## [SOLVED] (not) About 'Almost locked sets'.

Everything about Sudoku that doesn't fit in one of the other sections

### Re: About 'Almost locked sets'.

bat999 wrote:But I haven't thought this one through yet.
3) neither option forces a value into X, but each candidate in X is eliminated for at least one of the options.

I did not see that before, because we cross posted, when you added it.
This just means, that one candidate leaves X empty - a contradiction, so it must be false, and the 4 must be true there.
(It can't be possible in a valid sudoku, that both candidates lead to such a contradiction.)
In all 3 cases no common candidate is left in X.
eleven

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### Re: About 'Almost locked sets'.

eleven wrote:
bat999 wrote:But I haven't thought this one through yet.
3) neither option forces a value into X, but each candidate in X is eliminated for at least one of the options.

I did not see that before, because we cross posted, when you added it.
This just means, that one candidate leaves X empty - a contradiction, so it must be false, and the 4 must be true there.
(It can't be possible in a valid sudoku, that both candidates lead to such a contradiction.)

I wasn't trying to say that "for at least one of the options, each candidate in X is eliminated", which seems to be what you've inferred.

An example would be: if A is not a 4, then X must be a 1 or a 2 (although you can't say which), and if B is not a 4, then X must be a 3 or a 4 (and you can't say which). With no overlap in the possibilities, one of {A,B} must be a 4.

eleven wrote:In all 3 cases no common candidate is left in X.

Yes
blue

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### Re: About 'Almost locked sets'.

blue wrote:I wasn't trying to say that "for at least one of the options, each candidate in X is eliminated", which seems to be what you've inferred.

An example would be: if A is not a 4, then X must be a 1 or a 2 (although you can't say which), and if B is not a 4, then X must be a 3 or a 4 (and you can't say which). With no overlap in the possibilities, one of {A,B} must be a 4.

Ah, this is what you meant. I also had that in mind yesterday, it is not needed, that a candidate is forced.
To show a possible scenario.

Code: Select all
` 14  47 .  | 7+  .   .  .   .  .  |  .  .   .  .   .  .  |  . 123  . .   .  .  |  .  .   . ---------------------- .  457 .  | 7+ 2345 457  .   .  .  |  .  .   . .   .  .  |  . 123  .---------------------- 1+  .  .  |  .  1+  . `

If there are strong links for 1 in r7 and 7 in c4, the 1 in r1c1 eliminates 23, and 7r1c2 eliminates 45 in X=r4c5.
eleven

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### Re: About 'Almost locked sets'.

@ blue
"3) neither option forces a value into X, but each candidate in X is eliminated for at least one of the options."
"I wasn't trying to say that "for at least one of the options, each candidate in X is eliminated", which seems to be what you've inferred."

Yes, I understand now.
If one result says X must be 1 or 2 and the other result says X must be 3 or 4.
No overlap.
It's impossible.
That's the contradiction.

Also from your previous post, I realize now that A and/or B don't have to be bi-value squares.
They can have any number of candidates.
We're not interested what the candidates are, so long as A and B have one or more common candidates to work with.
After all, for this test we're not interested in setting any values for A and B. We only want to set them to 'not' 4 or whatever common candidate we fancy.

@ eleven
I can see what you've done in your example post.
"the 1 in r1c1 eliminates 23, and 7r1c2 eliminates 45 in X=r4c5"
Nothing left.

By the way
I'm not the sharpest knife in the drawer, so I don't think I have 'discovered' this test.
Is it the best-kept secret in the world of Sudoku, or is this test documented somewhere?
bat999
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### Re: About 'Almost locked sets'.

bat999 wrote:Is it the best-kept secret in the world of Sudoku, or is this test documented somewhere?

As far as i now, Instinct was the first one to formulate this kind of elimination this way. Here i guess it is seeen as an unnnamed AIC (chain elimination). The AIC for my scenario would be
(4=1)r1c1 - r7c1 = r7c5 -(1=23)r36c5-(23=45)r4c5-(45=7)r4c26-(7=4)r1c2 => r12c1=4
eleven

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### Re: About 'Almost locked sets'.

gurth wrote:
Leren wrote:This is logically correct, but most experienced solvers regard proof by contradiction as inelegant, if a constructive proof of similar complexity is available.

