## Double XY-Wing Triangle

Advanced methods and approaches for solving Sudoku puzzles
aran wrote:Re'born
I agree with what you write about independent statements etc and the
collection of short stories
metaphor is absolutely fine.
What you then go on to say is this I believe :
if it is a true statement : 46r7c49 not a pair =>r7c9 is 6
ie ~46r7c49=>6r7c9
then the logical converse must hold :
~6r7c9=>~~46r7c49.
There can be no disagreement so far.

Other than your use of the word converse instead of contrapositive, I agree so far.
aran wrote:The double "not" above is crucial because in "language" we might say that "not not a pair" means a pair...but does it ?
A pair here means 4,6 or 6,4 in r7c49 (and both remain possible)
Examine : not not (4,6 or 6,4).
not (4,6 or 6,4)=>neither.
not neither : is satisfied by the existence of either, without implying that both remain possible.

In language, perhaps. In logic, definitely not.
not not (4,6 or 6,4) is logically equivalent to (4,6 or 6,4) and the word 'or' does allow for both to be true. If I tell two students that "it is not the case that neither or you will pass, this implies that one will definitely pass and that both could pass.
re'born

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Joined: 31 May 2007

Re'born
(Just to recall the issue : I say that
if a chain starts ~a in cell C, then one cannot subsequently in the same logic stream refer to C as part of a pair ab. Because this latter leaves alive the possibility that C may be a, which is excluded).

aran wrote:

The double "not" above is crucial because in "language" we might say that "not not a pair" means a pair...but does it ?
A pair here means 4,6 or 6,4 in r7c49 (and both remain possible)
Examine : not not (4,6 or 6,4).
not (4,6 or 6,4)=>neither.
not neither : is satisfied by the existence of either, without implying that both remain possible.

In language, perhaps. In logic, definitely not.
not not (4,6 or 6,4) is logically equivalent to (4,6 or 6,4) and the word 'or' does allow for both to be true. If I tell two students that "it is not the case that neither or you will pass, this implies that one will definitely pass and that both could pass.

Our "or" here is an exclusive "xor" so the student analogy doesn't really work because both students can potentially pass whereas only one of the pair possibilities - (4,6) (6,4) - can "pass".
Original statement S : (4,6) xor (6,4)
~S => neither (4,6) nor (6,4)
~~S => one of (4,6) or (6,4).
This latter doesn't imply the pair, just that one of the two must hold.
And if one of the two is excluded, as I maintain, then it is the other which holds.
aran

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Joined: 02 March 2007

aran wrote:~~S => one of (4,6) or (6,4).
This latter doesn't imply the pair, just that one of the two must hold.
And if one of the two is excluded, as I maintain, then it is the other which holds.

Sounds like you now also see nothing wrong with the derived strong inference ... (6)r7c9 = (46)r7c49.

You must have been a victim of capillary fissure (your term, as I recall).
ronk
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Location: Southeastern USA

Perhaps I am confused but I can't see anywhere in the chain
Code: Select all
`(6)r7c9 = (6)r9c9 - (6=37)r9c56 - (7=46)r7c49`

where it is explicitly declared that r7c9<>6. To me it reads as r7c9=6 OR r9c9=6,r9c56={37},r7c49={46}
Glyn

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Joined: 26 April 2007

Ronk
aran wrote:
~~S => one of (4,6) or (6,4).
This latter doesn't imply the pair, just that one of the two must hold.
And if one of the two is excluded, as I maintain, then it is the other which holds.

Sounds like you now also see nothing wrong with the derived strong inference ... (6)r7c9 = (46)r7c49.

You must have been a victim of capillary fissure (your term, as I recall)

We must be at cross-purposes :
= (46)r7c49
as meaning a pair with all the potentiality of a pair, and on that basis can't agree. I wish I could to end the fissure.
aran

Posts: 334
Joined: 02 March 2007

### re: logic

aran wrote:Our "or" here is an exclusive "xor" so the student analogy doesn't really work because both students can potentially pass whereas only one of the pair possibilities - (4,6) (6,4) - can "pass".

Original statement S : (4,6) xor (6,4)
~S => neither (4,6) nor (6,4)
~~S => one of (4,6) or (6,4).

This latter doesn't imply the pair, just that one of the two must hold.

And if one of the two is excluded, as I maintain, then it is the other which holds.

sorry but you can't have it both ways

where S uses an (inclusive) OR,
indeed ~S becomes "neither" as you stated

but since your S uses a XOR,
your ~S would have to be "both-or-neither"

Pat

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Joined: 18 July 2005

Pat wrote

sorry but you can't have it both ways

where S uses an (inclusive) OR,
indeed ~S becomes "neither" as you stated

but since your S uses a XOR,
your ~S would have to be "both-or-neither"

What ? I can`t have my cake and eat it ?
I agree that "both" should have been included in logic, but since "both" is impossible, that brings us right back to "neither" => still have my gateau.

Another way of looking at this :
one cannot deduce (46)r7c4 then look back to r7c9 and see (46) and then say "aha ! a pair", because it is a mirage : there is no longer (46) in r7c9 but 4.
Obviously this 4 combines with (37r9c56) to reduce r7c4 to 6 etc.
The logic in all of that I maintain is that it is a second inference stream starting from 4r7c9 which is at work and not a fictitious pair within the first stream...
aran

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Joined: 02 March 2007

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