# A Trivial Pursuit #12 (Sports and Leisure 2/4): Logic Boat

This is part 12 of A Trivial Pursuit, a 25-part puzzle hunt. Each part is solvable on its own, with the exception of the meta-puzzle at the end.

This is a variation on a Battleships puzzle, where the custom fleet of ships to be placed includes some that are irregularly-shaped. In the spirit of Japanese grid-deduction puzzle naming conventions, I call this variant 'Logic Boat'. Once solved, a 6-letter word can be extracted from the puzzle - what is this word?

Battleships rules (adapted from Grandmaster Puzzles):

Locate the indicated fleet in the grid. Each segment of a ship occupies a single cell. Ships can be rotated. Different ships cannot be placed in adjacent cells that share an edge or corner. Some ship segments, or sea cells without any ship segments, are given in the grid. The numbers on the edges of the grid reveal the number of ship segments in that row or column.

Note: This puzzle can be solved entirely through logical deduction, without the need for guesswork. Please do not post answers consisting of only a screenshot of the solution without any logical explanation - to earn the checkmark, (as well as identifying the final 6-letter solution word) your answer should explain the logical steps required to reach the solution, so that somebody reading it can follow along and understand.

• Are we allowed to mirror the angle-shaped boat? Sep 26 at 18:38
• @The_spider That would be equivalent to a rotation, so yes
– Stiv
Sep 26 at 18:42
• @Stiv technically, mirroing is not reachable only with rotations, you cannot rotate $L$ to get $\Gamma$. But we take good note that mirroring is allowed ! Sep 28 at 9:28

The 6-letter word that can be extracted is:

SIPHON

This can be read by dividing the completed grid into 5x3 sections and highlighting the sea cells:

The solution path:

Let's start by filling in all the sea in the 0- row and column.

That leaves us with only 8 free cells in the "7" column, so all but one of them must contain ships.

We cannot have seven consecutive ship cells, so the isolated cell at the top must be ship.

That still leaves six ship cells (and one sea cell) in the lower section. The longest (linear) ship is four cells, so it must be split 4:2 or 3:3. So the single sea cell must be in the blue section, making all the green cells ship.

And wherever those ships are, all the red cells must be sea.

Now look at the rightmost column. A single ship cell in any of the green cells would create a small L-shaped or T-shaped ship that is not in the fleet. So we must have a pair of adjacent ship cells.

This tells us the 2x2 ship must use one of the green blocks, so the blue cell must be sea.

We also know that the other ship in the 7-column is the 4-ship, and the red cell must be part of it.

And, as the single 1-ship has been placed, the yellow cells must be sea. (A ship cell at the intersection of a 1-row and 1-column would be surrounded by sea giving us another 1-ship.)

Now, we still need another four ship cells in the 5-row. With the 4-ship elsewhere, this can only be 3:1 or 2:2. But purely horizontal ships here would make the entire row below sea. So we cannot have 2:2 and all the green cells must be ship.

(Apologies if the explanation here gets hard to follow, I'd be interested in suggestions for making it clearer.)
That makes all the blue cells in the 2-row below sea. There are now only three unknown cells in that row, two of which must be ship, and all of which are below known ship in the 5-row. So there are two ships oriented vertically here. The only way for that to work is if the V-ship is (partly) in the 5-row.

Finally for this section, we don't know precisely where the ship cells in that 2-row are, but they force two more (red) sea cells in the the 6-row below.

Now we can start making some real progress:
That leaves only five unknown (green) cells in the 6-row, all of which must be ship.

And that sets the location of of the O- and 4- ships on the right, filling in all the cells on that side (blue).

we also have several sea cells (red) around the 6-row. And that forces the location of the V-ship, completing the 5-row (yellow).

That gives us all the ship cells we require in both 1-columns and two of the 3-columns, making lots of cells sea (blue).

That means all remaining cells in the 3-rows must be ship (green).

And that means the incomplete 3-column must contain the last 3-ship (as it can't be 2:1) (red).

And finally the last two 2-ships are forced in the 6-column, completing the grid

• Great explanation, thanks :) The part you found trickiest to explain is indeed the trickiest part to explain overall, in my opinion. Well done!
– Stiv
Sep 26 at 18:44
• (My approach to that part was to note what R5C1&3 must be, then R4C2, then consider the options for R3C2...)
– Stiv
Sep 26 at 18:46
• I managed to mostly follow the explanations, but it's not clear to me how you managed to deduce that the 2 yellow cells in step 2 must be sea. I'd greatly appreciate if you could flesh out the reasoning. Sep 27 at 8:59
• @MatthieuM They are both at the intersection of a 1-row and 1-column, so a ship cell there would be surrounded by water giving us another 1-ship.
– fljx
Sep 27 at 9:19
• @fljx: Indeed, now I see it, thanks :) Sep 27 at 11:46