Leren

This statement is correct logically and any otherwise you can think of, only it is an understatement, most experienced solvers seem to have a phobia about contradictions. Which is absolutely absurd. There is nothing inelegant about contradictions. "Inelegant" is a purely subjective judgement, containing no logic whatsoever. Perhaps these people who hate contradictions so much, have a low tolerance for confrontations and disharmony of any sort.

I will be the first to concede that a proof, for any given case, that is simpler and/or shorter than a contradiction, is more elegant. But not if it is the same length and equally complex.

.

What is meant by the term 'contradication' here? The most common construct used by experienced solvers, the discontinuous AIC, gives its result(s) at the point of a contradiction so to say that 'most experienced solvers seem to have a phobia about contradictions would seem to be an oversimplification. I do know that many experienced solvers don't like bifurcation/forking or similar highly assumptive guessing methods.
DonM
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### Re: About 'Almost locked sets'.

eleven wrote:As far as i now, Instinct was the first one to formulate this kind of elimination this way....

Yes.
He has chosen two squares in the same row that have the common candidate 4.
Then he's worked out what happens when both squares are set to not 4.

In his example, the 'contradiction' was "resulting in two 1s in box 2".

In this thread here, I've been considering incompatible results in an object square as the 'contradiction'.
bat999
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### Re: About 'Almost locked sets'.

bat999 wrote:
eleven wrote:... the "result" for the candidates in X is different, but still both are possible, because X could resolve...

I've tightened up the question.
"Different" isn't strong enough.
"Contradictory" nails it.

Suppose that two bi-value squares A(47) and B(14) are in the same row or column or box.
Work out what happens to an object square X when...
A is not 4
and
B is not 4

If the result in X is contradictory, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect.
So one of the squares A or B must be 4.
And the 4's can be eliminated from squares that can see both A and B.

Is this true?

PS
I've just read blue's post again.
"It might be better to think of it as that "the results in X are incompatible".
I think me and blue are on the same wavelength.
--------------------------------------------------
These are my two examples of 'contradictory' results.
"The value of square X is 7 when A is not 4. The value of square X is 8 when B is not 4"
and
"The value of square X is 7 when A is not 4. The value of square X is not 7 when B is not 4"
--------------------------------------------------
And this was how blue expressed it...
1) one option forces some value into X, and the other option forces a different value into X.
2) one option forces X to be 'd', and the other eliminates 'd' from X.
But I haven't thought this one through yet.
3) neither option forces a value into X, but each candidate in X is eliminated for at least one of the options.

Yes! The way you have expressed it in your examples is now convincing.
Clearly, I am recanting what I said in my first reply!
gurth

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### Re: (not) About 'Almost locked sets'.

Leren wrote: This is logically correct, but most experienced solvers regard proof by contradiction as inelegant, if a constructive proof of similar complexity is available.

I knew I wasn't just imagining this. Read daj95376's comments on my Forcing Chain Contradiction move for the Network puzzle of 6th June 2014 and pjb's reference to "howls of indignation" to his post on the 10 June 2014 puzzle.

http://forum.enjoysudoku.com/network-6-15-2014-t31888.html?hilit=forcing%20contradiction

Since then I have been reversing Forcing Chain Contradiction moves so that they appear as Kraken moves, to avoid any controversy.

Personally I have no opinions on this matter but others apparently do.

Leren
Leren

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### Re: (not) About 'Almost locked sets'.

bat999 wrote:Suppose that two bi-value squares A(47) and B(14) are in the same row or column or box.
Work out what happens to an object square X when...
A is not 4
and
B is not 4

If the result in X is the not the same, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect.
So one of the squares A or B must be 4.
And the 4's can be eliminated from squares that can see both A and B.

Is this true?

I'm not going to plow through all of the replies to see if this topic has been resolved to your satisfaction. Instead, I'll just add another reply.

blue's first reply lists the conditions under which your logic works. This scenario gives a simple example where it doesn't work.

