## January 3, 2015

Post puzzles for others to solve here.

### Re: January 3, 2015

David, please check out my post here:
january-5-2015-t32254.html
DonM
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### Re: January 3, 2015

gurth wrote:I say it should mean "if we remove all values (a) and (b) from cells12, then we can deduce (and context and other cells can play a part in this deduction), that cells12 must contain (c) and (d). "

This would mean, that you interpret (ab)cells12 as (a or b) in cells12 (because the negation of "a and b" is "not a or not b") and (cd)cells12 as (c and d) in cells12, which is inconsistent.

Added: please don't miss the posts by Danny and David on the last page.
eleven

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### Re: January 3, 2015

eleven wrote:
gurth wrote:I say it should mean "if we remove all values (a) and (b) from cells12, then we can deduce (and context and other cells can play a part in this deduction), that cells12 must contain (c) and (d). "

This would mean, that you interpret (ab)cells12 as (a or b) in cells12 (because the negation of "a and b" is "not a or not b") and (cd)cells12 as (c and d) in cells12, which is inconsistent.

Added: please don't miss the posts by Danny and David on the last page.

At least you understand what I am saying, David apparently does not. Yes, to spell it out in even more detail, I am saying:
If ( NOT (a OR b) in cell1) AND ( NOT (a OR b) in cell2) then
((c in cell1) AND (d in cell2)) OR ((d in cell1) AND (c in cell2)).
There is no earthly reason why the arrangement before the 'then' should duplicate the arrangement after the 'then' . Neither arrangement has any influence on the other.
You should not read this as saying that in all cases this will automatically be true. In each case, the statement above would have to be proved by the context.
If we were to agree on your definition, (which is "If ( NOT (a AND b) in cell1) AND ( NOT (a AND b) in cell2) then..." (or do you mean "If ( NOT (a AND b) in cell1) OR ( NOT (a AND b) in cell2) then..)"?
then I hardly see how you could make any practical use of it. As you pointed out in your first post, there are multiple combinations possible, making it difficult to deduce anything.

Example of my usage: if we have 2 cells (5678) in a sector (say r1c12), then we can write (57)=(68)r1c12 (just as Steve did originally: I wish to state that he was 100% correct in this, whatever he may think now.
gurth

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Location: Cape Town, South Africa

### Re: January 3, 2015

gurth wrote:There is no earthly reason why the arrangement before the 'then' should duplicate the arrangement after the 'then' . Neither arrangement has any influence on the other.
You should not read this as saying that in all cases this will automatically be true. In each case, the statement above would have to be proved by the context.

Gurth,
as David said, your proposal, how to interpret the link notation is not according to the AIC definition.
One reason is, that an AIC is symmetric (can be read the same way from left to right and from right to left), which your interpretation is not, if the left side means (a or b) and the right one (c and d).

But let us forget the notation for a moment, and spell out the logic, how (i think) you see the elimination of 46 from r8c4:
- if none of the cells is 4 or 6, then only 38 remains for the 2 cells, and r8c5 nust be 3, r8c6=8, and r23c4 must be 46 (from Steves chain) => r8c4<>46
- if both would be 46 (which by the way is impossible, because one must be 8) => r8c4<>46
- if one of them is 4 or 6, the other one must be 8, and due to the strong link for 3 then r8c4=3 <>46

Then what you need, are the strong links for 3 and 8 and they should be part of the notation.
But if you use them, through the common cell r8c4 you get the link 3r8c4=8r8c6 and you don't need the 46 link at all.
eleven

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Joined: 10 February 2008

### Re: January 3, 2015

Eleven, Gurth works in his own ways and I don't rule out that it could be possible for his proposal to be worked into an assumptive solving scheme if it was the primary assumption.

In my previous post I was trying to establish what his ground rules were, and if he accepted if there was a difference in applying one or more Boolean operators and applying an inference that covered every possible situation without exception.
David P Bird
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### Re: January 3, 2015

Here's how I see it. Of course, that'd be wrong!

