I have an interesting suggestion.

To generate a sampling, you just repeat the following steps:

1. set $sum to 0, $count to 0
2. generate a random number $r between 1 and m
3. add this random number to the $sum += $r
4. if $sum <= 158, $count ++, else quit the loop

After this process, the variable $count contains the number of random number you used to reach 158, which can be considered as number of selections.

Repeat above process and you will get the satistics. Refine this a little bit and pick the right m, will give you soem good enough simulation. You have to refine this.

Here is some code to illustrate this:

use strict;
use warnings;

my @counts;

while (1) {
    my $count = 0;
    my $sum = 0;
    while (1) {
        $sum += rand(rand(rand(rand(158))));#need to refine
        print "$sum\n";
        ($sum <= 158) ? $count ++ : last; 
    }
    $counts[$count] ++;
    print "choose $count\n";
    sleep 1;
}
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For these kinds of problems, I always consult my favorite online book, Combinatorial Generation by Frank Ruskey. On page 87 of the PDF, it derives convenient ranking and unranking algorithms for combinations. Here they are in Perl:

use Math::Pari qw[:int binomial];
use Memoize;
use List::Util 'sum';

memoize 'binomial';

sub rank {
    my @comb = sort { $a <=> $b } @_;
    sum map { binomial( $comb[$_]-1, $_+1 ) } 0 .. $#comb;
}

# $k is the comination size
# $r is the rank

sub unrank {
    my ($k, $r) = @_;

    my @comb;
    for my $i (reverse 1 .. $k) {
        my $p = $i-1;
        $p++ until binomial($p,$i) > $r;
        $r -= binomial($p-1, $i);
        push @comb, $p;
    }
    return @comb;
}
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Since it makes a lot of use of computing binomial coefficients, I've used Math::Pari's version and also memoized it for speed. However, I haven't gotten the algorithm to work with large Pari numbers (I can't remember how to work with Pari objects, and it's getting late).

Now, assuming you can get these algorithms to work on the large numbers involved, and you can sample from the appropriately large range of integers, you can perform any sampling method you want on {0 .. C(N,k)-1} and apply it to the set of combinations.

....

Having directly addressed your question, let me say that I find it hard to imagine that the naïve sampling approach (repeatedly choosing a uniformly random k-subset) would significantly skew any statistical properties. The sample space is so large and the chance of collision is so small. Are you absolutely sure that all this effort to avoid repeated samples isn't overkill?

You can get a random subset without replacement using List::Util::shuffle().

use List::Util qw/shuffle/;
sub pick {
    return if @_ != 2;
    my ($num, $count) = @_;
    (shuffle 1 .. $num)[0 .. $count-1];
}
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my ($top, $ct, @samples) = (158, 8);
my $size = 50,000;

$#samples = $size;
$_ = [pick( $top, $ct)] for @samples;
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my %sample;
$sample{join ',', pick($top, $ct)} = undef
    while $size > keys %sample;
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Does order matter in a sample? If not, then sort them to get a canonical order before storage.

blokhead's approach is certainly the most general, and I'm glad he posted it. My naive approach for this case, however, would have been to use a simple modification of the Fisher-Yates shuffling algorithm (untested and not optimized):

sub random_pick {
  my ( $n, $arr ) = @_;
  my @idx = 0..$#$arr;
  my $i = 0;
  while ( $i < $n ) {
    my $j = $i + rand( @idx - $i );
    @idx[ $i, $j ] = @idx[ $j, $i ];
    ++$i;
  }
  return @{ $arr }[ sort { $a <=> $b } @idx[ 0 .. $n-1 ] ];
}
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Zaxo's approach also uses FY, but doing a full shuffle each time. On the one hand this is more work than is necessary, but on the other List::Util::shuffle is a compiled extension, so it may end up being faster than the solution above.

Repeatedly randomly sampling without replacement is equivalent to shuffling the list and just splitting the result into groups of the desired size. So do that but don't start over after that...

First, shuffle the list.

Now you can group the list into groups of 5 (or 8 or whatever) items and you've got int(158/5) unique samples. Pick a number that is relatively prime with your list size and you've got another int(158/5) samples, none of which match any of your other samples. Repeat as needed. If you run out of relatively prime numbers before you get enough samples, then you can reshuffle the list at that point (risking repeating a sampling but with even lower odds than those you theorized would be acceptable).

#!/usr/bin/perl -w
use strict;

# Here are some stubs so you can test my code.

# Replace this with a Fisher Yates shuffle:
sub PermuteList {
    my $av= shift @_;
    print "Not shuffling ( $av->[0] .. $av->[-1] )\n";
}

# Replace these with your own code:
sub GetPopulation { return( 100..(99+$ARGV[0]) ); }
sub ProcessSample { print "( @_ )\n"; }

sub Lcd
{
    my( $x, $y )= @_;
    while( 1 ) {
        if(  $y < $x  ) {
            ( $x, $y )= ( $y, $x );
        }
        if(  $x == 0  ) {
            return $y;
        }
        $y %= $x;
    }
}

sub GenNextSample
{
    my( $size, @list )= @_;
    my $step = @list;
    my( $start, $offset );
    return sub {
        while( 1 ) {
            if(  @list/2 <= $step  ) {
                $step= 1;
                PermuteList( \@list );
                $offset= $start= int rand @list;
            }
            if(  1 == Lcd( $step, 0+@list )  ) {
                my @sample;
                do {
                    return @sample
                        if  $size == @sample;
                    push @sample, $list[$offset];
                    $offset= ( $offset + $step ) % @list;
                } while(  $offset != $start  );
            }
            $step++;
        }
    };
}

my $size= $ARGV[1];
my $iter= GenNextSample( $size, GetPopulation() );
my $samples = 0;
while(  $samples++ < 64000  ) {
    my @sample = $iter->();
    ProcessSample( @sample );
}
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You can test the code like:

 > perl lcd.pl 16 3 | more
Not shuffling ( 100 .. 115 )
( 112 113 114 )
( 115 100 101 )
( 102 103 104 )
( 105 106 107 )
( 108 109 110 )
( 112 115 102 )
( 105 108 111 )
( 114 101 104 )
( 107 110 113 )
( 100 103 106 )
( 112 101 106 )
( 111 100 105 )
( 110 115 104 )
( 109 114 103 )
( 108 113 102 )
( 112 103 110 )
( 101 108 115 )
( 106 113 104 )
( 111 102 109 )
( 100 107 114 )
Not shuffling ( 100 .. 115 )
( 102 103 104 )
...
[download]

Repeatedly randomly sampling without replacement is equivalent to shuffling the list and just splitting the result into groups of the desired size. ...

I am afraid such a method would seriously alter the outcome of any statistical analysis. For instance, if the number of elements in each combination is a divisor of the number of elements you pick from (in this case 158), then all (158) elements would appear the exactly same number of times in your sample.


Cheers

Antonio


The stupider the astronaut, the easier it is to win the trip to Vega - A. Tucket