View previous topic :: View next topic 
Author 
Message 
keith
Joined: 19 Sep 2005 Posts: 3150 Location: near Detroit, Michigan, USA

Posted: Sun Oct 21, 2012 2:34 pm Post subject: Sudopedia: Eureka notation 


The following is copied from Sudopedia:
Eureka
The Eureka notation system is a compact method to write Alternating Inference Chains.
Cells are mainly written in the rncn notation, but other methods like k9 can also be used.
Candidate digits are written as a prefix to the cell, with the digit number in parenthesis. This is the way to write candidate for digit 1 in the cell at row 5 and column 4.
(1)r5c4
Weak links are represented by a dash.
(1)r5c4(1)r5c8
Strong links are represented by an equal sign.
(1)r5c8=(1)r9c8
There is no need to repeat the cell name when multiple candidates of that cell are used in the chain. A link between 2 candidates in a single cell is placed inside the parenthesis.
(14)r5c4=(4)r5c8
When groups of cells are used in the chain, they are named in the most efficient way.
All cells belong to a single row
(1)r5c456
All cells belong to a single column
(1)r45c4
Unordered group, in a box
(1)r5c4r6c45
When all candidates represent the same digit, this digit can be placed before the entire chain as a prefix.
(1): r5c4r5c8=r9c8
When an embedded ALS is part of the chain, the digit linked to the previous node is isolated from the remaining digits with a strong link symbol. The remaining digits are placed in such an order that the digit linked to the next node is the last one.
(1)r5c4(1=264)r5c789(4)r4c8
When read from left to right, the chain must contain alternating strong and weak links. There can be no 2 adjacent dashes or equal signs, whether or not they are placed inside the parenthesis.
The result is placed after the chain, separated by a double arrow.
(1)r5c4(1=4)r5c8(4)r9c8=(41)r7c7=(1)r7c4(1)r5c4 => r5c4<>1
Category: Chains and Loops
Last edited by keith on Wed Nov 21, 2012 7:50 am; edited 1 time in total 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5121 Location: Rochester, NY, USA

Posted: Fri Nov 02, 2012 12:31 am Post subject: 


I've been trying to learn notation and, among other problems, I'm having difficulty understanding inferences. In fact, in some cases, I can write notation without understanding the inferences that it depicts.
I have seen a number of examples that contain something similar to the first step shown above. R5c4 is a bivalue cell and I would've thought that all bivalue cells have a strong inference between the two candidates.
So what's the explanation behind a weak inference in a bivalue cell? (Why do I have the feeling that I'm stepping into a pile of deep doodoo)? 

Back to top 


arkietech
Joined: 31 Jul 2008 Posts: 1708 Location: Northwest Arkansas USA

Posted: Fri Nov 02, 2012 2:23 am Post subject: 


Marty R. wrote: 
I've been trying to learn notation and, among other problems, I'm having difficulty understanding inferences. In fact, in some cases, I can write notation without understanding the inferences that it depicts.
I have seen a number of examples that contain something similar to the first step shown above. R5c4 is a bivalue cell and I would've thought that all bivalue cells have a strong inference between the two candidates.
So what's the explanation behind a weak inference in a bivalue cell? (Why do I have the feeling that I'm stepping into a pile of deep doodoo)? 
A bivalue cell can be either a strong or weak link. Many mwings have a bivalue cell providing a weak link after an initial strong bivalue cell.
(a=b)b=(ba)=a
Note: eureka notation should always begin and end with a strong link 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3150 Location: near Detroit, Michigan, USA

Posted: Fri Nov 02, 2012 2:44 am Post subject: 


Marty R. wrote: 
I've been trying to learn notation and, among other problems, I'm having difficulty understanding inferences. In fact, in some cases, I can write notation without understanding the inferences that it depicts.
I have seen a number of examples that contain something similar to the first step shown above. R5c4 is a bivalue cell and I would've thought that all bivalue cells have a strong inference between the two candidates.
So what's the explanation behind a weak inference in a bivalue cell? (Why do I have the feeling that I'm stepping into a pile of deep doodoo)? 
Marty,
I will say about Eureka Notation, I am not going to worry about it at this stage of my life. I believe Winston Churchill said something similar about Cambodia.
Marty R. wrote:  (14)r5c4... 
I think this means 1 and 4 are weakly linked in R5C4, which means R5C4 is a tri (or more) valued cell.
14 means 1 is possible and excludes 4, or 4 is possible and excludes 1. Other possibilities are present.
1=4 means 1 or 4, no other possibilities are present.
Keith 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5121 Location: Rochester, NY, USA

