Puzzle 25: Polyamory - The Zotmeister
solving the puzzle of life one entry at a time
Jan. 25th, 2006
04:29 pm - Puzzle 25: Polyamory
Nikoli calls these simply "Block Puzzle". That won't do. That could describe hundreds of puzzle types. Besides, this involves polyominoes, not rectangles. I've been wanting to use Polyamory as a puzzle title for some time, and it may as well be used here.
The usual 5×5 sample puzzle is on the left; its unique solution is on the right rendered in indigo, with the cyan lines being just a solving aid. Yes, I just repeated a color scheme. What can I say, it fits.
Concise pedantism follows: The object is to take the given grid and divide it into polyominoes such that a permutation of the symbols in each polyomino matches one of the given permutations beneath the grid and that the given quantities of polyomino for each permutation are matched exactly.
Yep - another one-sentence ruleset. If you don't know what a permutation is, here's the numbered multi-line version:
1) Except for borders, the puzzle grid shows only the corners of the cells of the grid. (Black "cells" are simply holes in the grid pattern.) To solve the puzzle, draw in the needed borders between cells so as to divide the grid up into sections (polyominoes) following the remaining rules.
2) Each section will contain a quantity of symbols - letters, digits, whatever. It doesn't matter what order they are in in the grid, but for each section, those symbols must be able to exactly match one of the arrangements listed under the grid if those symbols were lined up and ordered properly.
3) The numbers next to each arrangement under the grid show how often the arrangement is used in the solution - that is, how many times a section matching those symbols appears in the solution. Divvy up the entire grid to match those numbers, and the puzzle is solved!
In the sample puzzle, each section has a single 'A', a single 'B', a single 'C', a single 'D', and a single 'E'. No other combinations are allowed, since "ABCDE" is the only one provided for under the grid. As a quick glance at the solution shows, the letters don't need to be in a nice, neat, intuitive alphabetical order inside of each section.
I always try to make my sample puzzles instructional, such that they can't really be solved by means other than one that enforces the points I'm trying to get across. Sometimes this results in the sample puzzle really being quite nasty for a sample. I believe this happened here. Don't feel ashamed to read behind the cut:
(Start of sample puzzle solution)
These can get tough - tough to the point of intuition being more useful than deduction for solving them, even. However, this one can be tackled entirely deductively if done so cautiously and observantly. The way to start this puzzle is to note that no polyomino in the solution can have a duplicate letter - the only combo we have under the grid is "one of each" - so borders can confidently be drawn in separating all pairs of orthogonally adjacent letters. There are nine such edges to be drawn in.
From there, I note the 'C' on the bottom row. It has 'D's on both sides. That 'C' needs to belong to a polyomino - every cell in the grid does - and that polyomino must ultimately read "ABCDE". Since it can have only one 'D', that polyomino needs to pick just one direction to expand in from there. It also needs to include one of each letter, and examining the right-hand side of the puzzle I see that there's no 'B' this polyomino could reach in that direction: those other two 'C's right above the one in question serve as a wall, so if the 'C' on the bottom row were to include the 'D' to its right in its section, it can get the 'E' and 'A' straight up from the corner but then have no space left to expand, remaining 'B'-less. This can't happen, so the 'C' on the bottom row gets the 'D' to its left instead. I indicate this by drawing a line connecting them; this is optional, but I find it helpful.
This 'C'-'D' section-in-the-making still needs an 'A', a 'B', and an 'E' to be completed. The 'A' and 'E' on the bottom row are the only ones this section can legally reach, but there are two choices of 'B' on the row above that could finish the section. Which is correct? Well, I wouldn't know yet, so I'll come back to it later. I still have my line linking the other four letters so I know where I left off.
I turn my attention to the other 'D' on the bottom row, the one in the corner. Since it needs to be a part of a pentomino (5-omino, or five-cell section), it needs to be linked to the letter right above it - up is the only direction it has left. From there, it needs an 'A', a 'B', and a 'C'. The only 'A' within reach is one cell further up, but it seems there are two possibilities for acquiring 'B' and 'C': both from the center row, or both from the row below it. However, I can see which is correct: between the walls separating the four 'B's in a square, the section I'm building, and the first partial section I was building, I'm surrounding the 'C' that's one cell diagonally in from the bottom-right corner. If I use the center-row 'B' and 'C' to complete this section, it'll be trapped! Since I can't leave a man behind in this puzzle, I need to use that 'C' here instead, and consequently the 'B' to its left as well. But hey - that resolves the decision I had to make with the first section! I now have two of the five polyominoes finished off.
The hardest parts are over - it's fairly straightforward from here on out. The 'E' in the leftmost column has only one means of getting the rest of the letters it needs (straight up to the corner, and the 'B' to the side), so that's easy to wall off. To finish off the last two sections, I note that the remaining 'A' in the rightmost column can only reach one of the two remaining 'D's, which means the other remaining 'A' on the top row has to use the other; its section is defined from there, and of course the remaining five cells must make up the last section, which they do legally.
(End of sample puzzle solution)