The New Sudoku Players' Forum

[Kozo Kataya]

Depend on [ Yogi's posts and so kind of Leren's replies ], i can enjoy this forum so much, many thanks for your posts.
On this forum to me, the most confusing matter is to get what means abbreviations and/or terms ?, recently [lclste].
Anyway on Eureka-AIC, i found ??? below.

Re: Chain Notation Mon Jul 25, 2016 10:17 am
The notation convention most commonly used on this forum is Eureka notation.
A reasonably compact description can be found on the Sudopedia site here.(EUREKA)
Leren

Accoding to EUREKA Grouped AIC example as follows.

This AIC can be formed (shown in condensed Eureka notation):
(1)r1c6=r5c6=(3-7)r6c6=grp(7)r6c79-r4c8-(5=6)r4c9-r2c9=(6-4)r2c7=grp(4)r2c46 => r1c6<>4

Code: Select all

+---+---+---+      +------------------+------------------+------------------+
|.6.|3..|...|      |2459  6     24579 |3     1245  1249  |2457  257   8     |
|3.1|.8.|.9.|      |3     27    1     |245   8     24    |24567 9     567   |
|8..|67.|..1|      |8     29    2459  |6     7     249   |2345  235   1     |
+---+---+---+      +------------------+------------------+------------------+
|.3.|...|1..|      |249   3     2489  |24589 2456  247   |1     57    567   |
|75.|...|.82|      |7     5     49    |149   146   134   |369   8     2     |     
|..6|...|.4.|      |129   1289  6     |2589  25    237   |3579  4     357   |
+---+---+---+      +------------------+------------------+------------------+
|6..|.98|..4|      |6     127   2357  |127   9     8     |2357  12357 4     |
|.4.|.3.|8.9|      |125   4     257   |127   3     6     |8     1257  9     |
|...|..5|.6.|      |129   12789 23789 |1247  124   5     |237   6     37    |
+---+---+---+      +------------------+------------------+------------------+


Wonder why Eureka express such a long link-chain just to elminate r1c6<>4.
Since r4c6=7 is false then r6c6=7 is true, then stte.

Regards kozo

[David P Bird]

Kozo
Sadly many contributors only think of themselves when they notate their solutions and don't consider that learners as well as experienced players will be trying to follow them. The condensed notation you quoted breaks a basic rule in writing Alternating Inference Chains that the link types in the chain must alternate.

Here is the chain showing all the links.
(1)r1c6 = (1-3)r5c6 = (3-7)r6c6 = (7)r6c79 - (7=5)r4c8 - (5=6)r4c9 - (6)r2c9 = (6-4)r2c7 = (4)r2c46 => r1c6 <> 4
As it isn't usual to use the grp term any more I've underlined where they occur.

Reading this it is easy to track the path link by link and to check that the logic is sound if you understand that a weak link connects Boolean arguments that can't both be true and a strong link connects them when they can't both be false.

AICs can be read left to right or right to left. They start from an observation usually that two candidates that are strongly linked. Following alternating inferences onwards you won't always prove what you were hoping to find but often will prove something else so you will made at least one step in the right direction. If you start by assuming one candidate is true and follow two weak links away from it the only thing you can prove is that your first assumption is true or false - nothing else. This makes following alternating inferences more productive especially for the harder puzzles.

The chain given has been provided to show how group links are used NOT to show the best way to solve the puzzle.

But now you fall into the same trap as the others I complain about when you write:

Wonder why Eureka express such a long link-chain just to elminate r1c6<>4.
Since r4c6=7 is false then r6c6=7 is true, then stte.

Please explain to a beginner how you prove (7)r4c6 is false!

DPB

[Kozo Kataya]

But now you fall into the same trap as the others I complain about when you write:

Since r4c6=7 is false then r6c6=7 is true, then stte.

Please explain to a beginner how you prove (7)r4c6 is false!
DPB

Thank you for your kind explanation.
Though i am not good at Eureka notation, please refer the followings

Code: Select all

if r4c6=7 then 134r5c6 is void, because   
   7r4c6=>5r4c8,6r4c9,37r69c9,5r2c9
   5r2c9=>24r2c46=>9r3c6,1r1c6,5r1c5
   5r1c5=>2r6c5,3r6c6 and 4r4c5
then 1r1c6+4r4c5+3r6c6(marked *) => void [134]r5c6: contradiction   
Thus, r6c6=7 then [ single to the end ]   


Code: Select all

*------------------+------------------+------------------*
|2459  6     24579 |3     1245  1249  |2457  257   8     |
|3     27    1     |245   8     24    |24567 9     567   |
|8     29    2459  |6     7     249   |2345  235   1     |
*------------------+------------------+------------------*
|249   3     2489  |24589 2456  247   |1     57    567   |
|7     5     49    149    146   134   |369   8     2     |
|129   1289  6     |2589  25    237   |3579  4     357   |
*------------------+------------------+------------------*
|6     127   2357  |127   9     8     |2357  12357 4     |
|125   4     257   |127   3     6     |8     1257  9     |
|129   12789 23789 |1247  124   5     |237   6     37    |
*------------------+------------------+------------------*


Code: Select all

  if r4c6=7 then 1r1c6+3r6c6+4r4c5 cause r5c6 is void
*------------------+------------------+------------------*
|2459  6     24579 |3     5     1*    |2457  257   8     |
|3     27    1     |24    8     24    |24567 9     5     |
|8     29    2459  |6     7     9     |2345  235   1     |
*------------------+------------------+------------------*
|249   3     2489  |24589 4*    7     |1     5     6     |
|7     5     49    149    146  [134]  |369   8     2     |
|129   1289  6     |2589  2     3*    |3579  4     37    |
*------------------+------------------+------------------*
|6     127   2357  |127   9     8     |2357  12357 4     |
|125   4     257   |127   3     6     |8     1257  9     |
|129   12789 23789 |1247  124   5     |237   6     37    |
*------------------+------------------+------------------*


Regards kozo

[David P Bird]

Kozo, unless you are using a computer to show you the effects of your assumptions, your proof is far from simple as you implied in your earlier post - so that was rather an unfair criticism.

As it seems you use assumptions and branched chains and I enjoy the challenge of avoiding these approaches, other members will be able to help you far better than me.

DPB