## August 1, 2020

Post puzzles for others to solve here.

### Re: August 1, 2020

eleven wrote:oops, thanks. This should repair it:
249r189c9 = 16r1c69 - (1|6=58)r1c15 - (5|8=964)r8c159 => -9r8c7, -4r23c9

Nice save! I love the first strong link. In fact, the whole thing could be written with just two:

(942)r981c9 = (1685)r1c9651 - (8|5=694)r8c519 => -9r8c7, -4r23c9; stte

SpAce

Posts: 2504
Joined: 22 May 2017

### Re: August 1, 2020

Hi SteveG48
It is strange that you ask for the sequence of the singles, however, here it is:
r3c9=1=>(r2c9=7;r3c4=7)=> (r1c4=4;r5c5=7)=>r4c5=2=>r8c9=4=>r4c7=9=>r8c7=3

Paolo
Ajò Dimonios

Posts: 196
Joined: 07 November 2019

### Re: August 1, 2020

SteveG48 wrote:Then write that chain.

That would be fun to see. (Or not.) Here's how I see the intended logic, using implication chains because it seems much more natural for this purpose:

Code: Select all
`-3r8c7 -> 3r8c3 -> 2r7c3 -> 1r9c3 -> -1r3c3 -> +1r3c9+3r8c7 --------------- singles --------------> +1r3c9=> +1 r3c9; stte`

(Or the same the other way proving +3r8c7, but just one or the other is necessary.) That is indeed valid logic, because it proves that whether 3r8c7 is true or not, 1r3c9 must be true. Or rather, it proves one part (-3r8c7 -> +1r3c9) but simply states the other as a fact (leaving it up to the reader to verify it's true). The original presentation doesn't depict that logic at all. The idea would have been clear if it had been written as a pseudo-AIC (or pseudo-DL Type 2) like this:

Code: Select all
`(1)r3c9 = r3c3 - (1=2)r9c3 - (2=3)r7c3 - r7c7 = (3)r8c7 -singles-> (1)r3c9 => +1r3c9; stte`

Either way, the biggest problem is that the original presentation attempts to hide the inherent complexity and inelegance of the full logic by presenting a simple AIC as the main piece. That's misleading the audience. The hardest part is obviously proving (rather than just stating) that 3r8c7 -> 1r3c9, and that is nowhere to be seen (*). Here's the full logic written as a matrix (I wouldn't even consider any other options to write it):

15x15 TM:

Code: Select all
` 1r3c9 1r3c3       1r9c3 2r9c3             2r7c3 3r7c3                   3r7c7 3r8c7                         3r7c7 2r7c7                               2r9c9 9r9c9                                     9r8c9 4r8c9                                     9r6c9 ..... 6r6c9                                                 6r6c2 5r6c2                                                       5r9c2 5r9c6                                                 6r4c7 ........... 6r4c4                                                             5r8c4 6r8c4 8r8c4                                                                   6r1c4 8r1c4 4r1c4                                                                   6r3c4 ..... 4r3c4 7r3c4 1r3c9 ................................... 4r3c9 6r3c9 ............................. 7r3c9==========================================================================================+1r3c9 `

Simple and elegant? Good luck writing that as an AIC. (The original simple AIC is in the first five rows. The other ten rows are the hard part.)

For comparison, here's the matrix for my solution proving two distant eliminations simultaneously:

7x7 TM:

Code: Select all
` 9r4c7,9r8c1 9r8c79             9r9c9  2r9c9                    2r9c3 1r9c3                          1r3c3 1r3c9                                7r3c9 7r2c9                    2r7c7 ................. 2r1c7 9r2c5,5r1c1 ........................ 7r2c5 5r1c7=================================================-9r4c5,5r8c1`

--
(*) Added. I see that an attempt to write the singles chain (for the corresponding +3r8c7 proof) has been made while I was writing this. The fact that it had to be written as an implication chain (leaving out a lot of details) proves my point. The original AIC presentation is misleading, because it's the simplest and the smallest part of the full logic. If the hard part is included, it's practically impossible to write as an AIC, as my matrix demonstrates.
--

