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Simple Sudoku and Colors
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angusj
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PostPosted: Tue Aug 30, 2005 3:14 am    Post subject: Re: Coloring Reply with quote

lennyh wrote:
Would you suggest that your helpful explanation in todays posting regarding multicolors001.ss is an example or particular case of either rule 1 or rule 2, or would you rather suggest that it is a separate rule -- perhaps specific to the multi-color consideration?

Yes, it's a separate multi-color rule, not covered by simple colors.
It's not on the hints page as I'm concerned these rather esoteric techniques are likely to overwhelm and deter people from enjoying sudoku (and also why SS currently doesn't make these puzzles in the released version). I've just included a few multi-colors (& xy-wings) in the install package to tempt the most adventurous.
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lennyh

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PostPosted: Tue Aug 30, 2005 7:28 am    Post subject: Reply with quote

Ah yes, this makes sense. Thank you.
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Nick70

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PostPosted: Tue Aug 30, 2005 9:20 am    Post subject: Reply with quote

In the above problem, also note there is a Turbot Fish in r3c4, r3c7, r2c8, r5c8, r5c4 that allows to remove candidate 7 from r3c4 and r5c8.
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kog

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PostPosted: Tue Aug 30, 2005 1:27 pm    Post subject: Reply with quote

Isn't simple coloring enough, like this?

partially colored i get here:

*** *** ***
*** *7* *7*
*** G7* 7**

**G *** 77*
*** B** *G*
**B *G* ***

*** *** ***
*** *** ***
*** *** ***
In column 7 r4c7 can't be G, so it must be B. Then the remaining cells can be colored:

*** *** ***
*** *G* *B*
*** GB* G**

**G *** BX*
*** B** *G*
**B *G* ***

*** *** ***
*** *** ***
*** *** ***

Since B is the right color, all missnig 7's can be placed.
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Nick70

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PostPosted: Tue Aug 30, 2005 2:08 pm    Post subject: Reply with quote

kog wrote:
Isn't simple coloring enough, like this?

partially colored i get here:

*** *** ***
*** *7* *7*
*** G7* 7**

**G *** 77*
*** B** *G*
**B *G* ***

*** *** ***
*** *** ***
*** *** ***
In column 7 r4c7 can't be G, so it must be B.

You can't say that.
However, you can say that in column 7 both placement for 7 exclude G, so G must be false.
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kog

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PostPosted: Tue Aug 30, 2005 6:00 pm    Post subject: Reply with quote

Nick70 wrote:
kog wrote:
In column 7 r4c7 can't be G, so it must be B.

You can't say that.
However, you can say that in column 7 both placement for 7 exclude G, so G must be false.


I was thinking that if r5c8=G, then r3c7=G, then r4c7=B. I see now that it is wrong continue colouring after an contradiction is found.

I can still put 7 in r5c4 and r6c3.

If i start with r2c5=G > r2c8=B > r3c7=G, is it safe to assume r5c4=G since r2c5=G then r3c4 is not? after that continue to see if one color is wrong.
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bart

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PostPosted: Wed Aug 31, 2005 1:02 pm    Post subject: Reply with quote

from copy/paste
Code:

 *-----------*
 |.72|...|854|
 |..5|7..|6..|
 |...|..4|2.7|
 |---+---+---|
 |...|..8|.63|
 |..6|5.7|4..|
 |52.|4..|...|
 |---+---+---|
 |6.4|8..|...|
 |..9|..6|1..|
 |257|...|98.|
 *-----------*


 *-----------*
 |.72|6.9|854|
 |4.5|782|6..|
 |.6.|.54|2.7|
 |---+---+---|
 |741|928|563|
 |..6|5.7|42.|
 |52.|46.|7..|
 |---+---+---|
 |614|895|372|
 |..9|276|145|
 |257|.4.|986|
 *-----------*


