Given an NxN chessboard and a Knight at position (x,y). The Knight has to take exactly K steps, where at each step it chooses any of the 8 directions uniformly at random. What is the probability that the Knight remains in the chessboard after taking K steps, with the condition that it can’t enter the board again once it leaves it.

Examples:

Let's take: 8x8 chessboard, initial position of the knight : (0, 0), number of steps : 1 At each step, the Knight has 8 different positions to choose from.

If it starts from (0, 0), after taking one step it will lie inside the board only at 2 out of 8 positions, and will lie outside at other positions. So, the probability is 2/8 = 0.25

One thing that we can observe is that at every step the Knight has 8 choices to choose from. Suppose, the Knight has to take k steps and after taking the Kth step the knight reaches (x,y). There are 8 different positions from where the Knight can reach to (x,y) in one step, and they are: (x+1,y+2), (x+2,y+1), (x+2,y-1), (x+1,y-2), (x-1,y-2), (x-2,y-1), (x-2,y+1), (x-1,y+2).

What if we already knew the probabilities of reaching these 8 positions after K-1 steps? Then, the final probability after K steps will simply be equal to the (Σ probability of reaching each of these 8 positions after K-1 steps)/8;

Here we are dividing by 8 because each of these 8 positions have 8 choices and position (x,y) is one of the choice.

For the positions that lie outside the board, we will either take their probabilities as 0 or simply neglect it.

Since, we need to keep track of the probabilities at each position for every number of steps, we need Dynamic Programming to solve this problem.

We are going to take an array dp[x][y][steps] which will store the probability of reaching (x,y) after (steps) number of moves.

Base case : if number of steps is 0, then the probability that the Knight will remain inside the board is 1.

Here is the implementation :

## C++

// C++ program to find the probability of the // Knight to remain inside the chessboard after // taking exactly K number of steps #include <bits/stdc++.h> using namespace std; // size of the chessboard #define N 8 // direction vector for the Knight int dx[] = { 1, 2, 2, 1, -1, -2, -2, -1 }; int dy[] = { 2, 1, -1, -2, -2, -1, 1, 2 }; // returns true if the knight is inside the chessboard bool inside(int x, int y) { return (x >= 0 and x < N and y >= 0 and y < N); } // Bottom up approach for finding the probability to // go out of chessboard. double findProb(int start_x, int start_y, int steps) { // dp array double dp1[N][N][N]; // for 0 number of steps, each position // will have probability 1 for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) dp1[i][j][0] = 1; // for every number of steps s for (int s = 1; s <= steps; ++s) { // for every position (x,y) after // s number of steps for (int x = 0; x < N; ++x) { for (int y = 0; y < N; ++y) { double prob = 0.0; // for every position reachable from (x,y) for (int i = 0; i < 8; ++i) { int nx = x + dx[i]; int ny = y + dy[i]; // if this position lie inside the board if (inside(nx, ny)) prob += dp1[nx][ny][s-1] / 8.0; } // store the result dp1[x][y][s] = prob; } } } // return the result return dp1[start_x][start_y][steps]; } // Driver program int main() { // number of steps int K = 3; cout << findProb(0, 0, K) << endl; return 0; }

## Java

// Java program to find the probability // of the Knight to remain inside the // chessboard after taking exactly K // number of steps class GFG { // size of the chessboard static final int N = 8; // direction vector for the Knight static int dx[] = { 1, 2, 2, 1, -1, -2, -2, -1 }; static int dy[] = { 2, 1, -1, -2, -2, -1, 1, 2 }; // returns true if the knight is // inside the chessboard static boolean inside(int x, int y) { return (x >= 0 && x < N && y >= 0 && y < N); } // Bottom up approach for finding // the probability to go out of // chessboard. static double findProb(int start_x, int start_y, int steps) { // dp array double dp1[][][] = new double[N][N][N]; // for 0 number of steps, each position // will have probability 1 for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) dp1[i][j][0] = 1; // for every number of steps s for (int s = 1; s <= steps; ++s) { // for every position (x, y) after // s number of steps for (int x = 0; x < N; ++x) { for (int y = 0; y < N; ++y) { double prob = 0.0; // for every position reachable // from (x, y) for (int i = 0; i < 8; ++i) { int nx = x + dx[i]; int ny = y + dy[i]; // if this position lie // inside the board if (inside(nx, ny)) prob += dp1[nx][ny][s - 1] / 8.0; } // store the result dp1[x][y][s] = prob; } } } // return the result return dp1[start_x][start_y][steps]; } // Driver code public static void main(String[] args) { // number of steps int K = 3; System.out.println(findProb(0, 0, K)); } } // This code is contributed by Anant Agarwal.

## Python3

# Python3 program to find the probability of # the Knight to remain inside the chessboard # after taking exactly K number of steps # size of the chessboard N = 8 # Direction vector for the Knight dx = [ 1, 2, 2, 1, -1, -2, -2, -1 ] dy = [ 2, 1, -1, -2, -2, -1, 1, 2 ] # returns true if the knight # is inside the chessboard def inside(x, y): return (x >= 0 and x < N and y >= 0 and y < N) # Bottom up approach for finding the # probability to go out of chessboard. def findProb(start_x, start_y, steps): # dp array dp1 = [[[0 for i in range(N + 1)] for j in range(N + 1)] for k in range(N + 1)] # For 0 number of steps, each # position will have probability 1 for i in range(N): for j in range(N): dp1[i][j][0] = 1 # for every number of steps s for s in range(1, steps + 1): # for every position (x,y) after # s number of steps for x in range(N): for y in range(N): prob = 0.0 # For every position reachable from (x,y) for i in range(8): nx = x + dx[i] ny = y + dy[i] # if this position lie inside the board if (inside(nx, ny)): prob += dp1[nx][ny][s-1] / 8.0 # store the result dp1[x][y][s] = prob # return the result return dp1[start_x][start_y][steps] # Driver code # number of steps K = 3 print(findProb(0, 0, K)) # This code is contributed by Anant Agarwal.

Output:

0.125

**Time Complexity**: O(NxNxKx8) which is O(NxNxK), where N is the size of the board and K is the number of steps.

**Space Complexity**: O(NxNxK)

This article is contributed by **Avinash Kumar Saw**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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