The New Sudoku Players' Forum

[Ajò Dimonios]

Eleven wrote:
Of course you can express the same with anti-tracks, but the concept is different. Here you define a fixed starting situation and look, what you can get out of it, i.e. if you can find a common outcome (which also can be that one side is wrong), basically the same concept as for double forcing chains.

What you say is the exact same thing as an AIC. Let's see the AIC and the anti-track of this scheme. I can say that the elimination of this AIC is obtained by forcing two chains. The first is (9 = 1) r2c8 - r6c8 = (1-7) r6c6 = 7r2c6-9r2c6 which is what is generally expressed and the second is the very simple one made up of 9r2c8-9r2c6 which is generally not highlighted. 9r2c6 is eliminated because it is wrong for the two chains. The same can be said for P '(9r2c8): (-9r2c8) => 1r2c8-> 1r6c6-> 7r2c6 => -9r2c6 with the theorem “if a candidate Z sees both E and P' (E), Z can be eliminated. In this statement And can be a candidate or a set of candidates. " The same can be said for AICs that create loops.

Paolo

[SteveG48]

Code: Select all

 *--------------------------------------------------*
 | 8    29   5    | 1    46   49   | 23   7    369  |
 | 3    6    4    | 8-7  2   a79   | 18  a19   5    |
 | 1    29   7    | 3    568  59   | 28   4    689  |
 *----------------+----------------+----------------|
 | 6    458  1    | 2    458  3    | 7    59   89   |
 | 7    458  9    | 6    458  145  | 138  2    38   |
 | 2   b58   3    |b78   9    15-7 | 6   b15   4    |
 *----------------+----------------+----------------|
 | 9    7    2    | 5    3    8    | 4    6    1    |
 | 4    1    8    | 9    7    6    | 5    3    2    |
 | 5    3    6    | 4    1    2    | 9    8    7    |
 *--------------------------------------------------*


(7=91)r2c68 - (1=587)r6c248 => -7 r2c4,r6c6 ; stte

[eleven]

9r2c6 is eliminated because it is wrong for the two chains.

This is exactly, what the creators and some defenders of the AIC wanted to avoid for any cost: to run into a contradiction.
In an AIC always at least one side is true, so the AIC never can be contradictory.

[Ajò Dimonios]

Eleven wrote:
This is exactly, what the creators and some defenders of the AIC wanted to avoid for any cost: to run into a contradiction.
In an AIC always at least one side is true, so the AIC never can be contradictory.

What you say is true. At least one of the two hypotheses is true. But I see it on the side of the candidate or of the eliminated candidates (which is in short the purpose I want to achieve) I wonder why it comes and / or are they eliminated? There is no other possible answer, because if it was true it would create a contradiction.

Paolo

[eleven]

If you want to see it that way, yes. (Generally any candidate you can eliminate is a contradiction to the solution.)
The AIC argument is:
One of the endpoints is true => that candidate is false
No contradiction here, it is just an implication. (if you want to argue, the candidate is a contradiction to the implication, it is your choice of seing things)

[Ajò Dimonios]

Eleven whrote:
If you want to see it that way, yes. (Generally any candidate you can eliminate is a contradiction to the solution.)
The AIC argument is:
One of the endpoints is true => that candidate is false
No contradiction here, it is just an implication. (if you want to argue, the candidate is a contradiction to the implication, it is your choice of seing things)

I think we are discussing “the sex of angels”. A truth or a non-truth is the same from the logical point of view. It only depends on the attitude of the person who mistakenly differentiates a positive implication other than a negative implication. It is like seeing a glass half-filled as half full or half empty.

Paolo

[eleven]

Maybe you don't know, that i have nothing against contradictions.
But i understand the argument, who would call the eliminations from a pair a contradiction elimination ? You ? And would you take each digit and show, that it leads to an empty cell, where the pair is ?

[Ajò Dimonios]

Eleven wrote:
Maybe you don't know, that i have nothing against contradictions.
But i understand the argument, who would call the eliminations from a pair a contradiction elimination ? You ? And would you take each digit and show, that it leads to an empty cell, where the pair is ?

