有必要
编写函数
- 要绘制棋盘，功能
drowChess (8,8) - 对于以随机顺序在其上的分型图.function addFigures（
['black']=>['King']，'white'=>['Rook','Officer']）。首先是两个对立的人物。例如，白黑之王 Rook.so，以至于每次调用此函数时，它们都不会与国际象棋定律相矛盾。 - 可能的功能必须是通用的，即如果设置第三个弄清楚它没有坏的功能。
简而言之，我认为是这样 - 我们必须尝试计算他们位置的算法。我已经开始解决问题，但算法有点难。
$x = 8;
$y = 8;
$xb = ['A','B','C','D','E','F','G','H'];
function drowChess($horisonal, $vertical, $xb)
{
$style = "width:30px;height:30px;";
$white = "background:white;color:black;text-align:center";
$black = "background:black;color:white;text-align:center";
$color = $black;
echo "<table border='1' style='margin:0 auto' cellpadding=0 cellspacing=0>";
echo "<tr>";
echo "<th style='width:30px;height:30px;'>$horisonal/$vertical</th>";
foreach($xb as $word){
echo "<th style='width:30px;height:30px;'>".$word."</th>";
}
echo "</tr>";
for($y = 1; $y <= $vertical; $y++)
{
if($color == $black)
$color = $white;
else
$color = $black;
echo "<tr>";
for($x = 1; $x <= $horisonal; $x++)
{
if($color == $black)
$color = $white;
else
$color = $black;
if($x == 1)
{
echo "<th style='$style'>".$y ."</th>";
}
echo "<td style='$style $color'>".$xb[$x-1] .$y."</td>";
}
echo "</tr>";
}
echo "</table>";
}
drowChess($x, $y,$xb);
我开始如此
$figures = [
'King' => function($x,$y)
{
//Logic + algorithm...
return [[$x,$y],[$x,$y],[$x,$y]];//Relevant possible positions , given a predetermined position
},
'Rook' => function($x,$y)
{
//Logic + algorithm ...
return [[$x,$y],[$x,$y],[$x,$y]];//given a predetermined position
},
//And so on for all figures
];
后
function addFigures(array $figures,array $algoritm)
{
foreach($figures as $figure)
{
if(!isset($algoritm[$figure])) return ['success'=>false,'message'=>'Such a figure does not exist'];
return $algoritm[$figure];
}
}
addFigures(['King','Rook']);
例如：
如果国王在位置（B2），Rook 它不能在位置（A1，B1，C1，C2，C3，B3，A3，A2和A8，H2...），需要一个算法，和适当的战术对齐
需要编写函数
function getFigurePositions($x,y)
{
//logic for figure - King
return ['A1','A2','A3','B3','C3','C2','C1','B1'];//available positions(I mean x,y coordinates )
}
getFigurePositions(['figure'=>'king','positions'=>'2,2']);
getFigurePositions(['figure'=>'Rock','positions'=>'3,7']);
您只能编写以下步骤的算法和策略。您认为如何实现它，有没有更好的方法？谢谢。
我为国王和车推导出一个算法。对于其他数字，移动相同的方案。
drowChess($x, $y,$xb);
$figures = [
'King' => function ($x,$y)
{
if((!$x or $x < 0) or (!$y or $y < 0))
return "Poziciya $x $y naxoditsya za polem";
$xb = ['A','B','C','D','E','F','G','H'];
$currentPosition = $xb[$x-1].$y;
$x1 = $x-1;
$y1 = $y;
$x2 = $x-1;
$y2 = $y-1;
$x3 = $x-1;
$y3 = $y+1;
$x4 = $x+1;
$y4 = $y;
$x5 = $x+1;
$y5 = $y-1;
$x6 = $x+1;
$y6 = $y+1;
$x7 = $x;
$y7 = $y-1;
$x8 = $x;
$y8 = $y+1;
$positions = [];
if(isset($xb[$x1-1]) && $y1 && $y1 <= 8 )
$positions[$xb[$x1-1].$y1] = ['x'=>$x1,'y'=>$y1];
if(isset($xb[$x2-1]) && $y2 && $y2 <= 8 )
$positions[$xb[$x2-1].$y2] = ['x'=>$x2,'y'=>$y2];
if(isset($xb[$x3-1]) && $y3 && $y3 <= 8 )
$positions[$xb[$x3-1].$y3] = ['x'=>$x3,'y'=>$y3];
if(isset($xb[$x4-1]) && $y4 && $y4 <= 8 )
$positions[$xb[$x4-1].$y4] = ['x'=>$x4,'y'=>$y4];
if(isset($xb[$x5-1]) && $y5 && $y5 <= 8 )
$positions[$xb[$x5-1].$y5] = ['x'=>$x5,'y'=>$y5];
if(isset($xb[$x6-1]) && $y6 && $y6 <= 8 )
$positions[$xb[$x6-1].$y6] = ['x'=>$x6,'y'=>$y6];
if(isset($xb[$x7-1]) && $y7 && $y7 <= 8 )
$positions[$xb[$x7-1].$y7] = ['x'=>$x7,'y'=>$y7];
if(isset($xb[$x8-1]) && $y8 && $y8 <= 8 )
$positions[$xb[$x8-1].$y8] = ['x'=>$x8,'y'=>$y8];
return [
'Korol na' => $currentPosition,
'Dostupnie emu xodi'=>$positions
];
},
'Rook' => function($x,$y)
{
if((!$x or $x < 0) or (!$y or $y < 0))
return "Poziciya $x $y naxoditsya za polem";
$positions = [];
$xb = ['A','B','C','D','E','F','G','H'];
$currentPosition = $xb[$x-1].$y;
//V pravo vniz i vverx
$yrb = $yrt = $y;
for( $x1 = $x+1; $x1 <= 8 ; $x1++ )
{
//Diaganal to rigth bottom
if((++$yrb) <= 8)
$positions[$xb[$x1-1].($yrb)] = ['x' => $x1, 'y' => $yrb];
//Diaganal to rigth top
if((--$yrt) > 0)
$positions[$xb[$x1-1].($yrt)] = ['x' => $x1, 'y' => $yrt];
}
//V pravo vniz i vverx
$ylb = $ylt = $y;
for( $x2 = $x-1; $x2 >= 1 ; $x2-- )
{
//Diaganal to left bottom
if((++$ylb) <= 8){
$positions[$xb[$x2-1].($ylb)] = ['x' => $x2, 'y' => $ylb];
}
//Diaganal to left top
if((--$ylt) > 0){
$positions[$xb[$x2-1].($ylt)] = ['x' => $x2, 'y' => $ylt];
}
}
return [
'Rook na' => $currentPosition,
'Dostupnie emu xodi'=>$positions
];
},
];
echo "<pre>";
print_r($figures['Rook'](5,5));
好吧，如果我们有可用的步骤，我们可以编写一个函数，该函数将使用不同的设置调用它们，直到它们可用为止 步骤没有不同。