Greetings PatmaxDaddy and kjellfp
As a "mere mortal" I have been trying to follow this thread and I can see the difficulty that many will have experienced with a thread of this length......and I have just looked at Wipekedia !
http://en.wikipedia.org/wiki/Mathematics_of_SudokuSuffice to say
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The number of grids has been calculated for the following grids
2×2 288
2×3 28200960
2×4 29136487207403520
2×5 1903816047972624930994913280000
3×3 6670903752021072936960 = c. 6.7×1021
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We have reliable estimations for
3×4 unknown, estimated c. 8.1064×10^46 Pettersen n/a
3×5 unknown, estimated c. 3.5086×10^84 Silver n/a
4×4 unknown, estimated c. 5.9584×10^98 Silver n/a
4×5 unknown, estimated c. 3.1764×10^175 Silver n/a
5×5 unknown, estimated c. 4.3648×10^308 Silver/Pettersen n/a
You have now worked out a formula for the 3xC and 4xC and now the 5x5 and 5x6 bands. I presume this relates to the first row of boxes.
A long time ago.............before Bertram calculated 6.7×10^21, various others were attempting to work it out long hand ! Estimations at the time needed were in years.
But there was a Monte Carlo estimation [unreferenced] which actually gave a very accurate result. Interestingly it is on page 2 of this thread [!]
josh wrote:I have heard of someone using a Monte Carlo algorithm on a supercomputer for 20 days and coming up with an answer in the region of 6x10^21
As I understood someone worked out the average number of grid solutions from 25 different B1B5B9 grid fillings from
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xxx......
xxx......
xxx......
...xxx...
...xxx...
...xxx...
......xxx
......xxx
......xxx
and multiplied the result by [9!]^3
The thought of working out the 4x4 horrified many....and still does !
Im guessing that the large number of solutions precludes this approximation......so how near to an exact count for the 4x4 will we get. ?
Regards