Post the puzzle or solving technique that's causing you trouble and someone will help
Sped wrote:
I don't see that as a closed chain. I see it as a regular XY chain starting on r7c6 and meandering to r1c7. It has loose 9s on each end so it allows the elimination of 9s in cells that see both r7c6 and r1c7.. in this case r1c6 and r7c7.

I'd write it like this:

9-(r7c6)-6-(r8c4)-9-(r8c9)-3-(r7c8)-4-(r3c8)-3-(r1c7)-9

I see. It is actually a matter of perspective. The difference in our perspectives is that yours is useful, practical and easy, whereas mine is inflexible and limited. Other than that, they are identical. I had always looked for xy-chains by finding the bivalue cells and then searching for loops that contradicted the pilot cell. If I didn't get a contradiction, then I knew that I would have the opportunity to make some deductions outside of the loop. What your post tells me is that it is more effective to start with a bivalue cell with candidates (XY) and follow a chain until you reach a cell with candidates (ZX). Now make eliminations of X in any cell seeing the pilot cell and the terminal cell. But you don't have to stop there. If you haven't enough eliminations, try extending the chain. If you keep your eyes open, you might find a subchain of use.

This seems to have the advantage of being shorter, more likely to spot deductions in polyvalued cells and, as my example shows, makes it easier to see additional eliminations.

Thanks for the insight!
re'born

Posts: 551
Joined: 31 May 2007

Sped wrote: rep'nA wrote: [...]What I have in mind is the chain
(1,6)9 > (7,6)6 > (8,4)9 > (8,9)3 > (7,8)4 > (3,8)3 > (1,7)9 => (1,6)!9.
[...]

I don't see that as a closed chain. I see it as a regular XY chain starting on r7c6 and meandering to r1c7. It has loose 9s on each end so it allows the elimination of 9s in cells that see both r7c6 and r1c7.. in this case r1c6 and r7c7.

I'd write it like this:

9-(r7c6)-6-(r8c4)-9-(r8c9)-3-(r7c8)-4-(r3c8)-3-(r1c7)-9

Coincidentally, I was explaining the same idea recently and used this picture ...

(in <this post>), recommending to visualise the "end-digits" (the 'a's) as "annihilators" that eliminate any other 'a' visible to both of them. (The picture is relatively free of specialised notation, because many people don't want to bother with it, wanting just to know how to apply xy-chains as a solving-method.)

BTW ... In that post, I erred in explaining the simple underlying logic (it was late! yadayada) -- and ironically, that error inadvertently underscores another point: One can effectively apply xy-chains as a solving-method without even considering why they work (although it's not recommended ).
r.e.s.

Posts: 337
Joined: 31 August 2005

Still Another Solution

Code: Select all
` *--------------------------------------------------------------------*  | 4      6      5      | 2      8      39     | 39     7      1      |  | 139    389    138    | 7      4      369    | 2      5      369    |  | 2      379    37     | 69     1      5      | 3469   34     8      |  |----------------------+----------------------+----------------------|  | 39     389    38     | 4      6      1      | 7      2      5      |  | 167    5      16     | 3      2      78     | 68     9      4      |  | 76     2      4      | 5      9      78     | 368    1      36     |  |----------------------+----------------------+----------------------|  | 8      37     367    | 1      5      69     | 349    34     2      |  | 36     1      2      | 69     7      4      | 5      8      39     |  | 5      4      9      | 8      3      2      | 1      6      7      |  *--------------------------------------------------------------------* `

[r6c7]=3|1=[r5c1]-1-[r5c3]-6-[r7c3]=6=[r7c6]=9=[r8c4]-9-[r8c9]-3-[r6c9](-6-[r6c7])-6-[r5c7]-8-[r6c7],

which implies r6c7<>6,8 => r6c7=3 and the puzzle is solved.

Carcul
Last edited by Carcul on Sat Apr 29, 2006 11:25 am, edited 1 time in total.
Carcul

Posts: 724
Joined: 04 November 2005

A 2-step solution:

1. x-wing in r17c67: r2c6<> 9 and r3c7<>9
Code: Select all
` 4     6     5     | 2     8    *39    |*39    7     1 139   389   138   | 7     4     36    | 2     5     369 2     379   37    |*69    1     5     |*346  *34    8-------------------+-------------------+------------------ 39    389   38    | 4     6     1     | 7     2     5 167   5     16    | 3     2     78    | 68    9     4 76    2     4     | 5     9     78    | 368   1     36-------------------+-------------------+------------------ 8     37    367   | 1     5     69    | 349   34    2 36    1     2     | 69    7     4     | 5     8     39 5     4     9     | 8     3     2     | 1     6     7`

2. Discontinuous ALS loop:

r1c6-9-r3c4-6-(ALS:r3c78=6|9=r1c7)-9-r1c6,

which implies r1c6<>9 and the puzzle is solved.
ronk
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