{13}   {7}    {2}    {6}    {13}   {9}    {8}    {5}    {4}   
{4}    {39}   {5}    {7}    {8}    {2}    {6}    {139}  {19}   
{1389} {6}    {38}   {13}   {5}    {4}    {2}    {39}   {7}   
{7}    {4}    {1}    {9}    {2}    {8}    {5}    {6}    {3}   
{389}  {389}  {6}    {5}    {13}   {7}    {4}    {2}    {18}   
{5}    {2}    {38}   {4}    {6}    {13}   {7}    {19}   {189} 
{6}    {1}    {4}    {8}    {9}    {5}    {3}    {7}    {2}   
{38}   {38}   {9}    {2}    {7}    {6}    {1}    {4}    {5}   
{2}    {5}    {7}    {13}   {4}    {13}   {9}    {8}    {6}   


from ss-file
Code:

 *-----------*
 |.72|...|854|
 |..5|7..|6..|
 |...|..4|2.7|
 |---+---+---|
 |...|..8|.63|
 |..6|5.7|4..|
 |52.|4..|...|
 |---+---+---|
 |6.4|8..|...|
 |..9|..6|1..|
 |257|...|98.|
 *-----------*

I806
I291
I517
I335
I603
I551
I617
I704
I662
I715
I622
I677
I764
I432
I312
I142
I225
I589
I595
I309
I284
I277
I094
I496
I196
I036
I138
I059
E44009
E25001



screenshot



let pretend we didn't see the 2 swordfishes
(r1c1,r1c5),(r5c1,r5c2,r5c5),(r8c1,r8c2)
or
(r3c3,r6c3),(r3c4,r9c4),(r6c6,r9c9)

and we did do coloring
in block 1 there's a blue and a green, so the 2 pink ones can be deleted for that reason?
just to make sure

btw, with the first swordfish you can delete the candidates in the same cells
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angusj
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PostPosted: Wed Aug 31, 2005 1:33 pm    Post subject: Reply with quote

bart wrote:
and we did do coloring
in block 1 there's a blue and a green, so the 2 pink ones can be deleted for that reason?

Yes.
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lennyh

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PostPosted: Fri Sep 02, 2005 9:14 am    Post subject: Coloring Question Reply with quote

In the following, I don't understand why I can't exclude the candidate in all the blue squares since cells r6c2 and r6c6 appear to be in a conjugate chain and are the same color and share the same group. Can you explain what I'm overlooking here?

Thank you.

*** B*G ***
*** *** ***
*** *** ***

*** *** ***
*G* *** ***
*B* G*B ***

*** *** ***
*** *** ***
*** *** ***
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angusj
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PostPosted: Fri Sep 02, 2005 1:11 pm    Post subject: Re: Coloring Question Reply with quote

lennyh wrote:
In the following, I don't understand why I can't exclude the candidate in all the blue squares since cells r6c2 and r6c6 appear to be in a conjugate chain

For a start cells r6c2 & r6c4 can't be conjugates because there's another cell containing the candidate in the same row (r6c6).
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lennyh

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PostPosted: Fri Sep 02, 2005 4:52 pm    Post subject: Reply with quote

I know I'm missing something here, but not sure what it is. I was trying to invoke rule 1 of the "Solving with Colors" writeup associated with Simple Sudoku. Apparently I am not clear on what constitutes a conjugate chain. Can you tell me why, in the above example, R1C4 - R1C6 - R6C6 - R6C4 - R6C2 - R5C2, is not a conjugate chain? Or maybe it is a conjugate chain and there are other reasons why this is not a sequence covered by rule 1.

Thank you.
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DHallman

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PostPosted: Fri Sep 02, 2005 5:53 pm    Post subject: Reply with quote

LennyH




R1C4 - R1C6 - R6C6 - R6C4 is a conjugate chain.
R6C2 - R5C2 is another. but it is not in the same chain. (P,A)

However R6C2 can be eliminated since R6C6 - R6C4 is a conjugate pair.
Thus solving R5C2.
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lennyh

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PostPosted: Fri Sep 02, 2005 6:52 pm    Post subject: Reply with quote

DHallman, thanks for your respopnse.

