## January 3, 2015

Post puzzles for others to solve here.

### Re: January 3, 2015

daj95376 wrote:What is the AIC/Eureka notation for the following?
Code: Select all
`(-v)r56c2 -> (wx)r56c2 -> (yz)r89c2 -> (u)r7c1 -> (r)r7c8 -> (v)r4c8  =>  (-v) r4c123`

Mine would be
(v=wx)r56c2 - (w|x=yz)r89c2 - (yz=u)r7c1 - (u=r)r7c8 - (r=v)r4c8 => r4c123 <> v
wxr56c2 can only mean w or x in the cell
yzr89c2 means both y and z in the 2 cells (more commonly used in this sense)
(w|x)r89c2 means w or x in the 2 cells
vr4c123 can only mean v in one of the 3 cells
[as stated (wx=yz)r89c2 meaning "both" on both sides is not possible, but then you don't know, on which side there should be the "or"]

Note, that this is only my suggestion, i am not interested to engage in clearer AIC definitions, after the inventors and defenders have missed that. After all it is just my secondary notation.
eleven

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### Re: January 3, 2015

eleven wrote:...Note, that this is only my suggestion, i am not interested to engage in clearer AIC definitions, after the inventors and defenders have missed that.

That's just not true, but understandable that it would come from someone who repeatedly said years ago that the notation was useless. While I fall into the category of 'defender' -and am one of the very few left who followed the development of the Eureka notation from its inception- I don't pretend to have retained the wealth of knowledge that accumulated regarding the Eureka AIC notation on the Eureka forum over a period of 5 years.

If that forum had not been totally obliterated by hacking and if we still had access to the original 'inventors' such as Ruud, Myth Jellies et al, we would be able to get more clarity regarding what the best notation is. Not to mention that during the very active period of manual solving of the UK Extreme puzzles from 2008-2011 a lot of fine tuning of the notation occurred as solvers came up with new constructs; all that is lost.

So, even though the Eureka notation could have continued to evolve if the Eureka forum had survived, I can't say that I've seen any solving methods or constructs on this forum over that last few years that the Eureka notation wouldn't already have addressed pretty well, if only solvers here had the ability to refer to archived Eureka forum information.
DonM
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### Re: January 3, 2015

DonM wrote:So, even though the Eureka notation could have continued to evolve if the Eureka forum had survived, I can't say that I've seen any solving methods or constructs on this forum over that last few years that the Eureka notation wouldn't already have addressed pretty well, if only solvers here had the ability to refer to archived Eureka forum information.

Who owns the archived information? Can it possibly be made available?
Last edited by SteveG48 on Wed Jan 21, 2015 3:48 am, edited 1 time in total.
Steve

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### Re: January 3, 2015

My understanding is that it is very throughly lost at this point. There are backups of tiny bit and pieces here and there, but the original copy is long gone from this world.

JasonLion
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### Re: January 3, 2015

JasonLion wrote:My understanding is that it is very throughly lost at this point. There are backups of tiny bit and pieces here and there, but the original copy is long gone from this world.

DonM
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### Re: January 3, 2015

SteveG48 wrote:Who owns the archived [Eureka notation] information? Can it possibly be made available?

Steve, in a Eureka notation every term must be a stand-alone Boolean independent of the truth state of any other term in the chain. This is what supports the bidirectional logic that AICs rely on to prove the elimination theorem. It also makes it easy to check if they are logically sound.

In forcing chains there are no such limitations and the requirements for a term to be considered true or false can be inferred from its context in a left to right reading of the chain.

Today's contributors are more interested in making their notations as terse as possible, each in their own way, and rely on their readers to sort out any ambiguities. As Don has said this is a self-serving approach. I also think that when they are asked for clarification, they spend more time responding and justifying their approach than their shorthand notations have saved. In this anarchic state discussions about notation are therefore a waste of time and space.

In practice posts from contributors who deviate too far from some evolving standard won't get read and slowly some popular norm will be reached but, like slang, it will be subject to the fashion of the day.
David P Bird
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### Re: January 3, 2015

David P Bird wrote:Today's contributors are more interested in making their notations as terse as possible, each in their own way, and rely on their readers to sort out any ambiguities. As Don has said this is a self-serving approach. I also think that when they are asked for clarification, they spend more time responding and justifying their approach than their shorthand notations have saved. In this anarchic state discussions about notation are therefore a waste of time and space.

