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Comments added... import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; /** * * @author prmsh */ public class SaveMatrix { public static void main(String args[]) { //declaring matrix.. String[][] data = new String[4][4]; int row = 0; // opening and reading file File file = new File(\"matrix.txt\"); try { Scanner sc = new Scanner(file); while (sc.hasNextLine()) { String line = sc.nextLine(); //splitting each row by spaces String[] lineArray = line.split(\" \"); //storing into array for(int i=0; i Solution Comments added... import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; /** * * @author prmsh */ public class SaveMatrix { public static void main(String args[]) { //declaring matrix.. String[][] data = new String[4][4]; int row = 0; // opening and reading file File file = new File(\"matrix.txt\"); try { Scanner sc = new Scanner(file); while (sc.hasNextLine()) { String line = sc.nextLine(); //splitting each row by spaces String[] lineArray = line.split(\" \"); //storing into array for(int i=0; i.
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import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
/**
*
* @author prmsh
*/
public class SaveMatrix {
public static void main(String args[]) {
//declaring matrix..
String[][] data = new String[4][4];
int row = 0;
// opening and reading file
File file = new File("matrix.txt");
try {
Scanner sc = new Scanner(file);
while (sc.hasNextLine()) {
String line = sc.nextLine();
//splitting each row by spaces
String[] lineArray = line.split(" ");
//storing into array
for(int i=0; i
Solution
Comments added...
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
/**
*](https://image.slidesharecdn.com/commentsadded-230412101704-a257cc08/85/Comments-added-import-java-io-File-import-java-io-FileNotF-pdf-1-320.jpg)
![* @author prmsh
*/
public class SaveMatrix {
public static void main(String args[]) {
//declaring matrix..
String[][] data = new String[4][4];
int row = 0;
// opening and reading file
File file = new File("matrix.txt");
try {
Scanner sc = new Scanner(file);
while (sc.hasNextLine()) {
String line = sc.nextLine();
//splitting each row by spaces
String[] lineArray = line.split(" ");
//storing into array
for(int i=0; i](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
i am finishing a project for my cs 0401 java course. I have complete.pdf
i am finishing a project for my cs 0401 java course. I have completed everything for the project
that requires us to sort and then using a binary search display a txt file read through the command
prompt. For some reason when the program treads through the file the first element isnt sorted
and just ends up being printed in front of all the other sorted elements.
my code is:
import java.util.*;
import java.io.*;
public class Project4
{
static final int INITIAL_CAPACITY = 5;
public static void main( String args[] ) throws Exception
{
// ALWAYS TEST FIRST TO VERIFY USER PUT REQUIRED INPUT FILE NAME ON
THE COMMAND LINE
if (args.length < 1 )
{
System.out.println(\"\ usage: C:\\\\> java Project4 \ \ \"); // i.e. C:\\> java Project4
P4input.txt
System.exit(0);
}
// LOAD FILE INTO ARR USING INSERINORDER
Scanner infile = new Scanner( new File( args[0] ) );
int[] arr = new int[INITIAL_CAPACITY];
int count= 0;
while (infile.hasNextInt())
{
if ( count==arr.length )
arr = upSizeArr(arr);
insertInOrder( arr, count++, infile.nextInt() );
}
infile.close();
arr=trimArr(arr,count); // Now count == .length
System.out.println( \"Sorted array of \" + arr.length + \" ints from \" + args[0] + \" after
insertInOrder:\" );
printArray( arr ); // we trimmed it thus count == length so we don\'t bother to pass in count
} // END MAIN
//
#############################################################################
###############################
// USE AS IS - DO NOT MODIFY
static void printArray( int[] arr )
{
for( int i=0 ; i= index + 1; i--)
{
arr[i] = arr[i - 1];
}
arr[index]=key; // LEAVE THIS HERE
}
}
static int bSearch(int[] a, int count, int key)
{
int hi = count-1;
int lo = 0;
int mid = (hi +lo)/2;
if(hi == -1)
{
return mid;
}
else
{
mid = (hi+lo)/2;
}
while (lo <= hi)
{
if (key==a[mid])
{
return (mid+1);
}
else if (key < a[mid])
{
hi = mid-1;
mid = (hi+lo)/2;
}
else
{
lo = mid +1;
mid = (hi+lo)/2;
}
}
return (mid +1);
}
}
and the .txt file has the elements:
Solution
HI, I have fixed the issue.
import java.util.*;
import java.io.*;
public class Project4
{
static final int INITIAL_CAPACITY = 5;
public static void main( String args[] ) throws Exception
{
// ALWAYS TEST FIRST TO VERIFY USER PUT REQUIRED INPUT FILE NAME ON
THE COMMAND LINE
if (args.length < 1 )
{
System.out.println(\"\ usage: C:\\\\> java Project4 \ \ \"); // i.e. C:\\> java Project4
P4input.txt
System.exit(0);
}
// LOAD FILE INTO ARR USING INSERINORDER
Scanner infile = new Scanner( new File( args[0] ) );
int[] arr = new int[INITIAL_CAPACITY];
int count= 0;
while (infile.hasNextInt())
{
if ( count==arr.length )
arr = upSizeArr(arr);
insertInOrder( arr, count++, infile.nextInt() );
}
infile.close();
arr=trimArr(arr,count); // Now count == .length
System.out.println( \"Sorted array of \" + arr.length + \" ints from \" + args[0] + \" after
insertInOrder:\" );
printArray( arr ); // we trimmed it thus count == length so we don\'t bother to pass in count
} // END MAIN
//
###################################.
Create a text file named lnfo.txt. Enter the following informatio.pdf
Create a text file named \"lnfo.txt\". Enter the following information using Linked List (any
method). Name rightarrow String Address rightarrow String Phone rightarrow String There
should be continues loop which will ask the above information and write down in the TEXT File.
When user inputs Exit in Name Field then program should be ended.
