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Hello BrowserUk,

A Stirling number of the second kind, S(n,k), gives the number of ways in which a set of n elements can be partitioned into exactly k non-empty subsets. It can be calculated by an explicit formula:

S(n,k) = 1/k!.SUM[j=0 to k]( (-1)^(k-j) . kCj . j^n )

where kCj is a binomial coefficient (“ k select j ”). But if you’re calculating successive values of S, it will be more efficient to use the recurrence relation:

S(n+1,k) = k.S(n,k) + S(n,k-1)

Here’s a simple implementation:

use strict; use warnings; my ($n, $k) = (20, 11); print "S($n, $k) = ", S_formulaic($n, $k), "\n"; print "S($n, $k) = ", S_recursive($n, $k), "\n"; sub S_recursive { my ($n, $k) = @_; return 1 if $n == 0 && $k == 0; return 0 if $k == 0; return 1 if $k == 1; return 1 if $n == $k; return S_recursive($n - 1, $k - 1) + $k * S_recursive($n - 1, $k); } sub S_formulaic { my ($n, $k) = @_; my $sum = 0; for my $j (0 .. $k) { $sum += (-1) ** ($k - $j) * C($k, $j) * $j ** $n; } return (1 / fact($k)) * $sum; } sub C { my ($n, $k) = @_; my $p = 1; $p *= $_ for ($n - $k + 1) .. $n; return $p / fact($k); } sub fact { my ($n) = @_; my $p = 1; $p *= $_ for 2 .. $n; return $p; }

Output:

17:17 >perl 1740_SoPW.pl S(20, 11) = 1900842429486 S(20, 11) = 1900842429486 17:17 >

I would say, “hope that helps,” but I’m sure you already know all of the above. So my best hope is that it will help you to clarify what you mean by “tangible explanation” and “transcribing the above description into English.” Then again, if what you’re really after is an explanation of the maths behind the formulae, you’ll need an actual mathematician. ;-)

Cheers,

Athanasius <°(((><contra mundum Iustus alius egestas vitae, eros Piratica,


In reply to Re: Can this be explained in layman's terms? by Athanasius
in thread [Answered; thanks.] Can this be explained in layman's terms? by BrowserUk

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