Is R6C2 not a part of the R1C4 - R1C6 - R6C6 - R6C4 chain because there are > 2 cells containing candidates in the row 6 group?
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Hakan

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PostPosted: Sun Oct 09, 2005 2:05 pm    Post subject: Multi- or Simple colors Reply with quote

angusj wrote:

Here's the layout at the critical step in multicolors1.ss :



I think this multi-colors example could be solved by the simple colors technique. Only one conjugate chain exists, I think. Actually, it looks to me as a good example of your first rule of simple colors:
"Whenever two cells in a conjugate chain have the same color and also share the same group, that color must be the 'false' color since each group can only have one of any value."

(Note: The 7s below are candidates):
*** *** ***
*** *7* *7*
*** 77* 7**

**7 *** 77*
*** 7** *7*
**7 *7* ***

*** *** ***
*** *** ***
*** *** ***

As I understand simple colors, the top right box belongs to your green/blue conjugate chain.

With simple colors (G= green, B=blue):
*** *** ***
*** *7* *7*
*** B7* B**

**B *** 77*
*** G** *B*
**G *B* ***

*** *** ***
*** *** ***
*** *** ***

Starting from r6c3=G and r4c3=B.
r6c3=G => r5c4=G.

r4c3=B => r5c8=B, causing three more cells to be marked B independently:
r6c5=B
r3c4=B
r3c7=B

Since r3c4 and r3c7 are two cells in a conjugate chain (green/blue), have the same color (blue) and also share the same group (row 3), that color (blue) must be the 'false' color.

Am I correct or not?
(I am trying to understand the coloring techniques.)


A follow-up question:
From the example above and using your first rule ...

"Whenever two cells in a conjugate chain have the same color and also share the same group, that color must be the 'false' color since each group can only have one of any value."

... I think you can NOT EXCLUDE candidate 7s with the FALSE color (r3c7 is blue, but could not be excluded). Instead, you could SOLVE the candidate 7s with the TRUE color (of course leading to some exclusion of candidate 7s).

If I am correct, you could clarify your first rule by adding the bold text:
"Whenever two cells in a conjugate chain have the same color and also share the same group, that color must be the 'false' color since each group can only have one of any value. Thus, the candidates having the 'true' color can be solved."
(or with something similar).

/Hakan (a fan of Simple Sudoku)
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Lummox JR

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PostPosted: Sun Oct 09, 2005 4:28 pm    Post subject: Reply with quote

I've redone this grid for readability, since using the asterisks as empty space wasn't helping that. I've changed those to periods instead and used asterisks to show the candidates.
Code:
...|...|...
...|.*.|.*.
...|**.|*..
-----------
..*|...|**.
...|*..|.*.
..*|.*.|...
-----------
...|...|...
...|...|...
...|...|...

Now, in this situation you don't actually have a single conjugate set of colors. I've always detested the blue-green notation for this reason: It only shows one conjugate pair. I much prefer to use a clearer notation of upper- and lowercase letters.
Code:
...|...|...
...|.a.|.A.
...|b*.|a..
-----------
..b|...|A*.
...|B..|.b.
..B|.b.|...
-----------
...|...|...
...|...|...
...|...|...

This notation shows you a much clearer picture of how the colors stand. And as you can see, the color labeled b appears alongside both a and A. Since either a or A must be true, b must be false, so the B's must be true. Therefore, you can place 7's in those locations and have this:
Code:
...|...|...
...|.a.|.A.
...|-A.|a..
-----------
..-|...|Aa.
...|7..|.-.
..7|.-.|...
-----------
...|...|...
...|...|...
...|...|...

There's no further elimination you can do from that point with the 7's.
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