Solution
package sample;
import java.io.*;
public class ReadWrite {
public static void main(String[] args) {
BufferedReader reader = null;
try {
reader = new BufferedReader(new InputStreamReader(System.in));
File file = new File(\"C:\\\\Temp\\\\Info.txt\");
// creates the file
file.createNewFile();
// creates a FileWriter Object
FileWriter writer = new FileWriter(file);
while (true) {
System.out.print(\"Enter Name : \");
String name = reader.readLine();
if (\"Exit\".equals(name)) {
System.out.println(\"Exit!\");
writer.flush();
writer.close();
System.exit(0);
}
System.out.print(\"Enter Address : \");
String address = reader.readLine();
System.out.print(\"Enter Phone : \");
String phone = reader.readLine();
// Writes the content to the file
writer.write(\"\ \ -----------\");
writer.write(\"\ Name : \"+name);
writer.write(\"\ Address : \"+address);
writer.write(\"\ Phone : \"+phone);
writer.write(\"\ -----------\");
System.out.println(\"Added Below Record Successfully \");
System.out.println(\"-----------\ \");
System.out.println(\"Name : \"+name);
System.out.println(\"Address : \"+address);
System.out.println(\"Phone : \"+phone);
System.out.println(\"-----------\ \");
}
} catch (IOException ex) {
ex.printStackTrace();
} finally {
if (reader != null) {
try {
reader.close();
} catch (IOException ex) {
ex.printStackTrace();
}
}
}
}
}
package sample;
import java.io.*;
public class ReadWrite {
public static void main(String[] args) {
BufferedReader reader = null;
try {
reader = new BufferedReader(new InputStreamReader(System.in));
File file = new File(\"C:\\\\Temp\\\\Info.txt\");
// creates the file
file.createNewFile();
// creates a FileWriter Object
FileWriter writer = new FileWriter(file);
while (true) {
System.out.print(\"Enter Name : \");
String name = reader.readLine();
if (\"Exit\".equals(name)) {
System.out.println(\"Exit!\");
writer.flush();
writer.close();
System.exit(0);
}
System.out.print(\"Enter Address : \");
String address = reader.readLine();
System.out.print(\"Enter Phone : \");
String phone = reader.readLine();
// Writes the content to the file
writer.write(\"\ \ -----------\");
writer.write(\"\ Name : \"+name);
writer.write(\"\ Address : \"+address);
writer.write(\"\ Phone : \"+phone);
writer.write(\"\ -----------\");
System.out.println(\"Added Below Record Successfully \");
System.out.println(\"-----------\ \");
System.out.println(\"Name : \"+name);
System.out.println(\"Address : \"+address);
System.out.println(\"Phone : \"+phone);
System.out.println(\"-----------\ \");
}
} catch (IOException ex) {
ex.printStackTrace();
} finally {
if (reader != null) {
try {
reader.close();
} catch (IOExcept.
Please I am trying to get this code to output in -txt file- I need you.pdf
Please I am trying to get this code to output in .txt file:
I need your help:
import java.io.File; // Import the File class
import java.io.FileNotFoundException; // Import this class to handle errors
// Import the Scanner class to read text files
import java.util.*;
import java.io.*;
public class KMeans {
List<double []> clusterCentroids;
Map<Integer,List<Double>> dataMatrix;
List<double[]> data;
Integer numClusters;
public void writeFile(List<Double> value, File filename) {
try {
FileWriter myWriter = new FileWriter(filename);
myWriter.write(value.toString());
myWriter.close();
System.out.println("Successfully wrote to the file.");
} catch (IOException e) {
System.out.println("An error occurred.");
e.printStackTrace();
}
}
public KMeans() {
data = new ArrayList<double []>();
dataMatrix = new LinkedHashMap<>();
}
void readData(String filePath, int n, List<Double> c) {
numClusters = n;
try {
BufferedReader br = new BufferedReader(new FileReader(new File(filePath)));
String line;
int i = 0;
while((line = br.readLine()) != null) {
String [] elem = line.split("\\s+");
int j = 0;
for(String d : elem) {
if(isCentroid(j, c)) {
this.clusterCentroids.get(i)[j] = ((Double.valueOf(d)));
}
//data.get(i)[j] = ((Double.valueOf(d)));
System.out.println(data.get(j)[i]);
}
System.out.println();
i++;
}
br.close();
} catch (IOException ex) {
ex.printStackTrace();
}
}
private boolean isCentroid(int j, List<Double> c) {
for(int i = 0; i < c.size(); i++) {
if(j == c.get(i)) {
return true;
}
}
return false;
}
public static void main(String [] args) {
Scanner sc = new Scanner(System.in);
List<Double> centroids = new ArrayList<>();
String filePath = ("C:\\Users\\mamdo\\eclipse-
workspace\\K_project\\src\\synthetic_control_data.txt");
int n = 6;
for(int i=0;i<n;i++) {
Random r = new Random();
centroids.add((double)r.nextInt(600)+1);
}
System.out.println("Initial Clusters Centroids : " + centroids);
KMeans km = new KMeans();
//km.readData(filePath,n,centroids);
//km.calc();
}
}
.
Java 3 Computer Science.pptx
Exceptions, I/O and Threads Input and Output in Java: The File Class, Standard Streams, Keyboard
Input, File I/O Using Byte Streams, Character Streams, File I/O Using Character Streams -
Buffered Streams, File I/O Using a Buffered Stream, Keyboard Input Using a Buffered Stream,Writing Text Files. Threads: Threads vs. Processes, Creating Threads by Extending Thread,
Creating Threads by Implementing Runnable, Advantages of Using Threads, Daemon Threads,
Thread States, Thread Problems, Synchronization. Exceptions: Exception Handling, The Exception
Hierarchy, throws statement, throw statement, Developing user defined Exception Classes- The
finally Block.