Or maybe they simply don't know any better. On the personal front, while I try to keep solutions as short as possible, I'm trying to learn the clearest notation I can. To that end, and as a result of this thread, I've been backing away from "abbreviated" notation. Unfortunately, it seems that the only way to really learn the notation is by following the example set by others (not always good!), and by occasional suggestions from other members. It's unfortunate that the archives to which Don refers have been lost. It's also unfortunate that there aren't that many members actually posting solutions on a regular basis. That limits the number of examples that we have to follow and the number of different patterns to which we're exposed. If order can't be imposed- and it really can't without ruining what is and should be a pleasurable pastime- then the only way out of anarchy is by setting the example.

One suggestion that I have is that someone who is knowledgeable about the notation and about what was done in the past and talked about in this thread might summarize it as much as possible. Then it could be posted in a "sticky" where it would be an easy reference.
Steve

SteveG48
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### Re: January 3, 2015

SteveG48 wrote:
David P Bird wrote:Today's contributors are more interested in making their notations as terse as possible, each in their own way, and rely on their readers to sort out any ambiguities. As Don has said this is a self-serving approach. I also think that when they are asked for clarification, they spend more time responding and justifying their approach than their shorthand notations have saved. In this anarchic state discussions about notation are therefore a waste of time and space.

Or maybe they simply don't know any better. On the personal front, while I try to keep solutions as short as possible, I'm trying to learn the clearest notation I can. To that end, and as a result of this thread, I've been backing away from "abbreviated" notation. Unfortunately, it seems that the only way to really learn the notation is by following the example set by others (not always good!), and by occasional suggestions from other members. It's unfortunate that the archives to which Don refers have been lost. It's also unfortunate that there aren't that many members actually posting solutions on a regular basis. That limits the number of examples that we have to follow and the number of different patterns to which we're exposed. If order can't be imposed- and it really can't without ruining what is and should be a pleasurable pastime- then the only way out of anarchy is by setting the example.

Good post and good thoughts on the subject. Here are some of mine:

The Good News:
Actually, there are 4 or 5 relatively regular solvers here, including yourself who do a fairly good job of using the Eureka notation for most basic chains and some of the more complex ones.

1. While I actually do appreciate the fun-factor (for others, but not really for me) of doing the typical daily puzzles here, a problem often arises in the attempt to keep the solutions to one line. When a simple chain won't do the job, questionably notated constructs are sometimes used in an attempt to shoe-horn everything into one chain resulting in a bastardization of the notation. (I hail from a solving era where a 2 chain simple chain solution was considered more elegant than a complex one-line net, but that's another subject.)

2. The over/inappropriate of computer solvers has affected the advance of good Eureka notation. A number of the computer solvers were programmed years ago and will spit out contructs and/or various terms that were defined/used in the first fews years of sudoku, but which were eventually no longer used as it was determined that either they could be incorporated into Eureka notation or they should be totally discarded as being too assumptive. I find it startling to periodically see some of these be resurrected on this forum as if we are back in 2006-7.

3. As already mentioned, the tendency for some solvers to insist on their shorthand, acronyms or other non-standard methods interferes with notation standardization. With so few overall solvers to begin with, that is a real problem.
DonM
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### Re: January 3, 2015

DonM wrote:2. The over/inappropriate of computer solvers has affected the advance of good Eureka notation. A number of the computer solvers were programmed years ago and will spit out contructs and/or various terms that were defined/used in the first fews years of sudoku, but which were eventually no longer used as it was determined that either they could be incorporated into Eureka notation or they should be totally discarded as being too assumptive. I find it startling to periodically see some of these be resurrected on this forum as if we are back in 2006-7.

Hi Don,

You've used the term "assumptive", and mentioned the idea that some things can be "too assumptive", and (perhaps) used the word in other ways, recently ... "least assumptive", I think (?) ... but I've never been able to grasp exactly what you mean.
Can I prevail upon you to do your best to clarify the meaning(s) ?
E.g. "more or less assumptive than something else" ... what might that mean ?
I'm guessing that techniques like singles, locked sets and unfinned fish, you'ld classify as being "non-assumptive" (?).

Also, would you shed some light on a threshold (or threshold area) where you perceive that things have become "too assumptive" ?
Couch it in terms of "IMHO", if you like. In fact, please do.
From what I remember about the Eureka forum, the conversations could become a little heated.
In fact, it was so bad (in that way) that after just a few visits, I tended to avoid it like the plague.

This is a side issue, but: I wonder if you think that there may have been advances in "Eureka notation" over the past (almost a full) decade, and that it might be worthwhile to view it as an "evolving standard" ?

I hope I'm OK in saying "thanks in advance" ...

