UR+2B/2SL

Advanced methods and approaches for solving Sudoku puzzles

Re: UR+2B/2SL

Postby ronk » Sun May 15, 2011 1:08 pm

David P Bird wrote:You've gone over to the dark side again daj!

(8a)r3c5 = (8)r3c4 - (8=96)r15c4 - (9b=27)r7c45 -[UR]- (27)r2c45 = (78c)r3c45
=> [ac] r3c5 <> 4, [bc] r3c4 <> 9 making (4)r1c5 single and (78)r2c45 a naked pair.

Based on just the cells in your AIC, this appears to be a counter-result.
Code: Select all
 .  .    .   | 9  .    .  | .     .    .
 .  .    .   |*7 *6    .  | .     .    .   
 .  .    .   | 8  4    .  | .     .    .
-------------+------------+-----------------
 .  .    .   | .  .    .  | .     .    .   
 .  .    .   | 6  .    .  | .     .    .
 .  .    .   | .  .    .  | .     .    .   
-------------+------------+-----------------
 .  .    .   |*2 *7    .  | .     .    .
 .  .    .   | .  .    .  | .     .    .
 .  .    .   | .  .    .  | .     .    . 
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Re: UR+2B/2SL

Postby David P Bird » Sun May 15, 2011 1:32 pm

Ronk, Yes you've shot me down in flames!
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Re: UR+2B/2SL

Postby daj95376 » Sun May 15, 2011 6:07 pm

David,

Working from here:

Code: Select all
 *-----------------------------------------------------------*
 | 2     469   1     | 69    456   59    | 3     7     8     |
 | 8     67    459   | 267   267   3     | 459   1     459   |
 | 3     479   459   | 789   478   1     | 459   6     2     |
 |-------------------+-------------------+-------------------|
 | 6     489   3     | 1     258   7     | 4589  289   459   |
 | 1     5     2     | 68    9     4     | 7     38    36    |
 | 7     489   49    | 3     2568  25    | 1     289   4569  |
 |-------------------+-------------------+-------------------|
 | 4     3     8     | 279   27    6     | 29    5     1     |
 | 9     1     6     | 5     3     28    | 28    4     7     |
 | 5     2     7     | 4     1     89    | 6     389   39    |
 *-----------------------------------------------------------*

And applying: r3c5<>8 r3c4=8 r5c4=6 r1c4=9 r1c6=5; I arrive at this grid:

Code: Select all
 *------------------------------------------------------------*
 | 2     46    1     |  9     46    5     | 3     7     8     |
 | 8     67    459   |  27    27+6  3     | 459   1     459   |
 | 3     479   459   |  8     47    1     | 459   6     2     |
 |-------------------+--------------------+-------------------|
 | 6     489   3     |  1     258   7     | 4589  289   459   |
 | 1     5     2     |  6     9     4     | 7     38    3     |
 | 7     489   49    |  3     258   2     | 1     289   4569  |
 |-------------------+--------------------+-------------------|
 | 4     3     8     |  27    27    6     | 29    5     1     |
 | 9     1     6     |  5     3     28    | 28    4     7     |
 | 5     2     7     |  4     1     89    | 6     389   39    |
 *------------------------------------------------------------*

Now, the UR Type 1 would force: r2c5=6 r1c5=4 r3c5=7

So, we can conclude that r3c5<>8 forces r3c5=7<>4 via the <27> UR. This leads to r1c5=4 cracking the puzzle -- as was part of your conclusion.

Regards, Danny
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Re: UR+2B/2SL

Postby David P Bird » Sun May 15, 2011 10:01 pm

Danny, As they say I've been "otherwise engaged" in recent weeks and have got rusty. I should have double checked before I posted. When I saw Ron's response I realised I was wrong and said so, but I didn’t have time then to look any further.

Check this alternative elimination using that UR in a single AIC I believe I've found that reduces the puzzle to singles:

(5=9)r1c4 - (9=8)r9c6 = (8-2)r7c6 = (2)x-wing:r27c45 -[UR]- (7)x-wing:r27c45 = (78-4)r3c45 = (4)r1c5 => r1c5 <> 5

(In your second grid you have already reduced the puzzles to singles too I think.)

Regards David
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Re: UR+2B/2SL

Postby ronk » Mon May 16, 2011 12:49 pm

David P Bird wrote:(5=9)r1c4 - (9=8)r9c6 = (8-2)r7c6 = (2)x-wing:r27c45 -[UR]- (7)x-wing:r27c45 = (78-4)r3c45 = (4)r1c5 => r1c5 <> 5

Adjacent strong inferences?
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Re: UR+2B/2SL

Postby David P Bird » Tue May 17, 2011 9:07 am

OH SUGAR! Yes, Ronk another kindergarten error.

