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		| immpy 
 
 
 Joined: 06 May 2017
 Posts: 574
 
 
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				|  Posted: Wed Dec 22, 2021 6:34 pm    Post subject: VH+ 122221 |   |  
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				| Hello all, enjoy the puzzle. Merry Christmas! 
 
  	  | Code: |  	  | +-------+-------+-------+
 | . 6 1 | 2 5 . | . . . |
 | . 9 7 | . 8 6 | . . . |
 | 3 . . | . . 7 | 9 6 . |
 +-------+-------+-------+
 | . . 9 | 8 2 . | . . . |
 | . 5 . | 6 3 1 | . 9 . |
 | . . . | . 9 4 | 5 . . |
 +-------+-------+-------+
 | . 7 3 | . . . | . . 9 |
 | . . . | . 6 . | 1 3 . |
 | . . . | . 7 3 | 4 5 . |
 +-------+-------+-------+
 
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 Play this puzzle online at the Daily Sudoku site
 
 cheers...immp
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Thu Dec 23, 2021 7:15 am    Post subject: |   |  
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				| Thanks Immp and Merry Christmas to you ! 
 I will leave after basics to others and white out my solution. Decided on a less conventional one.
 
 
  	  | Quote: |  	  | It's the march of the <28> pairs. Quite a few on the grid after basics and many can be linked. 
 What's interesting is being able to link the <28> pincer pairs in r5c9 with r8c1. The whole "bridge" consists only of six <28> pairs. What is nice is that they toggle each other so that if r5c9 is 2 then r9c1 must be 8 and of course vice versa.
 
 This means a double elimination of 2 and 8 at r5c1 leaving only 7 in that square and stte.
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 [Corrected x 2 - Diolch Tomos !]
 
 Last edited by Mogulmeister on Thu Dec 23, 2021 10:09 pm; edited 5 times in total
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		| TomC 
 
 
 Joined: 30 Oct 2020
 Posts: 359
 Location: Wales
 
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				|  Posted: Thu Dec 23, 2021 9:37 am    Post subject: |   |  
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				|  	  | Mogulmeister wrote: |  	  | Thanks Immp and Merry Christmas to you ! 
 I will leave after basics to others and white out my solution. Decided on a less conventional one.
 
 
  	  | Quote: |  	  | It's the march of the <28> pairs. Quite a few on the grid after basics and many can be linked. 
 What's interesting is being able to link the <28> pincer pairs in r5c9 with r8c1. The whole "bridge" consists only of six <28> pairs. What is nice is that they toggle each other so that if r5c9 is 2 then r9c1 must be 8 and of course vice versa.
 
 This means a double elimination of 2 and 8 at r1c5 leaving only 7 in that square and stte.
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 Very similar solution, started off with a <14> UR to then leave with a lot of <28>'s
 
 But I think you mean r5c1 in your post!
 
 Merry Christmas to all
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Thu Dec 23, 2021 11:46 am    Post subject: |   |  
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				| I did mean r5c1! Not r1c5. Too much Christmas cheer I suspect! The UR also comes first. |  | 
	
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		| TomC 
 
 
 Joined: 30 Oct 2020
 Posts: 359
 Location: Wales
 
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				|  Posted: Thu Dec 23, 2021 7:30 pm    Post subject: |   |  
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				| No problem, I found that r5c1<> 2 and this solves it, but didn't notice that it could also be not 8 
 Tomos
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Thu Dec 23, 2021 10:09 pm    Post subject: |   |  
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				| I’ve made the second correction ! Thanks! |  | 
	
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