Recommended
i am finishing a project for my cs 0401 java course. I have complete.pdf
i am finishing a project for my cs 0401 java course. I have completed everything for the project
that requires us to sort and then using a binary search display a txt file read through the command
prompt. For some reason when the program treads through the file the first element isnt sorted
and just ends up being printed in front of all the other sorted elements.
my code is:
import java.util.*;
import java.io.*;
public class Project4
{
static final int INITIAL_CAPACITY = 5;
public static void main( String args[] ) throws Exception
{
// ALWAYS TEST FIRST TO VERIFY USER PUT REQUIRED INPUT FILE NAME ON
THE COMMAND LINE
if (args.length < 1 )
{
System.out.println(\"\ usage: C:\\\\> java Project4 \ \ \"); // i.e. C:\\> java Project4
P4input.txt
System.exit(0);
}
// LOAD FILE INTO ARR USING INSERINORDER
Scanner infile = new Scanner( new File( args[0] ) );
int[] arr = new int[INITIAL_CAPACITY];
int count= 0;
while (infile.hasNextInt())
{
if ( count==arr.length )
arr = upSizeArr(arr);
insertInOrder( arr, count++, infile.nextInt() );
}
infile.close();
arr=trimArr(arr,count); // Now count == .length
System.out.println( \"Sorted array of \" + arr.length + \" ints from \" + args[0] + \" after
insertInOrder:\" );
printArray( arr ); // we trimmed it thus count == length so we don\'t bother to pass in count
} // END MAIN
//
#############################################################################
###############################
// USE AS IS - DO NOT MODIFY
static void printArray( int[] arr )
{
for( int i=0 ; i= index + 1; i--)
{
arr[i] = arr[i - 1];
}
arr[index]=key; // LEAVE THIS HERE
}
}
static int bSearch(int[] a, int count, int key)
{
int hi = count-1;
int lo = 0;
int mid = (hi +lo)/2;
if(hi == -1)
{
return mid;
}
else
{
mid = (hi+lo)/2;
}
while (lo <= hi)
{
if (key==a[mid])
{
return (mid+1);
}
else if (key < a[mid])
{
hi = mid-1;
mid = (hi+lo)/2;
}
else
{
lo = mid +1;
mid = (hi+lo)/2;
}
}
return (mid +1);
}
}
and the .txt file has the elements:
Solution
HI, I have fixed the issue.
import java.util.*;
import java.io.*;
public class Project4
{
static final int INITIAL_CAPACITY = 5;
public static void main( String args[] ) throws Exception
{
// ALWAYS TEST FIRST TO VERIFY USER PUT REQUIRED INPUT FILE NAME ON
THE COMMAND LINE
if (args.length < 1 )
{
System.out.println(\"\ usage: C:\\\\> java Project4 \ \ \"); // i.e. C:\\> java Project4
P4input.txt
System.exit(0);
}
// LOAD FILE INTO ARR USING INSERINORDER
Scanner infile = new Scanner( new File( args[0] ) );
int[] arr = new int[INITIAL_CAPACITY];
int count= 0;
while (infile.hasNextInt())
{
if ( count==arr.length )
arr = upSizeArr(arr);
insertInOrder( arr, count++, infile.nextInt() );
}
infile.close();
arr=trimArr(arr,count); // Now count == .length
System.out.println( \"Sorted array of \" + arr.length + \" ints from \" + args[0] + \" after
insertInOrder:\" );
printArray( arr ); // we trimmed it thus count == length so we don\'t bother to pass in count
} // END MAIN
//
###################################.
Create a text file named lnfo.txt. Enter the following informatio.pdf
Create a text file named \"lnfo.txt\". Enter the following information using Linked List (any
method). Name rightarrow String Address rightarrow String Phone rightarrow String There
should be continues loop which will ask the above information and write down in the TEXT File.
When user inputs Exit in Name Field then program should be ended.
Solution
package sample;
import java.io.*;
public class ReadWrite {
public static void main(String[] args) {
BufferedReader reader = null;
try {
reader = new BufferedReader(new InputStreamReader(System.in));
File file = new File(\"C:\\\\Temp\\\\Info.txt\");
// creates the file
file.createNewFile();
// creates a FileWriter Object
FileWriter writer = new FileWriter(file);
while (true) {
System.out.print(\"Enter Name : \");
String name = reader.readLine();
if (\"Exit\".equals(name)) {
System.out.println(\"Exit!\");
writer.flush();
writer.close();
System.exit(0);
}
System.out.print(\"Enter Address : \");
String address = reader.readLine();
System.out.print(\"Enter Phone : \");
String phone = reader.readLine();
// Writes the content to the file
writer.write(\"\ \ -----------\");
writer.write(\"\ Name : \"+name);
writer.write(\"\ Address : \"+address);
writer.write(\"\ Phone : \"+phone);
writer.write(\"\ -----------\");
System.out.println(\"Added Below Record Successfully \");
System.out.println(\"-----------\ \");
System.out.println(\"Name : \"+name);
System.out.println(\"Address : \"+address);
System.out.println(\"Phone : \"+phone);
System.out.println(\"-----------\ \");
}
} catch (IOException ex) {
ex.printStackTrace();
} finally {
if (reader != null) {
try {
reader.close();
} catch (IOException ex) {
ex.printStackTrace();
}
}
}
}
}
package sample;
import java.io.*;
public class ReadWrite {
public static void main(String[] args) {
BufferedReader reader = null;
try {
reader = new BufferedReader(new InputStreamReader(System.in));
File file = new File(\"C:\\\\Temp\\\\Info.txt\");
// creates the file
file.createNewFile();
// creates a FileWriter Object
FileWriter writer = new FileWriter(file);
while (true) {
System.out.print(\"Enter Name : \");
String name = reader.readLine();
if (\"Exit\".equals(name)) {
System.out.println(\"Exit!\");
writer.flush();
writer.close();
System.exit(0);
}
System.out.print(\"Enter Address : \");
String address = reader.readLine();
System.out.print(\"Enter Phone : \");
String phone = reader.readLine();
// Writes the content to the file
writer.write(\"\ \ -----------\");
writer.write(\"\ Name : \"+name);
writer.write(\"\ Address : \"+address);
writer.write(\"\ Phone : \"+phone);
writer.write(\"\ -----------\");
System.out.println(\"Added Below Record Successfully \");
System.out.println(\"-----------\ \");
System.out.println(\"Name : \"+name);
System.out.println(\"Address : \"+address);
System.out.println(\"Phone : \"+phone);
System.out.println(\"-----------\ \");
}
} catch (IOException ex) {
ex.printStackTrace();
} finally {
if (reader != null) {
try {
reader.close();
} catch (IOExcept.
