Knight Movement Problems: BFS and DP

Problem 1: Minimum Knight Steps (BFS)

Beginning from location (0,0) on an infinite board, what is the minimum number of steps needed for a knight to get to some other arbitrary location (x,y)?

Constraints: $1 \le x, y \le 8$

Why BFS?

This problem asks for the **minimum number of steps**, which immediately suggests a **Breadth-First Search (BFS)**.

BFS explores the graph layer by layer (level 1 moves, then level 2 moves, etc.).
Since all moves (edges) have a weight of 1, the first time BFS reaches the target $(x, y)$, it guarantees that it has found the shortest path.
The data structure used to manage the search is a **Queue (FIFO)**.

Knight's Move Set (The 8 Possible Jumps)

A knight moves in an 'L' shape: 2 squares in one direction (horizontal or vertical) and 1 square in the perpendicular direction. From $(0, 0)$, the possible moves are:

(1, 2)

(-1, 2)

(1, -2)

(-1, -2)

(2, 1)

(-2, 1)

(2, -1)

(-2, -1)

The BFS Algorithm Steps

Data Structures Needed

**Queue:** Stores the positions to visit. Each element is `(x, y, steps)`.
**Visited Set/Map:** Stores all `(x, y)` locations already processed to prevent cycles and redundant work. This is crucial for an infinite board.

Detailed Steps

**Initialization:** Add the starting position $\mathbf{(0, 0)}$ with $\mathbf{0}$ steps to the Queue and mark it as visited.
**Loop:** While the Queue is not empty, dequeue the front position $\mathbf{(x_c, y_c, s_c)}$.
**Check Goal:** If $\mathbf{(x_c, y_c)}$ is the target $\mathbf{(x, y)}$, return $\mathbf{s_c}$.
**Generate Next Moves:** Iterate through the 8 possible Knight moves (dx, dy). Calculate the new position $\mathbf{(x_n, y_n)}$.
**Enqueue and Mark:** If $\mathbf{(x_n, y_n)}$ has not been visited, mark it as visited and enqueue $\mathbf{(x_n, y_n, s_c + 1)}$.

Algorithm Complexity

For a general graph, Time is $O(V + E)$ (Vertices + Edges). On a chessboard, we can approximate the search space to be the area reachable within the minimum number of steps $M$.

Time: $O(M^2)$ (approximated search space)

Space: $O(M^2)$ (for the Visited Set and Queue)

Interactive BFS Demo (Start at 0, 0)

Target X (1-8):

Target Y (1-8):

Search Status

Waiting for target coordinates...

BFS Queue (Next to Visit)

Queue is empty.

Code Snippet (BFS Main Loop)

function bfs() {
    // Queue stores {x, y, steps}
    while (queue.length > 0) {
        let {x, y, steps} = queue.shift();
        if (x === targetX && y === targetY) {
            return steps; // Found shortest path!
        }
        
        // Explore 8 moves
        for (let i = 0; i < 8; i++) {
            let nextX = x + DX[i];
            let nextY = y + DY[i];
            let key = `${nextX},${nextY}`;

            if (!visited.has(key)) {
                visited.add(key);
                queue.push({x: nextX, y: nextY, steps: steps + 1});
            }
        }
    }
}

Problem 2: Knight Probability (Dynamic Programming)

Calculate the probability that a knight, starting at $(R, C)$ on an $N \times N$ board, remains on the board after exactly $K$ moves, where each move is random.

Constraints: $1 \le N \le 25, 0 \le K \le 100$.

The DP Approach

We use Dynamic Programming because the probability of the knight being at a certain square at step $k$ depends only on the probabilities of being at its neighboring squares at step $k-1$.

State Definition:

$DP[r][c]$ = Probability the knight is at $(r, c)$ after the current number of steps.

Transition:

The new probability $DP[r][c]$ is calculated by summing $\mathbf{1/8 \times \text{prevDP}[r'][c']}$ for all 8 positions $(r', c')$ from which the knight could jump to $(r, c)$ in one move.

Probability Calculator

Board Size N (1-25):

Total Moves K (0-100):

Start Row R (0-indexed):

Start Col C (0-indexed):

Result

Enter parameters and click 'Calculate' to see the result.

Code Snippet (DP Core Logic)

function calculateKnightProbability(n, k, startR, startC) {
    if (k === 0) { return 1.0; }
    
    // DP table: probability of being at (r, c)
    let prevDp = Array(n).fill(0).map(() => Array(n).fill(0));
    
    // Base case: 100% probability at start position
    if (startR >= 0 && startR < n && startC >= 0 && startC < n) {
        prevDp[startR][startC] = 1.0;
    } else {
        return 0.0;
    }
    
    // Knight's moves (same as before)
    const DX = [1, 1, 2, 2, -1, -1, -2, -2];
    const DY = [2, -2, 1, -1, 2, -2, 1, -1];

    for (let s = 1; s <= k; s++) {
        let currentDp = Array(n).fill(0).map(() => Array(n).fill(0));
        
        for (let r = 0; r < n; r++) {
            for (let c = 0; c < n; c++) {
                // Check all 8 possible previous positions (r_prev, c_prev)
                for (let i = 0; i < 8; i++) {
                    const prevR = r - DX[i];
                    const prevC = c - DY[i];
                    
                    // Check if the previous position was on the board
                    if (prevR >= 0 && prevR < n && prevC >= 0 && prevC < n) {
                        // The probability of reaching (r, c) is the sum of probabilities 
                        // from all valid previous spots, multiplied by the move chance (1/8)
                        currentDp[r][c] += prevDp[prevR][prevC] * 0.125;
                    }
                }
            }
        }
        prevDp = currentDp; // Move forward one step
    }

    // Sum all probabilities in the final DP table
    let totalProbability = 0;
    for (let r = 0; r < n; r++) {
        for (let c = 0; c < n; c++) {
            totalProbability += prevDp[r][c];
        }
    }

    return totalProbability;
}

♞ Knight Movement Problems: BFS and DP