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01/22/07 Daily Nightmare - Uniqueness Coloring

 
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Myth Jellies
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PostPosted: Tue Jan 23, 2007 8:24 am    Post subject: 01/22/07 Daily Nightmare - Uniqueness Coloring Reply with quote

Code:
 *-----------*
 |...|318|...|
 |1.6|...|3.5|
 |...|...|...|
 |---+---+---|
 |.4.|...|.1.|
 |...|8.3|...|
 |9..|.6.|..2|
 |---+---+---|
 |2..|...|..7|
 |.9.|...|.4.|
 |..3|954|2..|
 *-----------*

Basic methods get you here
Code:
 *--------------------------------------------------------------------*
 | 45     257    579    | 3      1      8      | 4679   2679   469    |
 | 1      278    6      | 247    2479   279    | 3      2789   5      |
 | 348    2378   789    |*56     279   *56     | 14789  2789   1489   |
 |----------------------+----------------------+----------------------|
 | 3568   4      2578   | 257    279    2579   | 56789  1      3689   |
 | 56     16     1257   | 8      2479   3      | 45679  5679   469    |
 | 9      3578   578    |*15+47  6     *15+7   | 4578   3578   2      |
 |----------------------+----------------------+----------------------|
 | 2      58     4      |*16     38    *16     | 59     359    7      |
 | 568    9      158    | 27     38     27     | 156    4      136    |
 | 7      16     3      | 9      5      4      | 2      68     168    |
 *--------------------------------------------------------------------*

There is a 56-61-15 bug-lite in r367c46. All digits are locked in place except for the 5's in r6, so r6c46 <> 5. A few more basics takes us to...
Code:
 *--------------------------------------------------------------------*
 | 45     257    579    | 3      1      8      | 4679   2679   469    |
 | 1      278    6      | 247    2479   279    | 3      2789   5      |
 | 348    2378   789    |B56     279   C56     | 14789  2789   1489   |
 |----------------------+----------------------+----------------------|
 | 368    4      278    |A257    279    2579   | 6789   1      3689   |
 | 56     16     1257   | 8      2479   3      | 45679  5679   469    |
 | 9      3578   578    | 147    6     E17     | 4578   3578   2      |
 |----------------------+----------------------+----------------------|
 | 2      58     4      | 16     38    D16     | 59     359    7      |
 | 568    9      158    | 27     38     27     | 156    4      136    |
 | 7      16     3      | 9      5      4      | 2      68     168    |
 *--------------------------------------------------------------------*
(5)r4c4 = (5-6)r3c4 = (6)r3c6 - (6=1)r7c6 - (1=7)r6c6 => r4c4 <> 7. This sets us up for a nice uniqueness deduction in multi-digit coloring.
Code:
 *--------------------------------------------------------------------*
 | 45     257    579    | 3      1      8      | 4679   2679   469    |
 | 1      278    6      | 247    2479   279    | 3      2789   5      |
 | 348    2378   789    |*5a6A   279   *5A6a   | 14789  2789   1489   |
 |----------------------+----------------------+----------------------|
 | 368    4      278    |*2a5A   279   *25a+79 | 6789   1      3689   |
 | 56     16     1257   | 8      2479   3      | 45679  5679   469    |
 | 9      3578   578    |*1a7+4   6     *1A7a   | 4578   3578   2      |
 |----------------------+----------------------+----------------------|
 | 2      58     4      |*1A6a   38    *1a6A   | 59     359    7      |
 | 568    9      158    |*27     38    *27     | 156    4      136    |
 | 7      16     3      | 9      5      4      | 2      68     168    |
 *--------------------------------------------------------------------*
The starred cells mark the 56-61-17-72-25 BUG-lite in r34678c46. Applying a little multi-digit coloring on the deadly pattern results in the conjugate markings 'a' and 'A'. Note that the only cells where the deadly pattern can be avoided are r4c6 and r6c4. Note also that in those cells, the 'a' color forces you into the deadly pattern digits. Therefore, by uniqueness, color 'a' must be false and color 'A' is true. You can also represent this as
(79=25)r4c6 -BUGLT- (17=4)r6c4 - (1)r6c4 = (1-6)r7c4 = (6)r7c6 - (6=5)r3c6 => r4c6 <> 5. This takes us here...
Code:
 *--------------------------------------------------------------------*
 |O45     257    579    | 3      1      8      | 4679   2679   469    |
 | 1     H278    6      |*247    2479  *29-7   | 3     J2789   5      |
 -A348    2378   789    | 6      279    5      | 14789  2789   1489   |
 |----------------------+----------------------+----------------------|
 |B368    4      278    | 5      279    279    | 6789   1    C3689   |
 |N56    M16     1257   | 8      2479   3      | 45679  5679   469    |
 | 9      3578   578    |*47     6      1      | 4578   3578   2      |
 |----------------------+----------------------+----------------------|
 | 2     G58     4      | 1     F38     6      | 59     359    7      |
 | 568    9      158    |*27    E38    *27     | 156    4     D136    |
 | 7     L16     3      | 9      5      4      | 2     K68     168    |
 *--------------------------------------------------------------------*

