# Puzzle 8: Magnetic Field

I just discovered this puzzle design today - sadly, I do not know its origin. The instances I found all had even-sided grid lengths, gave all numbers for all rows and columns, and always gave at least one domino filled in to start. I promptly did away with all that silliness in trying to compose my own, and I rather like the result. This is a cunning puzzle design, seemingly simple but possessing room for cleverness and a peculiar aesthetic. It is primarily a puzzle of parity.

Left side puzzle, right side unique solution - and it's not just the forest green "ink" that makes it unique.

The object is to determine the polarization - or lack thereof - of each domino in the grid: either positive in one half and negative in the other, or neutral (non-polar) in both. Cells with matching polarity cannot be orthogonally adjacent. Numbers to the left of each row and above each column must match the number of positive cells in that row or column, respectively; likewise, numbers to the right of each row and below each column must match the number of negative cells in that row or column, respectively.

And, for those ever-so-curious list lovers:

1) The puzzle grid is divided into dominoes - that is, pairs of cells. Some of these dominoes can be "polarized" - that is, just like a magnet, they have a positive end and a negative end: one of their two cells is "positive" and is to be marked with a plus sign, and the other is "negative" and to be marked with a minus sign. Any domino that isn't polarized is neutral: neither end is positive or negative, and the whole domino is to be shaded. If all dominoes are so marked without violating any of the following rules, the puzzle is solved. [Any "black cells" in the starting grid are actually just holes in the pattern - you can think of them as neutral cells without hurting anything.]
2) In much the same way two magnets will repel each other when ends of matching polarity are close together, the grid wouldn't be stable if polarized dominoes pushed each other. For this reason, two cells with the same polarity can't share a side. (So you can't have two minuses next to each other, although they can touch at corners. Same goes for plusses.)
3) See the plus sign outside the upper-left corner? Each of the numbers to its right - each above a column of the grid - is the number of plus signs that must be placed in its column, no more, no less. The numbers below that plus sign do the same thing for the rows they're in front of.
4) And yes, that's a minus sign outside the bottom-right corner, and the numbers above it and to its left must match the number of minus signs in their respective rows and columns.

(Start of sample puzzle solution)

The place to start is the top row. It needs three plusses and two minuses - and that's five marks total, covering the whole row. Since two matching plusses or minuses can't share a side, they will have to alternate across the row, and since there's one more plus than minus, the plusses will have go on the edges. So that's plus, minus, plus, minus, plus going from left to right across the top row. The domino in the upper-right corner is now only half filled in; the other half must be the opposite polarity, so the rightmost cell of the second row gets a minus.

Now look at the number above the rightmost column. There must be one plus in that column. We already have that plus; we can't have more. There's a second domino that lies vertically in that column. If it were polarized, one half of it - the top or bottom, it doesn't matter - would have to be another plus. Since we can't let that happen, that domino must be neutral instead. Shading it in, the bottom cell is all that remains in that column, and looking at the number just below it, it must be a minus, with a plus to its left.

There's another domino deducible from there, but first I'm going to the second column from the left for something useful. It needs two plusses and two minuses, and has five cells total. That means - with a little arithmetic - that there must be exactly one neutral cell in that column. But see that vertical domino in that column? Since it can't therefore be neutral in both cells, it must be polarized. Whether it's plus on top and minus on bottom or vice versa is not determined yet, but just knowing it's polarized is helpful. I like to mark both cells of such a domino with a dot, which I later cover with a plus and minus when I figure out which is really which.

Time to go back to that deducible domino I mentioned earlier. Looking at the fourth row from the top, I see the need for one plus, two minuses, and two neutrals. I already have one neutral on the far right. Using the same logic of the previous paragraph - albeit rotated ninety degrees - the horizontal domino in this row must be polarized, and since it touches one on the bottom row, the left of this one must be plus and the right minus. As for which of the other dominoes in the row provides the last minus, well, we wouldn't have been able to tell right away, but that dot makes it plain as day. So the bottom dot is a minus and the top dot a plus, and the domino in the bottom-left corner is neutral.

The second column from the left holds the last key. Doing the counting, the remaining two cells of that column must be a plus and a neutral. The plus can't be the cell in the second row, since it has a plus next to it already (the cell below); the last domino on the bottom row is therefore plus on the left and minus on the right.

So what of the last three dominoes? Easy - they can all be neutral! In fact, they have to be: the leftmost one can't be positive on either end, the rightmost one can't be negative on either end, and the last one can't be positive or negative on its left half. Shade in all three, and the puzzle is solved.

(End of sample puzzle solution)

I hope you find this entertaining. It was a fascinating exercise trying to take this discovery and make my own working prototype of it in such a short period, having to learn the rules and techniques myself - and quickly - and then apply them to build a non-trivial example worthy of sharing with others. Let me know what you think of it here, and of course, email me your solution if you solve it. - ZM

Tags:
Comments for this post were disabled by the author