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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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 Posted: Thu Aug 05, 2021 9:57 am Post subject: Aug 05 VH |
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| Code: |
+-------------------+-----------------+-------------------+
| 6 2 4 | 9 5 7 | 8 1 3 |
| 57 8 57 | 4 1 3 | 9 6 2 |
| 1 3 9 | 8 6 2 | 5 7 4 |
+-------------------+-----------------+-------------------+
| 8 a79 567 | 3 27 1 | 4 259 b69 |
| 2 1 6-7 | 5 4 9 |c67 3 8 |
| 59 4 3 | 6 27 8 | 27 59 1 |
+-------------------+-----------------+-------------------+
| 79 6 8 | 27 3 4 | 1 29 5 |
| 3 79 1 | 27 8 5 | 26 4 69 |
| 4 5 2 | 1 9 6 | 3 8 7 |
+-------------------+-----------------+-------------------+
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XY-Wing: (7=9)r4c2 - (9=6)r4c9 - (6=7)r5c7 => - 7r5c3; stte |
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TomC
Joined: 30 Oct 2020
Posts: 358
Location: Wales
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| Posted: Thu Aug 05, 2021 10:24 am Post subject: |
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| I used the very similar 679 XY to remove the 6's in r4c3 and r5c7 |
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TomC
Joined: 30 Oct 2020
Posts: 358
Location: Wales
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| Posted: Thu Aug 05, 2021 12:22 pm Post subject: |
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Another way, r4c8<>2
If you stick to the cells in row 4 and box 6 this would give either two 7's in box 6 or two 7's in row 4 |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Thu Aug 05, 2021 12:29 pm Post subject: |
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| TomC wrote: | Another way, r4c8<>2
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Or....via pincers for 2 at r4c5 and r7c8. |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Thu Aug 05, 2021 12:36 pm Post subject: |
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| (2=9)r7c8-(9=7)r7c1-(7=9)r8c2-(9=7)r4c2-(7=2)r4c5 => r4c8<>2 |
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TomC
Joined: 30 Oct 2020
Posts: 358
Location: Wales
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| Posted: Thu Aug 05, 2021 1:05 pm Post subject: |
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Or using the XY cells abc in Clement's grid
r6c7<> 7 as then no 7's box 5 |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Fri Aug 06, 2021 3:03 pm Post subject: |
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| Yes - Clement's xy wing layout at abc also shows that whatever the value in a(r4c2) the value of c (r5c7) is always 7 and so r6c7 must be <> 7. |
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