Create a C program that implements The Game of Life cellular auto.pdf
•
0 likes•2 views
Create a C program that implements \"The Game of Life\" cellular automata (CA) simulation. A few comments: You will want 2 two-dimensional arrays. One that holds the current state of the \"universe\" and second that is generated from the first using the Game of Life rules, which are given below. After the computations for filling the second array are complete, the contents of the second array can be copied back to the first. You can make the array as big as you want, but you should probably use 20 times 20 as a minimum. As discussed in class, the outside boundary of the \"universe\" can be a bit tricky. To keep it simple for this exercise, you can assume that the outside boundary cells are always \"dead\". You can use whatever character (or number) that you want to represent the \"living\" cells. The dead cells should probably be spaces. The rules for whether a particular cell lives or dies are determined by the 8 surrounding cells. If a particular cell is alive, then: If only 0 or 1 of the surrounding cells are alive, the cell dies. (It\'s lonely.) If 4 or more of the surrounding cells are alive, the cell dies. (Too much competition.) If 2 or 3 of the surrounds cells are alive, the cell stays alive. If a particular cell is dead, then it comes to life if exactly 3 of the surround cells are alive. You will have to pick some starting configuration for the universe. An easy thing to do is fill the array randomly, and then let things go and see what evolves. Write your program as described above and demo the operation to your TA. Then try some modifications. First, try using a different starting configurations. Look here for some ideas:. Gosper\'s Glider Gun is a famous one. Or try a simple spaceship or oscillator. Then try changing the rules determining life and death. A common variation is to include a \"birth\" when 6 neighboring cells are alive in addition to the \"3-cell\" birth. Finally, there will be a short quiz covering 2D arrays. This exercise should be very familiar. Solution #include #include #include #include #include #include #include #define ROWS 60 #define COLS 60 #define OFF 0 #define ON 1 #define BLACK 8, 0, 0, 0, 0 #define scr_width 600 #define scr_height 600 #define cell_width (scr_width / ROWS) #define cell_height (scr_height / COLS) We must hold two copies of the board so that we can analyze board and make * changes to temp when killing/creating cells so that we don\'t cause changes * to the board to affect the following cells. */ char board[ROWS][COLS]; char temp[ROWS][COLS]; SDL_Rect cells[ROWS][COLS]; /* Stores positions of each cell for blits. */ void randomize_board(void); void initialize_grid(SDL_Surface* screen); void blit_board(SDL_Surface* bcell, SDL_Surface* screen); int num_neighbours(int x, int y); void update_board(void); int clear_board(SDL_Surface* screen, Uint32 color); int clear_cell(SDL_Surface* screen, int x, int y, Uint32 color); void initialize_cells_array(void); int main(void) { SDL_Init(SDL_INIT_VIDEO).
Report
Share
Report
Share
Download to read offline
![Create a C program that implements "The Game of Life" cellular automata (CA) simulation. A
few comments: You will want 2 two-dimensional arrays. One that holds the current state of the
"universe" and second that is generated from the first using the Game of Life rules, which are
given below. After the computations for filling the second array are complete, the contents of the
second array can be copied back to the first. You can make the array as big as you want, but you
should probably use 20 times 20 as a minimum. As discussed in class, the outside boundary of
the "universe" can be a bit tricky. To keep it simple for this exercise, you can assume that the
outside boundary cells are always "dead". You can use whatever character (or number) that
you want to represent the "living" cells. The dead cells should probably be spaces. The rules
for whether a particular cell lives or dies are determined by the 8 surrounding cells. If a particular
cell is alive, then: If only 0 or 1 of the surrounding cells are alive, the cell dies. (It's lonely.) If
4 or more of the surrounding cells are alive, the cell dies. (Too much competition.) If 2 or 3 of
the surrounds cells are alive, the cell stays alive. If a particular cell is dead, then it comes to life
if exactly 3 of the surround cells are alive. You will have to pick some starting configuration for
the universe. An easy thing to do is fill the array randomly, and then let things go and see what
evolves. Write your program as described above and demo the operation to your TA. Then try
some modifications. First, try using a different starting configurations. Look here for some
ideas:. Gosper's Glider Gun is a famous one. Or try a simple spaceship or oscillator. Then try
changing the rules determining life and death. A common variation is to include a "birth" when
6 neighboring cells are alive in addition to the "3-cell" birth. Finally, there will be a short quiz
covering 2D arrays. This exercise should be very familiar.
Solution
#include #include #include #include #include #include #include #define ROWS 60 #define
COLS 60 #define OFF 0 #define ON 1 #define BLACK 8, 0, 0, 0, 0 #define scr_width 600
#define scr_height 600 #define cell_width (scr_width / ROWS) #define cell_height (scr_height /
COLS) We must hold two copies of the board so that we can analyze board and make *
changes to temp when killing/creating cells so that we don't cause changes * to the board to
affect the following cells. */ char board[ROWS][COLS]; char temp[ROWS][COLS];
SDL_Rect cells[ROWS][COLS]; /* Stores positions of each cell for blits. */ void
randomize_board(void); void initialize_grid(SDL_Surface* screen); void
blit_board(SDL_Surface* bcell, SDL_Surface* screen); int num_neighbours(int x, int y); void
update_board(void); int clear_board(SDL_Surface* screen, Uint32 color); int
clear_cell(SDL_Surface* screen, int x, int y, Uint32 color); void initialize_cells_array(void); int
main(void) { SDL_Init(SDL_INIT_VIDEO); SDL_WM_SetCaption("Conway's Game of
Life - Ryuurei", NULL); SDL_Event event; int breaker = 0; int paused = 1;](https://image.slidesharecdn.com/createacprogramthatimplementsthegameoflifecellularauto-230703010528-7c11bd0a/85/Create-a-C-program-that-implements-The-Game-of-Life-cellular-auto-pdf-1-320.jpg)
![SDL_Surface* screen = SDL_SetVideoMode( scr_width, scr_height, 0,
SDL_SWSURFACE); if (! screen) { perror("SDL_SetVideoMode");
return EXIT_FAILURE; } Uint32 bgcolor = SDL_MapRGB(screen->format, 0xFF, 0xFF,
0xFF); if (SDL_FillRect(screen, &(screen->clip_rect), bgcolor) == -1) {
perror("SDL_FillRect"); return EXIT_FAILURE; } initialize_cells_array();
initialize_grid(screen); SDL_Surface* bcell = SDL_CreateRGBSurface(
SDL_SWSURFACE, cell_width, cell_height, BLACK); if (! bcell) { perror("BLACK
SDL_CreateRGBSurface"); return EXIT_FAILURE; } SDL_Flip(screen); while (1)
{ /* Infinite game loop. */ /** The event loop handles any events in the poll/queue *
even while paused. */ while (SDL_PollEvent(&event)) { switch (event.type) {
case SDL_QUIT: breaker = 1; break; case
SDL_KEYDOWN: /* Pause on enter/return key press. */ if
(event.key.keysym.sym == SDLK_KP_ENTER || event.key.keysym.sym ==
SDLK_RETURN) { paused = !paused; /* Space causes the board to
be randomized. */ } else if (event.key.keysym.sym == SDLK_SPACE) {
SDL_FillRect(screen, &(screen->clip_rect), bgcolor); randomize_board();
/* Escape button also quits. */ } else if (event.key.keysym.sym ==
SDLK_ESCAPE) { breaker = 1; break; /* Either shift
button clears the screen. */ } else if (event.key.keysym.sym == SDLK_RSHIFT ||
event.key.keysym.sym == SDLK_LSHIFT) { if
(clear_board(screen, bgcolor) == -1) { perror("clear_board");
return EXIT_FAILURE; } } break; case
SDL_MOUSEBUTTONDOWN: /** The only mouse event we handle is the left
click. * If the cell being clicked is alive, kill it. * Otherwise, give it
life. */ if (event.button.button == SDL_BUTTON_LEFT) {
if (event.button.y > (ROWS * cell_height)) break; int tx = floor(event.button.x /
cell_width); int ty = floor(event.button.y / cell_height); if
(board[tx][ty] == OFF) { board[tx][ty] = ON; temp[tx][ty] =
ON; SDL_BlitSurface( bcell, &(bcell-
>clip_rect), screen, &cells[tx][ty]); } else
if (clear_cell(screen, tx, ty, bgcolor) == -1) { perror("clear_cell");
return EXIT_FAILURE; } } break;
default: continue; } if (breaker) break; } if (breaker)
break; /* Redraw the grid so it's not painted over by clear_cell. */
initialize_grid(screen); blit_board(bcell, screen); SDL_Flip(screen); if (!paused) {
update_board(); SDL_Delay(250); SDL_FillRect(screen, &(screen-](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
In Java using Eclipse, Im suppose to write a class that encapsulat.pdf
In Java using Eclipse, I\'m suppose to write a class that encapsulates a tic tac toe board using two
dimensional arrays. It should only involve the human player vs. the computer, and should
randomly select who should use \'X\' or \'O\' and whether the human player or the computer
should go first. Verify that all moves by the human player are to a valid space on the tic-tac-toe
board, and an incorrect choice should not halt or terminate the game. Below is my Java program
that is currently a work in progress. Can you help me remodify it? Thanks.
import java.util.Scanner;
public class LeavinesTicTacToe
{
public static Scanner sc = new Scanner(System.in);
public static void main(String[] args)
{
final int SIZE = 3;
char[][] board = new char[SIZE][SIZE]; // game board
resetBoard(board); // initialize the board (with \' \' for all cells)
// First, welcome message and display the board.
System.out.println(\"===== WELCOME TO THE TIC-TAC-TOE GAME!! =====\ \");
showBoard(board);
// Then ask the user which symbol (x or o) he/she wants to play.
System.out.print(\" Which symbol do you want to play, \\\"x\\\" or \\\"o\\\"? \");
char userSymbol = sc.next().toLowerCase().charAt(0);
char compSymbol = (userSymbol == \'x\') ? \'o\' : \'x\';
// Also ask whether or not the user wants to go first.
System.out.println();
System.out.print(\" Do you want to go first (y/n)? \");
char ans = sc.next().toLowerCase().charAt(0);
int turn; // 0 -- the user, 1 -- the computer
int remainCount = SIZE * SIZE; // empty cell count
// THE VERY FIRST MOVE.
if (ans == \'y\') {
turn = 0;
userPlay(board, userSymbol); // user puts his/her first tic
}
else {
turn = 1;
compPlay(board, compSymbol); // computer puts its first tic
}
// Show the board, and decrement the count of remaining cells.
showBoard(board);
remainCount--;
// Play the game until either one wins.
boolean done = false;
int winner = -1; // 0 -- the user, 1 -- the computer, -1 -- draw
while (!done && remainCount > 0) {
// If there is a winner at this time, set the winner and the done flag to true.
done = isGameWon(board, turn, userSymbol, compSymbol); // Did the turn won?
if (done)
winner = turn; // the one who made the last move won the game
else {
// No winner yet. Find the next turn and play.
turn = (turn + 1 ) % 2;
if (turn == 0)
userPlay(board, userSymbol);
else
compPlay(board, compSymbol);
// Show the board after one tic, and decrement the rem count.
showBoard(board);
remainCount--;
}
}
// Winner is found. Declare the winner.
if (winner == 0)
System.out.println(\"\ ** YOU WON. CONGRATULATIONS!! **\");
else if (winner == 1)
System.out.println(\"\ ** YOU LOST.. Maybe next time :) **\");
else
System.out.println(\"\ ** DRAW... **\");
}
public static void resetBoard(char[][] brd)
{
for (int i = 0; i < brd.length; i++)
for (int j = 0; j < brd[0].length; j++)
brd[i][j] = \' \';
}
public static void showBoard(char[][] brd)
{
int numRow = brd.length;
int numCol = brd[0].length;
System.out.println();
// First write the column he.
write the TODO part of the program.docx
write the //TODO part of the program:
/***************************************************************************
* scramble.c
*
* Problem Set 3
*
* Implements Scramble with CS50.
*
* Usage: scramble [#]
*
* where # is an optional grid number.
***************************************************************************/
#include [removed]
#include [removed]
#include [removed]
#include [removed]
#include [removed]
#include [removed]
// duration of a game in seconds
#define DURATION 30
// grid's dimensions
#define DIMENSION 4
// maximum number of words in any dictionary
#define WORDS 172806
// maximum number of letters in any word
#define LETTERS 29
// default dictionary
// http://www.becomeawordgameexpert.com/wordlists.htm
#define DICTIONARY "words"
// for logging
FILE* log;
// grid
char grid[DIMENSION][DIMENSION];
// flags with which we can mark grid's letters while searching for words
bool marks[DIMENSION][DIMENSION];
// defines a word as having an array of letters plus a flag
// indicating whether word has been found on grid
typedef struct
{
bool found;
char letters[LETTERS + 1];
}
word;
// defines a dictionary as having a size and an array of words
struct
{
int size;
word words[WORDS];
}
dictionary;
// prototypes
void clear(void);
bool crawl(string letters, int x, int y);
void draw(void);
bool find(string s);
void initialize(void);
bool load(string s);
bool lookup(string s);
void scramble(void);
// This is Scramble.
int main(int argc, string argv[])
{
// ensure proper usage
if (argc > 2)
{
printf("Usage: %s [#]\n", basename(argv[0]));
return 1;
}
// seed pseudorandom number generator
if (argc == 2)
{
int seed = atoi(argv[1]);
if (seed <= 0)
{
printf("Invalid grid.\n");
return 1;
}
srand(seed);
}
else
srand(time(NULL));
// determine path to dictionary
string directory = dirname(argv[0]);
char path[strlen(directory) + 1 + strlen(DICTIONARY) + 1];
sprintf(path, "%s/%s", directory, DICTIONARY);
// load dictionary
if (!load(path))
{
printf("Could not open dictionary.\n");
return 1;
}
// initialize the grid
initialize();
// initialize user's score
int score = 0;
// calculate time of game's end
int end = time(NULL) + DURATION;
// open log
log = fopen("log.txt", "w");
if (log == NULL)
{
printf("Could not open log.\n");
return 1;
}
// accept words until timer expires
while (true)
{
// clear the screen
clear();
// draw the current state of the grid
draw();
// log board
for (int row = 0; row < DIMENSION; row++)
{
for (int col = 0; col [removed]= end)
.
The following code, is a one player battleship game in JAVA. Im tryi.pdf
The following code, is a one player battleship game in JAVA. Im trying to Enhance the game to
be a two player game (were a player plays aggainst a computer opponent). With TWO boards
displayed side by side. The player and the computer must take turns selecting the coordinates on
their own boards. (the ships cannot be placed adjacent to each other).
BattleShipGame.java
package battleship;
import java.util.ArrayList;
import java.util.Scanner;
public class BattleshipGame {
private Ocean ocean;
private boolean[][] availableSpot;
private Scanner sc;
public BattleshipGame() {
// define a new ocean and a new 2D array to store available coordinates
ocean = new Ocean();
availableSpot = new boolean[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++){
availableSpot[i][j] = true;
}
}
}
/**
* prints the game menu and info
* //param select
*/
public void print(int select){
String info;
switch (select) {
case 1: info = \"Welcome to the World of Battleship!\";
break;
case 2: info = \"Enter coordinates to fire: \";
break;
case 3: info = \"Shots fired: \"+ocean.getShotsFired()+\", Ships sunk:
\"+ocean.getShipsSunk();
break;
case 4: info = \"Congratulations! You win!\";
break;
case 99: info = \"Invalid input. Please re-enter:\";
break;
case 100: info = \"--------------------------------------------\";
break;
case 101: info = \"\ ============================================\";
break;
default: info = \"Error selection\";
break;
}
System.out.println(info);
}
/**
* check if the input is valid
* //param input
* //return boolean
*/
public boolean checkValidInput(String input){
ArrayList numList = new ArrayList();
for (int i=0;i<10;i++){
numList.add(\"\"+i);
}
String[] coordinates = input.split(\" \");
//returns false if there are not 2 strings
if (coordinates.length!=2){
return false;
}
//returns false if any of the strings is not a single digit number
for (String str: coordinates){
if (numList.contains(str)==false){
return false;
}
}
//returns false if the coordinates have already been shot at
int row = Integer.parseInt(coordinates[0]);
int column = Integer.parseInt(coordinates[1]);
if (this.availableSpot[row][column]==false){
return false;
}
return true;
}
/**
* get the coordinates to shoot at from the String input
* //param input
* //return int[] coordinates
*/
public int[] getCoordinates(String input){
int[] coordinates = new int[2];
String[] strList = input.split(\" \");
int row = Integer.parseInt(strList[0]);
int column = Integer.parseInt(strList[1]);
coordinates[0] = row;
coordinates[1] = column;
return coordinates;
}
/**
* play the battleship game
*/
public void play(){
print(101);
print(1);
ocean.placeAllShipsRandomly();
boolean isGameOver = ocean.isGameOver();
sc = new Scanner(System.in);
//print the ocean and start the game
ocean.print();
print(3);
while (!isGameOver){
print(2);
String input = sc.nextLine();
//check if input is valid
while (!checkValidInput(input)){
print(99);
input = sc.nextLine();
}
//get coordinates and fire
int[] coord.
i have a code that runs, but it only lets the player to guess where .pdf
i have a code that runs, but it only lets the player to guess where the ships are(which is a one
player game with one board). is there a way some one can help turn this game into a player vs
computer game (with two boards one for player and one for computer) Please help me.
BattleShipGame.java
package battleship;
import java.util.ArrayList;
import java.util.Scanner;
public class BattleshipGame {
private Ocean ocean;
private boolean[][] availableSpot;
private Scanner sc;
public BattleshipGame() {
// define a new ocean and a new 2D array to store available coordinates
ocean = new Ocean();
availableSpot = new boolean[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++){
availableSpot[i][j] = true;
}
}
}
/**
* prints the game menu and info
* //param select
*/
public void print(int select){
String info;
switch (select) {
case 1: info = \"Welcome to the World of Battleship!\";
break;
case 2: info = \"Enter coordinates to fire: \";
break;
case 3: info = \"Shots fired: \"+ocean.getShotsFired()+\", Ships sunk:
\"+ocean.getShipsSunk();
break;
case 4: info = \"Congratulations! You win!\";
break;
case 99: info = \"Invalid input. Please re-enter:\";
break;
case 100: info = \"--------------------------------------------\";
break;
case 101: info = \"\ ============================================\";
break;
default: info = \"Error selection\";
break;
}
System.out.println(info);
}
/**
* check if the input is valid
* //param input
* //return boolean
*/
public boolean checkValidInput(String input){
ArrayList numList = new ArrayList();
for (int i=0;i<10;i++){
numList.add(\"\"+i);
}
String[] coordinates = input.split(\" \");
//returns false if there are not 2 strings
if (coordinates.length!=2){
return false;
}
//returns false if any of the strings is not a single digit number
for (String str: coordinates){
if (numList.contains(str)==false){
return false;
}
}
//returns false if the coordinates have already been shot at
int row = Integer.parseInt(coordinates[0]);
int column = Integer.parseInt(coordinates[1]);
if (this.availableSpot[row][column]==false){
return false;
}
return true;
}
/**
* get the coordinates to shoot at from the String input
* //param input
* //return int[] coordinates
*/
public int[] getCoordinates(String input){
int[] coordinates = new int[2];
String[] strList = input.split(\" \");
int row = Integer.parseInt(strList[0]);
int column = Integer.parseInt(strList[1]);
coordinates[0] = row;
coordinates[1] = column;
return coordinates;
}
/**
* play the battleship game
*/
public void play(){
print(101);
print(1);
ocean.placeAllShipsRandomly();
boolean isGameOver = ocean.isGameOver();
sc = new Scanner(System.in);
//print the ocean and start the game
ocean.print();
print(3);
while (!isGameOver){
print(2);
String input = sc.nextLine();
//check if input is valid
while (!checkValidInput(input)){
print(99);
input = sc.nextLine();
}
//get coordinates and fire
int[] coordinates = getCoordinates(input);
int row = coordinates[0];
int column = coord.
