The New Sudoku Players' Forum

[Maq777]

To begin with we would need to identify equivalence (conjugacy) classes. The only software for this is a group-theory package like GAP. The number of VPT's (Validity Preserving Transformations) involved, is, I believe, 2 x (12 ^ 10) = 123,834,728,448 for 16x16 Sudoku, compared with just 2 x 6^8 = 3,359,232 for 9x9 Sudoku.

With a group of that size, it's highly unlikely that any desktop PC is going to be able to accomplish even this first step! Some serious architecture would need to be thrown at the problem ...

? isnt VPT = 24^4,24^4,24^4,24^4 * 2 = 24^16 *2 = 24233149581890213117952 = 2e22

I think that the number of VPT's is actually 2 x (24 ^ 10) = 126,806,761,930,752

• band/stack permutations: 24 ^ 2

• rows/cols in band/stack: 24 ^ 8

• transposition: 2

[I see that I gave this count as 2 x 12^10 earlier on in this thread, unaccountably setting 4! = 12 (oops) Nobody picked up on this at the time ....]

Better all together, instead of just me.

Many brains think better than one

On the wikipedia page where these topics are discussed

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

Are a table in which they give the value of for all square sudoku puzzles, and for the 16x16 case they use the value of [(4!)^10 ]× 2 × 16! for "Size of VPT group".

The only thing to do to obtain that value in sudokus nxn is to multiply the Geometric Transformations by the relabels

Size of VPT group = [Geometric Transformations] * [Relabels]

or what i called

Maximum range: [2*(n!)^((2n)+2)] * [(n^2)!]

4x4
-------------------------------------------
Geometric Transformations: 128
Relabeling :24
Maximum range :3072 (VPT)

9x9
-------------------------------------------
Geometric Transformations: 3359232
Relabeling :362880
Maximum range :1218998108160 (VPT)

16x16
------------------------------------------
Geometric Transformations: 126806761930752
Relabeling: 20922789888000
Maximum range: 2,65315E+27 (VPT)

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

or am I wrong about something?

[EmilyVaughan]

No matter how much I searched I couldn't find anything suitable

[Maq777]

Hi all,

# Combinatorial Analysis of the First Band in 16x16 Sudoku (Hexadoku)

**Independent Researchers:**
* Miguel Ángel Quinteiro Fernández
* Miguel Ángel Quinteiro Piñero

**Research on Data Structure and Entropy Collapse**

## 1. Abstract
This study details the exact enumeration of the possible ways to complete the first band (rows 1 to 4) of a Sudoku of order n = 4. The analysis focuses on the interaction between box constraints (4x4 blocks) and row constraints, demonstrating a deterministic collapse of combinatorial freedom as the band is completed.

## 2. Counting Methodology
The concept of the Migration Matrix and Multinomial Coefficients is used to determine the distribution of number sets. To isolate the topology from the internal order, the structural divisor Sigma = (4!)^4 = 331,776 is employed.

## 3. Analysis by Regions and Rows
Due to the symmetry of the constraints, the count for completing the band region-by-region is isomorphic to the row-by-row count.

### Stage A: Region 1 / Row 1 (Initial Freedom)
* Total Ways: 20,922,789,888,000
* Structural Base: 63,063,000

### Stage B: Region 2 / Row 2 (Simple Exclusion)
* Total Ways: 248,341,303,296
* Structural Base: 748,521

### Stage C: Region 3 / Row 3 (Dual Exclusion)
* Total Ways: 600,514,560
* Structural Base: 1,810

### Stage D: Region 4 / Row 4 (Deterministic Closure)
* Total Ways: 331,776
* Structural Base: 1

## 4. The Master Number of the First Band
The total number of combinations to complete the first band is obtained by the product of the four stages (A x B x C x D):

1,035,230,499,039,714,288,129,983,395,009,658,880,000

## 5. Comparative Table of Results

| Step | Total Ways (Exact Figure) | Structural Base (/331,776) |
| :--- | :--- | :--- |
| 1 | 20,922,789,888,000 | 63,063,000 |
| 2 | 248,341,303,296 | 748,521 |
| 3 | 600,514,560 | 1,810 |
| 4 | 331,776 | 1 |

## 6. Conclusion
The analysis reveals that the fourth stage of a 16x16 Sudoku band is purely permutational at an internal level, with a structural base of 1. This implies that the configuration of sets for the band is "sealed" during the third stage, eliminating all uncertainty regarding which numbers must occupy the remaining spaces in the band's closure.

