## 3D Medusa Help on a specific puzzle

Post the puzzle or solving technique that's causing you trouble and someone will help

### Re: 3D Medusa Help on a specific puzzle

Also, there's no need to use a "grouped candidate" extension.
From this point already, both ' and " can see /.

Code: Select all
` *-------------------*-------------------*-------------------* | 2"7'  29    <8>   | 4     6     <1>   | 27"9  5     <3>   |  | 1"23/ 6     <5>   | 3\9/  <8>   <7>   | 129   19    <4>   |  | 1'7"  3\49  49    | <5>   <2>   3/9\  | <8>   1"7'  <6>   |  *-------------------*-------------------*-------------------* | 235   <1>   29    | 3/7'9 4     25    | <6>   37"9  <8>   |  | 235   2389  <6>   | 18    37"9  25    | <4>   1379  7'9"  |  | <4>   389   <7>   | 18    39    6     | 19    <2>   5     |  *-------------------*-------------------*-------------------* | <6>   5     <3>   | 7"9'  <1>   <8>   | 7'9"  4     2     |  | <9>   27    12    | <6>   <5>   4     | <3>   8     17    |  | <8>   47    14    | <2>   3/7'9 3\9/  | <5>   6     179'  |  *-------------------*-------------------*-------------------*`

1" can see 3/ in r2c1, and there are several places where ' can see /.
7' sees 3/ in r4c4, 7' sees 3/ in r9c5, 9' sees 9/ in c4, and 9' sees 9/ in r9.
blue

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### Re: 3D Medusa Help on a specific puzzle

Oops! An AIC with consecutive weak links! I got it wrong and I won't make excuses.

Thanks for alerting me, to this and also the other contradictions between the two colour pairs.
David P Bird
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### Re: 3D Medusa Help on a specific puzzle

[Edit: withdrawn.]
Last edited by daj95376 on Mon May 13, 2013 11:56 am, edited 1 time in total.
daj95376
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### Re: 3D Medusa Help on a specific puzzle

Hi Danny,

daj95376 wrote:
JasonLion wrote:(...)

In simple 3D Medusa (one pair of colors) all links are conjugate pairs, thus all links are both strong and weak.

(...)

(...)

I marked a statement above in red because I don't believe it is completely true. The definition for Y-Cycle does not require a strong link to exist between a value in two bivalue cells that see each other. Thus, the following weak inference might not be a strong link on <7> in [r2]:

Code: Select all
`  G B         G B (4=7)r2c2 - (7=5)r2c8`

... and still be included in a 3D Medusa solution.

I can see how someone might include a fragment like that in an AIC, if it makes for a shorter AIC.
In general, though, as long as the BG cluster has been constructed using the "only conjugate links" rule, then between B and any G candidate, an odd length chain of conjugate pairs will exist -- possibly several such chains -- and between any two "same color" candidates, an even length chain will exist.

Added: in your example above, the weak link between 7r2c2 and 7r2c8, would not have been used in constructing the BG-cluster.

Best Regards,
Blue.
blue

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### Re: 3D Medusa Help on a specific puzzle

[Edit: withdrawn.]
Last edited by daj95376 on Mon May 13, 2013 11:56 am, edited 1 time in total.
daj95376
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### Re: 3D Medusa Help on a specific puzzle

daj95376, Bob Hanson does say all that, but I don't believe that the implication that you can use true Y-Chains in a 3D Medusa actually works out. Y-Chains depend on maintaining an alternating strong - weak - strong - weak pattern. But the X-links break the pattern and allow you to get out of sync. For example you could end up with strong - weak - strong - strong - weak - strong, which doesn't work logically. It would work some of the time, depending on which order the links actually occurred in, but the simple color alternation rule no longer guarantees that it will work.

I have never seen Bob Hanson give an exact step by step procedure that makes sense. He says one thing, and then gives examples that do something slightly different. It really does seem like he understood what he was doing, the examples all work. But the explanations leave something to be desired. Rudd is the first person I can find to write it out as a set of rules that can be reliably followed. In Rudd's version there aren't true Y-Chains, no weak links in a group with a single pair of colors, and yet Rudd's rules allow you to solve all of the examples that Bob Hanson shows in the same way that Bob Hanson solves them.

JasonLion
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