The New Sudoku Players' Forum

[Ajò Dimonios]

Not all pairs of backdoors lead to this type of solution, only those in which it is possible to demonstrate the existence of a strong inference between them.
Space wrote:
That statement categorically denies the possibility of demonstration by any means. It doesn't say anywhere that demonstrations depending on bifurcation don't count. Adding that rule now means that you're moving the goal posts, because otherwise you'd have to admit that your claim was false. It's obvious that you'd do anything to avoid that horrible option.

t was clearly understood that I could not prove the existence of a strong inference with a single logical chain. To get the demonstration I need at least two steps.

Space wrore:
CODE: SELECT ALL
5,6r12c1 -> -1r12c1 \
-5r1c7 -> || -> -(1r12c1 & 1r3c3) -> 1r9c1|1r3c9
5,1r12c1 -> -1r3c3 /


1r12c1 -> -5,6r12c1 \
-(1r9c1|1r3c9) -> && -> -(5,6|5,1)r12c1 -> 5r1c7
1r3c3 -> -5,1r12c1 /

The passage between the results of the two inference "or" and the inference "and" is not permitted. The results of the two inferences "or" contradict each other. If it is true that r3c3 = 1 is false it can be concluded that it is false that r1c12 = 1 is false and if it is true that r1c12 = 1 is false then it can be concluded that it is false that r3c3 = 1 is false. This fact produces the result that the chain is valid only when r3c3 = 1 is false and not when r1c12 = 1 is false. Consequence of this deduction the two inferences "or" are not verified at the same time.

Paolo

[SpAce]

It was clearly understood that I could not prove the existence of a strong inference with a single logical chain. To get the demonstration I need at least two steps.

I have no problem with that as long as you keep speaking in first person. We can totally agree that you need at least two steps, if you say so. Just don't extend that claim to other people who don't share your limits. If you can't do something, it doesn't mean no one can.

The passage between the results of the two inference "or" and the inference "and" is not permitted.

??? I have no idea what you're saying. Perhaps the extra node in both implication chains got you confused. I added it behind the branching part to combine the results, because it was written that way in the AIC. I assure you, they're perfectly valid. It's not my fault if you can't see that. (Btw, if it wasn't clear, the two implication chains are one and the same chain, only demonstrating that the logic works both ways. It doesn't mean that two chains are needed. Both are separately equivalent to the AIC and to each other.)

The results of the two inferences "or" contradict each other. If it is true that r3c3 = 1 is false it can be concluded that it is false that r1c12 = 1 is false and if it is true that r1c12 = 1 is false then it can be concluded that it is false that r3c3 = 1 is false.

Can you say that in an even more complicated way? Please? If obfuscation is your main argumentation tactic, it's working. I almost didn't even bother to read that, because it's such an utterly horrible way of saying something very simple. Assuming I can follow that at all, you're saying that -1r3c3 -> +1r1c12 and -1r1c12 -> +1r3c3. Not so hard, is it? Or in your language: it is false that it is false that it is true that it is false that it is hard.

In any case, I don't know how you draw such conclusions, but they're neither valid nor have anything to do with my chain. In the solution both 1r3c3 and 1r1c12 are false, so a false value in one (a truth) can't imply a true value in the other (a falsehood), because truths can't lie. Thus such implications are pure nonsense. More importantly, I don't see any relation to my chain. Here's the full logic of my chain written in words:

(5)r1c7 = (5,6|5,1)r12c1 - (1)(r12c1 & r3c3) = (1)r9c1|r3c9 ->singles-> (5)r1c7 => +5r1c7; stte

First strong inference: If 5r1c7 is false then either (5,6)r12c1 OR (5,1)r12c1 must be true, and vice versa: if neither (5,6)r12c1 NOR (5,1)r12c1 is true, then 5r1c7 must be true. Thus, at least one side of the strong inference must be true. The inference is valid.

