php-计算置换的阶乘等级(N选择K)

我最近了解了CNSFNS,并且由于它们对我来说是如此优雅,因此我决定尝试并实现使用这些技术生成组合和置换的方法.我完成了将n个选择k组合转换为CSN等级的方法,反之亦然,但是我正在敲打墙壁,尝试对n个选择k(唯一)排列进行同样的操作.

多亏了@Joshua,我才可以使用未排序(从FNS到置换)的方法:

function Pr_Unrank($n, $k, $rank) { // rank starts at 1
    if ($n >= $k) {
        if (($rank > 0) && ($rank <= Pr($n, $k))) {
            $rank--;
            $result = array();
            $factoriadic = array();

            for ($i = 1; $i <= ($n - $k); ++$i) {
                $rank *= $i;
            }

            for ($j = 1; $j <= $n; ++$j) {
                $factoriadic[$n - $j] = ($rank % $j) + 1; $rank /= $j;
            }

            for ($i = $n - 1; $i >= 0; --$i) {
                $result[$i] = $factoriadic[$i];

                for ($j = $i + 1; $j < $n; ++$j) {
                    if ($result[$j] >= $result[$i]) {
                        ++$result[$j];
                    }
                }
            }

            return array_reverse(array_slice($result, 0 - $k));
        }
    }

    return false;
}

这是我当前尝试进行排名(向FNS排列)的方法:

function Pr_Rank($n, $k, $permutation) {
    if ($n >= $k) {
        $result = range(1, $n);
        $factoriadic = array();

        foreach ($permutation as $key => $value) {
            $factoriadic[$k - $key - 1] = array_search($value, $result);
            array_splice($result, $factoriadic[$k - $key - 1], 1);
        }

        $result = 1;

        foreach (array_filter($factoriadic) as $key => $value) {
            $result += F($key) * $value;
        }

        return $result;
    }

    return false;
}

这些是我正在使用的辅助函数:

function F($n) { // Factorial
    return array_product(range($n, 1));
}

function Pr($n, $k) { // Permutations (without Repetitions)
    return array_product(range($n - $k + 1, $n));
}

问题是,仅当n = k(demo)时,Pr_Rank()方法才返回正确的等级:

var_dump(Pr_Rank(5, 2, Pr_Unrank(5, 2, 10))); // 3, should be 10
var_dump(Pr_Rank(5, 3, Pr_Unrank(5, 3, 10))); // 4, should be 10
var_dump(Pr_Rank(5, 5, Pr_Unrank(5, 5, 10))); // 10, it's correct

我使用上面链接的Wikipedia文章和this MSDN article进行了指导,我知道他们都不打算考虑k大小的子集,但是我完全不知道这种逻辑是什么样的…

我还尝试了谷歌搜索并搜索现有问题/答案,但尚未找到任何相关内容.

解决方法:

睡个好觉后,笔& amp;的一点帮助纸,我想通了.如果有人感兴趣:

例如,第42个5选择3排列是4-2-5,但是如果查看Pr_Unrank(),调用array_slice()的话,您会注意到实际排列(按照字典顺序)实际上是4-2 -5 [-1-3],最后两个元素将被舍弃,因此最终只能得到k个元素.

这对于计算阶乘(3-1-2 [-0-0])的小数表示非常重要:

> 4-2-5 =(2!* 3)(1!* 1)(0!* 2)= 9
> 4-2-5-1-3 =(4!* 3)(3!* 1)(2!* 2)(1!* 0)(0!* 0)= 82

不过,82不是正确的答案.要获得它,我们必须将其除以以下结果:

> Pr(5,5)/ Pr(5,3)(=)(5-3)! = 120/60 = 2

所以82/2是41,我所要做的就是加1以使排名从1开始.

Array // 5 choose 3 permutations
(
    [1] => 1-2-3
    [2] => 1-2-4
    [3] => 1-2-5
    [4] => 1-3-2
    [5] => 1-3-4
    [6] => 1-3-5
    [7] => 1-4-2
    [8] => 1-4-3
    [9] => 1-4-5
    [10] => 1-5-2
    [11] => 1-5-3
    [12] => 1-5-4
    [13] => 2-1-3
    [14] => 2-1-4
    [15] => 2-1-5
    [16] => 2-3-1
    [17] => 2-3-4
    [18] => 2-3-5
    [19] => 2-4-1
    [20] => 2-4-3
    [21] => 2-4-5
    [22] => 2-5-1
    [23] => 2-5-3
    [24] => 2-5-4
    [25] => 3-1-2
    [26] => 3-1-4
    [27] => 3-1-5
    [28] => 3-2-1
    [29] => 3-2-4
    [30] => 3-2-5
    [31] => 3-4-1
    [32] => 3-4-2
    [33] => 3-4-5
    [34] => 3-5-1
    [35] => 3-5-2
    [36] => 3-5-4
    [37] => 4-1-2
    [38] => 4-1-3
    [39] => 4-1-5
    [40] => 4-2-1
    [41] => 4-2-3
    [42] => 4-2-5
    [43] => 4-3-1
    [44] => 4-3-2
    [45] => 4-3-5
    [46] => 4-5-1
    [47] => 4-5-2
    [48] => 4-5-3
    [49] => 5-1-2
    [50] => 5-1-3
    [51] => 5-1-4
    [52] => 5-2-1
    [53] => 5-2-3
    [54] => 5-2-4
    [55] => 5-3-1
    [56] => 5-3-2
    [57] => 5-3-4
    [58] => 5-4-1
    [59] => 5-4-2
    [60] => 5-4-3
)
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