排列-所有可能的数字集 [英] Permutations - all possible sets of numbers

问题描述

我有一个从0到8的数字.我希望得到这些数字的所有可能的集合，每个集合都应使用所有数字，每个数字在一个集合中只能出现一次.

I have numbers, from 0 to 8. I would like in result, all possible sets of those numbers, each set should use all numbers, each number can occur only once in a set.

我希望看到用PHP制作的解决方案可以打印出结果.或者，至少，我希望在组合理论上有所收获，因为我早已忘记了这一点.计算多少个排列的公式是什么?

I would like to see solution made in PHP that could print out result. Or, at least, I would like some refreshment in theory of combinatorics, as I have long forgotten it. What is the formula to calculate how many permutations will there be?

示例集:

• 0-1-2-3-4-5-6-7-8
• 0-1-2-3-4-5-6-8-7
• 0-1-2-3-4-5-8-6-7
• 0-1-2-3-4-8-5-6-7
• 0-1-2-3-8-4-5-6-7
• 0-1-2-8-3-4-5-6-7
• 依此类推...

推荐答案

您正在寻找置换公式:

nPk = n!/(n-k)!

在您的情况下，您有9个条目，并且想要选择所有条目，那就是9P9 = 9！ = 362880

In your case, you have 9 entries and you want to choose all of them, that's 9P9 = 9! = 362880

您可以在O'Reilly的"PHP Cookbook"的食谱4.26中找到一种可置换的PHP算法.

You can find a PHP algorithm to permutate in recipe 4.26 of O'Reilly's "PHP Cookbook".

pc_permute(array(0, 1, 2, 3, 4, 5, 7, 8));

从O'Reilly复制:

Copied in from O'Reilly:

function pc_permute($items, $perms = array( )) {
    if (empty($items)) { 
        print join(' ', $perms) . "\n";
    }  else {
        for ($i = count($items) - 1; $i >= 0; --$i) {
             $newitems = $items;
             $newperms = $perms;
             list($foo) = array_splice($newitems, $i, 1);
             array_unshift($newperms, $foo);
             pc_permute($newitems, $newperms);
         }
    }
}

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