The New Sudoku Players' Forum

[eleven]

As soon as you introduce some condition as "if none was a given", you are making some extra-logical assumption.

I am not "making some extra-logical assumption", i am just using the fact, that there is no given in the UR1.1 cells. If you are ignoring or rejecting this fact, this is not my problem.
There is no uniqueness assumption needed, it works for multi solution puzzles as well.

[creint]

And you can eliminate 1r1c1 with a kite (no 1 in the * cells)

Code: Select all

 12 . . | 12  . . | * * 1
 12 . . | 123 . . | * * *
 .  . . | .   . . | 1 * *
 --------------------------
 1  * * | *   * * | 1 * *

Is invalid because no valid sudoku exists with this property. (Start or end of chain must be true, and both ALS 12 12 must also be true which is impossible.)

Code: Select all

 12 . . | 12  . . | * * 1
 12 . . | 123 . . | * * *
 .  . . | .   . . | 1 * *
 --------------------------
 *  * * | 1   * * | 1 * *

Is possible.

[denis_berthier]

As soon as you introduce some condition as "if none was a given", you are making some extra-logical assumption.

I am not "making some extra-logical assumption", i am just using the fact, that there is no given in the UR1.1 cells. If you are ignoring or rejecting this fact, this is not my problem.
There is no uniqueness assumption needed, it works for multi solution puzzles as well.

Let me start with a counter-example to your last claim:

Code: Select all

..3456789
.......3.
9.....2..
...5678..
3.4.1....
.7.23....
74.1..3..
8......2.
........1

The UR1.1 cells are r1r4 x c1c2
Input this into Sudoku Explainer and you'll get the message "This Sudoku has multiple solutions." The views 1 and 2 give digits 1 and 9 has being in possible solutions, but if you add manually 2 in r4c2, the 2 is also in possible solutions. So, this is a puzzle with multiple solutions, with no given value in the UR1.1 cells, and each of the three values in the 3-value cell appears in different solutions.

After this, you may be open to the following more abstract considerations that explain the origin of the flaw in your reasoning.
There is no "fact" in logic". There are only axioms and theorems. And once a theorem has been proven, it is as good as an axiom.
The concrete problem is, there is no way to express that "rc=n is not a given".
You may want to state that rc=n is not an axiom, but one can only state axioms, not non-axioms.
You may also want to state rc≠n as an axiom, but stating that rc≠n is much stronger than stating that rc=n is not an axiom and it is not a way of defining a Sudoku puzzle. [You may try this in Sukaku.]

[Leren]

Just before I go to bed I'll just detail the 33 solutions found by my solver. This may aid further discussion by others.

Hidden Text: Show

123456789487921635956378214219567843364819572578234196745192368831645927692783451
123456789487921635956378214291567843364819572578234196742195368815643927639782451
123456789487921635956783214219567843364819572578234196745192368831645927692378451
123456789485729136967381254219567843354918672678234915742195368891643527536872491
123456789485729136967381254291567843354918672678234915742195368819643527536872491
123456789485792136967381254219567843354819672678234915746125398891643527532978461
123456789485792136967381254219567843354918672678234915746125398831649527592873461
123456789485792136967381254219567843354918672678234915746125398891643527532879461
123456789485792136967381254291567843354819672678234915746125398819643527532978461
123456789485792136967381254291567843354819672678234915749125368816943527532678491
123456789485792136967381254291567843354918672678234915746125398819643527532879461
123456789485972136967381254219567843354819672678234915742195368891643527536728491
123456789485972136967381254219567843354819672678234915746125398891643527532798461
123456789485972136967381254291567843354819672678234915742195368819643527536728491
123456789485972136967381254291567843354819672678234915746125398819643527532798461
123456789487921536965378214219567843354819672678234195746192358831645927592783461
123456789487921536965783214219567843354819672678234195746192358831645927592378461
213456789487921635956378214129567843364819572578234196745192368831645927692783451
213456789487921635956378214192567843364819572578234196745192368831645927629783451
213456789487921635956783214129567843364819572578234196745192368831645927692378451
213456789487921635956783214192567843364819572578234196745192368831645927629378451
213456789485729136967381254129567843354918672678234915742195368891643527536872491
213456789485792136967381254129567843354819672678234915746125398891643527532978461
213456789485792136967381254129567843354918672678234915746125398831649527592873461
213456789485792136967381254129567843354918672678234915746125398891643527532879461
213456789485792136967381254192567843354918672678234915746125398831649527529873461
213456789485792136967381254192567843354918672678234915749125368831649527526873491
213456789485972136967381254129567843354819672678234915742195368891643527536728491
213456789485972136967381254129567843354819672678234915746125398891643527532798461
213456789487921536965378214129567843354819672678234195746192358831645927592783461
213456789487921536965378214192567843354819672678234195746192358831645927529783461
213456789487921536965783214129567843354819672678234195746192358831645927592378461
213456789487921536965783214192567843354819672678234195746192358831645927529378461

Leren

[denis_berthier]

Just before I go to bed I'll just detail the 33 solutions found by my solver. This may aid further discussion by others.

