Given that the numbers can range from 1 to 20 (decimal) digits, a simple pack will always make an 8-byte string, but if values less than 10_000_000 happen to predominate (the OP suggests there's an upper bound of 10M elements to be tracked in the application), the result will be a net increase of memory usage.

Apart from that, there's a potential risk that some 64-bit int values, when packed into an 8-byte string, might collide with actual strings encountered in the data. There's a better way to pack integers into smaller strings.

The following "encode/decode" functions will always return a string of characters in the range "\x80" - "\xff", and the number of characters will be roughly half the number of decimal digits. (I'm assuming that BrowserUK's strings will always be in the ASCII range, so I'm using non-ASCII characters that happen to be stored internally as single bytes in perl): 

sub int2str {
    my $int = shift;
    my $str = '';
    while ( $int ) {
        $str .= chr(( $int % 128 ) + 0x80 );
        $int = int( $int / 128 );
    }
    $str;
}

sub str2int {
    my $mult = 1;
    my $int = 0;
    for ( split( //, shift )) {
        $int += ( ord() - 0x80 ) * $mult;
        $mult *= 128;
    }
    $int;
}

[download]
(There might be more efficient ways to implement that.)

Even so, the overall space saved seems a bit disappointing - on my (macports, 5.12) version of perl, encoding the numbers like that over the ~914K entries in BrowserUK's single-hash example (strings 'aaaa' .. 'zzzz', numbers incrementing from 1) yielded just 4.1% reduction in the total_size of the hash. (If I use random integers ranging up to 15 digits, it edges closer to 6% reduction.)

In reply to Re^2: Bidirectional lookup algorithm? (Updated: further info.) by graff
in thread Bidirectional lookup algorithm? (Updated: further info.) by BrowserUk

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