Code: Select all
` *--------------------------------------------------------------------* | 18     7      6      | 2      13     9      | 4      5      38     | | 12458  3      14589  | 47     1457   14     | 78     2789   6      | | 245    259    459    | 467    34567  8      | 37     279    1      | |----------------------+----------------------+----------------------| | 1458   1568   1458   | 3      9      46     | 1578   4678   2      | | 3      568    2      | 1      468    7      | 9      468    458    | | 7      1689   1489   | 468    2      5      | 138    468    348    | |----------------------+----------------------+----------------------| | 6      1258   1358   | 9      13478  1234   | 578    478    4578   | | 9      258    358    | 478    3478   234    | 6      1      4578   | | 18     4      7      | 5      168    16     | 2      3      9      | *--------------------------------------------------------------------*`

If we presume 7r2c4 and 1r2c6, then we have:

Code: Select all
`* 7r2c4                   -> -7r2c8* 1r2c6 -> 3r1c5 -> 8r1c9 -> -8r2c8`

We now have both initial values leading to different results in r2c8. However, the solution has r2c8=9. So, both -7r2c8 and -8r2c8 are true, and we can't deduce that one of r2c46 is 4.

_
daj95376
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### Re: (not) About 'Almost locked sets'.

bat999 wrote: If the result in X is the not the same, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect.

daj95376 wrote; We now have both initial values leading to different results in r2c8. However, the solution has r2c8=9. So, both -7r2c8 and -8r2c8 are true, and we can't deduce that one of r2c46 is 4.

Hi Danny, this doesn't look like a counterexample to me.

Firstly, when bat wrote "If the result in X is the not the same" you claim to have complied by showing that the results for r2c8 (X) are not 7 and not 8. This might seem to be linguistically in accordance with bat's assertion, but when I read it I assumed that the result in X would have to be a positive result ie something like r2c8 = 2 and r2c8 = 9. So it doesn't seem (to me) to address bat's assertion ie if X <> 7 and X <> 8, then X = 9 would satisfy both of your results but the positive result in X would be the same. But suppose I'm wrong and bat would be happy with -7 r2c8 and -8 r2c8 as satisfying his assertion. You then say that we can't deduce that one of r2c46 is 4. Well, in the solution to the puzzle, r2c4 = 4 !

So I'm forced to conclude that this example either doesn't address bat's assertion or if it does it is not a counterexample.

Leren
Leren

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### Re: (not) About 'Almost locked sets'.

I split your reply into two parts.

Leren wrote:
bat999 wrote: If the result in X is the not the same, then nothing is known about X, but it indicates that one of the values chosen for A or B were incorrect.

daj95376 wrote; We now have both initial values leading to different results in r2c8. However, the solution has r2c8=9. So, both -7r2c8 and -8r2c8 are true, and we can't deduce that one of r2c46 is 4.

Hi Danny, this doesn't look like a counterexample to me.

Firstly, when bat wrote "If the result in X is the not the same" you claim to have complied by showing that the results for r2c8 (X) are not 7 and not 8. This might seem to be linguistically in accordance with bat's assertion, but when I read it I assumed that the result in X would have to be a positive result ie something like r2c8 = 2 and r2c8 = 9. So it doesn't seem (to me) to address bat's assertion ie if X <> 7 and X <> 8, then X = 9 would satisfy both of your results but the positive result in X would be the same.

Never assume.

Leren wrote:But suppose I'm wrong and bat would be happy with -7 r2c8 and -8 r2c8 as satisfying his assertion. You then say that we can't deduce that one of r2c46 is 4. Well, in the solution to the puzzle, r2c4 = 4 !

So I'm forced to conclude that this example either doesn't address bat's assertion or if it does it is not a counterexample.

Just because you can't deduce something as true doesn't mean that it isn't true. It only means that you may have insufficient evidence (so far). Whew. How's that for a string of "...n't"s in a sentence. _ _

In my example, the evidence that's missing is r2c4=7 and ( r2c6=1 -> r1c9=8 ) would force r2c7 empty. However, the example still matches bat999's assertion.

_

[Edit: replaced first comment.]
Last edited by daj95376 on Mon Jan 12, 2015 11:57 pm, edited 1 time in total.
daj95376
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### Re: (not) About 'Almost locked sets'.

blue wrote :

Yes it's true, as long as you're careful about (sic) what is meant by "the result in X is the not the same".
It might be better to think of it as that "the results in X are incompatible".

