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

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";

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  • Comment on Re^2: Why is this code so much slower than the same algorithm in C?
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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]
      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 3*5*7*11*13*16*17*19"

      [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]
        You've got an extra factor of 3. The correct answer is 232792560. ;)
[reply]
        Why do that much analysis and then throw away so much speed by not using a lexical? Also why not finish the analysis? 
        perl -le 'my $i = 0;do { $i += 20 } while( $i%46558512 ); print $i'
232792560

        [download]
        Performance is essentially the same if you write 46558512 as (9*7*11*13*16*17*19).
[reply]
[d/l]
      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]

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