The New Sudoku Players' Forum

[SpAce]

Paolo,

Thank you for your answer. It was exactly what I expected.

(16)r4c56 = (16)r4c789 - r5c9 = r5c1 - r2c1 = r1c2 - (6=815)r139c8 - (5=7)r9c2 - r7c1 = r7c5 - r8c6 = (7)r4c6 => -2 r4c6, -7 r4c5; stte

Your chain is correct

No, it obviously isn't. Both of its end-points are false in the solution, which makes it possible to conclude incorrect (as well as correct) eliminations. (Note that only -2r4c6 is actually incorrect.)

Why does it have two false end-points? Because the first strong link is invalid. Since both of its options are false, it can "prove" anything, including total absurdities. It's worthless because it doesn't guarantee true conclusions.

(However, for that exact reason it works for you very well, because you obviously don't care about the truth. Many of your conclusions are just as absurd, for the same reasons.)

only the deletions that are not correct r4c6 = 2 and r4c5 = 7, because both r4c6 = 1 and r4c5 = 6 must be connected with weak inference to the candidates to be eliminated.

So you're denying either that A) the end-point (16)r4c56 is a locked pair, or B) locked pairs eliminate all of the other digits in their cells. Which one of those well-established facts do you want to deny this time to avoid admitting your mistake?

(What you're describing is (1|6)r4c56. That would indeed be a valid end-point for this AIC, though incapable of eliminating anything. Obviously that's not what's written in the AIC, so your argument is worthless.)

This AIC does not produce eliminations.

That is certainly true, but not because its end-points couldn't do it. They most certainly could -- if either one were true.

This also makes me review the Eleven solution with (16) r4c789 = 16r4c56, also in this case the elimination of r4C6 = 7 is not correct because it is not connected to r4c6 = 1 and r4c5 = 6.

So, you're now backtracking and returning to your original position that eleven's AIC was invalid? I thought you already understood his explanation:

there is a hidden pair 16 in r4c56, if both digits cannot be in r4c789. Then the only remaining cells for both are r4c56.

OK, now I understand

So, you didn't understand? Or did you now change your mind, because otherwise you'd have to admit how wrong you've been all this time?

(That said, eleven's wording is a bit ambiguous. Better: "there is a hidden pair 16 in r4c56, if neither digit can be in r4c789.")

Either way, that same hidden pair is the end-point in my invalid AIC, and it certainly works -- if it ever gets locked. Thus the problem must be somewhere else, and there's only one possibility for that.

--
Yet, none of that makes any difference, no matter how obvious the truth is. You will never admit it, because you simply can't. I knew that long before this discussion, but even I didn't realize how severe your problem is.

I hope someone will stop me if I ever make the mistake of trying to have a rational conversation with you again. You will simply invent your own facts and logic while you go, and armed with those you can obviously claim anything you like with total confidence. False premises allow that, just like my invalid AIC demonstrated. The two of you have a lot in common.

[Ajò Dimonios]

Space wrote:


So you're denying either that A) the end-point (16)r4c56 is a locked pair, or B) locked pairs eliminate all of the other digits in their cells. Which one of those well-established facts do you want to deny this time to avoid admitting your mistake?

(What you're describing is (1|6)r4c56. That would indeed be a valid end-point for this AIC, though incapable of eliminating anything. Obviously that's not what's written in the AIC, so your argument is worthless.)

No it does not. The final pair is formed by the pair 16r4c56, but only 1r4c6 is linked to 2r4c6, 6r4c5 is not linked to 2r4c6, this fact does not allow the elimination of candidate 2 in r4c6. In order to perform the elimination both (1 and 6) must have a weak link with 2 in r4c6. This is the reason why even in the Eleven solution, with the option I indicated (16) r4c789 = 16r4c56 does not allow the elimination of 7 in r4c6.

Paolo

[SpAce]

No it does not. The final pair is formed by the pair 16r4c56, but only 1r4c6 is linked to 2r4c6, 6r4c5 is not linked to 2r4c6, this fact does not allow the elimination of candidate 2 in r4c6. In order to perform the elimination both (1 and 6) must have a weak link with 2 in r2c5. This is the reason why even in the Eleven solution, with the option I indicated (16) r4c789 = 16r4c56 does not allow the elimination of 7 in r4c6.