PS. If one is going to skip writing the hard parts anyway, the logic could be written much shorter. This would do just as well:

Code: Select all
`(1)r3c9 = r3c3 - (1=2)r9c3 -singles-> (1)r3c9 => +1r3c9; stte`

Or this:

Code: Select all
`(2=1)r9c3 - r3c3 = (1)r3c9 -singles-> (2)r9c3 => +2r9c3; stte`

In other words, any derived strong link between two backdoors can be used to write such pseudo-DLs to "prove" either backdoor true, because backdoor candidates obviously imply each other. The logic is technically valid, but the value of such solutions is close to zero. If we started embracing such sloppy solutions, there would be no point in the whole exercise of publishing solutions anyway. Why would anyone bother to write complete chains anymore? Why would anyone bother to read anyone's solutions?

(Of course I could be called a hypocrite because I recently wrote a sloppy solution myself. Not only did it skip the details but it used a contradiction too. Neither is normally acceptable to me. My excuse was that there weren't a whole lot of OTP options in that puzzle, as evidenced by the lack of responses, and turning the one I found into a more complete presentation would have been very tedious with very little added value. It made more sense to highlight the relevant parts and leave the details out. In this puzzle, for example, there's no such excuse.)

SpAce

Posts: 2504
Joined: 22 May 2017

### Re: August 1, 2020

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. For this reason, the first part of the resolution, which is apparently the simplest, is essential to complete the resolution. In this example, the pairs (1r9c1; 1r3c9), (9r8c1; 1r3c9), (2r9c3; 1r3c9), (9r9c9; 1r3c9), (2r7c7; 1r3c9), (2r1c9; 1r3c9) also have this characteristic. note that it is always present the backdoor 1r3c9. Other backdoors (5r1c7; 4r2c7; 6r2c6; 1r2c3; 5r2c1; 6r1c1) do not have this property.

Paolo
Ajò Dimonios

Posts: 196
Joined: 07 November 2019

### Re: August 1, 2020

Ajò Dimonios wrote:Not all pairs of backdoors lead to this type of solution,

Sure they do.

only those in which it is possible to demonstrate the existence of a strong inference between them.

It's always possible. It may be more difficult for some than others, but it's always possible because two backdoors are strongly-linked by definition. Another way to see it is that if you assume a backdoor false it's a contradiction, and a contradiction can be forced to imply anything.

Other backdoors (5r1c7; 4r2c7; 6r2c6; 1r2c3; 5r2c1; 6r1c1) do not have this property.

A counter-example:

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

In fact, since we know that any backdoor is always strongly linked to any other backdoor, why bother to demonstrate it any more than the fact that any backdoor implies any other backdoor? We could skip that part too and write our solution simply:

(5)r1c7 == (1)r9c1 -> (5)r1c7 => +5r1c7; stte

Then again, since we know that a backdoor is a backdoor, we could simplify that even further:

+5r1c7; stte

Absurd, right? That's why this whole approach has the stink of circular reasoning. If you want to avoid it, you need to write the full logic for every part.

SpAce

Posts: 2504
Joined: 22 May 2017

### Re: August 1, 2020

Space wrote:
Ajò Dimonios wrote:
Not all pairs of backdoors lead to this type of solution,

Sure they do.

only those in which it is possible to demonstrate the existence of a strong inference between them.

It's always possible. It may be more difficult for some than others, but it's always possible because two backdoors are strongly-linked by definition. Another way to see it is that if you assume a backdoor false it's a contradiction, and a contradiction can be forced to imply anything.

Other backdoors (5r1c7; 4r2c7; 6r2c6; 1r2c3; 5r2c1; 6r1c1) do not have this property.

A counter-example:

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

In fact, since we know that any backdoor is always strongly linked to any other backdoor, why bother to demonstrate it any more than the fact that any backdoor implies any other backdoor? We could skip that part too and write our solution simply:

(5)r1c7 == (1)r9c1 -> (5)r1c7 => +5r1c7; stte

Then again, since we know that a backdoor is a backdoor, we could simplify that even further:

+5r1c7; stte

Absurd, right? That's why this whole approach has the stink of circular reasoning. If you want to avoid it, you need to write the full logic for every part.

