Someone asked me a question via e-mail about integer partitions the other day (as I had released a Perl module, Integer::Partition, to generate them), that I was unable to answer.

前几天有人通过电子邮件向我询问了有关整数分区的问题(因为我已经发布了一个Perl模块,Integer :: Partition来生成它们),我无法回答。

Background: here are all the integer partitions of 7 (the sum of each row equals 7).

背景:这里是7的所有整数分区(每行的总和等于7)。

7
6 1
5 2
5 1 1
4 3
4 2 1
4 1 1 1
3 3 1
3 2 2
3 2 1 1
3 1 1 1 1
2 2 2 1
2 2 1 1 1
2 1 1 1 1 1
1 1 1 1 1 1 1

Now if we look at the lengths of each partition and count how many there are of each length:

现在,如果我们查看每个分区的长度并计算每个长度的数量:

1 1
2 3
3 4
4 3
5 2
6 1
7 1

... we see one partition has a length of 1 (7), one has a length of 7 (1 1 1 1 1 1 1). There are 4 partitions that have a length of 3: (5 1 1), (4 2 1), (3 3 1), (3 2 2).

...我们看到一个分区的长度为1(7),一个分区的长度为7(1 1 1 1 1 1 1)。有4个分区长度为3:(5 1 1),(4 2 1),(3 3 1),(3 2 2)。

For larger numbers of N, if you graph the distribution of partition lengths, an asymetric curve emerges, skewed towards the origin. If you're curious, graph the following partition length counts for N=40.

对于较大数量的N,如果绘制分区长度的分布图,则会出现不对称曲线,偏向原点。如果您感到好奇,请将以下分区长度计算为N = 40。

1 20 133 478 1115 1945 2738 3319 3589 3590 3370 3036 2637 2241 1861 1530 1236 995 790 627 490 385 297 231 176 135 101 77 56 42 30 22 15 11 7 5 3 2 1 1

1 20 133 478 1115 1945 2738 3319 3589 3590 3370 3036 2637 2241 1861 1530 1236 995 790 627 490 385 297 231 176 135 101 77 56 42 30 22 15 11 7 5 3 2 1 1

If you're interested in generating these distribution counts, here's the code I used:

如果您对生成这些分发计数感兴趣,请使用以下代码:

#! /usr/local/bin/perl

use strict;
use warnings;

use Integer::Partition;

my $n = shift || 1;

while (1) {
    my $start = time;
    my $i = Integer::Partition->new($n);
    my %size;
    while (my $p = $i->next) {
        $size{scalar @$p}++;
    }

    open my $out, '>>', "bucket-count.out";
    for my $s (sort {$a <=> $b} keys %size) {
        print $out "$n\t$s\t$size{$s}\n";
    }
    close $out;
    my $delta = time - $start;
    print "$n\t$delta secs\n";
    ++$n;
}

(note: on my computer, N=90 takes about 10 minutes to generate).

(注意:在我的电脑上,N = 90大约需要10分钟才能生成)。

So my question is: what equation can be used to match the observed distribution curve? Is it a Gauss (can a Gaussian distribution be asymetric?) or Poisson distribution, or something else?

所以我的问题是:可以使用哪个方程来匹配观察到的分布曲线?它是高斯(高斯分布可以是不对称的吗?)还是泊松分布,还是别的什么?

How do I solve it for N? If I remember my maths from high-school, I can determine the peak by solving when the derivative intersects 0. How do I produce the derivative? I've searched the web but all I get back are abstruse mathematical papers. I just need some code :)

我如何为N解决它?如果我记得高中时的数学,我可以通过求解导数与0相交来确定峰值。如何产生导数?我在网上搜索过,但我得到的都是深奥的数学论文。我只需要一些代码:)

1 个解决方案