Code: Select all
`ALS strong link in Eureka notation:   (yz=x)cells  ->  if the pair yz is assumed/forced false, then x must be true                <-  if x is assumed/forced false, then the pair yz must be true`

Code: Select all
`AALS strong link in Eureka notation:   (wx=yz)cells  ->  if both w and x are assumed/forced false, then y and z must be true                 <-  if both y and z are assumed/forced false, then w and x must be true`

These definitions allow for the following mix of ALS and AALS strong links:

Code: Select all
`   ... (v=wx)cells_12 - (wx=yz)cells_34 - (yz=u)cells_56 ...`

_
daj95376
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### Re: January 3, 2015

daj95376 wrote:Here's how I see it. Of course, that'd be wrong!

Code: Select all
`ALS strong link in Eureka notation:   (yz=x)cells  ->  if the pair yz is assumed/forced false, then x must be true                <-  if x is assumed/forced false, then the pair yz must be true`

Code: Select all
`AALS strong link in Eureka notation:   (wx=yz)cells  ->  if both w and x are assumed/forced false, then y and z must be true                 <-  if both y and z are assumed/forced false, then w and x must be true`

These definitions allow for the following mix of ALS and AALS strong links:

Code: Select all
`   ... (v=wx)cells_12 - (wx=yz)cells_34 - (yz=u)cells_56 ...`

_

In your concluding chain you have 6 cells restricted to 6 candidates so (u) and (v) must both be true. This will allow the digits to be distributed like this:
(v)cell_1, (w)cell_2, (x)cell_3, (y)cell_4, (z)cell_5, (u)cell_6
where (xy)cells_34 is true. So in this case (wx=yx) is not a strong link.

I maintain that if you can find a scenario where that link must be strong then it can be made clear by appending #N to all the pairs to show the number of truths that are needed for them to be true. Therefore there is no need for any special interpretation of (ab=cd)cells_12.
David P Bird
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### Re: January 3, 2015

Danny's logic is of course valid here, which easily can be seen, if you write it as a simple implication chain:
If you have (vwx)cells_12, (wxyz)cells_34 (in a unit), then
Either (v)cells_12 or (wx)cells_12 => (yz)cells_34
which clearly says, that either v is in cells_12 or yz in cells_34.
(David: note that you only can have (xy)cells_34, if v in cells_12).

The problem is in the AIC notation again, because
(wx=yz)cells_34 in the notation means, either both cells are wx or both cells are yz, which is not true.
If not the ugly number signs are used, you could write (w|x=yz)cells_34 to indicate, either one of wx or both yz.
Note, that this is different from the (46=38) link we discussed, where you would have needed both 4 and 6 in the 2 cells to (directly ) give the conclusion.

Then (v=wx)cells_12 - (w|x=yz)cells_34 would have the same meaning as above:
If not v in cells_12, then the pair wx, then not one of wx in cells_34, if not one of wx, then both yz.
Just as Gurth wanted to have it (but in the wrong example).
From right to left: if not both yz in cells_34 , then (at least) one of wx, then not both of wx in cells_12, if not both wx, then v must be in there.

So be careful with this notation.
eleven

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Joined: 10 February 2008

### Re: January 3, 2015

Note that my chain is reversible: it can just as well be written (6=5)r1c5 - (345=12)r1c124 - (12=7)r23c1 - (7=6)r4c1, so eleven's suggestion that my method is not reversible is wrong.

Note also that cells r1c12 could also contain one or two 5s: e.g. if these cells are both 12345, then my chain would read exactly as written before.

Eleven, how would you write this?
gurth

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Location: Cape Town, South Africa

### Re: January 3, 2015

gurth wrote:Eleven, how would you write this?

Nice example, Gurth.