Please I am trying to get this code to output in -txt file- I need you.pdf
Please I am trying to get this code to output in .txt file:
I need your help:
import java.io.File; // Import the File class
import java.io.FileNotFoundException; // Import this class to handle errors
// Import the Scanner class to read text files
import java.util.*;
import java.io.*;
public class KMeans {
List<double []> clusterCentroids;
Map<Integer,List<Double>> dataMatrix;
List<double[]> data;
Integer numClusters;
public void writeFile(List<Double> value, File filename) {
try {
FileWriter myWriter = new FileWriter(filename);
myWriter.write(value.toString());
myWriter.close();
System.out.println("Successfully wrote to the file.");
} catch (IOException e) {
System.out.println("An error occurred.");
e.printStackTrace();
}
}
public KMeans() {
data = new ArrayList<double []>();
dataMatrix = new LinkedHashMap<>();
}
void readData(String filePath, int n, List<Double> c) {
numClusters = n;
try {
BufferedReader br = new BufferedReader(new FileReader(new File(filePath)));
String line;
int i = 0;
while((line = br.readLine()) != null) {
String [] elem = line.split("\\s+");
int j = 0;
for(String d : elem) {
if(isCentroid(j, c)) {
this.clusterCentroids.get(i)[j] = ((Double.valueOf(d)));
}
//data.get(i)[j] = ((Double.valueOf(d)));
System.out.println(data.get(j)[i]);
}
System.out.println();
i++;
}
br.close();
} catch (IOException ex) {
ex.printStackTrace();
}
}
private boolean isCentroid(int j, List<Double> c) {
for(int i = 0; i < c.size(); i++) {
if(j == c.get(i)) {
return true;
}
}
return false;
}
public static void main(String [] args) {
Scanner sc = new Scanner(System.in);
List<Double> centroids = new ArrayList<>();
String filePath = ("C:\\Users\\mamdo\\eclipse-
workspace\\K_project\\src\\synthetic_control_data.txt");
int n = 6;
for(int i=0;i<n;i++) {
Random r = new Random();
centroids.add((double)r.nextInt(600)+1);
}
System.out.println("Initial Clusters Centroids : " + centroids);
KMeans km = new KMeans();
//km.readData(filePath,n,centroids);
//km.calc();
}
}
.
Java 3 Computer Science.pptx
Exceptions, I/O and Threads Input and Output in Java: The File Class, Standard Streams, Keyboard
Input, File I/O Using Byte Streams, Character Streams, File I/O Using Character Streams -
Buffered Streams, File I/O Using a Buffered Stream, Keyboard Input Using a Buffered Stream,Writing Text Files. Threads: Threads vs. Processes, Creating Threads by Extending Thread,
Creating Threads by Implementing Runnable, Advantages of Using Threads, Daemon Threads,
Thread States, Thread Problems, Synchronization. Exceptions: Exception Handling, The Exception
Hierarchy, throws statement, throw statement, Developing user defined Exception Classes- The
finally Block.
Hi, I need help with a java programming project. specifically practi.pdf
Given 8 distinct integers from the set {1, 2, . . . , 16}, prove that there exists k such that ai aj = k
has at least 3 distinct solutions (ai , aj ) ?
Solution
For any k (unless the K is zero), at least 3 distinct solutions need not be existing.
Let\'s look at the next most favourable case
If k= 2;
We have to consider the case where the number of solutions of ai-aj=2 are minimum since we
have to show that at least 3 distinct solutions exist (which means even in the most unfavourable
case for a chosen k there will be 3 distinct solutions satisfying the equation ai-aj=2)
If k = 2; we start by taking numbers that are not at all 2 units apart from each other;
So we start with 1 and move 3 units successively ahead
1, 4, 7, 10,13, 16; Now we notice that these 6 numbers are of 3k+1 form; remaining numbers
can be any 2 ranomdly selected say 2 and 11; we get 4-2 = 2 and 13-11=2; Hence only 2 distinct
solutions are possible..
please navigate to cs112 webpage and go to assignments -- Trees. Th.pdf
please navigate to cs112 webpage and go to assignments --> Trees. This will give you the
directions for the assignment. Please do the last 2 methods in huffman.java called encode() and
decode()
Code is provided below:
HuffmanCoding.java
package huffman;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.util.ArrayList;
import java.util.Collections;
/**
* This class contains methods which, when used together, perform the
* entire Huffman Coding encoding and decoding process
*
* @author Ishaan Ivaturi
* @author Prince Rawal
*/
public class HuffmanCoding {
private String fileName;
private ArrayList sortedCharFreqList;
private TreeNode huffmanRoot;
private String[] encodings;
/**
* Constructor used by the driver, sets filename
* DO NOT EDIT
* @param f The file we want to encode
*/
public HuffmanCoding(String f) {
fileName = f;
}
/**
* Reads from filename character by character, and sets sortedCharFreqList
* to a new ArrayList of CharFreq objects with frequency > 0, sorted by frequency
*/
public void makeSortedList() {
StdIn.setFile(fileName);
/* Your code goes here */
}
/**
* Uses sortedCharFreqList to build a huffman coding tree, and stores its root
* in huffmanRoot
*/
public void makeTree() {
/* Your code goes here */
}
/**
* Uses huffmanRoot to create a string array of size 128, where each
* index in the array contains that ASCII character's bitstring encoding. Characters not
* present in the huffman coding tree should have their spots in the array left null.
* Set encodings to this array.
*/
public void makeEncodings() {
/* Your code goes here */
}
/**
* Using encodings and filename, this method makes use of the writeBitString method
* to write the final encoding of 1's and 0's to the encoded file.
*
* @param encodedFile The file name into which the text file is to be encoded
*/
public void encode(String encodedFile) {
StdIn.setFile(fileName);
/* Your code goes here */
}
/**
* Writes a given string of 1's and 0's to the given file byte by byte
* and NOT as characters of 1 and 0 which take up 8 bits each
* DO NOT EDIT
*
* @param filename The file to write to (doesn't need to exist yet)
* @param bitString The string of 1's and 0's to write to the file in bits
*/
public static void writeBitString(String filename, String bitString) {
byte[] bytes = new byte[bitString.length() / 8 + 1];
int bytesIndex = 0, byteIndex = 0, currentByte = 0;
// Pad the string with initial zeroes and then a one in order to bring
// its length to a multiple of 8. When reading, the 1 signifies the
// end of padding.
int padding = 8 - (bitString.length() % 8);
String pad = "";
for (int i = 0; i < padding-1; i++) pad = pad + "0";
pad = pad + "1";
bitString = pad + bitString;
// For every bit, add it to the right spot in the corresponding byte,
// and store bytes in the array when finished
for (char c : bitString.toCharArray()) {
if (c != '1' && c != '0') {
System.out.println("Invalid characters in bitstring");
return;
}
if (c .
