The New Sudoku Players' Forum

[Maq777]

When we want to factor in prime numbers the total amount of the sudoku4x4 solutions we obtain

288 = (2 ^ 5) * (3 ^ 2)

But when we want to do the same for sudoku9x9 we get

6,670,903,752,021,072,936,960 = (2 ^ 20) * (3 ^ 8) * 5 * 7 * 27,704,267,971

Which can also be seen as

6,670,903,752,021,072,936,960 = Relabeling * [Invariant Grids]

6,670,903,752,021,072,936,960 = 9! * [18,383,222,420,692,992]

And then obtaining the prime factors of 18,383,222,420,692,992 we obtain

18,383,222,420,692,992 = (2 ^ 13) × (3 ^ 4) × 27,704,267,971

Being able then to write the solutions as

6,670,903,752,021,072,936,960 = 9! * [(2 ^ 13) × (3 ^ 4) × 27,704,267,971]

The obvious question is what internal Sudoku mechanism creates or generates this very large prime number within the Invariant Grids?

The concept map of any size sudoku that I now have in my head tells me that:

n (is the base of sudoku),
--From n arise or emerge the number of cells (in my book "Las Casillas")
---- From the number of cells comes the number of templates (in my book "Las Grillas")
------ From the Templates the Invariants Grids arise (in my book "Los Modelos")
-------- Two things emerge from the Invariants Grids :
---------- 1.- The number of Essentially Different (in my book "Las Familias"), using for this the Geometric Permutations (in my book "Las Transformaciones Geometricas")
---------- 2.- The amount of Sudoku Solutions, Using for this the numerical transpositions or Relabeling (in my book "Las Transposiciones Númericas")
-------- From the multiplication of the Geometric Transformations by the Relabeling the VPT´s. (in my book "Alcance Máximo")

Following this conseptual map, I have an equation or mathematical formula for almost all the elements, but the Invariants Grids and the number of Solutions continue to elude me from a generalization and it is partly the fault of that huge prime number, whom I feel like a stone. inside my shoe.

Starting from:

6,670,903,752,021,072,936,960 = 9! * [(72 ^ 2) * (2 ^ 7) * 27.704.267.971]

Have I ever tried to generalize forms for all sudokus assuming this .....

https://github.com/MiguelQuinteiro/ImagenesSudoku/blob/master/SudokuSolutions.png

But it got me nowhere.

Someone will have a clearer idea of ​​this matter in the forum, what suduko mechanism may be generating such a large prime number involved in the number of invariant grids, if so please share it with this simple mortal.

Greetings

[Mathimagics]

Following this conseptual map, I have an equation or mathematical formula for almost all the elements, but the Invariants Grids and the number of Solutions continue to elude me from a generalization and it is partly the fault of that huge prime number, whom I feel like a stone. inside my shoe.

I think that prime number is simply evidence that there might not be a generalised formula of the type that you are looking for.

For both of these major counting problems (the counting of all grids, and the counting of ED grids) the only successful counting methods to date for 9x9 Sudoku involve:

• identification of equivalence classes

• physically counting the solutions in each class

The final count is then obtained by summing, for each equivalence class C, the number of solutions N(C) multiplied by the number of elements of C. In other words SUM {N(C) x |C|} over all C.

In the 9x9 countings (Felgenhauer, Jarvis, Russell etc) you can observe very much variance, both in the size of the equivalence classes, |C|, and in the number of solutions in a class, N(C).

So it really should come as no surprise, that the result of the summation contains a large prime factor.

Cheers
MM

[Maq777]

I think that prime number is simply evidence that there might not be a generalised formula of the type that you are looking for.

.... you can observe very much variance, both in the size of the equivalence classes, |C|, and in the number of solutions in a class, N(C).

So it really should come as no surprise, that the result of the summation contains a large prime factor.

Cheers
MM

Of course that's a possibility, the point is that I never give up.

I agree that there can be many variations, but they have to be in very discrete quantities and possible to determine.

You cannot have one (1) and only one isolated problem or load line.

