The New Sudoku Players' Forum

[eleven]

My 2 cents

Suppose there is a type 1 UR with one extra candidate and you can eliminate 3 candidates without using any uniqueness method to arrive at a UR1.1 (what of course is possible, though very rare).

I think that is impossible.

Why not ? I could with the first puzzle i tried, its by gsf from Patterns game.

Code: Select all

*-----------------------*
 | . . 5 | 7 . . | . . . |
 | . 8 . | . . 9 | 2 . . |
 | 9 . . | . . . | . 1 . |
 |-------+-------+-------|
 | 8 . . | 6 . . | . 7 . |
 | . . . | . . . | . . . |
 | . 9 . | . . 2 | . . 5 |
 |-------+-------+-------|
 | . 2 . | . . . | . . 6 |
 | . . 7 | 1 . . | . 3 . |
 | . . . | . . 3 | 4 . . |
 *-----------------------*
+----------------+----------------+----------------+
|*12  *13   5    | 7    28   4    | 68   69  *389  |
| 7    8    6    | 3    1    9    | 2    5    4    |
| 9    34   234  | 25   258  6    | 7    1    38   |
+----------------+----------------+----------------+
| 8    345  234  | 6    34   15   | 9    7    12   |
|*25   7    34   | 49   349  15   | 68  *26   128  |
| 6    9    1    | 8    7    2    | 3    4    5    |
+----------------+----------------+----------------+
| 3    2   #89   | 45   45   7    | 1   #89   6    |
| 4    6    7    | 1    29   8    | 5    3   *29   |
| 15   15  #89   | 29   6    3    | 4   #289  7    |
+----------------+----------------+----------------+

r7c3=8 -> r9c8=8 -> r5c8=2 -> r1c1=2 -> r1c2=1 -> r1c9=3 -> r8c9=9 -> r7c8=8
I.e. r7c3<>8, r7c3=9 -> r7c8=r9c3=8

[Edited typo and added *'s in the path]
[Edit 3: Added not to me misunderstood:]

What the proof says is:
If you arrive at a solution with the deadly pattern, you could exchange them to get another solution.

Which then proves that your solution path is flawed.

Thats what i meant. In this case the flaw would be to set r9c8 to 9.

[David P Bird]

We check if a particular two digit solution for a rectangle is only achievable in one way, we don’t assume it.

Well, so how do you check this?

Take my previous example:
abc ac
abc bc

We check to see how many ways we can fill these cells using digits (a) & (b) and find there is only one:
b a
a b

Now as we've been able to exclude (b) from the top-right cell, the rectangle isn't an isolated sub-puzzle. As it isn't an isolated sub-puzzle, it can't consist of only the digits (a) and (b), and one of the cells must contain (c).

In this example we can repeat this mental process for the other possible digit pairs too, but that won't always be the case.

The whole problem is: how could any logical or mathematical proof (of the solution) make a difference between an axiom (a given) and a consequence of an axiom?

I can't make sense out of this statement: Taking the words literally what it seems to say is that if we are provided with r1c1 = 9 as a given, we can't make use of that to say that r1c2 <> 9 because any given is only an axiom because it hasn't been proved beyond doubt by the composer. Perhaps that's so, but where does it lead us?

My stance is that if we can demonstrate a qualifying rectangle only has one way it can be resolved using two digits, we know just as certainly as we know r1c2<>9 above, that it isn't an isolated sub-puzzle - from which the other consequences follow.

[denis_berthier]

The whole problem is: how could any logical or mathematical proof (of the solution) make a difference between an axiom (a given) and a consequence of an axiom?

I can't make sense out of this statement: Taking the words literally what it seems to say is that if we are provided with r1c1 = 9 as a given, we can't make use of that to say that r1c2 <> 9 because any given is only an axiom because it hasn't been proved beyond doubt by the composer.

It means exactly the contrary: r1c1 = 9 has the same consequences, be it a given or a deducted value.
As a result, starting from the situation in Myth UR1.1
1 2
2 13
it can't change anything to add the condition that the first 3 values were not givens.

My stance is that if we can demonstrate a qualifying rectangle only has one way it can be resolved using two digits, we know just as certainly as we know r1c2<>9 above, that it isn't an isolated sub-puzzle - from which the other consequences follow.

"isolated sub-puzzle" is meaningless.


As what you want to prove is becoming more and more opaque, I can't see any point in discussing a 3rd pattern if you don't first answer by a mere yes or no each of the following questions:

1) are you claiming that standard UR1 doesn't depend on the axiom of uniqueness?
12 12
12 123

2) are you claiming that anything can be deduced from Myth UR1.1?
1 2
2 13

[David P Bird]

1) are you claiming that standard UR1 doesn't depend on the axiom of uniqueness?
12 12
12 123

No this does depend on the axiom of uniqueness as at this point it is un-resolvable with respect to (1) & (2) if (3) is found to be false. If this actually turns out to be the case, we have that isolated sub-puzzle for these four cells because, regardless of how the balance of the puzzle is completed, NOTHING is will be deducible about the arrangement of these digits.

2) are you claiming that anything can be deduced from Myth UR1.1?
1 2
2 13

Yes indeed I am!. Because we have been able to eliminate two instances of digits (1) & (2) we have proved that this rectangle isn't isolated from the rest of the puzzle. There is only one way to fill the four cells with (1) & (2). For that to be possible without a given being involved there must be at least 3 different digits in these four cells. (For a rectangle to be resolvable with respect to a pair of digits all that's needed is one of them to be eliminated from one of the cells.)

r1c1 = 9 has the same consequences, be it a given or a deducted value.
As a result, starting from the situation in Myth UR1.1
1 2
2 13
it can't change anything to add the condition that the first 3 values were not givens.

