Listener 4190: 2×2×2 by Oyler (or Holy Craps!)
Posted by Dave Hennings on 8 June 2012
Having recently gone back to a bit of part-time work after a few years being ‘retired’, I was reminded of why I never had time to do Listeners, Inquisitors, EVs, etc when I had a truly full-time job. OK, working overseas a lot of the time didn’t help, but even when I was back in London, knackeredness in the evenings (and, of course, golf at the weekends) meant that I had little time for crosswords. How members of our little group can hold down a job and spare the time for half a dozen serious crosswords a week (plus, no doubt, a couple of daily ones as well) is a source of admiration from me. And don’t get me started on those who set one or two handfuls of crosswords every year as well!
Consequently, the preamble of this puzzle from Oyler gave me cause for concern because it seemed as though it would entail quite a lot of work post grid … probably in addition to the grid itself. I think it fair to say that that is where the Listener series is headed these days. Generally I have no problem with that, but ‘mathematicals’ which encompass dominoes, dice and wrens (oops, sorry, that was a word puzzle) are a serious cause of panic! I wasn’t sure if I was really up to it.
I ended up dipping into and out of this puzzle quite a lot over the first few days of trying to ‘find my way in’. I found that it wasn’t the sort of thing that I could concentrate on on the train, either going into London or coming home in the evening. By the Friday, I was beginning to panic. Those of you who’ve read my previous mathematical blogs will recognise that this is a frequent feeling. Perhaps it’s a feeling you get yourselves.
So it was with great relief that I finally alit on 27dn. I suppose that in hindisght it should have been obvious to look at squares, but there were eight clues containing them, so which one to choose. 27dn came up trumps.
If a = 2, h = 5 gives 128 or h = 6 gives 256.
If a = 3, h = 5 gives 243.
Then 25dn gives various values of 3, 4, 5, 6 for o and then 32dn ANNA gives values of 3 or 4 for n
Moving on to MOON, it gives values of 5 and 6 for m and most result in c = 1. But from 34dn, c cannot be 1. In fact there are a couple of situations where m can be 1, leaving c to be 3 or 4.
In summary, with the blue lines giving the possibilities based on the final column being a length of 3:
Having got the down entries, the acrosses should have been a doddle. In fact they took about half an hour of further research:
So much for the easy part!
Splitting the grid up into the eight different ‘nets’ proved reasonably straightforward, although the top of the grid posed a bit of a problem until I realised that, just because a block of cells contained the numbers 1-6, didn’t mean that they were dice nets. In this case, the 6-2-3-5-1-4 in columns 2-5 could not form a cube, but the 6-2-3-5-1-4 in columns 2-6 could.
The next phase (and I was beginning to wonder how many more phases there were) needed me to find a different arrangement of the numbers to form a big cube. I listed them by using opposite faces. From the top clockwise I had:
The absence of 1,5 was perhaps a clue. However, at this stage I realised that dealing with the problem in purely mathematical terms was going to tax me to the limit. So I built some cubes of my own with which to experiment. A lot of printing and cutting and sticking later, and I had a set of eight dice, as shown in the picture to the left. I would like to say that the rest was easy and logical, but it must have taken another hour of trial and error to stumble across the correct solution, shown on the right.
I was pleased with the way that it all turned out, but it really was an uphill struggle. In all, I’m guessing it took me about six hours in total … actually, make that eight! Oyler, you owe me one, and if I’ve made a mistake, two!