That would cover cases where:
1) one option forces some value into X, and the other option forces a different value into X.
2) one option forces X to be 'd', and the other eliminates 'd' from X.
3) neither option forces a value into X, but each candidate in X is eliminated for at least one of the options

I think bat is right after all.

Paraphrasing 2) : A <> 4 => X = d; B <> 4 => X <> d. So A <> 4 and B <> 4 => X = d and not d. Contradiction => one of A and B must be 4.

Paraphrasing 3) : A <> 4 => via, say, 2 AIC's => X <> a,b; B <> 4 => via, say, 3 AIC's => X <> c,d,e; If X = abcde then A <> 4 and B <> 4 => X is empty. Contradiction => one of A and B must be 4.

So 2) and 3) are proofs by contradiction. Shock, horror, howl of indignation .

Paraphrasing 1) : A <> 4 ... some AIC 1 ... X = a; B <> 4 ... some other AIC 2 ... X = b. Rearranging we can say:

A <> 4 ... AIC 1 ... X = (a - b) ... reverse AIC 2 ... B = 4; and
B <> 4 ... AIC 2 ... X = (b - a) ... reverse AIC 1 ... A = 4. So shortening the notation:

A <> 4 ... AIC 1/2 ... B = 4; and
B <> 4 ... AIC 2/1 ... A = 4.

So 1) => A = 4 or B = 4, although I wouldn't say this means that the values that are forced into X (a and b) are incompatible. I think 1) was basically what bat was describing.

So what bat was describing, in a somewhat convoluted way, was simply a Type 1 AIC (ie AIC 1/2).

Of course this has nothing to do with Almost Locked Sets, and A and B don't have to share a house.

Leren
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### Re: About 'Almost locked sets'.

blue wrote:... or that A and B are bi-value cells, and it doesn't require that cells A and B can see each other.

Got it blue.

As you say, for this test, to find information about A and B, they don't need to be in the same row or column or box.
(Two 'not' tests give incompatible results in X)

And about the bi-value thing.
For this test we don't care about the value of the two squares A and B, just the 'state' of the squares.
With reference to the common candidate 4, the two squares each only have two states, no matter how many candidates they hold.
The two states are...
4 is off
and
4 is on
We use '4 is off' state for 'not'.

Also I think...
For the other type of test, to find information about X, the two squares A and B do have to be in the same row or column or box.
(Two 'not' tests give the same result (ie two compatible results) in X)

bat999
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### Re: About 'Almost locked sets'.

bat999 wrote:
blue wrote:... or that A and B are bi-value cells, and it doesn't require that cells A and B can see each other.

Got it blue.

As you say, for this test, to find information about A and B, they don't need to be in the same row or column or box.

In this sense e.g. w-wings are a special case of this method.

To give an example in Instincts grid, where A and B don't see each other:
Code: Select all
` *--------------------------------------------------------------------* |*18     7      6      | 2      13     9      | 4      5     *38     | |-12458  3      14589  |X47     1457  A14     |*78     2789   6      | | 245    259    459    | 467    34567  8      | 37     279    1      | |----------------------+----------------------+----------------------| | 1458   1568   1458   | 3      9      46     | 1578   4678   2      | | 3      568    2      | 1      468    7      | 9      468    458    | | 7      1689   1489   | 468    2      5      | 138    468    348    | |----------------------+----------------------+----------------------| | 6      1258   1358   | 9      13478  1234   | 578    478    4578   | | 9      258    358    | 478    3478   234    | 6      1      4578   | |B18     4      7      | 5      168   -16     | 2      3      9      | *--------------------------------------------------------------------*`

A=r2c6, B=r9c1, X=r2c4
A=4 => X=r2c4 = 7
B=8 => r1c9=8 -> r2c7=7 -> X=r2c4 = 4
=> A or B must be 1 => r2c1, r9c6<>1, stte

AIC: (1=4)r2c6-(4=7)r2c4-(7=8)r2c7-r1c9=r1c1-(8=1)r9c1
All AIC's with the same digit (in single cells) at the ends can be written (and found!) this way
eleven

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