## Network 6/15/2014

Post puzzles for others to solve here.

### Re: Network 6/15/2014

David P Bird wrote:
Code: Select all
` *-----------------*-----------------*-----------------* | <2>  3    9     | 7    <5>  1     | <8>  46   46    |  | 1    <4>  <8>   | 9    <2>  <6>   | 3    <7>  <5>   |  | 7    <5>  <6>   | 4    3    <8>   | 9    2    1     |  *-----------------*-----------------*-----------------* | 36   12   123   | <8>  4    <5>   | 267  <9>  237   |  | <9>  <8>  234#  | 6    <1>  7     | 24#  <5>  23    |  | 46#  <7>  <5>   | <2>  9    <3>   | <1>  <8>  46    |  *-----------------*-----------------*-----------------* | <5>  12   127   | 13   8    <4>   | 267  36   9     |  | 348# <9>  347@  | <5>  <6>  <2>   | 47#  <1>  78    |  | 48@  <6>  124   | 13   7    9     | 5    34   28    |  *-----------------*-----------------*-----------------*`

(2a=4)r5c7 - (4b=7)r8c7 - (7c=8)r8c9 - (8)r9c9 = (8-4)r9c1 =[(4)Skyscraper:r5c7,r6c19,r8c17]= (4d)r8c3 - (4=7e)r8c7 - (7=8)r8c9 - (8=2f)r9c9
=> [af] r7c7 <> 2, [bd]r8c1 <> 4, [ce] r7c7 <> 7 ste

Blue, looking at your skyscraper, I couldn't see a strong link between it and just (4)r9c1.
To disrupt it (without using memory/networking) either (4)r9c1 or (4)r8c3 must be true deriving a strong link between them. This allows (4)r8c1 and (4)r9c3 to be directly eliminated, but these aren't enough to solve the puzzle without one further chain - the simple approach.

Playing with it the skyscraper, I found I could locate continuous loops to pick off individual candidates but not enough of them together. I then found the weird extended chain above that traverses the same cells front and back to combine the essential eliminations.

Although it's weird, it's perfectly legitimate I think.

Hi David,

This is what I had in mind:

Code: Select all
`                                         +-----------------------+T can be either 4r5c3 or 4r6c9           | Skyscraper (4)c17\r8  |                                         |                       |                                         | 4r6c1 ------------------- T                                         |   ||                  | /                                         | 4r8c1 - r8c7 = 4r5c7 ---                                         |   ||                  |                                         +-----------------------+                                             ||T - (4=2)r5c7 - r7c7 = (2-8)r9c9 = 8r9c1 - 4r9c1`

I probably should have included the Kraken (column) equivalent:

Code: Select all
`Kraken column 4c1:4r6c1 - T  ||4r8c1 - r8c7 = 4r5c7 - T  ||4r9c1 - 8r9c1 = (8-2)r9c9 = r7c7 - (2=4)r5c7 - T`

You'll note that 4r8c3 doesn't play a role.

Your chain looks interesting, but I'm having the same trouble that you did -- I can't seem to fit the skyscraper part into the picture. It looks like maybe you're thinking about one where the strong links are in r6 and r8, and 4r8c3 is an "extra candiate" -- an "almost skyscraper". I can see where that one has potential, since it would eliminate the 4r5c7 in the initial (2=4)r5c7 link, but I can't tie it in with the (8-4)r9c1 link, that preceeds it in your chain.
blue

Posts: 683
Joined: 11 March 2013

### Re: Network 6/15/2014

Blue, yes as I thought, you were using a network not an AIC as it seemed to be notated.

I'm sorry for the confusion I've caused you, but I was considering your skyscraper as an oddagon but kept your description for simplicity.

At the risk of teaching grandma to suck eggs, this is the (4) map with the oddagon (or broken wing) is shown by the '#' cells:
Code: Select all
` *--------------*--------------*--------------* | .   .   .    | .   .   .    | .   4   4    |  | .   4   .    | .   .   .    | .   .   .    |  | .   .   .    | 4   .   .    | .   .   .    |  *--------------*--------------*--------------* | .   .   .    | .   4   .    | .   .   .    |  | .   .   4#   | .   .   .    | 4#  .   .    |  | 4#  .   .    | .   .   .    | .   .   4    |  *--------------*--------------*--------------* | .   .   .    | .   .   4    | .   .   .    |  | 4#  .  (4)   | .   .   .    | 4#  .   .    |  |(4)  .   4    | .   .   .    | .   4   .    |  *--------------*--------------*--------------*`

If r8c3 and r9c1 were both false these cells would form a ring of five conjugate links in b4,r5,c7,r8,c1. As a simple colouring of this pattern would produce a contradiction, it must be disrupted by one of those cells being true. The oddogon pattern therefore derives a strong link between these disruptors. Building the rest of the chain away from this strong link is then tortuous as there are multiple paths possible.

