Thursday, July 25, 2013

Solving the TopSpin Puzzle using GAP

TopSpin is an oval-track permutation puzzle that was made by Binary Arts; similar puzzles are made and sold by other manufacturers. Here's the Binary Arts TopSpin.

It's not a difficult puzzle to solve if you play around with it for a few hours and figure out how to generate various permutations. It's more interesting (and difficult) if you observe that the turntable has a distinguishable top and bottom. This suggests an interesting question - can you invert the turntable while keeping the numbers in the track in the same order?

The answer is, perhaps surprisingly, yes. Here's one way to find a sequence of operations that produces precisely this outcome.

GAP (Groups, Algorithms and Programming) is a freely available programming language that specializes in computational group theory, and it's perfect for solving permutation puzzles. Here's my GAP code for TopSpin. Label the top of the turntable with 21 and the bottom with 22. Flipping the turntable generates the permutation \[(1,4)(2,3)(21,22);\] rotating the oval to the left generates the permutation \[ (2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,1). \] Denoting these two permutations as \(x\) and \(y\), we can now simply ask GAP to find a sequence of operations on the free group generated by \(x,y\) that results in the pre-images \((1,2)\), flipping two adjacent numbers on the track, leaving everything else the same (including turntable parity); and \((21,22)\), flipping turntable parity, leaving everything else the same. This corresponds to the operation of flipping the turntable while keeping the order of the numbers in the track the same.

Running the code, we get these lovely results:
(1,2) = y*x^-1*y^-1*x^-1*y*x^-1*y*x^-1*y^-1*x^-1*y^-1*x^-1*y^2*x^-1*y^-1*x^-1*y^-1*x^-1*y^4*x^-1*y^-1*x^-1*y^2*x^-1*y^-2*x^-1*y*x^-1*y^2*x^-1*y^-3*x^-1*y^-3*x*y^-4*x*y*x*y^-1*x*y^4*x^-1*y^-5*x^-1*y^-1*x^-1*y^5*x^-1*y^-6*x^-1*y^2*x^-1*y^-1*x^-1*y^5*x*y*x*y*x*y*x*y*x*y^5*x*y*x*y^-1*x*y^-5*x^-1*y^-1*x^-1*y^-1*x^-1*y^-2*x*y*x*y^-1*x*y^4*x*y^-1*x*y^3*x^-1*y^-1*x^-1*y^6*x^-1*y^-1*x^-2*y^-4*x^-3*y^-3*x^-2*y^-1*x^-1*y*x^-1*y^2*x^-1*y^-2*x^-1*y^-1*x^-1*y*x*y*x^-1*y*x^-1*y^-1*x^-1*y^-2 
(21,22) = y*x^-1*y^2*x*y^-1*x*y*x*y^-1*x*y^-2*x*y^2*x*y^-1*x*y*x*y*x*y^-2*x^-1*y*x^-1*y^-2*x^-1
Since there are sequences of operations that allow us to flip any two adjacent numbers or the parity of the turntable, it follows that all possible configurations are both solvable and achievable.

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