Re: breaking an array into nearly equal parts

[revdiablo]

Did you notice a pattern in the table you pasted? It looks like it contains all the information you need. It's basically like this:

If $x%3 is 0, the arrays should all be of size $x/3. If $x%3 is not 0, there should be $x-($x%3) arrays of size $x/3 and $x%3 of size $x/3+1.

I'm not positive this will continue to hold true, but it looks good from what you've shown. There may be a simpler algorithm too, so I would be happy to see other replies. :-)

[spiritway]

You're right - and the pattern holds true. A positive integer can either be exactly divisible by three, have a remainder of 1, or a remainder of 2. Your idea is probably one of the simplest.

[Roy Johnson]

It is quite simple. Here's an implementation. The apportion figures out chunk sizes; multi_slice returns slices of those sizes. The for block runs a few examples.

sub apportion {
  my ($elements, $pieces) = @_;
  my $small_chunk = int $elements / $pieces;
  my $oversized_count = $elements % $pieces;
  ((1 + $small_chunk) x ($oversized_count), ($small_chunk) x ($pieces 
+- $oversized_count));
}

sub multi_slice {
  my ($aref, @chunk_sizes) = @_;
  my $hi_i = -1;
  map {
    my $lo_i = $hi_i + 1;
    $hi_i += $_;
    [@$aref[$lo_i..$hi_i]]
  } @chunk_sizes;
}

for my $try ([16,3], [17,4], [19,3]) {
  print "$try->[0] elements into $try->[1] pieces:\n";
  print "Sizes: ", join(', ', apportion(@$try)), "\n";
  print "@$_\n" for multi_slice([1..$try->[0]], apportion(@$try));
}