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
因此，所有这些都是 1+1+27+27=56 种可能性。
Interestingly there was an estimation of the number of possible sudokus posted in an internet forum before the actual value was calculated and published by Felgenhauer & Jarvis. The author of the post points out that there are some unproven assumptions in his guess. But the estimated value differs by 0.2% from the actual value published later.
In this Wiki you can find some estimation of other types of sudoku based on similar guesses.
Step A: Pretend that the only 'rule' was the 'block' rule, and that the row and column rules did not exist. Then each block could be arranged 9! ways, or 9!^9 ways to populate the puzzle (1.0911*10^50 'solutions').
Step B1: If we then say 'let us add a rule about unique values in a row', then the top three blocks can be filled as follows: Block 1: 9! ways Block 2: 56 ways to select which values go in each 3-cell row, and 3! ways to arrange them (remember that we haven't invented a column rule yet). Block 3: with 1 and 2 filled, the values that go in each row is now defined, but each row can be arranged 3! ways. Therefore, we have 9! * 56 * 3!^6 ways to fill the top three blocks, and this value cubed to fill all nine blocks. (or 8.5227*10^35 solutions). Note that this represents a 'reduction ratio' (denoted as R) of 1.2802*10^14, by adding this one new rule.
Step B2: But we could have just as easily added a 'unique in columns' rule, and achieved the same results downward instead of across, with the same value of R.
Step C: (and here is where my solution is not rigorous) What if we assume that each of these rules would constrain the number of valid solutions by exactly the same ratio? Then there would be a combined reduction ratio of R^2. So the intitial value of 1.0911*10^50 solutions would reduce by a factor of R^2, or 1.639*10^28, leaving 6.6571*10^21 valid solutions.
This post and the account are attributed to Kevin Kinfoil (Felgenhauer & Jarvis).
Additional notes
Assume the Block 1 is
1 2 3
4 5 6
7 8 9
Then we have the following possibilities for Block2, if we ignore the order of the rows
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
So all in all these are 1+1+27+27=56 possibilities.
"the number of valid Sudoku solution grids for the standard 9×9 grid was calculated by Bertram Felgenhauer and Frazer Jarvis in 2005 to be 6,670,903,752,021,072,936,960 . This number is equal to 9! × 722 × 27 × 27,704,267,971, the last factor of which is prime. The result was derived through logic and brute force computation."
You can read the most recent rewrite of the original publication by Bertram Felgenhauer and Frazer Jarvis : Mathematics of Sudoku, it details the computation over 7 pages. The calculation actually isn't trivial (the idea being to enumerate distinct and valid Sudoku grids, rather than all possible arrangements of digits over a 9x9 grid).
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有趣的是，在费尔根豪尔 &贾维斯。
帖子作者指出，他的猜测存在一些未经证实的假设。但估算值与后来公布的实际值存在0.2%的差异。
在此 Wiki 中，您可以根据类似的猜测找到对其他类型数独的一些估计。
以下是新数独玩家论坛：
这篇文章和帐户归属于 Kevin Kinfoil (Felgenhauer & Jarvis< /a>）。
附加说明
假设块 1 是
那么如果我们忽略行的顺序，那么对于块 2，我们有以下可能性
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因此，所有这些都是 1+1+27+27=56 种可能性。
Interestingly there was an estimation of the number of possible sudokus posted in an internet forum before the actual value was calculated and published by Felgenhauer & Jarvis.
The author of the post points out that there are some unproven assumptions in his guess. But the estimated value differs by 0.2% from the actual value published later.
In this Wiki you can find some estimation of other types of sudoku based on similar guesses.
Here is the full post from The New Sudoku Players' Forum:
This post and the account are attributed to Kevin Kinfoil (Felgenhauer & Jarvis).
Additional notes
Assume the Block 1 is
Then we have the following possibilities for Block2, if we ignore the order of the rows
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=27So all in all these are 1+1+27+27=56 possibilities.
您可以在这个 Wiki 中找到所有相关信息：http://en.wikipedia.org/wiki/Mathematics_of_Sudoku。
“Bertram Felgenhauer 和 Frazer Jarvis 在 2005 年计算出标准 9×9 网格的有效数独解网格数为 6,670,903,752,021,072,936,960 。这个数字等于 9！ × 722 × 27 × 27,704,267,971，最后一个因子是质数，结果是通过逻辑和强力计算得出的。”
You can find all about it in this Wiki: http://en.wikipedia.org/wiki/Mathematics_of_Sudoku.
"the number of valid Sudoku solution grids for the standard 9×9 grid was calculated by Bertram Felgenhauer and Frazer Jarvis in 2005 to be 6,670,903,752,021,072,936,960 . This number is equal to 9! × 722 × 27 × 27,704,267,971, the last factor of which is prime. The result was derived through logic and brute force computation."
您可以阅读 Bertram Felgenhauer 和 Frazer Jarvis 对原始出版物的最新重写：数独数学，7页详细介绍了计算过程。计算实际上并不简单（其想法是枚举不同和有效数独网格，而不是 9x9 网格上所有可能的数字排列）。
You can read the most recent rewrite of the original publication by Bertram Felgenhauer and Frazer Jarvis : Mathematics of Sudoku, it details the computation over 7 pages. The calculation actually isn't trivial (the idea being to enumerate distinct and valid Sudoku grids, rather than all possible arrangements of digits over a 9x9 grid).