# Puzzle 7: Polyominous - The Zotmeister

#### solving the puzzle of life one entry at a time

## Apr. 22nd, 2005

### 04:03 pm - *Puzzle 7: **Polyominous*

*Polyominous*

The pun is so obvious, I'm surprised no one else has used it as a name for this puzzle. But then again, maybe that's exactly *why* they haven't.

A polyomino is a tile made of a group of squares. The video game *Tetris* gets its name from the pieces being tetrominoes - that is, four-unit polyominoes. A two-unit polyomino is called a domino, a term most should be familiar with; a one-unit polyomino is called a monomino. For those who want to hum along without learning the words, a polyomino of *n* units can be called an *n*-omino, so a tetromino can be called a 4-omino, a polyomino with fourteen cells is a 14-omino, and so on.

On the left is an unsolved Polyominous puzzle. On the right is what its unique solution would look like if solved with purple ink.

The object is to take the given grid and divide it into polyominoes such that each given number

Yes, that's the entirety of the rules in one sentence. It's a simple puzzle. But for the checklist lovers out there:

1) Except for the outside border, the puzzle grid shows only the corners of the cells of the grid. To solve the puzzle, draw in the needed borders between cells so as to divide the grid up into sections following the remaining rules.

2) The number of squares in each section created by the borders must match the number in any cells of that section. (So if there's a 2 in the grid, the section with that 2 must have exactly one other cell with it, making two cells total.)

3) Two sections with the same number of cells cannot be orthogonally adjacent - that is, they may not touch except at corners. It is strongly recommended that as you learn what cells belong to what sections, you write in numbers in the blank cells - just like those already in the grid - so that you don't violate this rule. (So if you have a three-cell section, you can't have another three-cell section sharing a cell side with it. If you write a '3' into each cell of the section that needs it, you'll know not to write another '3' next to it as part of another section.)

Note that this

(Start of sample puzzle solution)

I like to start these by drawing in all the obvious borders - surrounding '1's and separating different numbers that start next to each other. In the sample, there are eleven edges you can draw in this way. After that, I look for numbers whose polyomino can only legally be in one place. Here, I start in the upper-left corner where the '2's are. Obviously, if the upper-left corner cell itself were to be given a '2' - that is, if it were part of a domino - then there would be two dominoes next to each other, which is against the rules. I find it very convenient to think in terms of giving each non-numbered cell its number; the rule then boils down to "you can't have the same number on both sides of a border". Since those '2's have no other direction to get their second cells from, the '2' on top gets another '2' to its right and the '2' on the left side gets another '2' below it. Surrounding them with borders, I see the upper-left corner is now isolated, without a number in it... but in this puzzle, that is okay, and often necessary. I put a '1' in it to mark it as solved and press on.

Next is the '4' along the left side. Note that it has borders on three sides already (including the outer border of the whole puzzle). It needs three more cells with '4' in them connected to it, and it has only one direction to go in, so the cell in the bottom-left corner gets a '4'. Now this just happens to be next to

In order to finish the puzzle, I now think backwards, as it were - rather than look at numbers and see where they must expand, I look at empty cells and see what numbers they must contain. I look in the upper-right corner and see the still-empty cell there. It can't have a '1', since there's a monomino on its left. It could be a '3', since it would link to the two '3's below it just fine, but any other whole number '2' or greater simply won't fit - it's a single cell. Therefore, '3' is the only option for it, and that completes a triomino. After placing its borders, I look at the cell in the center of the second row from the top. Let's see - there's a '1' to its left, a '2' above it, and a '3' to its right, so all those are off-limits. There is enough blank space for it to be a '5' or even '6', but then the remaining '3' and '4' couldn't be completed. It must therefore be a '4', and we can expand from that '4' just like it were a given number: the cell below it must be a '4'. Rather than finish that tetromino, I stop and look at the remaining '3' - it has exactly two spaces for expansion left! Since it has only one possible placement left, I fill it out - and with

(End of sample puzzle solution)

This was an interesting one to put together. I had some fun with patterns in the given numbers. It starts out easy enough, but I recommend caution as it goes along. Comments are welcome as always, and emailing me your solution might get you something. Note that a simple grid of numbers, if you fill in all the blanks, is sufficient to express the answer; the borders are trivially deducible from that. - ZM

On the left is an unsolved Polyominous puzzle. On the right is what its unique solution would look like if solved with purple ink.