Posted: Fri Nov 02, 2012 3:54 am Post subject: 


Keith, I have seen the weak link notation on more than one occasion for bivalue cells.
Here's an example from Ted which uses it twice with bivalue cells.
Code:  **
 6 1 59  8 45 349  7 2 34 
 78 4 3  19 17 2  189 5 6 
 2 59 78  6 1457 349  189 49 134 
++
 9 6 2  7 3 5  4 1 8 
 57 3 57  4 8 1  69 69 2 
 1 8 4  2 9 6  3 7 5 
++
 458 25 158  3 1246 7  1256 46 9 
 3 7 19  5 26 49  26 8 14 
 45 259 6  19 124 8  125 3 7 
**
9r9c2=9r3c2(9=4)r3c8r7c8=(41)r8c9=(19)r8c3=9r9c2 => 9r9c2 
Quote:  A bivalue cell can be either a strong or weak link. Many mwings have a bivalue cell providing a weak link after an initial strong bivalue cell. 
Thanks Dan, but I'm trying to find out why that is. 

Back to top 


daj95376
Joined: 23 Aug 2008 Posts: 3855

Posted: Fri Nov 02, 2012 3:28 pm Post subject: 


Here's my "working definition" of strong inference and weak inference.
Strong Inference: When you have two possible outcomes and you assume that one outcome isn't true, then you must conclude that the other outcome is true.
If you assume that one candidate in a bivalue cell isn't true, then you must conclude that the other candidate is true. Example: (5=9)r1c3 in the grid above. The assumption of <5> not being true forces the conclusion of <9> being true.
Weak Inference: When you have multiple, disjoint possible outcomes and you assume that one outcome is true, then you must conclude that all other outcomes aren't true.
A bivalue cell has multiple, disjoint possible outcomes  specifically, two. When you assume one candidate is true, then you must conclude that the other candidate isn't true. Example: (59)r1c3 in the grid above. The assumption of <5> being true forces the conclusion of <9> not being true.
Addendum:
Quote:  Note: eureka notation should always begin and end with a strong link.

Not true. Eureka notation does not have this constraint. In Myth Jellies definition of an AIC, it must start and end with a strong inference, but an AIC loop starts with a strong inference and ends with a weak inference. In addition, ronk would occasionally write discontinuous loops in Eureka notation. Some of his loops started with a weak inference and ended with a weak inference. It was a good way to show that the starting assumption of true resulted in a final conclusion of not true.
[Edit: removed last comment.]
Last edited by daj95376 on Fri Nov 02, 2012 4:41 pm; edited 1 time in total 

Back to top 


ronk
Joined: 07 May 2006 Posts: 397

Posted: Fri Nov 02, 2012 4:37 pm Post subject: 


daj95376 wrote:  In addition, ronk would occasionally write discontinuous loops in Eureka notation. Some of his loops started with a weak inference and ended with a weak inference. It was a good way to show that the starting assumption of true resulted in a final conclusion of not true. Of course, ronk would then ruin it by repeating the conclusion  similar to others adding a conclusion at the end of an AIC. _ _ 
I don't think that last sentence is very accurate. 

Back to top 


daj95376
Joined: 23 Aug 2008 Posts: 3855

Posted: Fri Nov 02, 2012 4:44 pm Post subject: 


ronk wrote:  I don't think that last sentence is very accurate. 
It was a failed attempt at "whimsey". Someone use to repeat the conclusion of discontinuous loops ... and I thought it was you. Sorry! 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5121 Location: Rochester, NY, USA

Posted: Fri Nov 02, 2012 5:13 pm Post subject: 


Thanks Danny, I will try and digest your working definitions, but I'm having trouble seeing how a bivalue can be both.
Thus far the definitions I've seen are both can't be false for strong and both can't be true for weak.
When I see a chain such as Remote Pairs it just looks like every inference fits the definition of both strong and weak, even though they alternate, and I just haven't been able to get that weak inference through my thick skull.
I can notate such things as M and WWings, XYChains and simpler AICs, but it's all mechanical, writing and following the rules without really understanding the inferences I'm writing.
Code: 
++++
 . . .  . . .  . . . 
 . . .  . . .  . 35 . 
 . . .  . . 35  35 . . 
++++
 . . .  . . .  . . . 
 . . .  . . .  . . . 
 . . .  . . .  . . . 
++++
 . . .  . . .  . 35 . 
 . . .  . . .  . . . 
 . . .  . . .  . . . 
++++

