我是靠谱客的博主 机智帅哥,这篇文章主要介绍PHP实现的数独求解问题示例,现在分享给大家,希望可以做个参考。

本文实例讲述了PHP实现的数独求解问题。分享给大家供大家参考,具体如下:

一、数独问题描述:

对于给出的数字二维数组,要求每行每列的数字不能重复。

二、实现代码:

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<?php /* 数独求解程序 * Created on 2017-4-18 * */ class Sudoku { var $matrix; function __construct($arr = null) { if ($arr == null) { $this->clear(); } else { $this->matrix = $arr; } } function clear() { for($i=0; $i<9; $i++) { for($j=0; $j<9; $j++) { $this->matrix[$i][$j] = array(); for ($k = 1; $k <= 9; $k++) { $this->matrix[$i][$j][$k] = $k; } } } } function setCell($row, $col, $value){ $this->matrix[$row][$col] = array($value => $value); //row for($i = 0; $i < 9; $i++){ if($i != $col){ if(! $this->removeValue($row, $i, $value)) { return false; } } } //col for($i = 0; $i < 9; $i++){ if($i != $row){ if(! $this->removeValue($i, $col, $value)) { return false; } } } //square $rs=intval($row / 3) * 3; $cs=intval($col / 3) * 3; for($i = $rs; $i < $rs + 3; $i++){ for($j = $cs; $j < $cs + 3; $j++){ if($i != $row && $j != $col){ if(! $this->removeValue($i, $j, $value)) return false; } } } return true; } function removeValue($row, $col, $value) { $count = count($this->matrix[$row][$col]); if($count == 1){ $ret = !isset($this->matrix[$row][$col][$value]); return $ret; } if (isset($this->matrix[$row][$col][$value])) { unset($this->matrix[$row][$col][$value]); if($count - 1 == 1) { return $this->setCell($row, $col, current($this->matrix[$row][$col])); } } return true; } function set($arr) { for ($i = 0; $i < 9; $i++) { for ($j = 0; $j < 9; $j++) { if ($arr[$i][$j] > 0) { $this->setCell($i, $j, $arr[$i][$j]); } } } } function dump() { for($i = 0; $i < 9; $i++){ for($j = 0; $j < 9; $j++){ $c = count($this->matrix[$i][$j]); if($c == 1){ echo " ".current($this->matrix[$i][$j])." "; } else { echo "(".$c.")"; } } echo "\n"; } echo "\n"; } function dumpAll() { for($i = 0; $i < 9; $i++){ for($j = 0; $j < 9; $j++){ echo implode('', $this->matrix[$i][$j]), "\t"; } echo "\n"; } echo "\n"; } function calc($data) { $this->clear(); $this->set($data); $this->_calc(); $this->dump(); } function _calc() { for($i = 0; $i < 9; $i++){ for($j = 0; $j < 9; $j++){ if(count($this->matrix[$i][$j]) == 1) { continue; } foreach($this->matrix[$i][$j] as $v){ $flag = false; $t = new Sudoku($this->matrix); if(!$t->setCell($i, $j, $v)){ continue; } if(!$t->_calc()){ continue; } $this->matrix = $t->matrix; return true; } return false; } } return true; } } $sd=new Sudoku; $sd->calc(array( array(0,5,0,0,0,6,0,9,0), array(0,4,7,0,8,2,6,0,0), array(0,8,0,0,0,7,0,5,2), array(7,0,1,0,3,4,0,0,6), array(0,3,0,0,2,0,0,8,0), array(2,0,0,0,0,1,9,0,4), array(4,7,0,1,0,0,0,6,0), array(0,0,9,4,6,0,3,7,0), array(0,1,0,2,0,0,0,4,0), )); $sd->calc(array( array(1,0,0,0,0,6,9,0,0), array(0,0,0,9,0,0,0,0,5), array(2,0,0,1,0,0,0,0,3), array(0,0,5,3,0,7,0,2,0), array(3,0,0,6,0,0,0,0,1), array(0,1,0,4,0,0,8,0,0), array(9,0,0,0,0,2,0,0,7), array(5,0,0,0,0,9,0,0,0), array(0,0,3,7,0,0,0,0,4), )); $sd->calc(array( array(7,0,0,1,0,0,0,0,5), array(0,0,6,0,4,0,0,8,0), array(0,0,1,0,0,0,0,0,0), array(0,6,0,0,8,0,0,0,3), array(0,8,0,0,0,9,0,7,0), array(1,0,0,0,0,0,0,5,0), array(0,0,0,0,0,0,9,0,0), array(0,4,0,0,3,0,1,0,0), array(9,0,0,0,0,7,0,0,2), )); $sd->calc(array( array(0,5,0,0,0,0,0,2,0), array(0,0,3,1,0,0,5,0,0), array(0,0,6,0,0,8,0,0,0), array(6,0,0,0,0,0,0,1,0), array(8,0,0,6,0,0,0,0,4), array(0,3,0,0,0,9,0,0,7), array(0,0,0,5,0,0,3,0,0), array(0,0,8,0,0,6,9,0,0), array(0,9,0,0,0,0,0,7,0), )); ?>

运行结果如下:

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1 5 2 3 4 6 7 9 8 9 4 7 5 8 2 6 1 3 3 8 6 9 1 7 4 5 2 7 9 1 8 3 4 5 2 6 5 3 4 6 2 9 1 8 7 2 6 8 7 5 1 9 3 4 4 7 3 1 9 8 2 6 5 8 2 9 4 6 5 3 7 1 6 1 5 2 7 3 8 4 9 1 3 7 2 5 6 9 4 8 4 6 8 9 7 3 2 1 5 2 5 9 1 8 4 6 7 3 6 8 5 3 1 7 4 2 9 3 9 4 6 2 8 7 5 1 7 1 2 4 9 5 8 3 6 9 4 6 5 3 2 1 8 7 5 7 1 8 4 9 3 6 2 8 2 3 7 6 1 5 9 4 7 3 8 1 9 6 4 2 5 2 9 6 3 4 5 7 8 1 4 5 1 2 7 8 3 9 6 5 6 9 7 8 4 2 1 3 3 8 2 5 1 9 6 7 4 1 7 4 6 2 3 8 5 9 6 2 7 4 5 1 9 3 8 8 4 5 9 3 2 1 6 7 9 1 3 8 6 7 5 4 2 9 5 1 3 6 7 4 2 8 7 8 3 1 4 2 5 6 9 2 4 6 9 5 8 7 3 1 6 2 9 4 7 5 8 1 3 8 7 5 6 1 3 2 9 4 1 3 4 2 8 9 6 5 7 4 6 7 5 9 1 3 8 2 3 1 8 7 2 6 9 4 5 5 9 2 8 3 4 1 7 6

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希望本文所述对大家PHP程序设计有所帮助。

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