Best regards,
Blue.
blue

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### Re: January 3, 2015

Eleven
The chain then has to be interpreted different from right to left

The logic that AICs are reversible is not in question
All that happens is that the reversibilty may require branching.
But that is true of virtually any non-basic chain.

daj
aran appears to be the only one willing to accept

Yes I would not have written that in any other way

Eleven
Mine would be
(v=wx)r56c2 - (w|x=yz)r89c2 - (yz=u)r7c1 - (u=r)r7c8 - (r=v)r4c8 => r4c123 <> v

wx are excluded from r89c2 every bit as much as yz from r7c1
therefore there is no need to adopt a theoretical wIx which I think confuses rather than enlightens
but if you do wish for wIx then do you not need yIz ?

David P Bird
in a Eureka notation every term must be a stand-alone Boolean independent of the truth state of any other term in the chain. This is what supports the bidirectional logic that AICs rely on to prove the elimination theorem. It also makes it easy to check if they are logically sound

You know well that every AIC chain is bidirectional in the sense that if a=>b then ~b=>~a
It may be that the route back has a different "topology" but that is a different point.
Take a simple case
a=b-b=XYwing-z=a
Now reverse that :
a=z then what ? Necessarily -z in two different cells, so it's no longer a basic chain.
Of course the initial chain only appears basic because of the acceptance of the pattern.
The point is this :
many interesting AIC chains are non-basic and reverse non basically.
The quest for concision which I do not fear will I think continue
blue
"assumptive"
It will be interesting to see what DonM may reply.

I think there can be no satisfactory answer.
Take an AIC.
It is most unlikely that the solver just looked at the board and saw both ends.
More likely he tried...and erred (ie got nowhere : the word "error" in trial and error is a misnomer...there is no error just failure to find something useful)...with a few starting points. Then found a chain a=b.....=a =><a>.
If we resume, he assumed a few x=y, tried and "erred" with those, ; then assumed a=b, and ended up with a useful AIC, that is a useful truth (all the others were valid truths just useless).
So at the end of the process there is a useful truth a=...=a =><a> but it was forged through trial, "error", and assumption.

Most solvers I think adhere to the idea that a pattern is non-assumptive.
But once again patterns are most often not seen by a quick glance at the board. The solver goes looking for a jellyfish. He tries with a digit, errs, tries with another digit and so on. Once he finds the pattern, he knows the logic is solid : he doesn't need to go through the steps (which proceed by contradiction - often otherwise frowned upon - that is advancing by assumption of the form : if this true, then contradiction).
So even here, trial, "error, and assumption are shadows in the background.

Yet who would argue that a jellyfish inside an AIC is not a pleasant thing to behold.
aran

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### Re: January 3, 2015

Right from the early days of 2006 I have found the phobias about bifurcation, assumption, nets and T&E to be highly objectionable, irrational, idiosyncratic and, frankly, disgusting and ridiculous. Most of these bloody and nauseating battles were fought in the Eureka battlefield, where I started my sudoku experience.
All logic is equally assumptive; M. Juillerat commences all his explanations with "If we assume that..."; the word "if" is essentially completely assumptive; to speak of "unwarranted assumptions" as Myth Jellies used to do is without sense.
Blue and Aran expose the difficulties arising from these stupid prejudices. Without "if" there can be no thought or logic in mathematics or in sudoku.
gurth

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### Re: January 3, 2015

aran wrote:Yet who would argue that a jellyfish inside an AIC is not a pleasant thing to behold.

The crux of the problem: how do you show a pattern in Eureka notation?
dan

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### Re: January 3, 2015

gurth wrote:Right from the early days of 2006 I have found the phobias about bifurcation, assumption, nets and T&E to be highly objectionable, irrational, idiosyncratic and, frankly, disgusting and ridiculous. Most of these bloody and nauseating battles were fought in the Eureka battlefield, where I started my sudoku experience.
All logic is equally assumptive; M. Juillerat commences all his explanations with "If we assume that..."; the word "if" is essentially completely assumptive; to speak of "unwarranted assumptions" as Myth Jellies used to do is without sense.
Blue and Aran expose the difficulties arising from these stupid prejudices. Without "if" there can be no thought or logic in mathematics or in sudoku.

Actually, a post laced with a series of negative adjectives reminds me of posts from the 2 or 3 individuals who tended to stir the pot over at the Eureka forum. Thankfully, they were far outnumbered by a lot of mild-mannered, extremely talented people who helped give rise to the real art of sudoku solving -and one of them was Myth Jellies.

