Listen With Others

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Listener 4334: Magic Cuboid by Brimstone

Posted by Jaguar on 13 Mar 2015

Hello again after a prolonged absence. I’ve been, shall we say, fairly busy, or distracted, or just buried up to my neck in the wonderful world of cruciverbal politics. Not a pleasant experience, but never mind. I have at least been keeping up with the Listener puzzles. So far in 2015 we’ve had a run of reasonably doable puzzles, albeit with one or two pitfalls here and there (apparently Stick Insect’s blank grid tripped up more than a few!), but at least that’s been balanced by a couple of puzzles marked generously.

This week sees the first of the year’s numerical efforts, something I still look forward to as that’s where I started getting into Listeners. On the other hand, Brimstone’s offering arrived at the start of one of the busiest weeks I’ve had, and so in fact I only properly got a chance to look at it on the Friday a week after it came out. By chance, next week’s puzzle proved to be not too difficult, so it was possible to catch up quickly.

I’d normally release a fully-worked solution but this time I don’t have one to offer as I solved this one essentially while half-asleep most of the time, and my method for solving the endgame was certainly not the most elegant solution! I ended up generating all 729 possible solutions to the first row that moved all single-digit cells apart from 1 and 9 (1, we were told, was fixed, and it became apparent that 9 had to stay still as well no matter what, along with the middle row), and then finding the handful of solutions that summed to 182 (= the middle row). Followed by checking what this made the third row sum to, and finding only one solution to that meant that the rest of the grid could be filled and voila! Somewhere along the way I must have deleted the “wrong” solution that doesn’t meet that “cells originally filled by 1-9 now sum to 200” condition, which is more luck than judgement, but there you go.

To go back to the beginning, I picked up that F had to be 2 and V was 3, and then it was sort of steady-ish progress to find the rest, which makes it sound rather a lot more trivial than it actually was, but I suppose the point is that I wasn’t paying enough attention to work out the solving path. All I can say is that, despite the intimidating preamble, this was a fairly doable numerical puzzle even if you didn’t hit on the intended solution path, and hopefully most people found it that way.

Now on the right side of a busier period, perhaps I’ll be able to keep up with the solving blogs more regularly, so until next time thanks, Brimstone, for marking the start of the numerical year with a fairly gentle ride.

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