For the Shuffle Method, Write the Steps for Its Algorithm for the Shuffle Method Elevents
Given an array, write a program to generate a random permutation of array elements. This question is also asked as "shuffle a deck of cards" or "randomize a given array". Here shuffle means that every permutation of array element should equally likely.
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Let the given array be arr[]. A simple solution is to create an auxiliary array temp[] which is initially a copy of arr[]. Randomly select an element from temp[], copy the randomly selected element to arr[0] and remove the selected element from temp[]. Repeat the same process n times and keep copying elements to arr[1], arr[2], … . The time complexity of this solution will be O(n^2).
Fisher–Yates shuffle Algorithm works in O(n) time complexity. The assumption here is, we are given a function rand() that generates random number in O(1) time.
The idea is to start from the last element, swap it with a randomly selected element from the whole array (including last). Now consider the array from 0 to n-2 (size reduced by 1), and repeat the process till we hit the first element.
Following is the detailed algorithm
To shuffle an array a of n elements (indices 0..n-1): for i from n - 1 downto 1 do j = random integer with 0 <= j <= i exchange a[j] and a[i]
Following is implementation of this algorithm.
C++
#include<bits/stdc++.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
void swap ( int *a, int *b)
{
int temp = *a;
*a = *b;
*b = temp;
}
void printArray ( int arr[], int n)
{
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
cout << "\n" ;
}
void randomize ( int arr[], int n)
{
srand ( time (NULL));
for ( int i = n - 1; i > 0; i--)
{
int j = rand () % (i + 1);
swap(&arr[i], &arr[j]);
}
}
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};
int n = sizeof (arr) / sizeof (arr[0]);
randomize (arr, n);
printArray(arr, n);
return 0;
}
C
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
void swap ( int *a, int *b)
{
int temp = *a;
*a = *b;
*b = temp;
}
void printArray ( int arr[], int n)
{
for ( int i = 0; i < n; i++)
printf ( "%d " , arr[i]);
printf ( "\n" );
}
void randomize ( int arr[], int n )
{
srand ( time (NULL) );
for ( int i = n-1; i > 0; i--)
{
int j = rand () % (i+1);
swap(&arr[i], &arr[j]);
}
}
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};
int n = sizeof (arr)/ sizeof (arr[0]);
randomize (arr, n);
printArray(arr, n);
return 0;
}
Java
import java.util.Random;
import java.util.Arrays;
public class ShuffleRand
{
static void randomize( int arr[], int n)
{
Random r = new Random();
for ( int i = n- 1 ; i > 0 ; i--) {
int j = r.nextInt(i+ 1 );
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
System.out.println(Arrays.toString(arr));
}
public static void main(String[] args)
{
int [] arr = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 };
int n = arr.length;
randomize (arr, n);
}
}
Python
import random
def randomize (arr, n):
for i in range (n - 1 , 0 , - 1 ):
j = random.randint( 0 ,i + 1 )
arr[i],arr[j] = arr[j],arr[i]
return arr
arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 ]
n = len (arr)
print (randomize(arr, n))
C#
using System;
class GFG
{
static void randomize( int []arr, int n)
{
Random r = new Random();
for ( int i = n - 1; i > 0; i--)
{
int j = r.Next(0, i+1);
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
for ( int i = 0; i < n; i++)
Console.Write(arr[i] + " " );
}
static void Main()
{
int [] arr = {1, 2, 3, 4,
5, 6, 7, 8};
int n = arr.Length;
randomize (arr, n);
}
}
PHP
<?php
function randomize ( $arr , $n )
{
for ( $i = $n - 1; $i >= 0; $i --)
{
$j = rand(0, $i +1);
$tmp = $arr [ $i ];
$arr [ $i ] = $arr [ $j ];
$arr [ $j ] = $tmp ;
}
for ( $i = 0; $i < $n ; $i ++)
echo $arr [ $i ]. " " ;
}
$arr = array (1, 2, 3, 4,
5, 6, 7, 8);
$n = count ( $arr );
randomize( $arr , $n );
?>
Javascript
<script>
let printArray = (arr, n)=>
{
ans = '' ;
for (let i = 0; i < n; i++)
{
ans += arr[i] + " " ;
}
console.log(ans);
}
let randomize = (arr, n) =>
{
for (let i = n - 1; i > 0; i--)
{
let j = Math.floor(Math.random() * (i + 1));
[arr[i], arr[j]] = [arr[j], arr[i]];
}
}
let arr = [1, 2, 3, 4, 5, 6, 7, 8];
let n = arr.length;
randomize (arr, n);
printArray(arr, n);
</script>
Output :
7 8 4 6 3 1 2 5
The above function assumes that rand() generates a random number.
Time Complexity: O(n), assuming that the function rand() takes O(1) time.
Auxiliary Space: O(1)
How does this work?
The probability that ith element (including the last one) goes to last position is 1/n, because we randomly pick an element in first iteration.
The probability that ith element goes to second last position can be proved to be 1/n by dividing it in two cases.
Case 1: i = n-1 (index of last element):
The probability of last element going to second last position is = (probability that last element doesn't stay at its original position) x (probability that the index picked in previous step is picked again so that the last element is swapped)
So the probability = ((n-1)/n) x (1/(n-1)) = 1/n
Case 2: 0 < i < n-1 (index of non-last):
The probability of ith element going to second position = (probability that ith element is not picked in previous iteration) x (probability that ith element is picked in this iteration)
So the probability = ((n-1)/n) x (1/(n-1)) = 1/n
We can easily generalize above proof for any other position.
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For the Shuffle Method, Write the Steps for Its Algorithm for the Shuffle Method Elevents
Source: https://www.geeksforgeeks.org/shuffle-a-given-array-using-fisher-yates-shuffle-algorithm/