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| Pep
| Joined: 27 Nov 2005 | Posts: 10 | : | Location: Bath, UK | Items |
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Posted: Sat Dec 03, 2005 8:35 pm Post subject: Medusa chain strength |
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If I understand Medusa chains correctly, there are strong chains where all the implications from one node to the next also work in the reverse direction, and weak ones where they work only one way.
Consider a cell C where r meets c meet, and therein we have either a 9 or not. And suppose that there is one other cell in the box, B, containing C which has a 9, say at D. Then there is a strong inference between C and D for 9.
Now look at another case of a B where D and E both contain 9 as well as C. Both D and E are in the same box and a row, say R. And in columns K,L respectively. And suppose each of R, K and L contain only one other 9.
Now the links between C and D, and between C and E are weak. But the link between the 9 in R and the two cells D and E _together_ is strong, and so is the one through B because the two cells D and E can be treated as a single entity in this case. Either exactly one is a 9 or none is.
Do I understand the situation properly? |
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| Bob Hanson
| Joined: 05 Oct 2005 | Posts: 187 | : | Location: St. Olaf College | Items |
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Posted: Sat Dec 03, 2005 9:17 pm Post subject: |
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OK, I think I'm following you. Yes, I think that's pretty much the idea.
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Consider a cell C where r meets c meet, and therein we have either a 9 or not. And suppose that there is one other cell in the box, B, containing C which has a 9, say at D. Then there is a strong inference between C and D for 9.
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Code: |
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c
B :
\:
..9C.....r..
:\
: 9D
: \
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Yes, that's correct.
Quote: |
Now look at another case of a B where D and E both contain 9 as well as C. Both D and E are in the same box and a row, say R. And in columns K,L respectively. And suppose each of R, K and L contain only one other 9.
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Code: |
K L
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R---9D----9E---
|. .|
| . . |
| 9C |
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9 9
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Quote: |
Now the links between C and D, and between C and E are weak. But the link between the 9 in R and the two cells D and E _together_ is strong, and so is the one through B because the two cells D and E can be treated as a single entity in this case. Either exactly one is a 9 or none is.
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I would NOT recommend thinking about 9D and 9E as one entity. They have different properties with respect to their OTHER candidates in their same cells. One will be TRUE and the other FALSE.
By the way, because 9C has a weak link to both parities of this chain (call them A and a), it can be eliminated.
The Medusa idea focuses on "marks" or "possibilities" rather than "cells."
The strong links are D(F)<==>E(T) and E(F)<==>D(T) and the weak ones are only D(T)==>C(F).
What makes it useful is that a FALSE can drive a TRUE -- that's the value of a strong connection.
My first attempt at discussion of X-cycles, Y-cycles, and the relation to 3D Medusa is at
http://www.stolaf.edu/people/hansonr/sudoku/top95-analysis.htm#medusa _________________ Bob Hanson
Professor of Chemistry
St. Olaf College
Northfield, MN
http://www.stolaf.edu/people/hansonr |
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