First (and only) weak inference: If either (5,6)r12c1 OR (5,1)r12c1 is true, then both 1r12c1 AND 1r3c3 can't be true (at least one of them must be false), and vice versa: if both 1r12c1 AND 1r3c3 are true, then neither (5,6)r12c1 NOR (5,1)r12c1 can be true. Thus, at least one side of the weak inference must be false. The inference is valid. (It makes no difference that 1r12c1 AND 1r3c3 can't be true together in reality. AICs don't care about such internal contradictions, as long as the links work. It just means that the chain could be written more shortly as a contradiction chain, as I already demonstrated much earlier.)

Second strong inference: If both 1r12c1 AND 1r3c3 aren't true (at least one is false), then either 1r9c1 OR 1r3c9 must be true, and vice versa: if neither 1r9c1 NOR 1r3c9 is true, then both 1r12c1 AND 1r3c3 must be true. Thus, at least one side of the strong inference must be true. The inference is valid. (Again, the AIC doesn't care about the obvious contradiction. We could use it, however, to cut the chain and prove directly that either 1r9c1 OR 1r3c9 must be true, but that would defeat the purpose of the exercise. Besides, if you don't like that, the other AIC I wrote doesn't have such a direct contradiction.)

That concludes the relevant part of the chain which proves the derived strong link between 5r1c7 and (1r9c1 OR 1r3c9). In other words, it proves that at least one of the three must be true. If one doesn't like the ORed end-point, it's easy to turn into just one or the other, because (as backdoors) the two are equivalent: 1r9c1 <-singles-> 1r3c9, but that just complicates things unnecessarily. We already have the proof we need to proceed. (And consequently, your claim that it wasn't possible for 5r1c7 is proven false. Sorry. I don't expect you to ever admit it, but that's how it is.)

The ->singles-> implication: If either 1r9c1 OR 1r3c9 is true, then it can be proved in both cases with singles that 5r1c7 is true. Both 1r9c1 and 1r3c9 are backdoors just like 5r1c7, so all three imply each other through singles. The implication is valid. QED.

That logic is bulletproof, unless I've suddenly turned braindead. I'm sure you can find holes in it anyway, but it doesn't make them true.

This fact produces the result that the chain is valid only when r3c3 = 1 is false and not when r1c12 = 1 is false. Consequence of this deduction the two inferences "or" are not verified at the same time.

The only valid conclusion here is that, once again, none of what you wrote makes sense. Are you trolling on purpose, or are you serious? I can't tell.

If you're serious, I'm wondering what makes you think you're such an expert in branching AICs? I don't think I've seen a single chain from you that even attempts to use ANDed and ORed nodes, and I highly doubt they'd be correct if you did write them. With a non-existent track record, what gives you such unlimited confidence to claim my chains are incorrect? It's absurd, considering my very existent track record. All evidence indicates that you have no idea what you're talking about, yet you keep lecturing me like there's no tomorrow. It's pretty funny, actually, but it's not a good attitude for learning. For your own sake, I recommend a change of course.

Writing branching logic as correct AICs is a skill that doesn't come without a thorough understanding of Boolean logic and a lot of practice. Most people's intuition produces totally incorrect results, as we can see regularly from those who haven't bothered to learn it. It doesn't really matter as long as the underlying logic is correct and the chain is understandable, but it isn't a matter of opinion whether an AIC is correct or not. The rules are crystal-clear and easily verified, because AICs can always be turned into pure Boolean logic. As such they can be checked with truth tables or Boolean calculators, if need be. There's zero ambiguity at all for those who understand the rules.

My first branched chains were also incorrect, for the exact reason that my intuition failed me like so many others. Fortunately I got good help early on (most notably from eleven and Steve), and unlike you, I didn't spit on it. I respected the fact that they obviously knew more than me, so I listened and learned and practiced instead. That's why my mistakes are extremely rare these days, no matter how complex the logic. (Yeah, I just made one, but it makes no dent in those probabilities. Besides, I eventually found and corrected it myself.)

