Strong Corner Marker for X-Wing AIC Deduction

Advanced methods and approaches for solving Sudoku puzzles

Postby Allan Barker » Mon Oct 20, 2008 12:04 am

An example of a classic weak-corner that removes 9r9c79 can be found in a sligtly different grid where 5r9c7 is removed. The same logic works with candidate 5r9c7 but the relation to nice loops is clearer without it.

Either r9c6=9
Or r9c6=5 & (r9c6=5 -> r6c4=5 -> r3c7=5 -> r9c7=8) -> r9c1=9
Thus r9c79<>9.
Code: Select all
New grid after eleven's grid
  +-----------------------------------------------------------+
  | 589   579   78    | 6     79    1     | 2     3     4     |
  | 4     2     3     | 8     5     79    | 79    6     1     |
  | 6     79    1     | 2     3     4     | 79(58)  5789 579  |
  +-----------------------------------------------------------+
  | 7     3     59    | 1     8     6     | 4     2     59    |
  | 2     8     59    | 4     79    79#5  | 6     1     3     |
  | 1     4     6     | 9(5)  2     3     | 79(5) 579   8     |
  +-----------------------------------------------------------+
  | 59    6     4     | 7     1     8     | 3     59    2     |
  | 3     579   78    | 59    4     2     | 1     589   6     |
  | #5(89) 1     2    | 3     6    #5(9)  | 7-9(8)  4   57-9  |
  +-----------------------------------------------------------+

9N6:    5r9c6A==========================9r9c6
        /    \                            |
5B5:   |    5r5c6==5r6c4                  |
       |             |                    |   
5C7:   |           5r6c7==5r3c7           |
       |                    |             |
8C7:   |                  8r3c7==8r9c7    |
       |                           |      |   
9N1: 5r9c1=======================8r9c1==9r9c1
                                          |
                                         X=9r9c79


Edit: clarify that this grid is derived from eleven's earlier grid.
Last edited by Allan Barker on Tue Oct 21, 2008 3:47 am, edited 2 times in total.
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Postby aran » Mon Oct 20, 2008 3:40 am

Allan

Code: Select all
+-----------------------------------------------------------+
  | 589   579   78    | 6     79    1     | 2     3     4     |
  | 4     2     3     | 8     5     79    | 79    6     1     |
  | 6     79    1     | 2     3     4     | 79(58)  5789 579  |
  +-----------------------------------------------------------+
  | 7     3     59    | 1     8     6     | 4     2     59    |
  | 2     8     59    | 4     79    79#5  | 6     1     3     |
  | 1     4     6     | 9(5)  2     3     | 79(5) 579   8     |
  +-----------------------------------------------------------+
  | 59    6     4     | 7     1     8     | 3     59    2     |
  | 3     579   78    | 59    4     2     | 1     589   6     |
  | #5(89) 1     2    | 3     6    #5(9)  | 7-9(8)  4   57-9  |
  +-----------------------------------------------------------+


Or
hp59r78c8=8r8c8-(8=79)r9c7-(79=5)r6c7-(5=9)r6c4=(5-9)r8c4=9r9c6
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Postby ronk » Mon Oct 20, 2008 3:51 am

Allan Barker wrote:
Code: Select all
9N6:    5r9c6A==========================9r9c6
        /    \                            |
5B5:   |    5r5c6==5r6c4                  |
       |             |                    |   
5C7:   |           5r6c7==5r3c7           |
       |                    |             |
8C7:   |                  8r3c7==8r9c7    |
       |                           |      |   
9N1: 5r9c1=======================8r9c1==9r9c1
                                          |
                                         X=9r9c79

Either r9c6=9
Or r9c6=5 & (r9c6=5 -> r6c4=5 -> r3c7=5 -> r9c7=8) -> r9c1=9
Thus r9c79<>9.

The relationship between a simple multi-inference nice loop and the diagram above would be better understood if, in this case, you "lopped off" the lower-left corner. Then ...

Code: Select all
Either r9c1=5:
  r9c1 -5- r9c6 -9- r9c79
Or r9c1<>5:
  r9c79 -9- r9c6 -5- r5c6 =5= r6c4 -5- r6c7 =5= r3c7 =8= r9c7 -8- r9c1 -9- r9c79
In both cases:
  r9c79<>9

aran wrote:hp59r78c8=8r8c8-(8=79)r9c7-(79=5)r6c7-(5=9)r6c4=(5-9)r8c4=9r9c6

There are two errors ... the worst highlighted in red.
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Postby Allan Barker » Mon Oct 20, 2008 6:14 am

ronk wrote:The relationship between a simple multi-inference nice loop and the diagram above would be better understood if, in this case, you "lopped off" the lower-left corner. Then ...

That's interesting. It could then be thought of as two continuous nice loops that partially overlap, the outer rectangle and the upper right triangle. Only the part that overlaps can cause eliminations.

