6x6 Enumerations

Everything about Sudoku that doesn't fit in one of the other sections

6x6 Enumerations

Postby SianKJones31 » Sat Jan 14, 2006 1:16 am

Not complete yet:

Given that the first cell is labelled canonically that is:

1 2 3
4 5 6

The number of Roku Doku can be reduced by a factor of 6!.

The number of combinations that can be formed in the second half of the band are fixed by the three numbers in the first column.
{1,2,3} must be followed by {4,5,6} in some order (3! ways)
{4,5,6} must be followed by {1,2,3} in some order (3! ways)

That is 3!^2 x 6! Combinations across the top row = 25920

The first stack is independent of the (1,2)mg in the ways mentioned above. As such the number of combinations down the first stack can be seen to have the following combinations.

If (1,1)mg is in canonical form, then there is one way to put both 1 and 2 on the second column. There are 4 ways to put 1 on the second and 2 on the third, and another 4 ways to do the opposite. And finally 1 way to put 1 and 2 both in the third column. That gives 1+8+1. Then there are 23=8 ways to choose the order of elements in each column, and so we have 8+64+8=80 choices for (1,2)mg.

It's correct also that (1,3)mg can always be chosen in 4 ways when (1,1)mg, (2,1)mg and (1,2)mg is chosen.

It's also correct that B2 as the last box can be chosen in either 0 or 4 ways. But the two cases do not contribute equally, The correct answer is 28.200.960.
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