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Marty R.
Joined: 12 Feb 2006 Posts: 5718 Location: Rochester, NY, USA

Posted: Wed Aug 20, 2014 5:13 am Post subject: August 20 VH 


Code: 
++++
 3 4 8  6 27 27  9 5 1 
 5 1 6  39 39 8  7 2 4 
 2 9 7  5 4 1  6 3 8 
++++
 6 57 39  27 8 23579  4 1 259 
 4 8 39  1 25 2359  235 6 7 
 1 57 2  4 6 3579  35 8 59 
++++
 7 3 4  8 1 25  25 9 6 
 8 2 5  79 79 6  1 4 3 
 9 6 1  23 235 4  8 7 25 
++++

Play this puzzle online at the Daily Sudoku site
Quote: 
1) XWing (2), c49=>r9c5<>2
2) XYWing (352), pivot r9c5, =>r4c4<>2 
(Select within quote to see solution) 

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arkietech
Joined: 31 Jul 2008 Posts: 1822 Location: Northwest Arkansas USA

Posted: Wed Aug 20, 2014 11:08 am Post subject: 


25wwing in c56 removes a gob of 5's and solves the puzzle 

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rmireland
Joined: 21 Sep 2013 Posts: 32 Location: New Orleans

Posted: Wed Aug 20, 2014 4:34 pm Post subject: 


Hi guys,
I desperately want to get to that Eureka moment, so here is my attempt at starting a Rosetta Stone...
First in english, using Marty's grid, in this puzzle r7c6 must be 5. The key is that r5c5 is either 2 or 5. If the value is 2 then r1c5 is 7 and r1c6 is 2 so the value in r7c6 cannot be 2, so must be 5. If the r5c5 value is 5 then r456c6 cannot be 5 so r7c6 must be 5 since we need a 5 in this column and this is the only remaining possibility.
So I have two paths to represent in one Eureka string. One path might be:
(2)r5c5(7)r1c5(2)r1c6=>2r7c6
The other path might be:
(5)r5c5(5)r456c6=>5r7c6
So, how do I combine these into one Eureka string? Like this?
(2=5)r5c5(7=2)r1c5(2=7)r1c6 [meanwhile on the other path] (5)r456c6=>2r7c6
Is that close? I have no idea what a = or  mean, other than I remember something about strong and weak links. I'm just pretending like I know.
If it's easier, could someone just show Marty's XY in Eureka? It would be much appreciated.
Rick 

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Marty R.
Joined: 12 Feb 2006 Posts: 5718 Location: Rochester, NY, USA

Posted: Wed Aug 20, 2014 7:06 pm Post subject: 


Hi Rick,
An XYWing is actually an XYChain, at three cells the shortest XYChain possible, so the notation would start not with the pivot cell, but at either end of the chain.
My wing would be notated as follows.
(2=3)r9c4(3=5)r9c5(5=2)r5c5=>r4c4<>2
In practical use, if I was describing an XYWing as part of a solution, I wouldn't notate it because by saying the three numbers and the pivot cell, the reader can easily see it and I don't think notating it would clarify anything.
If it was slightly more complicated, such as involving a pincer transport, then I would notate it.
Hope that helps. 

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rmireland
Joined: 21 Sep 2013 Posts: 32 Location: New Orleans

Posted: Wed Aug 20, 2014 10:22 pm Post subject: 


Eureka!
I was trying to do too much with this Eureka notation. I thought it was a means to discovery, but instead it's only a method to show what was discovered some other way.
I had mistakenly thought that it would start with the pivot, so I was confused how to show both paths from the pivot to the pincers. I now see that it starts and ends with the pincers (2=3) (3=5) (5=2) which for this wing are 2. The 3s pair and the 5s pair. so we have a 2 in r9c4 and a 2 in r5c5, and therefore any cell that "sees" both of these cannot be a 2, so =>r4c4<>2.
Because the string simply starts and ends at the pincers, I can read it in either direction and starting at either end, or in the middle somewhere, like at the pivot.
This particular string is setting values. I will next take some time to explore Eureka strings that eliminate possibilities.
Thanks, Marty and Don. I think I may be getting this, finally.
Rick 