Another starred uniqueness deduction...
(9=2&7)r28c6 -UR- (2&7 = 4)r28c4 - (2)r2c4 = (2)r8c4 - (2=7)r8c6 =>r2c6 <> 7
...then the long AIC marked with the letters above
(3)r3c1 = (3)r4c1 - (3)r4c9 = (3)r8c9 - (3=8)r8c5 - (8)r7c5 = (8)r7c2 - (8)r2c2 = (8)r2c8 - (8=6)r9c8 - (6)r9c2 = (6)r5c2 - (6=5)r5c1 - (5=4)r1c1 => r3c1 <> 4

This takes us most of the way.
Code:
 *-----------------------------------------------------------*
 | 4     257   579   | 3     1     8     | 67    279   69    |
 | 1     78    6     |*27    4    *2+9   | 3     789   5     |
 | 38    2378  789   | 6     79    5     | 14    2789  14    |
 |-------------------+-------------------+-------------------|
 | 38    4     2     | 5     79    79    | 68    1     36    |
 | 5     6     1     | 8     2     3     | 47    79    49    |
 | 9     378   78    | 4     6     1     | 58    35    2     |
 |-------------------+-------------------+-------------------|
 | 2     58    4     | 1     38    6     | 9     35    7     |
 | 6     9     58    |*27    38   *27    | 15    4     13    |
 | 7     1     3     | 9     5     4     | 2     6     8     |
 *-----------------------------------------------------------*
...we still have that potential 27-deadly UR pattern in r28c46. Now, the only way to avoid it is to exclude 2 from r2c6

This takes us here...
Code:
 *--------------------------------------------------*
 | 4    257  579  | 3    1    8    | 67   279  69   |
 | 1    78   6    | 2    4    9    | 3    78   5    |
 | 38   238  89   | 6    7    5    | 14   289  14   |
 |----------------+----------------+----------------|
 | 38   4    2    | 5    9    7    | 68   1    36   |
 | 5    6    1    | 8    2    3    | 47   79   49   |
 | 9    378  78   | 4    6    1    | 58   35   2    |
 |----------------+----------------+----------------|
 | 2    58   4    | 1    38   6    | 9    35   7    |
 | 6    9    58   | 7    38   2    | 15   4    13   |
 | 7    1    3    | 9    5    4    | 2    6    8    |
 *--------------------------------------------------*

A lot of ways to go from here. I had
(2)r1c2 = (2-3)r3c2 = (3)r6c2 - (3)r6c8 = (3)r7c8 - (3=8)r7c5 - (8=5)r7c2 => r1c2 <> 5 and you are done.