Reply if you have any doubts.public class Life { public stati.pdf
Reply if you have any doubts.
public class Life {
public static final int DEAD = 0;
public static final int LIVING = 1;
public static final int DYING = 2;
public static final int BORNING = 3;
private int size;
private int[][] cells;
/* Constructor is done. */
public Life(int size) {
this.size = size;
cells = new int[size][size];
}
/* getSize is a stub. */
public int getSize() {
return this.size;
}
/* setState is a stub. */
public void setState(int row, int col, int state)
{
cells[row][col] = state;
}
/* getState is a stub. */
public int getState(int row, int col) {
return this.cells[row][col];
}
/* evolve is a stub. */
public void evolve()
{
//Creating tempArray
//only after full cycle cells should evolve
int[][] TempCells = new int[size][size];
//looping through all the cells
for(int i=0;i 3)
{
TempCells[i][j] = DEAD;
}
}
else if(CurrentState == DEAD)
{
if(NeighbourCount == 3)
{
TempCells[i][j] = LIVING;
}
}
}
}
//set the cells to nex gen states
this.cells = TempCells;
}
/* getNumberOfNeighbors is a stub; note that it\'s private. */
private int getNumberOfNeighbors(int row, int col)
{
int neighbourX[] = new int[8];
int neighbourY[] = new int[8];
neighbourX[0] = row - 1; neighbourY[0] = col - 1;
neighbourX[1] = row - 1; neighbourY[1] = col;
neighbourX[2] = row - 1; neighbourY[2] = col + 1;
neighbourX[3] = row; neighbourY[3] = col - 1;
neighbourX[4] = row; neighbourY[4] = col + 1;
neighbourX[5] = row + 1; neighbourY[5] = col - 1;
neighbourX[6] = row + 1; neighbourY[6] = col;
neighbourX[7] = row + 1; neighbourY[7] = col + 1;
for(int i=0;i<8;i++)
{
if(neighbourX[i] < 0)
neighbourX[i] = size - 1;
if(neighbourY[i] < 0)
neighbourY[i] = size - 1;
if(neighbourX[i] >= size)
neighbourX[i] = 0;
if(neighbourY[i] >= size)
neighbourY[i] = 0;
}
int count = 0;
for(int i=0;i<8;i++)
{
count += this.cells[neighbourX[i]][neighbourY[i]];
}
return count;
}
/* The main only instantiates the objects and sets them working. */
public static void main(String[] args) {
Life life = new Life(25);
CellWorld world = new CellWorld(life);
}
}
Solution
Reply if you have any doubts.
public class Life {
public static final int DEAD = 0;
public static final int LIVING = 1;
public static final int DYING = 2;
public static final int BORNING = 3;
private int size;
private int[][] cells;
/* Constructor is done. */
public Life(int size) {
this.size = size;
cells = new int[size][size];
}
/* getSize is a stub. */
public int getSize() {
return this.size;
}
/* setState is a stub. */
public void setState(int row, int col, int state)
{
cells[row][col] = state;
}
/* getState is a stub. */
public int getState(int row, int col) {
return this.cells[row][col];
}
/* evolve is a stub. */
public void evolve()
{
//Creating tempArray
//only after full cycle cells should evolve
int[][] TempCells = new int[size][size];
//looping through all the cells
for(int i=0;i 3)
{
TempCells[i][j] = DEAD;
}
}
else if(CurrentState == DEAD)
{
if(NeighbourCount == 3)
{
TempCells[i][j] = LIVING;
}
}
}
}
//set the cells.
Here is the code for youimport java.util.Scanner; import java.u.pdf
Here is the code for you:
import java.util.Scanner;
import java.util.Random;
public class TicTacToeGame {
static char[] [] board = new char[3][3];
static Scanner input=new Scanner(System.in);
//Object of Stats class to maintain statistics
static Stats stat = new Stats();
/**
* Prints the TicTacToe board
* @param arr: The board so far
*/
public static void printBoard(char [][] arr){
System.out.println();
for (int i=0; i<3; i++)
{
for (int j=0; j<3; j++)
{
System.out.print(arr[i][j]);
if(j!=2)
//Print the | for readable output
System.out.print(\" \" + \"|\" + \" \");
}
System.out.println();
if(i!=2) {
System.out.print(\"_ _ _ \"); // Print _ for readability
System.out.println();;
}
}
}
/**
* Clear the TicTacToe board before starting a new game
* @param arr: The board so far
*/
public static void clearBoard(char [][] arr){
for (int i=0; i<3; i++)
{
for (int j=0; j<3; j++)
{
arr[i][j]=\' \';
}
}
}
/** Determines if the player with the specified token wins
*
* @param symbol: Specifies whether the player is X or O
* @return true if player has won, false otherwise
*/
public static boolean isWon(char symbol) {
for (int i = 0; i < 3; i++) //horizontal
if (board[i][0] == symbol
&& board[i][1] == symbol
&& board[i][2] == symbol) {
return true;
}
//TODO!!! Also check for vertical and the two diagonals
for (int i = 0; i < 3; i++) //vertical
if (board[0][i] == symbol
&& board[1][i] == symbol
&& board[2][i] == symbol) {
return true;
}
//Leading diagonal
if (board[0][0] == symbol
&& board[1][1] == symbol
&& board[2][2] == symbol) {
return true;
}
//Trailing diagonal
if (board[0][2] == symbol
&& board[1][1] == symbol
&& board[2][0] == symbol) {
return true;
}
return false;
}
/** Determines if the cell is occupied
*
* @param row: Row of the cell to be checked
* @param col: Column of the cell to be checked
* @return true if the cell is occupied, false otherwise
*/
public static boolean isOccupied(int row, int col){
if (board[row][col]!=\' \') return false;
else return true;
}
/** Determines who starts the game
*/
public static int whoStarts(){
//TODO: Randomly chooses between 0 and 1 and returns the choice
return (int)(Math.random() + 0.5 );
}
/** takes care of the human\'s move
* 1. Prompt for a cell, then column
* 2. Puts a symbol (X or O) on the board
* 3. Prints the updated board
* 4. If a human wins: prints, updates stats and returns true
* 5. If not a win yet, returns false */
public static boolean humanTurn(char symbol){
//Prompt for a cell. User must enter
//row and column with a space in between.
System.out.print(\"\ \ Enter your move: (row column): \" );
int row = input.nextInt();
int col = input.nextInt();
//TODO!!! Mark user move in the board, print
//the board and check if user has won!
board[row][col] = symbol;
printBoard(board);
if(isWon(symbol))
return true;
return false;
}
/** takes care of the computer\'s move
* 1. Generates numbers until finds an empty cell
* 2. Puts a symbol (X or O) on the board
* 3. Prints the updated board
* 4. If a comp .
Introduction You implemented a Deck class in Activity 2. This cla.pdf
The document provides instructions for completing the implementation of a Deck class. It includes:
1) The Deck class which represents a deck of cards, with methods to initialize, shuffle, deal cards, and check if empty. The shuffle method needs to be implemented using an efficient selection shuffle algorithm.
2) A DeckTester class that tests the Deck class methods. Additional code needs to be added to test shuffling.
3) Sample code for a Shuffler class that implements efficient selection shuffle which can be used for the Deck shuffle method.
implemement the game.cpp by the header file given. And create main.c.pdf
implemement the game.cpp by the header file given. And create main.cpp to make the program
work.
The game is called the game of life, https://en.wikipedia.org/wiki/Conway\'s_Game_of_Life
game.h:
#include // Provides ostream
#include // String operations
#include // Randomizer
namespace csci2312
{
using std::string;
using std::ostream;
using std::istream;
class Cell
{
friend class GameOfLife;
public:
static const char alive =\'o\'; // alive image
static const char dead = \'-\'; // dead image
// Default constructor sets the cell\'s state to false
Cell();
// Custom constructor sets the cell\'s state as per argument
Cell(bool state);
// Empty destructor
~Cell();
// Accessors have no intention to modify the object, so it is a good practice to make them
\'const\' functions
bool getState() const;
// Mutator to change cell\'s state
void setState(bool newState);
// Accessor to see the \'face\'
char getFace() const;
private:
bool state;
char face;
};
class GameOfLife
{
public:
static const unsigned int MAX_BOARD = 30;
GameOfLife();
GameOfLife(size_t boardSize);
~GameOfLife();
int seedBoard(string fileName);
void seedBoard(size_t seeds);
void run();
void run(unsigned int numberOfIterations);
// ADVANCED
// A const(!) accessor method that returns a handle to the private currentLife array.
// The return type must also be \'const\' because we return a pointer to a static array, and
these are fixed
// It is just an example. It is not needed if we have a friend operator.
const Cell(*getCurrentLife() const )[MAX_BOARD+2] { return currentLife; };
///////////////////////////////////////////////////////
// friend operator can access private members of GameOfLife
friend ostream& operator << (ostream& out, const GameOfLife& board);
friend istream& operator >> (istream& in, const GameOfLife& board);
private:
bool executeRules(unsigned int countAlive, bool currentState);
// With \"Halo\" approach we need a bigger board
Cell currentLife[MAX_BOARD + 2][MAX_BOARD + 2];
Cell nextLife[MAX_BOARD + 2][MAX_BOARD + 2];
// ADVANCED
// Example how to declare variable cl as a pointer/handle to our array of Cells of size
HALO_BOARD
// The accessor method getCurrentLife() above uses the same syntax for the return type
const Cell(*cl)[MAX_BOARD + 2] = currentLife;
////////////////////////////////////////////////////////
size_t boardSize; // Board size requested in the constructor
};
// NON-MEMBER OUTPUT FUNCTIONS
// Display cell\'s state with alive/dead face
ostream& operator << (ostream& out, const Cell& cell);
}
Solution
Answer:
#include
#include
#include
#include
using namespace std;
void copy(int input1[52][102], int input2[52][102])
{
for(int j = 0; j < 52; j++)
{
for(int i = 0; i < 102; i++)
input2[j][i] = input1[j][i];
}
}
void lifegame(int input[52][102], char option)
{
int support[52][102];
copy(input, support);
for(int j = 1; j < 51; j++)
{
for(int i = 1; i < 101; i++)
{
if(option == \'m\')
{
int increment = 0;
increment = input[j-1][i] +
input[j-1][i-1] +
input[j][i-1] .
Recommended
In Java using Eclipse, Im suppose to write a class that encapsulat.pdf
In Java using Eclipse, I\'m suppose to write a class that encapsulates a tic tac toe board using two
dimensional arrays. It should only involve the human player vs. the computer, and should
randomly select who should use \'X\' or \'O\' and whether the human player or the computer
should go first. Verify that all moves by the human player are to a valid space on the tic-tac-toe
board, and an incorrect choice should not halt or terminate the game. Below is my Java program
that is currently a work in progress. Can you help me remodify it? Thanks.
import java.util.Scanner;
public class LeavinesTicTacToe
{
public static Scanner sc = new Scanner(System.in);
public static void main(String[] args)
{
final int SIZE = 3;
char[][] board = new char[SIZE][SIZE]; // game board
resetBoard(board); // initialize the board (with \' \' for all cells)
// First, welcome message and display the board.
System.out.println(\"===== WELCOME TO THE TIC-TAC-TOE GAME!! =====\ \");
showBoard(board);
// Then ask the user which symbol (x or o) he/she wants to play.
System.out.print(\" Which symbol do you want to play, \\\"x\\\" or \\\"o\\\"? \");
char userSymbol = sc.next().toLowerCase().charAt(0);
char compSymbol = (userSymbol == \'x\') ? \'o\' : \'x\';
// Also ask whether or not the user wants to go first.
System.out.println();
System.out.print(\" Do you want to go first (y/n)? \");
char ans = sc.next().toLowerCase().charAt(0);
int turn; // 0 -- the user, 1 -- the computer
int remainCount = SIZE * SIZE; // empty cell count
// THE VERY FIRST MOVE.
if (ans == \'y\') {
turn = 0;
userPlay(board, userSymbol); // user puts his/her first tic
}
else {
turn = 1;
compPlay(board, compSymbol); // computer puts its first tic
}
// Show the board, and decrement the count of remaining cells.
showBoard(board);
remainCount--;
// Play the game until either one wins.
boolean done = false;
int winner = -1; // 0 -- the user, 1 -- the computer, -1 -- draw
while (!done && remainCount > 0) {
// If there is a winner at this time, set the winner and the done flag to true.
done = isGameWon(board, turn, userSymbol, compSymbol); // Did the turn won?
if (done)
winner = turn; // the one who made the last move won the game
else {
// No winner yet. Find the next turn and play.
turn = (turn + 1 ) % 2;
if (turn == 0)
userPlay(board, userSymbol);
else
compPlay(board, compSymbol);
// Show the board after one tic, and decrement the rem count.
showBoard(board);
remainCount--;
}
}
// Winner is found. Declare the winner.
if (winner == 0)
System.out.println(\"\ ** YOU WON. CONGRATULATIONS!! **\");
else if (winner == 1)
System.out.println(\"\ ** YOU LOST.. Maybe next time :) **\");
else
System.out.println(\"\ ** DRAW... **\");
}
public static void resetBoard(char[][] brd)
{
for (int i = 0; i < brd.length; i++)
for (int j = 0; j < brd[0].length; j++)
brd[i][j] = \' \';
}
public static void showBoard(char[][] brd)
{
int numRow = brd.length;
int numCol = brd[0].length;
System.out.println();
// First write the column he.
write the TODO part of the program.docx
write the //TODO part of the program:
/***************************************************************************
* scramble.c
*
* Problem Set 3
*
* Implements Scramble with CS50.
*
* Usage: scramble [#]
*
* where # is an optional grid number.
***************************************************************************/
#include [removed]
#include [removed]
#include [removed]
#include [removed]
#include [removed]
#include [removed]
// duration of a game in seconds
#define DURATION 30
// grid's dimensions
#define DIMENSION 4
// maximum number of words in any dictionary
#define WORDS 172806
// maximum number of letters in any word
#define LETTERS 29
// default dictionary
// http://www.becomeawordgameexpert.com/wordlists.htm
#define DICTIONARY "words"
// for logging
FILE* log;
// grid
char grid[DIMENSION][DIMENSION];
// flags with which we can mark grid's letters while searching for words
bool marks[DIMENSION][DIMENSION];
// defines a word as having an array of letters plus a flag
// indicating whether word has been found on grid
typedef struct
{
bool found;
char letters[LETTERS + 1];
}
word;
// defines a dictionary as having a size and an array of words
struct
{
int size;
word words[WORDS];
}
dictionary;
// prototypes
void clear(void);
bool crawl(string letters, int x, int y);
void draw(void);
bool find(string s);
void initialize(void);
bool load(string s);
bool lookup(string s);
void scramble(void);
// This is Scramble.
int main(int argc, string argv[])
{
// ensure proper usage
if (argc > 2)
{
printf("Usage: %s [#]\n", basename(argv[0]));
return 1;
}
// seed pseudorandom number generator
if (argc == 2)
{
int seed = atoi(argv[1]);
if (seed <= 0)
{
printf("Invalid grid.\n");
return 1;
}
srand(seed);
}
else
srand(time(NULL));
// determine path to dictionary
string directory = dirname(argv[0]);
char path[strlen(directory) + 1 + strlen(DICTIONARY) + 1];
sprintf(path, "%s/%s", directory, DICTIONARY);
// load dictionary
if (!load(path))
{
printf("Could not open dictionary.\n");
return 1;
}
// initialize the grid
initialize();
// initialize user's score
int score = 0;
// calculate time of game's end
int end = time(NULL) + DURATION;
// open log
log = fopen("log.txt", "w");
if (log == NULL)
{
printf("Could not open log.\n");
return 1;
}
// accept words until timer expires
while (true)
{
// clear the screen
clear();
// draw the current state of the grid
draw();
// log board
for (int row = 0; row < DIMENSION; row++)
{
for (int col = 0; col [removed]= end)
.
The following code, is a one player battleship game in JAVA. Im tryi.pdf
The following code, is a one player battleship game in JAVA. Im trying to Enhance the game to
be a two player game (were a player plays aggainst a computer opponent). With TWO boards
displayed side by side. The player and the computer must take turns selecting the coordinates on
their own boards. (the ships cannot be placed adjacent to each other).
BattleShipGame.java
package battleship;
import java.util.ArrayList;
import java.util.Scanner;
public class BattleshipGame {
private Ocean ocean;
private boolean[][] availableSpot;
private Scanner sc;
public BattleshipGame() {
// define a new ocean and a new 2D array to store available coordinates
ocean = new Ocean();
availableSpot = new boolean[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++){
availableSpot[i][j] = true;
}
}
}
/**
* prints the game menu and info
* //param select
*/
public void print(int select){
String info;
switch (select) {
case 1: info = \"Welcome to the World of Battleship!\";
break;
case 2: info = \"Enter coordinates to fire: \";
break;
case 3: info = \"Shots fired: \"+ocean.getShotsFired()+\", Ships sunk:
\"+ocean.getShipsSunk();
break;
case 4: info = \"Congratulations! You win!\";
break;
case 99: info = \"Invalid input. Please re-enter:\";
break;
case 100: info = \"--------------------------------------------\";
break;
case 101: info = \"\ ============================================\";
break;
default: info = \"Error selection\";
break;
}
System.out.println(info);
}
/**
* check if the input is valid
* //param input
* //return boolean
*/
public boolean checkValidInput(String input){
ArrayList numList = new ArrayList();
for (int i=0;i<10;i++){
numList.add(\"\"+i);
}
String[] coordinates = input.split(\" \");
//returns false if there are not 2 strings
if (coordinates.length!=2){
return false;
}
//returns false if any of the strings is not a single digit number
for (String str: coordinates){
if (numList.contains(str)==false){
return false;
}
}
//returns false if the coordinates have already been shot at
int row = Integer.parseInt(coordinates[0]);
int column = Integer.parseInt(coordinates[1]);
if (this.availableSpot[row][column]==false){
return false;
}
return true;
}
/**
* get the coordinates to shoot at from the String input
* //param input
* //return int[] coordinates
*/
public int[] getCoordinates(String input){
int[] coordinates = new int[2];
String[] strList = input.split(\" \");
int row = Integer.parseInt(strList[0]);
int column = Integer.parseInt(strList[1]);
coordinates[0] = row;
coordinates[1] = column;
return coordinates;
}
/**
* play the battleship game
*/
public void play(){
print(101);
print(1);
ocean.placeAllShipsRandomly();
boolean isGameOver = ocean.isGameOver();
sc = new Scanner(System.in);
//print the ocean and start the game
ocean.print();
print(3);
while (!isGameOver){
print(2);
String input = sc.nextLine();
//check if input is valid
while (!checkValidInput(input)){
print(99);
input = sc.nextLine();
}
//get coordinates and fire
int[] coord.
i have a code that runs, but it only lets the player to guess where .pdf
i have a code that runs, but it only lets the player to guess where the ships are(which is a one
player game with one board). is there a way some one can help turn this game into a player vs
computer game (with two boards one for player and one for computer) Please help me.
