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I learned something new ...

 
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Fri Apr 30, 2021 6:35 pm    Post subject: I learned something new ... Reply with quote
Code:
+-------+-------+-------+
| 5 . . | . . 8 | . 4 3 |
| . . . | 5 9 4 | . . . |
| 7 1 . | 3 . . | . . . |
+-------+-------+-------+
| . 7 5 | . 4 . | . . 8 |
| . . 6 | . . . | 1 . . |
| 9 . . | . 8 . | 5 2 . |
+-------+-------+-------+
| . . . | . . 3 | . 8 5 |
| . . . | 4 1 7 | . . . |
| 3 6 . | 8 . . | . . 2 |
+-------+-------+-------+

Keith
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dongrave



Joined: 06 Mar 2014
Posts: 568
PostPosted: Sat May 01, 2021 12:56 pm    Post subject: Reply with quote
After basics:

Code:
+------------+----------+--------+
| 5   29  29 | 1  7  8  | 6 4 3  |
| 6   38  38 | 5  9  4  | 2 7 1  |
| 7   1   4  | 3  26 26 | 8 5 9  |
+------------+----------+--------+
| 12  7   5  | 9  4  12 | 3 6 8  |
| 248 248 6  | 27 3  5  | 1 9 47 |
| 9   34  13 | 67 8  16 | 5 2 47 |
+------------+----------+--------+
| 14  49  19 | 26 26 3  | 7 8 5  |
| 28  5   28 | 4  1  7  | 9 3 6  |
| 3   6   7  | 8  5  9  | 4 1 2  |
+------------+----------+--------+

Then I used the following chain:

(1=2)r4c1 - r8c1 = r8c3 - (2=9)r1c3 - (9=1)r7c3 => r6c3 <> 1; stte.

What did you use to solve it Keith?
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Sat May 01, 2021 6:34 pm    Post subject: Reply with quote
After basics (see dongrave's grid above), this puzzle is a BUG+2.

In a BUG, every unsolved cell has two candidates, and each candidate occurs twice in each house (row, column, or box). A BUG does not have a unique solution: It has zero or two solutions (I believe).

Code:
+-----------------+-------------+-------------+
|  5     29   29  | 1   7   8   | 6   4   3   |
|  6     38   38  | 5   9   4   | 2   7   1   |
|  7     1    4   | 3   26  26  | 8   5   9   |
+-----------------+-------------+-------------+
|  12    7    5   | 9   4   12  | 3   6   8   |
| (2)48 2(4)8 6   | 27  3   5   | 1   9   47  |
|  9     34   13  | 67  8   16  | 5   2   47  |
+-----------------+-------------+-------------+
|  14    49   19  | 26  26  3   | 7   8   5   |
|  28    5    28  | 4   1   7   | 9   3   6   |
|  3     6    7   | 8   5   9   | 4   1   2   |
+-----------------+-------------+-------------+

The "+2" candidates are 2 in R5C1 and 4 in R5C1. You can see this by checking the candidates in C1 and then in C2, and noting then that the BUG condition is satisfied also in R5 and B4 if 2 is eliminated in R5C1 and 4 in R5C2.

Now, the correct statement, in this case, for a single unique solution is: R5C1 is 2 AND/OR R5C2 is 4.

(Here I was wrong. I thought the condition is always AND. In nearly 20 years of solving these things, I have never seen a case where the correct choice is OR.)

In either case, R5C1 is not 4 or R5C2 is not 2, and either of these eliminations solves the puzzle. We find R5C1=2 and R5C2=8.

Keith

(Thanks to Leren and Jamil for correcting me in another forum.)
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ZeroAssoluto



Joined: 05 Feb 2017
Posts: 933
Location: Rimini, Italy
PostPosted: Sun May 02, 2021 12:42 pm    Post subject: Reply with quote
Hi everyone,

WXYZ-Wing 1,2,3,8 in r268c3,r4c1 and -2 in r8c1

Ciao Gianni
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immpy



Joined: 06 May 2017
Posts: 571
PostPosted: Mon May 03, 2021 3:10 pm    Post subject: Reply with quote
Thanks Keith. Great clarification of the rare BUG+2 phenomenon. I had not been able to do much with them until now.

cheers...immp
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immpy



Joined: 06 May 2017
Posts: 571
PostPosted: Mon May 03, 2021 3:16 pm    Post subject: Reply with quote
Question: Will this only work when the BUG cells see each other, or can it be applied in any case?

Thanks in advance for any enlightenment.
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Mon May 03, 2021 7:52 pm    Post subject: Reply with quote
immpy wrote:
Question: Will this only work when the BUG cells see each other, or can it be applied in any case?

Thanks in advance for any enlightenment.

immpy,

The process is to identify the candidates which if eliminated will leave a BUG. That is, every unsolved cell is a bivalue, and each unsolved digit occurs twice in each row, column, and box. Then, one (or more) of those candidates must be true.

How that plays out I think depends on the particular puzzle. I'll try to find more examples.

Keith
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immpy



Joined: 06 May 2017
Posts: 571
PostPosted: Tue May 04, 2021 2:51 pm    Post subject: Reply with quote
I got it Keith. And it would depend on the particular puzzle. Thanks.

cheers...immp
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