You'd be better off too if you listened more and talked less until you get the hang of it. You can't improve with your current attitude because you can't even see or at least admit when you're dead wrong. It also makes discussions like this very painful for all parties, even though it might be entertaining for the audience.

[Ajò Dimonios]

Space wrote:
First strong inference: If 5r1c7 is false then either (5,6)r12c1 OR (5,1)r12c1 must be true, and vice versa: if neither (5,6)r12c1 NOR (5,1)r12c1 is true, then 5r1c7 must be true. Thus, at least one side of the strong inference must be true. The inference is valid.

Wrong. This is the fundamental error of all your reasoning. To ensure that there is a strong inference between 5r1c7 and (5,6 | 5,1) r12c1 then both {(5,6) r12c1 true and (5,1) r12c1 false} and {(5,6) r12c1 false and (5,1) r12c1 true} must be true, only in these conditions the strong inference is valid otherwise if only one of the two conclusions were to prove true then there would be the possibility that both hypotheses (5r1c7 and (5,6 | 5 , 1) r12c1) are false. When we build a strong inference it is not important that the conclusions of the inference are true or false or false by contradiction but only that the inference is really a strong inference. It is clear that it is a question of demonstrating a contradiction but this ensures that the inference is strong. It can also be thought that proving that this strong inference is valid implies that the initial hypothesis r1c7 ≠ 5 is certainly false, but this is another way of seeing the same thing with a contradictory chain.

In order not to bore the readers, I briefly summarize the terms of the discussion. I argue that two independent logical chains are needed to prove that between the backdoors r1c7 = 5 and the backdoor r3c9 = 1 there is a strong inference, while you argue that this is possible with a single chain. My assertion means that the two backdoors in question do not have the same behavior as other pairs of backdoors such as (r8c7 = 3; r3c9 = 1) whose strong inference 3r8c7 = 1r3c9 is easily proved by 3r8c7=c3r8 –(3=2)r7c3-(2=1)r9c3-1r3c3=1r3c9. To write a strong inference starting from 5r1c7 it is essential to start from the assertion that r1c7 = 5 is false. When I impose that r1c7 cannot contain 5, the sudoku evolves from:

Code: Select all

+-------------+-------------+--------------+
| 156 9  7    | 468 68  16  | 2456 3  1246 |
| 156 3  1456 | 2   79  169 | 456  8  1467 |
| 8   2  146  | 467 5   3   | 46   9  1467 |
+-------------+-------------+--------------+
| 7   1  8    | 56  29  4   | 69   25 3    |
| 3   4  569  | 567 279 569 | 1    25 8    |
| 2   56 569  | 1   3   8   | 7    4  69   |
+-------------+-------------+--------------+
| 4   8  23   | 9   1   7   | 23   6  5    |
| 569 7  356  | 568 68  2   | 349  1  49   |
| 19  56 12   | 3   4   56  | 8    7  29   |
+-------------+-------------+--------------+


to the following situation:

Code: Select all

+-----------+-------------+-------------+
| 5  9  7   | 468 68  16  | 246 3  1246 |
| 16 3  146 | 2   79  169 | 5   8  1467 |
| 8  2  146 | 467 5   3   | 46  9  1467 |
+-----------+-------------+-------------+
| 7  1  8   | 56  29  4   | 69  25 3    |
| 3  4  569 | 567 279 569 | 1   25 8    |
| 2  56 569 | 1   3   8   | 7   4  69   |
+-----------+-------------+-------------+
| 4  8  23  | 9   1   7   | 23  6  5    |
| 69 7  356 | 568 68  2   | 349 1  49   |
| 19 56 12  | 3   4   56  | 8   7  29   |
+-----------+-------------+-------------+