So you chose option B:

So you're denying either that A) the end-point (16)r4c56 is a locked pair, or B) locked pairs eliminate all of the other digits in their cells. Which one of those well-established facts do you want to deny this time to avoid admitting your mistake?

It's very interesting that you don't understand how Hidden Pairs work. One of the most basic techniques there is.

[Ajò Dimonios]

Hi Space

Space wrote:
Ajò Dimonios wrote:
No it does not. The final pair is formed by the pair 16r4c56, but only 1r4c6 is linked to 2r4c6, 6r4c5 is not linked to 2r4c6, this fact does not allow the elimination of candidate 2 in r4c6. In order to perform the elimination both (1 and 6) must have a weak link with 2 in r2c5. This is the reason why even in the Eleven solution, with the option I indicated (16) r4c789 = 16r4c56 does not allow the elimination of 7 in r4c6.

So you chose option B:

SpAce wrote:
So you're denying either that A) the end-point (16)r4c56 is a locked pair, or B) locked pairs eliminate all of the other digits in their cells. Which one of those well-established facts do you want to deny this time to avoid admitting your mistake?

It's very interesting that you don't understand how Hidden Pairs work. One of the most basic techniques there is.

When you built your chain you started from the hypothesis that (16) r4c56 was false up to the end of the chain to say that r4c6 = 7 is true, which allows you to say that r4c6 = 7 is weakly linked with r4c6 = 2. To get to the eliminations we have to establish, with Boolean logic, from ((16) r4c56 false) = A what (nor A) means. NorA is (1 true; 6 false), (1 false; 6 true) and (1 true; 6 true) in r4c56. At this point we must establish if all three possible hypotheses are linked with weak inference to r4c6 = 2, because only in this case we can proceed with the elimination. As can be seen, NorA in only two cases out of three shows a weak link with r4c6 = 2, in the case (1 false; 6 true) the link is not there, even though it is fully NorA. This shows that the deletion is not valid. In this way you can determine that your AIC does not lead to any elimination.
But let's go back to the main topic of our discussion which is certainly more interesting. Can you explain to me in terms of AIC how in the Eleven resolution (1 | 6) r4c789 = 16r4c56 leads to the elimination of r4c6 = 7 ?, given that at this point it is neither to you nor to me, for different reasons (16) r4c789 = 16r4c56 produces the elimination of r4c6 = 7. I remember that (1 | 6) r4c789 is equivalent to (1r4c789 or 6r4c789), since we have eliminated the last option (16r4c789).

P.S.
Or better still how can I conclude that 1r4c789 = 16r4c56 or 6r4c789 = 16r4c56?
Paolo

[totuan]

Paolo,
Thank you for your answer. It was exactly what I expected.

(16)r4c56 = (16)r4c789 - r5c9 = r5c1 - r2c1 = r1c2 - (6=815)r139c8 - (5=7)r9c2 - r7c1 = r7c5 - r8c6 = (7)r4c6 => -2 r4c6, -7 r4c5; stte

Your chain is correct

No, it obviously isn't.
........

I hope someone will stop me if I ever make the mistake of trying to have a rational conversation with you again.

Hahahaaaa..., you are quite clever
Yes and PLEASE, you can stop here - no need to explain more. Thank you.

totuan

[Ajò Dimonios]

Before laughing you should understand what it means the chain is correct?
Space wrote this chain to show that the first inference is not a strong inference and that consequently the conclusions are wrong (false deletions). Instead, I argue that the chain is formally correct and does not produce deletions as I demonstrated in my previous post.
You're smart enough too.

Paolo

[SpAce]

Before laughing you should understand what it means the chain is correct?

FYI, totuan is probably one of the best manual solvers ever, and his notational skills are legendary. Please don't make an even bigger fool of yourself by questioning his understanding of chains. It's on a whole different level than insulting me (which you shouldn't have done either).

