Need help writing Conways Game of Life. These are the instructions.pdf

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Need help writing Conway\'s Game of Life. These are the instructions: The game of Life is intended to model life in a society of organisms. Consider a rectangular array of cells, each of which may contain an organism. If the array is considered to extend indefinitely in both directions, then each cell has eight neighbors, the eight cells surrounding it. Births and deaths occur according to the following rules: An organism is born in any empty cell having exactly three neighbors. An organism dies from isolation if it has fewer than two neighbors. An organism dies from overcrowding if it has more than three neighbors. All other organisms survive to the next generation. Write a program to play the game of Life and investigate the patterns produced by various initial configurations. Some configurations die off rather rapidly; others repeat after a certain number of generations; others change shape and size and may move across the array. NOTE: For representing each organism, use \"@\" characters. For an array of cells, use 30x30 two-dimensional array. Each generation should be displayed on the screen and should be paused. (You can use cin.get( ) library function to pause running your program.) 4. The source file should be called life.cpp 5. All assignments are expected to be INDIVIDUAL work. All work handed in must be original. Duplicate or very similar programs receive zero points. Input A set of initial cells will be given using the interactive way using a keyboard. The followings are the prompt for the input: Please Enter the number of initial cells: 4 The position of cell is 10 10 The position of cell is 10 11 The position of cell is 10 12 The position of cell is 11 11 How many generations do you want to display? 3 Output The program then display each generation of organisms at a time. The 1-generation @@@ @ Press any key to continue!! (This picture should be displayed on the new screen.) The 2-generation @ @@@ @@@ Press any key to continue!! The 3-generation @@@ @ @ @ Do you want to do it again?(Yes/No) N Solution Note: User given template is used. Answer: #include #include #include #include #include using namespace std; int orAry[30][30]; void initilaize() { for(int kk=0;kk<30;kk++) { for(int bb=0;bb<30;bb++) { orAry[kk][bb]=0; } } } void disply() { cout<=0 ) { cc++; } if( orAry[kk+1][aa-1]==1 && kk+1<30 && aa-1>=0) { cc++; } if( orAry[kk+1][aa]==1 && kk+1<30) { cc++; } if( orAry[kk][aa+1]==1 && aa+1<30) { cc++; } if( orAry[kk-1][aa-1]==1 && kk-1>=0 && aa-1>=0) { cc++; } if(orAry[kk+1][aa+1]==1 && kk+1<30 && aa+1<30) { cc++; } if(orAry[kk-1][aa]==1 && kk-1>=0) { cc++; } if( orAry[kk-1][aa+1]==1 && kk-1>=0 && aa+1<30) { cc++; } return cc; } void runLife() { int bCnt=0; int dCnt=0; for(int kk=0;kk<30;kk++) { for(int bb=0;bb<30;bb++) { if(orAry[kk][bb]==0) { bCnt=ChechNeighBourCount(kk,bb); if(bCnt==3) { orAry[kk][bb]=1; } } else if(orAry[kk][bb]==1) { dCnt=ChechNeighBourCount(kk,bb); if(dCnt>3) { orAry[kk][bb]=0; } if(dCnt<2) { orAry[kk][bb]=0.

Education

(This picture should be displayed on the new screen.)
The 2-generation
@
@@@
@@@
Press any key to continue!!
The 3-generation
@@@
@ @
@
Do you want to do it again?(Yes/No) N
Solution
Note: User given template is used.
Answer:
#include
#include
#include
#include
#include
using namespace std;
int orAry[30][30];
void initilaize()
{
for(int kk=0;kk<30;kk++)
{
for(int bb=0;bb<30;bb++)
{
orAry[kk][bb]=0;
}
}
}
void disply()
{

cout<=0 )
{
cc++;
}
if( orAry[kk+1][aa-1]==1 && kk+1<30 && aa-1>=0)
{
cc++;
}
if( orAry[kk+1][aa]==1 && kk+1<30)
{
cc++;
}
if( orAry[kk][aa+1]==1 && aa+1<30)
{
cc++;
}
if( orAry[kk-1][aa-1]==1 && kk-1>=0 && aa-1>=0)
{
cc++;
}
if(orAry[kk+1][aa+1]==1 && kk+1<30 && aa+1<30)
{
cc++;
}
if(orAry[kk-1][aa]==1 && kk-1>=0)
{
cc++;
}
if( orAry[kk-1][aa+1]==1 && kk-1>=0 && aa+1<30)
{
cc++;
}
return cc;
}
void runLife()

