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Puzzle 18: To Each Their Own - The Zotmeister

Sep. 30th, 2005

04:36 pm - Puzzle 18: To Each Their Own

With the great popularity of Polyominous and my own Seeking Syren here, I figured it'd be worth presenting more variations on polyomino-divvying. I discovered this design in issue 100, the only one I own, of Puzzle Communication Nikoli; given that each letter in the grid needs its own polyomino, each character has its own congruence class, and Nikoli's title for the puzzle - NIKOJI - strikes me as rather pompous, I figure my title is a perfect fit. This design presents a first among my LiveJournal puzzles: no numbers! In many ways, this is a cipher Polyominous.

The grid on the left is an unsolved To Each Their Own puzzle; the grid on the right is the unique solution to that same puzzle, rendered in indigo. The cyan line segments are a solving aid I employ and are not part of the solution.

The object is to take the given grid and divide it into polyominoes such that each has exactly one given letter in it and that given any two such polyominoes arbitrarily, they otherwise match exactly - in size, shape, orientation, and the position of the letter within - if and only if the letters they contain match each other.

Still a simple puzzle, but the list is still here:

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) Every section the grid is split up into must have one letter in it - no more, no less.
3) If two sections have identical letters in them, then the sections they are in must be identical to each other. They must have matching cell counts; their shapes must be identical to each other; they must be lined up identically (if one section has to be rotated, flipped over, or both to look like the other, they do NOT match); they even need to have those matching letters in matching spots within.
4) If two sections do NOT have identical letters in them, then they must NOT be otherwise identical. At least one of the following must be true in that case: one has more cells than the other; their shapes don't match; one would have to be rotated, flipped over, or both in order to get the two to look like each other; their letters are not in matching cells with respect to the borders of the sections.

It is just as important to not use the word 'different' in puzzle instructions as it is to not use the word 'same'. Ambiguity is NOT a good thing in puzzle instructions. But if you're thinking "same letter = same section, different letter = different section", what you're thinking is hopefully correct. Look up at the sample solution: the two A-regions match each other, and the two E-regions match each other, but if you take an A-region and an E-region, they do not match... since although they have the same size, shape, and orientation, the letter is in the left-hand cell of the E-regions and the right-hand cell of the A-regions. I hope that quells any lingering doubts as to what I'm asking for, but if not - or if you can't seem to figure out how to approach this - here's the standard walkthrough:

(Start of sample puzzle solution)
The basic key to working through this type of puzzle is to note that whatever boundaries do or do not apply to any particular instance of a letter in the grid must apply to all of them. I'll start with 'B'. There's a 'B' in the top row, so all 'B's in the grid must have a border right above them; if one can't extend into the cell above it, none can. There's a 'B' in the rightmost column as well, so a border must always be to the right of a 'B'. Since each letter must be in its own section, the 'B' at the far-right end of the second row from the top tells me that borders must be drawn to the left and underside of every 'B' as well. That's right: all the 'B's are monominoes.

I next turn to 'A'. The 'A' in the rightmost column has borders above it and to its right; the other 'A' has a border below it. I transfer each of these elements. Now consider the 'A' on the bottom row. It can't be a monomino, since 'B' has already claimed that. Since it's only open on one side, this region will have to extend in that direction. I draw a line connecting that 'A' to the cell on its left, then draw a corresponding line from the other 'A'. Looking back at the bottom-row 'A', I see this newly-acquired second cell - the one in the bottom-left corner of the whole grid - is bordered on two sides, and with a letter above it, it gets the third border. This defines the A-region for this puzzle: it is a domino, oriented horizontally, with the 'A' in its right cell. I draw the rest of the border around the other A-region. Seven cells defined, eighteen to go.

My next move is in the bottom-right corner of the grid. That corner cell needs to be connected to a letter (remember, all regions must have a letter), and the only one accessible to it is the 'E' to its immediate left. I draw a line connecting those two cells. This is a small line that from the perspective of the 'E' travels one cell to the right, so the other 'E' in the left column gets a similar line. Now that 'E' has a border on its left; transferring that to the 'E' on the bottom row, I find I've completely surrounded it, defining the E-region as a domino, oriented horizontally, with the 'E' in its left cell. That makes it different from the A-region, so no problems there.

The center cell of the bottom row has nowhere to go but up now, and that runs it into the 'C' in the grid center. Just as whatever applies to one letter must apply to all its matches, whatever applies to anything connected to a letter must apply to its corresponding spot on matching letters' regions. I now have a line travelling two cells down from both 'C's and seven cell borders of a possible eight around each (incomplete) C-region. Similarly, the upper-right corner cell must travel left then down, hitting a 'D'. Transferring lines and borders between the 'D's and their connected cells, this defines the D-region as a bent triomino; the two cells that are still empty (no letter or line within) must therefore belong to the C-regions.
(End of sample puzzle solution)

I probably could have made a tougher one than the one below, but I felt like being cute with this first one. It's easy, but it's fun. Let me know what you think of this design in the comments; email me your solution if you finish, and maybe I'll reward you and maybe I won't.

On deck: two more designs I haven't presented here yet. - ZM

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