BattleShipGame.java
package battleship;
import java.util.ArrayList;
import java.util.Scanner;
public class BattleshipGame {
private Ocean ocean;
private boolean[][] availableSpot;
private Scanner sc;
public BattleshipGame() {
// define a new ocean and a new 2D array to store available coordinates
ocean = new Ocean();
availableSpot = new boolean[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++){
availableSpot[i][j] = true;
}
}
}
/**
* prints the game menu and info
* //param select
*/
public void print(int select){
String info;
switch (select) {
case 1: info = \"Welcome to the World of Battleship!\";
break;
case 2: info = \"Enter coordinates to fire: \";
break;
case 3: info = \"Shots fired: \"+ocean.getShotsFired()+\", Ships sunk:
\"+ocean.getShipsSunk();
break;
case 4: info = \"Congratulations! You win!\";
break;
case 99: info = \"Invalid input. Please re-enter:\";
break;
case 100: info = \"--------------------------------------------\";
break;
case 101: info = \"\ ============================================\";
break;
default: info = \"Error selection\";
break;
}
System.out.println(info);
}
/**
* check if the input is valid
* //param input
* //return boolean
*/
public boolean checkValidInput(String input){
ArrayList numList = new ArrayList();
for (int i=0;i<10;i++){
numList.add(\"\"+i);
}
String[] coordinates = input.split(\" \");
//returns false if there are not 2 strings
if (coordinates.length!=2){
return false;
}
//returns false if any of the strings is not a single digit number
for (String str: coordinates){
if (numList.contains(str)==false){
return false;
}
}
//returns false if the coordinates have already been shot at
int row = Integer.parseInt(coordinates[0]);
int column = Integer.parseInt(coordinates[1]);
if (this.availableSpot[row][column]==false){
return false;
}
return true;
}
/**
* get the coordinates to shoot at from the String input
* //param input
* //return int[] coordinates
*/
public int[] getCoordinates(String input){
int[] coordinates = new int[2];
String[] strList = input.split(\" \");
int row = Integer.parseInt(strList[0]);
int column = Integer.parseInt(strList[1]);
coordinates[0] = row;
coordinates[1] = column;
return coordinates;
}
/**
* play the battleship game
*/
public void play(){
print(101);
print(1);
ocean.placeAllShipsRandomly();
boolean isGameOver = ocean.isGameOver();
sc = new Scanner(System.in);
//print the ocean and start the game
ocean.print();
print(3);
while (!isGameOver){
print(2);
String input = sc.nextLine();
//check if input is valid
while (!checkValidInput(input)){
print(99);
input = sc.nextLine();
}
//get coordinates and fire
int[] coordinates = getCoordinates(input);
int row = coordinates[0];
int column = coord.
Reply if you have any doubts.public class Life { public stati.pdf
Reply if you have any doubts.
public class Life {
public static final int DEAD = 0;
public static final int LIVING = 1;
public static final int DYING = 2;
public static final int BORNING = 3;
private int size;
private int[][] cells;
/* Constructor is done. */
public Life(int size) {
this.size = size;
cells = new int[size][size];
}
/* getSize is a stub. */
public int getSize() {
return this.size;
}
/* setState is a stub. */
public void setState(int row, int col, int state)
{
cells[row][col] = state;
}
/* getState is a stub. */
public int getState(int row, int col) {
return this.cells[row][col];
}
/* evolve is a stub. */
public void evolve()
{
//Creating tempArray
//only after full cycle cells should evolve
int[][] TempCells = new int[size][size];
//looping through all the cells
for(int i=0;i 3)
{
TempCells[i][j] = DEAD;
}
}
else if(CurrentState == DEAD)
{
if(NeighbourCount == 3)
{
TempCells[i][j] = LIVING;
}
}
}
}
//set the cells to nex gen states
this.cells = TempCells;
}
/* getNumberOfNeighbors is a stub; note that it\'s private. */
private int getNumberOfNeighbors(int row, int col)
{
int neighbourX[] = new int[8];
int neighbourY[] = new int[8];
neighbourX[0] = row - 1; neighbourY[0] = col - 1;
neighbourX[1] = row - 1; neighbourY[1] = col;
neighbourX[2] = row - 1; neighbourY[2] = col + 1;
neighbourX[3] = row; neighbourY[3] = col - 1;
neighbourX[4] = row; neighbourY[4] = col + 1;
neighbourX[5] = row + 1; neighbourY[5] = col - 1;
neighbourX[6] = row + 1; neighbourY[6] = col;
neighbourX[7] = row + 1; neighbourY[7] = col + 1;
for(int i=0;i<8;i++)
{
if(neighbourX[i] < 0)
neighbourX[i] = size - 1;
if(neighbourY[i] < 0)
neighbourY[i] = size - 1;
if(neighbourX[i] >= size)
neighbourX[i] = 0;
if(neighbourY[i] >= size)
neighbourY[i] = 0;
}
int count = 0;
for(int i=0;i<8;i++)
{
count += this.cells[neighbourX[i]][neighbourY[i]];
}
return count;
}
/* The main only instantiates the objects and sets them working. */
public static void main(String[] args) {
Life life = new Life(25);
CellWorld world = new CellWorld(life);
}
}
Solution
Reply if you have any doubts.
public class Life {
public static final int DEAD = 0;
public static final int LIVING = 1;
public static final int DYING = 2;
public static final int BORNING = 3;
private int size;
private int[][] cells;
/* Constructor is done. */
public Life(int size) {
this.size = size;
cells = new int[size][size];
}
/* getSize is a stub. */
public int getSize() {
return this.size;
}
/* setState is a stub. */
public void setState(int row, int col, int state)
{
cells[row][col] = state;
}
/* getState is a stub. */
public int getState(int row, int col) {
return this.cells[row][col];
}
/* evolve is a stub. */
public void evolve()
{
//Creating tempArray
//only after full cycle cells should evolve
int[][] TempCells = new int[size][size];
//looping through all the cells
for(int i=0;i 3)
{
TempCells[i][j] = DEAD;
}
}
else if(CurrentState == DEAD)
{
if(NeighbourCount == 3)
{
TempCells[i][j] = LIVING;
}
}
}
}
//set the cells.
Here is the code for youimport java.util.Scanner; import java.u.pdf
Here is the code for you:
import java.util.Scanner;
import java.util.Random;
public class TicTacToeGame {
static char[] [] board = new char[3][3];
static Scanner input=new Scanner(System.in);
//Object of Stats class to maintain statistics
static Stats stat = new Stats();
/**
* Prints the TicTacToe board
* @param arr: The board so far
*/
public static void printBoard(char [][] arr){
System.out.println();
for (int i=0; i<3; i++)
{
for (int j=0; j<3; j++)
{
System.out.print(arr[i][j]);
if(j!=2)
//Print the | for readable output
System.out.print(\" \" + \"|\" + \" \");
}
System.out.println();
if(i!=2) {
System.out.print(\"_ _ _ \"); // Print _ for readability
System.out.println();;
}
}
}
/**
* Clear the TicTacToe board before starting a new game
* @param arr: The board so far
*/
public static void clearBoard(char [][] arr){
for (int i=0; i<3; i++)
{
for (int j=0; j<3; j++)
{
arr[i][j]=\' \';
}
}
}
/** Determines if the player with the specified token wins
*
* @param symbol: Specifies whether the player is X or O
* @return true if player has won, false otherwise
*/
public static boolean isWon(char symbol) {
for (int i = 0; i < 3; i++) //horizontal
if (board[i][0] == symbol
&& board[i][1] == symbol
&& board[i][2] == symbol) {
return true;
}
//TODO!!! Also check for vertical and the two diagonals
for (int i = 0; i < 3; i++) //vertical
if (board[0][i] == symbol
&& board[1][i] == symbol
&& board[2][i] == symbol) {
return true;
}
//Leading diagonal
if (board[0][0] == symbol
&& board[1][1] == symbol
&& board[2][2] == symbol) {
return true;
}
//Trailing diagonal
if (board[0][2] == symbol
&& board[1][1] == symbol
&& board[2][0] == symbol) {
return true;
}
return false;
}
/** Determines if the cell is occupied
*
* @param row: Row of the cell to be checked
* @param col: Column of the cell to be checked
* @return true if the cell is occupied, false otherwise
*/
public static boolean isOccupied(int row, int col){
if (board[row][col]!=\' \') return false;
else return true;
}
/** Determines who starts the game
*/
public static int whoStarts(){
//TODO: Randomly chooses between 0 and 1 and returns the choice
return (int)(Math.random() + 0.5 );
}
/** takes care of the human\'s move
* 1. Prompt for a cell, then column
* 2. Puts a symbol (X or O) on the board
* 3. Prints the updated board
* 4. If a human wins: prints, updates stats and returns true
* 5. If not a win yet, returns false */
public static boolean humanTurn(char symbol){
//Prompt for a cell. User must enter
//row and column with a space in between.
System.out.print(\"\ \ Enter your move: (row column): \" );
int row = input.nextInt();
int col = input.nextInt();
//TODO!!! Mark user move in the board, print
//the board and check if user has won!
board[row][col] = symbol;
printBoard(board);
if(isWon(symbol))
return true;
return false;
}
/** takes care of the computer\'s move
* 1. Generates numbers until finds an empty cell
* 2. Puts a symbol (X or O) on the board
* 3. Prints the updated board
* 4. If a comp .
Introduction You implemented a Deck class in Activity 2. This cla.pdf
The document provides instructions for completing the implementation of a Deck class. It includes:
1) The Deck class which represents a deck of cards, with methods to initialize, shuffle, deal cards, and check if empty. The shuffle method needs to be implemented using an efficient selection shuffle algorithm.
2) A DeckTester class that tests the Deck class methods. Additional code needs to be added to test shuffling.
3) Sample code for a Shuffler class that implements efficient selection shuffle which can be used for the Deck shuffle method.
implemement the game.cpp by the header file given. And create main.c.pdf
implemement the game.cpp by the header file given. And create main.cpp to make the program
work.
The game is called the game of life, https://en.wikipedia.org/wiki/Conway\'s_Game_of_Life
game.h:
#include // Provides ostream
#include // String operations
#include // Randomizer
namespace csci2312
{
using std::string;
using std::ostream;
using std::istream;
class Cell
{
friend class GameOfLife;
public:
static const char alive =\'o\'; // alive image
static const char dead = \'-\'; // dead image
// Default constructor sets the cell\'s state to false
Cell();
// Custom constructor sets the cell\'s state as per argument
Cell(bool state);
// Empty destructor
~Cell();
// Accessors have no intention to modify the object, so it is a good practice to make them
\'const\' functions
bool getState() const;
// Mutator to change cell\'s state
void setState(bool newState);
// Accessor to see the \'face\'
char getFace() const;
private:
bool state;
char face;
};
class GameOfLife
{
public:
static const unsigned int MAX_BOARD = 30;
GameOfLife();
GameOfLife(size_t boardSize);
~GameOfLife();
int seedBoard(string fileName);
void seedBoard(size_t seeds);
void run();
void run(unsigned int numberOfIterations);
// ADVANCED
// A const(!) accessor method that returns a handle to the private currentLife array.
// The return type must also be \'const\' because we return a pointer to a static array, and
these are fixed
// It is just an example. It is not needed if we have a friend operator.
const Cell(*getCurrentLife() const )[MAX_BOARD+2] { return currentLife; };
///////////////////////////////////////////////////////
// friend operator can access private members of GameOfLife
friend ostream& operator << (ostream& out, const GameOfLife& board);
friend istream& operator >> (istream& in, const GameOfLife& board);
private:
bool executeRules(unsigned int countAlive, bool currentState);
// With \"Halo\" approach we need a bigger board
Cell currentLife[MAX_BOARD + 2][MAX_BOARD + 2];
Cell nextLife[MAX_BOARD + 2][MAX_BOARD + 2];
// ADVANCED
// Example how to declare variable cl as a pointer/handle to our array of Cells of size
HALO_BOARD
// The accessor method getCurrentLife() above uses the same syntax for the return type
const Cell(*cl)[MAX_BOARD + 2] = currentLife;
////////////////////////////////////////////////////////
size_t boardSize; // Board size requested in the constructor
};
// NON-MEMBER OUTPUT FUNCTIONS
// Display cell\'s state with alive/dead face
ostream& operator << (ostream& out, const Cell& cell);
}
Solution
Answer:
#include
#include
#include
#include
using namespace std;
void copy(int input1[52][102], int input2[52][102])
{
for(int j = 0; j < 52; j++)
{
for(int i = 0; i < 102; i++)
input2[j][i] = input1[j][i];
}
}
void lifegame(int input[52][102], char option)
{
int support[52][102];
copy(input, support);
for(int j = 1; j < 51; j++)
{
for(int i = 1; i < 101; i++)
{
if(option == \'m\')
{
int increment = 0;
increment = input[j-1][i] +
input[j-1][i-1] +
input[j][i-1] .
you will use cellular automata to create a 2D predator–prey simulati.pdf
you will use cellular automata to create a 2D predator–prey simulation in your program. The
preys are ants and the predators are doodlebugs. These critters live in a 20 * 20 grid of cells.
Only one critter may occupy a cell at a time. The grid is enclosed and no critter may move off
the grid. Time is simulated in steps. Each critter performs some action every time step. The ants
behave according to the following model: Move: For every time step, the ants randomly try to
move up, down, left, or right. If the neighboring cell in the selected direction is occupied or
would move the ant off the grid, then the ant stays in the current cell. Breed: If an ant survives
for three time steps (not been eaten by doodlebugs), at the end of the time step (i.e., after
moving) the ant will breed. This is simulated by creating a new ant in an adjacent (up, down, left,
or right) cell that is empty randomly. If there is no empty cell available, no breeding occurs.
Once an offspring is produced, an ant cannot produce an offspring again until it has survived
three more time steps.
The doodlebugs behave according to the following model:
Move: For every time step, the doodlebug will firstly try to move to an adjacent cell containing
an ant and eat the ant (you can decide if there are several ants in the adjacent cells, how the
doodlebug will choose to move). If there are no ants in adjacent cells, the doodlebug moves
according to the same rules as the ant. Note that a doodlebug cannot eat other doodlebugs.
Breed: If a doodlebug survives for eight time steps, at the end of the time step, it will spawn off a
new doodlebug in the same manner as the ant.
Starve: If a doodlebug has not eaten an ant within three time steps, at the end of the third time
step it will starve and die. The doodlebug should then be removed from the grid of cells.
Initialize the world with 5 doodlebugs and 100 ants. You will randomly place them. You should
see a cyclical pattern between the population of predators and prey, although random
perturbations may lead to the elimination of one or both species. You will prompt the user to
enter the number of time steps to run.
Create a class named Critter that contains data and functions common to ants and doodlebugs.
This class should have a virtual function named move that is defined in the derived classes of
Ant and Doodlebug. Each class will be in its own source file.
For each time step, do the following in your program: after moves, when breeding, eating, and
starving are resolved, display the resulting grid. Draw the world using ASCII characters of “O”
for an ant, “X” for a doodlebug and “E” for an empty space (the characters should be arranged to
look like a grid). The doodlebugs will move before the ants in each time step. When you reach
the time steps entered by the user, ask them to enter another number and start to run the
simulation again or to exit the program. You must maintain the state of the current grid while
creating the next displa.
Quiz using C++
This document contains source code for an online quiz system created by Sushil Kumar Mishra of ECE-I. It includes classes for drawing lines and boxes, managing menus, and controlling quiz functions like adding, deleting, and modifying questions in different subject databases. The main menu allows users to play quizzes, add questions, edit questions, or quit. Playing a quiz displays questions one by one with a time limit to select the correct answer. Scores are tracked and displayed at the end.
Include
This document contains code for a Tic-Tac-Toe game written in C++ using the Allegro library. It includes functions to initialize the game board, place markers for X and O turns, check for a winner after each turn, and allow player movement and input using keyboard arrows and enter. The main game loop handles drawing the board, player input, placing markers, checking for a winner, and continues until the player presses escape to quit.
game.c Written by Peter Sutton. Game board da.docx
The document describes the code for a falling block game. It includes functions for initializing the game board, attempting to move or rotate the current block, fixing blocks to the board, and adding new blocks. It stores the board state in two representations - a bitfield for each row, and a display structure including colors. It also defines internal functions for checking for completed rows, adding new blocks, and collision detection between blocks.
tetris
This document contains code for a Tetris game implementation. It includes functions for:
- Drawing the game grid and pieces
- Detecting collisions and removing full rows
- Tracking the score, level, and current/next pieces
- Handling game events like pausing, restarting, and detecting a game over state
The core game logic involves randomly generating and moving falling pieces, removing full rows to increase the score, and checking for collisions to end the game.
i have a runable code below that works with just guessing where the .pdf
i have a runable code below that works with just guessing where the ships are. i need help editing
this code to be able to have a player play against computer and taking turns playing battleship.
BattleShipGame.java
package battleship;
import java.util.ArrayList;
import java.util.Scanner;
public class BattleshipGame {
private Ocean ocean;
private boolean[][] availableSpot;
private Scanner sc;
public BattleshipGame() {
// define a new ocean and a new 2D array to store available coordinates
ocean = new Ocean();
availableSpot = new boolean[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++){
availableSpot[i][j] = true;
}
}
}
/**
* prints the game menu and info
* //param select
*/
public void print(int select){
String info;
switch (select) {
case 1: info = \"Welcome to the World of Battleship!\";
break;
case 2: info = \"Enter coordinates to fire: \";
break;
case 3: info = \"Shots fired: \"+ocean.getShotsFired()+\", Ships sunk:
\"+ocean.getShipsSunk();
break;
case 4: info = \"Congratulations! You win!\";
break;
case 99: info = \"Invalid input. Please re-enter:\";
break;
case 100: info = \"--------------------------------------------\";
break;
case 101: info = \"\ ============================================\";
break;
default: info = \"Error selection\";
break;
}
System.out.println(info);
}
/**
* check if the input is valid
* //param input
* //return boolean
*/
public boolean checkValidInput(String input){
ArrayList numList = new ArrayList();
for (int i=0;i<10;i++){
numList.add(\"\"+i);
}
String[] coordinates = input.split(\" \");
//returns false if there are not 2 strings
if (coordinates.length!=2){
return false;
}
//returns false if any of the strings is not a single digit number
for (String str: coordinates){
if (numList.contains(str)==false){
return false;
}
}
//returns false if the coordinates have already been shot at
int row = Integer.parseInt(coordinates[0]);
int column = Integer.parseInt(coordinates[1]);
if (this.availableSpot[row][column]==false){
return false;
}
return true;
}
/**
* get the coordinates to shoot at from the String input
* //param input
* //return int[] coordinates
*/
public int[] getCoordinates(String input){
int[] coordinates = new int[2];
String[] strList = input.split(\" \");
int row = Integer.parseInt(strList[0]);
int column = Integer.parseInt(strList[1]);
coordinates[0] = row;
coordinates[1] = column;
return coordinates;
}
/**
* play the battleship game
*/
public void play(){
print(101);
print(1);
ocean.placeAllShipsRandomly();
boolean isGameOver = ocean.isGameOver();
sc = new Scanner(System.in);
//print the ocean and start the game
ocean.print();
print(3);
while (!isGameOver){
print(2);
String input = sc.nextLine();
//check if input is valid
while (!checkValidInput(input)){
print(99);
input = sc.nextLine();
}
//get coordinates and fire
int[] coordinates = getCoordinates(input);
int row = coordinates[0];
int column = coordinates[1];
ocean.shootAt(row, column);
availableSpot[row][column] = false;
isGa.