The existence of (4)r9c3 adds an interesting complexity as it looks as if it could also be a disruptor. I don't remember seeing this being discussed before, but I've always considered it a rule that a disruptor must be a member of one of the houses that cover the oddagon so (4)r9c3 wouldn't qualify. In fact if it's true it just creates another contradiction as it would eliminate r5c3 directly and r5c7 because r8c7 would be forced, which would leave r5 without a (4).
David P Bird
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### Re: Network 6/15/2014

daj95376 wrote:
pjb wrote:There were howls of indignation at my post on Jun 10: is Leren's post above any different?

The "howls of indignation" was yet another example of disagreement over definitions. As for your solution, it could have been written as a discontinuous loop or an XY-Chain: (note: typo in your solution corrected)

As for Leren's post, who knows what others are calling a Forcing Chain (w/Contradiction) today. To me, it's an acceptable description of his solution. As I mentioned above, if he were to reverse the streams, then he could call it a Kraken Cell solution. Either way, his solution is still a network.

The 'howls of indigination' had nothing to do with a disagreement over definitions. It had to do with the use of a 'forking' methods long thrown out in the dustbin when it comes to elegant (read: as non-assumptive as possible) manual solving, especially when it comes to simple puzzles. The methods used by pjb and Leren are close cousins (which is, presumably, why pjb raised the question).

A lot of the discussion about elegant manual solving -using as few assumptions as possible and in as few chains as possible- which occurred in the past led to many of the advanced techniques we have today. I can't help but feel that much of your perspective on the subject is colored by the fact that a computer solver doesn't care very much about solving elegance (ie. bifurcation, forking, forcing chains, Nishio, Ariadne's thread, networks, AICs- who cares which one as long as an elimination is arrived at).
DonM
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### Re: Network 6/15/2014

Don, the 'howls of indignation" was a piece of spin doctoring. The original question that was answered was what to call a method based on a single opening assumption.

The two principle rules for AICs are: 1) All nodes are Boolean and 2) Weak and strong links must alternate. If either of these disciplines is broken, it's simply self-delusional to claim the notated solution is an AIC.

That's not to say that other methods are logically invalid though, and indeed, experience shows they are required for the hardest puzzles. But used at the first opportunity they take the challenge out of the moderate puzzles. That's why so many authors have used use the term 'measures of last resort' for them.

The challenge to find a single step solution for a particular starting position isn't an unworthy one, but unfortunately it favours the less disciplined approaches when often two short AICs might use fewer inferences and be far more 'elegant'. It could be that counting the inferences required would make for an interesting secondary challenge?

A Sudoku puzzle is similar to a game of patience that you can try to solve following the rules or you can choose to cheat when the right cards don't turn up. Even if you permit yourself to cheat, you must then come to some sort of decision about how much and how often you'll allow yourself to bend the rules and still claim success.
David P Bird
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Location: Middle England

### Re: Network 6/15/2014

JC Van Hay wrote:UR(12)r47c23=*[3r4c1=*7r7c3-7r7c7=XYWing(24-6 )r57c7,r6c9-6r7c7=6r4c1] :=> -3r4c1

Ha! That's a cool one

Luke
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Location: Southern Northern California

### Re: Network 6/15/2014

David P Bird wrote: ...The challenge to find a single step solution for a particular starting position isn't an unworthy one, but unfortunately it favours the less disciplined approaches when often two short AICs might use fewer inferences and be far more 'elegant'. It could be that counting the inferences required would make for an interesting secondary challenge?

A Sudoku puzzle is similar to a game of patience that you can try to solve following the rules or you can choose to cheat when the right cards don't turn up. Even if you permit yourself to cheat, you must then come to some sort of decision about how much and how often you'll allow yourself to bend the rules and still claim success.

I think that the core of what we are both talking about is 'the art of sudoku solving' which you so eloquently describe above. One can fool oneself that the end justifies the means by the use of a computer solver or the use of sledgehammer methods that bring down simple puzzles, but in doing so one is missing the real fun of the solving process.

That said, it's likely that these highly assumptive methods are sometimes resurrected innocently, because, given the destruction of the 2 major sudoku forums, there isn't the written history and guidance that was previously available. Which is why 'we' pop in to comment every once in awhile.

Digressing a bit, I would like to be more of a solving presence here, but there's no challenge to solving what is close to the same puzzle over and over. I understand that one-steppers are easy to do on a daily basis when time is limited, but, really, almost by definition, they are puzzles that have one or more backdoors and it's far too easy to just find out where they are and reverse engineer. It would be nice to see, at the very least, a puzzle introduced once a week that requires at least 2 steps. That might bring in a few more regular solvers.
DonM
2013 Supporter

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