RogerBW's Blog

Perl Weekly Challenge 84: Reverse squares 28 October 2020

I’ve been doing the Perl Weekly Challenges. The latest involved reversing integers and a two-dimensional search. (Note that this is open until 1 November 2020.)

Task #1 › Reverse Integer

You are given an integer $N.

Write a script to reverse the given integer and print the result. Print 0 if the result doesn’t fit in 32-bit signed integer.

The first part is easy, of course. The second is rather harder: all four of the languages I'm using have larger standard integer types than 32 bits.

So in Perl we have:

  my $r=join('',reverse split '',$s);
  if ($r =~ /([0-9]+)-$/) {

Raku, with more complicated regexp syntax:

  my $r=$s.comb.reverse.join('');
  if ($r ~~ /(<[0..9]>+)\-$/) {


    if (a[-1] == '-'):
        a='-' + a[0:-1]


    if (a[-1] == '-'):
        a='-' + a[0:-1]

But how do we test whether the thing will fit into 32-bit signed?

Well, these languages may not have 32-bit data types, but they can encode/decode them for use with packed formats. So in Perl:

  if (unpack('l',pack('l',$r)) != $r) {
    return 0;
  return $r;

Python (needs a cast to integer):

    except struct.error as err:
        return 0
    return a

Ruby (ditto):

  if b != r
    return 0
  return r

But what about Raku? Well, its pack is still very primitive. But unlike the other languages here it does have an explicitly 32-bit-wide integer type. So I cast to that (which, perversely, doesn't raise an error if it doesn't fit) and see if it worked.

  my int32 $b=Int($r);
  if ($b != $r) {
    return 0;
  return $r;

Task #2 › Find Square

You are given matrix of size m × n with only 1 and 0.

Write a script to find the count of squares having all four corners set as 1.

So it's basically a two-dimensional search, but the delta-x and delta-y terms are always the same. Pretty straightforward really (using Perl for the example): find the array size;

  my $t=0;
  my $maxx=$#{$s};
  my $maxy=$#{$s->[0]};

iterate over each possible starting point;

  foreach my $x (0..$maxx-1) {
    foreach my $y (0..$maxy-1) {

see if that point is a potential anchor;

      if ($s->[$x][$y]==1) {

if so, iterate over all the sizes of square that might fit;

        foreach my $d (1..min($maxx-$x,$maxy-$y)) {

and check the other three corners.

          if ($s->[$x+$d][$y]==1 &&
                $s->[$x][$y+$d]==1 &&
                  ) {

And the other languages basically work the same way. Python wants a range operator, but otherwise they're all identical apart from minor details of syntax.

One could I suppose build a table of potential opposite corners but I'm not convinced that there's much optimisation to be had.

Full code on github.

  1. Posted by RogerBW at 02:35pm on 02 November 2020

    Only nine other blog posts were registered this week, and four of them are on unresponsive hosts. (And I really can't be bothered to read through 100+ examples of code.)

    As expected, the 32-bit check was the hard part of problem 1; some people ignored it, some invented very baroque ways of verifying it. Of the posts I could read, I seem to have been the only person to think of using internal 32-bit types or operators.

    For part 2, the usual optimisations seem to be the same ones I used – drop out if the starting corner isn't a 1, and stop checking at once if any of the other corners isn't.

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