Considering the purpose of this thread I was trying to weave the UR inference into a killing AIC. Using GEM, all the eliminations are shown in one mark-up, but I can't string them together into a single AIC.

From the UR inference the best I can get is:

Code: Select all
 *-----------------------------------------------------------*
 | 2     469   1     | 69    456   59    | 3     7     8     |
 | 8     67    459   | 267   267   3     | 459   1     459   |
 | 3     479   459   | 78-9  478   1     | 459   6     2     |
 |-------------------+-------------------+-------------------|
 | 6     489   3     | 1     258   7     | 4589  289   459   |
 | 1     5     2     | 68    9     4     | 7     38    36    |
 | 7     489   49    | 3     2568  25    | 1     289   4569  |
 |-------------------+-------------------+-------------------|
 | 4     3     8     | 279   27    6     | 29    5     1     |
 | 9     1     6     | 5     3     28    | 28    4     7     |
 | 5     2     7     | 4     1     89    | 6     389   39    |
 *-----------------------------------------------------------*
 
(78)r3c45 = (27)r2c45 -[UR]- (27=9)r7c45 - (9)r9c6 = (9)r1c6 => r3c4 <> 9
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Postby ronk » Tue May 17, 2011 5:01 pm

David P Bird wrote:(78)r3c45 = (27)r2c45 -[UR]- (27=9)r7c45 - (9)r9c6 = (9)r1c6 => r3c4 <> 9 [/code]

Much better. Since candidate <9> doesn't exist in r7c5, that can be shortened to ...

(78)r3c45 = (27)r2c45 -[UR]- (27=9)r7c45 => r3c4 <> 9

There's also a continuous loop using AUR(27)r27c45 and the <69> bivalue in r1c4.
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Re: UR+2B/2SL

Postby daj95376 » Sat May 21, 2011 11:00 pm

[Edit: withdrawn because of an interpretation error. Corrected post below.]
Last edited by daj95376 on Sun May 22, 2011 6:27 am, edited 1 time in total.
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Re: UR+2B/2SL

Postby David P Bird » Sat May 21, 2011 11:21 pm

Danny, Perhaps it will be clearer if I notated as

(78)HiddenPair:r3c45 = (27)HiddenPair:r2c45 -[UR]- (27=9)NakedPair:r7c45 - (9)r9c6 = (9)r1c6 => r3c4 <> 9

ie if r3c4 = 9 then r3c5 = 8 is forced and so (27)r2c45 must be true.
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Re: UR+2B/2SL

Postby daj95376 » Sun May 22, 2011 6:20 am

David,

After a long and tiring day returning from a trip, I hope to set things straight with a discontinuous chain:

Code: Select all
(98-7)r3c45 = (27)r2c45 -UR- (27=9)r7c54 - (98)r3c45  =>  r3c4<>9
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Re: UR+2B/2SL

Postby daj95376 » Fri Aug 05, 2011 12:50 am

This thread seems as good as any for my query.

I'm wondering if this qualifies as one of Mike Barker's UR patterns ???

Code: Select all
 +--------------------------------------------------------------+
 |  14    7     56    |  14    35    8     |  36    9     2     |
 | *15+4  9     2     |  6     7     134   |  38    148  *15+4  |
 | *15+46 3     8     |  145   9     2     |  7     14   *15+46 |
 |--------------------+--------------------+--------------------|
 |  26    26    9     |  7     4     5     |  1     3     8     |
 |  8     5     4     |  3     1     9     |  2     6     7     |
 |  3     1     7     |  8     2     6     |  4     5     9     |
 |--------------------+--------------------+--------------------|
 |  7     268   1     |  9     38    34    |  5     248   346   |
 |  25    28    3     |  145   6     7     |  9     1248  14    |
 |  9     4     56    |  2     58    13    |  68    7     13    |
 +--------------------------------------------------------------+
 # 46 eliminations remain

r2c9=1 => ( r2c1=5 and r3c9=5 ) => the following grid:

Code: Select all
 *-----------------------------------------------------------*
 | 14    7    (6)    | 14    35    8     | 36    9     2     |
 | 5     9     2     | 6     7     34    | 38    48    1     |
 | 146   3     8     | 14    9     2     | 7    (4)    5     |
 |-------------------+-------------------+-------------------|
 | 26    26    9     | 7     4     5     | 1     3     8     |
 | 8     5     4     | 3     1     9     | 2     6     7     |
 | 3     1     7     | 8     2     6     | 4     5     9     |
 |-------------------+-------------------+-------------------|
 | 7     268   1     | 9     38    34    | 5     248   346   |
 | 2     28    3     | 145   6     7     | 9     1248  4     |
 | 9     4     56    | 2     58    13    | 68    7     3     |
 *-----------------------------------------------------------*

The exposed Naked Singles r1c3=6 and r3c8=4 => r3c1=1 ... and we have a <15> DP.