Fractal Rendering in Developer C++ - 2012-11-06
Computer Graphics Seminar presentation of Aritra Sarkar at Indian Institute of Space Science and Technology
ECMAScript 6 and beyond
This document provides an overview of new features introduced in ECMAScript 6 (ES6), including arrow functions, constants and let declarations, template literals, classes, modules, promises, symbols, and more. It explains these features through examples and comparisons to ES5. Key points covered include arrow function syntax, block vs expression bodies, let and const block scoping, default parameters, destructuring, rest and spread operators, tagged template literals, class syntax and inheritance, module imports and exports, symbol use cases for unique keys and constants, and for...of loop iteration vs for...in enumeration.
adders/subtractors, multiplexers, intro to ISA
This document discusses binary and BCD adders and subtractors, including half-adders, full-adders, 4-bit binary adders, BCD adders, half/full subtractors, and multiplexers. It also discusses instruction set architecture (ISA), comparing RISC and CISC architectures, and the Intel x86 programming model.
c++ code#include iostream#include string#include stdlib.h.pdf
The document contains C++ and Java code that implements a minesweeper game. The C++ code defines a Grid class to represent the game board and includes functions to initialize the board with randomly placed mines, draw the board, and check for wins/losses when a player selects a tile. The Java code uses arrays to represent the game zones and sprites to render the board. It includes functions to initialize the board, place mines and numbers randomly, and handle player clicks on zones.
Javascript: The Important Bits
This document provides an introduction to JavaScript fundamentals and common patterns, jQuery optimization techniques, and an introduction to Node.js. It discusses JavaScript data types and variables, scoping, objects and classes. It also covers immediate execution functions, private patterns, event attaching in jQuery, and dependency management in Node.js projects. The document includes code examples for selecting elements and caching selectors in jQuery as well as creating an Express server and using middleware in Node.js applications.
AI CHALLENGE ADMIN
This document contains the code for a Connect Four game. It defines constants, global variables, and functions for tracking the game state (board, current player), validating and executing moves, checking for wins, and handling I/O like drawing the board. The main function runs the game loop, continuously prompting for player moves until the board is full or a player wins. Special move types like clearing rows/columns are supported in addition to normal drops.
EcmaScript 6
This contains all the slides used in Silicon Valley Code Camp presentation on Sunday Oct 4, 10:45 session on "Amazing new features in JavaScript". At the end ut also includes the last year presentation covering ES 5
Hidden loops
How to identify, when to remove and how to remove a common pattern in code: duplication that can be refactored towards loops and more complex data structures. Also with code examples.
Write a program that randomly generates a maze in javaSolution.pdf
Write a program that randomly generates a maze in java
Solution
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Random;
public class MyMaze {
private int dimensionX, dimensionY; // dimension of maze
private int gridDimensionX, gridDimensionY; // dimension of output grid
private char[][] grid; // output grid
private Cell[][] cells; // 2d array of Cells
private Random random = new Random(); // The random object
// initialize with x and y the same
public MyMaze(int aDimension) {
// Initialize
this(aDimension, aDimension);
}
// constructor
public MyMaze(int xDimension, int yDimension) {
dimensionX = xDimension;
dimensionY = yDimension;
gridDimensionX = xDimension * 4 + 1;
gridDimensionY = yDimension * 2 + 1;
grid = new char[gridDimensionX][gridDimensionY];
init();
generateMaze();
}
private void init() {
// create cells
cells = new Cell[dimensionX][dimensionY];
for (int x = 0; x < dimensionX; x++) {
for (int y = 0; y < dimensionY; y++) {
cells[x][y] = new Cell(x, y, false); // create cell (see Cell constructor)
}
}
}
// inner class to represent a cell
private class Cell {
int x, y; // coordinates
// cells this cell is connected to
ArrayList neighbors = new ArrayList<>();
// solver: if already used
boolean visited = false;
// solver: the Cell before this one in the path
Cell parent = null;
// solver: if used in last attempt to solve path
boolean inPath = false;
// solver: distance travelled this far
double travelled;
// solver: projected distance to end
double projectedDist;
// impassable cell
boolean wall = true;
// if true, has yet to be used in generation
boolean open = true;
// construct Cell at x, y
Cell(int x, int y) {
this(x, y, true);
}
// construct Cell at x, y and with whether it isWall
Cell(int x, int y, boolean isWall) {
this.x = x;
this.y = y;
this.wall = isWall;
}
// add a neighbor to this cell, and this cell as a neighbor to the other
void addNeighbor(Cell other) {
if (!this.neighbors.contains(other)) { // avoid duplicates
this.neighbors.add(other);
}
if (!other.neighbors.contains(this)) { // avoid duplicates
other.neighbors.add(this);
}
}
// used in updateGrid()
boolean isCellBelowNeighbor() {
return this.neighbors.contains(new Cell(this.x, this.y + 1));
}
// used in updateGrid()
boolean isCellRightNeighbor() {
return this.neighbors.contains(new Cell(this.x + 1, this.y));
}
// useful Cell representation
@Override
public String toString() {
return String.format(\"Cell(%s, %s)\", x, y);
}
// useful Cell equivalence
@Override
public boolean equals(Object other) {
if (!(other instanceof Cell)) return false;
Cell otherCell = (Cell) other;
return (this.x == otherCell.x && this.y == otherCell.y);
}
// should be overridden with equals
@Override
public int hashCode() {
// random hash code method designed to be usually unique
return this.x + this.y * 256;
}
}
// generate from upper left (In computing the y increases down often)
private void generateMaze() {
generateMaze(0, 0);
}
// gener.
TASK #1In the domain class you will create a loop that will prompt.pdf
TASK #1
In the domain class you will create a loop that will prompt the user to enter a value of 1, 2, or 3,
which will in turn, be translated to a floor number in the game. Make sure the user only picks a
selection you are expecting (1, 2, or 3) with a while-loop. Then you will create a switch or if-else
statement. If you are more comfortable with if-else then do switch, or if you are more
comfortable with switch then do if-else. Based on the number the user chooses you will set the
floor variable to the appropriate value. When they select 1 floor should be set to 3, when they
select 2 floor should be set to 6, and when they select 3 floor should be set to 10.
TASK #2
In the domain class you will create a constructor with 6 parameters, representing all the data
loaded from the input file that was saved from a previous adventure. This constructor receives 6
parameters: aName, anAttack, aDefense, aHealth, aCurrentFloor, & aMaxFloor.
TASK #3
In the domain class you will create a save method that will allow the user to save their progress
using a PrintWriter object. This file will overwrite whatever was there before, so no need to use
FileWriter, only PrintWriter. There are 6 attributes that you need to write to the file:
name, attack, defense, health, currentFloor, & maxFloor
public void saveFile()
{
String filename = “game.txt”;
PrintWriter pw = new PrintWriter(filename);
pw.println(name);
pw.println(attack);
pw.println(defense);
pw.println(health);
pw.println(currentFloor);
pw.println(maxFloor);
pw.close();
}
TASK #4
In the driver class you will create a load method that will allow the user to pick up from where
they left off. You will do this using a File object and a Scanner object. Remember to use the File
and Scanner classes, and to close the file object after you’re done. There are 6 attributes you will
need to load from the file to successfully continue an Adventure:
name, attack, defense, health, currentFloor, & maxFloor
After you read the record from the file, and store the data in these 6 variables, you can create a
new Adventure object called JavaQuest with those 6 variables. Remember JavaQuest is a global
variable defined at the beginning of the driver class. Then, within the load method, invoke the
startAdventure() method for the newly created Adventure object.
public static void load()
{
String filename = “load.txt”;
File myFile = new File(filename);
Scanner myScan = new Scanner(myFile);
String name;
int attack, defense, health, currentFloor, maxFloor;
name = myScan.nextLine();
attack = myScan.nextInt();
myScan.nextLine();
defense = myScan.nextInt();
myScan.nextLine();
…
javaQuest = new Adventure(name, attack, defense, health, currentFloor, maxFloor);
}
javaQuest.startAdventure();
The Files
package Adventure;
import java.io.BufferedWriter;
import java.io.File;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Random;
import java.util.Scanner;
public class Adventure
{
int health, defense, attack;
String mons.
This is the Java code i have for a Battleship project i am working o.pdf
This is the Java code i have for a Battleship project i am working on, i am now wanting to
convert this to a fully graphical Gui interface, but i just cant seem to get anywhere with it, so if i
could get some help with this, i would apperciate it....(just a reminder its in JAVA)
public class BattleshipGame {
private Ocean ocean;
private boolean[][] availableSpot;
private Scanner sc;
public BattleshipGame() {
// define a new ocean and a new 2D array to store available coordinates
ocean = new Ocean();
availableSpot = new boolean[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++){
availableSpot[i][j] = true;
}
}
}
/**
* prints the game menu and info
* //param select
*/
public void print(int select){
String info;
switch (select) {
case 1: info = \"Welcome to the World of Battleship!\";
break;
case 2: info = \"Enter coordinates to fire: \";
break;
case 3: info = \"Shots fired: \"+ocean.getShotsFired()+\", Ships sunk:
\"+ocean.getShipsSunk();
break;
case 4: info = \"Congratulations! You win!\";
break;
case 99: info = \"Invalid input. Please re-enter:\";
break;
case 100: info = \"--------------------------------------------\";
break;
case 101: info = \"\ ============================================\";
break;
default: info = \"Error selection\";
break;
}
System.out.println(info);
}
/**
* check if the input is valid
* //param input
* //return boolean
*/
public boolean checkValidInput(String input){
ArrayList numList = new ArrayList();
for (int i=0;i<10;i++){
numList.add(\"\"+i);
}
String[] coordinates = input.split(\" \");
//returns false if there are not 2 strings
if (coordinates.length!=2){
return false;
}
//returns false if any of the strings is not a single digit number
for (String str: coordinates){
if (numList.contains(str)==false){
return false;
}
}
//returns false if the coordinates have already been shot at
int row = Integer.parseInt(coordinates[0]);
int column = Integer.parseInt(coordinates[1]);
if (this.availableSpot[row][column]==false){
return false;
}
return true;
}
/**
* get the coordinates to shoot at from the String input
* //param input
* //return int[] coordinates
*/
public int[] getCoordinates(String input){
int[] coordinates = new int[2];
String[] strList = input.split(\" \");
int row = Integer.parseInt(strList[0]);
int column = Integer.parseInt(strList[1]);
coordinates[0] = row;
coordinates[1] = column;
return coordinates;
}
/**
* play the battleship game
*/
public void play(){
print(101);
print(1);
ocean.placeAllShipsRandomly();
boolean isGameOver = ocean.isGameOver();
sc = new Scanner(System.in);
//print the ocean and start the game
ocean.print();
print(3);
while (!isGameOver){
print(2);
String input = sc.nextLine();
//check if input is valid
while (!checkValidInput(input)){
print(99);
input = sc.nextLine();
}
//get coordinates and fire
int[] coordinates = getCoordinates(input);
int row = coordinates[0];
int column = coordinates[1];
ocean.shootAt(row, column);
availableSpot[row][column] = false;
isGameOver = o.
import javautilLinkedList import javautilQueue import .pdf
import java.util.LinkedList; import java.util.Queue; import java.util.Stack; class
Sudoku_Solver_question { public static int type = 0; public static int height = 0; // Node exploration
holds the sudoku board in every iteration private static class Exploration { char[][] board; int curCell
= 0; // constructor method public Exploration(char[][] board, int curCell) { this.board =
cloneBoard(board); this.curCell = curCell; } // clones the board for every state private char[][]
cloneBoard(char[][] board) { char[][] newBoard = new char[board.length][board[0].length];
copyIntoBoard(board, newBoard); return newBoard; } } public static boolean BFS(char[][] board) {
// TODO write BFS search algorithm } public static boolean DFS(char[][] board) { // TODO write
DFS search algorithm } /** * parameter: index * return: index * Returns the index of a row based on
type(6x6 / 9x9) of the board */ private static int getRowForIndex(int idx) { return idx / type; } /** *
parameter: index * return: index * Returns the index of a column based on type(6x6 / 9x9) of the
board */ private static int getColumnForIndex(int idx) { return idx % type; } /** * Parameters: source
board & destination board * return: null * Copies the source board to destination board in every
state */ protected static void copyIntoBoard(char[][] src, char[][] dst) { for (int i = 0; i < src.length;
i++) { System.arraycopy(src[i], 0, dst[i], 0, src[i].length); } } /** * Parameters: Board & box index
number * returns: true / false * Validates the box, returns true if there is no duplicate number in the
box else false */ private static boolean isValidSquare(char[][] board, int s) { boolean[] isPresent =
new boolean[board.length+1]; for (int r = (s / 3) * height; r < ((s / 3) + 1) * height; r++) { for (int c =
(s % 3) * height; c < ((s % 3) + 1) * height; c++) { char value = board[r][c]; if (value == '.') {
continue; } int intVal = Character.getNumericValue(value); if (isPresent[intVal]) { return false; }
isPresent[intVal] = true; } } return true; } /** * Parameters: Board & column index number * returns:
true / false * Validates the box, returns true if there is no duplicate number in the column else false
*/ private static boolean isValidColumn(char[][] board, int c) { boolean[] isPresent = new
boolean[board.length+1]; for (int i = 0; i < board.length; i++) { char value = board[i][c]; if (value ==
'.') { continue; } int intVal = Character.getNumericValue(value); if (isPresent[intVal]) { return false; }
isPresent[intVal] = true; } return true; } /** * Parameters: Board & row index number * returns: true /
false * Validates the box, returns true if there is no duplicate number in the row else false */ private
static boolean isValidRow(char[][] board, int r) { boolean[] isPresent = new
boolean[board.length+1]; for (int i = 0; i < board.length; i++) { char value = board[r][i]; if (value ==
'.') { continue; } int intVal = Character.getNumericValue(value); if (isPresent[intVal]) { return false; }
isPre.
D3.js workshop
This document provides an overview of key concepts for working with D3, including:
- D3 uses standard web technologies like HTML, SVG, and CSS rather than introducing new representations. Learning D3 largely means learning web standards.
- Visualization with D3 requires mapping data to visual elements using scales. Scales are functions that map from data values to visual values like pixel positions.
- Selections in D3 correspond to elements in the DOM. Data joins allow binding data to selections to drive attribute updates. The enter, update, exit pattern is used to handle new, existing and removed data.
- Common scale types include linear, log, quantize and quantile for quantitative data, and
i need an input of this program. anything good or bad. what could .docx
i need an input of this program. anything good or bad. what could be changed, what you may like or dislike about it. only one paragraph.
// newTetris.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include
#include
#include
#include
#define GAME_INTERVAL 20
#define DIR_DOWN 2
#define DIR_LEFT 4
#define DIR_RIGHT 6
#define DIR_ROTATE 5
using namespace std;
class TetrisShape {
public:
char shapeArray[4][4];
int shapeTopLeftX = 6;
int shapeTopLeftY = 0;
void populateShapeArray(int shape);
void rotate();
template
void setShape(char(&shape)[rows][cols]);
TetrisShape(int shape) { populateShapeArray(shape); };
TetrisShape() {};
};
void TetrisShape::rotate() {
char _shapeArray[4][4];
_shapeArray[0][0] = shapeArray[0][3]; _shapeArray[1][0] = shapeArray[0][2]; _shapeArray[2][0] = shapeArray[0][1]; _shapeArray[3][0] = shapeArray[0][0];
_shapeArray[0][1] = shapeArray[1][3]; _shapeArray[1][1] = shapeArray[1][2]; _shapeArray[2][1] = shapeArray[1][1]; _shapeArray[3][1] = shapeArray[1][0];
_shapeArray[0][2] = shapeArray[2][3]; _shapeArray[1][2] = shapeArray[2][2]; _shapeArray[2][2] = shapeArray[2][1]; _shapeArray[3][2] = shapeArray[2][0];
_shapeArray[0][3] = shapeArray[3][3]; _shapeArray[1][3] = shapeArray[3][2]; _shapeArray[2][3] = shapeArray[3][1]; _shapeArray[3][3] = shapeArray[3][0];
for (int _x = 0; _x < 4; _x++) {
for (int _y = 0; _y < 4; _y++) {
shapeArray[_x][_y] = _shapeArray[_x][_y];
}
}
}
void TetrisShape::populateShapeArray(int shape) {
switch (shape) {
case 1:
shapeArray[0][0] = ' '; shapeArray[1][0] = ' '; shapeArray[2][0] = ' '; shapeArray[3][0] = ' ';
shapeArray[0][1] = ' '; shapeArray[1][1] = 'X'; shapeArray[2][1] = ' '; shapeArray[3][1] = ' ';
shapeArray[0][2] = ' '; shapeArray[1][2] = 'X'; shapeArray[2][2] = ' '; shapeArray[3][2] = ' ';
shapeArray[0][3] = ' '; shapeArray[1][3] = 'X'; shapeArray[2][3] = 'X'; shapeArray[3][3] = ' ';
break;
case 2:
shapeArray[0][0] = ' '; shapeArray[1][0] = 'X'; shapeArray[2][0] = ' '; shapeArray[3][0] = ' ';
shapeArray[0][1] = ' '; shapeArray[1][1] = 'X'; shapeArray[2][1] = ' '; shapeArray[3][1] = ' ';
shapeArray[0][2] = ' '; shapeArray[1][2] = 'X'; shapeArray[2][2] = ' '; shapeArray[3][2] = ' ';
shapeArray[0][3] = ' '; shapeArray[1][3] = 'X'; shapeArray[2][3] = ' '; shapeArray[3][3] = ' ';
break;
case 3:
shapeArray[0][0] = ' '; shapeArray[1][0] = ' '; shapeArray[2][0] = ' '; shapeArray[3][0] = ' ';
shapeArray[0][1] = ' '; shapeArray[1][1] = 'X'; shapeArray[2][1] = ' '; shapeArray[3][1] = ' ';
shapeArray[0][2] = ' '; shapeArray[1][2] = 'X'; shapeArray[2][2] = 'X'; shapeArray[3][2] = ' ';
shapeArray[0][3] = ' '; shapeArray[1][3] = ' '; shapeArray[2][3] = 'X'; shapeArray[3][3] = ' ';
break;
case 4:
shapeArray[0][0] = ' '; shapeArray[1][0] = ' '; shapeArray[2][0] = ' '; shapeArray[3][0] = ' ';
shapeArray[0][1] = ' '; shapeArray[1][1] = ' '; shapeArray[2][1] = 'X'; shapeArray[3][1] = ' ';
shapeArray[0][2] = ' '; shapeArray[1][2] = 'X'; shapeArray[2][2] = 'X.
The age of the Earth is estimated from the amount of uranium isotope .pdf
The age of the Earth is estimated from the amount of uranium isotope 238 remaining.
Radioisotopes like U^238 slowly decay over time turning into either another isotope or another
element when they release radiation. Estimates are that only half of the U^238 that was
originally present in the Earth still exists. Since the half life (time it takes a radioisotope to lose
1/2 its material) is 4 5 times 10^9 that puts the age of the Earth at 4 5 billion years U^238 decays
into Th^234, which decays into Pa^234, which decays into U^234. The earliest fossil evidence
of bacteria is 3 5 billion years old. Vascular plants began to appear 400 million years ago.
