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Counting sudokus (Posted on 2018-07-13) Difficulty: 3 of 5
You are given a 9x9 grid of numbers such that:
a. each row,
b. column,
c. each of the nine 3x3 grids
has each number 1 through 9 exactly once.
Lets call it Sudoku puzzle's solution or SPS.
A grid with condition c. removed is called a Latin Square, LS.

1. (d2) how many 9x9 LS's are there?
2. (d4) And how many 9x9 SPS's ?

No Solution Yet Submitted by Ady TZIDON    
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Thoughts on Part 2 | Comment 2 of 6 |

This may be a dead end, but here goes: If you ignore the corner 3x3's and call the remaining five 3x3s "the cross" I think I have calculated how many legal crosses there are. N = (9!) [ 2 (3^3) (3!)^6 ]^2

Here is the logic: There are 9! possible centers. (All are equally represented in the solution set since the digits 1 to 9 are interchangeable symbols).  For a particular central 3x3 (C), how many valid left (L) 3x3s are there? Let's consider C to have row contents: C_Row 1:(123), C_Row2:(456), C_Row 3:(789), with the order in each row unspecified. Let's find all L forms allowed (again, with row element order unspecified). We can redistribute the C_Row 1 digits 1,2,3 into L_Rows 2 and 3 by placing either A: 1 element in L_Row 2 and 2 elements in L_Row 3, or B: vice versa.  If we do it the A way, then we _must_ redistribute  C_Row 2 by placing 1 element into L_Row 3 and 2 elements into L_Row 1, and we _also must_ redistribute C_Row 3 elements by placing 1 element into L_Row 1 and 2 into L_Row 2. One possible such "A" mapping makes L as (457), (189), (236) (the order of elements in a row are again unspecified. There are 3^3 ways to implement A since we have the choice of which one of the three elements to move singly out of each C_Row.

Each of these "A" L's makes for a unique right "R" 3x3 so long as we do not specify the element order in a row. To use our example:

(4 5 7)        (1 2 3)        (6 7 9)

(1 8 9)        (4 5 6)        (2 3 7)

(2 3 6)        (7 8 9)        (1 4 5)

Once we have made L and R, we can permute the elements in each of the six rows 3! ways, yielding (3!)^6 legal results. <o:p></o:p>

Mapping B has a parallel description. By redistributing the C_Row 1 digits 1,2,3 by placing 2 elements in L_Row 2 and 1 element in L_Row 3, the redistribution out if C_Rows 2, and 3 are determined. So Mapping A and B together put a "2" in the formula above for N.   Producing all legal L and Rs (with elements specified) from a unique C   (with elements specified) give the term in brackets [2 (3^3) (3!)^6].

We then square this term to include all the valid 3x3s located directly above and below  C. 

Now, is there a way to get the number of legal corners from this analysis? The problem is there may be many or none for a specific legal cross....

I am afraid that this may be a dead end, since for one C (that's all we need to solve all the corners for) there are 6 10^12 legal crosses, which does little to help narrow-in on valid corners. 

Edited on July 14, 2018, 7:09 pm
  Posted by Steven Lord on 2018-07-14 07:23:21

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