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immpy
Joined: 06 May 2017
Posts: 571
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 Posted: Tue Jul 27, 2021 2:22 pm Post subject: VH+ 072721 |
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Hello all, enjoy the puzzle...
| Code: |
+-------+-------+-------+
| 6 . . | . . . | 7 . . |
| . 8 . | . 3 . | 6 . . |
| . 1 . | . . 4 | . 5 9 |
+-------+-------+-------+
| 8 . 5 | . . . | 4 . . |
| . . . | . 2 . | . 9 . |
| . . 9 | . . . | . . . |
+-------+-------+-------+
| 5 . 2 | 4 . 3 | . . . |
| 3 . . | . . . | 9 7 . |
| . 9 8 | . . 2 | . . 3 |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
cheers...immp |
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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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| Posted: Tue Jul 27, 2021 4:35 pm Post subject: VH+072721 |
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| Code: |
+---------------+---------------------+--------------------+
| 6 5 34 | 129 18 19 | 7 1238 148 |
| 9 8 47 | 1257 3 157 | 6 12 14 |
| 2 1 37 | 67* 8-67* 4 | 38 5 9 |
+---------------+---------------------+--------------------+
| 8 2 5 | 1379 17 1679 | 4 136 167 |
| 4 36 1 | 3578 2 5678 | 38 9 5678 |
| 7 36 9 | 1358 4 1568 | 2 1368 1568 |
+---------------+---------------------+--------------------+
| 5 7 2 | 4 9 3 | 1 68 68 |
| 3 46 46 | 18 5 18 | 9 7 2 |
| 1 9 8 | 67* 67* 2 | 5 4 3 |
+---------------+---------------------+--------------------+
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Type 1 UR(67)r39c45, r3c5<>67; stte |
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TomC
Joined: 30 Oct 2020
Posts: 358
Location: Wales
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| Posted: Tue Jul 27, 2021 6:26 pm Post subject: |
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As an alternative to the 67 UR
r2c8<> 1 as if you stick to rows 2 and 3 you have a chain 1 - 4 - 7 - 3 - 8 which means r1c9 has nothing in it |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Fri Jul 30, 2021 3:42 pm Post subject: |
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That's a decent contradiction.
(1)r2c8-(1=4)r8c9-(4=7)r2c3-(7=3)r3c3-(3=8)r3c7-(8=14)r12c9-(1)r2c8 |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Tue Aug 03, 2021 1:37 pm Post subject: |
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As a post script there is another way using an ALS (Almost Locked Set) or to be accurate an AALS (Almost Almost Locked Set).
We have two ALS interacting:
Green {1,3,6,8} in (n-1=) 3 squares
Beige {1,2,3,4,8} in (n-1=) 4 squares
The restricted common should be 3 with the common candidate being 1 which can be removed from r2c8 because it can "see" all the other 1's in both ALS.
There's a wrinkle though.
The restricted common definition is bust because there's another 3 in r3c7. Otherwise it would fit and all would be well.
So logically we have a situation:
Either the ALS xz is true or the 3 is true in r3c7.
IF ALS is true then r2c8 <>1
If 3 is true (3)r3c7-(3=7)r3c3-(7=4)r2c3-(4=1)r2c9-r2c8 so r2c8 <>1
Makes the same elimination so all good. |
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