----------
I wonder if anyone can verify this.

Best regards

[Maq777]

Hi all,

Another factor to consider in the search for 16x16 solutions is the possibility that the following levels exist:

Code: Select all

Level | Stabilizer (Si) | Factor (sigma = 18432) | Family Size (Exact)    | Level Characteristics
------|-----------------|------------------------|------------------------|--------------------------------------
1     | 18,432          | 1      x sigma         | 6,879,707,136          | The Core (Seed): Total Octeract symm.
2     | 9,216           | 2      x sigma         | 13,759,414,272         | Loss of Transposition (Chirality).
3     | 6,144           | 3      x sigma         | 20,639,121,408         | Ternary Block Symmetry.
4     | 4,608           | 4      x sigma         | 27,518,828,544         | 4D Band Rotation.
5     | 3,072           | 6      x sigma         | 41,278,242,816         | Block Harmonic (24x128).
6     | 2,304           | 8      x sigma         | 55,037,657,088         | Block Duality (2 x 1152).
7     | 1,536           | 12     x sigma         | 82,556,485,632         | Axis Symmetry 2^9 * 3
8     | 1,152           | 16     x sigma         | 110,075,314,176        | The Master Block: 4x4 block stab.
9     | 768             | 24     x sigma         | 165,112,971,264        | Band Fragmentation.
10    | 576             | 32     x sigma         | 220,150,628,352        | Block Semi-symmetry (1152 / 2).
11    | 384             | 48     x sigma         | 330,225,942,528        | Hamming Network Level.
12    | 288             | 64     x sigma         | 440,301,256,704        | Quarter Block (1152 / 4).
13    | 192             | 96     x sigma         | 660,451,885,056        | Higher Binary Resonance.
14    | 144             | 128    x sigma         | 880,602,513,408        | Internal Row/Column Invariant.
15    | 96              | 192    x sigma         | 1,320,903,770,112      | The "Functional Minimum" of the 16x16
16    | 1               | 18,432 x sigma         | 126,806,761,930,752    | Total Chaos: Asymmetric models.


Of these, we are only certain that the first level (with the largest stabilizer) contains a single sub-family of 6,879,707,136 models (the equivalent of 5,184 in the 9x9 grid). It remains to be determined how many sub-families exist for the other levels and whether these are the only possible levels.

We are sharing this data for anyone who wishes to collaborate on the search.

Best regards.

[coloin]

Good that you are working on this, unfortunately Jim [RIP] [Mathimagics] is no longer with up but he would be pleased that you confirmed the Row 2 completions....

[blue]

(...)

## 4. The Master Number of the First Band
The total number of combinations to complete the first band is obtained by the product of the four stages (A x B x C x D):

1,035,230,499,039,714,288,129,983,395,009,658,880,000

(...)
----------
I wonder if anyone can verify this.

This isn't the right number.
The actual value is 16! * 1273431960 * (4!)^12, or 973,038,982,740,573,238,251,518,542,982,676,480,000 (~9.73 * 10^38).