I know very well that two backdoors are linked by a strong inference, however this must be demonstrated because I have not used the knowledge that the two hypotheses are backdoors in the resolution. When I speak of proving that two backdoors are linked by a strong link, I am referring exclusively to what the theory demonstrates about an alternation of strong and weak inferences of single hypotheses http://brunogreco.free.fr/bazar/sudoku/ ... hains1.htm (Theorem 3). As for other types of chains with multiple hypotheses (or; and and others) I don't know if the theory leads to the same proof as theorem 3. The fact that you say that everything can be proved and the opposite of everything seems to me a good indication that theorem 3 is not valid for this type of chains.

Paolo
Ajò Dimonios

Posts: 196
Joined: 07 November 2019

### Re: August 1, 2020

Ajò Dimonios wrote:I know very well that two backdoors are linked by a strong inference

I figured you did. My point was just that proving that link is also possible for all such pairs, though it might be quite complicated for some.

however this must be demonstrated because I have not used the knowledge that the two hypotheses are backdoors in the resolution. When I speak of proving that two backdoors are linked by a strong link, I am referring exclusively to what the theory demonstrates about an alternation of strong and weak inferences of single hypotheses http://brunogreco.free.fr/bazar/sudoku/ ... hains1.htm (Theorem 3). As for other types of chains with multiple hypotheses (or; and and others) I don't know if the theory leads to the same proof as theorem 3. The fact that you say that everything can be proved and the opposite of everything seems to me a good indication that theorem 3 is not valid for this type of chains.

I presume you're referring to this theorem:

Theorem 3 : when assumptions A,B,C,...X,Y,Z verify A==B--C...X--W==Z, where thick and thin links alternate, then A==Z.

That is certainly valid for any valid AIC. It doesn't matter how complex the individual nodes (A, B, C, ...) are, as long as they can be computed to have a single boolean truth value which is linked to the adjacent nodes (which can be simple or complex). An ORed node (a|b) is true if either a or b is true, otherwise false. An ANDed node (a&b) is true if both a and b are true, false otherwise.

That said, one might notice that my previous counter-example could be simplified a bit, because one of the nodes has a known truth value:

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

The node (1)(r12c1 & r3c3) is always false, because it's not possible for two 1s to exist in the same box. It means that its strongly linked partner node, (1)r9c1|r3c9, must be true. Thus, the first part of the chain is redundant, and we could write the logic more shortly like this (for example):

[1r12c1 - 1r3c3] -> (1r9c1|1r3c9) ->singles-> 5r1c7 => +5r1c7; stte

Of course we could replace 5r1c7 with any backdoor (or any value in the solution), and the logic would still hold. Technically valid, but not a very interesting proof.

SpAce

Posts: 2504
Joined: 22 May 2017

### Re: August 1, 2020

Space wrote:
That is certainly valid for any valid AIC. It doesn't matter how complex the individual nodes (A, B, C, ...) are, as long as they can be computed to have a single boolean truth value which is linked to the adjacent nodes (which can be simple or complex). An ORed node (a|b) is true if either a or b is true, otherwise false. An ANDed node (a&b) is true if both a and b are true, false otherwise.

That said, one might notice that my previous counter-example could be simplified a bit, because one of the nodes has a known truth value:

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

The node (1)(r12c1 & r3c3) is always false, because it's not possible for two 1s to exist in the same box. It means that its strongly linked partner node, (1)r9c1|r3c9, must be true. Thus, the first part of the chain is redundant, and we could write the logic more shortly like this (for example):

[1r12c1 - 1r3c3] -> (1r9c1|1r3c9) ->singles-> 5r1c7 => +5r1c7; stte

Of course we could replace 5r1c7 with any backdoor (or any value in the solution), and the logic would still hold. Technically valid, but not a very interesting proof.