Yes, thanks; we can clearly see that each of the 3 values in r4c2 does appear in several solutions.

[m_b_metcalf]

Just before I go to bed I'll just detail the 33 solutions found by my solver. This may aid further discussion by others.

FWIW:

Code: Select all

Common clues (40):
 . . 3 4 5 6 7 8 9
 4 8 . . . . . 3 .
 9 . . . . . 2 . 4
 . . . 5 6 7 8 4 3
 3 . 4 . 1 . . 7 2
 . 7 8 2 3 4 . . .
 7 4 . 1 . . 3 . 8
 8 . . . 4 . . 2 7
 . . . . . . 4 . 1

[eleven]

And you can eliminate 1r1c1 with a kite (no 1 in the * cells)

Code: Select all

 12 . . | 12  . . | * * 1
 12 . . | 123 . . | * * *
 .  . . | .   . . | 1 * *
 --------------------------
 1  * * | *   * * | 1 * *

Is invalid because no valid sudoku exists with this property. (Start or end of chain must be true, and both ALS 12 12 must also be true which is impossible.)

Code: Select all

 12 . . | 12  . . | * * 1
 12 . . | 123 . . | * * *
 .  . . | .   . . | 1 * *
 --------------------------
 *  * * | 1   * * | 1 * *

Is possible.

You are very right, thanks.

[eleven]

Let me start with a counter-example to your last claim:

Code: Select all

..3456789
.......3.no
9.....2..
...5678..
3.4.1....
.7.23....
74.1..3..
8......2.
........1

The UR1.1 cells are r1r4 x c1c2

There is no UR1.1 in these cells, but a UR type 1. Sorry, but if you still don't know the difference, you should not make any claims about the one or other.

After this, you may be open to the following more abstract considerations that explain the origin of the flaw in your reasoning.
There is no "fact" in logic". There are only axioms and theorems. And once a theorem has been proven, it is as good as an axiom.
The concrete problem is, there is no way to express that "rc=n is not a given".
You may want to state that rc=n is not an axiom, but one can only state axioms, not non-axioms.
You may also want to state rc≠n as an axiom, but stating that rc≠n is much stronger than stating that rc=n is not an axiom and it is not a way of defining a Sudoku puzzle. [You may try this in Sukaku.]

This may sound very clever, but is of no worth in this context. YOU DO NOT MAKE THE RULES, how a puzzle is allowed to be solved. You cannot forbid me to look, where givens are or not. And if there are none, it is a fact, that there are none, and logically i am allowed to use this fact with a logically proved rule. If you want or not.

[denis_berthier]

After this, you may be open to the following more abstract considerations that explain the origin of the flaw in your reasoning.
There is no "fact" in logic". There are only axioms and theorems. And once a theorem has been proven, it is as good as an axiom.
The concrete problem is, there is no way to express that "rc=n is not a given".
You may want to state that rc=n is not an axiom, but one can only state axioms, not non-axioms.
You may also want to state rc≠n as an axiom, but stating that rc≠n is much stronger than stating that rc=n is not an axiom and it is not a way of defining a Sudoku puzzle. [You may try this in Sukaku.]

This may sound very clever,

Thanks. Do you have any rational counter-argument?
Once more, what I've proven is, UR1.1 requires some extra-logical argument - and the assumption of uniqueness seems to be the required one. Do you have any counter-example to this, i.e. a multi-solution puzzle with the UR1.1, with no given in the 4 UR cells, in which the only value for the 3-value cell is as the rule says in all the solutions?

but is of no worth in this context. YOU DO NOT MAKE THE RULES, how a puzzle is allowed to be solved.

I meant rational.
I don't make the rules. The rules of FOL have been in existence for more than a century. And they make no difference between an axiom and a theorem for further poofs.

You cannot forbid me to look, where givens are or not. And if there are none, it is a fact, that there are none, and i am allowed to use this fact. If you want or not.

Once more, you're trying to turn it into something personal. You may do whatever you want. You may claim that the Earth is flat and refuse to consider any rational argument to the contrary. You may refuse to solve according to the rules of Logic. You may prove that some cell has value 1 and then pretend that it has some other value in order to conclude something else: "the Earth is round, but if it was flat it would have two shapes. Therefore, it must be square."

