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immpy
Joined: 06 May 2017
Posts: 571
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 Posted: Wed Dec 16, 2020 4:48 pm Post subject: VH+ 121620 |
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| Code: |
+-------+-------+-------+
| . 1 . | . . . | 2 . . |
| . . 7 | . . . | 5 . 4 |
| 6 5 . | . 9 7 | . 1 . |
+-------+-------+-------+
| . . 9 | 4 8 6 | . . . |
| . . 5 | 1 2 3 | 9 . . |
| . . . | 9 7 5 | 1 . . |
+-------+-------+-------+
| . 7 . | 6 4 . | 3 8 1 |
| 4 . 8 | . . . | 6 . . |
| . . 1 | . . . | . 2 . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Enjoy the puzzle, cheers...immp |
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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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| Posted: Wed Dec 16, 2020 8:53 pm Post subject: VH + 121620 |
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| Code: |
+-------------------+-----------------+-------------------+
|c8-9 1 3 | 58 a56 4 | 2 a679 a679 |
|c289 d289 7 | 38 36 1 | 5 6-9 4 |
| 6 5 4 | 2 9 7 | 8 1 3 |
+-------------------+-----------------+-------------------+
| 1 23 9 | 4 8 6 | 7 35 25 |
| 7 48 5 | 1 2 3 | 9 46 68 |
| 238 2348 6 | 9 7 5 | 1 34 28 |
+-------------------+-----------------+-------------------+
| 5 7 2 | 6 4 9 | 3 8 1 |
| 4 39 8 | 357 1 2 | 6 579 579 |
|b39 6 1 | 357 b35 8 | 4 2 579 |
+-------------------+-----------------+-------------------+
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(9=7|65)r1c985 - (5=39)r9c51 - r12c1 = (9)r2c2 => - 9r2c9,r1c1; stte |
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TomC
Joined: 30 Oct 2020
Posts: 358
Location: Wales
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| Posted: Thu Dec 17, 2020 8:43 am Post subject: |
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Sorry I have great trouble with notation
You can eliminate 5's in r1c5 and r89c4
using r1c1 89, r1c9 39, r1c4 58 and r9c5 35
maybe a XYZ wing not sure
Tom |
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dongrave
Joined: 06 Mar 2014
Posts: 568
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| Posted: Thu Dec 17, 2020 12:57 pm Post subject: |
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Mine looks a lot like like the other solutions.
(9=3)r9c1 - (3=5)r9c5 - (5=6)r1c5 - (6=7|9)r1c89 => r1c1 <> 9. |
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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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| Posted: Thu Dec 17, 2020 5:02 pm Post subject: |
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| TomC wrote: | Sorry I have great trouble with notation
You can eliminate 5's in r1c5 and r89c4
using r1c1 89, r1c9 39, r1c4 58 and r9c5 35
maybe a XYZ wing not sure
Tom |
Very Good solution Tom. There are many notations k9, box position(bp)etc, for me the notation which I find better is the Eureka Notation, it shows what is happening in the cell, e.g to write candidate 5 in row 9 column 5 is (5)r9c5 or for a bi value cell (8=5)r1c4 means 5 is the solution in r1c4. Your solution is an XY Alternating Inference Chain XY-AIC and can be written in Eureka Notation as follows:
(5=3)r9c5 - (3=9)r9c1 - (9=8)r1c1 - (8=5)r1c4 => - 5r1c5,r89c4; stte(singles to the end). When 5 is not in r9c5 you end with 5 in r1c4
The assumption in the first termin the chain is that 3 is the solution in r9c5, it is eliminated - in r9c1, 9 remains (3=9)r9c1 and so on. You can read more on this notation in Sudopedia: Eureka notation(I like it myself). |
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TomC
Joined: 30 Oct 2020
Posts: 358
Location: Wales
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| Posted: Thu Dec 17, 2020 8:12 pm Post subject: |
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Thanks Clement,
Will have to do some reading up on this.
My logic didn't start with r9c5, but used the 58 in r1c4 as a sort of pivot.
Another way is look at it is r1c4 <>8 because you would have 3 in r9c1 and 3 in r2c4 this would lead to no 3's in box 8
Tom |
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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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| Posted: Fri Dec 18, 2020 6:05 am Post subject: |
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| TomC wrote: | Thanks Clement,
Will have to do some reading up on this.
My logic didn't start with r9c5, but used the 58 in r1c4 as a sort of pivot.
Another way is look at it is r1c4 <>8 because you would have 3 in r9c1 and 3 in r2c4 this would lead to no 3's in box 8
Tom |
I forgot to mention that Chains are reversible, you can start with r1c4 and get the same solution:
(5=8)r1c4 - (8=9)r1c1 - (9=3)r9c1 - (3=5)r9c5 => - 5r1c5,r89c4.
If 5 is not in r1c4, it will be in r9c5. |
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immpy
Joined: 06 May 2017
Posts: 571
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| Posted: Fri Dec 18, 2020 4:12 pm Post subject: |
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Yes, lots of chain material in this one.
cheers...immp |
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