Modern humans are estimated to have appeared about 200,000 years ago. At which of these
times would the most original U^238 remain? The time bacteria first became fossilized The
time vascular plants appeared The time modem humans appeared At which of these times
would the least original U^238 remain? The time bacteria first became fossilized The time
vascular plants appeared The time modem humans appeared Fossils found in rock suggest that:
Compare these time spans by changing the scale from YEARS to SECOND: let 1 year = 1 sec.
Thus, in this scale the average biology student is 20 seconds old. At this scale, how many days
ago did humans appear? 1.7 hours 1 day 1.7 days 2 days 2.7 days How many years ago did the
dinosaurs vanish? 2 years 2.1 years 2.2 days 2.3 days 2.4 days How many years ago did
flowering plants begin to dominate? 1.5 years 2 years 2.5 years 3 years 3.5 years How many
years ago did plants start to live on land? 10 years 11 years 12 years 13 years 14 years How
many years ago did cells first fossilize? 91 years 111 years 121 years 131 years 141 years
Solution
a) Most U238 existed when the bacteria first bacame fossilised. The earliest fossil evidence of
bacteria is 3.5 billion years ago
b) Least U238 remains at time when first modern human appeared 200,000 years ago
2a) 180,000 years = 180,000 sec = 2.08 days or 2 days. Ans iv
b) 2.06 years
c)2.5 years
d) 12.6 years
e)111 years.
Suppose Michael creates an RSA cryptosystem with a very large modulu.pdf
Suppose Michael creates an RSA cryptosystem with a very large mod ulus n for which the
factorization of n cannot be found in a reasonable amount of time. Suppose that Nikita sends
messages to Michael by representing each alphabetic character as an integer between 0 and 26
(A corresponds to 1, B to 2, etc., and a space to 0), then encrypts each number separately using
Michael\'s RSA cryptosystem. Is this method secure? Explain your answer.
Solution
This method is not secure, it is just equal to a \"Ceaser Cipher\", that is a permutation of letters of
alphabet, which is well-known to be easily broken using a frequency analysis..
More Related Content
Similar to Create a C program that implements The Game of Life cellular auto.pdf
you will use cellular automata to create a 2D predator–prey simulati.pdf
you will use cellular automata to create a 2D predator–prey simulation in your program. The
preys are ants and the predators are doodlebugs. These critters live in a 20 * 20 grid of cells.
Only one critter may occupy a cell at a time. The grid is enclosed and no critter may move off
the grid. Time is simulated in steps. Each critter performs some action every time step. The ants
behave according to the following model: Move: For every time step, the ants randomly try to
move up, down, left, or right. If the neighboring cell in the selected direction is occupied or
would move the ant off the grid, then the ant stays in the current cell. Breed: If an ant survives
for three time steps (not been eaten by doodlebugs), at the end of the time step (i.e., after
moving) the ant will breed. This is simulated by creating a new ant in an adjacent (up, down, left,
or right) cell that is empty randomly. If there is no empty cell available, no breeding occurs.
Once an offspring is produced, an ant cannot produce an offspring again until it has survived
three more time steps.
The doodlebugs behave according to the following model:
Move: For every time step, the doodlebug will firstly try to move to an adjacent cell containing
an ant and eat the ant (you can decide if there are several ants in the adjacent cells, how the
doodlebug will choose to move). If there are no ants in adjacent cells, the doodlebug moves
according to the same rules as the ant. Note that a doodlebug cannot eat other doodlebugs.
Breed: If a doodlebug survives for eight time steps, at the end of the time step, it will spawn off a
new doodlebug in the same manner as the ant.
Starve: If a doodlebug has not eaten an ant within three time steps, at the end of the third time
step it will starve and die. The doodlebug should then be removed from the grid of cells.
Initialize the world with 5 doodlebugs and 100 ants. You will randomly place them. You should
see a cyclical pattern between the population of predators and prey, although random
perturbations may lead to the elimination of one or both species. You will prompt the user to
enter the number of time steps to run.
Create a class named Critter that contains data and functions common to ants and doodlebugs.
This class should have a virtual function named move that is defined in the derived classes of
Ant and Doodlebug. Each class will be in its own source file.
For each time step, do the following in your program: after moves, when breeding, eating, and
starving are resolved, display the resulting grid. Draw the world using ASCII characters of “O”
for an ant, “X” for a doodlebug and “E” for an empty space (the characters should be arranged to
look like a grid). The doodlebugs will move before the ants in each time step. When you reach
the time steps entered by the user, ask them to enter another number and start to run the
simulation again or to exit the program. You must maintain the state of the current grid while
creating the next displa.
Quiz using C++
This document contains source code for an online quiz system created by Sushil Kumar Mishra of ECE-I. It includes classes for drawing lines and boxes, managing menus, and controlling quiz functions like adding, deleting, and modifying questions in different subject databases. The main menu allows users to play quizzes, add questions, edit questions, or quit. Playing a quiz displays questions one by one with a time limit to select the correct answer. Scores are tracked and displayed at the end.
Include
This document contains code for a Tic-Tac-Toe game written in C++ using the Allegro library. It includes functions to initialize the game board, place markers for X and O turns, check for a winner after each turn, and allow player movement and input using keyboard arrows and enter. The main game loop handles drawing the board, player input, placing markers, checking for a winner, and continues until the player presses escape to quit.
game.c Written by Peter Sutton. Game board da.docx
The document describes the code for a falling block game. It includes functions for initializing the game board, attempting to move or rotate the current block, fixing blocks to the board, and adding new blocks. It stores the board state in two representations - a bitfield for each row, and a display structure including colors. It also defines internal functions for checking for completed rows, adding new blocks, and collision detection between blocks.
tetris
This document contains code for a Tetris game implementation. It includes functions for:
- Drawing the game grid and pieces
- Detecting collisions and removing full rows
- Tracking the score, level, and current/next pieces
- Handling game events like pausing, restarting, and detecting a game over state
The core game logic involves randomly generating and moving falling pieces, removing full rows to increase the score, and checking for collisions to end the game.
i have a runable code below that works with just guessing where the .pdf
i have a runable code below that works with just guessing where the ships are. i need help editing
this code to be able to have a player play against computer and taking turns playing battleship.
BattleShipGame.java
package battleship;
import java.util.ArrayList;
import java.util.Scanner;
public class BattleshipGame {
private Ocean ocean;
private boolean[][] availableSpot;
private Scanner sc;
public BattleshipGame() {
// define a new ocean and a new 2D array to store available coordinates
ocean = new Ocean();
availableSpot = new boolean[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++){
availableSpot[i][j] = true;
}
}
}
/**
* prints the game menu and info
* //param select
*/
public void print(int select){
String info;
switch (select) {
case 1: info = \"Welcome to the World of Battleship!\";
break;
case 2: info = \"Enter coordinates to fire: \";
break;
case 3: info = \"Shots fired: \"+ocean.getShotsFired()+\", Ships sunk:
\"+ocean.getShipsSunk();
break;
case 4: info = \"Congratulations! You win!\";
break;
case 99: info = \"Invalid input. Please re-enter:\";
break;
case 100: info = \"--------------------------------------------\";
break;
case 101: info = \"\ ============================================\";
break;
default: info = \"Error selection\";
break;
}
System.out.println(info);
}
/**
* check if the input is valid
* //param input
* //return boolean
*/
public boolean checkValidInput(String input){
ArrayList numList = new ArrayList();
for (int i=0;i<10;i++){
numList.add(\"\"+i);
}
String[] coordinates = input.split(\" \");
//returns false if there are not 2 strings
if (coordinates.length!=2){
return false;
}
//returns false if any of the strings is not a single digit number
for (String str: coordinates){
if (numList.contains(str)==false){
return false;
}
}
//returns false if the coordinates have already been shot at
int row = Integer.parseInt(coordinates[0]);
int column = Integer.parseInt(coordinates[1]);
if (this.availableSpot[row][column]==false){
return false;
}
return true;
}
/**
* get the coordinates to shoot at from the String input
* //param input
* //return int[] coordinates
*/
public int[] getCoordinates(String input){
int[] coordinates = new int[2];
String[] strList = input.split(\" \");
int row = Integer.parseInt(strList[0]);
int column = Integer.parseInt(strList[1]);
coordinates[0] = row;
coordinates[1] = column;
return coordinates;
}
/**
* play the battleship game
*/
public void play(){
print(101);
print(1);
ocean.placeAllShipsRandomly();
boolean isGameOver = ocean.isGameOver();
sc = new Scanner(System.in);
//print the ocean and start the game
ocean.print();
print(3);
while (!isGameOver){
print(2);
String input = sc.nextLine();
//check if input is valid
while (!checkValidInput(input)){
print(99);
input = sc.nextLine();
}
//get coordinates and fire
int[] coordinates = getCoordinates(input);
int row = coordinates[0];
int column = coordinates[1];
ocean.shootAt(row, column);
availableSpot[row][column] = false;
isGa.
Fractal Rendering in Developer C++ - 2012-11-06
Computer Graphics Seminar presentation of Aritra Sarkar at Indian Institute of Space Science and Technology
ECMAScript 6 and beyond
This document provides an overview of new features introduced in ECMAScript 6 (ES6), including arrow functions, constants and let declarations, template literals, classes, modules, promises, symbols, and more. It explains these features through examples and comparisons to ES5. Key points covered include arrow function syntax, block vs expression bodies, let and const block scoping, default parameters, destructuring, rest and spread operators, tagged template literals, class syntax and inheritance, module imports and exports, symbol use cases for unique keys and constants, and for...of loop iteration vs for...in enumeration.
adders/subtractors, multiplexers, intro to ISA
This document discusses binary and BCD adders and subtractors, including half-adders, full-adders, 4-bit binary adders, BCD adders, half/full subtractors, and multiplexers. It also discusses instruction set architecture (ISA), comparing RISC and CISC architectures, and the Intel x86 programming model.
c++ code#include iostream#include string#include stdlib.h.pdf
The document contains C++ and Java code that implements a minesweeper game. The C++ code defines a Grid class to represent the game board and includes functions to initialize the board with randomly placed mines, draw the board, and check for wins/losses when a player selects a tile. The Java code uses arrays to represent the game zones and sprites to render the board. It includes functions to initialize the board, place mines and numbers randomly, and handle player clicks on zones.
Javascript: The Important Bits
This document provides an introduction to JavaScript fundamentals and common patterns, jQuery optimization techniques, and an introduction to Node.js. It discusses JavaScript data types and variables, scoping, objects and classes. It also covers immediate execution functions, private patterns, event attaching in jQuery, and dependency management in Node.js projects. The document includes code examples for selecting elements and caching selectors in jQuery as well as creating an Express server and using middleware in Node.js applications.
AI CHALLENGE ADMIN
This document contains the code for a Connect Four game. It defines constants, global variables, and functions for tracking the game state (board, current player), validating and executing moves, checking for wins, and handling I/O like drawing the board. The main function runs the game loop, continuously prompting for player moves until the board is full or a player wins. Special move types like clearing rows/columns are supported in addition to normal drops.
EcmaScript 6
This contains all the slides used in Silicon Valley Code Camp presentation on Sunday Oct 4, 10:45 session on "Amazing new features in JavaScript". At the end ut also includes the last year presentation covering ES 5
Hidden loops
How to identify, when to remove and how to remove a common pattern in code: duplication that can be refactored towards loops and more complex data structures. Also with code examples.
Write a program that randomly generates a maze in javaSolution.pdf
Write a program that randomly generates a maze in java
Solution
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Random;
public class MyMaze {
private int dimensionX, dimensionY; // dimension of maze
private int gridDimensionX, gridDimensionY; // dimension of output grid
private char[][] grid; // output grid
private Cell[][] cells; // 2d array of Cells
private Random random = new Random(); // The random object
// initialize with x and y the same
public MyMaze(int aDimension) {
// Initialize
this(aDimension, aDimension);
}
// constructor
public MyMaze(int xDimension, int yDimension) {
dimensionX = xDimension;
dimensionY = yDimension;
gridDimensionX = xDimension * 4 + 1;
gridDimensionY = yDimension * 2 + 1;
grid = new char[gridDimensionX][gridDimensionY];
init();
generateMaze();
}
private void init() {
// create cells
cells = new Cell[dimensionX][dimensionY];
for (int x = 0; x < dimensionX; x++) {
for (int y = 0; y < dimensionY; y++) {
cells[x][y] = new Cell(x, y, false); // create cell (see Cell constructor)
}
}
}
// inner class to represent a cell
private class Cell {
int x, y; // coordinates
// cells this cell is connected to
ArrayList neighbors = new ArrayList<>();
// solver: if already used
boolean visited = false;
// solver: the Cell before this one in the path
Cell parent = null;
// solver: if used in last attempt to solve path
boolean inPath = false;
// solver: distance travelled this far
double travelled;
// solver: projected distance to end
double projectedDist;
// impassable cell
boolean wall = true;
// if true, has yet to be used in generation
boolean open = true;
// construct Cell at x, y
Cell(int x, int y) {
this(x, y, true);
}
// construct Cell at x, y and with whether it isWall
Cell(int x, int y, boolean isWall) {
this.x = x;
this.y = y;
this.wall = isWall;
}
// add a neighbor to this cell, and this cell as a neighbor to the other
void addNeighbor(Cell other) {
if (!this.neighbors.contains(other)) { // avoid duplicates
this.neighbors.add(other);
}
if (!other.neighbors.contains(this)) { // avoid duplicates
other.neighbors.add(this);
}
}
// used in updateGrid()
boolean isCellBelowNeighbor() {
return this.neighbors.contains(new Cell(this.x, this.y + 1));
}
// used in updateGrid()
boolean isCellRightNeighbor() {
return this.neighbors.contains(new Cell(this.x + 1, this.y));
}
// useful Cell representation
@Override
public String toString() {
return String.format(\"Cell(%s, %s)\", x, y);
}
// useful Cell equivalence
@Override
public boolean equals(Object other) {
if (!(other instanceof Cell)) return false;
Cell otherCell = (Cell) other;
return (this.x == otherCell.x && this.y == otherCell.y);
}
// should be overridden with equals
@Override
public int hashCode() {
// random hash code method designed to be usually unique
return this.x + this.y * 256;
}
}
// generate from upper left (In computing the y increases down often)
private void generateMaze() {
generateMaze(0, 0);
}
// gener.
TASK #1In the domain class you will create a loop that will prompt.pdf
TASK #1
In the domain class you will create a loop that will prompt the user to enter a value of 1, 2, or 3,
which will in turn, be translated to a floor number in the game. Make sure the user only picks a
selection you are expecting (1, 2, or 3) with a while-loop. Then you will create a switch or if-else
statement. If you are more comfortable with if-else then do switch, or if you are more
comfortable with switch then do if-else. Based on the number the user chooses you will set the
floor variable to the appropriate value. When they select 1 floor should be set to 3, when they
select 2 floor should be set to 6, and when they select 3 floor should be set to 10.
TASK #2
In the domain class you will create a constructor with 6 parameters, representing all the data
loaded from the input file that was saved from a previous adventure. This constructor receives 6
parameters: aName, anAttack, aDefense, aHealth, aCurrentFloor, & aMaxFloor.
TASK #3
In the domain class you will create a save method that will allow the user to save their progress
using a PrintWriter object. This file will overwrite whatever was there before, so no need to use
FileWriter, only PrintWriter. There are 6 attributes that you need to write to the file:
name, attack, defense, health, currentFloor, & maxFloor
public void saveFile()
{
String filename = “game.txt”;
PrintWriter pw = new PrintWriter(filename);
pw.println(name);
pw.println(attack);
pw.println(defense);
pw.println(health);
pw.println(currentFloor);
pw.println(maxFloor);
pw.close();
}
TASK #4
In the driver class you will create a load method that will allow the user to pick up from where
they left off. You will do this using a File object and a Scanner object. Remember to use the File
and Scanner classes, and to close the file object after you’re done. There are 6 attributes you will
need to load from the file to successfully continue an Adventure:
name, attack, defense, health, currentFloor, & maxFloor
After you read the record from the file, and store the data in these 6 variables, you can create a
new Adventure object called JavaQuest with those 6 variables. Remember JavaQuest is a global
variable defined at the beginning of the driver class. Then, within the load method, invoke the
startAdventure() method for the newly created Adventure object.
public static void load()
{
String filename = “load.txt”;
File myFile = new File(filename);
Scanner myScan = new Scanner(myFile);
String name;
int attack, defense, health, currentFloor, maxFloor;
name = myScan.nextLine();
attack = myScan.nextInt();
myScan.nextLine();
defense = myScan.nextInt();
myScan.nextLine();
…
javaQuest = new Adventure(name, attack, defense, health, currentFloor, maxFloor);
}
javaQuest.startAdventure();
The Files
package Adventure;
import java.io.BufferedWriter;
import java.io.File;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Random;
import java.util.Scanner;
public class Adventure
{
int health, defense, attack;
String mons.