In Java- Input is from a file- The input is all doubles- You are to in.pdf
In Java;
Input is from a file. The input is all doubles. You are to input these doubles 2 at a time (the
length and width), create a Rectangle and determine if it has the largest area. Output the
Rectangle with the largest area.
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Lab8Num2 {
public static void main(String[] args) {
File inFile = new File("lab8.in");
Scanner fileInput = null;
try {
fileInput = new Scanner(inFile);
} catch (FileNotFoundException ex) {
}
//get first Rectangle and make it the biggest;
//do NOT put into an arrayList and sort.
while ()
{ //get more data from file
//make Rectangle
//see if it is bigger than biggest so far
//if so, it is the new biggest
}
System.out.println("The biggest rectangle was " + biggest);
}
}
public class Rectangle
{
private double length, width;
public Rectangle()
{
length=0;
width=0;
}
public Rectangle(double len, double wid)
{
length=len;
width=wid;
}
public double getLength() {
return length;
}
public void setLength(double length) {
this.length = length;
}
public double getWidth() {
return width;
}
public void setWidth(double width) {
this.width = width;
}
public double area()
{
return length*width;
}
public double perimeter()
{
return 2*(length+width);
}
public String toString()
{
return "Length: " + length + " Width: " + width;
}
public int compareTo(Rectangle r)
{
}
}
.
Input is from a file- The input is all doubles- You are to input these.pdf
Input is from a file. The input is all doubles. You are to input these doubles 2 at a time (the
length and width), create a Rectangle and determine if it has the largest area. Output the
Rectangle with the largest area.
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Lab8Num2 {
public static void main(String[] args) {
File inFile = new File("lab8.in");
Scanner fileInput = null;
try {
fileInput = new Scanner(inFile);
} catch (FileNotFoundException ex) {
}
//get first Rectangle and make it the biggest;
//do NOT put into an arrayList and sort.
while ()
{ //get more data from file
//make Rectangle
//see if it is bigger than biggest so far
//if so, it is the new biggest
}
System.out.println("The biggest rectangle was " + biggest);
}
}
public class Rectangle
{
private double length, width;
public Rectangle()
{
length=0;
width=0;
}
public Rectangle(double len, double wid)
{
length=len;
width=wid;
}
public double getLength() {
return length;
}
public void setLength(double length) {
this.length = length;
}
public double getWidth() {
return width;
}
public void setWidth(double width) {
this.width = width;
}
public double area()
{
return length*width;
}
public double perimeter()
{
return 2*(length+width);
}
public String toString()
{
return "Length: " + length + " Width: " + width;
}
public int compareTo(Rectangle r)
{
}
}
.
Session 23 - JDBC
In this session you will learn:
1. File IO Continued
2. Intro to JDBC (Java Database Connectivity)
Complete the following Java code such that it opens a file named Fruit.docx
Complete the following Java code such that it opens a file named FruitList.txt, reads the first line from the file, and then prints the first line on console screen. Your Java code should handle possible exception(s). Assume that the file contains the name of a fruit in each line. You don\'t need to provide comments.
// complete import statement(s) here
public class FileRead
{
public static void main(String[] args)
{
// complete the code here
}
}
Solution
import java.io.File;
import java.util.Scanner;
/**
* @author
*
*/
public class FileRead {
/**
* @param args
*/
public static void main(String[] args) {
// complete the code here
Scanner scanner = null;
try {
File file = new File(\"FruitList.txt\");
scanner = new Scanner(file);
// checks the file contains data
if (scanner.hasNext()) {
// prints first line
System.out.println(scanner.nextLine());
}
} catch (Exception e) {
// TODO: handle exception
}
}
}
FruitList.txt
Apple
Orange
Papaya
Banana
Jack Fruit
Watermelon
KIWI
Grapes
Mango
Pineapple
OUTPUT:
Apple
.
5. Ввод-вывод, доступ к файловой системе
5.1 Доступ к файловой системе
5.2 Потоки байт
5.3 Потоки символов
5.4 Продвинутые возможности
phosphate group and deoxyribose The groups are .pdf
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose
sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the
\"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder.
Solution
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose
sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the
\"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder..
More Related Content
Similar to Comments added... import java.io.File; import java.io.FileNotF.pdf
Hi, I need help with a java programming project. specifically practi.pdf
Given 8 distinct integers from the set {1, 2, . . . , 16}, prove that there exists k such that ai aj = k
has at least 3 distinct solutions (ai , aj ) ?
Solution
For any k (unless the K is zero), at least 3 distinct solutions need not be existing.
Let\'s look at the next most favourable case
If k= 2;
We have to consider the case where the number of solutions of ai-aj=2 are minimum since we
have to show that at least 3 distinct solutions exist (which means even in the most unfavourable
case for a chosen k there will be 3 distinct solutions satisfying the equation ai-aj=2)
If k = 2; we start by taking numbers that are not at all 2 units apart from each other;
So we start with 1 and move 3 units successively ahead
1, 4, 7, 10,13, 16; Now we notice that these 6 numbers are of 3k+1 form; remaining numbers
can be any 2 ranomdly selected say 2 and 11; we get 4-2 = 2 and 13-11=2; Hence only 2 distinct
solutions are possible..
please navigate to cs112 webpage and go to assignments -- Trees. Th.pdf
please navigate to cs112 webpage and go to assignments --> Trees. This will give you the
directions for the assignment. Please do the last 2 methods in huffman.java called encode() and
decode()
Code is provided below:
HuffmanCoding.java
package huffman;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.util.ArrayList;
import java.util.Collections;
/**
* This class contains methods which, when used together, perform the
* entire Huffman Coding encoding and decoding process
*
* @author Ishaan Ivaturi
* @author Prince Rawal
*/
public class HuffmanCoding {
private String fileName;
private ArrayList sortedCharFreqList;
private TreeNode huffmanRoot;
private String[] encodings;
/**
* Constructor used by the driver, sets filename
* DO NOT EDIT
* @param f The file we want to encode
*/
public HuffmanCoding(String f) {
fileName = f;
}
/**
* Reads from filename character by character, and sets sortedCharFreqList
* to a new ArrayList of CharFreq objects with frequency > 0, sorted by frequency
*/
public void makeSortedList() {
StdIn.setFile(fileName);
/* Your code goes here */
}
/**
* Uses sortedCharFreqList to build a huffman coding tree, and stores its root
* in huffmanRoot
*/
public void makeTree() {
/* Your code goes here */
}
/**
* Uses huffmanRoot to create a string array of size 128, where each
* index in the array contains that ASCII character's bitstring encoding. Characters not
* present in the huffman coding tree should have their spots in the array left null.
