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Oct 22 VH

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ZeroAssoluto

Joined: 05 Feb 2017
Posts: 942
Location: Rimini, Italy

Posted: Sat Oct 22, 2022 9:55 am Post subject: Oct 22 VH

Hi everyone,

Code:

+---------------+-------------+-----------+
| 2 9 45 | 7 34 35 | 8 1 6 |
| 58 1 6 | 259 48 259 | 459 3 7 |
| 578 3 4578 | 569 468 1 | 459 2 59 |
+---------------+-------------+-----------+
| 358 4 1 | 69 2 689 | 59 7 359 |
| 9 6 2 | 3 5 7 | 1 4 8 |
| 3578 57 578 | 4 1 89 | 2 6 359 |
+---------------+-------------+-----------+
| 1 257 57 | 8 37 235 | 6 9 4 |
| 4 257 9 | 256 67 256 | 3 8 1 |
| 6 8 3 | 1 9 4 | 7 5 2 |
+---------------+-------------+-----------+

Play this puzzle online at the Daily Sudoku site

Quote:

XY-Wing 4,5,8 in r1c3,r2c15 and -4 in r1c5

Ciao Gianni

TomC

Joined: 30 Oct 2020
Posts: 358
Location: Wales

Posted: Sat Oct 22, 2022 11:44 am Post subject:

Sometimes eliminating the impossible can solve a puzzle

Remove <4> from r3c3 to solve

Code:

+----------------+----------------+-------------+
| 2 9 45 | 7 34 35 | 8 1 6 |
| 58 1 6 | 259 48 259 | 459 3 7 |
| 578 3 x4578 | 569 468 1 | 459 2 59 |
+----------------+----------------+-------------+
| 358 4 1 | c69 2 689 | 59 7 359 |
| 9 6 2 | 3 5 7 | 1 4 8 |
| 3578 57 a578 | 4 1 b89 | 2 6 359 |
+----------------+----------------+-------------+
| 1 257 57 | 8 37 235 | 6 9 4 |
| 4 257 9 | 256 67 256 | 3 8 1 |
| 6 8 3 | 1 9 4 | 7 5 2 |
+----------------+----------------+-------------+

Quote:

If x=4, a=8, b=9, c=6 leads to a <59> pair in three cells r3c479

Greenvilleguy

Joined: 01 Jul 2022
Posts: 24
Location: US

Posted: Wed Oct 26, 2022 12:40 pm Post subject: Oct 22 VH

I need more help with this one. I don't see why you can eliminate the 4 in R3C3.

dongrave

Joined: 06 Mar 2014
Posts: 568

Posted: Wed Oct 26, 2022 2:28 pm Post subject: Re: Oct 22 VH

Greenvilleguy wrote:

I need more help with this one. I don't see why you can eliminate the 4 in R3C3.

Tom's solution goes as follows:

If r3c3 is 4 then it's not 8 so r6c3 is 8, so r6c6 is not 8 so it's 9, so r4c4 is not 9 so it's 6,
so r3c4 is not 6 so it's 5 or 9, but r3c7 is also 5 or 9 (because it's not 4 because r3c3 is 4), but r3c9 is also 5 or 9!
But we can't have 3 cells in row 3 with just the candidates 5 and 9 in them therefore r3c3 cannot be 4.

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