XY Chains in Eureka

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XY Chains in Eureka

Postby Yogi » Mon Jan 18, 2021 9:37 pm

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XYCHain.png
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If r7c3 <> 8, then r6c6 = 8: If r6c6 <> 8, then r7c3 = 8. So r7c6 <> 8.
How to describe this in Eureka, remembering that in an XY Chain the strong links are within bi-value cells and the links between the cells are all weak links, or acting as weak links? How about:
(8=7)r7c3 – (7=3)r5c3 – (3=7)r6c2 – (7=6)r6c7 – (6=8)r6c6 => -8r7c6
A simple elimination which doesn’t immediately solve the puzzle, but the next one does:
(1=2)r5c4 – (2=7)r5c8 – (7=6)r6c7 – (6=8)r6c6 – (8=1)r6c1 => -1r5c1, r6c4
Any thoughts on the notation?
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Re: XY Chains in Eureka

Postby Leren » Mon Jan 18, 2021 10:11 pm

Code: Select all
*----------------------------------------*
| 9  1   2  | 58  7    58  | 3   6   4   |
| 4  5   38 | 6   39   1   | 789 27  29  |
| 7  38  6  | 4   39   2   | 89  5   1   |
|-----------+--------------+-------------|
| 2  6   9  | 7   5    3   | 1   4   8   |
| 18 4   37 | 12 a68   9   | 5   27 b236 |
| 18 37  5  | 12  4    8-6 | 67  9   236 |
|-----------+--------------+-------------|
| 3  789 78 | 58  2   d568 | 4   1  c69  |
| 6  29  4  | 3   1    7   | 29  8   5   |
| 5  28  1  | 9   8-6  4   | 26  3   7   |
*----------------------------------------*

Skyscraper : (6) r5c5 = r5c9 - r7c9 = (6) r7c6 => - 6 r6c7, r9c5; stte

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Re: XY Chains in Eureka

Postby Yogi » Wed Jan 20, 2021 9:10 pm

Well that was probably a simpler solution than the XY Chains, but I was trying to get some feedback on whether the notation I wrote for the XY Chains described them accurately.
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Re: XY Chains in Eureka

Postby Leren » Wed Jan 20, 2021 10:38 pm

Code: Select all
*----------------------------------------*
| 9   1   2  | 58 7    58  | 3    6  4   |
| 4   5   38 | 6  39   1   | 789  27 29  |
| 7   38  6  | 4  39   2   | 89   5  1   |
|------------+-------------+-------------|
| 2   6   9  | 7  5    3   | 1    4  8   |
|a18  4   37 |b12 6-8  9   | 5   c27 236 |
| 1-8 37  5  | 12 4   e68  |d67   9  236 |
|------------+-------------+-------------|
| 3   789 78 | 58 2    568 | 4    1  69  |
| 6   29  4  | 3  1    7   | 29   8  5   |
| 5   28  1  | 9  68   4   | 26   3  7   |
*----------------------------------------*

XY Chain : (8=1) r5c1 - (1=2) r5c4 - (2=7) r5c8 - (7=6) r6c7 - (6=8) r6c6 => - 8 r5c5, r6c1; stte

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Re: XY Chains in Eureka

Postby jco » Sun Feb 14, 2021 2:53 pm

Hello Yogi,

Yogi wrote:(8=7)r7c3 – (7=3)r5c3 – (3=7)r6c2 – (7=6)r6c7 – (6=8)r6c6 => -8r7c6
A simple elimination which doesn’t immediately solve the puzzle, but the next one does:
(1=2)r5c4 – (2=7)r5c8 – (7=6)r6c7 – (6=8)r6c6 – (8=1)r6c1 => -1r5c1, r6c4
Any thoughts on the notation?


I read from it that if 8 is not in r7c3, then 7 must be there, so we cannot have 7 in r5c3, and so 3 must be there, which in turn forbids 3 in r6c2, thus implying that 7 must be there and so on, leading to r6c6=8. Since r7c6<>8 follows immediately from r7c3=8, the written XY-chain implies that either we have r7c3=8 or r6c6=8, so 8 is false in r7c6. The notation describes the move (and the resulting elimination) and we can follow it looking at the board. The same happens with the other written XY-chain.
Regards,
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Re: XY Chains in Eureka

Postby Yogi » Tue Feb 23, 2021 7:43 am

Thank you jco! You have absolutely confirmed that I am on the right track. The objective was to ensure that the notation reflected the narrative.
A simpler description can jump straight through the forced conclusions. Say we worked the reverse of the first chain, remembering that with strong links we start off with the assumption that the first candidate is false. So if r6c6 is NOT 8, it must be 6. THEN r6c7 must be 7, r6c2 must be 3, r5c3 must be 7 and r7c3 must be 8, proving that r7c6 cannot be 8.
Great!
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