Halibut Towers has always been a location of the mind, not tied to anything as mundane as actual physical bricks and mortar, concrete and steel, wattle and daub, cowshit and bits of twig and what have you. And just as well, as our recent house move means that I'm currently sitting in the fourth physical manifestation of Halibut Towers since the birth of this blog back in late 2006 (one in Bristol, three in Newport). More on the move and the new house later (well, maybe) but here's a specific thing that caught my eye when we moved in.
Our predecessor left quite a considerable quantity of what you might call "bonus house contents" - or less charitably "random stuff", or less charitably still "shit" - in the house when he vacated it. Overall that's been a pain in the arse, though there have been a few things that we decided we might keep. Anyway, this item is really neither of these things but as it's just sitting quietly on a mantelpiece minding its own business and staying out of the way I'm fairly neutral about it.
So, it's a calendar. Big fucking deal, you might say, and you'd be mostly right, but the detail of its construction caught my eye. As you can see it's basically two cubes which you have to juggle around to make the correct number for the day of the month. Again, big fat hairy deal, you might say, but I was prompted to wonder about the distribution of the numbers across the two cubes, as you can't just randomly distribute the digits across the twelve available spaces and assume it'll work, as there may be numbers you need that you won't be able to make. Before we get into any theorising, here are the numbers on the first cube (0, 1, 2, 6, 7, 8):
And here are the numbers on the second (0, 1, 2, 3, 4, 5):
The secret with these things is not to try and bite off more than you can chew and come up with some all-encompassing Theory of Everything right off the bat, but instead make some obvious statements and see where they lead you. Here's a couple:
- You need a 1 and a 2 on both cubes, for two reasons: firstly that they need to pair with every other number to make 10-19 and 20-29 and there isn't enough space to store all the second digits on a single cube, and more obviously that you'll need to be able to make 11 and 22.
- You need a 0 on both cubes, for the first reason above (but not the second, as the zeroth of May is not a thing).
So each cube has three faces already spoken for by the digits 0, 1, 2. That leaves us with six faces as yet unoccupied, and the digits 3-9 to accommodate. Well, that's us fucked, then, you may be saying, because that's seven digits to fit in six spaces. And indeed we would be EXCEPT for the saving grace that you don't need two separate digits for 6 and 9 because you can just turn one upside-down to get the other (and the additional saving grace that you don't need a 69th of May, still less a 96th of May).
It is my assertion that it doesn't particularly matter how you distribute the remaining digits and the arrangement we have here of 3, 4, 5 on one cube and 6, 7, 8 (and by rotation 9) on the other is purely arbitrary. I guess the way to convince yourself of this is that none of the digits 3-9 ever have to be paired with another from the same range and there'll always be a 0, 1 or 2 on the "other" cube to pair with regardless of where they end up.
As always, needless to say I'm not the first person to ponder this problem: others including the great Martin Gardner have kicked it around as well.
