# Probability of Knight to remain in the chessboard

• Difficulty Level : Hard
• Last Updated : 19 Apr, 2021

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```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Approach:

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 has 8 choices and position (x,y) is one of the choices.

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 the number of steps is 0, then the probability that the Knight will remain inside the board is 1.
Below is the implementation of the above approach:

## 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][steps + 1];

// 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
int main()
{
// number of steps
int K = 3;

// Function Call
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][steps + 1];

// 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;

// Function Call
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+5)]
for j in range(N+5)]
for k in range(steps + 5)]

# 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

# Function Call
print(findProb(0, 0, K))

# This code is contributed by Anant Agarwal.
```

## C#

```// C# program to find the
// probability of the Knight
// to remain inside the
// chessboard after taking
// exactly K number of steps
using System;

class GFG {

// size of the chessboard
static 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 bool 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, steps+1];

// 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
static void Main()
{
// number of steps
int K = 3;

// Function Call
Console.WriteLine(findProb(0, 0, K));
}
}

// This code is contributed
// by Sam007```

## PHP

```<?php
// PHP 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 = array(1, 2, 2, 1, -1, -2, -2, -1 );
\$dy = array(2, 1, -1, -2, -2, -1, 1, 2 );

// returns true if the knight is
// inside the chessboard
function inside(\$x, \$y)
{
global \$N;
return (\$x >= 0 and \$x < \$N and
\$y >= 0 and \$y < \$N);
}

// Bottom up approach for finding the
// probability to go out of chessboard.
function findProb(\$start_x, \$start_y, \$steps)
{
global \$N, \$dx, \$dy;

// dp array
\$dp1 = array_fill(0, \$N,
array_fill(0, \$N,
array_fill(0, \$steps+1, NULL)));

// for 0 number of steps, each
// position will have probability 1
for (\$i = 0; \$i < \$N; ++\$i)
for (\$j = 0; \$j < \$N; ++\$j)
\$dp1[\$i][\$j][0] = 1;

// for every number of steps s
for (\$s = 1; \$s <= \$steps; ++\$s)
{
// for every position (x,y) after
// s number of steps
for (\$x = 0; \$x < \$N; ++\$x)
{
for (\$y = 0; \$y < \$N; ++\$y)
{
\$prob = 0.0;

// for every position
// reachable from (x,y)
for (\$i = 0; \$i < 8; ++\$i)
{
\$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;

// Function Call
echo findProb(0, 0, \$K) . "\n";

// This code is contributed by ita_c
?>

```

## Javascript

```<script>

// Javascript program to find the probability
// of the Knight to remain inside the
// chessboard after taking exactly K
// number of steps

// size of the chessboard
let N = 8;

// direction vector for the Knight
let dx = [ 1, 2, 2, 1, -1, -2, -2, -1 ];

let dy = [2, 1, -1, -2, -2, -1, 1, 2];

// returns true if the knight is
// inside the chessboard
function inside(x,y)
{
return (x >= 0 && x < N && y >= 0 && y < N);
}

// Bottom up approach for finding
// the probability to go out of
// chessboard.
function findProb(start_x, start_y, steps)
{
// dp array
let dp1 = new Array(N);
for(let i = 0; i < N; i++)
{
dp1[i] = new Array(N);
for(let j = 0; j < N; j++)
{
dp1[i][j] = new Array(steps + 1);
for(let k = 0; k < steps + 1; k++)
{
dp1[i][j][k] = 0;
}
}
}

// for 0 number of steps, each position
// will have probability 1
for (let i = 0; i < N; ++i)
for (let j = 0; j < N; ++j)
dp1[i][j][0] = 1;

// for every number of steps s
for (let s = 1; s <= steps; ++s)
{

// for every position (x, y) after
// s number of steps
for (let x = 0; x < N; ++x)
{

for (let y = 0; y < N; ++y)
{

let prob = 0.0;

// for every position reachable
// from (x, y)
for (let i = 0; i < 8; ++i)
{
let nx = x + dx[i];
let 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
let K = 3;

// Function Call
document.write(findProb(0, 0, K));

// This code is contributed by rag2127
</script> ```
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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.