Code: Select all
ooooooooo       ooooooooo
o..     o       ooo     o
o ...   o   +   . ooo   o
o   ... o       .   ooo o
ooooooooo       ......ooo
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Postby aran » Mon Oct 20, 2008 7:15 am

Ronk wrote
There are two errors ... the worst highlighted in red

What you probably mean is a little too succinct.
I take the view that the very obvious can be assumed without encumbering the notation in the interests of "legibility". Fair enough if you disagree...but
"dotting" the "i"s ain't the same thing as confusing grammar or maybe you make no such distinction.
:)
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Postby Allan Barker » Mon Oct 20, 2008 8:32 am

aran wrote:Or
hp59r78c8=8r8c8-(8=79)r9c7-(79=5)r6c7-(5=9)r6c4=(5-9)r8c4=9r9c6

Aran,

I'm not so strong on notation but usually get by. Try as I might, I could not get this to eliminate anything as far as I understand it. I assume you are after 9r9c79 but see one of these in your chain. Can you write it another way?

Thanks
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Postby aran » Mon Oct 20, 2008 9:42 am

Allan
Aran,

I'm not so strong on notation but usually get by. Try as I might, I could not get this to eliminate anything as far as I understand it. I assume you are after 9r9c79 but see one of these in your chain. Can you write it another way?


I guess I was too concise.
Objective : estabish the strong inference set : pair 59 r78c8 and 9r9c6.
At least one truth in there => r9c79 <9>.
I`ll use words to begin with :
in typical AIC type fashion (to establish a strong link) begin chain with : ~(hidden pair 59 r79c8) (not a hidden pair)
This=>r8c8=8 (the only way for there not to be a hidden pair)
=>r9c7~8=>r9c7=79=>(combined with r2c7=79) r6c7~79,and so=5.
=>r6c4=9=>r8c4=5=>r9c6=9. Hence the conclusion.
I think those words will be clear (well I hope so) the problem is how to transcribe that into eureka notation :
which IMO has great difficulty in dealing with the obvious (without labouring the presentation, I mean) in particular of the sort "which combined with this pre-existing thing on the grid=>" e.g as above for the pre-existing 79 at r2c7 to form the naked pair 79 so forcing r6c7=5.
Indeed I think footnotes for that sort of situation should be permitted :
So here is an amended version complete with footnote
hp59r78c8=8r8c8-(8=79)r9c7*-(79=5)r6c7-(5=9)r6c4=(5-9)r8c4=9r9c6 : =>r9c79 <9>
* which combined with r2c7
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Postby DonM » Mon Oct 20, 2008 10:21 am

aran wrote:Ronk wrote
There are two errors ... the worst highlighted in red

What you probably mean is a little too succinct.
I take the view that the very obvious can be assumed without encumbering the notation in the interests of "legibility". Fair enough if you disagree...but
"dotting" the "i"s ain't the same thing as confusing grammar or maybe you make no such distinction.
:)


The problem with the 'very obvious [being] assumed' is that sometimes what is being assumed is a mistake. Also, it is assuming that what is obvious to one person is obvious to everyone else. I don't mean any harm here- but as I read this thread and all the shorthand & non-standard presentations being used, my bet is that, more than not, more than they would care to admit, people are hardly understanding anyone's solutions or the points they are making. Still, I am always tempted to put down solutions in a way that confuses error-checking ronk as much as possible.:D

Again, no offense to anyone, but speaking of shorthand:?: :

Code: Select all
SIN:  9r9c7  5r9c6  5r6c4  [c7]~7  => [r9c7]<>9
SIN:  9r9c9  7r9c7  9r2c7  [b6]~5  => [r9c9]<>9


Thanks eleven for putting up that puzzle- it gave this thread a whole new life:) . It has an ER=7.8 -that plus the look of it at the point you presented it, one would think that it should go down with one chain...but obviously it doesn't. Anyway, here's 4 basic AICs that solve it- with the self-imposed objective being that the solution not make use of any of the methods already mentioned including W-wings & coloring.

Code: Select all
(9=7)r2c7-r2c6=(7-9)r1c5=r5c5-r5c3=r4c3-(9=5)r4c9-(5)r3c9=hp(58) => r3c78<>9

(9=5)r4c9-als(5=79)r2c7/r3c9-(7)r3c8=(7)r6c8  => r6c8<>9

(9)r7c8=r8c8-(9=5)r8c4-(5)r9c6=hp(578)r9c179  => r9c79<>9

(9)r4c9=r3c9-(9=7)r3c2-als(7=59)r7c1/r8c2-(9)r9c1=(9-5)r9c6=r5c6-r5c3=(5)r4c3  => r4c9<>5=9
Last edited by DonM on Mon Oct 20, 2008 3:34 pm, edited 7 times in total.
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Postby ronk » Mon Oct 20, 2008 10:38 am

aran wrote:So here is an amended version complete with footnote
hp59r78c8=8r8c8-(8=79)r9c7*-(79=5)r6c7-(5=9)r6c4=(5-9)r8c4=9r9c6 : =>r9c79 <9>
* which combined with r2c7

No need for a footnote ...

r9c79 -9- als:r78c8 -8- als:r29c7 -79- r6c7 -5- r6c4 -9- r8c4 =9= r9c6 -9- r9c79 ==> r9c79<>9

... and in AIC notation, it helps to have the inferences actually alternate.

