CODE CHALLENGE: Sudoku: Fastest Solver?

[(X)]

A fun challenge...

Who can come up with the fastest Sudoku Solver using GFA BASIC 32 for Windows?

You can use whatever source: human or machine, to help you.

You must submit your source code for others to compile, no external or 'hidden code' GL32 library, .BAT, .COM, .EXE or .DLL.

Post your code solution here, preferably with a screen capture and time to solve: (Timer_End - Timer_Beg) expressed in milliseconds.

Lets call the procedure: Sudoku_Solver_Calculation() that starts with a partially filled, 9 by 9, Byte typed, base 1, Global array called: Grid().

$Library "gfawinx"
$Library "UpdateRT"
UpdateRuntime      ' Patches GfaWin23.Ocx
Option Base 1
Global Grid(1 .. 9, 1 .. 9) As Byte

FullW 1
Sudoku_Solver_Calculations
Do : Sleep : Until Me Is Nothing

Procedure Sudoku_Solver_Calculations()
 Dim t0# = Timer
 
 '
 ' Calcs
 '
 
 Trace Timer - t0
EndProc



[scalion]

Yes it's a good challenge.

[(X)]

Ok. Try to beat this first attempt...

The start state is a diagonal set of numbers from 1 to 9, upper left to lower right.

Attachments:

Demo Sudoku Solver.G32 (17.02 KB)

[(X)]

This version is setup with a "difficult" puzzle and the original given numbers are in red.

Attachments:

Demo Sudoku Solver.G32 (17.64 KB)

[scalion]

I make one too

That result is the same :

My code : Sudoku Chalenge.G32 (4.31 KB)

[scalion]

Aug 9, 2023 18:14:39 GMT 1 (X) said:

This version is setup with a "difficult" puzzle and the original given numbers are in red.

View Attachment

i have an "array bound execed" in you demo...

[(X)]

Demo Sudoku Solver 2.G32 (18.04 KB)

Verification:

Dim i, j
For i = 1 To 9
 j = i - 1
 trace 1 + 3 * (j \ 3)
Next

TRACE:(1):1 + 3 * (j \ 3) = 1
TRACE:(1):1 + 3 * (j \ 3) = 1
TRACE:(1):1 + 3 * (j \ 3) = 1
TRACE:(1):1 + 3 * (j \ 3) = 4
TRACE:(1):1 + 3 * (j \ 3) = 4
TRACE:(1):1 + 3 * (j \ 3) = 4
TRACE:(1):1 + 3 * (j \ 3) = 7
TRACE:(1):1 + 3 * (j \ 3) = 7
TRACE:(1):1 + 3 * (j \ 3) = 7

This should fix the problem...

Local startRow = 1 + 3 * ((row-1) \ 3)Local startCol = 1 + 3 * ((col-1) \ 3)

[scalion]

I read this (FR) : www.lemonde.fr/passeurdesciences/article/2012/01/08/17-est-le-nombre-de-dieu-au-sudoku_5986165_5470970.html
They said the minimum of indices is 17 for a all sudoku grid combinations.

I want to go further in the challenge. Coding a grid reducer to find the minimum of 17 indices ! Who is not affraid ?

Good Luck.

To do this, i will make a function to count numbers of solutions for grid, but not great than 2.

If number of solutions is 0 : The indices are false

If number of solutions is 1 : The grid is good to be playable

If number is 2 (or more) : Indices missing !

I will product a random grid and I will remove each number one by one and test the number of solutions. There should only be 17 left!

[(X)]

En effet, Seldon, eh, scalion...

BingGPT répond:

Le nombre minimum d’indices pour qu’un puzzle Sudoku ait une solution unique est de 17. Cela signifie qu’au moins 17 chiffres doivent être donnés dans le puzzle initial pour qu’il n’ait qu’une seule solution possible. Les puzzles avec 16 indices ou moins doivent avoir plusieurs solutions.

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According to BingGPT:

The minimum number of clues for a Sudoku puzzle to have a unique solution is 17. This means that at least 17 numbers must be given in the initial puzzle for it to have only one possible solution. Puzzles with 16 clues or fewer must have multiple solutions.

[(X)]

Lets start this in another post: "CODE CHALLENGE: Sudoku: Find 17..."

gb32.proboards.com/thread/653/code-challenge-sudoku-find-17