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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Wed Oct 11, 2006 11:02 pm Post subject: Nightmare, September 20 |
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This is my second attempt and the second is just as big a mess as the first. What am I missing? All I've been able to do is a next-to-useless Finned X-Wing on "6."
| Code: | ----------------------------------------------------------
|2459 17 245 |357 8 1257 |1247 23479 6 |
|259 3 6 |57 125 4 |127 8 12579 |
|245 17 8 |9 12356 12567 |1247 2347 12457 |
----------------------------------------------------------
|23568 268 235 |1 4 2567 |9 267 278 |
|1 246 7 |8 269 269 |3 5 24 |
|2458 2468 9 |567 256 3 |12478 2467 12478 |
----------------------------------------------------------
|468 5 1 |2 69 689 |478 479 3 |
|7 24689 234 |346 1369 1689 |5 249 2489 |
|2348 2489 234 |345 7 589 |6 1 2489 |
---------------------------------------------------------- |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Thu Oct 12, 2006 3:35 am Post subject: Go Tigers! |
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Marty,
Sudoku Susser says you need chains:
5 x Comprehensive Forcing Chains
3 x Simple Forcing Chains
1 x Comprehensive Naked Sets
1 x Unique Rectangles and Loops
3 x Intersection Removal
2 x Simple Hidden Sets
2 x Simple Naked Sets
9 x Pinned Squares
Go Tigers!
Keith |
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Myth Jellies
Joined: 27 Jun 2006
Posts: 64
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| Posted: Thu Oct 12, 2006 8:36 am Post subject: |
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| Code: | *-----------*
|...|.8.|..6|
|.36|..4|.8.|
|..8|9..|...|
|---+---+---|
|...|14.|9..|
|1.7|8..|.5.|
|..9|..3|...|
|---+---+---|
|.51|2..|..3|
|7..|...|5..|
|...|.7.|61.|
*-----------*
*--------------------------------------------------------------------*
| 2459 17 245 | 357 8 1257 | 1247 23479 6 |
| 259 3 6 | 57 125 4 | 127 8 12579 |
| 245 17 8 | 9 12356 12567 | 1247 2347 12457 |
|----------------------+----------------------+----------------------|
|-23568 *268 235 |*1 4 2567 | 9 *267 278 |
| 1 #246 7 | 8 269 269 | 3 5 24 |
|-24568 *2468 9 |*567 256 3 | 12478 *2467 12478 |
|----------------------+----------------------+----------------------|
| 468 5 1 | 2 69 689 | 478 479 3 |
| 7 *24689 234 |*346 1369 1689 | 5 *249 2489 |
| 2348 2489 234 | 345 7 589 | 6 1 2489 |
*--------------------------------------------------------------------*
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Marty, you just stopped a tad too soon. The finned swordfish (r468c248 fin in r5c2) on the sixes enables you to solve for the 6 in r7c1. From there it slowly reduces via basics to...
| Code: | *--------------------------------------------------------------------*
| 59 17 24 | 35 8 127 |-1247 39 6 |
| 259 3 6 |%57 125 4 |%127 %8 #579 |
|*245 17 8 | 9 12356 1267 |-1247 -347 *457 |
|----------------------+----------------------+----------------------|
| 3 8 5 | 1 4 267 | 9 267 27 |
|*1 -246 7 | 8 26 9 | 3 5 *24 |
|#24 246 9 |%567 256 3 |%8 %2467 1 |
|----------------------+----------------------+----------------------|
| 6 5 1 |%2 9 8 |%47 %47 3 |
| 7 249 234 | 346 136 16 | 5 29 8 |
| 8 249 234 | 34 7 5 | 6 1 29 |
*--------------------------------------------------------------------*
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Then the fours (*) have an r35c19 finned (r6c1) sashimi x-wing.
And the sevens (%) have an r267c478 finned (r2c9) swordfish.
Now it is time for a little multidigit coloring. Start with the bivalue cell in r1c1 and give each digit the conjugate colors, 'B' and 'b'. Extend those conjugate colors as far as possible and you end up with something like this.
| Code: | *----------------------------------------------------------------------*
| 5B9b 17 24 |*3B5b 8 127 | 124 3b9B 6 |
| 259B 3 6 |*57 125 4 |*127b 8 579b |
| 245 17 8 | 9 123b56 1267 | 124 3B4b 57 |
|-----------------------+-----------------------+----------------------|
| 3 8 5 | 1 4 267 | 9 267 2B7b |
| 1 26 7 | 8 26 9 | 3 5 4 |
| 24 246 9 | 567 256 3 | 8 267 1 |
|-----------------------+-----------------------+----------------------|
| 6 5 1 | 2 9 8 | 4b7B 4B7b 3 |
| 7 249B 234 | 346 13B6 16 | 5 2B9b 8 |
| 8 249b 234 | 34 7 5 | 6 1 2b9B |
*----------------------------------------------------------------------*
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Note that color 'b' inhibits both the five and the seven from r2c4, therefore color 'b' cannot be true.
