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Carcul

Joined: 29 Dec 2005
Posts: 50
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Location: Coimbra, Portugal

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PostPosted: Sun Feb 12, 2006 9:47 pm    Post subject: Reply with quote

I have a complicated solution for this puzzle. Let's consider the grid after the basic steps:

Code:
 3      16     1456   | 245    8      125    | 25     9      7     
 589    2      158    | 6      7      1359   | 4      18     138   
 45789  179    14578  | 23459  3459   12359  | 25     6      138   
----------------------+----------------------+----------------------
 789    4      378    | 1      359    3589   | 39     2      6     
 89     5      368    | 2389   369    23689  | 1      7      4     
 1      69     2      | 7      369    4      | 389    5      89     
----------------------+----------------------+----------------------
 2      17     1457   | 34589  34569  35689  | 789    148    189   
 45     8      9      | 45     1      7      | 6      3      2     
 6      3      147    | 489    2      89     | 789    148    5     

and we have:

[r7c5](=6=[r7c6]=3=[r7c4]-3-[r3c4]){-4-[r8c4](-5-[r3c4])=4=[r8c1]=5=[r7c3]-5-[r2c3|r2c8]-1,8-[r2c9]-3-[r3c9]}-4-[r7c8]=4=[r9c8]=1=[r9c3]-1-[r2c3]-8-[r2c8]-1-[r3c9]-8-[r6c9](-9-[r4c7]-3-[r4c6])-9-[r6c2]{-6-[r1c2](-1-[r1c6])=6=[r1c3]=4=[r1c4]-4-[r3c4]}=9=[r3c2]-9-[r2c1]=9=[r2c6](-9-[r3c4])(-9-[r4c6])-9-[r9c6]-8-[r4c6]-5-[r1c6]-2-[r3c4].

This means that, if r7c5=4 then r3c4 would have no candidate. So, r7c5<>4 and that solve the puzzle.

Carcul
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tarek

Joined: 31 Dec 2005
Posts: 153
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Location: London, UK

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PostPosted: Tue Feb 14, 2006 4:52 pm    Post subject: Reply with quote

I had to use a discontinuous simple double implication chain only once.

Code:
*-----------------------------------------------------------------*
| 3      16     46    | 24     8      125   | 25     9      7     |
| 59     2      18    | 6      7      59    | 4      18     3     |
| 579    179    478   | 2349   3459   12359 | 25     6      18    |
|---------------------+---------------------+---------------------|
| 789    4      37    | 1      359    358   | 39     2      6     |
| 89     5      36    | 2389   369    236   | 1      7      4     |
| 1      69     2     | 7      369    4     | 389    5      89    |
|---------------------+---------------------+---------------------|
| 2      17     5     | 3489   346    3689  | 789    48     189   |
| 4      8      9     | 5      1      7     | 6      3      2     |
| 6      3      17    | 489    2      89    | 789    148    5     |
*-----------------------------------------------------------------*
Candidates in r3c9 will force r4c6 to have only 58 as valid Candidates (Level 1 Poly Implication chains)
r3c9=1: r3c9=1 => r2c8=8 => r2c3=1 => r9c3=7 => r4c3=3 => r4c6<>3 => r4c6=58
r3c9=8: r3c9=8 => r6c9=9 => r4c7=3 => r4c6<>3 => r4c6=58
Threfore r4c6=58


Tarek
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foxglove

Joined: 04 Feb 2006
Posts: 42
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Location: Portugal

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PostPosted: Thu Feb 16, 2006 1:14 am    Post subject: me too, I have a solution Reply with quote

#9 has this to say:



humans beware!

After the basic steps and some not so basic we have:
Code:

3   .   .   .   8   .   .   9   7
.   2   .   6   7   .   4   .   .
.   .   .   .   .   .   .   6   .
.   4   .   1   .   .   .   2   6
.   5   .   .   .   .   1   7   4
1   .   2   7   .   4   .   5   .
2   .   5   .   .   .   .   .   .
4   8   9   5   1   7   6   3   2
6   3   .   .   2   .   .   .   5


then:
remove the 6 @ 7/5



One little step...

(in these pictures a blue background is for a candidate that we suppose IS NOT, over the Yellow circles the candidate IS supposed. They form a chain of implications. Connected to the victim, the one surrounded by a blue circle, is a IS NOT. If he WAS our victim would be dead on the spot. That IS NOT will imply (=>) another candidate connected to the victim to be a IS (yellow). Pof! Dead!)


An X-wing removes a few 1s

Then a fine structure that I don't know what to call (Two Turbots having sex?) that removes the 8 @ 5/6 and then does this:



A single Turbot passes by and gobbles a 9 @ 7/5

A quaternion of zwitterions reacts with a pair of 8's


Another one gobbles 2's and places a 2:


Finally, the coup the grace:
A Hammer™ squashes the 1@ 3/6 and that will result in an avalanche of singles.


All this was cherry picked from the solutions given by #9.

If anyone can put a more meaningful name on the patterns I would be very grateful.
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