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arkietech
Joined: 31 Jul 2008 Posts: 1822 Location: Northwest Arkansas USA

Posted: Wed Aug 20, 2014 11:57 pm Post subject: 


Code:  **
 3 4 8  6 b27 c27  9 5 1 
 5 1 6  39 39 8  7 2 4 
 2 9 7  5 4 1  6 3 8 
++
 6 57 39  27 8 23795  4 1 259 
 4 8 39  1 a25 2395  235 6 7 
 1 57 2  4 6 3795  35 8 59 
++
 7 3 4  8 1 d25  25 9 6 
 8 2 5  79 79 6  1 4 3 
 9 6 1  23 235 4  8 7 25 
**
The wwing expressed in Eureka;
(5=2)r5c52r1c5=2r1c6(2=5)r7c6 => 5r9c5,r456c6; ste
explained: if r5c5 is a 5 then 5r9c5,r456c6 or r9c5,r456c6 cannot be a 5
is not 5 then it is 2 and r1c5 is not 2 and r1c6 is a 2
making r7c6 not 2 and there a 5 then 5r9c5,r456c6 


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Marty R.
Joined: 12 Feb 2006 Posts: 5718 Location: Rochester, NY, USA

Posted: Thu Aug 21, 2014 4:07 am Post subject: 


arkietech wrote:  Code:  **
 3 4 8  6 b27 c27  9 5 1 
 5 1 6  39 39 8  7 2 4 
 2 9 7  5 4 1  6 3 8 
++
 6 57 39  27 8 23795  4 1 259 
 4 8 39  1 a25 2395  235 6 7 
 1 57 2  4 6 3795  35 8 59 
++
 7 3 4  8 1 d25  25 9 6 
 8 2 5  79 79 6  1 4 3 
 9 6 1  23 235 4  8 7 25 
**
The wwing expressed in Eureka;
(5=2)r5c52r1c5=2r1c6(2=5)r7c6 => 5r9c5,r456c6; ste
explained: if r5c5 is a 5 then 5r9c5,r456c6 or r9c5,r456c6 cannot be a 5
is not 5 then it is 2 and r1c5 is not 2 and r1c6 is a 2
making r7c6 not 2 and there a 5 then 5r9c5,r456c6 

Dan, may I offer up an alternative?
(5=2)r7c6r9c4=2r4c4(2=5)r5c5=>same eliminations.
If r4c4=2, then r5c5=5. If r9c4=2, then r7c6=5. Any cell seeing both of r5c5 and r7c6 cannot contain 5. 

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rmireland
Joined: 21 Sep 2013 Posts: 32 Location: New Orleans

Posted: Thu Aug 21, 2014 2:57 pm Post subject: 


Wow! I am the last guy who should be mediating a Eureka discussion between you two. But I have noticed that you are both using the same ending cells, the only difference being the interim path, which we see all the time. There are often many intermediate paths between two pincer cells.
The difference seems to be that Dan is toggling bivalue cell r5c5, while Marty is keying on the fact that there can be only one 2 in c4. Without any criteria for judging paths, I cannot say that either path is better than the other, but Dan's has the advantage that I can see his logic by starting at the beginning of his string, while I have to analyze more to see Marty's starting point. But until I start speedreading Eureka, that's not a big difference.
I do have an observation, though, that I can trace Dan's string in either direction, from either end. With Marty's, I cannot start at 5r7c6 and get to 2r4c4 by following his path thru r9c4 (unless I do the X wing to get 2r9c5 first).
Would it make any sense in these strings to indicate the starting point by using italics or underline to denote the key cell or cells?
And finally Marty, are you missing a 2, as in 2r9c4? Is that just a different dialect of Eureka?
Thanks guys. This has been really helpful, at least to me.
Rick 

