The New Sudoku Players' Forum

[mith]

One more thing, regarding chains of three links:

A = B - C = D => A = D (strong/weak/strong implies strong link between ends; this is why a valid AIC works)
A - B = C - D => A - D (weak/strong/weak implies weak link between the ends)
A ^ B ^ C ^ D => A ^ D (xor/xor/xor implies xor link between the ends, for an odd number of links)

[Ajò Dimonios]

In fact it can be said that ¬A - ¬B is equivalent to A = B. The nature of the link is the same. They say virtually the same thing.

¬A - A = B - ¬B => ¬A - ¬B
A-¬A=¬B-B=>A-B

[mith]

Yes, exactly. If you have a strong link, you have a weak link between the negations, but not necessarily a weak link between the original statements.

While you can turn any strong link into a weak link by negation, what you can’t do is replace a strong link with a weak link in a chain (you can’t negate statements in the middle of a chain and expect the other links to remain). If it both strong and weak, then you get to choose how to use it.

[mith]

To put that last another way: you can’t go from

A = B = C = D
to
A = ¬B - ¬C = D

Unless the remaining strong links are still valid.

[Ajò Dimonios]

Hi Mith
you agree with what I wrote about the resolution
(2r4c6 = r6c12 = 2r5c1. (2 = 4,5,1,6) r5c3479- (1 | 6) r4c789 = 16r4c56 => - 7r4c6

My answer to question 3 is that the elimination of 7 in r4c6 is not legal. The reason is linked in the logic of eliminating any AIC. Our AIC shows that 2r4c6 = 16r4c56, for the triangle theorem any candidate that is linked with a weak inference to 2r4c6 and 16r4c56 can be eliminated. 7r4c6 is linked to 2r4c6 and 1r4c6 but not to 6r4c5, so it cannot be legitimately eliminated . But why is it important that 7r4c6 is linked to both candidates (1 and 6) in the tail of the chain? The reason is simple we must be sure that all possible combinations between 16r4c56 and 2r4c6 are tested because it is not known which is the correct combination that shows head and tail of the chain both true that see the false 7r4c6. The possible combinations are (r4c6 = 2; r4c5 = 1); (r4c6 = 2; r4c5 = 6; r4c6 = 1); (r4c6 = 2; r4c56 ≠ 16) and the opposites (r4c6 ≠ 2; r4c5 ≠ 1); (r4c6 ≠ 2; r4c5 ≠ 6; r4c6 ≠ 1); (r4c6 ≠ 2; r4c6 = 1; r4c5 = 6). Red combinations do not see the 7r4c6. In an AIC where head and tail are made up of only one candidate, all of this is immediate and you can proceed directly with the elimination.

Paolo

[mith]

No, I don't.

(16)r4c56 is the same as
1r4c56 AND 6r4c56, which is the same as
(1r4c5 AND 6r4c6) OR (1r4c6 AND 6r4c5)

If 7r4c6 is weakly linked to 1r4c6 (and obviously it is), then 7r4c6 is weakly linked to (1r4c6 AND 6r4c5).
7r4c6 is also weakly linked to 6r4c6, and therefore is weakly linked to (1r4c5 AND 6r4c6).
So 7r4c6 is weakly linked to (1r4c5 AND 6r4c6) OR (1r4c6 AND 6r4c5).

We happen to know from the puzzle that 6r4c6 is False, but this doesn't affect the links. What matters is that (16)r4c56 implies -7r4c56.

To say it another way: If both cells (r4c5 and r4c6) are occupied by 1 and 6 (in some order), then r4c6 must be occupied by 1 or 6, and therefore it cannot be occupied by 7.

[mith]

The error is here:

7r4c6 is linked to 2r4c6 and 1r4c6 but not to 6r4c5

7r4c6 does not need to be linked to 6r4c5. It is weakly linked to 1r4c6, and therefore (1r4c6 AND 6r4c5).

A - B => A - (B AND C)
A = B => A = (B OR C)

And here:

(r4c6 ≠ 2; r4c5 ≠ 1)

What the chain shows is that r4c6 ≠ 2 => (16)r4c56, so if r4c5 ≠ 1 then 6r4c5 and 1r4c6.

