I keep forgetting that this forum does not retain the username!
That was me again.
Anonymous wrote:IJ wrote:Could you give an example of how two manipulations could be the same? I can't quite see it, but I'm willing to believe it may be true.
It is not that the manipulations are the same. What I was saying is that the final and initial grids may 'look' the same, for some grids.
Consider:
123 456 789
456 789 123
789 123 456
231 564 897
564 897 231
897 231 564
312 645 978
645 978 312
978 312 645
There's a host of row / column / block by row or column 'swaps' that result in the starting point.
For example, ROLL blocks 147 to blocks 258, and ROLL blocks 258 to 369. Now roll the rows so that rows 123,456,789 become 312,645,978. Since each of 9 occurances of each digit are the same, you are back where you started.
tinfoil wrote:I keep forgetting that this forum does not retain the username!
That was me again.
IJ wrote:Yes - I started with 8, but then I managed to convince myself that other combinations were possible too. It was late!
Any comments on the rest?
afj wrote:I agree that we are agreed on 9! x 56 x 3!^6 for any "chute". This leads to 9! x (56 x 3!^6)^2 for any cross shape (L, T, +, etc., where we have one vertical chute and one horizontal chute). [Did you miss out blocks 2 and 8 in your new upper bound, by the way?] So, this gives the possible ways of filling in blocks 1, 2, 3, 4 and 7. We can fill in block 5 and block 9 in at most 448 ways each. Then blocks 6 and 8 can be filled in at most uniquely. This gives the upper bound: 9! x (56 x 3!^6)^2 x 448^2, as IJ already gave, I think.