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pdxsudo
Rookie

Joined: 05 Dec 2006
Posts: 6

 Posted: Thu Feb 22, 2007 10:16 pm    Post subject: XY-Chain bonus? The number "5" in R3C7 and R5C5 are in a XY-chain pattern. But I cannot eliminate any other "5"s in cells 2 and 6. However, there are only two "5"s in each of cell 3 and 5, that means both R3C7 and R5C5 must be "5"! Once I put these two 5s down in R3C7 and R5C5, the puzzle was solved. Is this a valid solving technique?
Ruud
Site Owner

Joined: 30 Dec 2005
Posts: 601

 Posted: Fri Feb 23, 2007 7:49 pm    Post subject: [brainfart deleted] Ruud Note to self: Think before you post a reply!Last edited by Ruud on Fri Feb 23, 2007 10:17 pm; edited 1 time in total
pdxsudo
Rookie

Joined: 05 Dec 2006
Posts: 6

 Posted: Fri Feb 23, 2007 8:45 pm    Post subject: Oops, I meant Box 2 and Box 6. In your example, R2C7's 7 is blue in color. Where did it chain to? I used my XY-chain short cut and it solved all the Unfair puzzles in SudoCue easily. Very handy.
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Fri Feb 23, 2007 9:11 pm    Post subject: Hi Ruud I am not sure how this xy-chain works but i can't seem to be able to recreate it. If you start in R2C7 and say R2C7 = 7(dark blue) it doesn't conclude that R3C8 = 8 and so don't finish the chain to R6C6 = 5. And starting anywhere else R3C7 = 2(light blue) doesn't lead to R2C7 = 5. If i am incorrect could you show me the logic behind these conclusions? Just solved the puzzle and that 5 is actually correct. So it shouldn't be eliminated. Maybe i see xy-chains incorrectly but bivalue-cells don't automatically colour differently in an xy-chain, right? greetings Para p.s. extending would eliminate the 2 in R1C9. From R6C6 = 5 -->> R1C6 = 2. 2's on both end see R1C9.
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Fri Feb 23, 2007 9:40 pm    Post subject: Hi Pdx Could you explain what you mean exactly? I don't really (or really don't) understand your logic. It just seems a bit like a coincidence, but if it is correct it should be very handy. Because it seems like you are using the fact that you can't eliminate anything with these digits as a fact that this would mean that these numbers can't be eliminated and should be correct. greetings Para
pdxsudo
Rookie

Joined: 05 Dec 2006
Posts: 6

 Posted: Fri Feb 23, 2007 10:11 pm    Post subject: I don't quite understand this either. But it always work for me when the XY-chain happens in this configuration: 1. There are only 2 possible cells in the beginning and ending X-Y chain boxes. In this example, both box 3 and box 5 have only 2 cells that contain the number "5". 2. In the above mentioned 2 cells in that box, each cell has only 2 numbers in it. In this example: Box3: R2C7 = 5,7 R3C7 = 5,8 Box5: R5C5 = 1,5 R6C6 = 1,5 3. Start the X-Y chain from the suspect number and end the chain in the suspect number. In this case, I start from the number "5" in R3C7 and end at R5C5. What I don't understand is... If you look at Ruud's diagram. The "5" in R3C7 is blue and R5C5 is green. But to solve this puzzle, both cells have to be "5" even though one is blue, the other is green. I guess there is something I don't fully understand about XY chain. I am as confused as ever. That is why I'm posting this question here. I'm thinking, it could be just a fluke that it happens... I'll do more testing!
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Sat Feb 24, 2007 2:28 am    Post subject: Hi Well the idea of the xy-chain is that you work from a certain bi-value cell and see what the implications are in other bi-value cells. This way you get 2 chains of which you know 1 has to be true. So if you get for example a 2 in each chain any cell that sees both these 2's can't contain a 2. This doesn't mean though that if one chain is true, the other has to be completely false because the 2 chains don't have to use the same cells. In this example, say you start from R5C8. A 4 in R5C8 excludes the 5 in R5C5 and places the 5 in R3C7. A 6 in R5C8 places the 5 in R5C5 but doesn't exclude the 5 in R3C7 so when there is a 6 in R5C8 there could still be a 5 in R3C7. These two 5's will have a different colour when you colour from R5C8. But as i said before this doesn't mean that of these coloured chains one is completely true and the other completely false. Here an xy-chain elimination would be for example. R3C8 is either 2 or 8. So R3C2 = 8 or R3C2 = 2 -->> R8C8 = 4 -->> R5C8 = 6 -->> R4C9 = 1 -->> R6C9 = 8 Now both chains have an 8 and both these 8's see R1C9 so the 8 can be eliminated from R1C9. Which then gives you a hidden single 8 in R1C4. For your theory i don't know if it is a fluke or a logic step. But we'll hear back from you and if you got more examples i'd like to see cause if it works it seems pretty handy. greetings Para
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