To keep with my suggestion above, an "or" sign between 1 and 2 would clarify it:
(6=7)r4c1 - (7=12)r23c1 -(1|2=345)r1c124 - (5=6)r1c5 => -6r4c5
If not on this forum, i would have written something like
If r4c1 not 6, then r4c1=7 -> r23c1=12 -> r1c12=34 -> r1c4=5 -> r1c5=6 => r4c5<>6

The crucial part in words:
If there is a pair 12 in r23c1, then none (not one) of 12 in r1c124, if not 1 or 2 in r1c124, then a triple 345
From the other side:
If there is no triple 345, then it contains 1 or 2, then 12 is not a pair in r23c1

(When i said, your definition of (ab=cd)xy is not reversable, i thought you would want to read the reverse side then as if not c OR d then a AND b)
eleven

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Joined: 10 February 2008

### Re: January 3, 2015

What a useless discussion – only one person has admitted to altering his opinion (but that didn't seem to last), and the conclusion is as clear as mud.

So that my position is clear:
1) In an AIC "(ab=cd)cells12" is invalid and needs to be qualified so that one of the arguments is considered true with one truth and the other considered true with two truths.
2) Ways to achieve this are to use (ab#1=cd) or (a|b=cd)
David P Bird
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### Re: January 3, 2015

Ok, if you think so.

For my part i am happy, that some misunderstandings, mis- and alternate use and pitfallls of the AIC notation could be clarified.
eleven

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Joined: 10 February 2008

### Re: January 3, 2015

On ab=cd
There is a fundamental difference between
ab=cd at the start of a chain (or subchain) eg SK-loop
and ab=cd within a chain eg x=ab-(ab=cd)
In this latter case, the meaning is clear : if both ab here, then neither a nor b there.
Because the meaning is clear, there is no need to seek alternative notation.
Put another way, the other theoretical possibilities (a but not b, b but not a) are ruled out by the preceding "both".

On the SK-loop
This demonstrated (to me) the superior logic provided by a base/cover approach.
With AIC logic one has to satisfy oneself that not only the "neither 2 nor 7" loop but also the "either 2 or 7 but not both" loop produce the same eliminations.

With base cover logic, there is a symmetrical 16 set rank-0 structure :
base = 16 cells (the cross cells in b1397)
r2c13 r13c2
r2c79 r13c8
r8c79 r79c8
r8c13 r79c2
cover = 8 box, 4 row 4 column
16b3b7
27b1b9
48c2 38r2
45r8 39c8
aran

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Joined: 02 March 2007

### Re: January 3, 2015

aran wrote:On ab=cd
There is a fundamental difference between
ab=cd at the start of a chain (or subchain) eg SK-loop
and ab=cd within a chain eg x=ab-(ab=cd)
In this latter case, the meaning is clear : if both ab here, then neither a nor b there.

But this does not make it very clear too. The chain then has to be interpreted different from right to left.
So i stick to my opinion, that in this case an "or" sign should be added, and that for the SK loop's pairs link a special notation should be used.
Otherwise you can notate it anyway and say, the meaning comes from the context ...
eleven

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Joined: 10 February 2008

### Re: January 3, 2015

What is the AIC/Eureka notation for the following?

Code: Select all
` +-----------------------------------------------+ |   .   .   .   |   .   .   .   |   .   .   .   | |   .   .   .   |   .   .   .   |   .   .   .   | |   .   .   .   |   .   .   .   |   .   .   .   | |---------------+---------------+---------------| |  -v  -v  -v   |   .   .   .   |   .  vr   .   | |   .  vwx  .   |   .   .   .   |   .   .   .   | |   .  vwx  .   |   .   .   .   |   .   .   .   | |---------------+---------------+---------------| |  uyz  .   .   |   .   .   .   |   .  ur   .   | |   .  wxy  .   |   .   .   .   |   .   .   .   | |   .  wxz  .   |   .   .   .   |   .   .   .   | +-----------------------------------------------+ (-v)r56c2 -> (wx)r56c2 -> (yz)r89c2 -> (u)r7c1 -> (r)r7c8 -> (v)r4c8  =>  (-v) r4c123`

aran appears to be the only one willing to accept:

Code: Select all
` (v=wx)r56c2 - (wx=yz)r89c2 - (yz=u)r7c1 - (u=r)r7c8 - (r=v)r4c8  =>  (-v) r4c123`

Maybe?

Code: Select all
` (v=wxyz)r5689c2 - (yz=u)r7c1 - (u=r)r7c8 - (r=v)r4c8  =>  (-v) r4c123`

_
daj95376
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