RogerBW's Blog

Perl Weekly Challenge 87: Consecutive Rectangles 19 November 2020

I’ve been doing the Perl Weekly Challenges. The latest involved list searches and matrices. (Note that this is open until 22 November 2020.)

TASK #1 › Longest Consecutive Sequence

You are given an unsorted array of integers @N.

Write a script to find the longest consecutive sequence. Print 0 if none sequence found.

The algorithm I used is: sort the list. Then for each list member, if it's one higher than the previous one (or it's the first one) push it onto the current sequence. Otherwise store the current sequence (if it's longer than one previously found), and push the list member onto the new current sequence. (And check at the end for a sequence in progress; I added another test case for that. There are doubtless prettier ways of doing this without duplicating code, but this is a quick hack that works.)

sub lcs {
  my $ns=shift;
  my @l=sort {$a <=> $b} @{$ns};
  my @w;
  my @r;
  while (my $n = shift @l) {
    if ((scalar @w == 0) || $n==$w[-1]+1) {
      push @w,$n;
    } else {
      if (scalar @w > scalar @r) {
        @r=@w;
      }
      @w=($n);
    }
  }
  if (scalar @w > scalar @r) {
    @r=@w;
  }
  if (scalar @r > 1) {
    return \@r;
  } else {
    return 0;
  }
}

Raku is basically identical except that I get proper parameter passing.

For Python, since I'm taking things off the list one at a time and Python lists are ineffecient if you do that at the front, I reverse the list before entering the algorithm and take things off the end.

def lcs(n):
    l=list(n)
    l.sort(reverse=True)
    w=list()
    r=list()
    while (len(l)):
        n=l.pop()
        if (len(w)==0 or n==w[len(w)-1]+1):
            w.append(n)
        else:
            if (len(w) > len(r)):
                r=w
            w=[n]
    if (len(w) > len(r)):
        r=w
    if (len(r)>1):
        return r
    return 0

Ruby works basically like the Perl/Raku version.

And Rust is more like Python in that I reverse the sorted list first. Defining the potential capacity of a vector can lead to a time saving since it doesn't need to be reallocated as it grows.

Also I changed the definition slightly: if no sequence is found, I return [0] rather than 0, because that way I have a single return type.

A slight complication in that pop() returns an Option and I have to test that – but I can't stick that in the while loop condition. (while 1 would probably work as well.)

fn lcs(ns: Vec<i32>) -> Vec<i32> {
    let mut l=ns;
    l.sort();
    l.reverse();
    let mut w=Vec::<i32>::with_capacity(l.len());
    let mut r=vec![0];
    while l.len()>0 {
            let nn=l.pop();
            let n: i32;
            match nn {
                None => break,
                Some(x) => n = x,
            }
            if w.len()==0 || n == w[w.len()-1]+1 {
                        w.push(n);
                    } else {
                        if w.len() > r.len() {
                                r = w;
                            }
                           w=Vec::<i32>::with_capacity(l.len());
                           w.push(n);
                    }
        }
          if w.len() > r.len() {
                  r = w;
              }
             return r;
}

TASK #2 › Largest Rectangle

You are given matrix m x n with 0 and 1.

Write a script to find the largest rectangle containing only 1. Print 0 if none found.

This is similar to the one in #84, but with more fiddliness. I'm sure this isn't the best algorithm, but it was one I could code up quickly: loop over each possible starting point, then loop over each possible rectangle size, then loop over each actual entry. This means a lot of repetition; probably a better approach would be, from the starting point, to extend each row until one hit a zero, each column ditto, then work out possible rectangles from there. But I didn't; and at least it should be fairly cheap to check values in a two-dimensional array.

"Largest" isn't defined; I chose "maximum area".

The real oddity here is that the problem wants the actual rectangle, rather than its dimensions or location. Since that's intrinsically a bunch of 1 values, I build it at the end from the dimensions rather than copying out the actual data.

sub lr {
  my $s=shift;
  my $maxx=$#{$s};
  my $maxy=$#{$s->[0]};
  my $maxa=0;
  my @c;

First loops: check all starting points, skip onward if it isn't a 1.

  foreach my $x (0..$maxx-1) {
    foreach my $y (0..$maxy-1) {
      if ($s->[$x][$y]==1) {

Second loops: check each possible rectangle size (minimum 2×2)

        foreach my $tx ($x+1..$maxx) {
          foreach my $ty ($y+1..$maxy) {

And within that rectangle, check each cell, aborting the check if we find a single zero.

