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### Re^3: Fast - Compact That String

by BrowserUk (Pope)
 on Feb 10, 2012 at 20:06 UTC ( #953120=note: print w/replies, xml ) Need Help??

in reply to Re^2: Fast - Compact That String
in thread Fast - Compact That String

An explanation of the code as requested:

```## Map the numbers, 0 .. 36 to the symbols we use
## to represent the number in base37
my @c1 = (' ', '0'..'9', 'A'..'Z' );

sub fromB37 {
my \$n = shift;      ## Get the number to convert
## Allocate space for the Base37 representation
## Initialise it to the representation of 0 (six spaces)
my \$s = '      ';

## For each position in the string
for( 0 .. 5 ) {
## extract the next base37 digit value
## look up its representaion character
## and assign it to the 'right place' i the string.
substr( \$s, \$_, 1 ) = \$c1[ \$n%37 ] );

## dividing by 37 effectively right-shifts
## the last digit's value out of the number
\$n /= 37;
}
\$s;
}

my @c2;
## Map the ordinal values of the symbols
## to their numeric values (0 .. 37)
## The sparse array is faster than a hash
\$c2[  ord( \$c1[ \$_ ] ) ] = \$_ for 0 .. 36;

sub toB37 {
my \$n = 0; ## initialise our return value to 0

## split the base37 representation
## into a list of the ordinal values of the symbols
## and reverse their order to match that produced by fromB37()
for( reverse unpack 'C*', \$_[0] ) {
## multiple the running total by 37
## (effectively left-shifting the accumulator
## to accommodate the next digit.)
## and add value of the next base37 digit
## by looking it up in the mapping array
\$n = \$n * 37 + \$c2[ \$_ ];
}
## return the accumulated value.
\$n;
}

As mbethke points out, these treat the base37 number in 'little-endian' fashion. This because you emphasised compression and speed, with no mention of needing to manipulate the compressed values numerically (sorting).

To get the sortable, big-endian representation, you could use:

```my @c1 = (' ', '0'..'9', 'A'..'Z' );
sub fromB37 {
my \$n = shift;
my \$s = '      ';
substr( \$s, \$_, 1, \$c1[ \$n%37 ] ), \$n /= 37 for 5, 4, 3, 2, 1, 0;
\$s;
}

my @c2;
\$c2[  ord( \$c1[ \$_ ] ) ] = \$_ for 0 .. 36;
sub toB37 {
my \$n = 0;
\$n = \$n * 37 + \$c2[\$_] for unpack 'C*', \$_[0];
\$n;
}

Which actually works out a bit quicker still, but not as fast as mbethke's unrolled version.

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