All the chains that you have written start from this situation. This situation is blocked and consequently in order to lead to conclusions you choose in the logic you wrote to verify the two possibilities in r2c1 (r2c1 = 1 or r2c1 = 6). Of these two possibilities, only r2c1 = 1, through 1r2c1-1r2c3 = 1r3c9, demonstrates the strong link between 5r1c7 = 1r3c9. The other possibility r2c1 = 6 triggers a chain of contradiction which as a first step leads to:

Code: Select all

+-------------+---------------+-------------+
| 5  9  7     | 468 68   16   | 246 3  1246 |
| 6  3  14    | 2   79   179  | 5   8  147  |
| 8  2  14    | 467 5    3    | 46  9  1467 |
+-------------+---------------+-------------+
| 7  1  8     | 56  269  4    | 269 25 3    |
| 3  4  569   | 567 2679 5679 | 1   25 8    |
| 2  56 569   | 1   3    8    | 7   4  69   |
+-------------+---------------+-------------+
| 4  8  23    | 9   1    7    | 23  6  5    |
| 9  7  356   | 568 68   2    | 349 1  49   |
| 19 56 12    | 3   4    56   | 8   7  29   |
+-------------+---------------+-------------+


As can be seen in the column 3 in block 1, candidate 1 is present both in r2c3 and in r3c3. This fact does not prove that r3c9 = 1. If we continue to insert other certain candidates into this contradictory net or chain it will lead us to a contradiction long before proving that r3c9 = 1. Depending on the path we used in the insertions, since the chain is contradictory, we can prove that r3c9 = 1 or that r3c9 = 7. All this logical speech that I wrote leads me to conclude that my thesis is correct: there is no AIC or AIC modified with memory that makes you conclude with a single passage that 5r1c7 = 1r3c9. I frankly do not understand your resistance to such a blatant fact.

Paolo

[SpAce]

First strong inference: If 5r1c7 is false then either (5,6)r12c1 OR (5,1)r12c1 must be true, and vice versa: if neither (5,6)r12c1 NOR (5,1)r12c1 is true, then 5r1c7 must be true. Thus, at least one side of the strong inference must be true. The inference is valid.

Wrong. This is the fundamental error of all your reasoning.

Your hubris has no limits, doesn't it? The fundamental error of my reasoning is that I'm still talking to you, despite it having been obvious for a long time that you can't be persuaded with reason. Such a gargantuan waste of time.

incomprehensible crap: Show

To ensure that there is a strong inference between 5r1c7 and (5,6 | 5,1) r12c1 then both {(5,6) r12c1 true and (5,1) r12c1 false} and {(5,6) r12c1 false and (5,1) r12c1 true} must be true, only in these conditions the strong inference is valid otherwise if only one of the two conclusions were to prove true then there would be the possibility that both hypotheses (5r1c7 and (5,6 | 5 , 1) r12c1) are false. When we build a strong inference it is not important that the conclusions of the inference are true or false or false by contradiction but only that the inference is really a strong inference. It is clear that it is a question of demonstrating a contradiction but this ensures that the inference is strong. It can also be thought that proving that this strong inference is valid implies that the initial hypothesis r1c7 ≠ 5 is certainly false, but this is another way of seeing the same thing with a contradictory chain.

The fundamental error in your reasoning is probably due to your TDP background (and other things best left unsaid). You're looking at too big chunks of the grid, storing memories, and predicting the future, when thinking of the validity of inferences.

That approach is completely invalid with AICs. With them the full context is irrelevant. In fact, it must not be used. There's no history or future or the rest of the grid in an AIC. An inference between two nodes is considered in isolation of all that, which greatly simplifies things.

That fact makes all of your arguments irrelevant and blatantly false. All they prove is that you don't really understand AICs at all, but that's been obvious all along -- probably to everyone except you.

In order not to bore the readers

No fear of that. You're making this very entertaining for the audience, unless their feelings of second-hand embarrassment overshadow that.