I'm sorry that this discussion bores someone. From what you have written I seem to understand that you define as contradiction what you get when the rules of the game are precisely contradicted, while in your previous post you said that also the contradiction that is obtained with the anti-track related to the theorem "if a candidate Z sees both E and P'(E), Z can be eliminated. In this statement E can be a candidate or a set of candidates. into this category." At this point you should define all the contradictions that are not good, clearly not considering the one linked to the "non-truth" of an AIC that I have shown you can be defined as obtained from two chains, one expressed and the other hidden.

Paolo

[SpAce]

In an AIC always at least one side is true, so the AIC never can be contradictory.

Strictly speaking that's only true for idiomatically written AICs, i.e. with a strong link at both ends. Of course I don't think any other kinds should be called AICs, so the definition could just as well have that as a requirement. Unfortunately it doesn't.

In fact, I once suggested that to David, but as expected, he didn't comment because it was me. That leaves two other end link combos possible: both weak or mixed strong and weak. Of those the latter is definitely the worst, and it's not even mentioned as a possibility in David's AIC Primer, though unfortunately it's not explicitly ruled out either. (I think, however, that we should start interpreting it as being ruled out to avoid idiotic debates about it. It would be much easier to just say that it's an invalid AIC, period, instead of having to explain (and fail) why it's merely an incredibly stupid way to write an AIC. Wasting time on debating such nonsense pisses me off.)

Anyway, as long as either of those other options is technically allowed, we can't say that AICs can never be contradictory. Whenever there's a weak link at either end or both (and it's not a closed loop), it becomes a contradiction chain (as per my definition) if it actually eliminates the weakly linked end node (otherwise it's just a chain fragment). Only if both are strong, as they should, then it's a verity chain. Edit: Even then it can actually be considered an implied contradiction chain in the special case when one end point is known to be false, i.e. a deadly pattern. See next post.

PS. AICs can also have (sometimes rather obvious) internal contradictions, but it's ok if the other rules are satisfied and the contradiction is not used as such. It also doesn't make it a contradiction chain, as long as it has a strong link at both ends.

--
Edit. Corrected an oversight.

[SpAce]

I'm only going to clarify what I mean by a chain of contradictions, because obviously we don't have the same definitions and we don't give the same meaning to the terms used.

Obviously. I don't mind you using whatever terms and definitions you choose with your TDP chains, but it would be nice if you accepted our normal definitions when you talk about AICs. In particular, please don't call AICs contradiction chains when they're not (according to our definition). Whenever there's a strong link (=) at both ends, i.e. the normal way of writing them, it can't be a contradiction chain (it's a verity chain). When there's a weak link (-) at either end or both, it's a contradiction chain. Simple.

That said, I don't think your definition (which I think I now understood) makes any sense for your own chains either, but deciding that is up to you. (Or perhaps you can explain to me why it does make sense.)

I say that the following chain (-9r2c8) => 1r2c8->1r6c6->7r2c6 => -9r2c6 is a chain of contradiction because 7r2c6 implies +9r2c8 which is a contradiction with -9r2c8. This contradiction is clearly visible.

No, it's not visible in the written chain at all, so it should not be a property of it either. To me that chain is clearly a verity chain, and it eliminates like a verity chain: an external candidate, using the proven verity (9r2c8 OR 7r2c6) which forbids 9r2c6. It makes zero difference what other unwritten logic could be read from the grid. Only the written chain matters. Thus, if you really want that mentioned extra implication and the resulting contradiction to count for something, you should write them out:

-9r2c8 -> 1r2c8 -> 1r6c6 -> 7r2c6 -> 9r2c8 (contradiction) => --9r2c8 <=> +9r2c8

Now, that's a real contradiction chain, and it eliminates like one: the initial assumption (turning it into a double-negation, i.e. a placement). What do you call that anyway if you already call the previous chain a contradiction chain? I hope you have different names for them because they use completely different logic.