Your analysis is assuming that the number of ways to fill R3 (or B3) against a fixed R12 (or B12), is the same for each R12/B12 -- 1810*(4!)^4.
That isn't true. Ten different values are possible:

Code: Select all

+------------------+--------------------+------------+
| R3/B3 base count | # R2/B2 base cases |    product |
+------------------+--------------------+------------+
|             4900 |                  3 |      14700 |
|             2450 |                192 |     470400 |
|             2050 |               1968 |    4034400 |
|             2000 |               3840 |    7680000 |
|             1850 |               7104 |   13142400 |
|             1810 |               5430 |    9828300 | (*)
|             1800 |              62208 |  111974400 |
|             1720 |             165120 |  284006400 |
|             1696 |             183168 |  310652928 |
|             1664 |             319488 |  531628032 |
+------------------+--------------------+------------+
| sums             |             748521 | 1273431960 |
+------------------+--------------------+------------+

See also, this post, and this one.
The "exact value of q(1,3)" number mentioned there -- "C" for you -- is "5215977308160 / 9241", which is: 1273431960 * (4!)^4 / 748521.

Best regards,
Blue.

[coloin]

i had forgotton all about those posts !!
however a "fruitless" discovery in the 16x16 area ....
This 55 clue puzzle by m_b_metcalf

Code: Select all

  .  .  .  9  .  .  .  .  .  3  .  .  .  .  .  2
  .  .  .  . 15  .  . 12 16  .  .  .  . 10  .  8
  .  4  .  5  .  .  .  .  .  9  .  .  .  .  .  .
  .  .  .  .  .  .  . 10  .  . 13  .  .  .  . 15
  .  .  8  .  .  .  .  .  .  .  .  .  .  .  . 16
  .  .  .  .  .  5  .  .  .  .  .  .  .  .  .  .
 10  . 15  .  .  .  .  .  .  .  .  .  .  .  . 12
  .  .  .  .  . 13  9  .  .  4  .  .  .  .  7  .
  .  .  .  . 16  .  . 14  .  .  .  .  .  .  .  .
  .  5  .  4  .  .  .  .  .  7  . 11  1 13  9  .
  .  .  .  3  .  .  .  .  .  1  .  .  5  .  4  .
  .  .  .  . 10  .  . 15  .  .  .  .  .  .  .  .
 15  . 16  .  .  .  .  .  8  . 10  .  .  .  . 14
  .  .  .  .  .  1  4  .  .  .  .  .  2  .  5  .
  8  .  .  .  .  .  .  . 12  . 16  .  .  .  .  .
  .  .  .  .  .  9  7  3  .  .  .  .  .  .  1  .


the solution is

Hidden Text: Show

Code: Select all

13 15 12 9 5 8 16 4 6 3 14 10 7 1 11 2
1 14 7 2 15 6 13 12 16 11 4 5 9 10 3 8
6 4 10 5 7 11 3 1 15 9 2 8 14 12 16 13
3 16 11 8 9 14 2 10 1 12 13 7 4 5 6 15
5 7 8 11 4 15 12 6 3 10 1 2 13 9 14 16
4 13 9 12 2 5 10 11 7 16 6 14 8 3 15 1
10 1 15 6 3 16 14 7 5 8 9 13 11 4 2 12
14 2 3 16 1 13 9 8 11 4 12 15 10 6 7 5
7 8 6 15 16 3 1 14 4 13 5 9 12 2 10 11
16 5 14 4 6 12 8 2 10 7 15 11 1 13 9 3
12 10 2 3 13 7 11 9 14 1 8 16 5 15 4 6
9 11 1 13 10 4 5 15 2 6 3 12 16 14 8 7
15 9 16 1 11 2 6 13 8 5 10 4 3 7 12 14
11 6 13 14 12 1 4 16 9 15 7 3 2 8 5 10
8 3 4 7 14 10 15 5 12 2 16 1 6 11 13 9
2 12 5 10 8 9 7 3 13 14 11 6 15 16 1 4


Counting up the clues.. there is only one 11 and no 6 ...

