Conjecturing number solutions in sudoku squares of any size

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Conjecturing number solutions in sudoku squares of any size

Postby Maq777 » Thu Mar 25, 2021 10:55 am

Conjecturing a mathematical equation to calculate the number of solutions in sudoku squares of any size

We know that based on "n" we can calculate certain interesting values ​​about the sudokus, what I have never been able to understand is why is there not an equation that tells us based on n the total number of solutions that exist?

here is a possible way to do it, or at least a line of research to follow.

In my post "Equations that work for any sudoku of the form n * n" I show some equations that can help us, especially the one for calculating the templates

http://forum.enjoysudoku.com/equations-that-work-for-any-sudoku-of-the-form-n-n-t38848.html

and in my post on "Shidoku Study with Graphs" I show the relationship that exists between Cells, templates and models "

http://forum.enjoysudoku.com/shidoku-study-with-graphs-t38842.html

That said we have:

[Total number of Sudoku Solutions] = [Number of Models] * [Relabeling]

The problem with this formula is that we did not have an equation or a way to determine the [Number of Models]

By representing the models on a graph of the family of hypercubes we can have some clue of how to solve.

Hypercubes are graphs and therefore have some properties, for example, number of vertices, number of edges, number of faces, etc.

The important point is that we have to find the relationship that exists between n and the Number of Models, making use of the knowledge of the value of the Number of Faces of the Hypercube that supports all the Templates of the particular case.

For example, for the 4x4 sudoku that has a value of n = 2 we have that the hypercube that supports its 16 templates in its vertices is the tesseract, and that the number of faces is 24. So we can determine that:

[Number of models] = [Number of Faces of the Tesseract] / [n]
[Number of models] = 24/2
[Number of models] = 12

and therefore the number of solutions is

[Total number of Sudoku Solutions] = [Number of Models] * [Relabeling]
[Total number of Sudoku Solutions] = 12 * 4!
[Total number of Sudoku Solutions] = 12 * 24
[Total number of Sudoku Solutions] = 288

Now for the 9x9 we know that the Number of models is 18,383,222,420,692,992 and that we must work with a Hypercube that has 46,656 Vertices, therefore we would have to know the relationship between the Number of Faces of the Hypercube (46656) and the number of 9x9 models (18,383,222,420,692,992) to see if we can express this relationship in terms of n = 3

We already know from Russell and Jarvis that:
[Total number of Sudoku 9x9 Solutions] = [Number of Models] * [Relabeling]
[Total number of Sudoku 9x9 Solutions] = 18,383,222,420,692,992 * 9!
[Total number of 9x9 Sudoku Solutions] = 18,383,222,420,692,992 * 362880
[Total number of 9x9 Sudoku Solutions] = 6,670,903,752,021,072,936,960

So what I'm needing your help for is so that a real mathematician, who deals with things like the hypercube family and its characteristics, can give me insight into how to calculate the number of faces that exist in hypercubes with 16 vertices, 46,656 vertices and 110,075,314,176 vertices. To see if by generalizing from these three cases we can find the relationship between the value of n, the number of templates, the number of faces of the hypercube that contains them, to finally be able to calculate the number of models that will be generated and in this way also the amount of solutions that these sodokus have.

Best Regards
Maq777
 
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Re: Conjecturing number solutions in sudoku squares of any s

Postby Maq777 » Thu Mar 25, 2021 8:54 pm

Basically I need the help of a real mathematician to help me understand this topic.

https://en.wikipedia.org/wiki/Regular_polytope

https://en.wikipedia.org/wiki/List_of_regular_polytopes_and_compounds
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Re: Conjecturing number solutions in sudoku squares of any s

Postby Maq777 » Fri Apr 02, 2021 11:03 pm

It seems to me that the matter comes more or less here.

.- For n = 2 we are in the case of sudoku 4x4, 16 vertices are needed for its 16 Templates, the way to build it is:
2 * 2 * 2 * 2 = 2 ^ 4 = 16 that is why the 4x4 sudoku puzzle is located at R4 and at P = 2
With this we can embed the 4x4 sudoku in the first tesseract (the usual Y (2-4)), for that we already know that it has 24 faces of which are used
only 12 to place the possible models or invariant grids of that sudoku puzzle

.- For n = 3 we are in the case of sudoku 9x9 (the typical one of the newspapers) 46656 vertices are needed for its 46656 Templates, the way to build it is:
6 * 6 * 6 * 6 * 6 * 6 = 6 ^ 6 = 46656 that is why the 9x9 sudoku puzzle is located at R6 and at P = 6
With this we can embed the sudoku 9x9 in the generalized hypercube Y (6-6), for that we do not know how many 9-faces it has,
but we know that there must be at least 18,383,222,420,692,992 which are used to place the possible models or invariant grids of the 9x9 sudoku

.- For n = 4 we are in the case of sudoku 16x16, 110075314176 vertices are needed for its 110075314176 Templates, the way to build it is:
24 * 24 * 24 * 24 * 24 * 24 * 24 * 24 = 24 ^ 8 = 110075314176 that is why the 16x16 sudoku is located at R8 and at P = 24
With this we can embed the 16x16 sudoku in the generalized hypercube Y (24-8), for that we do not know how many 16-faces it has,
nor do we know the order of magnitude of the possible 16x16 sudoku models

We can continue like this successively Embedding the larger and larger sudokus into bigger and bigger Politopes .