One could also find the following related AIC
(7=5)r2c4 - (5=3)r1c4 - (3)r1c8 = (3-4)r3c8 = (4)r7c8 - (4=7)r7c7 =>
chain implies that r2c4 = 7, r7c7 = 7, or both. Since r2c7 sees both r2c4 and r7c7, it cannot contain a 7, and the puzzle is unlocked.
[edited to fix the swordfish marked with %] |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Oct 12, 2006 4:47 pm Post subject: |
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Keith/MJ,
Both of you used some unfamiliar terms which I'll have to research. Thanks for taking the time to help.
Marty |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Thu Oct 12, 2006 7:06 pm Post subject: |
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Hi Marty,
suppose you are more familiar with these methods:
| Code: | *--------------------------------------------------------------------*
| 2459 17 245 | 357 8 1257 | 1247 23479 6 |
| 259 3 6 | 57 125 4 | 127 8 12579 |
| 245 17 8 | 9 12356 12567 | 1247 2347 12457 |
|----------------------+----------------------+----------------------|
| 23568 268 235 | 1 4 2567 | 9 267 278 |
| 1 246 7 | 8 269 269 | 3 5 24 |
| 2458 2468 9 | 567 256 3 | 12478 2467 12478 |
|----------------------+----------------------+----------------------|
| 468 5 1 | 2 69 689 | 478 479 3 |
| 7 24689 234 | 346 1369 1689 | 5 249 2489 |
| 2348 2489 234 | 345 7 589 | 6 1 2489 |
*--------------------------------------------------------------------*
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With 3 strong links you can eliminate 6 from r4c1:
r7c1=r8c2-r8c4=r6c4-r6c8=r4c8
(if 6 not in r7c1 then in r8c2,r6c4 and r4c8, i.e. either r7c1 or r4c8 is 6 => r4c1<>6)
=> r7c1=6
With quadruple and turbot fish (2 strong links) in 4 you come here:
| Code: | *--------------------------------------------------------------------*
| 59 17 24 | 35 8 127 | 1247 39 6 |
| 259 3 6 | 57 125 4 | 127 8 579 |
| 245 17 8 | 9 12356 1267 | 1247 347 57 |
|----------------------+----------------------+----------------------|
| 3 8 5 | 1 4 267 | 9 267 27 |
| 1 26 7 | 8 26 9 | 3 5 4 |
| 24 246 9 | 567 256 3 | 8 267 1 |
|----------------------+----------------------+----------------------|
| 6 5 1 | 2 9 8 | 47 47 3 |
| 7 249 234 | 346 136 16 | 5 29 8 |
| 8 249 234 | 34 7 5 | 6 1 29 |
*--------------------------------------------------------------------*
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r3c1=5 => r1c1=9 => r2c9=9 => r3c9=5
=>r3c9<>5
Then with a pair and triple you get r2c57=12 and a turbot fish in 7, which solves the puzzle. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Oct 12, 2006 10:52 pm Post subject: |
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| Quote: | With 3 strong links you can eliminate 6 from r4c1:
r7c1=r8c2-r8c4=r6c4-r6c8=r4c8
(if 6 not in r7c1 then in r8c2,r6c4 and r4c8, i.e. either r7c1 or r4c8 is 6 => r4c1<>6)
=> r7c1=6 |
Ravel,
I don't understand the notation, but I can follow along and see where r4c8=6 if r7c1 doesn't. However, I don't see where three strong links come into play.
| Quote: | | With quadruple and turbot fish (2 strong links) in 4 you come here: |
I'm afraid I don't understand this either.
Sorry. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Fri Oct 13, 2006 7:35 am Post subject: |
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Marty,
see here a good description for strong links.
The notation r7c1=r8c2-r8c4=r6c4-r6c8=r4c8 is my own (therefore i explained the logic), "=" is for a strong link (a conjugated pair for the candidate), "-" for a "weak" one, i.e. the two cells share a unit (row, column, box).
So you have: if 6 ist not in r7c1, then it must be in r8c2, then it cannot be in r8c4, then it must be in r6c4 and so on.
If you have such alternating strong and weak links (which also could be strong ones), starting and ending with a strong one, then you can eliminate the number from cells, that see both the first and the last cell in the chain.
This is nothing else than advanced coloring (if all links are strong it is simple coloring), also called x-chains, i.e. simple forcing chains only using one number, nothing difficult for you, so forgive me, when the notation confused you.
You can write it as:
r7c1=6 => r6c1<>6
r7c1<>6 => r8c2=6 => ....=> r4c8=6 => r6c1<>6
| Quote: | | With quadruple and turbot fish (2 strong links) in 4 you come here |
You can find 2 quads (if i remember right) then and can place the 4 in r5c9 after using 2 strong links in 4. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Oct 13, 2006 4:33 pm Post subject: |
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| Quote: | You can write it as:
r7c1=6 => r6c1<>6
r7c1<>6 => r8c2=6 => ....=> r4c8=6 => r6c1<>6 |
This type of logic is something I'm well familiar with.
"Advanced coloring" is a new term to me, presumably something different from "multi-coloring", a technique that I've had trouble grasping. So advanced is a form of simple coloring except for the alternating weak and strong links. But the basic principle appears to be the same; a single chain with an even number of cells so the starting and ending cells are of opposite polarity, so to speak.
I'll add this to the repertoire. Thanks. |
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