Let

KSP(N,M) be the expected number of NxM Sudoku grids, derived from the KSP formula, and let

G(N,M) be the actual number of grids. You ask whether

KSP(N,M) <

G(N,M) for all N,M > 1. I see reasons for why it should hold in general, and I have an explicit proof for the case M=2 (or N=2).

For general N,M, a vague idea: Given two (NM)x(NM) grids divided into boxes of size NxM, the first obeys the row and box constraints, the other the column and box constraints. Let

C(b) be the condition that the two grids are equal in a box

b. The idea behind the KSP-formula is the assumption that the

C(b) are independent for the different

b. This is not entirely true; if

C(b) holds for a (large) set of

b, it seems reasonable to assume that the elements of the grids are more likely to be "forced" into the same positions for the remaining

b. So it seems that the two grids are equal a little bit more than 1 out of (NM)!^(NM) times, making KSP an underestimate.

For M=2Let C_1, ..., C_T be the column configurations of Nx2 Sudoku bands. There are Ch(2N,N) ways to choose each Nx2 box configuration (Ch=Binomial), with N boxes we have T=Ch(2N,N)^N.

Each configuration C_i is subject to B(C_i) bands. Let B_0 be the average value of the B(C_i), and define e_i such that B(C_i) = B_0+e_i. Then the sum of all e_i is 0.

The number of Nx2 bands is B_(N,2) = sum_i B(C_i) = T * B_0.

The number of 2xN bands is known to be B_(2,N) = (2N)! * N!^2. Then

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` B_(N,2)^2 * B_(2,N)^N`

KSP(N,2) = --------------------- = T * B_0^2

(2N)!^(2N)

For the actual number, remember that when the configuration of the upper band in a grid is given, then so is the configuration of the lower band (swap columns in each box), and they each have the same number of bands. Therefore

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`G(N,2) = sum_i B(C_i)^2 = sum_i B_0^2 + 2*sum_i B_0*e_i + sum_i e_i^2`

= T * B_0^2 + 0 + sum_i e_i^2 = KSP(N,2) + sum_i e_i^2.

So

G(N,2) >=

KSP(N,2) with equality if and only if all B(C_i) are egual. If C_0 is a configuration where all boxes are equal, then B(C_0) = L^2 where L is the number of NxN Latin squares. On the other hand, let C_1 be a configuration where N-1 boxes are equal, while the last differs from the others by swapping a and b, one element from each column. Then you can show that the bands of C_1 correspond to those of C_0 where a and b are on the same row in the last box. Therefore B(C_1) = L^2/N < B(C_0), and so

G(N,2) >

KSP(N,2).

I have the time logs for 6x2, I'll come back to that later.