It seems to me that this chain is redundant from the first inference. The first inference has two possibilities: if r1c7 = 5 is false => (r1c1 = 5; r2c1 = 6) true or r1c1 = 5 false => (r1c1 = 5; r2c1 = 1) true, then the second inference, (5 , 6 | 1) r12c1 - (1) (r12c1 & r3c3), is valid only for the latter possibility. If the first were true, (5,6 | 1) r12c1 - (1) (r12c1 & r3c3), it would not be correct and consequently it could not be concluded with (1) (r12c1 & r3c3) = (1) r9c1 | r3c9 -> single-> (5) r1c7 => + 5r1c7; STTE.

Paolo
Ajò Dimonios

Posts: 196
Joined: 07 November 2019

### Re: August 1, 2020

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

The node (1)(r12c1 & r3c3) is always false, because it's not possible for two 1s to exist in the same box. It means that its strongly linked partner node, (1)r9c1|r3c9, must be true. Thus, the first part of the chain is redundant

It seems to me that this chain is redundant from the first inference. The first inference has two possibilities: if r1c7 = 5 is false => (r1c1 = 5; r2c1 = 6) true or r1c1 = 5 false => (r1c1 = 5; r2c1 = 1) true, then the second inference, (5 , 6 | 1) r12c1 - (1) (r12c1 & r3c3), is valid only for the latter possibility. If the first were true, (5,6 | 1) r12c1 - (1) (r12c1 & r3c3), it would not be correct and consequently it could not be concluded with (1) (r12c1 & r3c3) = (1) r9c1 | r3c9 -> single-> (5) r1c7 => + 5r1c7; STTE.

Wrong. If not logic, then basic probability calculation should tell you that answer

SpAce

Posts: 2504
Joined: 22 May 2017

### Re: August 1, 2020

Space wrote:
Wrong. If not logic, then basic probability calculation should tell you that answer

So from what I understand from your answer (5) r1c7 = (5,6 | 1) r12c1 is true because it is a consequence of the probability calculation. Unfortunately for you this is not written in the first inference. It would be good for each of us to report the proof of strong inferences when the latter are not immediate. The first strong inference you wrote is true only if (r1c1 = 5 and r2c1 = 6) is always false. The consequence of this is that you should have reported the proof that (r1c1 = 5 and r2c1 = 6) is always false. In practice you have to report a chain of contradiction (r1c1 = 5 and r2c1 = 6) => singles => contradiction. This is certainly true but not immediate. In practice, it all comes down to a chain of contradiction. But hadn't he told himself not to use chains that show a contradiction?

Paolo
Ajò Dimonios

Posts: 196
Joined: 07 November 2019

### Re: August 1, 2020

Ajò Dimonios wrote:So from what I understand from your answer (5) r1c7 = (5,6 | 1) r12c1 is true

That inference is not questionable in any way. The first strong link is valid no matter how you look at it. Edit 1: Wrong! See the added part of this post. Edit 2: In fact, there's nothing wrong with the strong link, just like I originally said. I shouldn't have started second-guessing myself. Other ways to write it would be:

(5)r1c7 = (5,6|5,1)r12c1

or:

(5)r1c7 = (5,(6|1))r12c1

or:

(5)r1c7 = ((5)r1c1 & (6|1)r2c1)

They all mean the same thing. (Edit: Yes, they do, and they're correct. The original doesn't mean the same thing, though.) If 5r1c7 is false, then there's a 5 in r1c1 and 6 or 1 in r2c1, and vice versa. Do you have a problem with that?

The following weak link is the only questionable part (because the third node is self-contradictory), but even that doesn't make any difference. The logic works and the AIC is still valid, even though it forces one end-point to be true if the nodes are looked more closely. It would be incorrect only if neither end-point were guaranteed to be true.

As a matrix (4x4 TM):

Code: Select all
` 5r1c7 5r1c1 1r3c9       1r3c3       5r2c1 1r2c1 6r2c1  1r9c1 1r1c1       1r2c1=========================> (5r1c7 == 1r3c9|1r9c1)`

because it is a consequence of the probability calculation.

You missed the joke, as I expected. What I meant was that the probability of me making mistakes with AICs is almost non-existent, rare typos not counted. You should know that if you'd been paying any attention, but it's obvious that you haven't. In any case, I'd recommend looking for mistakes in your own thinking before assuming them in mine.

