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Another question about Eureka notation

 
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Sun Sep 06, 2015 7:13 pm    Post subject: Another question about Eureka notation Reply with quote

I have another question about Eureka notation. Below is a grid that I arrived at after performing the basics of problem #513416 from 1sudoku.net (rated Evil).
Code:

1sudoku.net: # 513146 Evil (after performing the basics:)
 +------------+------------+--------------+
 | 7  8   3   | 4  19  6   | 5    19  2   |
 | 4  5   1   | 2  37  379 | 6    8   79  |
 | 2  6   9   | 8  5   17  | 3    4   17  |
 +------------+------------+--------------+
 | 59 1   457 | 79 6   8   | 479  2   3   |
 | 6  349 47  | 5  37  2   | 8    179 149 |
 | 8  39  2   | 1  4   379 | 79   5   6   |
 +------------+------------+--------------+
 | 59 7   45  | 3  19  149 | 2    6   8   |
 | 1  2   6   | 79 8   5   | 479  3   49  |
 | 3  49  8   | 6  2   479 | 1    79  5   |
 +------------+------------+--------------+

Since the Evil rated puzzles I 'd previously done from this site required chains, I began looking for one and I eventually found the following:
if r7c1=5, then r9c2=9 so r9c6=4 so r8c4=7 so r4c4=9 so r4c1=5;
Whenever I find a chain, I always try to write it out in Eureka notation - but when I tried to write this one out, I couldn't seem to do it without breaking the following rule listed in Sudopedia:

(From Sudopedia's definition of Eureka Notation:) '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.'

Is this rule generally accepted by everyone? The reason I ask is because as I was reading about Eureka on various sites, I discovered some examples that completely broke this rule! I noticed that I was able to represent the chain above in Eureka if I added some extra steps but I was wondering if the chain could be represented exactly as it's presented above?

I also eventually noticed that I could slightly alter (and simplify) the chain to begin and end with pincers (which I recently learned about from Marty and Bat) that goes as follows:
(5=4)r7c3-(4=19)r7c56-(19=7)r8c4-(7=9)r4c4-(9=5)r4c1 => r7c3=5
But I'm still wondering how you represent a chain in Eureka when it doesn't alternate perfectly between strong and weak links. Or is that something that is not done? Thanks in advance, Don.
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Sun Sep 06, 2015 8:32 pm    Post subject: Reply with quote

Yes, it's generally accepted. Don't forget, AIC = Alternating inference chain, so the = and - represent strong and weak inferences, respectively.

In your "if then" example. I'd have probably started with (5-9)=9r9c2 In your notation line, I'd have written the 3rd term as (9=7). How about an example of that chain that doesn't alternate between strong and weak?

I rarely (if ever) see chains that don't alternate.
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Mon Sep 07, 2015 3:22 pm    Post subject: Reply with quote

Marty R. wrote:
How about an example of that chain that doesn't alternate between strong and weak?
I rarely (if ever) see chains that don't alternate.

Hi Marty! I must be really dense when it comes to Eureka. I thought the 'if then' chain that I listed WAS an example that didn't alternate between strong and weak links! The only way I saw to represent it with alternating strong/weak links was to do the following: (9=5)r7c1-(5=4)r7c3-(4=9)r9c2-(9=7)r9c8-(79=4)r9c6-(4=19)r7c56-(9=7)r8c4-(7=9)r4c4-(9=5)r4c1. As you can see, I had to add a number of steps that referenced additional cells because I didn't see how to express the chain with the alternating strong/weak links that only included the 6 cells referenced in the 'if then' chain - even when I began with your (5-9)=9r9c2! Could you show me the entire expression so it might sink in? I'm sure I'm missing something obvious again. Hopefully this will click soon. Thanks.
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Mon Sep 07, 2015 4:16 pm    Post subject: Reply with quote

Don, I didn't think about the inferences when reading the "if then". I'm not saying I'm doing the right thing, but I go ahead and notate without giving much thought to how the numbers fit the definitions of strong and weak.

To notate your "if then", it's necessary to use ALS's (almost locked sets), an incredibly powerful tool. In this case, we're using ALS's as a way of notating what we know is right by doing the steps in our heads. Eureka doesn't seem to lend itself well when the solution gets into locked sets. ALS's can be used creatively to alter (or affect) the solution. Here's my interpretation of the "if then" example.


if r7c1=5, then r9c2=9 so r9c6=4 so r8c4=7 so r4c4=9 so r4c1=5


(5-9)r7c1=r9c2-(9=74)r9c86-(4=197)b8p234-(7=9)r4c4=> -9r4c1

Note the 3rd term. The b8p234 is a way of identifying cells when there's an ALS confined to a box. The "p" stands for "point" and refers to the cells' position in the box, 1-9 going from left to right and down. It's much simpler and easier to follow than listing all the individual cells.
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Mon Sep 07, 2015 11:29 pm    Post subject: Reply with quote

Wow! Thanks Marty! That is interesting! (And not obvious!) I really appreciate your help! Don.
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