To write a strong inference starting from 5r1c7 it is essential to start from the assertion that r1c7 = 5 is false.
When I impose that r1c7 cannot contain 5, the sudoku evolves from:

grid: Show

Code: Select all

+-------------+-------------+--------------+
| 156 9  7    | 468 68  16  | 2456 3  1246 |
| 156 3  1456 | 2   79  169 | 456  8  1467 |
| 8   2  146  | 467 5   3   | 46   9  1467 |
+-------------+-------------+--------------+
| 7   1  8    | 56  29  4   | 69   25 3    |
| 3   4  569  | 567 279 569 | 1    25 8    |
| 2   56 569  | 1   3   8   | 7    4  69   |
+-------------+-------------+--------------+
| 4   8  23   | 9   1   7   | 23   6  5    |
| 569 7  356  | 568 68  2   | 349  1  49   |
| 19  56 12   | 3   4   56  | 8    7  29   |
+-------------+-------------+--------------+


to the following situation:

grid: Show

Code: Select all

+-----------+-------------+-------------+
| 5  9  7   | 468 68  16  | 246 3  1246 |
| 16 3  146 | 2   79  169 | 5   8  1467 |
| 8  2  146 | 467 5   3   | 46  9  1467 |
+-----------+-------------+-------------+
| 7  1  8   | 56  29  4   | 69  25 3    |
| 3  4  569 | 567 279 569 | 1   25 8    |
| 2  56 569 | 1   3   8   | 7   4  69   |
+-----------+-------------+-------------+
| 4  8  23  | 9   1   7   | 23  6  5    |
| 69 7  356 | 568 68  2   | 349 1  49   |
| 19 56 12  | 3   4   56  | 8   7  29   |
+-----------+-------------+-------------+


All the chains that you have written start from this situation.

No. They "start" from the previous situation, and it never "evolves" from that into anything else, because AICs are not memory chains or TDP. Every link is verified against the relevant parts of the starting grid only, because there's no history or future or any other context to consider. Thinking otherwise is your fundamental error. It's a huge one, too.

Besides, AICs don't really "start" at all, because they're static Boolean-logic structures that don't depend on any assertions. It's a conceptual error to see them as directional implication chains, though that doesn't break anything unlike your fundamental error.

This situation is blocked and consequently in order to lead to conclusions you choose in the logic you wrote to verify the two possibilities in r2c1 (r2c1 = 1 or r2c1 = 6). Of these two possibilities, only r2c1 = 1, through 1r2c1-1r3c3 = 1r3c9, demonstrates the strong link between 5r1c7 = 1r3c9.

Indeed. The other possibility (5,6)r12c1 demonstrates the strong inference between 5r1c7 == 1r9c1. Did you really miss that?

The other possibility r2c1 = 6 triggers a chain of contradiction

Only if you miss the obvious fact that it was never meant to prove 1r3c9, and if you then consider the whole grid, store the history, and predict the future, which are all forbidden with AICs. Like I said, your TDP background betrays you miserably, and consequently you should stop talking about AICs as if you understood them at all -- until you actually do. Of course neither will ever happen because you simply can't understand how wrong you are.

When considered in isolation, as AIC nodes and links should be, there's no immediate contradiction with the (5,6) possibility any more than with the (5,1) possibility. (In full context both are obviously contradictory, because they're not in the solution.) Even if there were, it wouldn't matter one bit. Contradictions can be ignored in AICs as long as the links work.

