The New Sudoku Players' Forum

[mith]

Paolo: If you can't conclude they are two boys from my answer, then you admit that there is not a strong link between (two girls) and (two boys), yes? It is not the case that at least one is true. This is precisely equivalent to the fact that there is no strong link between (16)r4c789 and (16)r4c56, just a weak link. They can't both be true, but they can certainly both be false (and are - surprise, I have one of each).

The FF case is the negation of (1|6)r4c789. That's why it works. You reach the same conclusion because you are using the same node that eleven is correctly using.

Basically what you are doing is:

1. Ok, there's a weak link between (1r5c7, 6r5c9) and (16)r4c789. They can't both be true! (Fine, that is a valid weak link.)
2. It sure would be nice if at least one of (16)r4c789 and (16)r4c56 had to be true. Then I'd have a strong link!
3. Oh, wait, actually neither of them can be in r4c789 in this case, so they must both be in (16)r4c56. (Because you are using eleven's strong link.)
4. We reached the same conclusion, so it doesn't matter whether I write (1|6)r4c789 or (16)r4c789! They both work! Yay!

(I know, SpAce, but sometimes you need to scream into the void. )

[SpAce]

(I know, SpAce, but sometimes you need to scream into the void. )

I'm glad you do! That boy/girl analogy was perfect! I might actually give you a non-zero chance of getting through eventually, which was unthinkable for me. Even if it doesn't happen, you're not actually screaming into the void. I can see that I have a lot to learn from your argumentation style, so I'm following with great interest.

The most frustrating part is that when you think you've finally found the critical point and a perfect argument for it, like you just did, a new (or resurrected) mole-to-whack surfaces. It's like the previous point never mattered at all. That just happened now too:

I certainly can't conclude by saying that they are two boys because it could be a girl and a boy. Our case is different in the Eleven AIC we already have the information in the previous step that r5c7 = 1 and r5c9 = 6 therefore the only option that negates 16r4c789 usable in the next inference is FF.

A moment ago the question was about pure Boolean logic, as per Paolo's own words. Apparently he figured he was going to lose that argument, so he switched back to AICs. The last AIC in the stack was mine, but I guess he knew that he'd lose that argument too (obviously the flawed excuse above wouldn't work with it at all), so he went all the way back to eleven's AIC where he saw an opening. So here we are.

Of course that opening is a mirage too. Once that has been argued extensively and he sees there's no way out, he'll invent yet another excuse, and so it goes ad infinitum. I will be very impressed if someone can actually break that cycle.

[Ajò Dimonios]

I would like to bring some order to the discussion before the barrage of written messages takes us to other ports

1) Space claims that the AIC written by Eleven with (1 | 6) r4c789 is correct while with (16) r4c789 it is wrong. This is clearly false as it can be deduced from the definitions of "AND" and "OR". The two chains produce the same result.
2) Space claims that the chain with (16) r4c789 is wrong (it is not clear why only this one and not the one with (1 | 6) r4c789 see point 1) because to a check, after solving the puzzle, the two arguments A ( 16) r4c789 and B (16) r4c56 of the chain are false and contradict the definition of strong inference. This thesis can be countered by the fact that the definition of inference strong is redundant. In fact, given that A = ¬A; B = ¬B and ¬A = ¬B are true, A = B will also be true since A = ¬A = ¬B = B => A = B.
3) Space claims that the inference (16) r4c789 = 16r4c56 is not valid because there are AICs built with (16) r4c789 = 16r4c56 that produce false deletions. This has not been proven because the examples presented by Space are false due to a bad chain construction or bad candidate elimination but not because (16) r4c789 = 16r4c56 is not a strong inference.

Paolo

[SpAce]

I might actually give you [mith] a non-zero chance of getting through eventually.

I'm afraid I have to take that back There's no cure for this disease.

[mith]

Oh boy.

Paolo, I am only responding to point 2, because this is where your entire argument fails. In the process, I am going to ask a few simple yes/no questions, and if you change the subject again or otherwise don't answer, I'm done. If anyone who is not Paolo is confused and thinks he is right at this point, feel free to shoot me a message for more lessons on propositional calculus.
Question 1: If A is false, and B is false, is A = B a strong link/inference?

A is I have two girls.
¬A is I do not have two girls.
B is I have two boys.
¬B is I do not have two boys.

A = ¬A (any proposition and its negation have a strong link; at least one must be true)
B = ¬B (likewise)
¬A = ¬B (since I only have two children, at least one of these negations must be true; I cannot simultaneously have two children, two girls, and two boys)

Clearly it is incorrect to say A = B, though. We have already established this. It is possible for me to not have two girls and also not have two boys. And in fact that is the situation, I have one girl and one boy.
Question 2: Do you agree that there is no strong link between "both girls" and "both boys" in this example?

Where does the reasoning fail? It fails because A = B = C = D → A = D is not true in general for strong links defined as "at least one is true", because B and C can both be true (FTTF). You yourself already proved this and called it a contradiction. Why do you keep using something you know to be false?
Question 3: Do you agree that A = B = C = D → A = D, where = is defined as "at least one is true" is not a true implication in general?