Unfortunately for you this is not written in the first inference. It would be good for each of us to report the proof of strong inferences when the latter are not immediate. The first strong inference you wrote is true only if (r1c1 = 5 and r2c1 = 6) is always false. The consequence of this is that you should have reported the proof that (r1c1 = 5 and r2c1 = 6) is always false. In practice you have to report a chain of contradiction (r1c1 = 5 and r2c1 = 6) => singles => contradiction. This is certainly true but not immediate. In practice, it all comes down to a chain of contradiction. But hadn't he told himself not to use chains that show a contradiction?

None of that makes any sense. Sorry if I don't really feel like you're in a position to lecture me about writing valid AICs.

The only actually questionable part in my AIC was the third node (1)(r12c1 & r3c3), for the reason I already mentioned. It's self-contradictory and thus always false, but all that does is force the right-side end-point to be always true, which is enough for the AIC to do its job. However, it's a bit questionable if the first weak link is valid. There's no problem if you read it from left to right:

Code: Select all
`(5,6|1)r12c1 - (1)(r12c1 & r3c3)=>(5,6)r12c1 -> -1r12c1 => -(1r12c1 & 1r3c3) // OK||(5,1)r12c1 -> -1r3c3  => -(1r12c1 & 1r3c3) // OK`

There's a slight problem if flipped around, however:

Code: Select all
`(1)(r12c1 & r3c3) - (5,6|1)r12c1=>1r12c1 & 1r3c3 -> -(5,6)r12c1 // OK&1r12c1 & 1r3c3 -> -1r12c1     // ??`

The reason for the question marks is that if both 1r12c1 and 1r3c3 are true then they imply opposite truth values for 1r12c1. So, there's a contradiction, but it's because 1r12c1 and 1r3c3 are contradictory in the first place. I choose to consider the weak link valid, but I accept that it's not clear-cut. Nevertheless, the full AIC works. Edit: I no longer consider the weak link nor the strong link nor the AIC valid. See posts below.

--
Edit 2. Rolled back some of the previous edits with strikethroughs.
(All edits are colored additions. No part of the original text has been changed or removed.)
Last edited by SpAce on Tue Aug 11, 2020 8:01 pm, edited 2 times in total.

SpAce

Posts: 2504
Joined: 22 May 2017

### Re: August 1, 2020

Space wrote:
That inference is not questionable in any way. The first strong link is valid no matter how you look at it. Other ways to write it would be:

(5)r1c7 = (5,6|5,1)r12c1

or:

(5)r1c7 = (5,(6|1))r12c1

or:

(5)r1c7 = ((5)r1c1 & (6|1)r2c1)

They all mean the same thing. If 5r1c7 is false, then there's a 5 in r1c1 and 6 or 1 in r2c1, and vice versa. Do you have a problem with that?

What you say is obvious. If r1c7 = 5 is false, the inference (5) r1c7 = (5,6 | 1) r12c1 is equivalent to two contradicting inferences between them 5r1c7 => (r1c1 = 5; r2c2 = 1) "or non-inclusive" 5r1c7 = > (r1c1 = 5; r2c2 = 6), consequently the conclusion is an obvious one, it is true by definition because it includes the universe of all possible solutions that are in contradiction with each other, one valid inference cannot be defined but two contradicting inferences between them. The serious and incorrect fact is that you consider in the second weak inference only the conclusion that is to your liking because it leads you to the solution you want and completely ignores the one that forces you to the contradictory chain (r1c1 = 5 and r2c1 = 6) = > singles => contradiction.

Paolo
Ajò Dimonios

Posts: 196
Joined: 07 November 2019

### Re: August 1, 2020

Ajò Dimonios wrote:What you say is obvious.

Yes, it should be obvious, but just as obviously it wasn't for you since you claimed (and seem to continue to claim) that the first strong inference wasn't correct. Try to decide whether you think it's invalid or obvious.

