TL;DR

On with TASK #2 from the Perl Weekly Challenge #084. Enjoy!

# The challenge

You are given matrix of size m x n with only 1 and 0. Write a script to find the count of squares having all four corners set as 1.

# The questions

I guess that most questions I would ask about this script is what the input format is: is it an array-of-arrays? A string to be parsed? What about invalid inputs (like other stuff than 1 and 0)?

Another question is whether a lone 1 counts as a square or not. It seems to not count as a square, considering the examples.

# The solution

I guess these challenges don’t really get me create particularly interesting solutions, anyway… here’s my take.

 1 sub find_square (@matrix) {
2    my $m = @matrix; 3 my$n     = $matrix->@*; 4 my$count = 0;
5    for my $i (0 ..$m - 2) {    # no point in scanning last line
6       for my $j (0 ..$n - 2) {    # same for last column
7          next unless $matrix[$i][$j]; # only consider "1"s in upper left 8 my$k = 1;
9          while (($i +$k < $m) && ($j + $k <$n)) {
10             ++$count 11 if$matrix[$i][$j + $k] 12 &&$matrix[$i +$k][$j] 13 &&$matrix[$i +$k][$j +$k];
14             ++$k; 15 } ## end while (($i + $k <$m) && ...)
16       } ## end for my $j (0 ..$n - 2)
17    } ## end for my $i (0 ..$m - 2)
18    return $count; 19 } ## end sub find_square (@matrix)  We iterate through the matrix looking for suitable upper-left corners. This means that: • we will be searching for squares whose other vertices are either right or down (or both) from our starting corner; • we will skip the last column and the last row, because there would not be candidates for other corners. This accounts for the nested iterations in lines 5 and 6, where our iteration variables scan all but the list row and column. In this loop, we look for candidate upper-left corners, so we need that to be 1 and skip 0 (line 7, the string 0 is false in Perl). At this point, we’re looking for squares and we iterate different possible sizes, starting from 1 (line 8). We do another sub-iteration here, up to the point where the candidate square does not fit in the matrix any more (line 9). In this third loop, we already know that the upper-left corner is a 1, so we have to check the other three corners (lines 11 through 13) and increment our count accordingly (line 10). Line 14 increments the candidate square size, and line 18 eventually returns the count. Easy, uh!?! As always, here’s the full script should you want to play with it: #!/usr/bin/env perl use 5.024; use warnings; use experimental qw< postderef signatures >; no warnings qw< experimental::postderef experimental::signatures >; sub find_square (@matrix) { my$m     = @matrix;
my $n =$matrix->@*;
my $count = 0; for my$i (0 .. $m - 2) { # no point in scanning last line for my$j (0 .. $n - 2) { # same for last column next unless$matrix[$i][$j];    # only consider "1"s in upper left
my $k = 1; while (($i + $k <$m) && ($j +$k < $n)) { ++$count
if $matrix[$i][$j +$k]
&& $matrix[$i + $k][$j]
&& $matrix[$i + $k][$j + $k]; ++$k;
} ## end while (($i +$k < $m) && ...) } ## end for my$j (0 .. $n - 2) } ## end for my$i (0 .. $m - 2) return$count;
} ## end sub find_square (@matrix)

sub string2matrix ($string) { map { [split m{}mxs] } split m{\D+}mxs,$string;
}

sub print_matrix (@matrix) {
map { say join ' ', $_->@*;$_ } @matrix;
}

my $matrix = shift || "1101 1100 0111 1011"; say find_square(print_matrix(string2matrix($matrix)));


Comments? Octodon, , GitHub, Reddit, or drop me a line!