# ETOOBUSY π minimal blogging for the impatient

# PWC109 - Four Squares Puzzle

**TL;DR**

On with TASK #2 from the Perl Weekly Challenge #109. Enjoy!

# The challenge

You are given four squares as below with numbers named a,b,c,d,e,f,g.

`(1) (3) ββββββββββββββββ ββββββββββββββββ β β β β β a β β e β β β (2) β β (4) β βββββ«βββββββ«ββββ βββββ«ββββββββββ β β β β β β β β β β b β β d β β f β β β β β β β β β β β β β β β β β β ββββββββββββͺββββ βββββͺβββββββͺββββ β β c β β g β β β β β β β β β ββββββββββββββββ βββββββββββββββ`

Write a script to place the given unique numbers in the square box so that sum of numbers in each box is the same.

# The questions

Wellβ¦ a few:

- What to do if there is no solution?
- letβs assume that a nice message is enough

- What to do if there are multiple solutions?
- any will do, additional points if itβs not always the same

- Can numbers be negative?
- sure, why not?

- Can numbers be floating point? Can we assume that we will not be
hit by weird approximation errors that make
`0.3 == 0.2 + 0.1`

*false*in most languages (Perl included)?- interesting, but for this puzzle letβs stick to integers

- Can I assume that sums are within the limits of my platform?
- yes, if your platform is capable of holding integers between
`-128`

and`127`

.

- yes, if your platform is capable of holding integers between

# The solution

This time we go *brute force*, leveraging previous post Iterator-based
implementation of Permutations.

In particular, we will take any possible permutations of the input list, and check whether it leads to a solution or not. Hence, the core of the solution is this:

```
sub four_squares_puzzle (@values) {
my $it = permutations_iterator(items => [@values]);
while (my @S = $it->()) {
my $sum = $S[0] + $S[1];
next if $sum != $S[1] + $S[2] + $S[3];
next if $sum != $S[3] + $S[4] + $S[5];
next if $sum != $S[5] + $S[6];
my @keys = 'a' .. 'g';
my %retval;
@retval{@keys} = @S;
return %retval;
}
return;
}
```

I knowβ¦ itβs extremely lazy, but also extremely easy to code!

Using an iterator here is handy because it allows us to avoid computing
*all* permutations, and get out as soon as we find a suitable one.

I initially thought that there might be some magic way to figure out what the sum within each square would be by simply inspecting the numbers linearly, e.g. by summing them.

It turns out I was wrong.

Itβs possible to find solutions to the example input where the sum in each square can be 9:

```
a = 4
b = 5
c = 3
d = 1
e = 6
f = 2
g = 7
```

or 10:

```
a = 7
b = 3
c = 2
d = 5
e = 1
f = 4
g = 6
```

as well as 11:

```
a = 5
b = 6
c = 2
d = 3
e = 1
f = 7
g = 4
```

I doubt there are others.

This prompted me to *randomize* the input array, so that different
possible solutions should pop up; to this extent, List::Utilβs
`shuffle`

is invaluable.

The full solution, should you be curious:

```
#!/usr/bin/env perl
use 5.024;
use warnings;
use experimental qw< postderef signatures >;
no warnings qw< experimental::postderef experimental::signatures >;
use List::Util 'shuffle';
sub four_squares_puzzle (@values) {
my $it = permutations_iterator(items => [@values]);
while (my @S = $it->()) {
my $sum = $S[0] + $S[1];
next if $sum != $S[1] + $S[2] + $S[3];
next if $sum != $S[3] + $S[4] + $S[5];
next if $sum != $S[5] + $S[6];
my @keys = 'a' .. 'g';
my %retval;
@retval{@keys} = @S;
return %retval;
}
return;
}
my @input = @ARGV == 7 ? @ARGV : 1 .. 7;
my %solution = four_squares_puzzle(shuffle @input);
if (! scalar keys %solution) {
say 'no solution, sooooorry!';
}
else {
for my $key (sort keys %solution) {
say "$key = $solution{$key}";
}
}
sub permutations_iterator {
my %args = (@_ && ref($_[0])) ? %{$_[0]} : @_;
my $items = $args{items} || die "invalid or missing parameter 'items'";
my $filter = $args{filter} || sub { wantarray ? @_ : [@_] };
my @indexes = 0 .. $#$items;
my @stack = (0) x @indexes;
my $sp = undef;
return sub {
if (! defined $sp) { $sp = 0 }
else {
while ($sp < @indexes) {
if ($stack[$sp] < $sp) {
my $other = $sp % 2 ? $stack[$sp] : 0;
@indexes[$sp, $other] = @indexes[$other, $sp];
$stack[$sp]++;
$sp = 0;
last;
}
else {
$stack[$sp++] = 0;
}
}
}
return $filter->(@{$items}[@indexes]) if $sp < @indexes;
return;
}
}
```

Stay safe!

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