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我们在 BSV 区块链上实现了一个数独游戏智能合约,利用之前介绍过的一种合约范式可以将游戏中寻找解题方案的过程外包上链。因为求解数独问题的计算工作量会随着其行列数快速增长,实际上它也是一个 NP-完全 问题。不过我们可以借由 sCrypt 智能合约巧妙地寻求答案,只需要验证答案提供者所给出的解答是否满足要求即可,这样即可以将复杂的求解计算过程实现链下外包。
sCrypt 合约代码如下:
import "array.scrypt"; import "arrayUtil.scrypt"; contract Sudoku { bytes board; static const int N = 9; static const bytes EMPTY = b'00'; constructor(bytes board) { this.board = board; } function merge(bytes solution) : bytes { bytes newBoard = this.board; loop (N) : i { loop (N) : j { int value = readValue(newBoard, i, j); int inputValue = readValue(solution, i, j); if (value == 0) { require(inputValue <= 9); newBoard = setValue(newBoard, i, j, inputValue); } else { require(value == inputValue); } } } return newBoard; } public function solve(bytes solution) { require(len(solution) == Sudoku.N * Sudoku.N); bytes newBord = this.merge(solution); Array rowArray = new Array(); Array colArray = new Array(); Array squareArray = new Array(); loop (N) : i { loop (N) : j { // check for duplicate // in a row int rowElem = readValue(newBord, i, j); require(rowArray.indexOf(rowElem) == -1); rowArray.push(rowElem); // in a column int colElem = readValue(newBord, j, i); require(colArray.indexOf(colElem) == -1); colArray.push(colElem); // in a subgrid int squareElem = readSquareValue(newBord, i, j); require(squareArray.indexOf(squareElem) == -1); squareArray.push(squareElem); } rowArray.clear(); colArray.clear(); squareArray.clear(); } require(true); } static function readValue(bytes board, int i, int j) : int { return Utils.fromLEUnsigned(ArrayUtil.getElemAt(board, Sudoku.index(i, j))); } static function setValue(bytes board, int i, int j, int value) : bytes { return ArrayUtil.setElemAt(board, index(i, j), Utils.toLEUnsigned(value, 1)); } static function readSquareValue(bytes board, int i, int j) : int { return Utils.fromLEUnsigned(ArrayUtil.getElemAt(board, Sudoku.indexSquare(i, j))); } static function index(int row, int col) : int { return row * Sudoku.N + col; } static function indexSquare(int i, int j) : int { int row = i / 3 * 3 + j / 3; int col = i % 3 * 3 + j % 3; return Sudoku.index(row, col); } }
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