Even that implication chain can actually be seen more logically as a verity chain, because -9r2c8 -> 9r2c8 is the same as (9r2c8 OR 9r2c8), i.e. 9r2c8 must be true. Written in Eureka it's not a contradiction chain at all because it has a strong link at both ends (though they link to the same candidate). Thus it can be seen as both a valid AIC and a Discontinuous Nice Loop Type 2 (starts and ends with a strong link to the same candidate):

(9=1)r2c8 - r6c8 = (1-7)r6c6 = (7-9)r2c6 = (9)r2c8 => +9 r2c8

We just don't normally write AICs like that because they can always be replaced with a shorter elimination chain (which gives the same placement after the elimination).

On the other hand, the chain (-1r6c6) => 1r6c8->9r2c8->7r2c6 => -7r6c6 is not (for me) a chain of contradiction because 7r2c6 does not imply +1r6c6.

I understand the difference but I don't see how it's relevant in any way. Why is this distinction significant to you in the first place? The difference between real contradiction chains and verity chains (using our definitions) is extremely significant, but I don't see how this is.

In the same way, for me Denis Berthier's chains (whip, braid) are chains of contradiction.
For example, whip [3]: c2n3{r1 r5} - c2n8{r5 r3} - r1c3{n8 .} ==> r1c2 ≠ 7 expresses that the placement of 8r1c3 implies a contradiction in the puzzle.

I'm happy to hear that you're familiar with Denis' chains. I'm not but I'd like to learn. I understand the basic bivalue-chains easily (because they're basically AICs), but I don't know anything about whips or braids. I'd be happy if you could help with them (especially whips for now). I tried to ask Denis but he basically told me to look it up. Reading the freely available but extremely theoretical PBCS is not really my cup of tea for the same reasons I found your original TDP documentation tedious. Similar to that, I'm pretty sure the actual concepts (or how they relate to sudoku) are quite simple to understand if only they were explained in terms I already know well. So, I'd be very grateful if someone could help with that

So, you're saying whips are simply contradiction chains? Ok, that should be easy. I still don't understand your example, though. If you place 8r1c3 and it causes a contradiction (where? what?), shouldn't it eliminate 8r1c3 instead of 7r1c2? If I write that chain with Eureka, it doesn't make any sense to me:

Code: Select all

(3)r1c2 = (3-8)r5c2 = r3c2 - (8)r1c3 => -7 r1c2 ?

So, how should it really be interpreted?

Added 1. I think I might have figured it out (maybe). It should probably rather be something like:

Code: Select all

(7-3)r1c2 = (3-8)r5c2 = r3c2 - (8=!)r1c3 => -7 r1c2

In other words, it's not the placement of 8r1c3 but 7r1c2 that yields the contradiction (and the '.' represents that, similar to my '!'). Apparently the whip assumes a placement of the target and any global eliminations (not depicted in my chain) it gives before proceeding with the chain that leads to a contradiction. So, I think you're right that it's indeed a contradiction chain which eliminates the initial assumption. It's just not written explicitly as such, which makes it pretty unintuitive to me.

Added 2. Stupid me. There's actually no need to assume anything. Even with Eureka we can simply write:

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(3)r1c2 = (3-8)r5c2 = r3c2 - (8=!)r1c3 => +3r1c2

That's actually how I would write it too (and have), though I'd probably reverse it (and possibly replace the '!' with 'DP'). It's a rare example of an AIC with strong links at both ends but which is still a sort of contradiction chain because one of the end points is a contradiction (which means the other must be true). The normal AIC logic is valid, i.e. (3r1c2 OR !) => +3r1c2, so it can be seen as a verity chain like any AIC. Yet, because one of the two options is a contradiction forcing the other to be true, it could be seen as an implicit contradiction chain as well. Because of that it's actually an exception to my earlier rule that an AIC with strong links at both ends can't be a contradiction chain -- it can if one of them links to a contradiction.

Anyway, would you agree with that whip interpretation?

--
Edits. Added the new whip interpretation. Amended it with another one.

