Interestingly, before Felgenhauer & Jarvis calculated and published the actual price, an estimate of the possible amount of sudoku was published on the online forum. The author of the post indicates that there are unproven assumptions in his guess. But the calculated value differs by 0.2% from the actual value published later.
On this wiki, you can find some estimates of other types of sudoku based on similar assumptions.
Here is the full post from the new sudoku players forum :
by Guest "Fri Apr 22, 2005 1:27 p.m.
Let's try it from a completely different direction:
Step A:
Imagine that the only “rule” was the “block” rule, and the row and column rules did not exist. Then each block can be arranged 9! ways, or 9! ^ 9 ways to complete the puzzle (1.0911 * 10 ^ 50 "solutions").
Step B1:
If then we say “let's add a rule about unique values in a row”, then the top three blocks can be filled as follows:
Block 1: 9! the way
Block 2: 56 ways to select values for each row from 3 cells and 3! ways to place them (remember that we have not yet invented a column rule).
Block 3: with filled in 1 and 2 values that go on each line, are now defined, but each line can be organized 3! the way.
Therefore, we have 9! * 56 * 3! ^ 6 ways to fill the top three blocks, and this value is cubed to fill all nine blocks. (or 8.5227 * 10 ^ 35 solutions). Note that this represents a “reduction factor” (denoted by R) of 1.2802 * 10 ^ 14 by adding this new rule.
Step B2: But we could just as easily add the “uniqueness in columns” rule and achieve the same results down, not across, with the same value of R.
Step C: (and here my decision is not strict). What if we assume that each of these rules will limit the number of valid decisions in exactly the same ratio? Then there will be a combined reduction coefficient of R ^ 2. Thus, the initial value of 1.0911 * 10 ^ 50 solutions would decrease in R ^ 2 or 1.639 * 10 ^ 28, leaving 6.6571 * 10 ^ 21 valid solutions.
This post and account is owned by Kevin Kinfoilom (Felgenhauer & Jarvis ).
Additional notes
Assume Block 1
1 2 3
4 5 6
7 8 9
Then we have the following options for Block2, if we ignore the row order
1 2 3 4 5 6
4 5 6 7 8 9
7 8 9 1 2 3
this is 1 possibility
1 2 3 7 8 9
4 5 6 1 2 3
7 8 9 4 5 6
this is 1 possibility
1 2 3 two of 4,5,6, one of 7,8,9 3 * 3
4 5 6 the two remaining of 7,8,9, one of 1,2,3 3
7 8 9 the two remaining of 1,2,3, the remaining of (two of 4,5,6) 1
these are (3 * 3) * 3 * 1 = 27 possibilities
1 2 3 two of 7,8,9, one of 4,5,6 3 * 3
4 5 6 two of 1,2,3, the remaining of 7,8,9 3
7 8 9 the two remaining of 4,5,6, the remaining of two of 1,2,3 1
these are (3 * 3) * 3 * 1 = 27
Thus, in general, this is 1 + 1 + 27 + 27 = 56 possibilities.