I had never before combined coloring and BUG-Lite like I did in this puzzle.
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Sudtyro
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PostPosted: Sun Jan 28, 2007 8:32 pm    Post subject: Re: 01/22/07 Daily Nightmare - Uniqueness Coloring Reply with quote

Myth Jellies wrote:
A few more basics takes us to...
Code:
 *--------------------------------------------------------------------*
 | 45     257    579    | 3      1      8      | 4679   2679   469    |
 | 1      278    6      | 247    2479   279    | 3      2789   5      |
 | 348    2378   789    |B56     279   C56     | 14789  2789   1489   |
 |----------------------+----------------------+----------------------|
 | 368    4      278    |A257    279    2579   | 6789   1      3689   |
 | 56     16     1257   | 8      2479   3      | 45679  5679   469    |
 | 9      3578   578    | 147    6     E17     | 4578   3578   2      |
 |----------------------+----------------------+----------------------|
 | 2      58     4      | 16     38    D16     | 59     359    7      |
 | 568    9      158    | 27     38     27     | 156    4      136    |
 | 7      16     3      | 9      5      4      | 2      68     168    |
 *--------------------------------------------------------------------*
(5)r4c4 = (5-6)r3c4 = (6)r3c6 - (6=1)r7c6 - (1=7)r6c6 => r4c4 <> 7.


Sorry to be slow on the chain...is it a discontinuous nice loop with the discontinuity at r4c4 due to the two (unlabeled) weak links to digit 7? And if so, how would a learner spot this loop?
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Myth Jellies
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PostPosted: Mon Jan 29, 2007 7:48 am    Post subject: Re: 01/22/07 Daily Nightmare - Uniqueness Coloring Reply with quote

Sudtyro wrote:
Myth Jellies wrote:
A few more basics takes us to...
Code:
 
 *--------------------------------------------------------------------*
 | 45     257    579    | 3      1      8      | 4679   2679   469    |
 | 1      278    6      | 247    2479   279    | 3      2789   5      |
 | 348    2378   789    |B56     279   C56     | 14789  2789   1489   |
 |----------------------+----------------------+----------------------|
 | 368    4      278    |A257    279    2579   | 6789   1      3689   |
 | 56     16     1257   | 8      2479   3      | 45679  5679   469    |
 | 9      3578   578    | 147    6     E17     | 4578   3578   2      |
 |----------------------+----------------------+----------------------|
 | 2      58     4      | 16     38    D16     | 59     359    7      |
 | 568    9      158    | 27     38     27     | 156    4      136    |
 | 7      16     3      | 9      5      4      | 2      68     168    |
 *--------------------------------------------------------------------*
(5)r4c4 = (5-6)r3c4 = (6)r3c6 - (6=1)r7c6 - (1=7)r6c6 => r4c4 <> 7.


Sorry to be slow on the chain...is it a discontinuous nice loop with the discontinuity at r4c4 due to the two (unlabeled) weak links to digit 7? And if so, how would a learner spot this loop?


If you would rather label your linkages instead of noting their endpoints, then it is the discontinuous nice loop you mention. I like the candidate-to-candidate AIC way of looking at it better.

After a cursory look for shorter xy-wing-type chains, I tend to use a multidigit multicoloring method to help find my AICs. Applying multidigit multicoloring to the relevent parts of the above grid gives you...
Code:
 
 *--------------------------------------------------------------------*
 | 45     257    579    | 3      1      8      | 4679   2679   469    |
 | 1      278    6      | 247    2479   279    | 3      2789   5      |
 | 348    2378   789    | 5a6A   279    5A6a   | 14789  2789   1489   |
 |----------------------+----------------------+----------------------|
 | 368    4      278    | 25A7   279    25a79  | 6789   1      3689   |
 | 56     16     1257   | 8      2479   3      | 45679  5679   469    |
 | 9      3578   578    | 1a47   6      1A7a   | 4578   3578   2      |
 |----------------------+----------------------+----------------------|
 | 2      58     4      | 1A6a   38     1a6A   | 59     359    7      |
 | 568    9      158    | 2B7b   38     2b7B   | 156    4      136    |
 | 7      16     3      | 9      5      4      | 2      68     168    |
 *--------------------------------------------------------------------*

The candidate labels 'a' and 'A' represent a pair of conjugate colors, as do 'b' and 'B', etc. Simple multidigit coloring worked for this one as the 7 in r4c4 sees (is weakly linked to) both an 'a' and 'A', and so it can't be valid.

To show how multicoloring works, say you missed the fact that there was a 7a and 5A in the same box. We have a color AIC

A = a - B = b

Which tells us color A or b must be true. Thus we also have the 7 in r4c4 seeing both a 5A and a 7b, so it still can't be true.