Every approach you make on the sudoku grid is instantly accompanied by a well-determined consort from other boards. At the end of the day, the only thing we are allowed to do is Geometrically Transform it and then transform it with the relabeling.

Then as you say you have to add many partial results. But since no one came to me in isolation I wonder.

Where does such a large prime number come from?

[Mathimagics]

Where does such a large prime number come from?

It comes from pure chance, is what I am trying to say. Consider these known grid counts from the Wiki MoS page:

• Sudoku 6x6: total grids = 28,200,960 = 2^12 × 3^4 × 5 × 17

• Sudoku 8x8: total grids = 29,136,487,207,403,520 = 2^25 × 3^5 × 5 x 7 × 23^2 × 193

• Sudoku 6x6: ED grids = 49 = 7^2

• Sudoku 8x8: ED grids = 1,673,187 = 3 × 557729

• Sudoku 9x9: ED grids = 5,472,730,538 = 2 × 11^2 × 23 × 983243

This is simply what typically happens when you start adding up a series of numbers. The bigger the total, the more likely it is that you will get some large prime factor(s) ... even if the numbers being added are themselves smooth (i.e. have only small prime factors), it still happens, as you can see.

Counting total grids, and counting ED grids, are both essentially combinatorial problems, rather than geometric ones, I think ...

[JPF]

Recurrent discussion...

see Red Ed's answer in 2005!

JPF

[Maq777]

Recurrent discussion...

see Red Ed's answer in 2005!

JPF

I don't think it's a recurring discussion and let me explain why:

In the timeline of the discussion you mention, it seems to me that the person was doubting the result obtained by Jarvis and Russell due to the huge prime number found. That is not our case. In fact, my Father and I, with programs designed by us independently, went through almost all the calculations that Russell and Jarvis went through and got the same results. So we can guarantee that if they made any miscalculation we made it too.

We find Jarvis's intellectual work admirable as he has been able to reduce the scan size to just 44 equivalence classes and the programs produced by Russell also have their touch of computational genius.

Having said all this, my problem with the enormous prime number found comes more from the side that Mathimagic warned, simply the number hinders me in my objective of being able to find a generalization in an equation that allows us to find the number of invariant grids in sudoku of any size .

As a computer guy one of my obsections is prime numbers, being able to find and predict them is one of the holy grails of computing, since huge prime numbers are used as the basis of cryptography with their public key and private key encryption.

Assuming that each higher-order sudoku puzzle has a huge prime number associated with it and that we can, through some other method, determine where it comes from, we would not only be finding a formula to determine the solutions to any sudoku puzzle, but also being able to find a huge number prime that does not depend on the famous Mersen prime number search [(2 ^ n) + 1] or [(2 ^ n) - 1].

kind regards.

[coloin]

Well, like most things in sudoku combionotronics ... the answer is what the answer is ..

For example We have so far been able to find x number of ED 17 clue puzzles .... maybe this number x is a prime number ... but we havnt proved that we have found them all yet !!!
But 9*9 sudoku is a big finite space of ED puzzles and ED grid solutions ....

The number of grid solutions is approx 6e21 ...[6,670,903,752,021,072,936,960]
this can be reduced impirically by the inherent isomorphic transformations 9! and 6^8 *2 [9! and 3^8 and 2^9]
now

6,670,903,752,021,072,936,960 = (2 ^ 20) * (3 ^ 8) * 5 * 7 * 27,704,267,971

The chance that any random number above 2e10 is a prime number im told by my mathmatical son is around 6%

I wonder if you took any random number around 6e21 and factorized it - I reckon the biggest prime number in its factors might be quite a big number ... [ maybe ]

[JPF]

For those who love questions about large prime numbers in the sudoku field:

The number of different patterns of a 9x9 sudoku grid is = 2,417,851,639,229,258,349,412,352 = 2^81 (81 factors)

But...the number of esssentialy different patterns is 746,186,061,249,180,790 (see here):
which can be written with prime numbers:
746,186,061,249,180,790 = 2 × 5 × 7 × 16,033 × 664,866,268,009

JPF