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Re: How to generate restricted partitions of an integer

by tachyon (Chancellor)
on Nov 11, 2004 at 13:08 UTC ( #407021=note: print w/ replies, xml ) Need Help??


in reply to How to generate restricted partitions of an integer

This reduces the 8316 loops your code requires to 197 which is almost 2 orders of magnitude faster tahn your original and 3.5x faster than grinders example that needs 686 loops. Only 4:1 wastage on the test is pretty good as semi brute force goes and certainly beats the original 166:1. All it does different is to limit the looping to the limited subset of possible cases....

my ( $loops, $cnt ); printf "%4d %4d %4d %4d %4d\n", 100,50,20,10,5; for (my $e5=0; $e5<=100; $e5+=10 ) { for (my $e10=0; $e10<=100-$e5; $e10+=10 ) { for (my $e20=0; $e20<=100-$e10-$e5; $e20+=20 ) { for (my $e50=0; $e50<=100-$e20-$e10-$e5; $e50+=50 ) { for (my $e100=0; $e100<=100-$e50-$e20-$e10-$e5; $e100+=100 ) { $loops++; if ( $e5+$e10+$e20+$e50+$e100 == 100 ) { printf "%4d %4d %4d %4d %4d\n",$e100/100,$e50/50,$e20/20,$ +e10/10,$e5/5; $cnt++; } } } } } } print "\n$cnt possibilities in $loops loops\n"; __DATA__ 100 50 20 10 5 1 0 0 0 0 0 2 0 0 0 0 0 5 0 0 0 1 2 1 0 0 0 4 2 0 0 1 1 3 0 0 0 3 4 0 0 1 0 5 0 0 0 2 6 0 0 0 1 8 0 0 0 0 10 0 0 1 2 0 2 0 0 4 1 2 0 1 1 2 2 0 0 3 3 2 0 1 0 4 2 0 0 2 5 2 0 0 1 7 2 0 0 0 9 2 0 0 4 0 4 0 1 1 1 4 0 0 3 2 4 0 1 0 3 4 0 0 2 4 4 0 0 1 6 4 0 0 0 8 4 0 1 1 0 6 0 0 3 1 6 0 1 0 2 6 0 0 2 3 6 0 0 1 5 6 0 0 0 7 6 0 0 3 0 8 0 1 0 1 8 0 0 2 2 8 0 0 1 4 8 0 0 0 6 8 0 1 0 0 10 0 0 2 1 10 0 0 1 3 10 0 0 0 5 10 0 0 2 0 12 0 0 1 2 12 0 0 0 4 12 0 0 1 1 14 0 0 0 3 14 0 0 1 0 16 0 0 0 2 16 0 0 0 1 18 0 0 0 0 20 50 possibilities in 197 loops

cheers

tachyon


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