One can use semantics regarding the very broad subject of logic (ie. 'All logic is equally assumptive') to cloud/distort the real subject when applied to solving a puzzle game.
Last edited by DonM on Thu Jan 22, 2015 9:21 pm, edited 1 time in total.
DonM
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### Re: January 3, 2015

I see that Don has responded while I've been working on the (very) long post that follows.
He makes good points.

This is what I was working on.
I don't know if it's free of typos (and other errors) at this point, but I'll post it anyway ...

----------------

gurth wrote:(...) I have found the phobias about bifurcation, assumption, nets and T&E to be highly objectionable, irrational, idiosyncratic and, frankly, disgusting and ridiculous.

gurth wrote:Blue and Aran expose the difficulties arising from these stupid prejudices.

Such negativity, gurth !
I'm shocked, and frankly I'm feeling a bit "nauseated".

The absolute last thing I wanted to do with my post, was to ignite another Eureka-style firestorm.
I'm hoping, and indeed truly anticipating that after collecting his thoughts, DonM will respond thoughtfully, respectfully, and with a full understanding that this can be an emotional topic for many people.

May I (respectfully and sincerely) suggest that your post would have been more effetive, if you had left the "stupid" out of "stupid prejudices", and replaced the first part with something along the lines of:
(...) I have found the misgivings about bifurcation, assumption, nets and T&E to be highly questionable.

---

aran wrote:blue
"assumptive"
It will be interesting to see what DonM may reply.

My sentiments exactly.

gurth wrote:Without "if" there can be no thought or logic in mathematics or in sudoku.

I'm pretty sure -- almost positive -- that I understand what you mean.
But ... and as a way, perhaps, of getting at what Don might mean by "less assumptive" ... let me cover three different situations.

1) A naked pair elimination: (NP: <12>r1c12) => -2r1c9

• One way of looking at it is this: If, in a solution to the puzzle, r1c9 was a '2', then neither of {r1c1,r1c2} could contain a '2'. In that case, the situation would be that the only value that could appear in {r1c1,r1c2}, would be '1'. If r1c1 contained a '1', there would be nothing left to place in r1c2, and vica-versa. Hence r1c9 cannot be a '2' in any solution to the puzzle.
• Another way of looking at it is this: In any solution to the puzzle, r1c1 must contain a '1' or a '2'. If it contained a '2', then r1c9 could not contain a '2'. The alternative to it containing a '2', would be for it to contain a '1', but in that case, r1c2 would need to contain a '2', and again r1c9 could not contain a '2'. Either way, r1c9 cannot contain a '2'.
• A third way of looking at it would be this: The only digits that can go in r1c1 and r2c2, are '1' and '2'. In any solution to the puzzle, one cell would contain a '1' and the, other a '2'. With that, r1c9 cannot contain a '2' in any solution to the puzle.
The three interpretations, seem to be ordered from "most" to "least" assumptive -- but I have no idea if that's what DonM has in mind.

2) A skyscraper elimination. A "skyscraper" seems to sit right on the borderline between being a "chain elimination" and a "pattern elimination" -- some people see it one way, and some the other. For the sake of what follows, suppose that in the "chain view", it looks like this: 1r1c1 = 1r4c1 - 1r4c4 = 1r2c4 => -1r2c3. There are (at least) three way's of looking at the situation here, as well.

• One way of looking at it is this: If, in a solution to the puzzle, r2c3 contained a '1', then neither of {r1c1,r2c4} could contain a '1', and we would be faced with a situation where both r4c1 and r4c4 need to contain a '1'. According to the rules of Sudoku, though, they can't both contain a '1', and so it follows that r2c3 could not contain a '1' in any solution to the puzzle.
• Another way of looking at it is this: In c1, one of {r1c1,r4c1} must contain a '1' (in any solution to the puzzle). Ifr1c1 contained a '1', then r2c3 could not contain a '1'. The alternative is that r4c1 contains a '1', but that would mean that r4c4 cannot contain a '1', and since c4 must contain a '1', it would need to go in r2c4, and again r2c3 could not contain a '1'. In either case, r2c4 cannot contain a '1' in any solution to the puzzle.
• A third way of looking at it would be this: Since In any solution to the puzzle, r4c1 and r4c4 can't both be '1', and so one of {r4c1,r4c4} must be false. With that, and since 1's must appear in both columns, it follows that (in any solution to the puzzle) at least one of {r1c1,r2c4} would need to be a '1', and from that it follows that r2c3 cannot be a '1' in any solution to the puzzle.
In this case, the last option seems to be the "least assumptive" (in fact, "non-assumptive") ... and the first two, might seem to be "equally assumptive", since "If" occurs only once in each. In the other hand, the first case amounts to a "trial and error" elimination, and the second case seems amount to an argument that "if r1c1 didn't contain a '1', then r2c4 would need to contain a '1'", and a statement of the "follow on" consequence.