* Set encodings to this array.
*/
public void makeEncodings() {
/* Your code goes here */
}
/**
* Using encodings and filename, this method makes use of the writeBitString method
* to write the final encoding of 1's and 0's to the encoded file.
*
* @param encodedFile The file name into which the text file is to be encoded
*/
public void encode(String encodedFile) {
StdIn.setFile(fileName);
/* Your code goes here */
}
/**
* Writes a given string of 1's and 0's to the given file byte by byte
* and NOT as characters of 1 and 0 which take up 8 bits each
* DO NOT EDIT
*
* @param filename The file to write to (doesn't need to exist yet)
* @param bitString The string of 1's and 0's to write to the file in bits
*/
public static void writeBitString(String filename, String bitString) {
byte[] bytes = new byte[bitString.length() / 8 + 1];
int bytesIndex = 0, byteIndex = 0, currentByte = 0;
// Pad the string with initial zeroes and then a one in order to bring
// its length to a multiple of 8. When reading, the 1 signifies the
// end of padding.
int padding = 8 - (bitString.length() % 8);
String pad = "";
for (int i = 0; i < padding-1; i++) pad = pad + "0";
pad = pad + "1";
bitString = pad + bitString;
// For every bit, add it to the right spot in the corresponding byte,
// and store bytes in the array when finished
for (char c : bitString.toCharArray()) {
if (c != '1' && c != '0') {
System.out.println("Invalid characters in bitstring");
return;
}
if (c .
In Java- Input is from a file- The input is all doubles- You are to in.pdf
In Java;
Input is from a file. The input is all doubles. You are to input these doubles 2 at a time (the
length and width), create a Rectangle and determine if it has the largest area. Output the
Rectangle with the largest area.
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Lab8Num2 {
public static void main(String[] args) {
File inFile = new File("lab8.in");
Scanner fileInput = null;
try {
fileInput = new Scanner(inFile);
} catch (FileNotFoundException ex) {
}
//get first Rectangle and make it the biggest;
//do NOT put into an arrayList and sort.
while ()
{ //get more data from file
//make Rectangle
//see if it is bigger than biggest so far
//if so, it is the new biggest
}
System.out.println("The biggest rectangle was " + biggest);
}
}
public class Rectangle
{
private double length, width;
public Rectangle()
{
length=0;
width=0;
}
public Rectangle(double len, double wid)
{
length=len;
width=wid;
}
public double getLength() {
return length;
}
public void setLength(double length) {
this.length = length;
}
public double getWidth() {
return width;
}
public void setWidth(double width) {
this.width = width;
}
public double area()
{
return length*width;
}
public double perimeter()
{
return 2*(length+width);
}
public String toString()
{
return "Length: " + length + " Width: " + width;
}
public int compareTo(Rectangle r)
{
}
}
.
Input is from a file- The input is all doubles- You are to input these.pdf
Input is from a file. The input is all doubles. You are to input these doubles 2 at a time (the
length and width), create a Rectangle and determine if it has the largest area. Output the
Rectangle with the largest area.
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class Lab8Num2 {
public static void main(String[] args) {
File inFile = new File("lab8.in");
Scanner fileInput = null;
try {
fileInput = new Scanner(inFile);
} catch (FileNotFoundException ex) {
}
//get first Rectangle and make it the biggest;
//do NOT put into an arrayList and sort.
while ()
{ //get more data from file
//make Rectangle
//see if it is bigger than biggest so far
//if so, it is the new biggest
}
System.out.println("The biggest rectangle was " + biggest);
}
}
public class Rectangle
{
private double length, width;
public Rectangle()
{
length=0;
width=0;
}
public Rectangle(double len, double wid)
{
length=len;
width=wid;
}
public double getLength() {
return length;
}
public void setLength(double length) {
this.length = length;
}
public double getWidth() {
return width;
}
public void setWidth(double width) {
this.width = width;
}
public double area()
{
return length*width;
}
public double perimeter()
{
return 2*(length+width);
}
public String toString()
{
return "Length: " + length + " Width: " + width;
}
public int compareTo(Rectangle r)
{
}
}
.
Session 23 - JDBC
In this session you will learn:
1. File IO Continued
2. Intro to JDBC (Java Database Connectivity)
Complete the following Java code such that it opens a file named Fruit.docx
Complete the following Java code such that it opens a file named FruitList.txt, reads the first line from the file, and then prints the first line on console screen. Your Java code should handle possible exception(s). Assume that the file contains the name of a fruit in each line. You don\'t need to provide comments.
// complete import statement(s) here
public class FileRead
{
public static void main(String[] args)
{
// complete the code here
}
}
Solution
import java.io.File;
import java.util.Scanner;
/**
* @author
*
*/
public class FileRead {
/**
* @param args
*/
public static void main(String[] args) {
// complete the code here
Scanner scanner = null;
try {
File file = new File(\"FruitList.txt\");
scanner = new Scanner(file);
// checks the file contains data
if (scanner.hasNext()) {
// prints first line
System.out.println(scanner.nextLine());
}
} catch (Exception e) {
// TODO: handle exception
}
}
}
FruitList.txt
Apple
Orange
Papaya
Banana
Jack Fruit
Watermelon
KIWI
Grapes
Mango
Pineapple
OUTPUT:
Apple
.
5. Ввод-вывод, доступ к файловой системе
5.1 Доступ к файловой системе
5.2 Потоки байт
5.3 Потоки символов
5.4 Продвинутые возможности
Similar to Comments added... import java.io.File; import java.io.FileNotF.pdf (20)
Hi, I need help with a java programming project. specifically practi.pdf
Hi, I need help with a java programming project. specifically practi.pdf
please navigate to cs112 webpage and go to assignments -- Trees. Th.pdf
please navigate to cs112 webpage and go to assignments -- Trees. Th.pdf
In Java- Input is from a file- The input is all doubles- You are to in.pdf
In Java- Input is from a file- The input is all doubles- You are to in.pdf
Input is from a file- The input is all doubles- You are to input these.pdf
Input is from a file- The input is all doubles- You are to input these.pdf
Complete the following Java code such that it opens a file named Fruit.docx
Complete the following Java code such that it opens a file named Fruit.docx
More from anwarfoot
phosphate group and deoxyribose The groups are .pdf
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose
sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the
\"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder.