[Ajò Dimonios]

candidate 6 is not present i r4c6

[Ajò Dimonios]

When 16r4c56 is false one of the 3 possibilities is (r4c5 = 1; r4c5 ≠ 6; r4c6 ≠ 1) which are the candidates for the solution of the puzzle. In this case r4c6 = 1 is not present and does not see the candidate r4c6 = 7

[mith]

If (16)r4c56 is false, then at least one of (16)r4c56 and 7r4c6 is false.

¬A => (¬A OR ¬B)

And since 2r4c6 and (16)r4c56 are strongly linked, NOT((16)r4c56) => 2r4c6

(A OR B) AND ¬A => B (this is called Disjunctive Syllogism)

[Ajò Dimonios]

ok thanks i understood where i was wrong

[mith]

You’re welcome.

[Ajò Dimonios]

Hi Mith

One last question on the chain 2r4c6 =r6c6- r6c12 = 2r5c1- (2 = 4,5,1,6) r5c3479- (1 | 6) r4c789 = 16r4c56. Since the last link between (1| 6) r4c789 and 16r4c56 is a weak inference since they are both false, tying the chain from right to left, if 16r4c56 is false => (1 | 6) r4c789 is true (apparently a strong inference), the inference between 2r4c6 and 16r4c56 which link is? You can proceed to the elimination of r4c6 = 7?

Paolo

[SpAce]

Hi mith,

ok thanks i understood where i was wrong

You’re welcome.

And you're a miracle worker!! Am I really seeing the words 'I' and 'wrong' in the same sentence? And even 'thanks'?

Wow. That alone is worthy of recognition in the archives of the Jedi Order! Apparently I didn't misjudge you when I gave you a non-zero chance of pulling it off. Too bad I lost faith in it right after, but I'm glad to see you didn't.

I also must thank you, because this is the first significant debate in a long time that has achieved any kind of a positive result. For me a debate can end successfully in three ways (in this preference order):

1. Both sides see and admit that they were partly wrong (everyone learns)
2. I see and admit that I was wrong (I learn)
3. The other side sees and admits they were wrong (they learn)

On the other hand, all of these are failures (worst first):

1. Both are at least partly wrong but fail to see it (no one learns, everyone's frustrated)
2. I'm wrong but fail to see it (I don't learn, everyone's frustrated)
3. The other side is wrong but fails to see it (they don't learn, everyone's frustrated)

All of the recent AIC-debates have ended up as failures in that scale, without any kind of a mutually agreeable closure. It's been really frustrating and eating me up. Would you like to arbitrate those too? I think you've proven your skills and impartiality. (Those cases are a tad bit more difficult than this. Here the only failure option was obviously 3, but in those other cases there might be a small chance for other possibilities, too.)

Unlike some people undoubtedly think, I don't have an obsession of winning arguments at any cost. I don't even have an obsession of being right. What I want is the truth to win -- especially if it's not on my side. If someone can convince me that I'm wrong, I'm thankful and more than happy to admit it. What I hate the most is failing to see my own mistakes. If that has happened in any of the recent unclosed debates, I'd really like to know. If not, it would be nice to know that too.

[mith]

I am happy to weigh in on anything I understand (or think i understand).

(1 | 6) r4c789 = 16r4c56. Since the last link between (1| 6) r4c789 and 16r4c56 is a weak inference since they are both false

(1|6)r4c789 = (16)r4c56 is an XOR link, so it is both strong and weak. Exactly one of the following is true:

1r4c789 OR 6r4c789
1r4c56 AND 6r4c56

(And if exactly one is true, at least one is true, and at least one is false.)

(A XOR B) => (A OR B) (any xor link is also a strong link)
(A XOR B) => (¬A OR ¬B) (any xor link is also a weak link)

The resulting chain alternates = - = - = - =, so the ends are strongly linked.

(16)r4c789 - (16)r4c56 is only a valid weak link. They can't both be true, but they can both be false.