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| daj95376
| Joined: 05 Feb 2006 | Posts: 349 | : | | Items |
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Posted: Sun Sep 17, 2006 5:09 pm Post subject: Coloring a Double Implication Chain |
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Code: | ..8..4.1...3.9....7..1.......75.2.642..6.9..886.4.72.......5..7....6.8...3.7..4..
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Below is the partially solved puzzle. Although it's not necessary to eliminate <5> in [r1c1] to easily complete this puzzle, I'd like to discuss that elimination anyway. Skipping over Forcing Chains, there's a Double Implication Chain on <5> in [c7] that will perform the elimination. I'd like to consider the DIC in terms of coloring using the following labels.
Code: | B: Blue
G: Green
b: not Blue
g: not Green
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If I label cell [r1c7] Blue and (conjugate) cell [r5c7] Green, then this is as far as I can proceed with simple Coloring on <5>. However, there are now four cells that can be viewed as (g) -- not Green. This leaves [r6c3] in [b4] uncolored for <5>, and so I contend that it must be treated as Green. This then forces two additional cells in [c3] to be viewed as (g) -- not Green. This leaves [r9c1] in [b7] uncolored for <5>, and so I contend that it must be treated as Green. At this point, [r1c7] being Blue and [r9c1] being Green forces the desired elimination.
Code: | *-----------------------------------------------------------*
| 69-5 59 8 | 2 7 4 | 35B 1 3569 |
| 15 125 3 | 8 9 6 | 7 4 25 |
| 7 249 46 | 1 5 3 | 69 8 269 |
|-------------------+-------------------+-------------------|
| 3 19 7 | 5 8 2 | 19 6 4 |
| 2 145g 145 g | 6 13 9 | 35G 7 8 |
| 8 6 159 G | 4 13 7 | 2 359 g 1359g |
|-------------------+-------------------+-------------------|
| 19 8 1269 | 39 4 5 | 16 239 7 |
| 4 7 259 g | 39 6 1 | 8 2359 359 |
| 569G 3 1569g | 7 2 8 | 4 59 16 |
*-----------------------------------------------------------*
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Would someone please save me a lot of scanning through the forum and tell me if this coloring operation has a name. Also, if I were to include this elimination in a listing for solving this puzzle, then should it be referenced as DIC or as (???) Coloring?
Essentially, I'm wondering if a DIC that only tracks a single value should be viewed as Coloring. TIA!!! |
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| Steve
| Joined: 12 Apr 2006 | Posts: 12 | : | | Items |
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| Ruud Site Admin
| Joined: 17 Sep 2005 | Posts: 708 | : | Location: Netherlands | Items |
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Posted: Sun Sep 17, 2006 8:00 pm Post subject: |
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This type of coloring has become very popular lately.
Check this recent topic for detailed information.
Names used by different people are:
- Weak coloring
- X-Colors
- Equivalence Marking
It looks like a double Nishio trick, where you ask yourself the question:
"Are there any candidates that would be eliminated when I place either of these 2 conjugate candidates in this house?"
When you backtrack from the eliminated candidate, you can use it as a single Nishio starting location.
Ruud _________________ Meet me at sudocue.net |
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| rkral
| Joined: 21 Oct 2005 | Posts: 233 | : | | Items |
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Posted: Mon Sep 18, 2006 12:37 pm Post subject: Re: Coloring a Double Implication Chain |
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daj95376 wrote: | Would someone please save me a lot of scanning through the forum and tell me if this coloring operation has a name. Also, if I were to include this elimination in a listing for solving this puzzle, then should it be referenced as DIC or as (???) Coloring?
Essentially, I'm wondering if a DIC that only tracks a single value should be viewed as Coloring. TIA!!! |
"Coloring" ... and I would refer to your specific example as "Grouped Simple Coloring" because ...
- Most people familiar with coloring already know that 'Simple Coloring' refers to a single chain of conjugate links.
- The adjective 'Grouped' conveys the addition to the technique better than 'X' or 'Equivalence' or even 'Weak'.
I think you are confusing the technique name with the logical expression(s) of the deduction. Your specific example may be expressed with the nice loop:
r1c1-5-r1c7=5=r5c7-5-r5c23=5=r6c3-5-r89c3=5=r9c1-5-r1c1
... which has (as one possible set of) double implication chains (DICs) ...
r6c2=5 -> r89c3!=5 -> r9c1=5 -> r1c1!=5, and
r6c2!=5 -> r5c23=5 -> r5c7!=5 -> r1c7=5 -> r1c1!=5
... but many different techniques result in DICs.
