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VH 111821

 
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immpy



Joined: 06 May 2017
Posts: 479

PostPosted: Thu Nov 18, 2021 8:05 pm    Post subject: VH 111821 Reply with quote

Hello all, enjoy the puzzle...

Code:

+-------+-------+-------+
| . . . | . . . | 1 . 3 |
| . . . | . 9 1 | 8 . . |
| 7 . . | 4 . . | . 2 9 |
+-------+-------+-------+
| . 5 6 | . 2 . | . . . |
| . . . | . 8 . | . . 1 |
| . 3 8 | . 6 . | . . . |
+-------+-------+-------+
| 9 . . | 2 . . | . 1 8 |
| . . . | . 1 3 | 9 . . |
| . . . | . 4 . | 7 . 5 |
+-------+-------+-------+

Play this puzzle online at the Daily Sudoku site

cheers...immp
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Mogulmeister



Joined: 03 May 2007
Posts: 1017

PostPosted: Fri Nov 19, 2021 10:19 am    Post subject: One way Reply with quote

After basics:

Code:
+---------------+--------------+--------------+
| 568 4689 459  | 68  57  2    | 1   47  3    |
| 3   2    45   | 57  9   1    | 8   47  6    |
| 7   68   1    | 4   3   68   | 5   2   9    |
+---------------+--------------+--------------+
| 14  5    6    | 179 2   79   | 3   8   47   |
| 2   79   79   | 3   8   4    | 6   5   1    |
| 14  3    8    | 157 6   57   | 2   9   47   |
+---------------+--------------+--------------+
| 9   67   3    | 2   57  567  | 4   1   8    |
| 58  478  457  | 78  1   3    | 9   6   2    |
| 68  1    2    | 689 4   689  | 7   3   5    |
+---------------+--------------+--------------+


Quote:
Pincers for 7 at r7c2 and r8c4

Eliminates 7 from r8c23 and r7c56
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TomC



Joined: 30 Oct 2020
Posts: 169
Location: Wales

PostPosted: Fri Nov 19, 2021 12:25 pm    Post subject: Reply with quote

I think I was on the same lines as you Mogulmeister, but saw it differently

If r8c4=8 then r7c2=7 leaves no space for a <7> in box 8
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TomC



Joined: 30 Oct 2020
Posts: 169
Location: Wales

PostPosted: Fri Nov 19, 2021 3:26 pm    Post subject: Another way Reply with quote

There is a <68> in r1 and c1, so r1c1<>6 as then no <8> in box 8
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Mogulmeister



Joined: 03 May 2007
Posts: 1017

PostPosted: Fri Nov 19, 2021 10:23 pm    Post subject: Reply with quote

There is a pleasing symmetry about that pattern with a lovely simplicity to it. As you say Tom, a 6 in r1c1 triggers the 8 in both <68> outliers and box 8 is toast for 8s.

As always, you can plug 6 into r1c1 and follow a loop round and it removes itself from r1c1.

+6r1c1-(6=8)r9c1-(8)r9c46=(r8c4)-(8=6)r1c4 so r1c1 < > 6
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TomC



Joined: 30 Oct 2020
Posts: 169
Location: Wales

PostPosted: Sun Nov 21, 2021 9:58 am    Post subject: Reply with quote

I agree, this is a very good looking contradiction

It stems from the pattern of 8's in box 8

First, just give the two <68> pairs in r1 and c1 names, say a and b (not really needed but makes my point cleaner)

Look at box 8 and the position of the 8's in that box. They make an "L" shape which shows that wherever you put an 8 it will remove an 8 in a or b. But crucially, it also shows that a and b cannot both be 8's as this gives the contradiction.

With this information, the <568> in r1c1 pops up to force the contradiction and solve the puzzle.
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Mogulmeister



Joined: 03 May 2007
Posts: 1017

PostPosted: Sun Nov 21, 2021 4:21 pm    Post subject: Reply with quote

Nice.
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