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Re^4: Performance, Abstraction and HOP

by BrowserUk (Pope)
on Sep 01, 2005 at 18:31 UTC ( #488453=note: print w/ replies, xml ) Need Help??

in reply to Re^3: Performance, Abstraction and HOP
in thread Performance, Abstraction and HOP

I can't image trees implemented with arrays being that much more memory hungary than arrays alone.

Don't imagine--measure :)

P:\test>junk Array[ 1..10000]: 200056 Sum @array = 50005000 Tree[ 1..10000 ]: 1120016 Sum tree = 50005000 Rate tree array tree 17.1/s -- -82% array 95.1/s 455% --

I think that 6x bigger and 5x slower pretty much makes my point.

#!/usr/bin/perl -lw use strict; use List::Util qw[ sum shuffle ]; use Benchmark qw[ cmpthese ]; use Devel::Size qw[ total_size ]; my $MAX = 10000; our @array = 1 .. $MAX; our $tree = make_tree($MAX/2,undef,undef); $tree = insert( $_, $tree) for shuffle 1 .. $MAX; print "Array[ 1..$MAX]: ", total_size( \@array ); print "Sum \@array = ", sum( @array ); print "Tree[ 1..$MAX ]: ", total_size( $tree ); print "Sum tree = ", sum_tree( $tree ); cmpthese -1, { tree => q[ my $sum = sum_tree( $tree ); ], array=> q[ my $sum = sum @array ], }; exit; sub sum_tree { my $tree = shift; return 0 if not defined($tree); return node($tree) + sum_tree(right($tree)) + sum_tree(left($tree) +); } sub insert { (my $elem, my $tree) = @_; if(not defined($tree)){ return make_tree($elem, undef, undef); } my $curr = node($tree); if( $elem == $curr) { return $tree; } elsif($elem < $curr) { return make_tree($curr, insert($elem,left($tree)), right($tree)); } elsif($elem > $curr) { return make_tree($curr, left($tree), insert($elem,right($tree))); } } sub make_tree { [$_[0], $_[1], $_[2]] } sub node { $_[0]->[0] } sub left { $_[0]->[1] } sub right { $_[0]->[2] }

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal?
"Science is about questioning the status quo. Questioning authority".
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Comment on Re^4: Performance, Abstraction and HOP
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Re^5: Perl memory usage
by Anonymous Monk on Sep 01, 2005 at 20:02 UTC
    I think that 6x bigger and 5x slower pretty much makes my point.
    Holy crap, now I'm curious. Only 5x slower is faster than what I would have guessed. But 6x larger is crazy. Anyone know how much boxing Perl does? I would have thought that the rough estimate for the size of a scalar number would be say 16 bytes (4 bytes for a pointer, 4 bytes for type/reference-count information, 8 bytes for a double precision floating point number). And the overhead for an array at maybe 16 bytes (8 bytes for a length field, 4 bytes for type/reference-count info, 4 bytes for a pointer to the array of pointers). For the tree structure above (an array composed of one scalar and two arrays) that would be 16 (array overhead)+3*4(three elements in first array, 4 byte pointers (32 bit machine))+16(the scalar)+2*16(the left and right branches) = 76 bytes. I guess that's starting to add up, but it is still shy of the 112 bytes measured above. Perl must preallocate space for each array to make growing it faster (maybe 12 elements initially?). Does that sound about right? Any way to get a more slimed down data structure in pure Perl?
      Is 6x really that crazy? Doesn't a typical generic C tree use about 4x? (If you're storing integers in the tree. If you are storing something larger, the overhead is smaller.) The perl array includes at least a reference count and a length that aren't in the C Node struct, so 6x quickly becomes plausible.
        Well, on a 32bit machine, it is 3x larger...
        struct tree { int elem; struct tree *left, *right; }; int main(int argc, char **argv) { printf("sizeof(int)=%d,sizeof(struct tree)=%d\n", sizeof(int),sizeof(struct tree)); }
        ...of course, the 12 bytes from above is a tad bit slimmer than the apparently 112 byte structure on the perl side.

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