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Re^3: Who's a thief? (recursion limitations)

by sleepingsquirrel (Hermit)
on Jan 14, 2005 at 18:29 UTC ( #422338=note: print w/replies, xml ) Need Help??


in reply to Re^2: Who's a thief? -- the follow up
in thread Who's a thief? -- the follow up

...one problem that happens with larger systems like this is the systems tend to rely heavily on recursion and this is not Perl's strength :/
You're not giving perl enought credit for what it can do. The following program executes in under 9 seconds on my (1GB) machine.
#!/usr/bin/perl -w print deep(2_500_000); sub deep { $_[0]>0 ? deep($_[0]-1) : "done\n"; }
...and I can calculate the 36th fibonacci number using the simplistic algorithm below in about a minute. It ends up calling "fib" over 48 million times.
#!/usr/bin/perl -w $count = 0; my $n = ($ARGV[0] < 1) ? 1 : $ARGV[0]; print fib($n)."\ncount = $count\n"; sub fib { $count++; $_[0]<2 ? 1 : fib($_[0]-2) + fib($_[0]-1) }


-- All code is 100% tested and functional unless otherwise noted.

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Re^4: Who's a thief? (recursion limitations)
by Ovid (Cardinal) on Jan 14, 2005 at 18:47 UTC

    As soon as I tried your "deep" program, I immediately got a "deep recursion" error. I've a 1.6 GHz machine with 640 megs of RAM. I don't know how you could have run that. Every time Perl enters a sub (normally), it creates a new stack frame and this quickly eats memory.

    On my machine, it pretty much ground the box to a halt after about 1.5 million iterations.

    Cheers,
    Ovid

    New address of my CGI Course.

      If you can arrange your algorithm to be tail recursive, and use the magic form of goto, you can cut the memory growth to zero and achieve some very impressive figures.

      8 seconds for 10 million recursions and 83 seconds for 100 million with the memory consumption never over 1.5 MB.

      #!/usr/bin/perl -w use strict; $|++; my $count; deep( $ARGV[ 0 ] ); print $count; sub deep { ++$count and --$_[ 0 ] and goto \&deep; } __DATA__ [19:09:45.02] P:\test>422344 10000000 10000000 [19:09:53.02] P:\test> [19:10:03.21] P:\test>422344 100000000 100000000 [19:11:26.84] P:\test>

      It takes a little effort to do, but when you need it it is worth it.


      Examine what is said, not who speaks.
      Silence betokens consent.
      Love the truth but pardon error.

        You can avoid magic stuff with a more cumbersome notation:

        sub deep {SUB:{ ++$count and --$_[ 0 ] and redo SUB; }}

        Which is actually going to be faster afaik. (Havent tested though)

        ---
        demerphq

        I've never seen doing a goto to a reference of a named subroutine. Does this really get around the non-tail-recursive nature of goto &deep;?

        Being right, does not endow the right to be rude; politeness costs nothing.
        Being unknowing, is not the same as being stupid.
        Expressing a contrary opinion, whether to the individual or the group, is more often a sign of deeper thought than of cantankerous belligerence.
        Do not mistake your goals as the only goals; your opinion as the only opinion; your confidence as correctness. Saying you know better is not the same as explaining you know better.

      Ovid,
      Every time Perl enters a sub (normally), it creates a new stack frame and this quickly eats memory.

      Presumably, you are referring to &sub as the abnormal example where you could use goto &sub to bypass the recursion limit as seen in the following example:

      #!/usr/bin/perl use strict; use warnings; print sum( 1000 ); sub sum { my ($num, $tot) = @_; return $tot if ! $num; @_ = ($num - 1, $tot += $num); goto &sum; }
      According to TimToady, this actually does create a new stack, but only after the old one is stripped off. Of course you could always ignore the warning, turn off the 'recursion' warning, or recompile Perl to avoid the warning. In practice, I have never needed this kind of technique because it seems that iterative solutions aren't that difficult to come up with. In this case, the formula (n^2 + n) / 2 would do the trick.

      And as I am previewing this, I see BrowserUk has already said something similar - oh well.

      Cheers - L~R

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