#591, April 27 - May 3, 2024: The Weekly 'Unsolvable' Sudoku Puzzle
by Richard Kröger, guest compiler
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WARNING: This is the Weekly 'Unsolvable' Sudoku, rated above 'Extreme'.
This is currently unsolvable by my solver, except perhaps with trial and error strategies.
Using the solver will not help you. (much).
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Discussion...
Post an idea here...
... 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
Archive
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