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Find the probability of reaching all points after N moves from point N
  • Difficulty Level : Hard
  • Last Updated : 15 Oct, 2020

Given N which denotes the initial position of the person on the number line. Also given L which is the probability of the person of going left. Find the probability of reaching all points on the number line after N moves from point N. Each move can be either to the left or to the right. 

Examples: 

Input: n = 2, l = 0.5 
Output: 0.2500 0.0000 0.5000 0.0000 0.2500 
The person can’t reach n-1th position and n+1th position in 2 passes, hence the probability is 0. The person can reach 0th position by only moving 2 steps left from index 2, hence the probability of reaching 0th index is 05*0.5=0.25. Similarly for 2n index, the probability is 0.25. 

Input: n = 3, l = 0.1 
Output: 0.0010 0.0000 0.0270 0.0000 0.2430 0.0000 0.7290 
The person can reach n-1th in three ways, i.e., (llr, lrl, rll) where l denotes left and r denotes right. Hence the probability of n-1th index is 0.027. Similarly probabilities for all other points are also calculated. 

Approach: Construct an array arr[n+1][2n+1] where each row represents a pass and the columns represent the points on the line. The maximum a person can move from index N is to 0th index at left or to 2nth index at right. Initially the probabilities after one pass will be left for arr[1][n-1] and right for arr[1][n+1]. The n-1 moves which are left will be done, hence the two possible moves will either be n steps to the right or n steps to the left. So the recurrence relations for right and left moves for all will be: 



arr[i][j] += (arr[i – 1][j – 1] * right) 
arr[i][j] += (arr[i – 1][j + 1] * left)

The summation of probabilities for all possible moves for any index will be stored in arr[n][i]. 

Below is the implementation of the above approach: 

C++




// C++ program to calculate the
// probability of reaching all points
// after N moves from point N
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the probabilities
void printProbabilities(int n, double left)
{
    double right = 1 - left;
 
    // Array where row represent the pass and the
    // coloumn represents the points on the line
    double arr[n + 1][2 * n + 1] = {{0}};
 
    // Initially the person can reach left
    // or right with one move
    arr[1][n + 1] = right;
    arr[1][n - 1] = left;
 
    // Calculate probabilities for N-1 moves
    for (int i = 2; i <= n; i++)
    {
        // when the person moves from ith index in
        // right direction when i moves has been done
        for (int j = 1; j <= 2 * n; j++)
            arr[i][j] += (arr[i - 1][j - 1] * right);
 
        // when the person moves from ith index in
        // left direction when i moves has been done
        for (int j = 2 * n - 1; j >= 0; j--)
            arr[i][j] += (arr[i - 1][j + 1] * left);
    }
 
    // Print the arr
    for (int i = 0; i < 2*n+1; i++)
        printf("%5.4f ", arr[n][i]);
}
 
// Driver Code
int main()
{
    int n = 2;
    double left = 0.5;
    printProbabilities(n, left);
    return 0;
}
 
/* This code is contributed by SujanDutta */


Java




// Java program to calculate the
// probability of reaching all points
// after N moves from point N
import java.util.*;
class GFG {
 
    // Function to calculate the probabilities
    static void printProbabilities(int n, double left)
    {
        double right = 1 - left;
 
        // Array where row represent the pass and the
        // coloumn represents the points on the line
        double[][] arr = new double[n + 1][2 * n + 1];
 
        // Initially the person can reach left
        // or right with one move
        arr[1][n + 1] = right;
        arr[1][n - 1] = left;
 
        // Calculate probabilities for N-1 moves
        for (int i = 2; i <= n; i++) {
 
            // when the person moves from ith index in
            // right direction when i moves has been done
            for (int j = 1; j <= 2 * n; j++) {
                arr[i][j] += (arr[i - 1][j - 1] * right);
            }
 
            // when the person moves from ith index in
            // left direction when i moves has been done
            for (int j = 2 * n - 1; j >= 0; j--) {
                arr[i][j] += (arr[i - 1][j + 1] * left);
            }
        }
        // Calling function to print the array with probabilities
        printArray(arr, n);
    }
 
    // Function that prints the array
    static void printArray(double[][] arr, int n)
    {
        for (int i = 0; i < arr[0].length; i++) {
            System.out.printf("%5.4f ", arr[n][i]);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 2;
        double left = 0.5;
        printProbabilities(n, left);
    }
}


Python3




# Python3 program to calculate the
# probability of reaching all points
# after N moves from point N
 
# Function to calculate the probabilities
def printProbabilities(n, left):
     
    right = 1 - left;
  
    # Array where row represent the pass
    # and the coloumn represents the
    # points on the line
    arr = [[0 for j in range(2 * n + 1)]
              for i in range(n + 1)]
  
    # Initially the person can reach
    # left or right with one move
    arr[1][n + 1] = right;
    arr[1][n - 1] = left;
  
    # Calculate probabilities
    # for N-1 moves
    for i in range(2, n + 1):
     
        # When the person moves from ith
        # index in right direction when i
        # moves has been done
        for j in range(1, 2 * n + 1):
            arr[i][j] += (arr[i - 1][j - 1] * right);
  
        # When the person moves from ith
        # index in left direction when i
        # moves has been done
        for j in range(2 * n - 1, -1, -1):
            arr[i][j] += (arr[i - 1][j + 1] * left);
     
    # Print the arr
    for i in range(2 * n + 1):
        print("{:5.4f} ".format(arr[n][i]), end = ' ');
 
# Driver code   
if __name__=="__main__":
     
    n = 2;
    left = 0.5;
     
    printProbabilities(n, left);
     
# This code is contributed by rutvik_56


C#




// C# program to calculate the
// probability of reaching all points
// after N moves from point N
using System;
 
class GFG
{
 
    // Function to calculate the probabilities
    static void printProbabilities(int n, double left)
    {
        double right = 1 - left;
 
        // Array where row represent the pass and the
        // coloumn represents the points on the line
        double[,] arr = new double[n + 1,2 * n + 1];
 
        // Initially the person can reach left
        // or right with one move
        arr[1,n + 1] = right;
        arr[1,n - 1] = left;
 
        // Calculate probabilities for N-1 moves
        for (int i = 2; i <= n; i++)
        {
 
            // when the person moves from ith index in
            // right direction when i moves has been done
            for (int j = 1; j <= 2 * n; j++)
            {
                arr[i, j] += (arr[i - 1, j - 1] * right);
            }
 
            // when the person moves from ith index in
            // left direction when i moves has been done
            for (int j = 2 * n - 1; j >= 0; j--)
            {
                arr[i, j] += (arr[i - 1, j + 1] * left);
            }
        }
        // Calling function to print the array with probabilities
        printArray(arr, n);
    }
 
    // Function that prints the array
    static void printArray(double[,] arr, int n)
    {
        for (int i = 0; i < GetRow(arr,0).GetLength(0); i++)
        {
            Console.Write("{0:F4} ", arr[n,i]);
        }
    }
     
    public static double[] GetRow(double[,] matrix, int row)
    {
        var rowLength = matrix.GetLength(1);
        var rowVector = new double[rowLength];
 
        for (var i = 0; i < rowLength; i++)
            rowVector[i] = matrix[row, i];
 
        return rowVector;
    }
     
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 2;
        double left = 0.5;
        printProbabilities(n, left);
    }
}
 
/* This code contributed by PrinciRaj1992 */


Output: 

0.2500 0.0000 0.5000 0.0000 0.2500



 

Time Complexity: O(N2
Auxiliary Space: O(N2)
 

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