[edit - mixed up my colors]


Last edited by Myth Jellies on Sat Feb 03, 2007 12:09 am; edited 1 time in total
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Sudtyro
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PostPosted: Mon Jan 29, 2007 1:40 pm    Post subject: Reply with quote

Very "nice" explanation. Many thanks!
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Ron Moore
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PostPosted: Sun Feb 25, 2007 10:37 pm    Post subject: Two Grouped AIC's Reply with quote

MJ,

I'm sorry I've gotten a bit behind. First, I'll comment that the date of the Nightmare you show is 1/21/07 rather than 1/22/07.

In the first position you posted (shown below), I found a grouped AIC showing a strong link between (4)r1c1 and (3)r3c1, as does your long AIC later on.
Code:

 *--------------------------------------------------------------------*
 |A45     257    579    | 3      1      8      | 4679   2679   469    |
 | 1      278    6      | 247    2479   279    | 3      2789   5      |
 |F348    2378   789    | 56     279    56     | 14789  2789   1489   |
 |----------------------+----------------------+----------------------|
 |E3568   4      2578   | 257    279    2579   | 56789  1     D3689   |
 |A56     16     1257   | 8      2479   3      | 45679  5679   469    |
 | 9      3578   578    | 1457   6      157    | 4578  C3578   2      |
 |----------------------+----------------------+----------------------|
 | 2     B58     4      | 16     38     16     |B59    B359    7      |
 |A568    9      158    | 27     38     27     | 156    4      136    |
 | 7      16     3      | 9      5      4      | 2      68     168    |
 *--------------------------------------------------------------------*

The nodes in the chain are labelled with letters A through F in the diagram. Incidentally, in my posts I try to follow the conventions given on the Sudopedia page for Eureka notation, which differ a bit from your style. Following these conventions the chain is written as:
    (4=568)r158c1 - (8=593)r7c278 - (3)r6c8 = (3)r4c9 - (3)r4c1 = (3)r3c1 => r3c1 <> 4
After this elimination there is a hidden "14" pair in r3c79, and basic follow-up advances the solution (considerably) to this point:
Code:
.---------------.---------------.---------------.
| 4    257  579 | 3    1    8   | 67   279  69  |
| 1   A78   6   | 27   4    279 | 3    789  5   |
| 38   2378 789 | 6    79   5   | 14   2789 14  |
:---------------+---------------+---------------:
| 38   4    2   | 5    79   79  | 68   1    36  |
| 5    6    1   | 8    2    3   | 47   79   49  |
| 9   A378  78  | 4    6    1   | 58   5-3  2   |
:---------------+---------------+---------------:
| 2   A58   4   | 1    38   6   | 9   B35   7   |
| 6    9    58  | 27   38   27  | 15   4    13  |
| 7    1    3   | 9    5    4   | 2    6    8   |
'---------------'---------------'---------------'

In this position your "27" based UR in r28c46 is very evident, but (I later found that) it's not even needed since the ALS XZ elimination represented by this grouped chain is enough to complete the solution (ALS's marked with A and B in the diagram):
    (3=785)r267c2 - (5=3)r7c8 => r6c8 <> 3
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Sudtyro
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PostPosted: Tue Feb 27, 2007 3:57 pm    Post subject: Re: Two Grouped AIC's Reply with quote

Ron Moore wrote:
Code:

 *--------------------------------------------------------------------*
 |A45     257    579    | 3      1      8      | 4679   2679   469    |
 | 1      278    6      | 247    2479   279    | 3      2789   5      |
 |F348    2378   789    | 56     279    56     | 14789  2789   1489   |
 |----------------------+----------------------+----------------------|
 |E3568   4      2578   | 257    279    2579   | 56789  1     D3689   |
 |A56     16     1257   | 8      2479   3      | 45679  5679   469    |
 | 9      3578   578    | 1457   6      157    | 4578  C3578   2      |
 |----------------------+----------------------+----------------------|
 | 2     B58     4      | 16     38     16     |B59    B359    7      |
 |A568    9      158    | 27     38     27     | 156    4      136    |
 | 7      16     3      | 9      5      4      | 2      68     168    |
 *--------------------------------------------------------------------*


    (4=568)r158c1 - (8=593)r7c278 - (3)r6c8 = (3)r4c9 - (3)r4c1 = (3)r3c1 => r3c1 <> 4

Need help in understanding how inferences (and the Eureka notation) work for ALS’s and other groups. Ron used (8=593)r7c278 for the “B” nodes in r7. Those three cells form an ALS with an “extra” digit of either 3 or 8. So, to make the inferences come out right, does the Eureka notation require the first and last candidates (in the parentheses) to always be the extras, and therefore not repeated elsewhere in the set? Or is it less restrictive than that?