P.S. Apologies for using "!=" instead of "<>" but this site is treating "<>" as if it were HTML. |
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| daj95376
| Joined: 05 Feb 2006 | Posts: 349 | : | | Items |
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Posted: Mon Sep 18, 2006 4:37 pm Post subject: |
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Thanks steve, Ruud, and rkral for replying to my query!
rkral: Yes, the elimination I described can be obtained in several ways. I'm not comfortable with nice loops and need to rectify that shortcoming. My solver uses Templates and it's always popping out eliminations that I quickly recognize as simple DICs, so that's how I reference them. Since my Templates are based on a single value, in this case <5>, I started viewing the DICs as an extended form of Coloring. That's why I started this thread.
BTW, the DIC that I spotted for this puzzle is the one I showed as colored:
[r1c7]=5 => [r1c1]!=5
[r5c7]=5 => [r6c3]=5 => [r9c1]=5 => [r1c1]!=5 |
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| TheSpaniard
| Joined: 28 Aug 2006 | Posts: 20 | : | Location: Spain | Items |
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Posted: Fri Sep 22, 2006 9:22 am Post subject: |
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To daj95376:
Ruud had mentioned that you can eliminate the "5" in r1c1 using X-Colors.
Steve and rkral have mention different ways of eliminating it too.
I will complete Ruud's comment, showing how X-Colors easily solves this situation:
Step 1: r1c7-Blue and r5c7-Green.
Step 2 (the "Single colors" stuff): No new cells coloured, as they are not other conjugate cells in "5's".
Step 3 (the "X-Ray" stuff):
a) r6c3-Green (is the only cell in its box not been a peer of r5c7, that is Green, so r6c3 is Green).
b) r9c1-Green (is the only cell in its box not been a peer of r6c3, that is Green, so r9c1 is Green).
Step 4.1) You can eliminate r1c1's candidate 5, due to the fact that it is peer of two cells (r1c7 and r9c1) coloured with two different colors.
Note that X-Colors overrides in a single and well defined technique both "Single Colors" and "Multicolors", and solves some other situations (like the one you proposed) that cannot be solved nor with Colors neither with Multicolors, needing other kind of more complicated techniques (Weak Coloring, Grouped Coloring, MC+Hinge, Implication chains, Nishio subsets, etc).
It is not clear until now (at least it is not clear for me) if X-Colors can completely override some or all of this different techniques; I have never found any situation that can be solved with any of these techniques and not with X-Colors (the contradiction rule), but I have not proven that rule yet (for instance, that WeakColoring(x,z) implies X-Colors(x,z)) in a more mathematical way.
Best Regards. |
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| Carcul
| Joined: 29 Dec 2005 | Posts: 50 | : | Location: Coimbra, Portugal | Items |
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Posted: Mon Sep 25, 2006 5:46 pm Post subject: |
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Code: | *-----------------------------------------------------*
| 569 59 8 | 2 7 4 | 35 1 3569 |
| 15 125 3 | 8 9 6 | 7 4 25 |
| 7 249 469 | 1 5 3 | 69 8 269 |
|------------------+---------------+------------------|
| 3 19 7 | 5 8 2 | 19 6 4 |
| 2 145 145 | 6 13 9 | 35 7 8 |
| 8 6 159 | 4 13 7 | 2 359 1359 |
|------------------+---------------+------------------|
| 169 8 1269 | 39 4 5 | 16 239 7 |
| 4 7 259 | 39 6 1 | 8 2359 359 |
| 1569 3 1569 | 7 2 8 | 4 59 16 |
*-----------------------------------------------------*
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r9c3=1 or r2c2=2. But
[r9c3]=1|2=[r2c2](-2-[r3c2])-2-[r2c9](-5-[r8c9])-5-[r1c7]=5=[r5c7]-5-
-[r6c89]=5=[r6c3](-5-[r9c3])=9=[r4c2](-9-[r3c2]-4-[r3c3])-9-[r4c7]-1-
-[r6c9|r7c7]={Unique Pattern: r6c89|r7c48|r8c49}=2=[r7c8]-2-[r8c8]-
-{Unique Rectangle: r68c89}-5-[r9c8]-{TILA: r9c38|r37c7|r3c3}.
So, r9c3=1 which solves the puzzle.
Carcul |
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