This is the Java code i have for a Battleship project i am working o.pdf
This is the Java code i have for a Battleship project i am working on, i am now wanting to
convert this to a fully graphical Gui interface, but i just cant seem to get anywhere with it, so if i
could get some help with this, i would apperciate it....(just a reminder its in JAVA)
public class BattleshipGame {
private Ocean ocean;
private boolean[][] availableSpot;
private Scanner sc;
public BattleshipGame() {
// define a new ocean and a new 2D array to store available coordinates
ocean = new Ocean();
availableSpot = new boolean[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++){
availableSpot[i][j] = true;
}
}
}
/**
* prints the game menu and info
* //param select
*/
public void print(int select){
String info;
switch (select) {
case 1: info = \"Welcome to the World of Battleship!\";
break;
case 2: info = \"Enter coordinates to fire: \";
break;
case 3: info = \"Shots fired: \"+ocean.getShotsFired()+\", Ships sunk:
\"+ocean.getShipsSunk();
break;
case 4: info = \"Congratulations! You win!\";
break;
case 99: info = \"Invalid input. Please re-enter:\";
break;
case 100: info = \"--------------------------------------------\";
break;
case 101: info = \"\ ============================================\";
break;
default: info = \"Error selection\";
break;
}
System.out.println(info);
}
/**
* check if the input is valid
* //param input
* //return boolean
*/
public boolean checkValidInput(String input){
ArrayList numList = new ArrayList();
for (int i=0;i<10;i++){
numList.add(\"\"+i);
}
String[] coordinates = input.split(\" \");
//returns false if there are not 2 strings
if (coordinates.length!=2){
return false;
}
//returns false if any of the strings is not a single digit number
for (String str: coordinates){
if (numList.contains(str)==false){
return false;
}
}
//returns false if the coordinates have already been shot at
int row = Integer.parseInt(coordinates[0]);
int column = Integer.parseInt(coordinates[1]);
if (this.availableSpot[row][column]==false){
return false;
}
return true;
}
/**
* get the coordinates to shoot at from the String input
* //param input
* //return int[] coordinates
*/
public int[] getCoordinates(String input){
int[] coordinates = new int[2];
String[] strList = input.split(\" \");
int row = Integer.parseInt(strList[0]);
int column = Integer.parseInt(strList[1]);
coordinates[0] = row;
coordinates[1] = column;
return coordinates;
}
/**
* play the battleship game
*/
public void play(){
print(101);
print(1);
ocean.placeAllShipsRandomly();
boolean isGameOver = ocean.isGameOver();
sc = new Scanner(System.in);
//print the ocean and start the game
ocean.print();
print(3);
while (!isGameOver){
print(2);
String input = sc.nextLine();
//check if input is valid
while (!checkValidInput(input)){
print(99);
input = sc.nextLine();
}
//get coordinates and fire
int[] coordinates = getCoordinates(input);
int row = coordinates[0];
int column = coordinates[1];
ocean.shootAt(row, column);
availableSpot[row][column] = false;
isGameOver = o.
import javautilLinkedList import javautilQueue import .pdf
import java.util.LinkedList; import java.util.Queue; import java.util.Stack; class
Sudoku_Solver_question { public static int type = 0; public static int height = 0; // Node exploration
holds the sudoku board in every iteration private static class Exploration { char[][] board; int curCell
= 0; // constructor method public Exploration(char[][] board, int curCell) { this.board =
cloneBoard(board); this.curCell = curCell; } // clones the board for every state private char[][]
cloneBoard(char[][] board) { char[][] newBoard = new char[board.length][board[0].length];
copyIntoBoard(board, newBoard); return newBoard; } } public static boolean BFS(char[][] board) {
// TODO write BFS search algorithm } public static boolean DFS(char[][] board) { // TODO write
DFS search algorithm } /** * parameter: index * return: index * Returns the index of a row based on
type(6x6 / 9x9) of the board */ private static int getRowForIndex(int idx) { return idx / type; } /** *
parameter: index * return: index * Returns the index of a column based on type(6x6 / 9x9) of the
board */ private static int getColumnForIndex(int idx) { return idx % type; } /** * Parameters: source
board & destination board * return: null * Copies the source board to destination board in every
state */ protected static void copyIntoBoard(char[][] src, char[][] dst) { for (int i = 0; i < src.length;
i++) { System.arraycopy(src[i], 0, dst[i], 0, src[i].length); } } /** * Parameters: Board & box index
number * returns: true / false * Validates the box, returns true if there is no duplicate number in the
box else false */ private static boolean isValidSquare(char[][] board, int s) { boolean[] isPresent =
new boolean[board.length+1]; for (int r = (s / 3) * height; r < ((s / 3) + 1) * height; r++) { for (int c =
(s % 3) * height; c < ((s % 3) + 1) * height; c++) { char value = board[r][c]; if (value == '.') {
continue; } int intVal = Character.getNumericValue(value); if (isPresent[intVal]) { return false; }
isPresent[intVal] = true; } } return true; } /** * Parameters: Board & column index number * returns:
true / false * Validates the box, returns true if there is no duplicate number in the column else false
*/ private static boolean isValidColumn(char[][] board, int c) { boolean[] isPresent = new
boolean[board.length+1]; for (int i = 0; i < board.length; i++) { char value = board[i][c]; if (value ==
'.') { continue; } int intVal = Character.getNumericValue(value); if (isPresent[intVal]) { return false; }
isPresent[intVal] = true; } return true; } /** * Parameters: Board & row index number * returns: true /
false * Validates the box, returns true if there is no duplicate number in the row else false */ private
static boolean isValidRow(char[][] board, int r) { boolean[] isPresent = new
boolean[board.length+1]; for (int i = 0; i < board.length; i++) { char value = board[r][i]; if (value ==
'.') { continue; } int intVal = Character.getNumericValue(value); if (isPresent[intVal]) { return false; }
isPre.
D3.js workshop
This document provides an overview of key concepts for working with D3, including:
- D3 uses standard web technologies like HTML, SVG, and CSS rather than introducing new representations. Learning D3 largely means learning web standards.
- Visualization with D3 requires mapping data to visual elements using scales. Scales are functions that map from data values to visual values like pixel positions.
- Selections in D3 correspond to elements in the DOM. Data joins allow binding data to selections to drive attribute updates. The enter, update, exit pattern is used to handle new, existing and removed data.
- Common scale types include linear, log, quantize and quantile for quantitative data, and
i need an input of this program. anything good or bad. what could .docx
i need an input of this program. anything good or bad. what could be changed, what you may like or dislike about it. only one paragraph.
// newTetris.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include
#include
#include
#include
#define GAME_INTERVAL 20
#define DIR_DOWN 2
#define DIR_LEFT 4
#define DIR_RIGHT 6
#define DIR_ROTATE 5
using namespace std;
class TetrisShape {
public:
char shapeArray[4][4];
int shapeTopLeftX = 6;
int shapeTopLeftY = 0;
void populateShapeArray(int shape);
void rotate();
template
void setShape(char(&shape)[rows][cols]);
TetrisShape(int shape) { populateShapeArray(shape); };
TetrisShape() {};
};
void TetrisShape::rotate() {
char _shapeArray[4][4];
_shapeArray[0][0] = shapeArray[0][3]; _shapeArray[1][0] = shapeArray[0][2]; _shapeArray[2][0] = shapeArray[0][1]; _shapeArray[3][0] = shapeArray[0][0];
_shapeArray[0][1] = shapeArray[1][3]; _shapeArray[1][1] = shapeArray[1][2]; _shapeArray[2][1] = shapeArray[1][1]; _shapeArray[3][1] = shapeArray[1][0];
_shapeArray[0][2] = shapeArray[2][3]; _shapeArray[1][2] = shapeArray[2][2]; _shapeArray[2][2] = shapeArray[2][1]; _shapeArray[3][2] = shapeArray[2][0];
_shapeArray[0][3] = shapeArray[3][3]; _shapeArray[1][3] = shapeArray[3][2]; _shapeArray[2][3] = shapeArray[3][1]; _shapeArray[3][3] = shapeArray[3][0];
for (int _x = 0; _x < 4; _x++) {
for (int _y = 0; _y < 4; _y++) {
shapeArray[_x][_y] = _shapeArray[_x][_y];
}
}
}
void TetrisShape::populateShapeArray(int shape) {
switch (shape) {
case 1:
shapeArray[0][0] = ' '; shapeArray[1][0] = ' '; shapeArray[2][0] = ' '; shapeArray[3][0] = ' ';
shapeArray[0][1] = ' '; shapeArray[1][1] = 'X'; shapeArray[2][1] = ' '; shapeArray[3][1] = ' ';
shapeArray[0][2] = ' '; shapeArray[1][2] = 'X'; shapeArray[2][2] = ' '; shapeArray[3][2] = ' ';
shapeArray[0][3] = ' '; shapeArray[1][3] = 'X'; shapeArray[2][3] = 'X'; shapeArray[3][3] = ' ';
break;
case 2:
shapeArray[0][0] = ' '; shapeArray[1][0] = 'X'; shapeArray[2][0] = ' '; shapeArray[3][0] = ' ';
shapeArray[0][1] = ' '; shapeArray[1][1] = 'X'; shapeArray[2][1] = ' '; shapeArray[3][1] = ' ';
shapeArray[0][2] = ' '; shapeArray[1][2] = 'X'; shapeArray[2][2] = ' '; shapeArray[3][2] = ' ';
shapeArray[0][3] = ' '; shapeArray[1][3] = 'X'; shapeArray[2][3] = ' '; shapeArray[3][3] = ' ';
break;
case 3:
shapeArray[0][0] = ' '; shapeArray[1][0] = ' '; shapeArray[2][0] = ' '; shapeArray[3][0] = ' ';
shapeArray[0][1] = ' '; shapeArray[1][1] = 'X'; shapeArray[2][1] = ' '; shapeArray[3][1] = ' ';
shapeArray[0][2] = ' '; shapeArray[1][2] = 'X'; shapeArray[2][2] = 'X'; shapeArray[3][2] = ' ';
shapeArray[0][3] = ' '; shapeArray[1][3] = ' '; shapeArray[2][3] = 'X'; shapeArray[3][3] = ' ';
break;
case 4:
shapeArray[0][0] = ' '; shapeArray[1][0] = ' '; shapeArray[2][0] = ' '; shapeArray[3][0] = ' ';
shapeArray[0][1] = ' '; shapeArray[1][1] = ' '; shapeArray[2][1] = 'X'; shapeArray[3][1] = ' ';
shapeArray[0][2] = ' '; shapeArray[1][2] = 'X'; shapeArray[2][2] = 'X.
Similar to Create a C program that implements The Game of Life cellular auto.pdf (20)
you will use cellular automata to create a 2D predator–prey simulati.pdf
you will use cellular automata to create a 2D predator–prey simulati.pdf
game.c Written by Peter Sutton. Game board da.docx
game.c Written by Peter Sutton. Game board da.docx
i have a runable code below that works with just guessing where the .pdf
i have a runable code below that works with just guessing where the .pdf
c++ code#include iostream#include string#include stdlib.h.pdf
c++ code#include iostream#include string#include stdlib.h.pdf
Write a program that randomly generates a maze in javaSolution.pdf
Write a program that randomly generates a maze in javaSolution.pdf
TASK #1In the domain class you will create a loop that will prompt.pdf
TASK #1In the domain class you will create a loop that will prompt.pdf
This is the Java code i have for a Battleship project i am working o.pdf
This is the Java code i have for a Battleship project i am working o.pdf
import javautilLinkedList import javautilQueue import .pdf
import javautilLinkedList import javautilQueue import .pdf
i need an input of this program. anything good or bad. what could .docx
i need an input of this program. anything good or bad. what could .docx
More from arihantsherwani
The age of the Earth is estimated from the amount of uranium isotope .pdf
The age of the Earth is estimated from the amount of uranium isotope 238 remaining.
Radioisotopes like U^238 slowly decay over time turning into either another isotope or another
element when they release radiation. Estimates are that only half of the U^238 that was
originally present in the Earth still exists. Since the half life (time it takes a radioisotope to lose
1/2 its material) is 4 5 times 10^9 that puts the age of the Earth at 4 5 billion years U^238 decays
into Th^234, which decays into Pa^234, which decays into U^234. The earliest fossil evidence
of bacteria is 3 5 billion years old. Vascular plants began to appear 400 million years ago.
Modern humans are estimated to have appeared about 200,000 years ago. At which of these
times would the most original U^238 remain? The time bacteria first became fossilized The
time vascular plants appeared The time modem humans appeared At which of these times
would the least original U^238 remain? The time bacteria first became fossilized The time
vascular plants appeared The time modem humans appeared Fossils found in rock suggest that:
Compare these time spans by changing the scale from YEARS to SECOND: let 1 year = 1 sec.
Thus, in this scale the average biology student is 20 seconds old. At this scale, how many days
ago did humans appear? 1.7 hours 1 day 1.7 days 2 days 2.7 days How many years ago did the
dinosaurs vanish? 2 years 2.1 years 2.2 days 2.3 days 2.4 days How many years ago did
flowering plants begin to dominate? 1.5 years 2 years 2.5 years 3 years 3.5 years How many
years ago did plants start to live on land? 10 years 11 years 12 years 13 years 14 years How
many years ago did cells first fossilize? 91 years 111 years 121 years 131 years 141 years
Solution
a) Most U238 existed when the bacteria first bacame fossilised. The earliest fossil evidence of
bacteria is 3.5 billion years ago
b) Least U238 remains at time when first modern human appeared 200,000 years ago
2a) 180,000 years = 180,000 sec = 2.08 days or 2 days. Ans iv
b) 2.06 years
c)2.5 years
d) 12.6 years
e)111 years.
Suppose Michael creates an RSA cryptosystem with a very large modulu.pdf
Suppose Michael creates an RSA cryptosystem with a very large mod ulus n for which the
factorization of n cannot be found in a reasonable amount of time. Suppose that Nikita sends
messages to Michael by representing each alphabetic character as an integer between 0 and 26
(A corresponds to 1, B to 2, etc., and a space to 0), then encrypts each number separately using
Michael\'s RSA cryptosystem. Is this method secure? Explain your answer.
Solution
This method is not secure, it is just equal to a \"Ceaser Cipher\", that is a permutation of letters of
alphabet, which is well-known to be easily broken using a frequency analysis..
Match the terms to the definitions provided.Question 2 optionsd.pdf
Match the terms to the definitions provided.
Question 2 options:
dehydrogenations
anaerobic respiration
fermentation
digestion
oxidative phosphorylation
heterotrophs
aerobic respiration
ATP synthase
autotrophs
substrate-level phosphorylation
glycolysis
High-energy electrons from chemical bonds are transferred to a final electron acceptor, which is
oxygen
An enzyme that couples the reentry of protons into the matrix with the phosphorylation of ADP
to ATP
ATP synthetase makes ATP with energy from a proton gradient generated by an electron
transport chain.
Organisms that live on the organic compounds produced by other organisms.
The use of inorganic molecules as final electron acceptors for an electron transport chain.
A series of 10 reactions used to catabolize glucose and produce ATP by substrate levels
phosphorylation.
Reactions in which electrons lost are accompanied by protons.
ATP is formed by transfer of a phosphate group directly to ADP from a phosphate-bearing
intermediate.
The enzymatic breakdown of large molecules into smaller ones.
The electrons generated by glycolysis are donated to organic molecules and NAD+ is recycled
Organisms that harvest energy from the sun and convert it to chemical energy by photosynthesis.
dehydrogenations
anaerobic respiration
fermentation
digestion
oxidative phosphorylation
heterotrophs
aerobic respiration
ATP synthase
autotrophs
substrate-level phosphorylation
glycolysis1.
High-energy electrons from chemical bonds are transferred to a final electron acceptor, which is
oxygen2.
An enzyme that couples the reentry of protons into the matrix with the phosphorylation of ADP
to ATP3.
ATP synthetase makes ATP with energy from a proton gradient generated by an electron
transport chain.4.
Organisms that live on the organic compounds produced by other organisms.5.
The use of inorganic molecules as final electron acceptors for an electron transport chain.6.
A series of 10 reactions used to catabolize glucose and produce ATP by substrate levels
phosphorylation.7.
Reactions in which electrons lost are accompanied by protons.8.
ATP is formed by transfer of a phosphate group directly to ADP from a phosphate-bearing
intermediate.9.
The enzymatic breakdown of large molecules into smaller ones.10.
The electrons generated by glycolysis are donated to organic molecules and NAD+ is
recycled11.
Organisms that harvest energy from the sun and convert it to chemical energy by
photosynthesis.
Solution
Answer :
Dehydrogenations : Reactions in which electrons lost are accompanied by protons.
Anaerobic respiration : The use of inorganic molecules as final electron acceptors for an electron
transport chain.
Fermentation : The electrons generated by glycolysis are donated to organic molecules and
NAD+ is recycled
Digestion : The enzymatic breakdown of large molecules into smaller ones.
oxidative phosphorylation : ATP synthetase makes ATP with energy from a proton gradient
generated by an electron transport chain.
Heterotrophs : Organi.
Simple linear regression and multiple regression are both based on t.pdf
Simple linear regression and multiple regression are both based on the use of quantitative
variables to predict a single dependent variable. It is the number of independent variables that
changes.
Describe a regression relationship using multiple independent variables.
Solution
Consider Y: weight
Let X1: gender
X2: age
X3: height
X4: nature of job
X5: heredity factors
This is one example where multiple regression is used..
Sand, Mell, and Rand are partners who share incomes and losses in a .pdf
Sand, Mell, and Rand are partners who share incomes and losses in a 1:4:5 ratio. After lengthy
disagreements among the partners and several unprofitable periods, the partners decided to
liquidate the partnership. Before the liquidation, the partnership balance sheet showed the
following:
Cash $10,000
Total “other assets,” $106,000
Total liabilities, $88,000
Sand, Capital, $1,200
Mell, Capital, $11,700
Rand, Capital, $15,100
The “other assets” were sold for $ 85,000. Proceeds from the sale of other assets were used to
payoff existing liabilities.
Determine the following:
1)The gain (or loss) realized on the sale of the assets and recording of liabilities’ payment.
2)The balances in the partners’ capital accounts after the distribution of this gain or loss to the
capital accounts.
3)Assume that if any capital deficits exist, they are not made up and the deficient partner pays
down his or her deficit to zero. How much cash will each of the partners receive in the final
liquidation?
Solution
Balance Sheet Assets Amount Liabilities Amount Cash $ 10,000.00 Total
liabilities $ 88,000.00 Other Assets $ 106,000.00 Partner\'s Capital A/c Sand $
1,200.00 Mell $ 11,700.00 RAND $ 15,100.00 Total $ 116,000.00 Total $
116,000.00 Journal Entries Particular Amount(DR) Amount(CR) 1) Cash
A/c $ 85,000.00 Loss on realization $ 21,000.00 To Other Assets $
106,000.00 (Being amount realized on sale of assets) Sand=($21000*1/10) $
2,100.00 Mell=($21000*4/10) $ 8,400.00 Rand=($21000*5/10) $ 10,500.00
To Loss on realization $ 21,000.00 (Being loss on realization distribute to
partner\'s in their profit sharing ratio) Total Liabilities $ 88,000.00 To Cash
$ 88,000.00 (Being amount paid to total liabilities) 2) Sand\'s Capital
A/c Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $
2,100.00 By Balance B/d $ 1,200.00 By Cash A/c $ 900.00 Total $ 2,100.00
Total $ 2,100.00 Mell\'s Capital A/c Particular Amount(Dr) Particular
Amount(Cr) To Loss on realization $ 8,400.00 By Balance B/d $ 11,700.00 To
Cash $ 3,300.00 Total $ 11,700.00 Total $ 11,700.00 Rand\'s Capital A/c
Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $ 10,500.00 By
Balance B/d $ 15,100.00 To Cash $ 4,600.00 Total $ 15,100.00 Total $
15,100.00 3) Cash A/c Particular Amount(Dr) Particular Amount(Cr) To
Balance b/d $ 10,000.00 By Total Liabilities $ 88,000.00 To Other assets $ 85,000.00
By Mell\'s Capital A/c $ 3,300.00 To Sand\'s Capital A/c $ 900.00 By Rand\'s
Capital A/c $ 4,600.00 Total $ 95,900.00 Total $ 95,900.00.
PLEASE ANSWER ALL THE QUESTIONES BELOW AS DETAILED AS POSSIBLE. THAN.pdf
PLEASE ANSWER ALL THE QUESTIONES BELOW AS DETAILED AS POSSIBLE.
THANK YOU!
Why do all (-)RNA virus have an RNA polymerase in its virion?
How does vesicular stomatitis virus regulate its mRNA transcription?
How does vesicular stomatitis virus regulate Protein synthesis?
State numerous factors for Ebola virus being an emerging deadly pathogen for us.
Why can’t influenza virus replicate in a cell where the nucleus has been removed?
PLEASE ANSWER ALL THE QUESTIONES ABOVE AS DETAILED AS POSSIBLE.
THANK YOU!
Solution
1) Negative-sense RNA has a nucleotide sequence complementary to the mRNA that it encodes.
Like DNA, this RNA cannot be translated into protein directly. Instead, it must first be
transcribed into a positive-sense RNA that acts as an mRNA.
In order to get transcribed into a positive-sense RNA these viruses need a RNA Polymerase
enzyme. As soon as the virion enters the host cell the virus uses its own RNA replicase, known
as RNA dependent RNA Polymerase to form positive RNA template strands through
complementary base pairing. This positive sense RNA strand acts as mRNA and can be
translated into structural capsomere proteins and viral RNA polyemrase enzymes for further
synthesis of virions.