(I've never seen this bad case of the Dunning–Kruger effect, and I've seen pretty bad ones.)

I argue that the chain is formally correct and does not produce deletions as I demonstrated in my previous post.

For argument's sake, let's assume you're right (you're not). What's your excuse for this variant, then:

Code: Select all

.------------------.----------------------.----------------------.
|  5     g26  3    | 1     78        9    |  4    h68     2678   |
| f26     1   478  | 478   3478      5    |  278   9      23678  |
|  478    79  4789 | 2     3478      6    |  5    h18     1378   |
:------------------+----------------------+----------------------:
|  9      3   5    | 478  b14678   bn17-2 | c128  c1468  c12468  |
| e1246   8   14   | 45    9         3    |  12    7     d12456  |
|  12467  26  147  | 458   1468     a12   |  9     3      124568 |
:------------------+----------------------+----------------------:
| j17     59  2    | 3    k17        8    |  6     45     49     |
|  3      4   1789 | 6     5        m17   |  178   2      1789   |
|  178   i57  6    | 9     2         4    |  3    h158    178    |
'------------------'----------------------'----------------------'

(2=1)r6c6 - (16)r4c56 = (16)r4c789 - r5c9 = r5c1 - r2c1 = r1c2 - (6=815)r139c8 - (5=7)r9c2 - r7c1 = r7c5 - r8c6 = (7)r4c6 => -2 r4c6

(The only difference is the extra node at the start.)

--
Of course we could add a similarly redundant node to eleven's perfectly valid chain, but it just adds extra length and complexity:

2r4c6 = r6c6 - r6c12 = 2r5c1 - (2=4,5,1,6)r5c3479 - (1|6)r4c789 = 16r4c56 - (1|6=24587)b5p98741 => -7r4c6, stte

(Every hidden subset has a corresponding naked subset with the same eliminations. I guess you didn't know that either.)

[Ajò Dimonios]

Space wrote:
(2=1)r6c6 - (16)r4c56 = (16)r4c789 - r5c9 = r5c1 - r2c1 = r1c2 - (6=815)r139c8 - (5=7)r9c2 - r7c1 = r7c5 - r8c6 = (7)r4c6 => -2 r4c6

(The only difference is the extra node at the start.)

Can you explain to me how you manage to write 1r6c6- (16) r5c56? I inform you that r6c6 = 1 is a backdoor and so is r5c5 = 6. With only one weak inference you can prove that when 1 is true in r6c6 6 in r5c5 it is false. In practice, when one backdoor is true the other is false. It is truly original.

P.S. Perhaps it is better that you answer the other questions I asked in the previous post, clearly excluding that (16) r4c789 = 16r4c56 which for you is not a strong inference.


Paolo

[mith]

If 1 is in r6c6, there is not a 16 pair in r4c56. If there is a 16 pair in r4c56, there is not a 1 in r6c6. That's the weak link.

The reason there is no strong link as you write it is that it is possible for there to not be a 16 pair in r4c56 and also to not be a 16 pair in r4c789 - as is the case in the actual solution. On the other hand, it is not possible for both sides of eleven's link to be false (either there is a 16 pair in r4c789, or there is a 1 OR a 6 OR both in r4c56).

(1|6)r4c56 = (1|6)r4c789 would also not be a valid strong link, in this case because they can be (and are) both true. [edit]Revised later in the thread.[/edit]

[SpAce]

(2=1)r6c6 - (16)r4c56 = (16)r4c789 - r5c9 = r5c1 - r2c1 = r1c2 - (6=815)r139c8 - (5=7)r9c2 - r7c1 = r7c5 - r8c6 = (7)r4c6 => -2 r4c6

Can you explain to me how you manage to write 1r6c6- (16) r5c56?