The object is to take the given grid and divide it into polyominoes such that each given number

*n*in the grid is part of an*n*-omino and that no two polyominoes of matching size (quantity of cells) are orthogonally adjacent.Yes, that's the entirety of the rules in one sentence. It's a simple puzzle. But for the checklist lovers out there:

1) Except for the outside border, the puzzle grid shows only the corners of the cells of the grid. To solve the puzzle, draw in the needed borders between cells so as to divide the grid up into sections following the remaining rules.

2) The number of squares in each section created by the borders must match the number in any cells of that section. (So if there's a 2 in the grid, the section with that 2 must have exactly one other cell with it, making two cells total.)

3) Two sections with the same number of cells cannot be orthogonally adjacent - that is, they may not touch except at corners. It is strongly recommended that as you learn what cells belong to what sections, you write in numbers in the blank cells - just like those already in the grid - so that you don't violate this rule. (So if you have a three-cell section, you can't have another three-cell section sharing a cell side with it. If you write a '3' into each cell of the section that needs it, you'll know not to write another '3' next to it as part of another section.)

Note that this

*isn't**Islands in the Stream*. The numbers in the grid are not necessarily one-to-a-polyomino. You can have multiple given numbers in the same section, like the triomino (3-omino) in the upper right of the sample, or even no digits at all, like the monomino in the upper left of the sample. You'll need to pay careful attention as you're divvying up the grid. Don't settle for trial and error - stick to logical deduction.(Start of sample puzzle solution)

I like to start these by drawing in all the obvious borders - surrounding '1's and separating different numbers that start next to each other. In the sample, there are eleven edges you can draw in this way. After that, I look for numbers whose polyomino can only legally be in one place. Here, I start in the upper-left corner where the '2's are. Obviously, if the upper-left corner cell itself were to be given a '2' - that is, if it were part of a domino - then there would be two dominoes next to each other, which is against the rules. I find it very convenient to think in terms of giving each non-numbered cell its number; the rule then boils down to "you can't have the same number on both sides of a border". Since those '2's have no other direction to get their second cells from, the '2' on top gets another '2' to its right and the '2' on the left side gets another '2' below it. Surrounding them with borders, I see the upper-left corner is now isolated, without a number in it... but in this puzzle, that is okay, and often necessary. I put a '1' in it to mark it as solved and press on.

Next is the '4' along the left side. Note that it has borders on three sides already (including the outer border of the whole puzzle). It needs three more cells with '4' in them connected to it, and it has only one direction to go in, so the cell in the bottom-left corner gets a '4'. Now this just happens to be next to

*another*cell with a '4' in it, but again, that's okay in this puzzle. We have no choice but to include that '4' in this tetromino; we need one more cell, and the only one available is along the bottom. Surrounding this piece with borders, it is easy to see that the '3' along the bottom edge has only one legal option, following similar logic to the 4-omino just determined.In order to finish the puzzle, I now think backwards, as it were - rather than look at numbers and see where they must expand, I look at empty cells and see what numbers they must contain. I look in the upper-right corner and see the still-empty cell there. It can't have a '1', since there's a monomino on its left. It could be a '3', since it would link to the two '3's below it just fine, but any other whole number '2' or greater simply won't fit - it's a single cell. Therefore, '3' is the only option for it, and that completes a triomino. After placing its borders, I look at the cell in the center of the second row from the top. Let's see - there's a '1' to its left, a '2' above it, and a '3' to its right, so all those are off-limits. There is enough blank space for it to be a '5' or even '6', but then the remaining '3' and '4' couldn't be completed. It must therefore be a '4', and we can expand from that '4' just like it were a given number: the cell below it must be a '4'. Rather than finish that tetromino, I stop and look at the remaining '3' - it has exactly two spaces for expansion left! Since it has only one possible placement left, I fill it out - and with

*that*, the 4-omino I was working on has only one possibility left as well. Give the one remaining cell the monomino it needs, and that's it.(End of sample puzzle solution)

This was an interesting one to put together. I had some fun with patterns in the given numbers. It starts out easy enough, but I recommend caution as it goes along. Comments are welcome as always, and emailing me your solution might get you something. Note that a simple grid of numbers, if you fill in all the blanks, is sufficient to express the answer; the borders are trivially deducible from that. - ZM