All that matters is that the inferences are valid in isolation, and mine certainly are. All three parts of the strong inference (5)r1c7 = (5,1|5,6)r12c1 are valid possibilities when considered in isolation, and they're also the only possibilities, proving that at least one of them must be true. Thus the strong link is perfectly valid.

grid: Show

Code: Select all

+-------------+---------------+-------------+
| 5  9  7     | 468 68   16   | 246 3  1246 |
| 6  3  14    | 2   79   179  | 5   8  147  |
| 8  2  14    | 467 5    3    | 46  9  1467 |
+-------------+---------------+-------------+
| 7  1  8     | 56  269  4    | 269 25 3    |
| 3  4  569   | 567 2679 5679 | 1   25 8    |
| 2  56 569   | 1   3    8    | 7   4  69   |
+-------------+---------------+-------------+
| 4  8  23    | 9   1    7    | 23  6  5    |
| 9  7  356   | 568 68   2    | 349 1  49   |
| 19 56 12    | 3   4    56   | 8   7  29   |
+-------------+---------------+-------------+


As can be seen in the column 3 in block 1, candidate 1 is present both in r2c3 and in r3c3. This fact does not prove that r3c9 = 1.

As I already said (and what should have been obvious), the (5,6) option was never meant to prove 1r3c9. It proves 1r9c1, a different backdoor, which implies 5r1c7 just as well. In fact, that should have been so obvious that I'm beginning to think you must be trolling intentionally. The alternative is less flattering for you.

In any case, as a backdoor 1r9c1 implies (through singles) any other backdoor, including 1r3c9, so there's no problem proving that (5,6)r12c1 implies 1r3c9 too. As I mentioned in the last post, I could have written the chain like that, but it would have been an unnecessary complication. Instead I chose to prove that 5r1c7 is strongly linked to either 1r3c9 OR 1r9c1, which works just as well and more simply, because they both imply 5r1c7. But, since you have a problem understanding that (among other things), here's that redundant version too:

(5)r1c7 = (5,6|5,1)r12c1 - (1)(r12c1 & r3c3) = (1)r9c1|r3c9 ->singles-> (1)r3c9 ->singles-> (5)r1c7 => +5r1c7; stte

If we continue to insert other certain candidates into this contradictory net or chain it will lead us to a contradiction long before proving that r3c9 = 1.

Only in your TDP-corrupted mind (and only if (5,6)r12c1 were trying to to prove 1r3c9 in the first place, which it isn't). With AICs there's no such problem because any history is erased as soon as a link is applied and the chain moves to the next node. You're completely ignoring that fundamental AIC property, or perhaps you've never realized its existence.

AICs are not memory chains, unlike TDP and others that take into account the full history and the whole grid at every step. That's what makes AICs much simpler, more elegant, and easier to verify.

(My chain turns into an implied TDP-style memory chain once the ->singles-> implication is applied. Not before then. The first part that proves the derived strong inference 5r1c7 == 1r9c1|1r3c9 is a fully valid AIC without any memories.)

Depending on the path we used in the insertions, since the chain is contradictory, we can prove that r3c9 = 1 or that r3c9 = 7. All this logical speech that I wrote leads me to conclude that my thesis is correct: there is no AIC or AIC modified with memory that makes you conclude with a single passage that 5r1c7 = 1r3c9.

Indeed. It's obvious that your "logical speech" may lead you to conclude whatever the hell you want. It's the nature of false premises. They can imply anything at all.

My argumentation options are much more limited because I choose to deal with true premises only. They can only imply other truths, so I'm missing a whole world of weird and wonderful conclusions that seem natural to you. Of course your conclusions have nothing to do with reality, but that's just a minor detail.

I frankly do not understand your resistance to such a blatant fact.

I could tell you the reason why you don't understand it, but it would get me kicked out. Perhaps merely implying it will too, but I'm beyond caring at this point. It would be an acceptable price for not having to deal with you ever again. This is like playing chess with a pigeon.

[SpAce]

PS. To the audience I just want to say one thing: "Are you not entertained?" I hope you are, because I'm not. This is not why I'm here. That's why I won't be dignifying this idiotic discussion with any more comments, no matter what my opponent says. In fact, I'm positive that he'll be strutting around claiming victory.

That's ok. There was never any other outcome possible with this opponent. So, let's just pretend that I never had any valid arguments, just like he predicted from the get-go. I should have let him keep his illusions voluntarily, because I knew in advance that there was no way to make a dent in them anyway. Yet, I let myself get drawn into the quagmire, which I deeply regret. It ends now.

