Re: Why is this code so much slower than the same algorithm in C?

ptoulis's comment about stack-oriented loop structures got me thinking. What if the inner loop were replaced with individual lines of code? Here's what I came up with:

sub new {
    my $i = 0;
    while ($i += 20) {
        next if ($i % 1);
        next if ($i % 2);
        next if ($i % 3);
        next if ($i % 4);
        next if ($i % 5);
        next if ($i % 6);
        next if ($i % 7);
        next if ($i % 8);
        next if ($i % 9);
        next if ($i % 10);
        next if ($i % 11);
        next if ($i % 12);
        next if ($i % 13);
        next if ($i % 14);
        next if ($i % 15);
        next if ($i % 16);
        next if ($i % 17);
        next if ($i % 18);
        next if ($i % 19);
        print "Number: $i\n"; last;
    }
}
[download]

Then I benchmarked it against your original Perl code and got these results:

         s/iter original      new
original   25.7       --     -63%
new        9.57     168%       --
[download]

Admittedly, it doesn't scale very well and this example is very case specific, but I'll be paying closer attention to where and how I use loops from now on.

Comment on Re: Why is this code so much slower than the same algorithm in C? Select or Download Code

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Re^2: Why is this code so much slower than the same algorithm in C? by snowhare (Friar) on Dec 10, 2008 at 16:58 UTC
With a little analysis to omit test values that don't affect the answer and some improvement on the if..next..if..next logic, I was able to tighten that up slightly: `#!/usr/bin/perl use integer; my $i = 0; while ($i += 20) { last unless ($i % 3 or $i % 6 or $i % 7 or $i % 8 or $i % 9 or $i % 11 or $i % 12 or $i % 13 or $i % 14 or $i % 15 or $i % 16 or $i % 17 or $i % 18 or $i % 19 or $i % 20); } print "Number: $i\n";` [download]	[reply] [d/l]
Re^3: Why is this code so much slower than the same algorithm in C? by Limbic~Region (Chancellor) on Dec 10, 2008 at 20:08 UTC
snowhare, Can this be reduced even further? `for values of x > 20 if x % 12 == 0 then x % 6 == 0` [download] Is there a reason to have them both? Update: In other words, order the list in descending order removing any exact multiples of previous items. When you reach 16, one of two things will happen. Either there will be a remainder and it will exit the loop or there won't be a remainder and it will continue. There is no need to test 8 after 16. Cheers - L~R	[reply] [d/l]
Re^4: Why is this code so much slower than the same algorithm in C? (2s) by tye (Sage) on Dec 10, 2008 at 20:46 UTC
`perl -le "do { $i += 20 } while( $i%3\|\|$i%7\|\|$i%11\|\|$i%13\|\|$i%16\|\|$i%1 +7\|\|$i%19 ); print $i" 77597520` [download] or just `perl -le "print 3571113161719"` [download] Update:* Replace 3 with 9 in both samples above to get the correct answer, as noted and further explained in the replies below. - tye	[reply] [d/l] [select]
Re^5: Why is this code so much slower than the same algorithm in C? (2s) by snowhare (Friar) on Dec 11, 2008 at 19:11 UTC
Re^6: Why is this code so much slower than the same algorithm in C? (2s) by tye (Sage) on Dec 12, 2008 at 03:31 UTC
Re^5: Why is this code so much slower than the same algorithm in C? (2s) by tilly (Archbishop) on Dec 15, 2008 at 13:19 UTC
Re^6: Why is this code so much slower than the same algorithm in C? (why) by tye (Sage) on Dec 16, 2008 at 02:55 UTC
Some notes below your chosen depth have not been shown here
Re^4: Why is this code so much slower than the same algorithm in C? by snowhare (Friar) on Dec 11, 2008 at 19:06 UTC
Good catch. `#!/usr/bin/perl use integer; my $i = 0; while ($i += 20) { last unless ( $i % 19 or $i % 18 or $i % 17 or $i % 16 or $i % 15 or $i % 14 or $i % 13 or $i % 12 or $i % 11 ); } print "Number: $i\n";` [download] Is 17% faster than my original version.	[reply] [d/l]