[eleven]

Thanks. Do you have any rational counter-argument?
Once more, what I've proven is, UR1.1 requires some extra-logical argument - and the assumption of uniqueness seems to be the required one.

I am sorry for you, that your sudoku model (FOL or not) cannot distinguish between givens and derived numbers. So it neither can handle some uniqueness methods nor UR1.1. Your mistake is, that therefore you think, that UR1.1 somehow needs uniqueness, which is definitely not the case. It just needs this differentiation, you are not able to make.

Do you have any counter-example to this, i.e. a multi-solution puzzle with the UR1.1, with no given in the 4 UR cells, in which the only value for the 3-value cell is as the rule says in all the solutions?

Do you mean that honestly ? The proof (maybe you should read it again, Red Ed's or mine or the other 5 in the old threads) definitely says, that the rule is valid for any puzzles. If someone has to come up with a counter-example, it is you.

[denis_berthier]

Thanks. Do you have any rational counter-argument?
Once more, what I've proven is, UR1.1 requires some extra-logical argument - and the assumption of uniqueness seems to be the required one.

I am sorry for you, that your sudoku model (FOL or not) cannot distinguish between givens and derived numbers. So it neither can handle some uniqueness methods nor UR1.1. Your mistake is, that therefore you think, that UR1.1 somehow needs uniqueness, which is definitely not the case. It just needs this differentiation, you are not able to make.

Not I. No consistent logic can make any difference between an axiom and a theorem (a given and a derived number), as far as further proofs are concerned. If I do remember well, you have some mathematical background. Have you ever seen some new theorem proven on the condition that a theorem was not an axiom?

Do you have any counter-example to this, i.e. a multi-solution puzzle with the UR1.1, with no given in the 4 UR cells, in which the only value for the 3-value cell is as the rule says in all the solutions?

Do you mean that honestly ? The proof (maybe you should read it again, Red Ed's or mine or the other 5 in the old threads) definitely says, that the rule is valid for any puzzles. If someone has to come up with a counter-example, it is you.

Your old "proof" is precisely what my new argument proves to be flawed. You haven't yet understood this.
There is no "proof" if it can't be expressed in any logic way. Maybe you think repeating the same false arguments and refusing to discuss new ones will end in making them true, but that's not how it works.
Your answer makes it clear: you have neither any rational argument nor any counter-example.

[StrmCkr]

Eleven was pointing out the clear distinctions between the 81 cell grid string and the populated candidate string of 729 characters

The 81 grid string is used for the ur1.1 (and higher order avoidable rectangles) formation of cells that potentially house it then
Then one of the 4 cells cannot contain the 2 digits use the 729 pm string to identify which cell specifically

UR 2, and beyond focus on the 729 grid string only.

[eleven]

Denis,

words like FOL, consistent logic, axioms, and further proof don't make your incomplete sudoku model complete.
If you are able to understand the UR1.1 rule, you will realize, that in some situations it turns out, that givens and derived numbers have different logical impacts.
Your only argument against the proof is, that it does not fit into your model.

Again, if you still think, that the rule is not valid (which you always state), YOU have to give a counter-example.

[denis_berthier]

words like FOL, consistent logic, axioms, and further proof don't make your incomplete sudoku model complete.

Sudoku IS a logic puzzle. It has a straightforward logic formulation and it obeys the laws of logic, like it or not. This is a fact that doesn't depend on any sudoku model. You are free to prefer using crystal balls or prophecies or whatever you want; but I'm not very interested in discussing such methods.

If you are able to understand the UR1.1 rule, you will realize, that in some situations it turns out, that givens and derived numbers have different logical impacts.

There can be no logical difference between an axiom and a theorem, no different impact. Doesn't it print? You can try to replace the word consequence by impact in order to purposely blur the problem, but it doesn't change anything.

Your only argument against the proof is, that it does not fit into your model.

No. This is what you are trying to make believe. My argument is, your "proof" is not a proof at all because it contradicts the most basic laws of logic. But conversely, you have NO argument against my proof - except blathering against logic in general.

Again, if you still think, that the rule is not valid (which you always state),

No; you're distorting what I say. What I say is, the rule is not valid without any extra-logic assumption.

YOU have to give a counter-example.

That's really funny. Consider UR1.1 in case of multi-solution puzzles. You are claiming that it is valid. I show the logical flaws in what you call a "proof"; you're unable to find any flaw in my proof (except globally rejecting logic); moreover, you're unable to provide any example. And you ask me to provide a counter-example!

[eleven]

So what ? When i look, if there are givens in 4 cells, that is an extra-logic assumption ? Because your poor model does not allow a distinction between given and derived numbers ?
You determine, what logic is in sudoku ?
Please go away.