If r1c7 = 5 is false, the inference (5) r1c7 = (5,6 | 1) r12c1 is equivalent to two contradicting inferences between them 5r1c7 => (r1c1 = 5; r2c2 = 1) "or non-inclusive" 5r1c7 = > (r1c1 = 5; r2c2 = 6), consequently the conclusion is an obvious one, it is true by definition because it includes the universe of all possible solutions that are in contradiction with each other, one valid inference cannot be defined but two contradicting inferences between them. The serious and incorrect fact is that you consider in the second weak inference only the conclusion that is to your liking because it leads you to the solution you want and completely ignores the one that forces you to the contradictory chain (r1c1 = 5 and r2c1 = 6) = > singles => contradiction.

I cannot say that I do not disagree with you. (Translation.) The serious and correct fact is that nothing above makes any sense. Let me quote yourself back to you:

Ajò Dimonios wrote:I'm sorry you don't admit your mistake. You probably have no valid arguments to refute what I wrote.

And we get back to why I have zero interest in debating anything with you. I'm sorry to say but your understanding of AICs is so flawed that there's no way we could have a rational discussion about them. The worst part is that you don't realize it yourself, which gives you an unlimited amount of confidence to make absurd claims instead of listening and learning to fix your mistakes. Thus, the whole discussion is pointless. (The other possibility is that you're expressing yourself so dismally that I simply can't see the brilliance of your thinking. Either way, the conclusion is the same.)
--

Added. The questionable weak link is quite interesting, actually. Put into a truth table:

Code: Select all
`            a  b              b       c        (5)r1c7 = (5,6|1)r12c1 - (1)(r12c1 & r3c3) = (1)r9c1|r3c9 ->singles-> (5)r1c7 => +5r1c7; sttea b c (a|b) (b&c) ((a|b) NAND (b&c))------------------------------------T T T   T     T           F *T T F   T     F           T T F T   T     F           TT F F   T     F           TF T T   T     T           F *F T F   T     F           TF F T   F     F           TF F F   F     F           T`

What it says is that if both b and c (i.e. 1r12c1 and 1r3c3) are true, then the weak link (NAND) is false. In other words, if the chain is read from right to left then it can't propagate past that node, which is bad. However, since we know that b and c can't be true together (being natively weakly-linked), those false rows could be eliminated from the truth table leaving a technically valid but rather lop-sided weak link. The AIC can still do its main job, which is to prove one or the other end-point true (in this case always the right-side one). However, it fails the test of all of its links working both ways as intended. Thus, I've come to the conclusion that it's probably best to disqualify it.

Of course the problem is more or less academic, but it's also easy to avoid, so let's do it. Here's a slight variant that avoids any questionable nodes and links:

(5)r1c7 = (5,1|5,6)r12c1 - ((1)r3c3 & (59|69)r89c1) = (1)r3c9|r9c1 ->singles-> (5)r1c7 => +5r1c7; stte

SpAce

Posts: 2504
Joined: 22 May 2017

### Re: August 1, 2020

Since I am not at all interested in insulting and receiving free insults but only in discussing the methodologies for solving sudoku, I reply to:

Space wrote:

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

a b c (a|b) (b&c) ((a|b) NAND (b&c))
------------------------------------
T T T T T F *
T T F T F T
T F T T F T
T F F T F T
F T T T T F *
F T F T F T
F F T F F T
F F F F F T

What it says is that if both b and c (i.e. 1r12c1 and 1r3c3) are true, then the weak link (NAND) is false. In other words, if the chain is read from right to left then it can't propagate past that node, which is bad. However, since we know that b and c can't be true together (being natively weakly-linked), those false rows could be eliminated from the truth table leaving a technically valid but rather lop-sided weak link. The AIC can still do its main job, which is to prove one or the other end-point true (in this case always the right-side one). However, it fails the test of all of its links working both ways as intended. Thus, I've come to the conclusion that it's probably best to disqualify it.

Of course the problem is more or less academic, but it's also easy to avoid, so let's do it. Here's a slight variant that avoids any questionable nodes and links:

(5)r1c7 = (5,1|5,6)r12c1 - ((1)r3c3 & (59|69)r89c1) = (1)r3c9|r9c1 ->singles-> (5)r1c7 => +5r1c7; stte

The AIC you report is also incorrect.
You cannot write (5)r1c7 = (5,1|5,6)r12c1 - ((1)r3c3 & (59|69)r89c1) ) completely ignore that among the conclusions there may also be the possibility that 1r2c3 is false. You have to be more careful when writing about chains that have "or" and "and".
The fact that the other backdoors (5r1c7; 4r2c7; 6r2c6; 1r2c3; 5r2c1; 6r1c1) do not have the characteristic of the previous ones is also proved by the fact that to obtain a solution with these backdoors I am forced to use a method T & E = 2
While for the others a method T & E = 1 is sufficient.

