Find probability of selecting element from kth column after N iterations

Given a matrix M of order N*M where Mij represent the occurrence of j in row i. The task is to find the probability of occurrence of given k in last row after applying the given operation:

  • Starting from 1st row, we select one element from any column of the entire row, and add it to the same column of next row. We repeating this till the last row.

Examples:

Input: k = 1, M[][] = 
{{0, 1},
{1, 1}}   

Output:0.666667

Input: k = 1, M[][] = 
{{0, 1},
{1, 1}}  

Output: 0.333333

Approach :

  • Pre-calculate a sum[] array which stores the sum of all elements of 1st row.
  • Fill 1st row of dp[][] by the the first row of M[][] elements.
  • Calculate the probability of selecting an element from each column of a particular row and add the same to the corresponding column.
  • Also, update the Sum[] of the row while updating the value of dp[][] matrix.
  • Finally, the value of dp[n][k] for given k is the required value for the probability of selecting element k after all iterations.

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
#define n 4
#define m 4
using namespace std;
  
// Function to calculate probability
float calcProbability(int M[][m], int k)
{
    // declare dp[][] and sum[]
    float dp[m][n], sum[n];
  
    // precalculate the first row
    for (int j = 0; j < n; j++) {
        dp[0][j] = M[0][j];
        sum[0] += dp[0][j];
    }
  
    // calculate the probability for
    // each element and update dp table
    for (int i = 1; i < m; i++) {
        for (int j = 0; j < n; j++) {
            dp[i][j] += dp[i - 1][j] / sum[i - 1] + M[i][j];
            sum[i] += dp[i][j];
        }
    }
  
    // return result
    return dp[n - 1][k - 1] / sum[n - 1];
}
  
// Driver code
int main()
{
  
    int M[m][n] = { { 1, 1, 0, 3 },
                    { 2, 3, 2, 3 },
                    { 9, 3, 0, 2 },
                    { 2, 3, 2, 2 } };
    int k = 3;
  
    cout << calcProbability(M, k);
    return 0;
}

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Java

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// Java implementation of the above approach 
  
class GFG
{
    final static int n = 4
    final static int m = 4 ;
  
    // Function to calculate probability 
    static float calcProbability(int M[][], int k) 
    
        // declare dp[][] and sum[] 
        float dp[][] = new float[m][n] ;
        float sum[] = new float[n]; 
      
        // precalculate the first row 
        for (int j = 0; j < n; j++) 
        
            dp[0][j] = M[0][j]; 
            sum[0] = sum[0] + dp[0][j]; 
        
      
        // calculate the probability for 
        // each element and update dp table 
        for (int i = 1; i < m; i++) 
        
            for (int j = 0; j < n; j++)
            
                dp[i][j] += dp[i - 1][j] / sum[i - 1] +
                            M[i][j]; 
                sum[i] += dp[i][j]; 
            
        
      
        // return result 
        return dp[n - 1][k - 1] / sum[n - 1]; 
    
      
    // Driver code 
    public static void main(String []args)
    
        int M[][] = { { 1, 1, 0, 3 }, 
                      { 2, 3, 2, 3 }, 
                      { 9, 3, 0, 2 }, 
                      { 2, 3, 2, 2 } }; 
        int k = 3
        System.out.println(calcProbability(M, k));
          
    
}
  
// This code is contributed by Ryuga

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Python3

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# Python3 implementation of the 
# above approach
  
n = 4
m = 4
  
# Function to calculate probability
def calcProbability(M, k):
  
    # declare dp[][] and sum[]
    dp = [[0 for i in range(n)] 
             for i in range(m)]
    Sum = [0 for i in range(n)]
  
    # precalculate the first row
    for j in range(n): 
        dp[0][j] = M[0][j]
        Sum[0] += dp[0][j]
      
    # calculate the probability for
    # each element and update dp table
    for i in range(1, m): 
        for j in range(n):
            dp[i][j] += (dp[i - 1][j] / 
                         Sum[i - 1] + M[i][j])
            Sum[i] += dp[i][j]
          
    # return result
    return dp[n - 1][k - 1] / Sum[n - 1]
  
# Driver code
  
M = [[ 1, 1, 0, 3 ],
     [ 2, 3, 2, 3 ],
     [ 9, 3, 0, 2 ],
     [ 2, 3, 2, 2 ]] 
k = 3
  
print(calcProbability(M, k))
  
# This code is contributed
# by mohit kumar 29

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C#

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// C# implementation of the above approach 
using System;
  
class GFG
{
    static int n = 4 ; 
    static int m = 4 ;
  
    // Function to calculate probability 
    static float calcProbability(int[,] M, int k) 
    
        // declare dp[][] and sum[] 
        float[,] dp = new float[m,n] ;
        float[] sum = new float[n]; 
      
        // precalculate the first row 
        for (int j = 0; j < n; j++) 
        
            dp[0, j] = M[0, j]; 
            sum[0] = sum[0] + dp[0, j]; 
        
      
        // calculate the probability for 
        // each element and update dp table 
        for (int i = 1; i < m; i++) 
        
            for (int j = 0; j < n; j++)
            
                dp[i, j] += dp[i - 1,j] / sum[i - 1] +
                            M[i, j]; 
                sum[i] += dp[i, j]; 
            
        
      
        // return result 
        return dp[n - 1,k - 1] / sum[n - 1]; 
    
      
    // Driver code 
    public static void Main()
    
        int[,] M = { { 1, 1, 0, 3 }, 
                    { 2, 3, 2, 3 }, 
                    { 9, 3, 0, 2 }, 
                    { 2, 3, 2, 2 } }; 
        int k = 3; 
        Console.Write(calcProbability(M, k));
    
}
  
// This code is contributed by Ita_c.

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PHP

Output:

0.201212


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