Michal D. wrote: "I'm not sure why to prefer
doubling an entire array as opposed to dividing a single scalar?"
Doubling an integer provides an integer, while division transcends the realm of
integers. That's why. - Bo
--
For
H.]ere are Raul's two definitions, f and g and an explicit definition h
which is turned to tacit.
n=:2
d=:3
n(f=:4 : 'x>+:(?]) x#y')d
1 0 0
0 1 0
5!:4 <'f'
-- 4
-- : -+- ,:'x>+:(?]) x#y'
n(g=:[>[:+:[:(?])#)d
0 1 0
1 0 0
5!:4 <'g'
-- [
+
If J is operating correctly, you should get a stack error.
If J implemented tail recursion this stack error would instead be an
infinite loop (but one that you could break out of).
--
Raul
On Sat, Nov 3, 2012 at 11:37 PM, Linda Alvord wrote:
>3 :'y`:6
> y' 3 :'y`:6 y'`''
>
http://www.jsoftware.com/help/dictionary/dtdotu.htm
http://www.jsoftware.com/help/dictionary/dtdotm.htm
FYI,
--
Raul
On Sun, Nov 4, 2012 at 1:31 AM, Linda Alvord wrote:
> You explain what ^t. does, but what does t. do?
>
> t. i.6
> |syntax error
> | t.i.
>1 t. i.6
> |length
On my Mac with Lion 10.7.3 in the terrminal jconsole I get the following.
I don't know the interpretation of Segmentation fault #11.
server:~ brian$ /Users/brian/j64-701/bin/jconsole ; exit;
9!:12''
5
9!:14''
j701/2011-01-10/11:25 build: Feb 6 2011 16:16:29
3 :'y`:6 y' 3 :'y`:6 y'`''
S
"But what does t. do?"
Use the dictionary.
t. is an adverb.
The dictionary heading on the page says
u t. 0 0 0
This means (and it took me 2 years before I understood how to read
dictionary headings. I learned a lot of j via the "parts of speech"
dictionary access page)
The verb formed by
u
In
A=. 0= ? (2$n)$2 NB. generate random matrix of [0,1]
The 0= is unnecessary, and probably reflects a habit based on the
false idea that boolean algebra is does not have an integer domain.
Boolean rings have (subset of) integer domains, and [even after
redefinition] boolean algebra is a
> I take it that picking # over $ is a purely stylistic preference.
"#" returns a scalar and "$" returns a vector. Each of these applied
monadically to a vector will give the same number but applied to anything
of lower or higher dimension will return different answers entirely.
'a';'few';'wo
As far as I can understand the "revise" verb, you don't need the
adjacency matrix a, although it does help to have a 2-column matrix of
index pairs.
I think Raul has overlooked your use of the right argument ys within
that verb, though he's right about selecting with it. Here's a
derivation
On Sun, Nov 4, 2012 at 12:53 PM, Mike Day wrote:
> I think Raul has overlooked your use of the right argument ys within that
> verb, ...
You are right, I had not even tried to read 'revise'.
Looking at it now, it has a signature:
NB.* '(A;a;http://www.jsoftware.com/forums.htm
Actually...
ac is a wrapper for revise, which supplies a value for ys (which is
the same thing as xs) which initially takes the value i.#D which is
i.n but this changes over time... I need to think about this pattern.
But for the moment, since the right argument to revise is a pair, and
it's used
Your ($#: I.@,) can be replaced:
((4$. $.) -: $#: I.@,) 4 5$ 0 1 0
1
R.E. Boss
> -Oorspronkelijk bericht-
> Van: programming-boun...@forums.jsoftware.com
> [mailto:programming-boun...@forums.jsoftware.com] Namens Mike Day
> Verzonden: zondag 4 november 2012 18:53
> Aan: programm.
In revise, you are using a pair for your right argument.
It looks to me like this might be unnecessary -- that you could use
ys=. I.+./"1 D
But this does not exactly reproduce your algorithm. So if this change
breaks your algorithm, could you find an example situation which
illustrates this issu
I'm blathering a lot here, sorry about that, but I found a case, now I
just need to understand it:
A=: (|:++:) #:0 8 4 4
C=: (a:,];|:) #:0 6 11 1
D=: #:9 0 0 4
Now to see if I can understand what's happening here...
I will try to refrain from posting my incomplete thoughts -- I'll try
to come ba
Hi Linda,
Thanks for your input. I've added:
NB. boxed display for non-nouns.
9!:3 (2)
to my startup script so that lets me leave the parsing up to J which I'm
still taking advantage of quite a bit. I take it that's for the most part
what you're trying to show me in addition to the clever use
So maybe the best solution, at least from a performance perspective, is (<.
