fiendish puzzle from sudukozone

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fiendish puzzle from sudukozone

Postby steffo72 » Wed Feb 08, 2006 9:55 am

5 9 68 | 1 48 7 | 346 348 2
3 2 68 | 568 9 456 | 46 1 7
4 1 7 | 268 3 26 | 5 9 68
-----------------|--------------------------|-----------------------
2 367 19 | 35689 58 13569 | 367 357 4
16 346 5 | 2368 7 12346 | 9 38 68
8 3467 49 | 3569 45 34569 | 2 357 1
-----------------|------------------------- |------------------------
69 46 2 | 379 1 39 | 8 47 5
19 8 14 | 579 6 59 | 47 2 3
7 5 3 | 4 2 8 | 1 6 9


Can someone please give me a hint on how to solve this puzzle.
I'm fairly new at this so maybe there's some rule that applies that I don't know about.
The original puzzle can be found at www.sudukozone.com
Click at fiendish
Last edited by steffo72 on Thu Feb 09, 2006 6:46 am, edited 1 time in total.
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Postby Wolfgang » Wed Feb 08, 2006 10:31 am

Hi Steffo,

one possibility is to use an x-chain (or multicoloring) for the number 4 over the cells r1c5,r6c5,r6c3,r8c3,r8c7,r7c8, which says that one of the cells r1c5 or r7c8 must be 4 and allows you to eliminate 4 from r1c8. [edited typo r9c8]
To verify it, you can suppose r1c5 <> 4 or r9c8 <> 4 and walk through the chain.
Last edited by Wolfgang on Wed Feb 08, 2006 6:43 am, edited 1 time in total.
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Postby vidarino » Wed Feb 08, 2006 10:33 am

If you use a plain text editor (notepad) to lay out your puzzle, and stick it between [ code ] and [ /code ] tags when posting, it will look much nicer:
Code: Select all
      5      9     68  |      1     48      7  |    346    348      2
      3      2     68  |    568      9    456  |     46      1      7
      4      1      7  |    268      3     26  |      5      9     68
-----------------------+-----------------------+-----------------------
      2    367     19  |  35689     58  13569  |    367    357      4
     16    346      5  |   2368      7  12346  |      9     38     68
      8   3467     49  |   3569     45  34569  |      2    357      1
-----------------------+-----------------------+-----------------------
     69     46      2  |    379      1     39  |      8     47      5
     19      8     14  |    579      6     59  |     47      2      3
      7      5      3  |      4      2      8  |      1      6      9


First, you can do a couple of eliminations in the middle box due to some Locked Candidates. (E.g. the 8s in box 6 are confined to the middle row (row 5), so therefore there can be no 8 in the rest of that row.)

After those it gets a bit trickier, though. I have one solver that spots and X-Cycle in 4s which solves the puzzle. The other solves beats around the bush a bit with a couple of Nice Loops before solving the rest with singles.

I can post the solutions if you want, but perhaps you'd like to try to find them yourself first? That's the fun part, right?:)
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type-o

Postby steffo72 » Wed Feb 08, 2006 12:31 pm

Vidarino:
first of all i'm sorry, but i missed one 8 in r4c8 so the elimination of other eights don't apply:(
about the x-cycle you're talking about, i'm not really familiar with it so maybe you could "walk me through it" so i can get the hang of it and use it again later?
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Postby Wolfgang » Wed Feb 08, 2006 12:59 pm

On the second glance: at least on paper an xy-chain should be easier to spot:
if r2c7=6->r3c9=8->r5c9=6->r5c1=1->r8c1=9->r8c3=4
so either r2c7=4 or r8c3=4 and you can eliminate 4 fom r8c7.

To the x-chain again. It says
if r1c5<>4->r6c5=4->r6c3<>4->r8c3=4->r8c7<>4->r7c8=4
if r7c8<>4->r8c7=4->r8c3<>4->r6c3=4->r6c5<>4->r1c5=4
so either r1c5 or r7c8 must be 4
Note that for the deduction "A=4 -> B<>4" you can have more 4's in the unit of the cells A and B, but for "A<>4 -> B=4" A and B must be a conjugated pair, ie the only cells that contain a 4.

PS: see here for x-cycles. You get an x-cycle, when you add r1c8 (at the beginning and end) to the x-chain cells above.
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Postby tarek » Wed Feb 08, 2006 3:15 pm

Code: Select all
*-----------------------------------------------------------------*
| 5      9      68    | 1     *48     7     | 346   *348    2     |
| 3      2      68    | 568    9      456   | 46     1      7     |
| 4      1      7     | 268    3      26    | 5      9      68    |
|---------------------+---------------------+---------------------|
| 2      367    19    | 35689  58     13569 | 367    3578   4     |
| 16    #346    5     | 2368   7      12346 | 9      38     68    |
| 8     *3467  -49    | 3569  *45     34569 | 2      357    1     |
|---------------------+---------------------+---------------------|
| 69    *46     2     | 379    1      39    | 8     *47     5     |
| 19     8      14    | 579    6      59    | 47     2      3     |
| 7      5      3     | 4      2      8     | 1      6      9     |
*-----------------------------------------------------------------*
Eliminating 4 From r6c3 (Finned Swordfish in Columns 258)


Should seal the deal

Tarek
Last edited by tarek on Wed Feb 08, 2006 12:09 pm, edited 1 time in total.
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Postby TKiel » Wed Feb 08, 2006 3:41 pm

I'm a little bit confused. When I entered the puzzle as posted, Simple Sudoku indicated that it was asymmetrical but valid. If 8 is entered in r4c8 (which I understood from Steffo72's post at 1:31p.m. was part of the initial clue set) then SS says it's invalid with no solution. (The website listed as the source of the puzzle seems to be currently down as I tried to go there several times without success.)

The puzzle as origionally posted is also solvable by colouring on 4's with two conjugate chains, which use many of the same cells as Wolfgang's x-cycle, though the links between the cells in his x-cycle are not as apparent to me as the links in the chains I used. Must read up on x-cycle.

Tracy
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Postby tarek » Wed Feb 08, 2006 4:08 pm

I've just noticed the post about the missing 8 in r4c8, my post will be edited to include it, however it won't affect the outcome.
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Postby Crazy Girl » Wed Feb 08, 2006 10:17 pm

Tracy,

The Website stated above is spelt wrong, so it should read http://www.sudukozone.com/fiendish-suduko-puzzles.htm where you can find the fiendish puzzle.
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thanx

Postby steffo72 » Thu Feb 09, 2006 11:53 am

thanx everyone for the help. now "all" i have to do is sit down and figure out exactly how these rules work. it's all a bit mind-boggling at the moment, but it'll stick eventually i suppose. but then again, part of the fun is getting the ol' brain to shift up a gear now and again, right?
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