Therefore there is a 32 clue UA with no smaller UAs
changing the clues 6-->1 and 11--> 2

Code: Select all

+-------+--------+--------+--------+
|. . . .| . . . .| 1 . . .| . . 2 .|
|. . . .| . 1 . .| . 2 . .| . . . .|
|1 . . .| . 2 . .| . . . .| . . . .|
|. . 2 .| . . . .| . . . .| . . 1 .|
+-------+--------+--------+--------+
|. . . 2| . . . 1| . . . .| . . . .|
|. . . .| . . . 2| . . 1 .| . . . .|
|. . . 1| . . . .| . . . .| 2 . . .|
|. . . .| . . . .| 2 . . .| . 1 . .|
+-------+--------+--------+--------+
|. . 1 .| . . . .| . . . .| . . . 2|
|. . . .| 1 . . .| . . . 2| . . . .|
|. . . .| . . 2 .| . . . .| . . . 1|
|. 2 . .| . . . .| . 1 . .| . . . .|
+-------+--------+--------+--------+
|. . . .| 2 . 1 .| . . . .| . . . .|
|2 1 . .| . . . .| . . . .| . . . .|
|. . . .| . . . .| . . . .| 1 2 . .|
|. . . .| . . . .| . . 2 1| . . . .|
+-------+--------+--------+--------+


These are common in 9*9 ...as yet an unexplored area in 16*16 ....

[Maq777]

Blue,

You were right. I was overconfident because for the 9x9 we did:

Block 1: 9!
Block 2: [(1^3 + 3^3 + 3^3 + 1^3) * (3!^3)]
Block 3: (3!^3)

Calculation:
9! * [(1^3 + 3^3 + 3^3 + 1^3) * (3!^3)] * (3!^3)
362,880 * [56 * 216] * 216
362,880 * 12,096 * 216

Result:
948,109,639,680


We initially thought that for the 16x16 it would be:
Block 1: 16!
Block 2: 748,521 * (4!^4)
Block 3: [(1^4 + 4^4 + 6^4 + 4^4 + 1^4)] * (4!^4)
Block 4: 1 * (4!^4)

However, we failed to account for the fact that Block 2 creates a structural bifurcation for Block 3:

We have now processed 363 matrices under the "Invariance Herds" doctrine (the Quinteiro Family taxonomy) and discovered that they group into "exactly 10 families."

These are the ten (10) exact ways in which Block 3 can be filled, depending on the matrix herd it inherits from Block 2:

* 72 matrices - allow - 1,664 - ways to fill Block 3.
* 57 matrices - allow - 1,696 - ways to fill Block 3.
* 84 matrices - allow - 1,720 - ways to fill Block 3.
* 60 matrices - allow - 1,800 - ways to fill Block 3.
* 15 matrices - allow - 1,810 - ways to fill Block 3.
* 24 matrices - allow - 1,850 - ways to fill Block 3.
* 18 matrices - allow - 2,000 - ways to fill Block 3.
* 18 matrices - allow - 2,050 - ways to fill Block 3.
* 12 matrices - allow - 2,450 - ways to fill Block 3.
* 3 matrices - allow - 4,900 - ways to fill Block 3 (the most chaotic asymmetries leave more paths open).

Notice that 1,810 is exactly the number we deduced by adding: 1^4 + 4^4 + 6^4 + 4^4 + 1^4 = 1810.
Our geometric intuition with Pascal's Triangle did not fail; we simply calculated the exact survival rate for the circulating matrices herd.

Therefore, it is exactly as you pointed out:
Block 1: 63,063,000 (Ways to group 16 numbers into 4 sets of 4).
Block 2 and 3 together: 1,273,431,960 (The exact constant of cross-dimensional tension between matrices).
Block 4: 1 (Collapse by gravity).

Calculation:
[63,063,000 * (4!^4)] * [1,273,431,960 * (4!^4) * (4!^4)] * [1 * (4!^4)]
[63,063,000 * 1,273,431,960 * (4!^16)]
63,063,000 * 1,273,431,960 * 12,116,574,790,945,106,558,976

Final Result:
973,038,982,740,573,238,251,518,542,982,676,480,000

Best Regard