Play this puzzle online at the Daily Sudoku site 

Back to top 


arkietech
Joined: 31 Jul 2008 Posts: 1708 Location: Northwest Arkansas USA

Posted: Fri Nov 02, 2012 5:31 pm Post subject: 


Code:  ++++
 . . .  . . .  . . . 
 . . .  . . .  . 35 . 
 . . .  . . 35  35 . . 
++++
 . . .  . . .  . . . 
 . . .  . . .  . . . 
 . . .  . . .  . . . 
++++
 . . .  . .35  . 35 . 
 . . .  . . .  . . . 
 . . .  . . .  . . . 
++++
(3=5)r3c6(5=3)r3c7(3=5)r2c8(5=3)r7c8 => 35r7c6 


Back to top 


JC Van Hay
Joined: 13 Jun 2010 Posts: 364 Location: Charleroi, Belgium

Posted: Fri Nov 02, 2012 9:37 pm Post subject: 


arkietech wrote:  Code:  ++++
 . . .  . . .  . . . 
 . . .  . . .  . 35 . 
 . . .  . . 35  35 . . 
++++
 . . .  . . .  . . . 
 . . .  . . .  . . . 
 . . .  . . .  . . . 
++++
 . . .  . .35  . 35 . 
 . . .  . . .  . . . 
 . . .  . . .  . . . 
++++
(3=5)r3c6(5=3)r3c7(3=5)r2c8(5=3)r7c8 => 35r7c6 
 Dan, your chain implies only : 3r3c6=3r7c8 :=> 3r7c6
To get 5r7c6, one should also write : (5=3)r3c6(3=5)r3c7(5=3)r2c8(3=5)r7c8 => 5r3c6=5r7c8 :=> 5r7c6 !
That said, by comparison : as in a loop, the 4 cells have only 2 solutions giving either (r3c6=3 and r7c8=5) or (r3c6=5 and r7c8=3) forbidding r7c6=3 and r7c6=5 in both cases. 

Back to top 


arkietech
Joined: 31 Jul 2008 Posts: 1708 Location: Northwest Arkansas USA

Posted: Fri Nov 02, 2012 10:12 pm Post subject: 


JC Van Hay wrote:  arkietech wrote:  Code:  ++++
 . . .  . . .  . . . 
 . . .  . . .  . 35 . 
 . . .  . . 35  35 . . 
++++
 . . .  . . .  . . . 
 . . .  . . .  . . . 
 . . .  . . .  . . . 
++++
 . . .  . .35  . 35 . 
 . . .  . . .  . . . 
 . . .  . . .  . . . 
++++
(3=5)r3c6(5=3)r3c7(3=5)r2c8(5=3)r7c8 => 35r7c6 
 Dan, your chain implies only : 3r3c6=3r7c8 :=> 3r7c6
To get 5r7c6, one should also write : (5=3)r3c6(3=5)r3c7(5=3)r2c8(3=5)r7c8 => 5r3c6=5r7c8 :=> 5r7c6 !
That said, by comparison : as in a loop, the 4 cells have only 2 solutions giving either (r3c6=3 and r7c8=5) or (r3c6=5 and r7c8=3) forbidding r7c6=3 and r7c6=5 in both cases. 
In my way of looking at it the notation shows it both ways.
(3=5)r3c6(5=3)r3c7(3=5)r2c8(5=3)r7c8 => 35r7c6
but I'm not too smart. 

Back to top 


daj95376
Joined: 23 Aug 2008 Posts: 3855

Posted: Fri Nov 02, 2012 11:13 pm Post subject: 


Marty R. wrote:  Thanks Danny, I will try and digest your working definitions, but I'm having trouble seeing how a bivalue can be both.

A bivalue cell is neither a strong inference nor a weak inference. There's only how you use the bivalue cell that relates to strong inference and weak inference.
[Edit: dropped clunky analogy using a screwdriver.]
Last edited by daj95376 on Sun Nov 04, 2012 4:01 am; edited 1 time in total 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3150 Location: near Detroit, Michigan, USA

Posted: Sat Nov 03, 2012 12:35 am Post subject: 


daj95376 wrote:  Marty R. wrote:  Thanks Danny, I will try and digest your working definitions, but I'm having trouble seeing how a bivalue can be both.