The same exact way I wrote (16)r4c789 - 6r5c9, which you accepted. As mith already explained, it's a perfectly valid weak link, just like the other. Perhaps it's easier for you to understand from the other direction: (61)r4c56 - 1r6c6. Since AICs are bidirectional, so are all links obviously:

(7)r4c6 = r8c6 - r7c5 = r7c1 - (7=5)r9c2 - (5=18)r93c8 - (8=6)r1c8 - r1c2 = r2c1 - r5c1 = r5c9 - (61)r4c789 = (61)r4c56 - (1=2)r6c6 => -2 r4c6

Thus, both weak links surrounding the critical strong link are valid, as is everything else in the chain (which you already accepted). That leaves only one place where the error can be.

I'm afraid you've now run out of even remotely understandable excuses, and I will certainly not entertain any more of them.

P.S. Perhaps it is better that you answer the other questions I asked in the previous post, clearly excluding that (16) r4c789 = 16r4c56 which for you is not a strong inference.

I won't entertain any other questions either until you concede this one. There's no point in discussing anything else until you prove that you're actually capable of admitting a mistake and learning from it. Otherwise it's just a waste of time for both, and very frustrating at that.

The only smart move left at this point is to concede. Everyone can see that it's probably almost impossible for you under any circumstances, but you might be surprised if you tried. I respect people who can admit their mistakes, especially if it's obviously very hard for them. Not so much the other kind.

Sometimes when you innovate, you make mistakes. It is best to admit them quickly, and get on with improving your other innovations.

[Ajò Dimonios]

Hi Mith

Mith wrote:
If 1 is in r6c6, there is not a 16 pair in r4c56. If there is a 16 pair in r4c56, there is not a 1 in r6c6. That's the weak link.
The reason there is no strong link as you write it is that it is possible for there to not be a 16 pair in r4c56 and also to not be a 16 pair in r4c789 - as is the case in the actual solution. On the other hand, it is not possible for both sides of eleven's link to be false (either there is a 16 pair in r4c789, or there is a 1 OR a 6 OR both in r4c56).

(1|6)r4c56 = (1|6)r4c789 would also not be a valid strong link, in this case because they can be (and are) both true.

When 1 is in r6c6 the only certain thing is that in r4c56 there is no 1 and consequently no pair of 1s with 4; 8; 6; 7 and 2, it is obvious, but in writing it means that there are no candidates 1 and 6, otherwise we cannot proceed with strong inference = (16) r4c789 which means 1 and 6 are present in r4c789.

I have never written that (16) r4c789 = 16r4c56 means that the pair 16 is not present on either side of the strong inference. This result is given by the solution of the puzzle. What you are saying is Space's thesis which appeals to the definition of strong inference which exists when at least one hypothesis is true and consequently (16) r4c789 = 16r4c56 cannot be a strong inference. I instead appeal to cases in which both hypotheses are true as I wrote previously.

I'll give you another even better clarifying example. A and B are two backdoors of a puzzle and there is also a strong inference between them, as in the August 1 puzzle. So the inference A = B is certainly a strong inference, but it is also a strong inference Nor A = Nor B, since Nor A = A = B = Nor B. If you analyze the inference Nor A = Nor B you will realize that even being bound by strong inference Nor A and Nor B are both false. This case is practically identical to our case (16) r4c789 = 16r4c56.

Paolo

[mith]

I instead appeal to cases in which both hypotheses are true as I wrote previously.

Neither of them is true, though.

(16)r4c56 means, in the notation eleven/SpAce/etc. are using, that both 1 and 6 are present in r4c56. It is false if only one (or neither) is present. Likewise, (16)r4c789 means that both 1 and 6 are present in r4c789. Obviously these cannot be both true.

I think the misunderstanding here is simply in what is meant by (16). Though I guess I don't understand what you think it means, either.

[Ajò Dimonios]

Space wrote:
Ajò Dimonios wrote:
SpAce wrote:
(2=1)r6c6 - (16)r4c56 = (16)r4c789 - r5c9 = r5c1 - r2c1 = r1c2 - (6=815)r139c8 - (5=7)r9c2 - r7c1 = r7c5 - r8c6 = (7)r4c6 => -2 r4c6

Can you explain to me how you manage to write 1r6c6- (16) r5c56?