In fact, crap like this has made me lose my interest in the whole scene. So, I'll be going back to voluntary exile, effective immediately. That will surely come as a relief to many, and they should hail the Great Ajò Demonios as the One who finally beat me into submission. Hope you all like what you get in the trade.

That said, I don't make any promises that I won't come back some day (unless that choice is made for me, which is a distinct possibility), so don't lose your fear of the Dark Side in the mean time. Lol.

[Ajò Dimonios]

Hi Space

Despite your gratuitous insults, I am truly sorry that you decide to exile. I invite you to review your decision. This was certainly not what I wanted. However, I don't think the discussion is idiotic. Those who have followed will realize that writing an AIC or any chain is not so trivial. I apologize to you if I used harsh tones at certain times that may seem offensive perhaps to my bad English, I assure you that it was not my will. I renew the invitation again to review your decision. Your interventions, particular and intelligent and well documented, are certainly useful to stimulate interest in this game in all of us.
Paolo

[SteveG48]

SpAce, I wish you'd stay around. You argued. Neither of you convinced the other. Move on.

There are other things to discuss.

[tarek]

SpAce, I wish you'd stay around. You argued. Neither of you convinced the other. Move on.
There are other things to discuss.

ditto

[totuan]

SpAce, I wish you'd stay around. You argued. Neither of you convinced the other. Move on.
There are other things to discuss.

ditto

ditto

totuan

[Cenoman]

In order not to bore the reader

You're making this very entertaining for the audience

It was very very much boring, and definitely not entertaining.

I join other forum members to wish both of you to stay around.
The forum need people with your skills to stay alive. Stay and do your best to keep it friendly.

[SpAce]

Paolo,

Hi Space

While I appreciate the change in your tone in general, I would also appreciate if you spelled my handle correctly. I've mentioned it before. Seems like a small thing, but it's an obvious fact that some people misspell it for the exact purpose of disrespecting and irritating me. If that's not your intention, I suggest you spell it correctly to avoid such associations. Even small things have snowballing effects.

Despite your gratuitous insults, I am truly sorry that you decide to exile. I invite you to review your decision. This was certainly not what I wanted. However, I don't think the discussion is idiotic. Those who have followed will realize that writing an AIC or any chain is not so trivial. I apologize to you if I used harsh tones at certain times that may seem offensive perhaps to my bad English, I assure you that it was not my will. I renew the invitation again to review your decision. Your interventions, particular and intelligent and well documented, are certainly useful to stimulate interest in this game in all of us.

As I said, I appreciate that. However, the roots of my decision go much deeper than you and me, so it's not going to change any time soon. Our discussion just broke the camel's back, nothing more. For what it's worth, I can exonerate you from most of the responsibility. I also apologize for the generous insults on my part. When snowballs get rolling, things tend to get ugly.

[SpAce]

Steve, tarek, totuan, Cenoman,

SpAce, I wish you'd stay around. You argued. Neither of you convinced the other. Move on.
There are other things to discuss.

ditto

ditto

I join other forum members to wish both of you to stay around.
The forum need people with your skills to stay alive. Stay and do your best to keep it friendly.

Thanks, guys. I appreciate it. I've enjoyed and learned from our conversations very much, and will be missing all of you.

I wish it had been like that with everyone else. I've learned something from almost every discussion here, but I haven't always enjoyed them a whole lot. Despite appearances, I really don't like discussions that turn into mudslinging contests. I'd much rather have mutually respectful and fun conversations with factual and logical arguments only. This should be a particularly suitable arena for that, but for some reason things tend to get personal all too often. I'm sure I carry my share of fault for that, but I'm not the only one. Yet it seems I'm the one who gets to pay the price every time, both in the forms of administrative punishments and people's contempt. I'm losing the war despite (or because of) winning pretty much every battle. I'm tired of it.