A bivalue cell is neither a strong inference nor a weak inference. There's only how you use the bivalue cell that relates to strong inference and weak inference.
Think of a screwdriver. One day you use it to secure a wood screw. The next day you use it to open a can of paint. The next day you tell your son to get you the screwdriver. He asks if you are going to use it on a wood screw or a can of paint. The screwdriver doesn't dictate how it's used until you place it in context with your planned course of action. The same is true of a bivalue cell (screwdriver) and whether or not you plan to use it for a strong inference (wood screw) or for a weak inference (can of paint). 
I am missing the point here.
How is a bivalue cell a weak inference between the two candidates?
Your logic may only need a weak inference. That does not change the fact it is a strong inference.
Help me here:
(3=5)r3c6 says r3c6 is a bivalue cell. Right?
(35)r3c6 says what??
Keith 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5121 Location: Rochester, NY, USA

Posted: Sat Nov 03, 2012 12:41 am Post subject: 


Thanks again Danny. I had no idea that this thing was so looseygoosey. I hope that will make my learning process easier. 

Back to top 


ronk
Joined: 07 May 2006 Posts: 397

Posted: Sat Nov 03, 2012 2:18 am Post subject: 


keith wrote:  Help me here:
(3=5)r3c6 says r3c6 is a bivalue cell. Right?
(35)r3c6 says what?? 
That the two candidates share a cell, and we don't know, or even care, if the cell contains other candidates.
Honestly, I don't mean to sound flippant. 

Back to top 


daj95376
Joined: 23 Aug 2008 Posts: 3855

Posted: Sat Nov 03, 2012 6:24 am Post subject: 


keith wrote:  Help me here:
(3=5)r3c6 says r3c6 is a bivalue cell. Right?
(35)r3c6 says what??

Wrong!
(3=5)r3c6 says there's a strong inference between the candidate <3> and the candidate <5>, and that both candidates reside in r3c6. Coincidentally, the only way this can happen is if r3c6 is a bivalue cell.
(35)r3c6 says there's a weak inference between the candidate <3> and the candidate <5>, and that both candidates reside in r3c6. It's inconsequential that r3c6 is a bivalue cell.
Strong inference and weak inference specify relationships between candidates. It is not a property of the cell(s) containing the candidates. The cell(s) only indicate where the relationship occurs. 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3150 Location: near Detroit, Michigan, USA

Posted: Sat Nov 03, 2012 3:56 pm Post subject: 


daj95376 wrote:  keith wrote:  Help me here:
(3=5)r3c6 says r3c6 is a bivalue cell. Right?
(35)r3c6 says what??

Wrong!
(3=5)r3c6 says there's a strong inference between the candidate <3> and the candidate <5>, and that both candidates reside in r3c6. Coincidentally, the only way this can happen is if r3c6 is a bivalue cell.
(35)r3c6 says there's a weak inference between the candidate <3> and the candidate <5>, and that both candidates reside in r3c6. It's inconsequential that r3c6 is a bivalue cell.
Strong inference and weak inference specify relationships between candidates. It is not a property of the cell(s) containing the candidates. The cell(s) only indicate where the relationship occurs. 
I don't see what can be Wrong! with asking a question. Let me put it another way. If r3c6 contains only the two candidates <35>, why on earth would you write (35)r3c6?
Keith 

Back to top 


daj95376
Joined: 23 Aug 2008 Posts: 3855

Posted: Sat Nov 03, 2012 4:58 pm Post subject: 


keith wrote:  I don't see what can be Wrong! with asking a question. Let me put it another way. If r3c6 contains only the two candidates <35>, why on earth would you write (35)r3c6?

The Wrong! was in response to your Right?
Well, you would if you intend to use Eureka notation for this MWing. (see r2c6)
Code:  Puzzle: Ruud50k #39657 (1step)  from ronk's collection of MWings
..5..3.1...7.6....3.91.7..........25..8...9..74..........2.84.9....1.5...6.9..8..
++
 6 28 5  48 9 3  27 1 478 
 4 1 7  58 6 c25  23 9 38 
 3 28 9  1 248 7  26 5 468 
++
 9 3 16  4678 478 146  167 2 5 
 12 5 8  367 237 d126  9 4 367 
 7 4 126  356 235 9  136 8 36 
++
 15 7 13  2 35 8  4 6 9 
 8 9 34  3467 1 46  5 37 2 
 a25 6 234  9 3457 b45  8 37 1 
++
# 60 eliminations remain
*********
MWing 1A: (2=5)r9c1  r9c6 = (52)r2c6 = (2)r5c6 => r5c1<>2



Back to top 


