Minimum steps to reach target by a Knight | Set 1

Given a square chessboard of N x N size, the position of Knight and position of a target is given. We need to find out minimum steps a Knight will take to reach the target position.

Examples:



In above diagram Knight takes 3 step to reach 
from (4, 5) to (1, 1) (4, 5) -> (5, 3) -> (3, 2) 
-> (1, 1)  as shown in diagram

This problem can be seen as shortest path in unweighted graph. Therefore we use BFS to solve this problem. We try all 8 possible positions where a Knight can reach from its position. If reachable position is not already visited and is inside the board, we push this state into queue with distance 1 more than its parent state. Finally we return distance of target position, when it gets pop out from queue.
Below code implements BFS for searching through cells, where each cell contains its coordinate and distance from starting node. In worst case, below code visits all cells of board, making worst-case time complexity as O(N^2)

C++

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// C++ program to find minimum steps to reach to
// specific cell in minimum moves by Knight
#include <bits/stdc++.h>
using namespace std;
  
// structure for storing a cell's data
struct cell
{
    int x, y;
    int dis;
    cell() {}
    cell(int x, int y, int dis) : x(x), y(y), dis(dis) {}
};
  
// Utility method returns true if (x, y) lies
// inside Board
bool isInside(int x, int y, int N)
{
    if (x >= 1 && x <= N && y >= 1 && y <= N)
        return true;
    return false;
}
  
// Method returns minimum step to reach target position
int minStepToReachTarget(int knightPos[], int targetPos[],
                                                int N)
{
    // x and y direction, where a knight can move
    int dx[] = {-2, -1, 1, 2, -2, -1, 1, 2};
    int dy[] = {-1, -2, -2, -1, 1, 2, 2, 1};
  
    // queue for storing states of knight in board
    queue<cell> q;
  
    // push starting position of knight with 0 distance
    q.push(cell(knightPos[0], knightPos[1], 0));
  
    cell t;
    int x, y;
    bool visit[N + 1][N + 1];
  
    // make all cell unvisited
    for (int i = 1; i <= N; i++)
        for (int j = 1; j <= N; j++)
            visit[i][j] = false;
  
    // visit starting state
    visit[knightPos[0]][knightPos[1]] = true;
  
    // loop untill we have one element in queue
    while (!q.empty())
    {
        t = q.front();
        q.pop();
  
        // if current cell is equal to target cell,
        // return its distance
        if (t.x == targetPos[0] && t.y == targetPos[1])
            return t.dis;
  
        // loop for all reachable states
        for (int i = 0; i < 8; i++)
        {
            x = t.x + dx[i];
            y = t.y + dy[i];
  
            // If reachable state is not yet visited and
            // inside board, push that state into queue
            if (isInside(x, y, N) && !visit[x][y]) {
                visit[x][y] = true;
                q.push(cell(x, y, t.dis + 1));
            }
        }
    }
}
  
// Driver code to test above methods
int main()
{
    int N = 30;
    int knightPos[] = {1, 1};
    int targetPos[] = {30, 30};
    cout << minStepToReachTarget(knightPos, targetPos, N);
    return 0;
}

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Python3

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# Python3 code to find minimum steps to reach 
# to specific cell in minimum moves by Knight 
class cell:
      
    def __init__(self, x = 0, y = 0, dist = 0):
        self.x = x
        self.y = y
        self.dist = dist
          
# checks whether given position is 
# inside the board
def isInside(x, y, N):
    if (x >= 1 and x <= N and 
        y >= 1 and y <= N): 
        return True
    return False
      
# Method returns minimum step to reach
# target position 
def minStepToReachTarget(knightpos, 
                         targetpos, N):
      
    #all possible movments for the knight
    dx = [2, 2, -2, -2, 1, 1, -1, -1]
    dy = [1, -1, 1, -1, 2, -2, 2, -2]
      
    queue = []
      
    # push starting position of knight
    # with 0 distance
    queue.append(cell(knightpos[0], knightpos[1], 0))
      
    # make all cell unvisited 
    visited = [[False for i in range(N + 1)] 
                      for j in range(N + 1)]
      
    # visit starting state
    visited[knightpos[0]][knightpos[1]] = True
      
    # loop untill we have one element in queue 
    while(len(queue) > 0):
          
        t = queue[0]
        queue.pop(0)
          
        # if current cell is equal to target 
        # cell, return its distance 
        if(t.x == targetpos[0] and 
           t.y == targetpos[1]):
            return t.dist
              
        # iterate for all reachable states 
        for i in range(8):
              
            x = t.x + dx[i]
            y = t.y + dy[i]
              
            if(isInside(x, y, N) and not visited[x][y]):
                visited[x][y] = True
                queue.append(cell(x, y, t.dist + 1))
  
# Driver Code    
if __name__=='__main__'
    N = 30
    knightpos = [1, 1]
    targetpos = [30, 30]
    print(minStepToReachTarget(knightpos,
                               targetpos, N))
      
# This code is contributed by 
# Kaustav kumar Chanda

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Output:

20

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Improved By : Kaustav kumar Chanda