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The Formal definition of Unified Colors Technique
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Mike Barker

Joined: 03 Sep 2006
Posts: 6
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PostPosted: Tue Sep 05, 2006 7:13 pm    Post subject: Reply with quote

Three Strong Links: r3c4=3=r3c8-3-r8c8=3=r9c7-3-r9c5=3=r1c5-3-r3c4 => r12c8<>3,r9c2<>3
Code:
+----------------+--------------+---------------+
|  369  239    5 |    4 G39  12 |  368 -1389  7 |
|  369    4  239 |    8   7  12 |   36  -139  5 |
|    1    8    7 |  B39   5   6 |    4   G39  2 |
+----------------+--------------+---------------+
|    2   39   39 |    7   1   8 |    5     6  4 |
|    8    1    6 |    5   2   4 |    9     7  3 |
|    7    5    4 |   69  69   3 |   28    28  1 |
+----------------+--------------+---------------+
|   39    7  239 |   23   8   5 |    1     4  6 |
|    5  236    1 |  236   4   9 |    7   B23  8 |
|    4 -236    8 |    1 B36   7 |  G23     5  9 |
+----------------+--------------+---------------+
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Mike Barker

Joined: 03 Sep 2006
Posts: 6
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PostPosted: Tue Sep 05, 2006 7:24 pm    Post subject: Reply with quote

Code:
+-----------------+--------------------+-------------------+
|    2   7      6 |     9    3458 -358 |  345    348     1 |
|    8  45    459 |    24   B2345    1 |    7      6   G39 |
|  459   3      1 |    46    4568    7 |  459     48     2 |
+-----------------+--------------------+-------------------+
|  345   9  34578 |   147    4578    2 |   34   1347     6 |
|  345  12  34578 |  1467  456789  589 |  349  12347  3789 |
|    6  12    478 |     3    4789   89 |   12      5   789 |
+-----------------+--------------------+-------------------+
|    7  45   G345 |     8      12    6 |   12      9   B35 |
|  B39   6      2 |     5      17  G39 |    8     17     4 |
|    1   8    359 |    27     -39    4 |    6    237   357 |
+-----------------+--------------------+-------------------+

Three Strong Links: r3c4=3=r3c8-3-r8c8=3=r9c7-3-r9c5=3=r1c5-3-r3c4 => r12c8<>3,r9c2~=3.
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Mike Barker

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PostPosted: Tue Sep 05, 2006 7:25 pm    Post subject: Reply with quote

This puzzle can be solved with an Empty Column Rectangle: r29c9|r2c5 => r9c5~=3. Four submissions for one post - I don't know why.
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daj95376

Joined: 05 Feb 2006
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PostPosted: Tue Sep 05, 2006 8:03 pm    Post subject: Reply with quote

[Addendum to my previous entry] Actually, this could be considered on-topic because the DIC uses only the value <2> ... and I would think that a UCT should be able to handle the following relationship. In fact, I have this nagging feeling that I've run into this pattern in other puzzles that I was solving manually.

Code:
*--------------------------------------*
| .   .   .  | .   2g  .  | .   .   .  |
| .   .   .  | .   2g  .  | .   .   .  |
| .   .   .  | 2b  .   2b | .   2G  .  |
|------------+------------+------------|
| .   .   .  | .   .   .  | .   .   .  |
| .   .   .  | .   .   .  | .   .   .  |
| .   .   .  | .   .   .  | .   .   .  |
|------------+------------+------------|
| .   .   .  | .   .   .  | .   .   .  |
| .   .   .  | .   -2  .  | .   2B  .  |
| .   .   .  | .   .   .  | .   .   .  |
*--------------------------------------*


BTW: Am I the only person whose having trouble posting since the HTML/BBcode was updated? It's a nightmare at times!!!
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TheSpaniard

Joined: 28 Aug 2006
Posts: 20
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PostPosted: Tue Sep 05, 2006 9:55 pm    Post subject: Reply with quote

Hello to you all.

I think most of the doubts have been responded by Angus (thank you, AJ) and others.

Yes, as you have realised, you can extend the so-called weak links green to green and blue to blue when you apply Step 3 of the procedure, not green to blue, that can be only made in Step 2. Step 2 is in fact what you make in Simple Coloring.

But there are an exception: When both cells that form a conjugated pair have been coloured INDEPENDENTLY with opposite colors, then this conjugated pair of cells FORMS PART of the main chain of conjugates, and in this case (and only in this case) you can go back to step 2.

Cheers to everybody. Buenas noches.
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Starflyer Alien

Joined: 05 Sep 2006
Posts: 7
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Location: Rhyl, UK

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PostPosted: Wed Sep 06, 2006 8:32 am    Post subject: Reply with quote

daj95376 wrote:
Is there anything in the UCT to explain [r8c5]<>2 ???


Here's a non-UCT method... Embarassed
Code:
[r8c5]-2-[r8c8]=2=[r3c8]-2-[r3c46]=2=[r12c5]-2-[r8c5] => [r8c5]<>2

... I know it's not what you asked, but I didn't want a restless night to go totally to waste! Wink
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TheSpaniard

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Location: Spain

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PostPosted: Wed Sep 06, 2006 10:05 am    Post subject: Reply with quote

Good day to everybody.

To Mike Barber:

In the puzzle you posted, you can solve it using X-Colors as defined in the beginning of the post.

Looking at candidate "3", you can apply "Standard" Colors (that is, Steps 1 and 2 of the X-Colors Method). Then, you have:
R8C6-Blue, R1C6-Green, R9C5-Green, R8C1-Green.

Applying from this position Step 3 (the X-Ray technique), you will find that
R2C9 is Green (is the only cell in its box not been a peer of R1C6), and
R7C9 is Green (is the only cell in its box not been a peer of R9C5).

It originates two green cells in column 9, so, applying Step 4.2, you can conclude that BLUE CELLS ARE TRUE. R8C6 is therefore a "3".

The rest of the puzzle from here is obvious.

Saludos.
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TheSpaniard

Joined: 28 Aug 2006
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Location: Spain

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PostPosted: Wed Sep 06, 2006 10:41 am    Post subject: Reply with quote

Mike: I referred in my previous post to the Second Puzzle you posted, sorry for not mention it.

The first one is quite the same: Looking at "3's", after Steps 1,2:
R1C5-Green, R3C4-Blue, R3C8-Green, R9C5-Blue.

Applying Step 3 to this position, then R8C8 is Blue (it's the only cell of its box not been a peer of R9C5-Blue). Applying Step 4.1, you can then eliminate the victims: R1C8 and R2C8, as they both are peers (in column 8) with R3C8-Green and R8C8-Blue.

In this position, beginning again the complete process (well, in fact it is not needed, it is to formally use the method), you can reach with Steps 1-2 to a situation in which you can eliminate the candidate "3" in R9C2.

Then you will easily find an XY-Wing that eliminates "2" in R7C3, and the rest is easy.

You can, if you want, distinguish between what is commonly named Strong and Weak Conjugates, but, with a little bit of practice, you won't need it. Believe me.

And, BTW did you realize that X-Colors generalises too X-Wing?
That is, X-Wing(x,z) implies X-Colors(x,z). In any position in which X-Wing will find a solution, X-Colors will find the same solution, too.

Should anyone of us prove the same with, let's say, Swordfish? JellyFish?
Finned Fish? Whatever Fish? Maybe it needs to add some more steps to the definition, but it seems to be a good thing, Don't you think so??

Saludos de nuevo.
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