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| Carcul
| Joined: 29 Dec 2005 | Posts: 50 | : | Location: Coimbra, Portugal | Items |
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Posted: Sun Feb 12, 2006 9:47 pm Post subject: |
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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
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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 | : | Location: London, UK | Items |
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Posted: Tue Feb 14, 2006 4:52 pm Post subject: |
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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 | : | Location: Portugal | Items |
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Posted: Thu Feb 16, 2006 1:14 am Post subject: me too, I have a solution |
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#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
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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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