Paolo
Ajò Dimonios

Posts: 196
Joined: 07 November 2019

### Re: August 1, 2020

Ajò Dimonios wrote:Since I am not at all interested in insulting and receiving free insults

Lol. Nice hypocrisy. Many of your comments are freely insulting my well-earned AIC-expertise. Or rather, they would be if they weren't so absurd. I'm just returning the compliment, only with a bit better aim. It's not the way I want to discuss at all, but you leave no choice.

For that reason I wouldn't really want to discuss anything at all with you, because my patience for disrespectful nonsense is very thin. You'd have to be on a much higher level with AICs to earn the right to talk to me the way you do, though if you were, you probably wouldn't. The worst part is that if I let myself respond the way it deserves, I'll get banned again. It's possibly already happened, we'll see. (You can congratulate yourself in that case.)

SpAce wrote:Here's a slight variant that avoids any questionable nodes and links:

(5)r1c7 = (5,1|5,6)r12c1 - ((1)r3c3 & (59|69)r89c1) = (1)r3c9|r9c1 ->singles-> (5)r1c7 => +5r1c7; stte

The AIC you report is also incorrect.

Lol. No, it's not. It's almost certainly perfectly correct. Like I said, your limited and flawed understanding of AICs just doesn't let you see it, so it's pointless to debate it.

You cannot write (5)r1c7 = (5,1|5,6)r12c1 - ((1)r3c3 & (59|69)r89c1) ) completely ignore that among the conclusions there may also be the possibility that 1r2c3 is false.

What has 1r2c3 to do with anything? It's totally irrelevant. Comments like that prove my previous point.

You have to be more careful when writing about chains that have "or" and "and".

You keep thinking you have what it takes to lecture me about AICs? I'm sorry but the evidence doesn't point in that direction. Like I said, condescending comments like that would be very insulting if they weren't so absurd, so you really can't claim innocence in that department.

The fact that the other backdoors (5r1c7; 4r2c7; 6r2c6; 1r2c3; 5r2c1; 6r1c1) do not have the characteristic of the previous ones is also proved by the fact that to obtain a solution with these backdoors I am forced to use a method T & E = 2
While for the others a method T & E = 1 is sufficient.

I guess by T&E 2 you mean bifurcation. Where exactly did you originally state that it was forbidden? You said:

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.

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.
--

Added 1. There's actually a simpler way to fix my original chain:

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

Turns out that the real culprit was in fact the second node. The way I originally wrote it (5,6|1)r12c1 doesn't work as written, though the intended logic was correct. It just can't be shortened like that because it changes the semantics, which breaks both links surrounding the node the weak link. Thus it's not actually a synonym for the more explicit (and correct) variants I wrote before. It should have been obvious to me in the first place, but sometimes the improbable happens. So, in fact I did make a mistake. How refreshing! (Edit: The mistake was actually smaller than I feared. It did break the weak link, which was already known, but not the strong link. Thus the only new thing is the fix.)

With that little fix both links should work even with the original chain, as demonstrated by these implication chains:

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 /`

PS. Paolo, this would be a good time to claim that that was what you meant all along. However, I can't honestly see anything in what you've written that demonstrates my actual mistake, much less a way to fix it. Nevertheless, I did make a mistake and I have no problem admitting it. A bit embarrassing, perhaps, but it doesn't change the big picture in any way.
--

Added 2. I actually got a bit carried away with the self flagellation. There was nothing wrong with the strong link in the original, after all. Thus, everything was just like I said earlier: the strong link was good, but the weak link was bad. The fixed chain is good on both accounts, and so is the previously introduced variant, unless I'm really blind to something.

SpAce

Posts: 2504
Joined: 22 May 2017

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