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immpy
Joined: 06 May 2017
Posts: 574
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 Posted: Fri Oct 13, 2023 6:21 pm Post subject: VH++ 101323 |
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Hello all, enjoy the puzzle.
| Code: |
+-------+-------+-------+
| 3 2 . | . . . | . . 4 |
| 6 4 . | . . 8 | 7 5 . |
| . . 7 | . . . | 3 . . |
+-------+-------+-------+
| . . . | 6 . 5 | . . . |
| . . . | 1 2 9 | 5 . 6 |
| . . . | 7 . 4 | . . . |
+-------+-------+-------+
| . . 4 | . . . | 9 . . |
| 7 5 6 | . . 3 | 1 2 8 |
| 9 8 . | . . . | . . 5 |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
cheers...immp |
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dongrave
Joined: 06 Mar 2014
Posts: 572
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| Posted: Sat Oct 14, 2023 2:23 pm Post subject: |
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Thanks for the puzzle immp!
After basics: | Code: | +-----------------+---------------+-----------------+
| 3 2 c58 | d59 1679 167 | 68 19 4 |
| 6 4 f19 | 3 e19 8 | 7 5 2 |
| 58 g19 7 | 45 46 2 | 3 68 19 |
+-----------------+---------------+-----------------+
| 148 379 2389 | 6 38 5 | 248 134789 1379 |
| 48 37 b38 | 1 2 9 | 5 3478 6 |
| 158 6 23589 | 7 38 4 | 28 1389 139 |
+-----------------+---------------+-----------------+
| 2 3-1 4 | 8 5 167 | 9 367 37 |
| 7 5 6 | 49 49 3 | 1 2 8 |
| 9 8 a13 | 2 167 167 | 46 3467 5 |
+-----------------+---------------+-----------------+
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Then I used the chain (1=3)r9c3-(3=8)r5c3-(8=5)r1c3-(5=9)r1c4-(9=1)r2c5-r2c3=r3c2 => r7c2 <> 1; stte. |
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Clement
Joined: 24 Apr 2006
Posts: 1113
Location: Dar es Salaam Tanzania
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| Posted: Sun Oct 15, 2023 3:25 pm Post subject: VH++101323 |
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| Code: |
+------------------+-----------------+---------------------+
| 3 2 a58 |a59 1679 167 |a68 19 4 |
| 6 4 c19 | 3 b19 8 | 7 5 2 |
| 58 d19 7 | 45 46 2 | 3 68 19 |
+------------------+-----------------+---------------------+
| 148 379 2389 | 6 38 5 | 248 134789 1379 |
| 48 37 38 | 1 2 9 | 5 3478 6 |
| 158 6 259 | 7 38 4 | 28 1389 139 |
+------------------+-----------------+---------------------+
| 2 e13 4 | 8 5 167 | 9 f367 f37 |
| 7 5 6 | 49 49 3 | 1 2 8 |
| 9 8 13 | 2 167 167 |h4-6 g3467 5 |
+------------------+-----------------+---------------------+
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(6=859)r1c734 - (9=1)r2c5 - r2c3 = r3c2 - (1=3)r7c2 - r7c89 = (3-4)r9c8 = (4)r9c7 => -6r9c7; stte |
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glesco
Joined: 12 May 2022
Posts: 39
Location: North Vancouver, BC, Canada
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| Posted: Thu Oct 19, 2023 5:47 pm Post subject: |
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I used a colouring technique.
In r1c3, I set the 5 to Green & the 8 to Blue. As 8 is a strong link in r1 so the 8 in r1c7 is Green with the 6 being Blue. In c2 the 5 is also a strong link so in r6c3 the 5 is set to Blue. There are a lot more cells that can be coloured but it turns out not to be necessary to find the solution.
If Green is the solution and looking at the 28 bivalued cell in r6c7, the 2 would also be Green as the 8 sees a Green 8. That 2 is also, however, part of a strong link in r6 which would imply 2 in r6c3 is Blue. But r6c3 already has a Blue 5! So Green can not be the solution and thus solves for all the cells coloured Blue.
Trust this explanation is relatively clear! |
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