The effect this is having on the discussion is that you are taking it as a given that if the negations of two propositions have a strong link/inference, the two propositions themselves must also have a strong link/inference, despite the fact that you know both propositions are false.
Question 4: Do you agree that both (16)r4c789 and (16)r4c56 are false in the puzzle?

[totuan]

Hi All,
For me, the understanding simply about strong link/inference like this: A=B means "at least one must be true IN SOLUTION".
For SpAce's example, at start we can't know at least one of [ pair (16)r4c56 and pair (16)r4c789 ] must be true IN SOLUTION or not, so pair (16)r4c56 and pair (16)r4c789 is not strong link/inference at start.

totuan

[SpAce]

For me, the understanding simply about strong link/inference like this: A=B means "at least one must be true IN SOLUTION".

Exactly. That's the whole point of a strong link (or a larger SIS). If valid, it provides a 100% certainty that at least one of the options must be true in the solution, which gives it its solving power. If not valid, it can't prove anything. Any use of an invalid strong link amounts to guessing. Thus:

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valid (A=B) => (A|B) in solution; not (A|B) in solution => not valid (A=B)

Can't get any simpler than that.

Btw, consequently a puzzle with zero solutions doesn't have any strong links, by definition. It makes sense too, because there are no guaranteed truths in such a puzzle.

[Ajò Dimonios]

HI Mith

Paolo, I am only responding to point 2, because this is where your entire argument fails. In the process, I am going to ask a few simple yes/no questions, and if you change the subject again or otherwise don't answer, I'm done. If anyone who is not Paolo is confused and thinks he is right at this point, feel free to shoot me a message for more lessons on propositional calculus.
Question 1: If A is false, and B is false, is A = B a strong link/inference?

A is I have two girls.
¬A is I do not have two girls.
B is I have two boys.
¬B is I do not have two boys.

A = ¬A (any proposition and its negation have a strong link; at least one must be true)
B = ¬B (likewise)
¬A = ¬B (since I only have two children, at least one of these negations must be true; I cannot simultaneously have two children, two girls, and two boys)

Clearly it is incorrect to say A = B, though. We have already established this. It is possible for me to not have two girls and also not have two boys. And in fact that is the situation, I have one girl and one boy.
Question 2: Do you agree that there is no strong link between "both girls" and "both boys" in this example?

Where does the reasoning fail? It fails because A = B = C = D → A = D is not true in general for strong links defined as "at least one is true", because B and C can both be true (FTTF). You yourself already proved this and called it a contradiction. Why do you keep using something you know to be false?
Question 3: Do you agree that A = B = C = D → A = D, where = is defined as "at least one is true" is not a true implication in general?

The effect this is having on the discussion is that you are taking it as a given that if the negations of two propositions have a strong link/inference, the two propositions themselves must also have a strong link/inference, despite the fact that you know both propositions are false.
Question 4: Do you agree that both (16)r4c789 and (16)r4c56 are false in the puzzle?

Small introduction to questions 1 and 2
¬A ≠ (I do not have two gilrs) but = (I have a girl and a boy OR I have two boys)
¬A is the negation of A and has two solutions linked by an OR
¬B ≠ (I do not have two boys) but = (I have a girl and a boy OR I have two girls)
¬B is the negation of B and It has two solutions linked by an OR
Clearly, by OR we mean OR operator
Logical OR operator |: The | operator computes the logical OR of its operands. The result of x | y is true if either x or y evaluates to true. Otherwise, the result is false.

http://pages.di.unipi.it/corradini/Dida ... /main.html

Answer to question 1: Yes, because ¬B and ¬A are both true.
Answer to question 2:No because as I wrote previously A = ¬A = ¬B = B => A = B. The strong inference arises from the fact that ¬B and ¬A are strongly linked because both are true.
Answer 3 :No as per previous demonstration A = ¬A = ¬B = B => A = B. The definition of strong inference is redundant. From its definition, as a premise, it is possible in certain cases to deduce the negation of the premise itself. This fact is clearly redundant but the logic is correct. In my opinion it means that the definition is not complete.
Answer to question 4: Yes

Paolo

[mith]

Question 5: What is the definition of a strong inference/link?

[mith]

Aside, since this is not really relevant:

¬A ≠ (I do not have two gilrs) but = (I have a girl and a boy OR I have two boys)

This may be a language issue? "I do not have two girls" in English
is equivalent to NOT(I have two girls)
is equivalent to NOT(first child is a girl AND second child is a girl)
is equivalent to (NOT(first child is a girl) OR NOT(second child is a girl))
is equivalent to (I have a girl and a boy OR I have two boys).

Question 6: Do you understand that "=" is not the boolean operator for equivalence in this context?

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A == B == C == D → A == D
(A == B) && (B == C) && (C == D) → (A == D), to be more precise

is true; equivalence between A and B, B and C, and C and D does imply equivalence between A and D.