{
int bCnt=0;
int dCnt=0;
for(int kk=0;kk<30;kk++)
{
for(int bb=0;bb<30;bb++)
{
if(orAry[kk][bb]==0)
{
bCnt=ChechNeighBourCount(kk,bb);
if(bCnt==3)
{
orAry[kk][bb]=1;
}
}
else if(orAry[kk][bb]==1)
{
dCnt=ChechNeighBourCount(kk,bb);
if(dCnt>3)
{
orAry[kk][bb]=0;
}
if(dCnt<2)
{
orAry[kk][bb]=0;
}
}
}
}
}
int main()
{
int rr,cc;

int npos=0;
int ngen=0;
int xv,yv;
char re;
srand(time(NULL));
do
{
initilaize();
cout<<"Please Enter the number of initial cells:";
cin>>npos;
for(int kk=0;kk>xv>>yv;
if(orAry[xv][yv]==0)
{
orAry[xv][yv]=1;
}
}
cout<<"How many generations do you want to display?";
cin>>ngen;
for(int kk=0;kk>re;
}while(re!='N');
return 0;
}
Sample Output:
Please Enter the number of initial cells:4
The position of cell is
10 10
The position of cell is
10 11
The position of cell is
10 12
The position of cell is
11 11
How many generations do you want to display?3
1 - generation

@@@
@
Press any key to continue
2 - generation

@@
@ @
@
Press any key to continue
3 - generation

@@
@ @
@
Press any key to continue
Do you want to do it again?(Yes/No) N
Exit code: 0 (normal program termination)

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## Part B: Conway's Game of Life ### Introduction John Conway's Game of Life is a well known "game" with a very simple set of rules that can simulate incredibly sophisticated behavior (as sophisticated, in fact, as any computer program). Its setting -- which we'll refer to as the game's "world" -- is a two-dimensional grid of cells, each one either living or dead, and considered to be in contact with its eight neighbors (i.e., orthogonally or diagonally adjacent cells). Given the state of each cell in the world at time $t_n$, we can compute the state of each cell at time $t_{n+1}$ (i.e., one generation later) using the following rules: - **Birth rule**: a dead cell with exactly three living neighbors comes alive. - **Survival rule**: a living cell with two to three neighbors remains alive. - **Death rule**: a living cell with zero or one neighbors dies in isolation; a living cell with four or more neighbors dies of overcrowding The image below depicts a world before and after the rules are applied to a group of cells. <img src="images/game-of-life-1.png" alt="Game of Life example" style="display: block; margin: 0 auto; width:650px;"/> Many living cell patterns have been discovered that result in stable/non-changing shapes and repeatedly oscillating shapes, among other types (see the [example patterns in the Game of Life Wikipedia article](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life#Examples_of_patterns) for some). An interesting fact about the Game of Life is that there is no algorithm which can reliably predict whether or not a given pattern will appear in the world given some starting state (this is related to the halting problem, covered in CS 330). Since each world state (after the initial state) is a pure function of the preceding one, and the game world can be easily modeled using lists, implementing the Game of Life makes for another fun list processing exercise! ### Implementation in Haskell We will represent the game world using the type: `((Int,Int), [[Bool]])`. I.e., as a tuple where the first element is a tuple of two integers $(w,h)$ representing the width and height of the world, and the second element is a list of lists of Booleans representing the 2D grid, where each sublist of length $w$ represents the cells in a row of the world, with $h$ sublists total. A barren (i.e., containing only dead cells) world can be created with the following function, which takes as input the width and height: makeWorld :: (Int,Int) -> ((Int,Int), [[Bool]]) makeWorld dims@(w,h) = (dims, replicate h (replicate w False)) In "src/MP2b.hs" you will find declarations for functions you must complete to implement the Game of Life. The first two functions compute the number of live neighbors of a cell and the next world state: liveNeighbors :: ((Int,Int), [[Bool]]) -- world -> (Int,Int) -- cell -> Int -- number of living neighbors nextWorld :: ((Int,Int), [[Bool]]) -- world -> ((Int,Int), [[Bool]]) -- next world (one generation later) To animat.

