The puzzle seemed straight-forward: just arrange the six lowest-value dominoes into a square where each of the sides add up to the same value. A quick bit of brow-furrowing got me to the value - it has to be four, but actually solving the puzzle was rather trickier. I began to feel rather like the unfortunate contestant, who has also failed to solve the problem - and I had the considerable advantage of not having any time limit (plus I didn't have the other contestants shouting 'often less-than useful' advice at me all the time.
Eventually I found the solution, but it wasn't a very satisfying answer. The reason has to do with the way that I think about dominoes. I spent many of my formative Saturday nights at 'Domino Drives', mainly because my Dad was a seasoned card player who was pretty successful at the accompanying 'Whist Drives', and so transport wasn't a problem. As a result, I think of dominoes as being arranged with ends matching, and doubles rotated through ninety degrees. Now there were some local variations: Up North, where I lived at the time, they played with dominoes that went all the way up to Double Nines, and the One spots were not red, nor were they a different size. White dots all the way from none to nine was what I was brought up with, and it wasn't until many years later that I discover the many variations of domino that existed elsewhere...
Subconsciously, I was applying the 'match the ends' rule as a constraint to this problem. Not rotating the doubles so that they were across the flow wasn't a problem, because I had grown up with players who didn't cross doubles, and there were always people around the table who would 'tut-tut' and rotate any uncrossed double during play. But matching those ends was totally automatic, and so I quickly came up against the problem that wherever you placed the Double Two domino, the two dominoes either side immediately made those two sides add up to more than four!
Eventually it dawned on me that the only way to solve the problem was to ignore my self-imposed constraint and not to match the ends of dominoes. Once you do this, then the solution drops out quite quickly.
But, as frequent readers of this blog will tell you, my head doesn't let me stop there. My mind continuously looks beyond the obvious, and I now realised that actually, not all of the junctions between dominoes broke the rule/constraint - just some of them. Now I already knew that the 'no junctions break the rule/constraint' was not possible, so was it possible to break the rule/constraint at all the junctions?
It seems that you can't do this either. This was my best result, and here all but one of the junctions breaks the rule/constraint.
So, today's observation is that: 'Sometimes you unconsciously impose rules where there aren't any rules at all.' - plus the corollary that: 'breaking constraints sometimes produces interesting results', which gives s revised, and more difficult puzzle:
Can you arrange the six lowest-value dominoes in a square so that the sides all add up to the same number, and with the highest possible number of junctions between dominoes where they have different numbers of dots?
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