The same exact way I wrote (16)r4c789 - 6r5c9, which you accepted. As mith already explained, it's a perfectly valid weak link, just like the other. Perhaps it's easier for you to understand from the other direction: (61)r4c56 - 1r6c6. Since AICs are bidirectional, so are all links obviously:

(7)r4c6 = r8c6 - r7c5 = r7c1 - (7=5)r9c2 - (5=18)r93c8 - (8=6)r1c8 - r1c2 = r2c1 - r5c1 = r5c9 - (61)r4c789 = (61)r4c56 - (1=2)r6c6 => -2 r4c6

Thus, both weak links surrounding the critical strong link are valid, as is everything else in the chain (which you already accepted). That leaves only one place where the error can be.

I'm afraid you've now run out of even remotely understandable excuses, and I will certainly not entertain any more of them.

No you cannot write this logical chain because in this case (16) r4c789 = 16r4c56 cannot be a strong inference. The non-existence of a pair 16 in r4c56 does not imply that a pair 16 exists in r4c789 as long as it is possible that 6 is in r4c5, while the absence of 1 and 6 in r4c56 implies that 1 and 6 are present in r4c789. You understand that it is very different.

Space wrote:
P.S. Perhaps it is better that you answer the other questions I asked in the previous post, clearly excluding that (16) r4c789 = 16r4c56 which for you is not a strong inference.

I won't entertain any other questions either until you concede this one. There's no point in discussing anything else until you prove that you're actually capable of admitting a mistake and learning from it. Otherwise it's just a waste of time for both, and very frustrating at that.

The only smart move left at this point is to concede. Everyone can see that it's probably almost impossible for you under any circumstances, but you might be surprised if you tried. I respect people who can admit their mistakes, especially if it's obviously very hard for them. Not so much the other kind.

I understand the options are two or you don't know the answer or you don't want to tell us. I lean towards the first option. Bye

Paolo

[mith]

To get to the eliminations we have to establish, with Boolean logic, from ((16) r4c56 false) = A what (nor A) means. NorA is (1 true; 6 false), (1 false; 6 true) and (1 true; 6 true) in r4c56.

This seems to be the correct understanding of (16)r4c56 (other than the last "and" needing to be an "or"). I'm going switch to P and Q here, because it's what I'm used to:

P = (16)r4c56 = (1r4c56 AND 6r4c56)
¬P = NOT((16)r4c56) = NOT(1r4c56 AND 6r4c56) = (-1r4c56 OR -6r4c56) = (1r4c56 AND -6r4c56) OR (-1r4c56 AND 6r4c56) OR (-1r4c56 AND -6r4c56)

Q = (16)r4c789 = (1r4c789 AND 6r4c789)
¬Q = NOT((16)r4c789) = NOT(1r4c789 AND 6r4c789) = (-1r4c789 OR -6r4c789) = (1r4c789 AND -6r4c789) OR (-1r4c789 AND 6r4c789) OR (-1r4c789 AND -6r4c789)

P → ¬Q, and Q → ¬P, so they form a weak link. However, ¬P→ Q is false (as is ¬Q → P), so they do not form a strong link. At most one of (P, Q) is true. Since the AIC requires a strong link there, it is not a valid AIC.

On the other hand:

[edited to match the original chain]
Q' = (1|6)r4c789 = (1r4c789 OR 6r4c789) = (1r4c789 AND -6r4c789) OR (-1r4c789 AND 6r4c789) OR (1r4c789 AND 6r4c789)
¬Q' = NOT(1r4c789 OR 6r4c789) = (-1r4c789 AND -6r4c789)
And here P ↔ ¬Q' (and equivalently Q' ↔ ¬P), so there is a strong link. Exactly one of (P, Q') is true.

Can you point to what you disagree with here?

[mith]

As can be seen, NorA in only two cases out of three shows a weak link with r4c6 = 2, in the case (1 false; 6 true) the link is not there, even though it is fully NorA. This shows that the deletion is not valid. In this way you can determine that your AIC does not lead to any elimination.

This is precisely why the AIC itself is not valid. Because ¬P→ Q is false in this case (here both ¬P and ¬Q), the weak link from Q goes nowhere, and we don't ultimately get to 7r4c6.