Re^3: How to code this?

Oh, I missed that! Something like this then:

#!/usr/lib/perl
use strict;
use warnings;

my $kPoolSpan = 2;
my %hash      = (
    3  => [1 .. 2],
    4  => [2 .. 2],
    6  => [4 .. 5],
    7  => [2 .. 6],
    8  => [1 .. 3],
    11 => [5 .. 10],
    );
my @indexes = sort {$a <=> $b} keys %hash;
my @pool;

while (@indexes) {
    my $oldPoolSize = @pool;

    push @pool, shift @indexes if !@pool;
    push @pool, shift @indexes
        while @indexes && $pool[0] + $kPoolSpan >= $indexes[0];

    my $poolSum;
    my @new = @pool[$oldPoolSize .. $#pool];

    for my $poolEntry (@pool) {
        $poolSum += $_ for @{$hash{$poolEntry}};
    }

    print "Pool @pool sum (added @new): $poolSum\n";
    shift @pool while @indexes && @pool && $pool[0] + $kPoolSpan < $in
+dexes[0];
}
[download]

Prints:

Pool 3 4 sum (added 3 4): 5
Pool 4 6 sum (added 6): 11
Pool 6 7 8 sum (added 7 8): 35
Pool 11 sum (added 11): 45
[download]

True laziness is hard work

Comment on Re^3: How to code this? Select or Download Code

Replies are listed 'Best First'.
Re^4: How to code this? (Solved) by BrowserUk (Patriarch) on Nov 15, 2011 at 03:40 UTC
Thank you. With a few tweaks to deal with stuff not mentioned in the question, this is my addaption of your code: our $INT //= 4; my %hash; push @{ $hash{ calcScore( $_ ) } }, $_ while <>; my @grpKeys = sort{ $a <=> $b } keys %hash; my @oldGrp; while( my $next = shift @grpKeys ) { ## Remove any arrays from teh previously processed group ## that are below the interval of the next value - $INT shift @oldGrp while ( $next - $oldGrp[ 0 ] ) > $INT. ## Process items in the new array for my $i ( 0 .. $#{ $hash{ $next } } ) { ## against each of the (other) items in the new array for my $j ( $i+1 .. $#{ $hash{ $next } } ) { process( $i, $j ); } ## and all of the items in each of the arrays ## retained from the previous pass for my $g ( @oldGrp ) { for my $j ( 0 .. $#{ $hash{ $g } } ) { process( $i, $j ); } } } ## Add this array to the previously processed group push @oldGrp, $next; } [download] With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday' Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. "Science is about questioning the status quo. Questioning authority". In the absence of evidence, opinion is indistinguishable from prejudice.	[reply] [d/l]

Replies are listed 'Best First'.

Re^4: How to code this? (Solved)
by BrowserUk (Patriarch) on Nov 15, 2011 at 03:40 UTC

Thank you. With a few tweaks to deal with stuff not mentioned in the question, this is my addaption of your code:

our $INT //= 4;

my %hash;
push @{ $hash{ calcScore( $_ ) } }, $_ while <>;

my @grpKeys = sort{ $a <=> $b } keys %hash;

my @oldGrp;
while( my $next = shift @grpKeys ) {

    ## Remove any arrays from teh previously processed group
    ## that are below the interval of the next value - $INT
    shift @oldGrp while ( $next - $oldGrp[ 0 ] ) > $INT.

    ## Process items in the new array
    for my $i ( 0 .. $#{ $hash{ $next } } ) {

        ## against each of the (other) items in the new array
        for my $j ( $i+1 .. $#{ $hash{ $next } } ) {
            process( $i, $j );
        }

        ## and all of the items in each of the arrays 
        ## retained from the previous pass
        for my $g ( @oldGrp ) {
            for my $j ( 0 .. $#{ $hash{ $g } } ) { 
                process( $i, $j );
            }
        }
    }
    
    ## Add this array to the previously processed group
    push @oldGrp, $next;
}
[download]

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'

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

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

[reply]
[d/l]