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I need help with these two questions 4.12 and 4.13 dealing with MATLAB. Thank you in advance. Solution clear all; bigBal=2*10^5; endBal=bigBal; i=1; payment=1199.11; while endBal>0 A(i)=endBal*(1+0.5/100); B(i)=payment; C(i)=A(i)-B(i); endBal=C(i); i=i+1; end fileID = fopen(\'exp.txt\',\'w\'); x=1:1:(i-1); M=[x; A; B; C]; fprintf(fileID,\'%4s %12s %12s %12s\ \',\'month\',\'Beg Balance\',\'payment\', \'end Balance\'); fprintf(fileID,\'%6.2f %12.2f %8.2f %6.2f\ \',M); type exp.txt if endBal<0 fprintf(\' Final balance amount to be refunded is: %f\', -endBal); end %out put month Beg Balance payment end Balance 1.00 201000.00 1199.11 199800.89 2.00 200799.89 1199.11 199600.78 3.00 200598.79 1199.11 199399.68 4.00 200396.68 1199.11 199197.57 5.00 200193.55 1199.11 198994.44 6.00 199989.42 1199.11 198790.31 7.00 199784.26 1199.11 198585.15 8.00 199578.07 1199.11 198378.96 9.00 199370.86 1199.11 198171.75 10.00 199162.61 1199.11 197963.50 11.00 198953.32 1199.11 197754.21 12.00 198742.98 1199.11 197543.87 13.00 198531.59 1199.11 197332.48 14.00 198319.14 1199.11 197120.03 15.00 198105.63 1199.11 196906.52 16.00 197891.05 1199.11 196691.94 17.00 197675.40 1199.11 196476.29 18.00 197458.67 1199.11 196259.56 19.00 197240.86 1199.11 196041.75 20.00 197021.96 1199.11 195822.85 21.00 196801.96 1199.11 195602.85 22.00 196580.87 1199.11 195381.76 23.00 196358.67 1199.11 195159.56 24.00 196135.35 1199.11 194936.24 25.00 195910.92 1199.11 194711.81 26.00 195685.37 1199.11 194486.26 27.00 195458.70 1199.11 194259.59 28.00 195230.88 1199.11 194031.77 29.00 195001.93 1199.11 193802.82 30.00 194771.84 1199.11 193572.73 31.00 194540.59 1199.11 193341.48 32.00 194308.19 1199.11 193109.08 33.00 194074.62 1199.11 192875.51 34.00 193839.89 1199.11 192640.78 35.00 193603.98 1199.11 192404.87 36.00 193366.90 1199.11 192167.79 37.00 193128.63 1199.11 191929.52 38.00 192889.16 1199.11 191690.05 39.00 192648.50 1199.11 191449.39 40.00 192406.64 1199.11 191207.53 41.00 192163.57 1199.11 190964.46 42.00 191919.28 1199.11 190720.17 43.00 191673.77 1199.11 190474.66 44.00 191427.04 1199.11 190227.93 45.00 191179.07 1199.11 189979.96 46.00 190929.86 1199.11 189730.75 47.00 190679.40 1199.11 189480.29 48.00 190427.69 1199.11 189228.58 49.00 190174.72 1199.11 188975.61 50.00 189920.49 1199.11 188721.38 51.00 189664.99 1199.11 188465.88 52.00 189408.21 1199.11 188209.10 53.00 189150.14 1199.11 187951.03 54.00 188890.79 1199.11 187691.68 55.00 188630.14 1199.11 187431.03 56.00 188368.18 1199.11 187169.07 57.00 188104.92 1199.11 186905.81 58.00 187840.34 1199.11 186641.23 59.00 187574.43 1199.11 186375.32 60.00 187307.20 1199.11 186108.09 61.00 187038.63 1199.11 185839.52 62.00 186768.72 1199.11 185569.61 63.00 186497.46 1199.11 185298.35 64.00 186224.84 1199.11 185025.73 65.00 185950.86 1199.11 184751.75 66.00 185675.50 1199.11 184476.39 67.00 185398.78 1199.11 184199.67 68.00 185120.66 1199.11 183921.55 69.00 184841.16 1199.11 183642.05 70.00 184560.26 1199.11 183361.15 71.00 18427.