Solution
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose
sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the
\"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder..
in case of solid oxygen the atoms of oxygen are s.pdf
in case of solid oxygen the atoms of oxygen are sharing elctrons and have a
covalent bond between them so there is a dipole dipole bond
Solution
in case of solid oxygen the atoms of oxygen are sharing elctrons and have a
covalent bond between them so there is a dipole dipole bond.
Yeast is a microorganism and doesnt itself rise.pdf
Yeast is a microorganism and doesn\'t itself rise. Yeast eats carbohydrate and makes
Carbon Dioxide (CO2) - the gas in soda water - as a waste product and it is this gas that makes
the bread or cake \"rise\".
Solution
Yeast is a microorganism and doesn\'t itself rise. Yeast eats carbohydrate and makes
Carbon Dioxide (CO2) - the gas in soda water - as a waste product and it is this gas that makes
the bread or cake \"rise\"..
First, lets analyze what each component of the .pdf
First, let\'s analyze what each component of the solution is (strong base/acid, weak
base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of
your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH:
A strong base (once again, something you should know off the top of your head. Virtually any
ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH
we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as
previously mentioned, this is completely soluble in water and adds F- ions, which will affect the
equilibrium of HF). Like before, we can assume that however much NaF we have gives us the
same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are
happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m
leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==>
F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF
into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing
the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH.
Now, all we have is HF and F-. Recall the Henderson-Hasselbalch equation, which tells you how
to calculate pH based on the ratio of an acid and its salt. In the case of HF, it will be pH = pKa +
log([F-]/[HF]) You have [F-] in the form of [NaF], and you have [HF] which we just
determined. You can calculate pKa from the given Ka value. Thus, you should be able to
calculate pH.
Solution
First, let\'s analyze what each component of the solution is (strong base/acid, weak
base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of
your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH:
A strong base (once again, something you should know off the top of your head. Virtually any
ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH
we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as
previously mentioned, this is completely soluble in water and adds F- ions, which will affect the
equilibrium of HF). Like before, we can assume that however much NaF we have gives us the
same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are
happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m
leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==>
F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF
into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing
the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH.
Now,.
when they reach equilibrium te cell potential is .pdf
when they reach equilibrium te cell potential is zero
Solution
when they reach equilibrium te cell potential is zero.
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s.pdf
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s) 2UO2+(aq) + Sn2+(aq)
[UO22+] = [UO2+] =[Sn2+] = 1.1mM = 0.0011 M Q =[UO2+]2[Sn2+]/[UO22+]2 = 0.0011
E = E° - 0.0592/n logQ = 0.21 - 0.0592/2 log 0.0011 = 0.2976 V
Solution
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s) 2UO2+(aq) + Sn2+(aq)
[UO22+] = [UO2+] =[Sn2+] = 1.1mM = 0.0011 M Q =[UO2+]2[Sn2+]/[UO22+]2 = 0.0011
E = E° - 0.0592/n logQ = 0.21 - 0.0592/2 log 0.0011 = 0.2976 V.
D) Not B) because Cl-Benzene bond develops a doub.pdf
D) Not B) because Cl-Benzene bond develops a double bond character due to
resonance
Solution
D) Not B) because Cl-Benzene bond develops a double bond character due to
resonance.
YesBecause (1,1) is missing, it is not reflexive though (3,3) (2,2.pdf
Yes
Because (1,1) is missing, it is not reflexive though (3,3) (2,2) are there
Solution
Yes
Because (1,1) is missing, it is not reflexive though (3,3) (2,2) are there.
We know that from Nernst Equation ,E o cell = E o – ( 0.059 n ) .pdf
We know that from Nernst Equation ,
E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ])
Given Eo cell = 0.17 V
Eo = 0.24 V
[products ] = [Cd 2+ ] = ?
[reactants ] = [ Ni 2+ ] = 1.0 M
N = no . of electrons transferred = 2
Solution
We know that from Nernst Equation ,
E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ])
Given Eo cell = 0.17 V
Eo = 0.24 V
[products ] = [Cd 2+ ] = ?
[reactants ] = [ Ni 2+ ] = 1.0 M
N = no . of electrons transferred = 2.
lower.. think about it.. take a straw draw a line.pdf
lower.. think about it.. take a straw draw a line on it and fill it with water, then look
at it from different angles
Solution
lower.. think about it.. take a straw draw a line on it and fill it with water, then look
at it from different angles.
r= 1+ sin theta drd = cos = 0Therefore = 2Solutionr=.pdf
r= 1+ sin theta
dr/d = cos = 0
Therefore
= /2
Solution
r= 1+ sin theta
dr/d = cos = 0
Therefore
= /2.
From left to right iodocyclopropane; 1-bromo-3-m.pdf
From left to right: iodocyclopropane; 1-bromo-3-methylcyclopentane; 1-iodo-2,2-
dimethylpropane;
Solution
From left to right: iodocyclopropane; 1-bromo-3-methylcyclopentane; 1-iodo-2,2-
dimethylpropane;.
For inorganic compounds Chemical properties remai.pdf
For inorganic compounds Chemical properties remain same but colour or physical
appearance changes. CuSO4 .5H2O is blue and crystalline but CuSO4 anhydrous is
colourless.Colour change can be explained by theories of Co ordination compounds. In organic
compounds ; chemical properties also change as dehydration changes the chemical nature;
C2H5OH---> C2H4+H2O
Solution
For inorganic compounds Chemical properties remain same but colour or physical
appearance changes. CuSO4 .5H2O is blue and crystalline but CuSO4 anhydrous is
colourless.Colour change can be explained by theories of Co ordination compounds. In organic
compounds ; chemical properties also change as dehydration changes the chemical nature;
C2H5OH---> C2H4+H2O.
n=4 l=3 we know that L=h2(l(l+1))L=3.710-34Solutionn.pdf
n=4
l=3
we know that
L=h/2(l(l+1))
L=3.7*10-34
Solution
n=4
l=3
we know that
L=h/2(l(l+1))
L=3.7*10-34.