2) The RNA-dependent RNA polymerase of vesicular stomatitis virus, a nonsegmented negative-
strand RNA virus, directs two discrete RNA synthetic processes, transcription and replication.
The two short extragenic regions at the genomic termini, the 3 leader (Le) and the complement
of the 5 trailer (TrC), contain essential signals for these processes.
The viral genome comprises a 47-nt leader region (Le), five transcriptional units for the N, P, M
(matrix), G (glycoprotein), and L mRNAs, and a 59-nucleotide trailer region (Tr), arranged in the
order 3 Le-N-P-M-G-L-Tr 5. The 3 Le is thought to act as a promoter for transcription of the
viral mRNAs.
The promoter for mRNA transcription was shown to include specific sequence elements within
Le at positions 19 to 29 and 34 to 46, a separate element at nucleotides 47 to 50, the
nontranscribed leader-N gene junction.
Transcription process
3) Translation in Vesicular Stomatitis Virus
3) Ebola, commonly known as Ebola hemorrhagic fever, is a rare and deadly disease caused by
infection with one of the Ebola virus strains. Ebola can cause disease in humans and nonhuman
primates (monkeys, gorillas, and chimpanzees). Ebola viruses belong to a family of viruses
called Filoviridae, and there are five known species of Ebola virus
5) Influenza A virus belongs to the family of Orthomyxoviridae. Orthomyxoviridae viruses are
one of two RNA viruses that replicate in the nucleus (the other being retroviridae). The reason is
that the machinery of orthomyxo viruses cannot make their own mRNAs. The influenza virus
has a negative sense RNA genome. In the host nucleus, the virus does primary transcription to
produce necessary proteins for replication..
Livingstone is a marble freak and loves to paly at school. During th.pdf
Livingstone is a marble freak and loves to paly at school. During the first break, he doubled the
number of marbles he had. During the second break, he lost 4 marbles. During the third break, he
increased his stock of marbles by one-third. During lunch, however, he lost half of his marbles.
After school, livingston played again and tripled what he had. He went home with 72 marbles.
How much did he start with? work backword to answer this question.
Solution
consider he had x marblles
During the first break, he doubled the number of marbles he had. = 2x
During the second break, he lost 4 marbles =2x-4
During the third break, he increased his stock of marbles by one-third =2x-4 +1/3 (2x-4)
During lunch, however, he lost half of his marbles. = (2x-4 +1/3 (2x-4))/2
After school, livingston played again and tripled what he had = 3(2x-4 +1/3 (2x-4))/2
He went home with 72 marbles
3(2x-4 +1/3 (2x-4))/2 = 72
(3/2)(4/3)(2x-4) = 72
4x-8 =72
4x = 80
x=20
he start with 20 marbles.
Invertebrates colonized terrestrial environments independently of ve.pdf
Invertebrates colonized terrestrial environments independently of vertebrates. However, they
faced similar challenges adapting to land. Compare the adaptations of flatworms, mollusks and
arthropods in regards to: i) water conservation, 2) movement/body support and 3) reproduction
that allowed them to successfully colonize terrestrial environments.
Solution
Free-living flatworms are carnivores and scavengers that eat organic matter,which they acquire
through their pharynx.Tapeworms attach to their host’s intestine and absorb food through the
body wall.Flatworms move by contracting muscles against a hydrostatic skeleton.Flatworms
reproduce asexually by pinching in half;they can also reproduce sexually inside a host by
producing eggs that are dispersed.The flat body shape of a flatworm maximizes the surface area
in contact with the environment.
Mollusks have a mantle that secretes a shell in most species,a muscular foot that provides
movement,a visceral mass containing the internal organs,and a radula.Mollusks feed by filtering
organic particles out of the water;by scraping algae off rocks;or by hunting prey.Mollusks move
by contracting muscles against a hydrostatic skeleton.Many mollusks protect themselves with a
hard shells,others can evade predators by changing color and darting away,often in a cloud of
ink.
Arthropods use their jointed appendages for movement,defense,prey capture,sensing the
environment,ornaments,weapons,and copulation.Arthropods feed on almost every imaginable
inorganic substance,either by chewing or sucking.On land,arthropods exchange gases via tubules
called tracheae or via book lungs;arthropods that live in water respire with gills.Arthropods have
specialized organs that remove nitrogenous wastes,which are deposited into the end of the
digestive tract and ejected though the anus.Arthropods have specialized sense organs that are
sensitive to light,sound,scents,chemicals, potential mates,food proximity,touch,air currents,and
body position.Arthropods move by contracting muscles attached internally to the
exoskeleton.Although there are asexually reproducing species, most arthropods reproduce
sexually,using internal fertilization.The exoskeleton offers some protection against predators, but
many arthropods defend themselves by fleeing,biting,stinging,pinching,making noises,or giving
off foul odors.Camouflage is another defense,as are wing spots that may startle
predators.Aquatic arthropods respire with gills.Terrestrial forms rely on diffusion through tiny
tubes called trachea.Trachea are cuticle-lined air ducts that branch throughout the body,and open
in tiny holes called spiracles,located along the abdomen.Insects can open and close these
spiracles,to conserve water that would otherwise be lost to evaporation from the open tubes.Their
reliance on diffusion for respiration is one of the reasons that insects are small.Being small,an
arthropod has a larger relative surface area than a human, so it loses more water.The exoskele.
Is there a similarity transformation that maps Delta ABC to Delta DBF.pdf
Is there a similarity transformation that maps Delta ABC to Delta DBF? If so, identify the
similarity transformation and write a similarity statement. Explain your answer.
Solution
yes there is a similarity in both the triangle
since all the three sides are in same proportion to the corresponding sides in the other triangle,
therefore these triangle are similar
AC/DF=CB/FE=AB/DE=1/2
SO SSS similarity transformation.
Felix and Janet Green live in Swarthmore, PA. Janets father, Larry,.pdf
Felix and Janet Green live in Swarthmore, PA. Janet\'s father, Larry, lives in Sweden. For each
of the following transactions that occur in their lives, identify whether it is included in the
calculate investment (I), government purchases (G), exports (X), or imports (M). Check all that
apply.
Solution
1. It is part of investment (I)
2. Yes, It is part of imports (M)
3. Yes, It is part of consumption (C)
4. Yes, It is part of consumption (C)
5. Yes, It is part of government spendig (G).
Do the leaves of all plants have an abscission layer If not, what h.pdf
Do the leaves of all plants have an abscission layer? If not, what happens to the leaves on such
plants?
Solution
Plants exhibit the phenomenon of abscission when it no longer needs its parts like leaves, flowers
and fruits. Most of the plants undergo abscission to conserve nutrients and energy. Almost all
plants and trees of deciduous, coniferous and rain forests shed leaves by forming abscission
layer. Abscission layer formation shed the plant part with deposition of secondary structures
made up of parenchyma, suberin and lignin as a protective layer.
Abscission in plants take place under 3 major conditions- 1. to conserve water and
photosynthetic energy; 2. to resist the pests or as part defense mechanism and 3. to improve
photosynthetic efficiency of plants. Generally, the leaves that became older are shed through
abscission. This is because, the older leaves lose chlorophylls and may develop other pigments
like carotenes, xanthophylls and anthocyanins abundantly. If abscission does not take place, the
older leaves remaining on plant may not carry out photosynthesis efficiently as a chlorophyllous
leaf does. Due to this plant may exhibit poor growth and nutrient deficiency and may even die
due to inadequate nutrient synthesis..
Discuss the outcomes and challenges surrounding shortages in various.pdf
Discuss the outcomes and challenges surrounding shortages in various markets. Below are a few
examples from NBC learn. Pick one of these to discuss: Bees Tamiflu Workers Rice
Solution
The report notes that the record population can be viewed as a success because it means people
are living longer — average life expectancy has increased from about 48 years in the early 1950s
to about 68 in the first decade of the 21st century — and more children are surviving worldwide.
But not everyone has benefited from a higher quality of life.
In some of the poorest countries women are having more babies, stymieing development and
perpetuating poverty; in some of the wealthier countries low fertility rates and a shortage of
workers are raising concerns about the sustainability of economic growth and social programs.
World population by the billions
1 billion - 1804
2 billion - 1927
3 billion - 1959
4 billion - 1974
5 billion - 1987
6 billion - 1999*
7 billion - 2011
*Year when the milestone was observed by the United Nations.
Source: United Nations Population Fund
\"This report makes the case that with planning and the right investments in people now — to
empower them to make choices that are not only good for themselves but for our global
commons — our world of 7 billion can have thriving, sustainable cities, productive labor forces
that can fuel economic growth, youth populations that contribute to the well-being of economies
and societies, and a generation of older people who are healthy and actively engaged in the social
and economic affairs of their communities,\" writes Babatunde Osotimehin, executive director of
the UNFPA..
Copy a String in Reverse Order Write a program with a loop and ind.pdf
Copy a String in Reverse Order
Write a program with a loop and indirect addressing that copies a string from source to target,
reversing the character order in the process. Use the following variables:
source BYTE \"This is the source string\",0
target BYTE SIZEOF source DUP(\'#\')
and here is what I have got so far
;reverse
INCLUDE IRVINE32.INC
.DATA
SOURCE byte \"this is the source string\", 0
target BYTE SIZEOF source DUP(\'#\')
.code
main PROC
Call Clrscr
mov esi,0
mov edi,SIZEOF source-TYPE source
mov ecx,LENGTHOF source
L1: mov al, source[edi]
mov target[esi]
sub edi TYPE source
loop L1
exit
main ENDP
END main
Solution
program with a loop and indirect addressing that copies a string from source to target
MOV SI, source ; Get source address
MOV DI, (target + SIZEOF source) ; Get the ending address for target
LOOP:
MOV AL, [SI] ; Get a byte
MOV [DI], AL ; Store a byte
INC SI ; Move forward one in source
DEC DI ; Move back one in target
CMP AL, 0 ; Have we reached the zero termination?
JNZ LOOP
RET.
Compare and contrast prokaryotic cells and eukaryotic cells. Compare.pdf
Compare and contrast prokaryotic cells and eukaryotic cells. Compare and contrast passive
transport and active transport. Compare and contrast animal cells and plant cells. Summarize
the endosymbiosis theory including the evidence that supports it.
Solution
1. Similarities between prokaryotic cells and eukaryotic cells:
Differences between Prokaryotic cells and Eukaryotic cells:
2. Similarities between active and passive Transport:
Differentiate between Active and Passive Transport:Prokaryotic CellsEukaryotic cellsCells
don\'t have nucleus.Cells have nucleus.They don\'t have organelles.They have organelles.They
all are unicellular.They may be unicellular or multicellular..
Case Study 1Hotel worker Danny Ruiz was living with his wife and f.pdf
Case Study 1
Hotel worker Danny Ruiz was living with his wife and four children in a cramped New York
apartment when he saw a television ad promising the family a way out. “Why rent when you can
own your own home?” Penn- sylvania builder Gene Percudani asked. The company even offered
to pay his rent for a year, while he saved for a down payment. So the Ruiz family fled the cityhe
Pocono Mountains, where they bought a three- bedroom Cape Cod home for $171,000.
However, when they tried to refinance less than two years later, the home was valued at just
$125,000. “I just about flipped,” said Mr. Ruiz. Later Mrs. Ruiz remarked about her husband,
“He went nuts.” Percudani, a 51-year-old native of Queens, New York, built a thriv- ing
homebuilding business in this market, running folksy television ads offering New Yorkers new
homes in Pennsylvania. If they joined Percudani’s program, called “Why Rent,” homeowners
would find financing through another of his companies, Chapel Creek Mort- gage, which
brokered loans from J. P. Morgan Chase and the company’s Chase Manhattan Mortgage unit.
For years, the “Why Rent” program appealed to workers with modest salaries, such as Eberht
Rios, a truck driver for UPS. Rios bought a home in the Poco- nos for $140,000. This year, when
he tried to refinance, he was told the home was valued at only $100,000. One local appraiser,
Dominick Stranieri, signed off on most of the “Why Rent” deals that state officials now say were
overpriced, including the Rios and Ruiz homes. Percudani’s firm picked Stranieri as his
appraiser because of his quick work and low fee of $250, instead of the typical $300 to $400. In
exchange for a steady stream of work, Mr. Stranieri accepted without ques- tion valuations from
Percudani’s company. Other common methods of creating revenues include investors and others
buying distressed proper- ties and then, using inflated appraisals, selling them for a big profit. In
order to secure the efforts of a “dirty appraiser,” those involved with the fraud would pay up to
$1,500 under the table on top of the appraiser’s stan- dard fee of $400. Another unique twist to
the plot is that few of the people involved in making mortgage loans have a long- term interest in
them. Traditionally, bankers made loans directly and held them, giving the lenders a strong
incentive to find fair appraisals to protect their interest. Today, however, many appraisers are
picked by independent mortgage brokers, who are paid per transaction and have little stake in the
long-term health of the loans. Many lenders have also lost a long-term interest in their loans,
because they sell them off to investors. Appraisers increasingly fear that if they don’t go along
with higher valuations sought by bro- kers, their business will dry up. Do you think a county
appraiser would do a lot bet- ter than a private practitioner? Joel Marcus, a New York–based
attorney recently had his property valued at $2.2 million by a county appraiser, up fro.
Assignment Submission Assignment, you submit answers by question by .pdf
Assignment Submission Assignment, you submit answers by question by question parts. The
number of submissions remaining for each question part only changes if you submit or change
the answer. Assignment Scoring Assignment submission is used for your score. The complex
number z = -1 + i in polar form is z =. The complex number z = 2(cos(pi/6) + i sin(pi/6)) in
rectangular form is z = ___________. The complex number graphed below can be expressed in
rectangular form as z = or in polar form as z = _________.
Solution
a)given z =-1+ 1i
r =[(-1)2+12]
r=2
=tan-1(1/-1)
=-tan-1(1/1)
=-(/4)
=(3/4)
z=2 [cos(3/4) +i sin(3/4)]
====================================
b)z=2 [cos(/6) +i sin(/6)]
z=2 [(3 /2) +i(1/2)]
z=3 +i1
z=3 +i
c) z =(1+ 1i)
z=2 [cos(/4) +i sin(/4)].
A linear dynamical system can be created for two masses connected by.pdf
A linear dynamical system can be created for two masses connected by springs between one
another, and connected to opposing walls. The state vector at time t is a 4?×?1 vector consisting
of the displacement and velocity of each of the two masses. The transition matrix for this
dynamical system is given by the following matrix.
[?i
i
? 30
? ? 30]
Solution.
A 7 year old female with a history of previous UTI’s suddenly develo.pdf
A 7 year old female with a history of previous UTI’s suddenly develops painful, dark-orange
colored urine, with fever and back pain. Her urinalysis reveals dipstick tests that are positive for
protein (2+), blood (1+), leukocyte esterase (4+), and nitrate (4+). The microscopic examination
yields 20-25 WBC/HPF, 5-10 RBC/HPF, 5-10 WBC Casts/LPF, and numerous bacteria. The
patient has been taking Tylenol and vitamin C supplements daily for the past 24 hours, since the
condition began.
Based on the results 1) what diagnosis/condition would you surmise at this point? 2) what
dipstick/microscopic test support this? 3) is the patient’s history significant? 4) can the
medication or vitamin supplement have any impact on the test results?
Solution
Women often experience urinary tract infections that are often marked by pain in the abdomen or
lower back. The urine is dark-orange colored due to blood mixing with dark yellow urine. The
infection may be due to bacteria E. coli, (MayoClinic).
Leukocyte esterase test dipstick that indicates; the presence of white blood cells either as whole
cells or as lysed cells; supports this assumption and the microscopic examination yielding 20-25
WBC/HPF also backs this assumption.
Yes with all the medical history and the details of the tests done; one can proceed with the
treatment.
Yes according to researches; Vitamin C spiking of urine has shown to demonstrate false-negative
results at various concentrations, (Ko DH, 2015) in case of Dipstick Test Results..
5. Toxco, Inc. emits a noxious gaseous substance as a waste product o.pdf
5. Toxco, Inc. emits a noxious gaseous substance as a waste product of its operations. As part of
its manufacturing process, it also generates liquid solvent wastes such as toluene. Analyze which
if any, of these materials are regulated under RCRA. If there is any additional information that
would be needed to answer this question, identify what that is. (15 points)
Solution
Following the discovery of an illegal meth lab by law enforcement, officers collect
samples of the chemicals used in that lab as evidence for prosecution. After all forensic samples
are gathered, a Hazardous Materials (Hazmat) Team is summoned to clean up the remaining
chemicals and other by-products or wastes. Estimates indicate that for each pound of meth
produced, between five and six pounds of highly toxic waste is generated.1
The by-products or
wastes removed by the Hazmat Teams are generally treated as RCRA hazardous waste without
further analysis, and are sent to a RCRA permitted site for treatment and disposal.
State and Federal officials also have found that during the operation of the meth lab, the
residue from meth production may enter porous building materials such as wallboard, carpets
and furnishings.2
In addition to the potential contamination of the building and its contents, there
may be further contamination from the improper disposal of these by-products or wastes by the
producers of the meth, which is typically poured into the indoor plumbing drains that flow either
into the city sewer or individual sewage treatment systems (e.g., septic systems), or modified to
allow drainage directly onto the soil near the meth lab. The other disposal method typically seen
at functioning meth labs is a burn or burial pit. The primary environmental concern from
drainage of these materials into the soil or from disposal into a burn or burial pit is
contamination
of ground water by the chemicals, including volatile organic chemicals used in the production of
meth, as well as vapor intrusion which involves soil vapor gas from ground water plumes
infiltrating, and accumulating in, nearby buildings (residences and nonresidential sites) at
dangerous levels.3
In most States, the property owner is responsible for decontamination of the property. In
some States, the health department or another agency may evaluate the property for risks from
long-term exposure to residual chemicals, including the potential environmental impacts of
chemical spills into the environment or improper waste disposal.
V. How Does RCRA Regulate Hazardous Waste?
The determination of whether a particular waste is a Subtitle C RCRA hazardous waste is
a multi-step process that depends on a number of factors, including the chemical character and
composition of the waste. The regulations for making the hazardous waste determination are
found in 40 CFR part 261.4
The first step is to determine whether the waste in question is a solid waste as defined
under RCRA. The RCRA regulations require this determinatio.
21. Two fair coins are tossed. Let E be the event not more than one h.pdf
21. Two fair coins are tossed. Let E be the event not more than one head and F the event at least
one of each face. Define the sample space S = {HH, HT, TH, TT) and assign probability ¼ to
each simple event. a. Determine the P(E) b. Find the P(F) c. Find the P(E intersection F) d. Are E
and F independent or dependent? 23. Three fair coins are tossed. Let E be the event not more
than one head and F the event at least one of each face. Define the sample space S = {HHH,
HHT, ..., TTT} and assign probability 1/8 to each simple event. a. Determine the P(E) b. Find the
P(F) c. Find the P(E intersection F) d. Are E and F independent or dependent?
Solution
21.
a)
E be the Event = { Not more than one Head } = {(H,T) (T,H) (T,T) }
b)
F be the Event = { atleast one of each face } = { (H,T) (T,H)} =
Sample Space = { (H,T) (T,H) (T,T) (H,H) }
c.
P( E n F) = {(H,T) (T,H)}
d.
Dependent
Note : We are suppose to answer only 1 question in time.
