Re^2: Why does Perl get slower when building a larger hash? (Not due to the memory swapping)

[chialingh]

Yes, what I did in the code is measuring the time taken to compare every file with every other file. However, the time log I showed is the running time divided by the number of file pairs, i.e. $time/$x

In that case, the time is the unit time for each comparison and it shouldn't go up with the number of files, right?

Sorry for the misleading code, I'll revise it. :)

[tmharish]

Dividing by $x ( number of files ) does not help cause you still have a O( log(N) ) in there ( the inner loop )

As a demonstration, consider the following code that benchmarks hash access times:

use strict                    ;
use warnings                  ;
use Time::HiRes qw( time )    ;



my @bench_mark_points = ( 1_000, 10_000, 100_000, 1_000_000, 10_000_00
+0, 20_000_000, 30_000_000, 40_000_000, 50_000_000 ) ;
my $bench_mark_point  = 0                                     ;
my %large_hash ;
my @keys       ;
for( my $elem = 0 ; $elem < 50_000_001; $elem++ ) { 
    my $key = rand( $elem ) . "hello" . rand( $elem );
    $large_hash{ $key } = 1 ;
    push @keys, $key ;
    if( $elem == $bench_mark_points[ $bench_mark_point ] ) { 
    _bench_mark( $bench_mark_points[ $bench_mark_point ] ) ;
    $bench_mark_point++ ;
    }
}


sub _bench_mark { 

    my $benchmark_point = shift ;

    my @benchmark_keys  = map( { $keys[ int( rand( $benchmark_point ) 
+) ] } ( 0 .. 1_000_000 ) ) ;

    my $total_time = 0 ;
    foreach my $key ( @benchmark_keys ) { 
    my $start_time = time;
    my $val = $large_hash{ $key } ;
    my $end_time = time ;
    $total_time += ( $end_time - $start_time ) ;
    }

    print "Benchmarked Hash access of size $benchmark_point \t -> Acce
+ss time for 1_000_000 keys: " . $total_time . "\n";

    return ;

}

Output

Benchmarked Hash access of size 1000    	 -> Access time for 1_000_000 keys: 0.11689305305481
Benchmarked Hash access of size 10000   	 -> Access time for 1_000_000 keys: 0.121062278747559
Benchmarked Hash access of size 100000   	 -> Access time for 1_000_000 keys: 0.125393152236938
Benchmarked Hash access of size 1000000 	 -> Access time for 1_000_000 keys: 0.116819381713867
Benchmarked Hash access of size 10000000 	 -> Access time for 1_000_000 keys: 0.118601083755493
Benchmarked Hash access of size 20000000 	 -> Access time for 1_000_000 keys: 0.117170572280884

You will notice that the time to access a million keys is nearly always the same - The memory on the other hand shoots up dramatically reaching 7.8 GB at around 25 Million entries ( which is why my bench-marking stops there )

Perl is working really hard in the background to ensure that it builds structures that can ensure that finding an element ( which should be O(N) ) is closing constant time here.

UPDATE: Here is the output again with the CPU throttled so the difference can actually be seen:

Benchmarked Hash access of size 1000    	 -> Access time for 1_000_000 keys: 0.438439846038818
Benchmarked Hash access of size 10000   	 -> Access time for 1_000_000 keys: 0.467595815658569
Benchmarked Hash access of size 100000  	 -> Access time for 1_000_000 keys: 0.615071773529053
Benchmarked Hash access of size 1000000 	 -> Access time for 1_000_000 keys: 0.804181814193726
Benchmarked Hash access of size 10000000 	 -> Access time for 1_000_000 keys: 0.873048782348633
Benchmarked Hash access of size 20000000 	 -> Access time for 1_000_000 keys: 0.910530567169189

So the time does go up - and thats expected. You cannot find an element in less than O( logN ); The speed here is because Perl optimizes this at a low level, but eventually the effect adds up.