The New Sudoku Players' Forum

[SpAce]

The strong links are there, if the puzzle has a solution or not.

Did you read a word I just said? Your statement is true if you're using mith's practical definition only. I was clearly speaking about the consequences of totuan's definition, which I consider the official one. Stick to that if you want to criticize what I wrote.

I definitely can use them to prove, that this puzzle has no solution, because applying them (correctly) leads to a contradiction.

Yeah, because your (like everyone else's) process looks like this:

1. Assume the puzzle has at least one solution.
2. Apply apparent strong links correctly, by mith's definition.
3. Arrive at a contradiction.
4. Conclude that that puzzle has no solution.

What you're missing is a fifth step based on that conclusion:

5. Conclude that the apparent strong links weren't real, by totuan's definition.

Which part do you disagree with?

--
Personally I think we need both definitions: one globally (totuan's) and the other locally (mith's).

The former is sufficient for puzzles that have at least one solution, because any valid strong link guarantees that at least one of the options is in a solution. It also provides the theoretical meaning for a strong link, and a quick way to prove one's invalidity if neither option is found in a solution.

However, the latter is what we all use in practice, and it's also necessary for when there is no solution (to prove that). Thus, in the no-solution context this is not exactly true:

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.

That global definition of validity only applies if there's at least one solution. In the no-solution context we must use a local definition (apparent validity), or otherwise we can't prove that there's no solution.

[Ajò Dimonios]

At this point I too would like to ask questions.
1) Is the Eleven solution (2r4c6 = r6c12 = 2r5c1. (2 = 4,5,1,6) r5c3479- (1 | 6) r4c789 = 16r4c56 => - 7r4c6, stte) valid?
2) If it is valid, can you explain why?
3) If it is not valid can you explain the reason?

Paolo

[SpAce]

At this point I too would like to ask questions.

Hidden Text: Show

1) Is the Eleven solution (2r4c6 = r6c12 = 2r5c1. (2 = 4,5,1,6) r5c3479- (1 | 6) r4c789 = 16r4c56 => - 7r4c6, stte) valid?
2) If it is valid, can you explain why?
3) If it is not valid can you explain the reason?

1) eleven's solution is valid, but your sloppy copy of it is not.
2) Not to a troll.
3) N/A

[Ajò Dimonios]

Space can you explain to me why (2r4c6 = r6c12 = 2r5c1. (2 = 4,5,1,6) r5c3479- (1 6) r4c789 = 16r4c56 => - 7r4c6, stte) is not valid?
I remind you that (1 6) r4c789 and (1 | 6) r4c789 produce an identical result, perhaps this is a detail that you have not yet digested.

[SpAce]

can you explain to me [anything at all]?

No, unfortunately I can't perform miracles. See number 2 in my previous answers.

[Ajò Dimonios]

Hi All
However, I am not interested in arguing about an already acquired result. My interest is only on the answers to questions 2 or 3.
2) if it is valid, can you explain why?
3) if it is not valid can you explain the reason?
A detailed answer would be useful because in my opinion it answers most of the topics discussed so far.

Paolo

[totuan]

However, I am not interested in arguing about an already acquired result. My interest is only on the answers to questions 2 or 3.
2) if it is valid, can you explain why?
3) if it is not valid can you explain the reason?
A detailed answer would be useful because in my opinion it answers most of the topics discussed so far.

That is important and necessary for you to incease your skills on solving puzzles? I suggest two options.
Option 1: If it's important and necessary, you should read through the previous pages carefully to get your own answers.
Option 2: If it is not important and necessary, we can stop it here.

totuan

[mith]

Paolo, my questions 5 and 6 are really simple and should be easy to provide a yes or no answer to. You don't even have to add anything else. Just yes or no.

I don't think anyone has any interest in discussing this with you further if you can't provide that. The questions you asked have been answered numerous times.

[Ajò Dimonios]

Hi All

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]

Paolo, my questions 5 and 6 are really simple and should be easy to provide a yes or no answer to. You don't even have to add anything else. Just yes or no.

I don't think anyone has any interest in discussing this with you further if you can't provide that. The questions you asked have been answered numerous times.

[Ajò Dimonios]

Hi mith

Sorry if I didn't answer question 5. My bad English doesn't allow me to answer quickly. However, in a previous post I had already provided the definition of strong inference that is known in the literature and what in my opinion should be the generalization to include these particular cases as well.
Definition 1 : Let A, B be two assumptions. We say that A,B are thicklinked and note A == B when at least one of A,B is true . We say that A,B are thinlinked and note A--B when at most one of A,B is true.

Paolo

[mith]

And question 6?

[Ajò Dimonios]

Hi Mith

Mith Wrote:
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?

So it's a translation problem because in Italian "I don't have two girls" can very well mean that I have one or none. At this point your example has only two options. By transferring a cell, a row, a column, a block with only two candidates to the Sudoku puzzle. In this case we do not need to discuss because it is certain that if the sudoku is valid at least one candidate is true and the other is false. It will never happen that they can both be true or both are false. I was discussing more complex situations where, as two strongly linked Backdoors are both true. It can very well be shown that two backdoors are strongly linked. See the example of the puzzle of August 1 and even the definition allows it because we are talking about at least one must be true but the possibility that they are both is not excluded. In this case A = B (both true) but also A = ¬A and B = ¬B (¬A false and ¬B false). I write the chain that binds them all together (chain of strong inferences only) ¬A = A = B = ¬B => ¬A = ¬B. This tells me that the strong inference between the two backdoors determines an identical strong inference between their boolean negations. I think the answer to your question 6 is also the operator “NOT” is a Boolean operator while = is a mathematical operator it means equal to.


Paolo

[mith]

So, no, you do not understand what "=" means in this context.

¬A = A = B = ¬B => ¬A = ¬B

Ignoring for the moment that you've swapped the negations, "=" in this statement represents a strong link. It does not represent equality or equivalence.

Question 7: Do you understand that A ≠ ¬A for any statement A, where ≠ is the boolean operator for non-equivalence? (That is, A and ¬A always have different truth values.)
Question 8: Do you understand that A = ¬A (strong link) means the same thing as A || ¬A (boolean operator) and A ∨ B (logical disjunction)?
Question 9: Do you understand that A = ¬A (strong link) does not mean A == ¬A (boolean equivalence)?
Question 10: Is the following statement true for boolean operations? (A || B) && (B || C) && (C || D) => (A || D)

[mith]

And revisiting earlier answers:

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

Definition 1 : Let A, B be two assumptions. We say that A,B are thicklinked and note A == B when at least one of A,B is true

(Note, this should read "A = B", not "A == B"; "==" is the boolean operator for equivalence.)

Question 1: If A is false, and B is false, is A = B a strong link/inference?

Answer to question 1: Yes, because ¬B and ¬A are both true.

Premise 1: By your answer to question 5, A and B have a strong inference/link (are thicklinked) when at least one of A,B is true.
Premise 2: By your answer to question 1, A and B have a strong inference/link (are thicklinked) when they are both false, because ¬B and ¬A are true.
Conclusion: Therefore, all pairs of statements (A and B) have a strong inference/link (are thicklinked), because either (at least one of A,B is true) or (both A and B are false).

Question 11: Is this a logical conclusion? Are all pairs of statements strongly linked?