I need help with these two questions 4.12 and 4.13 dealing with MATL.pdf

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I don\'t understand the min and max Find the smallest positive value of t for which f(t) = 5 sin (t - pi/4). is the following. f(t) is zero. f(t)- attains a maximum. f(t) attains a minimum value. Solution we know that sine function attains maximum at pi/2 for first time and minimum at pifor first time. so, argument is x- pi/4 for maximunum x-pi/4 = pi/2 x = 3pi/4 for minimum, x-pi/4 = pi x = 5pi/4 zero when x-pi/4 = 0 x = pi/4.

I dont understand the min and max Find the smallest positive value.pdf

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How can the activation of effector molecules within cells be restricted to subcellular localities? Solution The effector proteins can be transported to subcellular localities through specialised protein secretion systems carried by bacteria which infect the host organism. These systems carry out precise localization of the effector molecules into the host cell thereby enahancing their effectiveness..

How can the activation of effector molecules within cells be restric.pdf

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Explain how the skin of reptiles adaptive for terrestial existence. (Is your skin also adaptive for a terrestial existence?) Solution Reptiles posess many adaptations that permits them to live in dry conditions. One of the important adaptation of reptiles is their skin. Reptiles have dry, cornified skin without glands and with scales. The thick, scaly skin helps conserve water inside their bodies and prevent water loss. This is an essential feature that helps them survive not only on land, but in dry, desert areas as well. The skin functions as water proof cover to conserve water inside, so they do not feel the need of water unlike other animals. This important adaptation has helped reptiles to survive in terestrial habitats..

Explain how the skin of reptiles adaptive for terrestial existence. .pdf

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Distinguish the roles of the four economic actors in economics Solution Following are the four economic actors: Consumer: Consumer buys the goods and service. He seeks to maximizes his utility. Consumer is regarded as rational consumer. Worker: Worker is most important human resource. worker provides the important service in the form of labor. Firm: firm seeks to maximizes its own profit. Like Microsoft tries to sell more software to earn more profits. Government. Govt aims to maximize the welfare of its citizens..

Distinguish the roles of the four economic actors in economicsSo.pdf

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Explain the term liquidity and how it relates to the classified balance sheet. Why is it important to classify accounts in order of their liquidity? Solution Liquidity means the cash conversion/equivalent value of the asset. The liquidity relates to the balance sheet towards the presentation of assets in order of their liquidity in the assets column in the balance sheet and in the order of the amount of time the asset would usually take to convert them into cash. Thus, in the defined cash is always presented first and then followed by other assets like marketable securities, accounts receivable, inventory, fixed assets and at the last the goodwill. The importance of presentation of the balance sheet in order of the liquidity nature of the assets is that the management and other user of the balance sheet would able to know how the firm is performing in terms of liquidity and in case of emergency will the firm will able to convert its assets into cash. It will provide confidence to the creditors towards the firm not going into liquidation and their investment is safe. =====================.

Explain the term liquidity and how it relates to the classified bala.pdf

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