Mike, I think this will work as an alternative to adj
A
0 0 1 0
0 0 0 0
2 0 0 1
0 0 2 0
adj
<@#&0 1 2 3@(0&<)
adj A
--TT---T-┐
│2││0 3│2│
L-++---+--
h
0 1 2 3 <@#~ 0 < ]
h A
--TT---T-┐
│2││0 3│2│
L-++---+--
Can anyone remove the final @ from h ?
Linda
-----Original Message-----
From: programming-boun...@forums.jsoftware.com
[mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Raul Miller
Sent: Saturday, November 10, 2012 12:44 PM
To: programm...@jsoftware.com
Subject: Re: [Jprogramming] Arc consistency in J
On Sat, Nov 10, 2012 at 12:16 PM, Michal D. <michal.dobrog...@gmail.com>
wrote:
> Here X is telling us to use the constraint c1 (presumably b/c C is not
> shown) between the variables 1 and 3 (0 based). Likewise, use the
> transpose going the other direction (3,1).
Ouch, you are correct, I did not specify C. On retesting, though, it looks
like my results stay the same when I use:
arccon=:3 :0
'D c1 X'=: y
'n d'=: $D
adj =: ((<@#)&(i.n)) @ (0&<)
A =: adj X
C=: a: , c1 ; (|:c1)
ac =: > @ (1&{) @ (revise^:_) @ ((i.n)&;)
ac D
)
For longer scripts like this, I really need to get into the habit of
restarting J for every test. So that probably means I should be using jhs.
> Given the structure of X, only variables 1 and 3 can possibly change.
> So if they are all changing something is definitely wrong.
This line of thinking does not make sense to me. I thought that the
requirement was that a 1 in D exists only when there is a valid relationship
along a relevant arc. If a 1 in D can also exist in the absence of any
relevant arc, I am back to needing a description of the algorithm.
> Unfortunately I've run out of time to read the rest of your response
> but hopefully I can get through it soon. I've also wanted to write a
> simpler version of the algorithm where the right argument of ac is
> only D and it runs through all the arcs in the problem instead of
> trying to be smart about which ones could have changed.
Yes... I am currently suspicious of the "AC-3 algorithm".
In the case of symmetric consistency, I think that it's unnecessary
complexity, because the system converges on the initial iteration.
In the case of asymmetric consistency, I think that the work involved in
maintaining the data structures needed for correctness will almost always
exceed the work saved.
But I could be wrong. I am not sure yet if I understand the underlying
algorithm!
--
Raul
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