Detailed explanation of PHP permutation recursion and permutation and combination example codes

1. Arrangement recursion

If P represents the full arrangement of n elements, and Pi represents the full arrangement of n elements that does not include element i, (i) Pi means adding in front of the arrangement Pi Arrangement with the prefix i, then the total arrangement of n elements can be recursively defined as:
① If n=1, then the arrangement P has only one element i;
② If n>1, then the total arrangement P is represented by the arrangement ( i) Pi composition;
According to the definition, it can be seen that if the arrangement Pi of (k-1) elements has been generated, then the arrangement of k elements can be generated by adding element i in front of each Pi.

Code:

function rank($base, $temp=null)
{
    $len = strlen($base);
    if($len <= 1)
    {
        echo $temp.$base.&#39;<br/>&#39;;
    }
    else
    {
        for($i=0; $i< $len; ++$i)
        {
            rank(substr($base, 0, $i).substr($base, $i+1, $len-$i-1), $temp.$base[$i]);
        }
    }
}
rank(&#39;123&#39;);

However, after multiple test runs, it was found that there is a problem: if there are the same elements, the entire arrangement will be repeated.
For example, there are only three situations for the full arrangement of '122': '122', '212', and '221'; but the above method is repeated.
Slightly modified, added a flag to determine duplication, solved the problem (the code is as follows):

function fsRank($base, $temp=null)
{
    static $ret = array();
    $len = strlen($base);
    if($len <= 1)
    {
        //echo $temp.$base.&#39;<br/>&#39;;
        $ret[] = $temp.$base;
    }
    else
    {
        for($i=0; $i< $len; ++$i)
        {
            $had_flag = false;
            for($j=0; $j<$i; ++$j)
            {
                if($base[$i] == $base[$j])
                {
                    $had_flag = true;
                    break;
                }
            }
            if($had_flag)
            {
                continue;
            }
            fsRank(substr($base, 0, $i).substr($base, $i+1, $len-$i-1), $temp.$base[$i]);
        }
    }
    return $ret;
}
print &#39;<pre class="brush:php;toolbar:false">&#39;;
print_r(fsRank(&#39;122&#39;));
print &#39;

';

2. Permutation and combination examples

<?php
/**
 * 要解决的数学问题    ：算出C(a,1) * C(b, 1) * ... * C(n, 1)的组合情况，其中C(n, 1)代表从n个元素里任意取一个元素
 *
 * 要解决的实际问题样例：某年级有m个班级，每个班的人数不同，现在要从每个班里抽选一个人组成一个小组，
 *                       由该小组来代表该年级参加学校的某次活动，请给出所有可能的组合
 */
/* ################################### 开始计算 ################################### */
/**
 * 需要进行排列组合的数组
 *
 * 数组说明：该数组是一个二维数组，第一维索引代表班级编号，第二维索引代表学生编号
 */
$CombinList = array(1 => array("Student10", "Student11"),
                    2 => array("Student20", "Student21", "Student22"),
                    3 => array("Student30"),
                    4 => array("Student40", "Student41", "Student42", "Student43"));
/* 计算C(a,1) * C(b, 1) * ... * C(n, 1)的值 */
$CombineCount = 1;
foreach($CombinList as $Key => $Value)
{
    $CombineCount *= count($Value);
}
$RepeatTime = $CombineCount;
foreach($CombinList as $ClassNo => $StudentList)
{
    // $StudentList中的元素在拆分成组合后纵向出现的最大重复次数
    $RepeatTime = $RepeatTime / count($StudentList);
    $StartPosition = 1;
    // 开始对每个班级的学生进行循环
    foreach($StudentList as $Student)
    {
        $TempStartPosition = $StartPosition;
        $SpaceCount = $CombineCount / count($StudentList) / $RepeatTime;
        for($J = 1; $J <= $SpaceCount; $J ++)
        {
            for($I = 0; $I < $RepeatTime; $I ++)
            {
               $Result[$TempStartPosition + $I][$ClassNo] = $Student;
            }
            $TempStartPosition += $RepeatTime * count($StudentList);
        }
        $StartPosition += $RepeatTime;
    }
}
/* 打印结果 */
echo "<pre class="brush:php;toolbar:false">";
print_r($Result);
?>

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