            my $valid=1;
          OUTER:
            foreach my $sx ($x..$tx) {
              foreach my $sy ($y..$ty) {
                if ($s->[$sx][$sy] != 1) {
                  $valid=0;
                  last OUTER;
                }
              }
            }

If we found a rectangle, and its area is larger than any we've previously found, store its width and height.

            if ($valid) {
              my @l=($tx-$x+1,$ty-$y+1);
              my $a=$l[0]*$l[1];
              if ($a > $maxa) {
                $maxa=$a;
                @c=@l;
              }
            }
          }
        }
      }
    }
  }

If we found a rectangle, generate a list of lists of 1s and return it.

  if (@c) {
    my @o;
    foreach (1..$c[0]) {
      push @o,[(1) x $c[1]];
    }
    return \@o;
  }
  return 0;
}

Raku is similar, with some tweaks to the list generation (it wants to produce a Seq rather than a List):

  if (@c) {
    my @o;
    for (1..@c[0]) {
      push @o,(1 xx @c[1]).List;
    }
    return @o;
  }

Python cannot, by design, break out of an inner loop in one step, so:

                        valid=1
                        for sx in range(x,tx+1):
                            for sy in range(y,ty+1):
                                if (s[sx][sy]!=1):
                                    valid=0
                                    break
                            if (valid==0):
                                break
                        if (valid):
                            l=[tx-x+1,ty-y+1]
                            a=l[0]*l[1]
                            if (a > maxa):
                                maxa=a
                                c=l

and at the end (note * for duplicating a list):

    if (len(c)):
        o=list()
        for i in range(0,c[0]):
            o.append([1] * c[1])
        return o
    else:
        return 0

Ruby sort of can break out of an inner loop but it gets fiddly, so I used the same pattern here.

            valid=true
            x.upto(tx) do |sx|
              y.upto(ty) do |sy|
                if (s[sx][sy] != 1)
                  valid=false
                  break
                end
              end
              if valid==false
                break
              end
            end

and the same * for list duplication:

  if (c.length > 0)
    o=Array.new
    1.upto(c[0]) do
      o.push([1] * c[1])
    end
    return o

Again with Rust I tweaked the rules to make the return type consistent. (Doubtless there is some way of fiddling this, but it's not what Rust wants you to do.) No break-from-inner here either, as far as I can tell from a superficial check. But oh, how beautiful the compilation of the return value:

    return vec![vec![1; c[1]]; c[0]];

Full code on github.

Comments on this post are now closed. If you have particular grounds for adding a late comment, comment on a more recent post quoting the URL of this one.

Search
Archive
Tags 1920s 1930s 1940s 1950s 1960s 1970s 1980s 1990s 2000s 2010s 3d printing action advent of code aeronautics aikakirja anecdote animation anime army astronomy audio audio tech base commerce battletech beer boardgaming book of the week bookmonth chain of command children chris chronicle church of no redeeming virtues cold war comedy computing contemporary cornish smuggler cosmic encounter coup covid-19 crime crystal cthulhu eternal cycling dead of winter doctor who documentary drama driving drone ecchi economics en garde espionage essen 2015 essen 2016 essen 2017 essen 2018 essen 2019 essen 2022 essen 2023 essen 2024 existential risk falklands war fandom fanfic fantasy feminism film firefly first world war flash point flight simulation food garmin drive gazebo genesys geocaching geodata gin gkp gurps gurps 101 gus harpoon historical history horror hugo 2014 hugo 2015 hugo 2016 hugo 2017 hugo 2018 hugo 2019 hugo 2020 hugo 2021 hugo 2022 hugo 2023 hugo 2024 hugo-nebula reread in brief avoid instrumented life javascript julian simpson julie enfield kickstarter kotlin learn to play leaving earth linux liquor lovecraftiana lua mecha men with beards mpd museum music mystery naval noir non-fiction one for the brow opera parody paul temple perl perl weekly challenge photography podcast politics postscript powers prediction privacy project woolsack pyracantha python quantum rail raku ranting raspberry pi reading reading boardgames social real life restaurant reviews romance rpg a day rpgs ruby rust scala science fiction scythe second world war security shipwreck simutrans smartphone south atlantic war squaddies stationery steampunk stuarts suburbia superheroes suspense television the resistance the weekly challenge thirsty meeples thriller tin soldier torg toys trailers travel type 26 type 31 type 45 vietnam war war wargaming weather wives and sweethearts writing about writing x-wing young adult
Special All book reviews, All film reviews
Produced by aikakirja v0.1