My own approach would have been to use single candidates and develop the equivalent chain for (8=593)r7c278 as:
(8)r7c2 = (8 – 3)r7c5 = (3)r7c8.

But, then following Ron’s lead, the 2-cell ALS in r7c78 suggests trying:
(8 = 5)r7c2 – (5 = 93)r7c78.
But, are the inferences correct in the ALS? The 3 is the extra, but the 5’s are not.

BTW, Ron’s chain also provides for:
(8=593)r7c278 – (3)r6c8 = (3)r6c2 => r6c2 <> 8.
That might help in another stage of the puzzle.
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Myth Jellies
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PostPosted: Thu Mar 01, 2007 4:58 pm    Post subject: Re: Two Grouped AIC's Reply with quote

Sudtyro wrote:
Ron Moore wrote:
Code:

 *--------------------------------------------------------------------*
 |A45     257    579    | 3      1      8      | 4679   2679   469    |
 | 1      278    6      | 247    2479   279    | 3      2789   5      |
 |F348    2378   789    | 56     279    56     | 14789  2789   1489   |
 |----------------------+----------------------+----------------------|
 |E3568   4      2578   | 257    279    2579   | 56789  1     D3689   |
 |A56     16     1257   | 8      2479   3      | 45679  5679   469    |
 | 9      3578   578    | 1457   6      157    | 4578  C3578   2      |
 |----------------------+----------------------+----------------------|
 | 2     B58     4      | 16     38     16     |B59    B359    7      |
 |A568    9      158    | 27     38     27     | 156    4      136    |
 | 7      16     3      | 9      5      4      | 2      68     168    |
 *--------------------------------------------------------------------*


    (4=568)r158c1 - (8=593)r7c278 - (3)r6c8 = (3)r4c9 - (3)r4c1 = (3)r3c1 => r3c1 <> 4

Need help in understanding how inferences (and the Eureka notation) work for ALS’s and other groups. Ron used (8=593)r7c278 for the “B” nodes in r7. Those three cells form an ALS with an “extra” digit of either 3 or 8. So, to make the inferences come out right, does the Eureka notation require the first and last candidates (in the parentheses) to always be the extras, and therefore not repeated elsewhere in the set? Or is it less restrictive than that?


The ALS digits can be ordered however you like, and the strong link symbol may be placed anywhere you like. They can all be potential extra digits. The trick is, if the digit appears in multiple places in the ALS then any weak link to that digit in the ALS must see all examples of that digit in the ALS.

It is convention to place the ALS digits being weakly linked closest to their weak link symbol.

I like placing ampersands between digits because I think...

(4=5&6&8)r158c1

...conveys better that r158c1 either contains a 4 or it contains 5 & 6 & 8 in a locked set. You could continue on with...

(4=5&6&8)r158c1 - (568=3)r4c1

or

(4=5&6&8)r158c1 - (5v6v8=3)r4c1

...where here, the 568 in r4c1 represents a choice rather than a locked set.

Of course you could have just started with

(4=5&6&8&3)r1458c1 in the first place. Smile
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Sudtyro
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Joined: 16 Jan 2007
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PostPosted: Thu Mar 01, 2007 10:47 pm    Post subject: Re: Two Grouped AIC's Reply with quote

Myth Jellies wrote:

They can all be potential extra digits. The trick is, if the digit appears in multiple places in the ALS then any weak link to that digit in the ALS must see all examples of that digit in the ALS.


Well, that really helped...the key is the "trick." It's hard for a newcomer to ferret out these gems! Once again...many thanks!
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