More from arihantsherwani (20)
The age of the Earth is estimated from the amount of uranium isotope .pdf
The age of the Earth is estimated from the amount of uranium isotope .pdf
Suppose Michael creates an RSA cryptosystem with a very large modulu.pdf
Suppose Michael creates an RSA cryptosystem with a very large modulu.pdf
Match the terms to the definitions provided.Question 2 optionsd.pdf
Match the terms to the definitions provided.Question 2 optionsd.pdf
Simple linear regression and multiple regression are both based on t.pdf
Simple linear regression and multiple regression are both based on t.pdf
Sand, Mell, and Rand are partners who share incomes and losses in a .pdf
Sand, Mell, and Rand are partners who share incomes and losses in a .pdf
PLEASE ANSWER ALL THE QUESTIONES BELOW AS DETAILED AS POSSIBLE. THAN.pdf
PLEASE ANSWER ALL THE QUESTIONES BELOW AS DETAILED AS POSSIBLE. THAN.pdf
Livingstone is a marble freak and loves to paly at school. During th.pdf
Livingstone is a marble freak and loves to paly at school. During th.pdf
Invertebrates colonized terrestrial environments independently of ve.pdf
Invertebrates colonized terrestrial environments independently of ve.pdf
Is there a similarity transformation that maps Delta ABC to Delta DBF.pdf
Is there a similarity transformation that maps Delta ABC to Delta DBF.pdf
Felix and Janet Green live in Swarthmore, PA. Janets father, Larry,.pdf
Felix and Janet Green live in Swarthmore, PA. Janets father, Larry,.pdf
Do the leaves of all plants have an abscission layer If not, what h.pdf
Do the leaves of all plants have an abscission layer If not, what h.pdf
Discuss the outcomes and challenges surrounding shortages in various.pdf
Discuss the outcomes and challenges surrounding shortages in various.pdf
Copy a String in Reverse Order Write a program with a loop and ind.pdf
Copy a String in Reverse Order Write a program with a loop and ind.pdf
Compare and contrast prokaryotic cells and eukaryotic cells. Compare.pdf
Compare and contrast prokaryotic cells and eukaryotic cells. Compare.pdf
Case Study 1Hotel worker Danny Ruiz was living with his wife and f.pdf
Case Study 1Hotel worker Danny Ruiz was living with his wife and f.pdf
Assignment Submission Assignment, you submit answers by question by .pdf
Assignment Submission Assignment, you submit answers by question by .pdf
A linear dynamical system can be created for two masses connected by.pdf
A linear dynamical system can be created for two masses connected by.pdf
A 7 year old female with a history of previous UTI’s suddenly develo.pdf
A 7 year old female with a history of previous UTI’s suddenly develo.pdf
5. Toxco, Inc. emits a noxious gaseous substance as a waste product o.pdf
5. Toxco, Inc. emits a noxious gaseous substance as a waste product o.pdf
21. Two fair coins are tossed. Let E be the event not more than one h.pdf
21. Two fair coins are tossed. Let E be the event not more than one h.pdf
Recently uploaded
clinical examination of hip joint (1).pdf
described clinical examination all orthopeadic conditions .
South African Journal of Science: Writing with integrity workshop (2024)
South African Journal of Science: Writing with integrity workshop (2024)Academy of Science of South Africa
A workshop hosted by the South African Journal of Science aimed at postgraduate students and early career researchers with little or no experience in writing and publishing journal articles.Executive Directors Chat Leveraging AI for Diversity, Equity, and Inclusion
Let’s explore the intersection of technology and equity in the final session of our DEI series. Discover how AI tools, like ChatGPT, can be used to support and enhance your nonprofit's DEI initiatives. Participants will gain insights into practical AI applications and get tips for leveraging technology to advance their DEI goals.
How to Manage Your Lost Opportunities in Odoo 17 CRM
Odoo 17 CRM allows us to track why we lose sales opportunities with "Lost Reasons." This helps analyze our sales process and identify areas for improvement. Here's how to configure lost reasons in Odoo 17 CRM
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...National Information Standards Organization (NISO)
This presentation was provided by Steph Pollock of The American Psychological Association’s Journals Program, and Damita Snow, of The American Society of Civil Engineers (ASCE), for the initial session of NISO's 2024 Training Series "DEIA in the Scholarly Landscape." Session One: 'Setting Expectations: a DEIA Primer,' was held June 6, 2024.The History of Stoke Newington Street Names
Presented at the Stoke Newington Literary Festival on 9th June 2024
www.StokeNewingtonHistory.com
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Aberdeen
Community pharmacy- Social and preventive pharmacy UNIT 5
Covered community pharmacy topic of the subject Social and preventive pharmacy for Diploma and Bachelor of pharmacy
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
His expertise extends across a diverse spectrum of reporting, database, and web development applications, underpinned by an exceptional grasp of data storage and virtualization technologies. His proficiency in application testing, database administration, and data cleansing ensures seamless execution of complex projects.
What sets Denis apart is his comprehensive understanding of Business and Systems Analysis technologies, honed through involvement in all phases of the Software Development Lifecycle (SDLC). From meticulous requirements gathering to precise analysis, innovative design, rigorous development, thorough testing, and successful implementation, he has consistently delivered exceptional results.
Throughout his career, he has taken on multifaceted roles, from leading technical project management teams to owning solutions that drive operational excellence. His conscientious and proactive approach is unwavering, whether he is working independently or collaboratively within a team. His ability to connect with colleagues on a personal level underscores his commitment to fostering a harmonious and productive workplace environment.
Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
-------------------------------------------------------------------------------
Find out more about ISO training and certification services
Training: ISO/IEC 27001 Information Security Management System - EN | PECB
ISO/IEC 42001 Artificial Intelligence Management System - EN | PECB
General Data Protection Regulation (GDPR) - Training Courses - EN | PECB
Webinars: https://pecb.com/webinars
Article: https://pecb.com/article
-------------------------------------------------------------------------------
For more information about PECB:
Website: https://pecb.com/
LinkedIn: https://www.linkedin.com/company/pecb/
Facebook: https://www.facebook.com/PECBInternational/
Slideshare: http://www.slideshare.net/PECBCERTIFICATION
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptx
How to Make a Field Mandatory in Odoo 17
In Odoo, making a field required can be done through both Python code and XML views. When you set the required attribute to True in Python code, it makes the field required across all views where it's used. Conversely, when you set the required attribute in XML views, it makes the field required only in the context of that particular view.
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Your Skill Boost Masterclass: Strategies for Effective UpskillingExcellence Foundation for South Sudan
Strategies for Effective Upskilling is a presentation by Chinwendu Peace in a Your Skill Boost Masterclass organisation by the Excellence Foundation for South Sudan on 08th and 09th June 2024 from 1 PM to 3 PM on each day.The simplified electron and muon model, Oscillating Spacetime: The Foundation...
Discover the Simplified Electron and Muon Model: A New Wave-Based Approach to Understanding Particles delves into a groundbreaking theory that presents electrons and muons as rotating soliton waves within oscillating spacetime. Geared towards students, researchers, and science buffs, this book breaks down complex ideas into simple explanations. It covers topics such as electron waves, temporal dynamics, and the implications of this model on particle physics. With clear illustrations and easy-to-follow explanations, readers will gain a new outlook on the universe's fundamental nature.
The Diamonds of 2023-2024 in the IGRA collection
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
How to Add Chatter in the odoo 17 ERP Module
In Odoo, the chatter is like a chat tool that helps you work together on records. You can leave notes and track things, making it easier to talk with your team and partners. Inside chatter, all communication history, activity, and changes will be displayed.
Recently uploaded (20)
South African Journal of Science: Writing with integrity workshop (2024)
South African Journal of Science: Writing with integrity workshop (2024)
Executive Directors Chat Leveraging AI for Diversity, Equity, and Inclusion
Executive Directors Chat Leveraging AI for Diversity, Equity, and Inclusion
How to Manage Your Lost Opportunities in Odoo 17 CRM
How to Manage Your Lost Opportunities in Odoo 17 CRM
Liberal Approach to the Study of Indian Politics.pdf
Liberal Approach to the Study of Indian Politics.pdf
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
What is Digital Literacy? A guest blog from Andy McLaughlin, University of Ab...
Community pharmacy- Social and preventive pharmacy UNIT 5
Community pharmacy- Social and preventive pharmacy UNIT 5
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Pengantar Penggunaan Flutter - Dart programming language1.pptx
Digital Artefact 1 - Tiny Home Environmental Design
Digital Artefact 1 - Tiny Home Environmental Design
Film vocab for eal 3 students: Australia the movie
Film vocab for eal 3 students: Australia the movie
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Your Skill Boost Masterclass: Strategies for Effective Upskilling
The simplified electron and muon model, Oscillating Spacetime: The Foundation...
The simplified electron and muon model, Oscillating Spacetime: The Foundation...
Create a C program that implements The Game of Life cellular auto.pdf
- 1. Create a C program that implements "The Game of Life" cellular automata (CA) simulation. A few comments: You will want 2 two-dimensional arrays. One that holds the current state of the "universe" and second that is generated from the first using the Game of Life rules, which are given below. After the computations for filling the second array are complete, the contents of the second array can be copied back to the first. You can make the array as big as you want, but you should probably use 20 times 20 as a minimum. As discussed in class, the outside boundary of the "universe" can be a bit tricky. To keep it simple for this exercise, you can assume that the outside boundary cells are always "dead". You can use whatever character (or number) that you want to represent the "living" cells. The dead cells should probably be spaces. The rules for whether a particular cell lives or dies are determined by the 8 surrounding cells. If a particular cell is alive, then: If only 0 or 1 of the surrounding cells are alive, the cell dies. (It's lonely.) If 4 or more of the surrounding cells are alive, the cell dies. (Too much competition.) If 2 or 3 of the surrounds cells are alive, the cell stays alive. If a particular cell is dead, then it comes to life if exactly 3 of the surround cells are alive. You will have to pick some starting configuration for the universe. An easy thing to do is fill the array randomly, and then let things go and see what evolves. Write your program as described above and demo the operation to your TA. Then try some modifications. First, try using a different starting configurations. Look here for some ideas:. Gosper's Glider Gun is a famous one. Or try a simple spaceship or oscillator. Then try changing the rules determining life and death. A common variation is to include a "birth" when 6 neighboring cells are alive in addition to the "3-cell" birth. Finally, there will be a short quiz covering 2D arrays. This exercise should be very familiar. Solution #include #include #include #include #include #include #include #define ROWS 60 #define COLS 60 #define OFF 0 #define ON 1 #define BLACK 8, 0, 0, 0, 0 #define scr_width 600 #define scr_height 600 #define cell_width (scr_width / ROWS) #define cell_height (scr_height / COLS) We must hold two copies of the board so that we can analyze board and make * changes to temp when killing/creating cells so that we don't cause changes * to the board to affect the following cells. */ char board[ROWS][COLS]; char temp[ROWS][COLS]; SDL_Rect cells[ROWS][COLS]; /* Stores positions of each cell for blits. */ void randomize_board(void); void initialize_grid(SDL_Surface* screen); void blit_board(SDL_Surface* bcell, SDL_Surface* screen); int num_neighbours(int x, int y); void update_board(void); int clear_board(SDL_Surface* screen, Uint32 color); int clear_cell(SDL_Surface* screen, int x, int y, Uint32 color); void initialize_cells_array(void); int main(void) { SDL_Init(SDL_INIT_VIDEO); SDL_WM_SetCaption("Conway's Game of Life - Ryuurei", NULL); SDL_Event event; int breaker = 0; int paused = 1;
- 2. SDL_Surface* screen = SDL_SetVideoMode( scr_width, scr_height, 0, SDL_SWSURFACE); if (! screen) { perror("SDL_SetVideoMode"); return EXIT_FAILURE; } Uint32 bgcolor = SDL_MapRGB(screen->format, 0xFF, 0xFF, 0xFF); if (SDL_FillRect(screen, &(screen->clip_rect), bgcolor) == -1) { perror("SDL_FillRect"); return EXIT_FAILURE; } initialize_cells_array(); initialize_grid(screen); SDL_Surface* bcell = SDL_CreateRGBSurface( SDL_SWSURFACE, cell_width, cell_height, BLACK); if (! bcell) { perror("BLACK SDL_CreateRGBSurface"); return EXIT_FAILURE; } SDL_Flip(screen); while (1) { /* Infinite game loop. */ /** The event loop handles any events in the poll/queue * even while paused. */ while (SDL_PollEvent(&event)) { switch (event.type) { case SDL_QUIT: breaker = 1; break; case SDL_KEYDOWN: /* Pause on enter/return key press. */ if (event.key.keysym.sym == SDLK_KP_ENTER || event.key.keysym.sym == SDLK_RETURN) { paused = !paused; /* Space causes the board to be randomized. */ } else if (event.key.keysym.sym == SDLK_SPACE) { SDL_FillRect(screen, &(screen->clip_rect), bgcolor); randomize_board(); /* Escape button also quits. */ } else if (event.key.keysym.sym == SDLK_ESCAPE) { breaker = 1; break; /* Either shift button clears the screen. */ } else if (event.key.keysym.sym == SDLK_RSHIFT || event.key.keysym.sym == SDLK_LSHIFT) { if (clear_board(screen, bgcolor) == -1) { perror("clear_board"); return EXIT_FAILURE; } } break; case SDL_MOUSEBUTTONDOWN: /** The only mouse event we handle is the left click. * If the cell being clicked is alive, kill it. * Otherwise, give it life. */ if (event.button.button == SDL_BUTTON_LEFT) { if (event.button.y > (ROWS * cell_height)) break; int tx = floor(event.button.x / cell_width); int ty = floor(event.button.y / cell_height); if (board[tx][ty] == OFF) { board[tx][ty] = ON; temp[tx][ty] = ON; SDL_BlitSurface( bcell, &(bcell- >clip_rect), screen, &cells[tx][ty]); } else if (clear_cell(screen, tx, ty, bgcolor) == -1) { perror("clear_cell"); return EXIT_FAILURE; } } break; default: continue; } if (breaker) break; } if (breaker) break; /* Redraw the grid so it's not painted over by clear_cell. */ initialize_grid(screen); blit_board(bcell, screen); SDL_Flip(screen); if (!paused) { update_board(); SDL_Delay(250); SDL_FillRect(screen, &(screen-
- 3. >clip_rect), bgcolor); } } return EXIT_SUCCESS; } void randomize_board(void) { /* Clear the board then randomly set ~1/5 of them to ON. */ memset((void *)board, OFF, ROWS * COLS); for (int y = 0; y < ROWS; y++) { for (int x = 0; x < COLS; x++) { if (rand() % 5 == 0) { board[x][y] = ON; } temp[x][y] = board[x][y]; } } } void initialize_grid(SDL_Surface* screen) { /** Position horizontal and veritcal lines to form a grid to * make it easier to count cell spaces when trying to draw things. */ SDL_Surface* linex = SDL_CreateRGBSurface( /* Vertical lines */ SDL_SWSURFACE, 1, scr_height, BLACK); SDL_Surface* liney = SDL_CreateRGBSurface( /* Horizontal lines */ SDL_SWSURFACE, scr_width, 1, BLACK); SDL_Rect pos_x; SDL_Rect pos_y; pos_x.y = pos_y.x = 0; for (int i = 0; i < scr_width / (cell_width); i++) { pos_x.x = cell_width + cell_width * i; SDL_BlitSurface(linex, &(linex->clip_rect), screen, &pos_x); } for (int i = 0; i < scr_height / (cell_height); i++) { pos_y.y = cell_height + cell_height * i; SDL_BlitSurface(liney, &(liney->clip_rect), screen, &pos_y); } } void blit_board(SDL_Surface* bcell, SDL_Surface* screen) { /* Blit all the live cells onto the board. */ for (int y = 0; y < ROWS; y++) { for (int x = 0; x < COLS; x++) { if (board[x][y] == ON) { SDL_BlitSurface( bcell, &(bcell->clip_rect), screen, &cells[x][y]); } } } } int num_neighbours(int x, int y) { /** Count the number of live cells all around the cell given by * the position (x, y). Of course, this includes all diagonal, vertical, * and horizontal living cells. The program works using a toroidal array, * thus we must check to see if it is necessary to wrap to the other side * of the array in order to find cells that must be taken into account. */ int num_adj = 0; int tmpy = y; int tmpx = x; if (y-1 < 0) tmpy = ROWS - 1; else tmpy = y - 1; if (board[x][tmpy] == ON) num_adj++; if (y+1 >= ROWS) tmpy = 0; else tmpy = y + 1; if (board[x][tmpy] == ON) num_adj++; if (x-1 < 0) tmpx = COLS - 1; else tmpx = x - 1; if (board[tmpx][y] == ON) num_adj++; if (x+1 >= COLS) tmpx = 0; else tmpx = x + 1; if (board[tmpx][y] == ON) num_adj++; if (y-1 < 0) tmpy = ROWS - 1; else tmpy = y - 1; if (x-1 < 0) tmpx = COLS - 1; else tmpx = x - 1; if (board[tmpx][tmpy] == ON) num_adj++; if (x+1 >= COLS) tmpx = 0; else tmpx = x + 1; if (board[tmpx][tmpy] == ON) num_adj++; if (y+1 >= ROWS) tmpy = 0; else tmpy = y + 1; if (x+1 >= COLS) tmpx = 0; else tmpx = x + 1; if (board[tmpx][tmpy] == ON) num_adj++; if (x-1 < 0) tmpx = COLS - 1; else tmpx = x - 1; if (board[tmpx][tmpy] == ON) num_adj++; return num_adj; } void update_board(void) { /** Detirmine which cells live and which die. Operate on temp so that * each change does not affect following changes, then copy temp into * the main board to be displayed. */ int neighbours = 0; for (int y = 0; y < ROWS; y++) { for (int x = 0; x
- 4. < COLS; x++) { neighbours = num_neighbours(x, y); if (neighbours < 2 && board[x][y] == ON) { temp[x][y] = OFF; /* Dies by underpopulation. */ } else if (neighbours > 3 && board[x][y] == ON) { temp[x][y] = OFF; /* Dies by overpopulation. */ } else if (neighbours == 3 && board[x][y] == OFF) { temp[x][y] = ON; /* Become alive because of reproduction. */ } /* Otherwise the cell lives with just the right company. */ } } for (int y = 0; y < ROWS; y++) { for (int x = 0; x < COLS; x++) { board[x][y] = temp[x][y]; } } } int clear_board(SDL_Surface* screen, Uint32 color) { /* Simply set both the main and temporary boards to completely off. */ for (int y = 0; y < ROWS; y++) { for (int x = 0; x < COLS; x++) { board[x][y] = OFF; temp[x][y] = OFF; } } return SDL_FillRect(screen, &(screen->clip_rect), color); } int clear_cell(SDL_Surface* screen, int x, int y, Uint32 color) { /* Clear a cell by coloring it white and setting the board pos. to off. */ SDL_Rect rect; rect.x = cell_width * x; rect.y = cell_height * y; board[x][y] = OFF; temp[x][y] = OFF; return SDL_FillRect(screen, &rect, color); } void initialize_cells_array(void) { /** Initialize the array of SDL_Rect structs that store the cell * coordinates by iteratively calculating their successive positions. */ for (int y = 0; y < ROWS; y++) { for (int x = 0; x < COLS; x++) { (cells[x][y]).x = (cell_width * x); (cells[x][y]).y = (cell_height * y); } } }