RogerBW's Blog

Perl Weekly Challenge 101: Spiral Origin 24 February 2021

I’ve been doing the Perl Weekly Challenges. The latest involved packing an array and a standard geometry problem. (Note that this is open until 28 February 2021.)

TASK #1 › Pack a Spiral

You are given an array @A of items (integers say, but they can be anything).

Your task is to pack that array into an MxN matrix spirally counterclockwise, as tightly as possible.

‘Tightly’ means the absolute value |M-N| of the difference has to be as small as possible.

I see two parts to this problem: first determine the appropriate dimensions for the array, then do the packing.

The factors of the number of items that are closest to equal will correspond to the shape closest to a square. So I solve part 1 by taking the highest integer less than or equal to the square root of the number of items, checking for divisibility into that number, then decrementing until that test succeeds. (In the limiting case it'll always work when the integer reaches 1. If this were a real-world problem I might do a prime factorisation instead and binary walk the result.)

  my $n=scalar @a;
  my $f2=int(sqrt($n));
  my $f1;
  while (1) {
    if ($n % $f2 == 0) {
      $f1=$n/$f2;
      last;
    }
    $f2--;
  }

Then fill the output array with zeroes, or whatever, just to get it to the right size. (In the languages that have types, there are ways of being generic about the array contents, but I didn't bother.)

  my @out;
  foreach (1..$f2) {
    push @out,[(0) x $f1];
  }

Then it's time to walk the array filling it up. Clearly there are four filling directions, and enough differs between them (which value I'm changing, in which direction, what's the limiting value and condition) that I ended up just writing the four separately, though it might be an amusing exercise in lambdas. (Each method breaks out of the overall loop if it runs out of list entries.)

  my ($x,$y)=(-1,0);
  my $maxx=$f1-1;
  my $maxy=$f2-1;
  my $minx=0;
  my $miny=1;
 ARR:
  while (1) {
    while ($x < $maxx) {
      $x++;
      $out[$y][$x]=pop @a;
      unless (@a) {
        last ARR;
      }
    }
    $maxx-=1;
    while ($y < $maxy) {
      $y++;
      $out[$y][$x]=pop @a;
      unless (@a) {
        last ARR;
      }
    }
    $maxy-=1;
    while ($x > $minx) {
      $x--;
      $out[$y][$x]=pop @a;
      unless (@a) {
        last ARR;
      }
    }
    $minx++;
    while ($y > $miny) {
      $y--;
      $out[$y][$x]=pop @a;
      unless (@a) {
        last ARR;
      }
    }
    $miny++;
  }

Other languages are basically similar. Python can't break out of an inner loop (by design apparently), and Ruby does it with throw/catch. Rust needs my subscripts to be of a usize type, which can't be set to -1 ever, so that needs a flag for the first pass incrementing x.

TASK #2 › Origin-containing Triangle

You are given three points in the plane, as a list of six co-ordinates: A=(x1,y1), B=(x2,y2) and C=(x3,y3).

Write a script to find out if the triangle formed by the given three co-ordinates contain origin (0,0).

Print 1 if found otherwise 0.

I felt free to declare that what I was getting was a list of three coordinate pairs, not a single flat list of six numbers, and code on that basis.

This is a standard geometry problem (well, it's usually to an arbitrary point rather than the origin), with two solutions I can think of: one is to build a new triangle from each line to the origin, measure the areas of each, and compare with the area of the original triangle, but that would potentially mean doing floating-point comparisons. So instead I used the other approach: take the cross products of the vectors (point A to origin) and (point A to point B) (actually just the z component of that cross product, since they're coplanar in x/y) which is effectively a function of the angle between them.

(What do you get if you cross a dog with a cat? |dog| |cat| sin theta.)

If all those cross products are positive (or zero in the case where the origin lies along one of the edges), or if they're all negative-or-zero depending on how the calculation was done, then the origin lies inside the triangle. If some cross products are positive and others are negative, it's outside.

Part two is usually meant to be the harder part of the pair of problems, but the hard bit has already been achieved in working out the above; the actual coding is relatively simple. In Raku:

sub ot(@points) {
  my @pp=@points;
  push @pp,@points[0];
  my @xp;
  for (0..2) -> $i {
    push @xp,(@pp[$i][0] *
                (@pp[$i+1][1]-@pp[$i][1]))
      -
      (@pp[$i][1] *
       (@pp[$i+1][0]-@pp[$i][0]));
  }
  @xp=sort @xp;
  if (@xp[0] <=0 && @xp[2] <=0) {
    return 1;
  }
  if (@xp[0] >=0 && @xp[2] >=0) {
    return 1;
  }
  return 0;
}

(This is a good example, I think, of when not to make a thing a function – the cross product calculation is an obvious candidate, but in this specific code I'm only using it in this one place so there's no particular virtue to splitting it out. In real life I'd probably be coding this up as part of a geometry library, in which case there would most certainly be a separate cross product function that would be called from many places.)

Full code on github.


  1. Posted by RogerBW at 11:30am on 02 March 2021

    Part 1 saw a split between the square-root approach and exhaustive search. (I think that square-root and counting down from there should minimise the number of calculations.) Most of the spiral-fillers worked like mine, though a couple tried to get more sophisticated.

    For part 2, cross product appears to have been the most popular approach.

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