The idea is to know if by construction or by generalization of the euler equation for regular polygons a quick calculation can be obtained
of the number of faces of the polytopes, and if with that result we can find a direct relationship with the value of "n" which is the base of each sudoku puzzle
and the value of the invariant models or grids of that sudoku puzzle.

That is what we are looking for.
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Re: Conjecturing number of solutions grids of size "n*n"

Postby coloin » Sat Apr 03, 2021 3:38 pm

Maq777 wrote:nor do we know the order of magnitude of the possible 16x16 sudoku models


Well we do have an idea ! - number-of-possible-16x16-sudoku-grids

There are other ironically termed "fruitless" threads I could point you to - which enumerate the template counting for the more manageable and finite 3*3 case [46556 and co] but probably you want to have a go yourself !!
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Re: Conjecturing number of solutions grids of size "n*n"

Postby Maq777 » Sat Apr 03, 2021 6:12 pm

coloin wrote:
Maq777 wrote:nor do we know the order of magnitude of the possible 16x16 sudoku models


Well we do have an idea ! - number-of-possible-16x16-sudoku-grids

There are other ironically termed "fruitless" threads I could point you to - which enumerate the template counting for the more manageable and finite 3*3 case [46556 and co] but probably you want to have a go yourself !!



Greetings Coloin ..

About the number of possible 16x16 sudoku grids I answered in the same post.

About working on my own, although I have been researching sudoku for about 10 years with my Father, I have been a reader of this forum for at least 7 years.

You have been a mandatory reference, and many of the topics discussed here are beyond my understanding.

I consider Russell, Jarvis, Gupta, Serg and others as true heroes, because figuring out what Sudoku is all about is not easy.

What I have been learning over the years I will try to share with you, I only ask you for some patience since I do not speak English, just a little. Many times to read or write to you I have to use the google translator.

Best regards
Maq777
 
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Re: Conjecturing number solutions in sudoku squares of any s

Postby Maq777 » Sat Apr 10, 2021 3:59 pm

If we have a graph with 46656 vertices and it is totally connected, that is, all the vertices touch through an edge with all the others and we want to form groups of nine vertices at a time, we can perform the combinatorial of (46656 taken from a 9 ). Which would give us an astonomically large number.

For the case in which the 46656 vertices are associated with a Sudoku9x9 template, the graph could not be one that is fully connected, since each template can be combined only with other 17972 templates, the graph would already suffer some limitation or restriction.

The question is, how then can you calculate the possible combinations of nine vertices at a time? when the graph is not fully connected.

Someone knows ?
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Re: Conjecturing number solutions in sudoku squares of any s

Postby Mathimagics » Sun Apr 11, 2021 7:58 am

.
Ok, let's suppose our graph has a vertex for each template.

If the graph has an edge between pairs of vertices (v[x], v[y]), where v[x] and v[y] correspond to templates T[x], T[y], then T[x] and T[y] must be mutually disjoint. That is, they have no cells in common.

For the 9x9 Sudoku case, we then want to count the number of 9-cliques (complete subgraphs of 9 vertices).

This is still a difficult problem.
"Most versions of the clique problem are hard"

(Wiki: Clique Problem)


The real question, I suppose, is this: is the clique-counting method of counting Sudoku's "easier" than other methods :?:
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Re: Conjecturing number solutions in sudoku squares of any s

Postby Maq777 » Sun Apr 11, 2021 6:25 pm

The main idea is not to count or enumerate, it is rather to be able to deduce on the other hand.

Well, the assumption must be made for sudoku9x9, for sudoku4x4 the work is already done, you can see it in the discussion of Studying Shidoku with graphs. Or in the videos of my youtube channel.

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

Once this adjacency matrix was created for templates in sudoku4x4, the graphs seen in the discussion could be generated, the KEY is that the polytope found here is a Hypercube (tesseract) and that for the tesseract we know in advance that it has a quantity of 24 [ 4-faces] in total, is one of its characteristics, we already have an equation that determines it, dividing the total number of [4-faces] by n = 2 gives us exactly 12, which exactly matches the number of grids invariants that we are going to have for that sudoku4x4. (This could all be pure chance !?).

My idea was to be able to make the graph of 46656 vertices, make its corresponding adjacency matrix and create some equivalent polytope for the sudoku9x9, and them by formulas or equations of the polytope itself we could know the total amount of [9-faces] that this may have.

The key is to be able to find a polytope that has already been studied enough to determine how many [9-Faces] it has in total, by equations of the characteristics of the polytope, and see if it is in a direct relationship with the value of n = 3 that is the basis of sudoku9x9, I had thought that this polytope could be a generalized hypercube of Y (6/6), the problem is that starting from a Y (6/6) I don't know how the [9-faces could be found there ].

The main idea is to NOT COUNT, but instead to calculate with a formula the total number of [9-faces] and to be able to compare the results with the number of Invariant Grid of sudoku9x9 determined by Jarvis and Russell, to see if they are in a direct relationship.

None of this is proven, is just a guess or a hunch. But I think it could be a line of research to explore.
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