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Dec 14 VH

 
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Clement



Joined: 24 Apr 2006
Posts: 969
Location: Dar es Salaam Tanzania

PostPosted: Thu Dec 13, 2018 9:47 pm    Post subject: Dec 14 VH Reply with quote

Code:

+-----------+-----------+---------+
| 4 8   3   | 19  6   7 | 5  2 19 |
| 5 169 16  | 129 12  8 | 7  4 3  |
| 7 19  2   | 4   3   5 | 1-96 8  |
+-----------+-----------+---------+
| 1 25  4   | 7   25  3 | 8  9 6  |
| 8 25  9   | 12  125 6 | 3  7 4  |
| 6 3   7   | 8   4   9 | 2  1 5  |
+-----------+-----------+---------+
| 9 167 168 | 5   78  4 | 16 3 2  |
| 2 4   68  | 3   78  1 | 69 5 79 |
| 3 17  5   | 6   9   2 | 4  8 17 |
+-----------+-----------+---------+

W-Wing 19 r3c2& r1c9 with SL on 1 in r9c29 => - 9r3c7; stte


Last edited by Clement on Thu Dec 13, 2018 9:52 pm; edited 1 time in total
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ZeroAssoluto



Joined: 05 Feb 2017
Posts: 360
Location: Rimini, Italy

PostPosted: Fri Dec 14, 2018 9:18 am    Post subject: Reply with quote

Hi everyone,

Skyscraper with number 1 in r3c27,r9c29 and -1 in r1c9,r7c7
or
Skyscraper with number 1 in r27c3,r37c7 and -1 in r3c2

Ciao Gianni
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Earl



Joined: 30 May 2007
Posts: 606
Location: Victoria, KS

PostPosted: Fri Dec 14, 2018 3:57 pm    Post subject: Dec14 VH Reply with quote

167 xyzwing @R7C2, -1R7C3.

Early Earl
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TerenceF



Joined: 21 Dec 2007
Posts: 26
Location: Takapuna, NZ

PostPosted: Sun Dec 16, 2018 10:11 am    Post subject: Re: Dec 14 VH Reply with quote

Is it a valid move to observe that if r5c4 is 1 then there is no longer a unique solution because there would be a 25 option in r4c25 and r5c25?

Hence r5c4 is 2?


Clement wrote:
Code:

+-----------+-----------+---------+
| 4 8   3   | 19  6   7 | 5  2 19 |
| 5 169 16  | 129 12  8 | 7  4 3  |
| 7 19  2   | 4   3   5 | 1-96 8  |
+-----------+-----------+---------+
| 1 25  4   | 7   25  3 | 8  9 6  |
| 8 25  9   | 12  125 6 | 3  7 4  |
| 6 3   7   | 8   4   9 | 2  1 5  |
+-----------+-----------+---------+
| 9 167 168 | 5   78  4 | 16 3 2  |
| 2 4   68  | 3   78  1 | 69 5 79 |
| 3 17  5   | 6   9   2 | 4  8 17 |
+-----------+-----------+---------+

W-Wing 19 r3c2& r1c9 with SL on 1 in r9c29 => - 9r3c7; stte
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ZeroAssoluto



Joined: 05 Feb 2017
Posts: 360
Location: Rimini, Italy

PostPosted: Sun Dec 16, 2018 2:06 pm    Post subject: Reply with quote

r5c4 must be 1 or the schema would not have a single solution.
If you have number 2 and 5 in r45c25 at the same time you have a double solution.

Ciao Gianni
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immpy



Joined: 06 May 2017
Posts: 231

PostPosted: Wed Dec 19, 2018 1:43 am    Post subject: Reply with quote

This is a fine example of a Uniqueness Test Type 1, sometimes referred to as a Unique Rectangle. A valid Sudoku Puzzle can have only ONE correct solution, so we can key on the appearance of (25) in r45c25 with an extra candidate of (1) in r5c5. Terence is correct in his assessment. The extra candidate, providing there is only ONE of such, in ALL cases of a Type 1 Uniqueness Test MUST be placed in its particular cell in order to avoid multiple solutions to the puzzle. This surely assumes one is attempting to solve a TRULY valid puzzle. I was exposed to many invalid puzzles in the past before learning to only accept puzzles that were from reliable websites (such as this one) and/or sources.

I would gently suggest to Terence a clearer description of the logic...to avoid the confusion of including the bi-value (12) cell @r5c4 in the explanation of this. It would be clearer this way:

Uniqueness Test 1: 2/5 in r4c25,r5c25 =>r5c5<>2, r5c5<>5(=1!).

One could also state simply that r5c5=1, which is true in this instance. But I have come across many cases where there were two (rarely more) extra candidates in the one cell of the Uniqueness Test 1, meaning one would have to be content with the eliminations only, at this stage.

cheers all...immp
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