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February 7

Puzzle degrees of Freedom question

Assume an octahedral puzzle. It has two modes. In one mode, the two halves of the puzzle may be rotated by each other on any of the three planes that connect four points through the center. They can rotate either 1/4, 1/2 or 3/4 turns relative to each other. (Sort of like a Rubix cube with rotational axes through the points.) In the other mode, each of the faces of the Octahedron rotate, with each face rotating in the opposite direction from the faces that it borders on an edge. This can rotate either 1/3 turn clockwise or counterclockwise. If the puzzle is taken apart and reassembled, the number of possible positions that it can have (for this purpose, orienting face A and its corners in the same position as the start is 7!*3^7 (the 7 possible positions for the other triangular faces times the orientations of those faces). Of those positions, how many can actually be achieved simply using the proper moves of the toy?Naraht (talk) 02:57, 7 February 2017 (UTC)[reply]

I don't think the turning of all the faces by 1/3 gives any extra moves, if so the puzzle is completely equivalent to the Pocket Cube with the corners mapped to the center of the sides,, so 7!*3^6 possibilities. Dmcq (talk) 11:57, 7 February 2017 (UTC)[reply]

Agreed, without the face rotation, it would be equivalent to the Pocket Cube, but I don't understand why the exponent is 6 in the 3^6. All of the other 7 should be independent, right?Naraht (talk) 17:43, 7 February 2017 (UTC)[reply]

Even with the gear type face rotation it is equivalent to the Pocket Cube. That move can be generated using a sequence of the first type of rotations. The rotations of the faces are not completely independent, the eight is determined by the other seven. Dmcq (talk) 18:14, 7 February 2017 (UTC)[reply]

What is the series of mode 1 moves equivalent to a mode 2 rotation? And how do you show that the 8th is determined by the other 7?Naraht (talk) 21:13, 8 February 2017 (UTC)[reply]

I think I'll leave the first question as an exercise :) Try figuring out for yourself a sequence which twists two adjacent faces in the opposite direction and leaves everything else unchanged. Then apply it four times.

For the second I think it is fairly straightforward but this is my proof from a think earlier this morning rather than one I read so there's a possibility of a mistake :) Put an A at two opposite corners and put ABC clockwise on every face so every face is labelled. Now consider a twist round A. Every ABC will move to another ABC so no renaming is necessary to fit the faces into the original labeling. If you twist on quarter turn around one of the side corners though you move BCA, CAB, BCA, CAB to CAB, BCA, CAB, BCA so one needs two 1./3 turns clockwise of the faces and two 1/3 turns anticlockwise to get back to the original labeling. The turns clockwise or anticlockwise will always add up to zero (or a factor of three times 1/3 which is one whole turn if you want to count -1/3 as the same as +2/3) But if you try relabelling where just one triangle has been turned a 1/3 round then one will get a total which is not zero or some multiple of 3 times 1/3, you just get 1/3 or -1/3. So there is no way of gtting to such a position with the moves you gave. Dmcq (talk) 13:22, 10 February 2017 (UTC)[reply]

If you could twist all the faces in the same direction you'd get another factor of three. Dmcq (talk) 12:09, 7 February 2017 (UTC)[reply]

Yes, but that's not one of the modes. In mode 2, 4 faces turn clockwise and 4 faces turn counterclockwise. (think replacing the sides with gears)Naraht (talk) 17:43, 7 February 2017 (UTC)[reply]