# Fighting Fish: An Aquarium-Star Battle Hybrid

This is an entry for Fortnightly Topic Challenge #44: Introduce a new grid deduction genre to the community.

UPDATE NOTE: The original image had the upper right indicator that there are two fish per row, column and shape to the left of the fish, which could be interpreted as being an Aquarium clue. It is not...I used the standard Star Battle indication for this unthinkingly. Many, MANY apologies!

This puzzle is a hybrid of two grid deduction puzzles that rely on the partition of a grid into shapes. Aquarium recently appeared on PSE in Stiv's post. The goal of Aquarium is to shade ("add water") to each shape, regarded as a fishbowl, so that the clues outside the grid indicate the number of shaded squares in a row/column. The catch is that in each shape, either every cell in a row is shaded or is not shaded, and no shaded row can appear above an unshaded row.

Star Battle also made a recent appearance on PSE. The rules of Star Battle, summarized from https://www.puzzle-star-battle.com are simple: place stars in the grid such that each row, column, and shape contains exactly some fixed number of stars, where stars cannot be adjacent, even diagonally.

This Fighting Fish hybrid puzzle asks you to both shade the squares of the grid below according to the rules of Aquarium, and then place two fish in each row, column and shape according to the rules of Star Battle. The extra requirement: fish have to be in water. Note that neither individual puzzle is uniquely determined, but there is a unique solution to the hybrid. I hope you enjoy!

Solver Notes I don't think the puzzle is particularly hard, but I like the need to go back and forth between the logic of the component puzzles. If you really want to get in the spirit, the Unicode for the fish is U+1F41F, 🐟. Also, I might be stretching the letter of the FTC a little bit, since both component puzzles have technically appeared recently on PSE, but both have appeared only a single time, and I feel this is still in the spirit.

• If a shape has two disconnected sides going down, can one of them have water at a level above the other side? So for the top-right shape, can there be water in R3C9-10 but no water in R3C12? Dec 5, 2020 at 16:26
• @bobble: That's a great question. According to official Aquarium rules, no. But I've been thinking of doing a variant with better physics. For this puzzle, assume that the level must be uniform, even across "disconnected" parts of a tank. Dec 5, 2020 at 17:17
• The cell height is 43 px but the width is 42 px. I do appreciate they are whole numbers, but ... still :P Dec 5, 2020 at 19:03

## 1 Answer

Apologies in advance for the length of this answer. Blue is confirmed shaded, green is confirmed unshaded. I did use the Unicode fish.

Step 1:

First off, there has to be at least some water in each region, or it can't have fish in it. Therefore the bottom row of each shape can be shaded. I also shaded some additional rows until there was a way to fit 2 non-touching fish into each shape.

Step 2:

The 8 needs 3 more shaded cells, and there is only one way to give it that.

Step 3:

That shape just completed has only one way to fit two non-touching fish in.

Step 4:

The 8 has two options for getting its last 4 shaded squares. One of them would mean the bottom-left shape would have only its bottom row shaded, so the bottom-right shape couldn't use its bottom row. However there's no way for the bottom-right corner shape to fit two non-touching fish in if it can't use its bottom row. Therefore the second row of the bottom-left shape must be shaded, completing the 8

Step 5:

There are two ways to give the 6 its last 5 shaded squares. If it doesn't use the squares in columns 1-5, then the shape third-from-bottom on the left side would only have its bottom row, and the shape directly to its right would only have the 2x3 at its bottom. Then both of third-from-bottom's fish would go in its bottom row, leaving no way to fit two fish into the shape to its right. Therefore Row 6 Columns 1-5 are shaded, completing the 6

Step 6:

To fit two fish into the newly-completed shape without touching the one in R9C9, one must go in R11C8. That leaves only one way to fit two non-touching fish into the bottom-right corner shape.

Step 7:

The bottom-left shape can't use its bottom row, and it can't put two fish into its second row (that would mean Row 11 would have 3 fish). Therefore it must have at least one fish in its third row, and there is only one place to put that fish.

Step 8 (in which I demonstrate a contradiction):

I put some "f"s in the grid - for each pair of "f"s, at least one must be a fish. I also centered the fish so they look nicer. There are two ways of satisfying the 5. If it doesn't use Rows 1-4, then this situation occurs. Now there is no way of placing two fish in the left-side shape. They would have to both be in Column 1 (to avoid touching R10C3) but can't use the bottom two rows (would make 3 fish in a row) or the top two (would touch both of the "f"s there). So these Rows 1-4 are shaded, satisfying the 5.

Step 9:

Now that middle shape has only two ways of placing fish into it to avoid touching the "f"s.

Step 10:

The left-side shape must have both its fish in its top 3 rows (to avoid making 3 fish in a row) so one fish and two "f"s can be placed.

Step 11:

No fish can go where the "n"s are to avoid touching or making 3 fish in a row. Only one fish can go in Column 3 since there is already a fish in that column. Therefore there must be a fish in R4C2 and some fs as well.

Step 12:

Placing those fish and "f"s, one of the "f"s in the left-side shape can be ruled out and a fish can be placed.

Step 13:

The 10 needs 4 more squares to be satisfied. If it doesn't use the 3 that are part of the top-right shape, then it doesn't have enough - so those square are used, and the 10 can be completed.

Step 14:

The 7 needs two more squares. If it uses the three that are part of the top-right shape then it has too many, so it doesn't use those and the 7 can be completed.

Step 15:

The top-right shape must have 2 fish. One must be at the bottom of Column 12; it will go in whatever row 6 or 7 doesn't have two fish after the other set of fs in those rows is decided. The other fish must go on the other side; I've made a grouping of 3 "f"s to demonstrate this. The rising-bar-chart top-right shape can't fit two fish in without using its bottom row; any attempt that doesn't do so will touch all the "f"s or make 3 fish in a column. Therefore the middle-top shape can't just use its bottom row (this would make 3 fish in Row 5) and the second row can be shaded in.

Step 16:

The only way to fit all the fish into the top-right shape and the rising-bar-chart is to use R1C11 as a fish and also use either both "a"s or both "b"s

Step 17:

The top-left shape can't use only its bottom row; that would make 3 fish in Row 2. Therefore its top row is shaded. Now we've completed the Aquarium, the rest is just Star Battle Logic.

Step 18:

Column 2 needs one more fish and there is only one place to put it. Row 2 needs one more fish and can't use the top-left shape (both its fish must be in Row 1) so the last fish must be placed in the skinny shape.

Step 19:

Column 7 needs one more fish and there is only one more place to put it. That shape now has only one place to put its last fish, as any other would make 3 fish in a column/row.

Step 20:

Columns 4 and 8 each need one more fish and there is only one place to put them.

Step 21 (and the solution!):

Now the one of the "f"s in Column 3 can be ruled out (no-touching), and then one of the "f"s in Column 12 (only 2 to a row), and I realized I goofed earlier - it's not either a or b, it has to be the "a"s because the "b"s touch a fish. And the puzzle is done!

• Awesome solution bobble! All of your deductions are of course correct. I think the solve path is a little shorter if you had made your deduction in step 17 sooner (it requires only your steps 1 and 14), but you definitely got there. +1 for use of Unicode fish :-) Dec 5, 2020 at 19:08