[eleven]

What you say is the exact same thing as an AIC. Let's see the AIC and the anti-track of this scheme. I can say that the elimination of this AIC is obtained by forcing two chains. The first is (9 = 1) r2c8 - r6c8 = (1-7) r6c6 = 7r2c6-9r2c6 which is what is generally expressed and the second is the very simple one made up of 9r2c8-9r2c6 which is generally not highlighted. 9r2c6 is eliminated because it is wrong for the two chains. The same can be said for P '(9r2c8): (-9r2c8) => 1r2c8-> 1r6c6-> 7r2c6 => -9r2c6 with the theorem “if a candidate Z sees both E and P' (E), Z can be eliminated. In this statement And can be a candidate or a set of candidates. " The same can be said for AICs that create loops.

Paolo, sorry, but i did not understand that weird post at all (please try to read it yourself and to find the Enter button on your keyboard). You talk about forcing two chains (??) and then there is an AIC ending with "-9r2c6" instead of "=> -9r2c6" (??) and then a weak inference (??) ...
So my assumption was, that you tried 9r2c6 (which is normal in TDP) and got a contradiction with 9r2c8.

Later you just insisted, that an implication elimination is a contradiction elimination.

[Ajò Dimonios]

Hi Eleven

Eleven wrote:

Paolo, sorry, but i did not understand that weird post at all (please try to read it yourself and to find the Enter button on your keyboard). You talk about forcing two chains (??) and then there is an AIC ending with "-9r2c6" instead of "=> -9r2c6" (??) and then a weak inference (??) ...
So my assumption was, that you tried 9r2c6 (which is normal in TDP) and got a contradiction with 9r2c8.

Later you just insisted, that an implication elimination is a contradiction elimination.

Hi Eleven I'm sorry I can't explain my point of view, certainly due to my bad English, I try to be clearer. I fully agree in the definition that you give of an AIC and also that at least one of the two AiC start and end statements is true. My discordance with you is in the fact that you interpret the eliminations as an implication of this simple truth. Would you like to explain to me what is the logical mechanism that justifies this implication? In practice I ask you simply why the truth of the AIC leads to eliminate candidates? My answer is that if it were not so, a contradiction would be created. This implication is justified only by this fact, there are no other reasons. It is the same reason that leads when you have a single candidate in a row (truth) implies the elimination of all the other candidates of the same value in the column and in the belonging box (it is a contradiction, they cannot be present in the same box or row or column more than one candidate of the same value). If you are disturbed by the way the anti-track is written you may very well not take into account the last weak inference of the chain, it may very well not write it, the elimination is always valid.
Paolo

[eleven]

Paolo,

if you have a single, you can eliminate the other candidates of the digit of the row/box/column, because according to the sudoku rules the digit can be there exactly once. No contradiction needed.
Maybe you never solved a puzzle without pencilmarks. There you do not even have any other digits of that candidate to eliminate. You have just filled a cell with a number. And obviously only the other digits can go to the other cells of these units.
To construct contradictions here for all the eliminated digits is really far-fetched.

The same with a pair (see above).

The same with an AIC. If a digit has to go to one of 2 cells, it cannot be in any cell, which sees both. dot. No contradiction needed.

I have nothing against the way, the tracks and anti-tracks are written e.g. by Robert, because it is the same as forcing chains are written by myself, with the addition, that digits are remembered (not clearly marked, but once i knew that, i had no problem to understand them).

And again, your text is hard to read in one line. When you paste the translation, please invest some seconds to structure it a bit using the Enter key.

[Ajò Dimonios]

Hi Eleven

if you have a single, you can eliminate the other candidates of the digit of the row/box/column, because according to the sudoku rules the digit can be there exactly once. No contradiction needed.

I'm really sorry that you don't understand what I wrote, you probably have a different concept of contradiction from mine. I looked for the simplest example that any program can run automatically and you answer me saying that breaking this rule is not a contradiction but it is an automatic mechanism not to break it. I frankly don't know what a contradiction is for you. Probably by dint of avoiding them, you call all the mechanisms to avoid them, implications. But then why do you avoid chains of contradiction, you could call them chains that avoid contradictions in order to respect the rules of sudoku, that is, of implication chains.

Paolo

[eleven]

Oh, stupid me, you are a program, and i have wasted so much time.