| View previous topic :: View next topic |
| Author |
Message |
immpy
Joined: 06 May 2017
Posts: 571
|
 Posted: Wed Dec 22, 2021 6:34 pm Post subject: VH+ 122221 |
 |
|
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 . |
+-------+-------+-------+
|
Play this puzzle online at the Daily Sudoku site
cheers...immp |
|
| Back to top |
|
 |
Mogulmeister
Joined: 03 May 2007
Posts: 1151
|
| Posted: Thu Dec 23, 2021 7:15 am Post subject: |
|
|
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. |
[Corrected x 2 - Diolch Tomos !]
Last edited by Mogulmeister on Thu Dec 23, 2021 10:09 pm; edited 5 times in total |
|
| Back to top |
|
|
TomC
Joined: 30 Oct 2020
Posts: 358
Location: Wales
|
| Posted: Thu Dec 23, 2021 9:37 am Post subject: |
|
|
| 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. |
|
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 |
|
| Back to top |
|
|
Mogulmeister
Joined: 03 May 2007
Posts: 1151
|
| Posted: Thu Dec 23, 2021 11:46 am Post subject: |
|
|
| I did mean r5c1! Not r1c5. Too much Christmas cheer I suspect! The UR also comes first. |
|
| Back to top |
|
|
TomC
Joined: 30 Oct 2020
Posts: 358
Location: Wales
|
| Posted: Thu Dec 23, 2021 7:30 pm Post subject: |
|
|
No problem, I found that r5c1<> 2 and this solves it, but didn't notice that it could also be not 8
Tomos |
|
| Back to top |
|
|
Mogulmeister
Joined: 03 May 2007
Posts: 1151
|
| Posted: Thu Dec 23, 2021 10:09 pm Post subject: |
|
|
| I’ve made the second correction ! Thanks! |
|
| Back to top |
|
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