Would someone please remind me why Mike Barker and I learned AIC notation in order to post on the Eureka! forum?
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Postby DonM » Mon Oct 20, 2008 10:47 am

ronk wrote:Would someone please remind me why Mike Barker and I learned AIC notation in order to post on the Eureka! forum?


Because it's better? (the notation that is).

P.S. There's still life in Eureka- it's just on hiatus for awhile... (and the hurricane season has ended. ;) )
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Postby Allan Barker » Mon Oct 20, 2008 5:48 pm

aran wrote:hp59r78c8=8r8c8-(8=79)r9c7*-(79=5)r6c7-(5=9)r6c4=(5-9)r8c4=9r9c6 : =>r9c79 <9>
* which combined with r2c7

Not sure how I could miss that r2c7 was involved, but it works fine now. If I consider
that 9r6c7 and 9r6c4 are also linked, this also eliminates 5r6c8. Following ronk:

r9c79 -9- als:r78c8 -8- als:r29c7 -79- r6c7 -5- r6c4 -9- r8c4 =9= r9c6 -9- r9c79 ==> r9c79<>9 r6c8<>5

or by itself,

als:r78c8 -89- r29c7 -79- r6c7 -59- r6c4 ==> r6c8<>5

One reason I sometimes post set disagrams is they make no assumptions, although they often look longer.
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Postby ronk » Mon Oct 20, 2008 6:37 pm

Allan Barker wrote:by itself,

als:r78c8 -89- r29c7 -79- r6c7 -59- r6c4 ==> r6c8<>5

Did you mean r6c8<>9 ... or r6c8<>59:?:

Code: Select all
 589  579  78   | 6    79   1    | 2    3    4
 4    2    3    | 8    5    79   |B79   6    1
 6    79   1    | 2    3    4    | 5789 5789 579
----------------+----------------+---------------
 7    3    59   | 1    8    6    | 4    2    59
 2    8    59   | 4    79   579  | 6    1    3
 1    4    6    |C59   2    3    |C579  7-59 8
----------------+----------------+---------------
 59   6    4    | 7    1    8    | 3   A59   2
 3    579  78   | 59   4    2    | 1   A589  6
 589  1    2    | 3    6    59   |B789  4    579

               A             B             C
r6c8 -59- als:r78c8 -8- als:r29c7 -7- als:r6c47 -59- r6c8 ==> r6c8<>59

The derived strong inference (in AIC notation) is: (59)r78c8 = (59)r6c47

Relative to eleven's pencilmarks here you dropped candidate 5 in r9c7. What was the basis for that elimination?:?:

[edit2: total rewrite]
Last edited by ronk on Tue Oct 21, 2008 1:08 am, edited 2 times in total.
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Postby aran » Tue Oct 21, 2008 2:50 am

Ronk wrote
No need for a footnote ...

r9c79 -9- als:r78c8 -8- als:r29c7 -79- r6c7 -5- r6c4 -9- r8c4 =9= r9c6 -9- r9c79 ==> r9c79<>9

... and in AIC notation, it helps to have the inferences actually alternate.

"no need for footnote"...and you then proceed that to write a different chain (discontinuous loop from r9c6)...but anyway !

I'm afraid you've resorted to concision with : "9- als:r78c8 -8- als:r29c7" which skips the step r8c8-->r9c7. I don't mind but by your own rules this is a serious error, deserving of the red-ink treatment.

Visually a long sequence of "-" doesn't really look well. And could confuse those accustomed to alternating "= -". Concision gone too far ?
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Postby Allan Barker » Tue Oct 21, 2008 7:37 am

Ronk wrote:
Allan Barker wrote:by itself,

als:r78c8 -89- r29c7 -79- r6c7 -59- r6c4 ==> r6c8<>5

Did you mean r6c8<>9 ... or r6c8<>59 ?

I really meant:

. als:r78c8 -89- r9c7 -79- r6c7 -59- r6c4 ==> r6c8<>5

which does not require r2c7. The extra '2' was left in after I trimmed the larger logic in my previous post. As you point out, this also has a nice symmetrical form with ALS on either end:

. als:r78c8 -89- r9c7 -79- als:r6c47 ==> r6c8<>5

Ronk wrote:Relative to eleven's pencilmarks here you dropped candidate 5 in r9c7. What was the basis for that elimination?

Only because it provided the nice clear weak-corner example. I have changed the post at the top to reflect this. Thanks, that was pretty unclear.
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Postby ronk » Wed Oct 22, 2008 11:33 am

Allan Barker wrote:As you point out, this also has a nice symmetrical form with ALS on either end:

. als:r78c8 -89- r9c7 -79- als:r6c47 ==> r6c8<>5

Wouldn't the following be the most accurate nice loop notation for the six linksets (weak links):?:

r6c8 -5- als:r78c8 -89- aals:r9c7 -7- als:r6c47 -59- r6c8

Many will view the "als:" and "aals:" labels as excess baggage.:)
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