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Re: Can this be explained in layman's terms?

by jaredor (Priest)
on Jan 14, 2017 at 07:27 UTC ( #1179559=note: print w/replies, xml ) Need Help??


in reply to [Answered; thanks.] Can this be explained in layman's terms?

Hi BrowserUk,

I'm a lackluster mathematics student, but perhaps I can volunteer an explanation that doesn't stray too far from the rigor.

First, I think I would like to recast your statement as "For any set of n elements the number of ways to partition the set into k non-empty subsets is denoted S(n,k) and these numbers are known as Stirling numbers of the second kind."

  • S(0,0) = 1 : There is only one way to make no partitions out of nothing.
  • S(n,0) = 0 if n>0 : If you have a set of something, you cannot have a partition with no subsets--the elements of the non-empty set have to go somewhere.
  • S(n,1) = S(n,n) = 1 : There is only one way to partition a non-empty set into one non-empty subset--the subset being the set itself. There is only one way to partition a set of n elements into n non-empty subsets--each element has to be in its own subset (a singleton set).
  • S(n,k) = S(n-1, k-1) + kS(n-1,k) : Every partition of a set of n elements into k non-empty subsets can be created from one of the partitions of a set of n-1 elements. The sum comes about by looking at the two ways a new, identified, element can be added to the partitions (thus making a new partition that is of a larger set of n elements)
    1. You can add the singleton set containing just the new element to any partition of n-1 elements into k-1 subsets, thus creating a partition of a set of n elements into k subsets. There are S(n-1, k-1) partitions that we can do this to.
    2. You can add the new element to any non-empty subset of a partition of n-1 elements into k non-empty subsets. There are S(n-1, k) partitions that we can do this to and since we have k subsets in any partition, we have k ways to modify each of the S(n-1, k) partitions, i.e., there are k*S(n-1, k) ways to get to a partition of a set of n elements into k non-empty subsets this way.
    Note that because we are talking about an identified element being included, then this sum counts each possible partition of a set of n elements into k non-empty subsets exactly once. The identified element must be in any partition of this set of n elements. The identified element is either in a subset by itself (case 1) or in a subset with at least one other element (case 2).

Sometimes the rule S(n,k) = 0 if k>n is given for completeness, i.e., you can't partition something up into more non-empty subsets than for which it has elements, but this is considered obvious and you won't run into the need for it if you start with any reasonable case where n>=k.

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Re^2: Can this be explained in layman's terms?
by huck (Parson) on Jan 14, 2017 at 08:19 UTC

    Sometimes the rule S(n,k) = 0 if k>n is given for completeness

    VERY important if you dont realize n elements ... k non-empty subsets and you just start with

    for $n (0..10) { for $k (0..10) { print "$n $k '.fs($n,$k)."\n"; } }

    use strict; use warnings; my $l='k '; $l.=sprintf "%3d ",0; for my $k (0..10){ $l.=sprintf " %10s",$k; } print $l."\n"; $l=~s/./-/g; print $l."\n"; for my $n (0..10){ printf "n %3d ",$n; for my $k (0..10){ printf " %10s",fs($n,$k); } print "\n"; } sub fs { my $n=shift; my $k=shift; if ( $n ==0 && $k == 0 ) {return 1}; if ( $k > $n ) {return 0}; # important if ( $n > 0 && $k == 0 ) {return 0} if ( $k == 1 ) {return 1} if ($n == $k ) {return 1} my $p1=fs($n-1,$k-1); my $p2=fs($n-1,$k ); return $p1 + ($k*$p2); }
    And i learned never to name my subroutines s
    use strict; use warnings; my $l='k '; $l.=sprintf "%3d ",0; for my $k (0..10){ $l.=sprintf " %10s",$k; } print $l."\n"; $l=~s/./-/g; print $l."\n"; for my $n (0..10){ printf "n %3d ",$n; for my $k (0..10){ printf " %10s",s($n,$k); } print "\n"; } sub s { my $n=shift; my $k=shift; if ( $n ==0 && $k == 0 ) {return 1}; if ( $k > $n ) {return 0}; # important if ( $n > 0 && $k == 0 ) {return 0} if ( $k == 1 ) {return 1} if ($n == $k ) {return 1} my $p1=s($n-1,$k-1); my $p2=s($n-1,$k ); return $p1 + ($k*$p2); }
    output
    Global symbol "$p2" requires explicit package name at buk-2.pl line 30 +. syntax error at buk-2.pl line 31, near "return" (Might be a runaway multi-line ;; string starting on line 29) Global symbol "$p1" requires explicit package name at buk-2.pl line 31 +. Global symbol "$p2" requires explicit package name at buk-2.pl line 31 +. Missing right curly or square bracket at buk-2.pl line 32, at end of l +ine syntax error at buk-2.pl line 32, at EOF Execution of buk-2.pl aborted due to compilation errors.

         ...And i learned never to name my subroutines s

      I wish I could give you more ++ for that !

              ...it is unhealthy to remain near things that are in the process of blowing up.     man page for WARP, by Larry Wall

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