Thus, we can conclude that r2c9<>1.

I'm aware that the Naked Single r3c8=4 (alone) also forces r3c4=1. That's part of my quandry.

Regards, Danny


An interesting companion chain: (5)r2c9 = r3c9 - (5=416)r3c841 - (6=5)r1c3 - r2c1 = (5)r2c9
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Re: UR+2B/2SL

Postby David P Bird » Fri Aug 05, 2011 8:03 am

Danny, as your UR cells contain the only instances of 6 in row 3, it's clearly impossible. How kosher it is to use its possibility as premise is therefore questionable but I guess that wouldn't concern you. However, I take such instances to indicate that there should be more telling exclusion chains about.

Still using uniqueness, there's this chain to show that if r3c1 contains 5 a (14) deadly pattern would be created in the top tier:

(6)r2c1 = (6)r2c9 - (36=8)r12c7 - (8=14)r23c8 -[DP]- (14)r12c1,r14c4 = (5)r2c1,r3c4 => r3c1 <> 5
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Re: UR+2B/2SL

Postby daj95376 » Fri Aug 05, 2011 10:53 am

David, I accept that a false premise may lead to multiple contradiction states. That's not the issue. The question still remains as to whether or not a Mike Barker UR pattern exists that would explain r2c9=1 leading to a DP contradiction using the bivalue cells r1c3 and r3c8?

The pattern would look something like:

Code: Select all
(a)d

abW  --------- abX
         a      |
                |a
                |
abcd           ab(Z) (ab)c

abX <> b

It's not that much different from this Mike Barker UR pattern.

UR+4x/2SL:

Code: Select all
abW-----abX
     a   |
         |a
         |
aby     ab(Z)  (ab)y

abX <> b
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Re: UR+2B/2SL

Postby David P Bird » Fri Aug 05, 2011 2:10 pm

Danny in Mikes figure I believe it's implied that there is a compund strong link between the b's involved as shown in the chain I give below, otherwise it would be possible to solve the cells as shown on the right.
Code: Select all
abW-----abX                          a-bW------b-aX         
     a   |                                 a    | 
         |a                                     |a 
         |                                      |
aby     ab(Z)  (ab)y                 y-ab      a-bZ                  b-ay           

abx <> b

(a)r1c4 = (a#2)r1c1,r4c3 -[UR]- (b#2)r1c4,r3c1 = (b)r3c46 => r1c4 <> b

 +--------------------------------------------------------------+
 |  14    7     56    |  14    35    8     |  36    9     2     |
 | *15+4  9     2     |  6     7     134   |  38    148  *15+4  |
 | *15+46 3     8     |  145   9     2     |  7     14   *15+46 |
 |--------------------+--------------------+--------------------|
 |  26    26    9     |  7     4     5     |  1     3     8     |
 |  8     5     4     |  3     1     9     |  2     6     7     |
 |  3     1     7     |  8     2     6     |  4     5     9     |
 |--------------------+--------------------+--------------------|
 |  7     268   1     |  9     38    34    |  5     248   346   |
 |  25    28    3     |  145   6     7     |  9     1248  14    |
 |  9     4     56    |  2     58    13    |  68    7     13    |
 +--------------------------------------------------------------+
 
(5)r2c9 = (5#2)r2c1,r3c9 -[UR]- (1#2)r2c9,r3c1 = ??

Transferring this logic to your grid we find there is no such compound strong link.

[Edit notation for the bottom row of the upper figure corrected from r4 to r3.]
Last edited by David P Bird on Sat Aug 06, 2011 3:59 pm, edited 1 time in total.
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Postby ronk » Fri Aug 05, 2011 5:07 pm

daj95376 wrote:... The question still remains as to whether or not a Mike Barker UR pattern exists that would explain r2c9=1 leading to a DP contradiction using the bivalue cells r1c3 and r3c8?

The pattern would look something like:

Code: Select all
(a)d

abW  --------- abX
         a      |
                |a
                |
abcd           ab(Z) (ab)c

abX <> b

Danny, I'm reasonably sure that Mike Barker did not define such a pattern. One has to draw the line somewhere, and I think that line is ... "include one ALS but not two."

David P Bird wrote:Danny in Mikes figure I believe it's implied that there is a compund strong link between the b's involved ...

No implicit strong links are required. There already is a set of seven explicit strong inferences: (1) the four cells of the UR, (2) the two conjugate links on 'a' as shown, and (3) the (a)(b)y cell that combines with the 'aby' cell of the UR to comprise an ALS.

Mike Barker wrote:UR+4x/2SL:

Code: Select all
abW-----abX
     a   |
         |a
         |
aby     ab(Z)  (ab)y

abX <> b
All seven may not actually be required.

[edit: add the 'aby' cell to the description]
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