--- id: 5a23c84252665b21eecc7ed5 title: Knight's tour challengeType: 5 forumTopicId: 302297 --- ## Description
Knight's TourProblem: You have an empty w * h chessboard, but for a single knight on some square. The knight must perform a sequence of legal moves that result in the knight visiting every square on the chessboard exactly once. Note that it is not a requirement that the tour be "closed"; that is, the knight need not end within a single move of its start position.
## Instructions
Write a function that takes w and h as parameters and returns the number of initial positions from where it is possible to achieve the task stated above.
## Tests
```yml tests: - text: knightTour should be a function. testString: assert(typeof knightTour == 'function'); - text: knightTour(6, 6) should return a number. testString: assert(typeof knightTour(6, 6) == 'number'); - text: knightTour(6, 6) should return 35. testString: assert.equal(knightTour(6, 6), 35); - text: knightTour(5, 6) should return 20. testString: assert.equal(knightTour(5, 6), 20); - text: knightTour(4, 6) should return 10. testString: assert.equal(knightTour(4, 6), 10); - text: knightTour(7, 3) should return 4. testString: assert.equal(knightTour(7, 3), 4); - text: knightTour(8, 6) should return 47. testString: assert.equal(knightTour(8, 6), 47); ```
## Challenge Seed
```js function knightTour(w, h) { } ```
## Solution
```js function knightTour(w, h) { var b, cnt = 0; var dx = [-2, -2, -1, 1, 2, 2, 1, -1]; var dy = [-1, 1, 2, 2, 1, -1, -2, -2]; function init_board() { var i, j, k, x, y; // * b is board; a is board with 2 rows padded at each side for (i = 0; i < h; i++) { for (j = 0; j < w; j++) { b[i][j] = 255; } } for (i = 0; i < h; i++) { for (j = 0; j < w; j++) { for (k = 0; k < 8; k++) { (x = j + dx[k]), (y = i + dy[k]); if (b[i][j] == 255) b[i][j] = 0; if (x >= 0 && x < w && y >= 0 && y < h) b[i][j]++; } } } } function walk_board(x, y) { var i, nx, ny, least; var steps = 0; // printf(E"H"E"J"E"%d;%dH"E"32m[]"E"m", y + 1, 1 + 2 * x); while (1) { // * occupy cell b[y][x] = 255; // * reduce all neighbors' neighbor count for (i = 0; i < 8; i++) if (y + dy[i] >= 0 && x + dx[i] >= 0 && y + dy[i] < h && x + dx[i] < w) b[y + dy[i]][x + dx[i]]--; // find neighbor with lowest neighbor count least = 255; for (i = 0; i < 8; i++) { if (y + dy[i] >= 0 && x + dx[i] >= 0 && y + dy[i] < h && x + dx[i] < w) if (b[y + dy[i]][x + dx[i]] < least) { nx = x + dx[i]; ny = y + dy[i]; least = b[ny][nx]; } } if (least > 7) { return steps == w * h - 1; } steps++; (x = nx), (y = ny); } } function solve(x, y) { b = new Array(h); for (var i = 0; i < h; i++) b[i] = new Array(w); init_board(); if (walk_board(x, y)) { cnt++; } } for (var i = 0; i < h; i++) { for (var j = 0; j < w; j++) { solve(j, i); } } return cnt; } ```