TL;DR

Itâ€™s time to start coding for aquarium, letâ€™s begin from the constraints that allow us to tell a good - if partial - solution from an evidently wrong one.

The code for this stage can be found in stage 3.

# Letâ€™s start from the rules

The rules for aquarium are simple, from the site:

• The puzzle is played on a rectangular grid divided into blocks called â€śaquariumsâ€ť
• You have to â€śfillâ€ť the aquariums with water up to a certain level or leave it empty.
• The water level in each aquarium is one and the same across its full width
• The numbers outside the grid show the number of filled cells horizontally and vertically.

The constraints are stated in the third and fourth bullet, letâ€™s address them individually. The following sections assume that youâ€™re comfortable with the data structure to represent the whole puzzle, you can take a refresher in the first post Aquarium - parse puzzle input.

# Water has one level only in one aquarium

This constraint can be translated into the following checks:

• vertically in one column, if two adjacent cells belong to the same aquarium, the upper one MUST have a value that is less than, or equal to, the lower one. This stems from the fact that empty (-1) is lighter than or equal to unknown (0), which is lighter than or equal to water (1);
• horizontally, whatever level we find for the leftmost cell of an aquarium, all other cells in the same aquarium MUST hold the same value.

This is the code for this constraint:

`````` 1  sub assert_water_level (\$puzzle) {
2     my (\$n, \$field, \$status) = \$puzzle->@{qw< n field status >};
3     for my \$i (0 .. \$n - 1) {    # iterate rows from top to bottom
4         my %expected;
5         for my \$j (0 .. \$n - 1) {
6             my \$id = \$field->[\$i][\$j];
7             my \$st = \$status->[\$i][\$j];
8
9             die "wrong vertical leveling for aquarium \$id\n"
10                if (\$i > 0)
11                && (\$id == \$field->[\$i - 1][\$j])
12                && (\$st < \$status->[\$i - 1][\$j]);
13
14             \$expected{\$id} //= \$st;
15             die "wrong horizontal leveling for aquarium \$id\n"
16                if \$expected{\$id} != \$st;
17
18         } ## end for my \$j (0 .. \$n - 1)
19     } ## end for my \$i (0 .. \$n - 1)
10     return \$puzzle;
21  } ## end sub assert_water_level (\$puzzle)
``````

It is basically a straight translation into code of the bullets above.

The vertical check is performed in lines 9 to 12; it can only be performed from the second row on, which is the reason of the test in line 10.

Hash `%expected` tracks the expected value for each aquarium at the same level (second bullet), initializing it with the first value found for each aquarium (line 14 `\$expected{\$id //= \$st`) and complaining if values differ (lines 15 and 16). This test is only for the single horizontal level, hence `%expected` is declared inside the outer loop so that it is reset for each new row.

# Row-level and Column-level constraints

Boundary conditions are easy to check: it suffices to count how many water- filled cells are there, and check that itâ€™s the right number. Put it like this, anyway, the check would be somehowâ€¦ strict because it can only be fulfilled by a complete solution. We will check that there is not too much water instead, as well as not too much emptiness:

`````` 1 sub assert_boundary_conditions (\$puzzle) {
2    my (\$n, \$status, \$items_by_row, \$items_by_col) =
3       \$puzzle->@{qw< n status items_by_row items_by_col >};
4
5    # the field is square and this is an advantage, \$i and \$j can be
6    # thought as either row-column or column-row
7    for my \$i (0 .. \$n - 1) {
8       my \$i1 = \$i + 1; # useful for the exception
9       my (\$water_row, \$empty_row, \$water_col, \$empty_col) = (0) x 4;
10       for my \$j (0 .. \$n - 1) {
11          \$water_row++ if \$status->[\$i][\$j] > 0;
12          \$empty_row++ if \$status->[\$i][\$j] < 0;
13          \$water_col++ if \$status->[\$j][\$i] > 0;
14          \$empty_col++ if \$status->[\$j][\$i] < 0;
15       }
16
17       die "too many filled cells in row \$i1\n"
18          if \$water_row > \$items_by_row->[\$i];
19
20       die "too many empty cells in row \$i1\n"
21          if \$empty_row > \$n - \$items_by_row->[\$i];
22
23       die "too many filled cells in col \$i1\n"
24          if \$water_col > \$items_by_col->[\$i];
25
26       die "too many empty cells in col \$i1\n"
27          if \$empty_col > \$n - \$items_by_col->[\$i];
28
29    }
30    return \$puzzle;
31 }
``````

In other terms, we allow for some cells to still be unknown, so our checks translate into ensuring that the count of empty or filled cells is within the expected bounds.

Iteration over variable `\$i` (outer iteration) does the trick for both rows and columns at the same time - the trick is to keep track of four different variables to count the amount of water and confirmed empty spaces for row `\$i` and column `\$i`. The actual counting is done in lines 10 to 15, where variable `\$j` iterates the other dimension.

Tests in lines 17 to 27 are straightforward: if the specific quantity is out of bounds, an exception is raised.

# Constraints in action!

The following asciinema recording shows examples of all the different ways to fail the constraints:

Until next timeâ€¦ happy coding!

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