This assertion is merely your own hypothesis which needs to be proved in all cases. However, to refute the contrary hypothesis that I've stated, all you need is one counter-example.

If you tried to compose a set of givens that would be sufficient to solve a particular puzzle you would find there are rules that must be followed - eg a minimum of eight digits must be in the set of givens. You seem oblivious to these requirements, but they're the basis of the logic here.

BTW I wouldn't label that situation as a UR as its resolution doesn't depend on uniqueness being assumed, just that there are one or more valid solutions.

[StrmCkr]

removed

[denis_berthier]

i did not sway your pov on it overly either.
you dont use them.
you feel its contrited information or lacking some logical value,
there is mathmatical reasons to use them as well.

I can't accept your systematic distortion of my POV.
I always said that standard UR is logically valid as a rule depending on the axiom of uniqueness. And I already said I have voded it in SudoRules.
I don't use it personally because I'm studying systematically homogeneous sets of rules and UR is not homogeneous with them.

[denis_berthier]

1) are you claiming that standard UR1 doesn't depend on the axiom of uniqueness?
12 12
12 123

No

Good. At least one thing we agree upon.

2) are you claiming that anything can be deduced from Myth UR1.1?
1 2
2 13

Yes indeed I am!. Because we have been able to eliminate two instances of digits (1) & (2) we have proved that this rectangle isn't isolated from the rest of the puzzle. For that to be possible without a given being involved...

Too bad. This shows how risky it is to use undefined notions ("isolated") as an argument of proof. You can repeat indefinitely the same flawed reasoning based on it, that won't make it true.
Allan has introduced this word without back thoughts. And now you are sticking to it as to a life belt. But read his post: he doesn't claim having proven anything.
Sudoku is a logic game. I'm dismayed to see how easily some players with no logical or mathematical background get stucked in incantations based on undefined notions.
If anything can be deduced from this pattern, it has to be deduced from the pattern itself, not from a reference to past states or to givens.

r1c1 = 9 has the same consequences, be it a given or a deducted value.
As a result, starting from the situation in Myth UR1.1
1 2
2 13
it can't change anything to add the condition that the first 3 values were not givens.

This assertion is merely your own hypothesis which needs to be proved in all cases.

Not my hypothesis but a general logical and mathematical fact: a logical formula (e.g. r1c1 = 9) has the same consequences independent of whether it is an axiom (e.g. a given) or it is a consequence of the axioms (e.g. a decided value).

[StrmCkr]

removed

[denis_berthier]

I don't use it personally because I'm studying systematically homogeneous sets of rules and UR is not homogeneous with them.

fair enough.
why are people still debatign it on here considering the above?

Because, I think, this thread is about resolution rules in general and what is discussed with this particular case of UR is not really uniquenees. It serves as an example of what shouldn't be done with resolution rules:
- putting conditions on patterns based on past values,
- making a difference between a given and a decided value,
- using undefined notions ("isolated sub-puzzle"), especially when their name ("isolated") has some intrinsic incantatory power,
- thinking that such notions make a proof in themselves,
- thinking that mathematical truth is a matter of belief,
- thinking that not having a counter example is a proof of truth,
- thinking that incantation is proof.


Let it be clear that:
- I'm not claiming that no conclusion can be drawn from the UR1.1 pattern; I don't know and nobody knows; some believe it, but they can't say they know it.
What I'm claiming is two things:
1) no conclusion has been proven;
2) a conclusion cannot depend on a reference to anything being given or deduced.
"Believers", to take David's very significant expression, should try to prove it.

- I'm not claiming that some notion of an "isolated sub-puzzle" couldn't be defined. I'm claiming that it has not been.



But you're right. If anyone "believes" that any conclusion can be drawn from UR1.1 or that UR1 doesn't depend on the assumption of uniqueness, that's worth a thread of its own.

[Red Ed]

I'm coming late to this, so my apologies to anyone that's already reeled out the following argument.

Fact: if a puzzle has a unique solution grid then it cannot contain the following pattern on four unclued cells:

Code: Select all

 .  .  . | .
 1  .  . | 2
 2  .  . | 1
---------+---
 .  .  . | .

Consequence: if, part way through solving a puzzle that has a unique solution, you find the following pattern on four unclued cells ...

Code: Select all

 .  .  . | .
 1  .  . | 2
 2  .  . | 13
---------+---
 .  .  . | .

... then that bivalue cell must resolve to "3".

Is that what the UR1.1 argument was all about? Not worth a thread of its own, if so.

[denis_berthier]

I'm coming late to this, so my apologies to anyone that's already reeled out the following argument.

Fact: if a puzzle has a unique solution grid then it cannot contain the following pattern on four unclued cells:

Code: Select all

 .  .  . | .
 1  .  . | 2
 2  .  . | 1
---------+---
 .  .  . | .

That's the whole point: you're presenting this as an obvious fact, which it is certainly not. How do you prove this?

[StrmCkr]

removed

[denis_berthier]

No, StrmCkr,
what I want is a proof of exactly what RedEd stated as an obvious fact, not of your re-statement.

[StrmCkr]

removed

[denis_berthier]

StrmCkr,

This is still claiming, but not proving.

Notice that, with the same hypothesis, RedEd invokes uniqueness of the solution and you invoke existence of a solution.
Very strange.

Billions of billions of examples can't prove a general rule.