A = B is a strong link in our context, and means A || B. To write a chain of strong links A = B = C = D with boolean operators, you would say:

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(A || B) && (B || C) && (C || D)

and this does not imply (A || D).

(I mean, it should be obvious that A is not equivalent to ¬A, but I am taking nothing for granted here.)

[eleven]

mith,

to keep the analogy, you indeed should use the bot's definitions of ¬A and ¬B.
¬A = ¬B says, that at least one of
- (I have a girl and a boy OR I have two boys)
- (I have a girl and a boy OR I have two girls)
must be true, which is correct (if you have 2 children with defined sex).

The problem is, that it cannot be used as a weak inference, because both are true, if you have a girl and a boy.

Thus the link ¬A - ¬B (if one is true, the other is false) is wrong and therefore
A = ¬A - ¬B = B
is wrong, which would say, that you only can have 2 boys or 2 girls (or both, but that would be 4 children, so we can exclude it here).
And of course
A = ¬A = ¬B = B => A=B is wrong

[edit: weak link description]

[eleven]

Btw, consequently a puzzle with zero solutions doesn't have any strong links, by definition. It makes sense too, because there are no guaranteed truths in such a puzzle.

That's not true. To get an example, i added a wrong clue to mith's digit symmetric puzzle:

Code: Select all

+-------------------+-------------------+-------------------+
| 9     3     8     | 7     16    4     | 256   12    156   |
| 6     7     1     | 2     9     5     | 38    4     38    |
| 25    25    4     | 8     16    3     | 69    7     169   |
+-------------------+-------------------+-------------------+
| 7     568   356   | 4     2     68    | 1     9     368   |
| 28    4     36    | 1     38    9     | 2368  5     7     |
| 1     268   9     | 56    358   7     | 2368  28    4     |
+-------------------+-------------------+-------------------+
| 58    5689  56    | 3     4     2     | 7     18    1589  |
| 3     589   2     | 59    7     1     | 4     6     589   |
| 4     1     7     | 569   58    68    | 59    3     2     |
+-------------------+-------------------+-------------------+

[mith]

I'm inclined to agree with eleven on this one - my personal definition of a strong link would be something along the lines of "a strong link exists between two propositions if the current state of the puzzle (givens and candidates) implies that at least one of the propositions is true". The problem is that ultimately you could then show a strong link between anything, because the current state of the puzzle is false, and "false implies X" is true for any X.

(The same is true for uniqueness tests, really. Valid logic, but fails if the puzzle does not have a unique solution.)

[SpAce]

Btw, consequently a puzzle with zero solutions doesn't have any strong links, by definition. It makes sense too, because there are no guaranteed truths in such a puzzle.

That's not true. To get an example, i added a wrong clue to mith's digit symmetric puzzle

Can you be a bit more specific? I'm not proficient with symmetries so you have to explain in more detail, if it's related to that property.

In any case I think my first sentence is definitely true, if we use totuan's definition of a strong link (and I do):

"at least one must be true IN SOLUTION"

Then "a puzzle with zero solutions doesn't have any strong links" is clearly true, because nothing can be true in a solution that doesn't exist. Do we agree on that triviality?

The second sentence is more problematic, and I realized it myself when I wrote it:

there are no guaranteed truths in such a puzzle

I know I didn't think that through, and it's clearly not true generally. I was mainly thinking of an otherwise single-solution puzzle where a false given would cause a global contradiction. Since that contradiction can be moved around freely, not much if anything is guaranteed to end up in the "solution". Did I go too far with that claim even in that context? (Of course one could imagine guaranteed complex truths in any context, but I was thinking of the basic variety here.)

In any case, the bigger problem I imagined even when I wrote it was with puzzles that would otherwise have multiple solutions. Their sub-puzzles would be isolated from the contradiction, and thus could have guaranteed and relatively simple truths (though with multiple solutions). Yet, even they wouldn't have strong links by the simple definition above, because there's no global solution.

my personal definition of a strong link would be something along the lines of "a strong link exists between two propositions if the current state of the puzzle (givens and candidates) implies that at least one of the propositions is true".

For all practical purposes I agree. All normal solving techniques are based on the assumption that the puzzle has at least one solution. Thus the player is obviously within their rights to assume the same, or otherwise there's no point in trying to solve the puzzle in the first place. So, from the player's point of view strong links can be considered valid locally even if they aren't that globally (due to the no-solution nature of the puzzle itself).

However, I don't think that's a good enough reason to change the actual definition of a strong link. Just like any other solving technique, it's meaningful only if there's at least one solution. (Or at least I can't imagine anything else.)

The problem is that ultimately you could then show a strong link between anything, because the current state of the puzzle is false, and "false implies X" is true for any X.

Exactly.

[eleven]

The strong links are there, if the puzzle has a solution or not.
I definitely can use them to prove, that this puzzle has no solution, because applying them (correctly) leads to a contradiction.