Minimum Knight Moves (BFS) Interactive Tutorial

Problem Statement (BFS Introduction)

Beginning from location (0,0), 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: $O(x \cdot y)$ or $O(M^2)$ (approximated search space)

Space: $O(x \cdot y)$ (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 (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});
            }
        }
    }
}

♞ Minimum Knight Moves using BFS