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Dec 17 VH

 
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ZeroAssoluto



Joined: 05 Feb 2017
Posts: 940
Location: Rimini, Italy
PostPosted: Fri Dec 17, 2021 9:44 am    Post subject: Dec 17 VH Reply with quote
Hi everyone,

Code:

+-------+-----------+------------+
| 2 4 7 | 369 1  59 | 358 569 89 |
| 1 5 3 | 69  8  2  | 4   69  7  |
| 6 8 9 | 7   35 4  | 35  1   2  |
+-------+-----------+------------+
| 7 3 1 | 2   6  59 | 58  4   89 |
| 5 9 8 | 4   7  1  | 2   3   6  |
| 4 2 6 | 39  35 8  | 7   59  1  |
+-------+-----------+------------+
| 9 7 5 | 1   2  3  | 6   8   4  |
| 3 1 2 | 8   4  6  | 9   7   5  |
| 8 6 4 | 5   9  7  | 1   2   3  |
+-------+-----------+------------+

Play this puzzle online at the Daily Sudoku site

Quote:
XYZ-Wing 5,6,9 in r1c68,r2c8 and -9 in r1c9
or
Skyscraper with number 5 in
r36c5,r16c8 and -5 in r1c6,r3c7
r3c57,r6c58 and -5 in r1c8,r4c7
r3c57,r4c67 and -5 in r1c6,r6c5
r14c6,r16c8 and -5 in r4c7,r6c5
or
W-Wing
5,9 in r1c6,r6c8 SL with number 9 in r2c48 and -5 in r1c8
3,5 in r3c7,r6c5 SL with number 5 in r4c67 and -3 in r3c5


Ciao Gianni
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TomC



Joined: 30 Oct 2020
Posts: 358
Location: Wales
PostPosted: Fri Dec 17, 2021 10:18 am    Post subject: Reply with quote
I noticed that the position of <3> in box 2 means that r3c7 and r6c4 can't both be <3> as this would cause a contradiction.

But, if r1c8 = 5 then r3c7=3 and also r6c8=9, r6c4=3, so r1c8 <>5
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Mogulmeister



Joined: 03 May 2007
Posts: 1151
PostPosted: Fri Dec 17, 2021 10:06 pm    Post subject: Reply with quote
Code:
+-------+-----------+------------+
| 2 4 7 | 369 1  59a| 358 569e 89|
| 1 5 3 | 69  8  2  | 4   69  7  |
| 6 8 9 | 7   35 4  | 35  1   2  |
+-------+-----------+------------+
| 7 3 1 | 2   6  59b| 58c 4    89|
| 5 9 8 | 4   7  1  | 2   3   6  |
| 4 2 6 | 39  35 8  | 7   59d  1 |
+-------+-----------+------------+
| 9 7 5 | 1   2  3  | 6   8   4  |
| 3 1 2 | 8   4  6  | 9   7   5  |
| 8 6 4 | 5   9  7  | 1   2   3  |
+-------+-----------+------------+


If we look at our chain of 5's running from a to e we can see the following:

1)If we make r1c6 =5 then follow the chain a to e we get r1c8 also = 5 = contradiction so r1c6 < > 5.

Go in reverse from e to a

2)If we make r1c8 = 5 then follow the chain e to a and we get r1c6 also = 5 = contradiction so r1c8 < > 5.

The only legit 5 left on row 1 must be r1c7.
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