Hi all-- Take a look at the following (partial) computation, and try to poke as many holes in it as you can
I've ruthlessly stolen some of the ideas in the first few pages, and redone the calculations so that they should be more correct: I found a few flaws in reasoning, some of which were emperically pointed out.
I'll start with the 5 block: 9! choices of numbers, to reordering
123
456
789
Then attack the 4 and 6 blocks:
We'll start (as before) with 4,5, and 6.
There are two main cases: 456 all in the same row, and 456 spread across the two rows.
Case 1) There are 2 choices of rows, and (3!)^2 reorderings within each row.
Ex.
546123___
___456___
___789645
Then the row-position of all the remaining numbers are fixed.
There are 3! ways to reorder the 4 remaining rows. Column doesn't matter here, because we can't duplicate numbers within boxes.
Total for case 1) 2*(3!)^6 = 2*6^6
Case 2) This one is more work.
There are 2 ways to choose which row (the upper or lower) will contain only one of 4,5,6 in box 4.
Then there are 3 ways to choose which of 4,5,6 is alone in that row.
There are 3 choices of position for that number in each of those 2 rows.
Similarly, there are 3! ways to position to the remaining two numbers in each of the 2 remaining rows.
Subtotal: 2*3*3^2*(3!)^2
Ex.
45_123_6_
___456___
6__7895_4
Now we have 3! ways to arrange 123 in the bottom row, and another 3! ways to arrange the 789 in the top row.
Ex.
457123968
___456___
631789524
The remaining 3 numbers in each block is determined. There are 3! ways to order the numbers in each of those blocks.
Total, Case 2) 2*3*3^2*(3!)^2 * (3!)^4 = 9 * 6^7
Total number of ways to arrange the 4 and 6 blocks:
9 * 6^7 + 2 * 6^6
We start to lose exactness if we try to say
"Similarly, there are 9 * 6^7 + 2 * 6^6 ways to arrange blocks 2 and 8" because there could (and will) be indirect constraints. If our goal, however, is to bound the solution, then we can make such a claim.
Total at this point: 9! (9 * 6^7 + 2 * 6^6)^2 = 3*10^17ish
Since we're committed to bounding at this point, let's look forward to the 1359 blocks.
There are 8 ways to place all the remaining 1's.
There are four allowable spots in each block, and there are only two configurations that work for each spot:
1# ## /// 1# ##
## 1# /// ## #1
#1 ## /// ## 1#
## #1 /// #1 ##
Thus the absolute maximum (though it should be much smaller) number of ways to fill the remaining blocks is 8^8. This bound can probably be reduced by a couple orders of magnitude with some cleverness. For instance, I want to say that once 6 numbers are filled in, the other 3 are determined, or else there is no solution: this would reduce this bound by a factor of 64.
= 4.16 * 10^24
I'd put money on the final answer being in the 10^20-10^21 range...