... by: Frans Goosens

#591

With trial and error

Combination A1=247 and I9=589

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A1=2 I9=5 Wrong, Undo calculation

All reset to initial position

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A1=2 I9=8 Wrong, Undo calculation

All reset to initial position

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A1=2 I9=9 ( C3=8 ) Solved,

Total solving time is : 36 sec.
Number of logical steps is : 8536

... by: Dieter

Thanks a lot @Serban.
I'll dig into it ;-)

... by: Serban

# 591 G2=6 Basics to 22

1. Box 1 ABC123 n=3 e=6 Nc=16 Ns=16 => B2=1 & C2=4 OR A1=2 & A3=7 solve

2. Box 3 GDJ123 n=5 e=4 Nc=4 Ns=3 => J3=5 OR G3=9 OR H2=2 basics to 34
--2a.Column 2 n=5 e=4 Nc=4 Ns=2 => D2=8 solve
-- Note: G1=7

@ Dieter Column 2
mC =
1 9 8 4 5 7 1 6 2 3 74
2 9 8 4 2 7 1 6 5 3 77
3 9 8 2 4 7 1 6 5 3 79
4 9 2 4 8 7 1 6 5 3 78
5 9 2 8 4 7 1 6 5 3 71
6 9 1 4 8 7 5 6 2 3 81 solve
--- ---
7 9 1 2 8 7 4 6 5 3 78
8 9 1 8 5 7 4 6 2 3 72
9 9 1 8 4 7 5 6 2 3 77
10 9 1 8 2 7 4 6 5 3 79

Solving algorithm
Each unsolved Sudoku game is a 9*9 'S' matrix with 81 partially occupied cells
N cells occupied, E=81-N unoccupied cells
Each solved Sudoku game- N=81 E=0
- each row, column and box contains all the numbers 1 to 9. n=9 e=0
Matrix S contains 36 matrices u with 9 partially, fully or unoccupied cells
( 9-row 1*9,9-column 9*1,9-box 3*3,9-number )
Program=Basic techniques+ X-wing ,X-chain, W-wing, XY-wing , XY-chain , XYZ-wing
The program determines the numbers that can occupy each free cell
In each free cell there are 2 or more numbers
%======================================
1. Choose a partially occupied 'u' matrix n<=7 e=9-n
2. From 'u' we obtain - more-Nc- matrices 'u1' fully occupied n=9 e=0
3. For each case, enter u1 (the fully occupied matrix) in the S matrix.
Result Nc matrices S1 with N+9-n =N+e occupied cells
4. The matrix S1 is -analyzed- solved with the same program, one of the cases Ns <= Nc
being the solution of the game
Note
Pmax=e! - permutations of e
if n=7 e=2 Nc = Pmax= 2
if n=6 e=3 Nc < Pmax= 6
if n=5 e=4 Nc < Pmax= 24

... by: Dieter

@Serban:
Hi ! I had another approach with GHJ 123 > 2 valid matrices only.
As I do not use "technical help" I came to my solution running the two possible matrices manually.
I worked out your one with column 2 getting 9 or 10 variations.
But how did you decide that B2=1 & C2=4 resolves?
Did you try all variations or is there a rule if you have so many possible combinations?
Thanks in advance for your feedback.

... by: Dieter

G2 = 6 with basics
GHJ 123 > 2 valid matrices only
A1 = (2,4) and F3 = (1,3,5)
A1 = 2 and F3 = 1 > solved

... by: Serban

# 591 G2=6 Basics to 22

1. Number 4 n=2 e=7 N=28 Nc=28 Ns=15 => D1=E4=4 solve
Note: G5=4
2. Column 2 n=4 e=5 Nc=10 Ns=6 => B2=1 & C2=4 solve