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Marty R.
Joined: 12 Feb 2006 Posts: 5718 Location: Rochester, NY, USA

Posted: Thu Aug 21, 2014 3:38 pm Post subject: 


Quote:  And finally Marty, are you missing a 2, as in 2r9c4? 
My second term is r9c4. Using it in conjunction with the first term, if r7c6=2 then r9c4 cannot be=2.
As to my English explanation, that follows how I view a WWing: two remote pairs connected by a strong link. There is a strong link on 2 in c4, each of which sees one of the 25 remote pairs, therefore one of the pairs must be=5 since one of the 2's in c4 must be true.
Hope that's understandable. I offered that up as an alternative to Dan's; it's not better, just different. Different people see things in different ways. 

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dongrave
Joined: 06 Mar 2014 Posts: 285

Posted: Thu Aug 21, 2014 5:05 pm Post subject: 


These discussions have been a huge help to me too! The more Eureka examples I see, the clearer it becomes. I have one question. I noticed that Arkitech's solution is almost identical to Rick's solution posted earlier on this page  but there is one subtle difference that seems to make Rick's a 'forcing chain' of either the 'Unit', 'Digit', or 'Cell' sort (which one I'm not sure) instead of a wwing. How do you represent Rick's 'Digit Forcing Chain' (or whatever type of chain it is) in Eureka? Or is this not the type of solution that one would typically write out in Eureka? Thanks, Don. 

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Marty R.
Joined: 12 Feb 2006 Posts: 5718 Location: Rochester, NY, USA

Posted: Thu Aug 21, 2014 7:38 pm Post subject: 


This is one possible way to notate Rick's "forcing chain."
(5=2)r7c6r1c6=r1c5(2=5)r5c5r456=5r7c6 Contradiction=>r7c6<>2 

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dongrave
Joined: 06 Mar 2014 Posts: 285

Posted: Fri Aug 22, 2014 10:35 pm Post subject: 


Hi Marty, Hey, that's surprisingly simple! I guess I was expecting to see two expressions (one for each path) delimited somehow, and then capped off with symbols for the conclusion. I've been under the impression that these expressions would only be 'correct' if you adhered to some strict set of rules  but now it's appearing to me that maybe it's okay to include embellishments (like adding the word 'Contradiction', or embedding an explanation to a step, etc.) as long as you use some common sense  and as long as others can understand it. Is this true? By the way, the reason I asked specifically about this example is because I only recently discovered these 'multiple path forcing chains' and at the same time I'm trying to understand Eureka better. I can read Eureka expressions that others post but I had no idea how to represent these 'multiple path' solutions. Thanks, Don. 

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Marty R.
Joined: 12 Feb 2006 Posts: 5718 Location: Rochester, NY, USA

Posted: Sat Aug 23, 2014 12:00 am Post subject: 


Don,
If I were doing a forcing chain I'd use two strings of notation to show a common outcome. But you already mentioned that either way the cell had to be=5. So 2 had to be invalid so I started with that to see when the notation would show the contradiction. 

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dongrave
Joined: 06 Mar 2014 Posts: 285

Posted: Sat Aug 23, 2014 12:36 am Post subject: 


Marty R. wrote:  Don,
If I were doing a forcing chain I'd use two strings of notation to show a common outcome. But you already mentioned that either way the cell had to be=5. So 2 had to be invalid so I started with that to see when the notation would show the contradiction. 
Oh, I see! Thanks for the clarification! 

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dongrave
Joined: 06 Mar 2014 Posts: 285

Posted: Sat Aug 30, 2014 1:57 am Post subject: 


I hope Marty's okay! He's normally right on top of removing those unrelated posts by Nina. Hey Marty, are you okay? We all hope so! (except maybe Nina.) Let us know if you're alright![/list] 

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