That's why I rather leave now on my own terms than wait for yet another ban to kick in. The seemingly arbitrary and definitely opaque admin policies on this forum have made sure that I can't stick around any longer. It's not very motivating to make any contributions when you constantly feel a knife on your back just waiting to be stabbed in without any explanations.

I've now been banned twice in a short time for "personal attacks". I actually don't even know exactly what the latest ban was for. Perhaps the bans were justified, but what about those who provoked my reactions leading to them? Did they even get warned? I just don't know, but if not, then I think there are clear double-standards. To avoid such questions, it'd be nice if all such admin actions were public and explained, so everyone could judge for themselves whether different people were treated the same way for similar offenses. Right now I have my doubts, but I could be wrong. Transparency would fix that (unless it would confirm my doubts).

I also find it interesting that the admin policies seem to have gotten stricter since I first got here. I never got even a slap on my wrist for that stupid initial confrontation with David. In retrospect, I probably should have. I can freely admit that it's the one fight I actually regret, because it was more or less unnecessary and eventually led to a loss of a valuable contributor. In that light, I think it's a bit weird that the administration has now taken a much more active role in policing the discussions. That's exactly what David wanted, but it's too late now for him. (It's also exactly what I don't want, especially when executed with zero transparency.)

Why wasn't David shielded the way some other people have been lately? It seems illogical to me, especially considering that those other people haven't been exactly angels themselves. It's probably a bit ironic coming from me, but I actually think David should have been treated with more respect than what he got in the end, considering his lasting contributions to the sudoku scene. It's quite obviously the real reason for why he left, even though I'm the one who gets eternally blamed for it.

In any case, I really have to take a break now. I don't feel like even solving puzzles at the moment, much less discussing them. The passion is gone. Perhaps it'll come back some day, we'll see.

--
PS. Obviously Paolo and I couldn't convince each other in this discussion, as Steve said. The all-important question remains open, though: did either of us convince anyone else? I think it would be kind of nice to get closure on whether my AICs were valid or not. To me they obviously were (except the first one), but I'm all ears if someone can actually argue the opposite in a way that I could understand. Despite my apparent arrogance, I always keep the option open that I've been blind to something. Was I? I think there's enough AIC expertise around to make that judgment.

[mith]

As far as I can tell, this is where the disagreement lies?

To ensure that there is a strong inference between 5r1c7 and (5,6 | 5,1) r12c1 then both {(5,6) r12c1 true and (5,1) r12c1 false} and {(5,6) r12c1 false and (5,1) r12c1 true} must be true, only in these conditions the strong inference is valid otherwise if only one of the two conclusions were to prove true then there would be the possibility that both hypotheses (5r1c7 and (5,6 | 5 , 1) r12c1) are false.

(5,6|5,1)r12c1 is the same as
(5,6)r12c1 OR (5,1)r12c1 which is the same as
(5r1c1 AND 6r2c1) OR (5r1c1 AND 1r2c1) which is the same as
5r1c1 AND (6r2c1 OR 1r2c1) by

Code: Select all

Distribution: (A AND B) OR (A AND C) is equivalent to A AND (B OR C)

Additionally:
-5r1c7 => 5r1c1 => (6r2c1 OR 1r2c1), so
-5r1c7 => (5r1c1 AND (6r2c1 OR 1r2c1)) by

Code: Select all

Hypothetical Syllogism: (A => B) AND (B => C); therefore A => C
Composition: (A => B) AND (A => C) is equivalent to A => (B AND C)

So either 5r1c7 is true, or (5r1c1 AND (6r2c1 OR 1r2c1)) is true, and the strong link is valid.

[Ajò Dimonios]

This solution is valid but it is certainly not sequential.It is always possible to write an AIC that justifies the correct deletions

[mith]

You'll have to define "not sequential" for me. I'm not sure what you mean as it relates to this chain.