TL;DR

On with TASK #2 from The Weekly Challenge #123. Enjoy!

The challenge

You are given coordinates of four points i.e. (x1, y1), (x2, y2), (x3, y3) and (x4, y4).

Write a script to find out if the given four points form a square.

Example

Input: x1 = 10, y1 = 20
       x2 = 20, y2 = 20
       x3 = 20, y3 = 10
       x4 = 10, y4 = 10
Output: 1 as the given coordinates form a square.

Input: x1 = 12, y1 = 24
       x2 = 16, y2 = 10
       x3 = 20, y3 = 12
       x4 = 18, y4 = 16
Output: 0 as the given coordinates doesn't form a square.

The questions

As we will be doing some potentially floating point maths, a first question would be what tolerance should the operations have, in particular what tolerance is there to consider a value to be the same as 0.

As nothing is said about the ordering of the points, we will assume they can be in any order and not necessarily assuming that close points in the list are also adjacent in the candidate square.

The examples seem to indicate that the points we consider are in a plane.

Last, I’d ask if this is meant to be a tricky question. The first example is about a square whose sides are parallel to the coordinate axes, but… squares might also be rotated in the plane!

The solution

We’ll use some vector maths here. Assuming that the input sequence of points $(P_0, P_1, P_2, P3)$ is ordered, i.e. that each consecutive pair is a side of the candidate polygon we want to check, we end up with the following vectors representing the four sides:

\[s_0 = P_1 - P_0 \\ s_1 = P_2 - P_1 \\ s_2 = P_3 - P_2 \\ s_3 = P_0 - P_3\]

Much like the points, these “vector sides” are represented by pairs of numbers, so we can “blur” the line and use the same representation for the two.

In a square, two consecutive sides $s_i$ and $s_{i + 1}$ MUST fulfil the following two conditions:

  • have the same length;
  • be orthogonal, i.e. form an angle of $\pm 90°$.

Fun fact: we only need to check the two conditions above for the first three sides $s_0$, $s_1$, and $s_2$. If the comply, the fourth side $s_3$ will comply too.

The length of a vector is calculated with Pythagora’s theorem:

\[L_v = \sqrt{v_x^2 + v_y^2}\]

In comparing two sides, though, we can equivalently look a the squares and avoid calculating the square root:

\[L_v^2 = v_x^2 + v_y^2\]

Checking for orthogonality can be done calculating their regular scalar (or dot) product:

\[v \cdot w = v_x w_x + v_y w_y\]

This is 0 if and only if the two vectors are orthogonal, so it’s exactly the condition we are after.

OK, enough theory now… show us the code!

Raku

Raku first, which also gets the nice commenting. We define a class to represent our points and vectors:

# a tiny class for handling a limited set of vector operations
class Vector {
   has @.cs is built is required;

   # "dot", i.e. scalar, product
   method dot (Vector $a) { return [+](self.cs »*« $a.cs) }

   # the *square* of the length is all we need in our solution
   method length_2 ()     { return self.dot(self) }
}

To make the implementation easier to read, we also override the difference operator (so that we can calculate vectorized sides by difference of two points):

multi sub infix:<->(Vector $a, Vector $b) {
   Vector.new(cs => [$a.cs »-« $b.cs]);
}

as well as the dot product, which relies on the dot method:

multi sub infix:<*>(Vector $a, Vector $b) { $a.dot($b) }

Our basic test function is the following:

sub is-sequence-a-square (@points is copy) {

   # comparing candidate sides means that we consider a "previous" side
   # and a "current" one. A side is defined as the vector resulting from
   # the difference of two consecutive points.
   my $previous = @points[1] - @points[0];

   # we just need to compare 3 sides, if they comply then the 4th will too
   for 1, 2 -> $i {
      my $current = @points[$i + 1] - @points[$i];

      # check if sides have the same length (squared)
      return False if $previous.length_2 != $current.length_2;

      # approximation might give surprises, we'll accept as orthogonal
      # sides whose scalar product is below our tolerance
      return False if $previous * $current > tolerance;

      # prepare for next iteration
      $previous = $current;
   }

   # three sides are compliant, it's a square!
   return True;
}

Now, of course, our input sequence of points might not be in the “right” order, so we wrap the test above to check different alternative orderings.

How many permutations should we consider? Out of 4 points, we have $4! = 24$ of them, but we don’t need to consider them all.

First, we can fix our point in the first position as our starting point, so in case we only have to consider permutations of the other three, i.e. $3! = 6$ of them.

Then, we can observe that two arrangements that have the same point as the opponent (i.e. non-adjacent) point to the starting point are actually the same candidate polygon, traversed in opposite directions. Hence, we can just consider one of these two.

In the end, we can just consider three possible permutations, like in the following function:

sub is-square (*@points) {

   # try out permutations of the inputs that can yield a square. We fix
   # point #0 and only consider one permutation for each of the other
   # points as the opposite, ignoring the other because symmetric.
   state @permutations = (
      [0, 2, 1, 3],  # 0 and 1 are opposite
      [0, 1, 2, 3],  # 0 and 2 are opposite
      [0, 2, 3, 1],  # 0 and 3 are opposite
   );
   for @permutations -> $permutation {
      my @arrangement = @points[@$permutation].map({Vector.new(cs => @$_)});
      return 1 if is-sequence-a-square(@arrangement);
   }
   return 0;
}

A couple of final remarks:

  • Math::Vector was of… great inspiration for getting the implementation right. I used it in the first place, but it takes ages to load and eventually re-implemented only the relevant parts;
  • inlining the class as I did means that the definition of the overloaded multi sub infix operators must appear outside the class definition. This took me a while to figure out.

Perl

The Perl translation is pretty much straightforward, also thanks to the overload module that allows us to overload a couple of operators. Here’s the complete program:

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';
use constant False => 0;
use constant True  => 1;

use constant tolerance => 1e-7;

package Vector2D {
   use overload
     '-' => sub ($u, $v, $x) { v([ map { $u->[$_] - $v->[$_] } 0, 1 ]) },
     '*' => sub ($u, $v, $x) { $u->dot($v) };

   sub dot ($S, $t)   { return $S->[0] * $t->[0] + $S->[1] * $t->[1] }
   sub length_2 ($S)  { return $S->dot($S) }
   sub v ($v)         { return bless [$v->@*], __PACKAGE__ }
}

sub is_sequence_a_square (@points) {
   my $previous = $points[1] - $points[0];
   for my $i (1 .. $#points - 1) {
      my $current = $points[$i + 1] - $points[$i];
      return False if $previous->length_2 != $current->length_2;
      return False if $previous * $current > tolerance;
      $previous = $current;
   }
   return True;
}

sub is_square (@points) {
   state $permutations = [
      [0, 2, 1, 3],
      [0, 1, 2, 3],
      [0, 2, 3, 1],
   ];
   for my $permutation ($permutations->@*) {
      my @arrangement = map { Vector2D::v($_) } @points[@$permutation];
      return 1 if is_sequence_a_square(@arrangement);
   }
   return 0;
}

say is_square([10, 20], [20, 20], [20, 10], [10, 10]);
say is_square([12, 24], [16, 10], [20, 12], [18, 16]);
say is_square([0, 0], [1, 1], [0, 2], [-1, 1]);

Thanks!

Thank you for reading this far and stay safe!