3) A naked triple elimination, in which all three cells contain all three candidates: (NT: 123r1c123) => -1r1c9. Again, there are (at least) three different ways of looking at the situation:

• The first way, like it was above, would views the elimination as a "trial and error" elimination, where the argument is that in the end, 1r1c9 cannot be "true" in any solution to the puzzle, since if it was true, then (...). The rest of the argument, could be done in a few ways. Easiest, might be to argue that in that situation {r1c1,r1c2} would reduce to a naked pair ... "(NP: <23>r1c12)", which would make it impossible for r1c3 to contain a '2' or a '3', and to conclude that since (from the initial assumption) it also couldn't contain a '1', and no digit could be placed there, it follows that r1c9 cannot contain a '1' (in any solution to the puzzle).
• A second way, would be: First, note that in any solution to the puzzle, r1c1 would need to contain a '1', '2' or '3'. If it contained a '1' then r1c9 could not contain a '1'. (And) if it contained a '2', then r1c23 would be reduced to a naked pair, <13>r1c23, and again r1c9 could not contain a '1'. (And finally) the remaining alternative, would be for r1c1 to contain a '3', in which case r1c23 would be reduced to the naked pair, <12>r1c23, and still r1c9 could not contain a '1'. Since all three options lead to it being impossible for r1c9 to contain a '1', r1c9 cannot be a '1' in any solution to the puzzle.
• The third way, would be similar to the third way in the naked pair case: {r1c1,r1c2,r1c3} can only contain digits '1', '2' and '3', and in a solution to the puzzle, each number would need to appear in exactly one of those cells. Since that means that a '1' would need to appear in at least one of the cells, it follows that r1c9 cannot contain a '1' in any solution to the puzzle.
In this case, the 3rd option seems to be "least assumptive", and again "non-assumptive" in that the word "if", doesn't appear in the argument. The first two options include "if". In the first, it's only used once, and in the 2nd, it's used twice. On the other hand, the first two arguments appeal to known facts (read "additional arguments") regarding naked pairs ... and if you covered those in all of thier detail, I don't know but it seems like the 2nd argument would have more occurences of the word "if". On the other hand, the first argument, is a "T&E" argument, and many people prefer to avoid such arguments. Would DonM think that the first one is "more assumptive" than the other, regardless of how many "if's" occur ? I really don't know.

Cheers,
Blue.
Last edited by blue on Thu Jan 22, 2015 7:47 pm, edited 1 time in total.
blue

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### Re: January 3, 2015

The kitchen sink seems to have been thrown into the mix here!

To answer Aran and ArkieTech on combining patterns and AICs; a pattern is something that is true or false and can therefore be used as a Boolean node. In the XXWing example if the pattern is made good by eliminating a digit in the middle of the three cells then the AIC would go (a) – (abc)XYWing:cells123 = (b)cells13. Aran, this topic was discussed on this sub-forum while you were off on your extended cruise and is an example of an 'evolving AIC standard' (Blue). The inferences from patterns can be proved by whatever logic is convenient including case by case listing, but they should be recognisable to a trained eye. Where the boundary lies on that score is questionable as we have contributors who are virtually pattern savants.

On assumptivity the argument to claim AICs aren't assumptive went like this. 'First we observe two nodes that are strongly linked and then we apply alternating inference rules to build an elimination chain. At no time did we need to consider which of the starting terms was true, and sometimes they both turn out to be true.' To my way of thinking "just applying the rules" was a bit of BS (Blue Smoke) as the vast majority of solvers won't follow an inference chain purely mechanically but will think of consequences. Therefore it largely boiled down to saying 'twin assumptions good, single assumptions bad'. Nevertheless at the higher difficulty levels there are also efficiency arguments to be made.

I have found that by avoiding 'assumptive' and 'brute force' methods I'm faced with the sort of challenge I want to exercise my brain. I'm not in it as a race – to solve a problem within 15 minutes of it being published - but more like a game of patience. My target is to see how far I extend the difficulty of the problems I can solve, but I appreciate I'm alone in this.

Now I think everyone here has their own self-set targets (eg Eleven's solving without pencil marks), and the easier they make them, the quicker they will lose interest and move on. What seems to the common denominator between solvers (rather than programmer/analysts) is aim to be able to describe and justify their solution paths (hence the desire for common terms and notation methods), otherwise we would all just use a Dancing Links program. We can learn from one another but there is little point in trying to determine who's right and who's wrong, or who's the better solver.
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