Over forty interstellar molecules have been found, ranging from simp.pdf
Over forty interstellar molecules have been found, ranging from simple OH radicals to molecules
with over a dozen different atoms, including a few polycyclic. Their significance lies in that
planets are formed from such nebulas( area of the cloud), so this tells us what may be found.
Solution
Over forty interstellar molecules have been found, ranging from simple OH radicals to molecules
with over a dozen different atoms, including a few polycyclic. Their significance lies in that
planets are formed from such nebulas( area of the cloud), so this tells us what may be found..
One of the four prominent symptoms of typhoid is coated tongue. Even.pdf
One of the four prominent symptoms of typhoid is coated tongue. Even before the onset of fever,
the tongue appears furred, with centrally coated and a red margin around. There are other illness
also which shows coated tongue, but with different appearance. In malaria, the tongue is
uniformly coated whereas in candidia thrush, a thick white layer of which patch is formed. In
typhoid, the patch is centrally located on the tongue surrounded by a red margin.
During the first week of typhoid, the dorsum of tongue is covered with a thin dirty white fur,
which occurs due to the toxins released by the pathogen. The coat gradually clears from the tip
and edges and completely disappears by the thid week of infection, leaving a glazed, dry or
smooth red tongue.
By observing the coat pattern, the Doctor would have diagnosed the fever well in advance.
Solution
One of the four prominent symptoms of typhoid is coated tongue. Even before the onset of fever,
the tongue appears furred, with centrally coated and a red margin around. There are other illness
also which shows coated tongue, but with different appearance. In malaria, the tongue is
uniformly coated whereas in candidia thrush, a thick white layer of which patch is formed. In
typhoid, the patch is centrally located on the tongue surrounded by a red margin.
During the first week of typhoid, the dorsum of tongue is covered with a thin dirty white fur,
which occurs due to the toxins released by the pathogen. The coat gradually clears from the tip
and edges and completely disappears by the thid week of infection, leaving a glazed, dry or
smooth red tongue.
By observing the coat pattern, the Doctor would have diagnosed the fever well in advance..
Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a
template for new DNA synthesis, a process called semi-conservative replication. At that time,
there were three proposed models for DNA replication put forward by the scientific community
after DNA structure had been discovered- Semi-conservative, conservative and dispersive
replication.
Semi-conservative replication: In this model, the two strands of DNA unwind from each other,
and each acts as a template for synthesis of a new, complementary strand. This results in two
DNA molecules with one original strand and one new strand.
Conservative replication. In this model, DNA replication results in one molecule that consists of
both original DNA strands and another molecule that consists of two new strands with exactly
the same sequences as the original molecule.
Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli
bacteria) to establish that DNA replicates using the semi-conservative model of replication as
follows:
Results of analysis:
Solution
Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a
template for new DNA synthesis, a process called semi-conservative replication. At that time,
there were three proposed models for DNA replication put forward by the scientific community
after DNA structure had been discovered- Semi-conservative, conservative and dispersive
replication.
Semi-conservative replication: In this model, the two strands of DNA unwind from each other,
and each acts as a template for synthesis of a new, complementary strand. This results in two
DNA molecules with one original strand and one new strand.
Conservative replication. In this model, DNA replication results in one molecule that consists of
both original DNA strands and another molecule that consists of two new strands with exactly
the same sequences as the original molecule.
Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli
bacteria) to establish that DNA replicates using the semi-conservative model of replication as
follows:
Results of analysis:.
Introduction One of the key goals for the Windows Subsystem for Li.pdf
Introduction
One of the key goals for the Windows Subsystem for Linux is to allow users to work with their
files as they would on Linux, while giving full interoperability with files the user already has on
their Windows machine. Unlike a virtual machine, where you have to use network shares or
other solutions to share files between the host and guest OS, WSL has direct access to all your
Windows drives to allow for easy interop.
Windows file systems differ substantially from Linux file systems, and this post looks into how
WSL bridges those two worlds.
File systems on Linux
Linux abstracts file systems operations through the Virtual File System (VFS), which provides
both an interface for user mode programs to interact with the file system (through system calls
such as open, read, chmod, stat, etc.) and an interface that file systems have to implement. This
allows multiple file systems to coexist, providing the same operations and semantics, with VFS
giving a single namespace view of all these file systems to the user.
File systems are mounted on different directories in this namespace. For example, on a typical
Linux system your hard drive may be mounted at the root, /, with directories such as /dev, /proc,
/sys, and /mnt/cdrom all mounting different file systems which may be on different devices.
Examples of file systems used on Linux include ext4, rfs, FAT, and others.
VFS implements the various system calls for file system operations by using a number of data
structures such as inodes, directory entries and files, and related callbacks that file systems must
implement.
Inodes
The inode is the central data structure used in VFS. It represents a file system object such as a
regular file, directory, symbolic link, etc. An inode contains information about the file type, size,
permissions, last modified time, and other attributes. For many common Linux disk file systems
such as ext4, the on-disk data structures used to represent file metadata directly correspond to the
inode structure used by the Linux kernel.
While an inode represents a file, it does not represent a file name. A single file may have
multiple names, or hard links, but only one inode.
File systems provide a lookup callback to VFS which is used to retrieve an inode for a particular
file, based on the parent inode and the child name. File systems must implement a number of
other inode operations such as chmod, stat, open, etc.
Directory entries
VFS uses a directory entry cache to represent your file system namespace. Directory entries only
exist in memory, and contain a pointer to the inode for the file. For example, if you have a path
like /home/user/foo, there is a directory entry for home, user, and foo, each with a pointer to an
inode. Directory entries are cached for fast lookup, but if an entry is not yet in the cache, the
inode lookup operation is used to retrieve the inode from the file system so a new directory entry
can be created.
File objects
When an inode is opened, .
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- 1. Comments added... import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; /** * * @author prmsh */ public class SaveMatrix { public static void main(String args[]) { //declaring matrix.. String[][] data = new String[4][4]; int row = 0; // opening and reading file File file = new File("matrix.txt"); try { Scanner sc = new Scanner(file); while (sc.hasNextLine()) { String line = sc.nextLine(); //splitting each row by spaces String[] lineArray = line.split(" "); //storing into array for(int i=0; i Solution Comments added... import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; /** *
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