in reply to Near equal partitions.

Another approach 

sub part {
  my ($n, $m) = @_;
  my $q = int($n / $m);
  my $r = $n % $m;
  return ($q+1) x $r, ($q) x ($m-$r);
}

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Re^2: Near equal partitions.
by ELISHEVA (Prior) on Jan 14, 2011 at 05:44 UTC

    That list generation is lovely!

    Just one question. Is the Perl interpreter smart enough to know that int($n/$m) should use integer division rather than floating point division? i.e. would there be any difference in performance between int($n/$m) vs. ($n-$r)/$m?

    In more strongly typed languages, the type of the variables determines the meaning of the '/' operator (integer division vs. floating point division) so ($n-$r)/$m would be faster. However in Perl, the operator determines the type of the operands, so maybe there is no difference in performance between the two expressions in Perl? Or maybe ($n-$r)/$m is slower because the int(...) gives the Perl interpreter a clue to context and hence the opportunity to use integer division when it sees such a simple expression as int( token / token )?

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      Just one question. Is the Perl interpreter smart enough to know that int($n/$m) should use integer division rather than floating point division?

      No.

      i.e. would there be any difference in performance betwee int($n/$m) vs. ($n-$r)/$m?

      Sorry, you asked to ask two questions. But to save you from having to ask to ask another question, I'll answer it anyway.

      Performance should be roughly the same (4 trivial ops in the former, 5 trivial ops in the latter). The latter is better from a numerical methods point of view, no possibility of floating point error.

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        btw, what about use integer? It looks it's little bit faster with that.
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