-: n).
On Sun, Nov 4, 2012 at 12:01 AM, Bo Jacoby wrote:
> Michal D. wrote: "I'm not sure why to
> prefer doubling an entire array as opposed to dividing a single scalar?"
>
> Doubling an integer provides an integer, w
Unfortunately, I do not know what "works" means. Maybe if this was
wired up to a sudoku solver, I could see it in action -- I know how to
test a sudoku solution for correctness.
Anyways, here's the case I am struggling with right now:
A=: (|:++:) #:0 8 4 4
C=: (a:,];|:) #:0 6 11 1
D=: #
It's a little hard for me to catch up with the flurry of messages, but I am
trying =) I'm also on Vancouver time so I'm guessing you guys had a head
start.
Good ideas, thank you:
removing 0=
adj=: < @ I. @: *
Devon for elaboration of differences b/w # and $.
The domain of the left argument of ar
The adjacency matrix A is important because it tells us which constraint to
look up in C. This also explains why the lower diagonal is doubled.
For example, take the less than constraint:
c =: |: (i.n) wrote:
> As far as I can understand the "revise" verb, you don't need the
> adjacency ma
On Sun, Nov 4, 2012 at 3:25 PM, Michal D. wrote:
> (0 1 4). I follow the argument for (arcs=: $ #: I.@) but my J is slow. We
> would have to be able to select an or of equals for example select from first
> column where the value is 0 or 1 or 4.
Here's select the rows of Y where number in the f
I think I am having a problem envisioning how a sudoku-style puzzle
can work with a relationship which is not commutative, and I am also
having a problem having to do with the transitive character of the
relationship.
As you have put it:
"(note: constraints are not symmetric, for example x wro
The solver can solve sudoku puzzles (hopefully =)) but it should also be
able to handle more general csp instances where the constraints are not
symmetric. In the case of a sudoku puzzle, I think a single symmetric
not-equal constraint would suffice.
n=: 81
d=: 9
] C=: a: , < (i.d) ~:/
On Sun, Nov 4, 2012 at 4:47 PM, Michal D. wrote:
> The solver can solve sudoku puzzles (hopefully =)) but it should also be
> able to handle more general csp instances where the constraints are not
> symmetric. In the case of a sudoku puzzle, I think a single symmetric
> not-equal constraint woul
On Sun, Nov 4, 2012 at 5:17 PM, I wrote:
> Here's a brute force implementation of A for sudoku:
>
>C=: |: R=: ,9#,:i.9 NB. identity within columns (C) and rows (R)
>B=: ,3#3#"1 i.3 3 NB. identity within boxes
>A=: ((2*>/~i.81)+/~i.81)+http://www.jsoftware.com/forums.htm
> Here's a brute force implementation of A for sudoku:
>
>C=: |: R=: ,9#,:i.9 NB. identity within columns (C) and rows (R)
>B=: ,3#3#"1 i.3 3 NB. identity within boxes
>A=: ((2*>/~i.81)+
> I left the extraneous right-most set of parenthesis in place because I
> liked the rhythm.
>
Nice
On Sun, Nov 4, 2012 at 5:32 PM, Michal D. wrote:
>> If D were instead a list of 81 by 9 bits, with extra copies of D being
>> introduced as necessary, the "multiple answer" case could be addressed
>> by having an extra item in D for every valid solution. Here, the
>> initial D would have shape 1
> As far as I can understand the "revise" verb, you don't need the
> adjacency matrix a, although it does help to have a 2-column matrix of
> index pairs.
>
> I think Raul has overlooked your use of the right argument ys within that
> verb, though he's right about selecting with it. Here's a deri
Arc consistency by itself will only be able to solve easy sudoku puzzles.
To solve any puzzle, the search component is needed which does not
currently exist.
Ok, but I am still stuck on making sure I understand what "arc
> consistent" means.
>
http://en.wikipedia.org/wiki/Local_consistency#Arc_c
i. :. (i. :. +) ^:_1 ^:_1 ]5
0 1 2 3 4
The obverse of i. :. (i. :. +) should be (i. :. +), and
the obverse of that should be + . I think. But it isn't.
Henry Rich
--
For information about J forums see http://www.jsoftware.c
Good one! I tend to overlook sparse arrays as I seem to encounter
nonce errors when trying to use them for real.
Mike
On 04/11/2012 7:03 PM, R.E. Boss wrote:
Your ($#: I.@,) can be replaced:
((4$. $.) -: $#: I.@,) 4 5$ 0 1 0
1
R.E. Boss
-Oorspronkelijk bericht-
Van: progr
Hello, Michal D - Mike Day here!
Sorry, I meant your lower case "a", the boxed array of to-nodes,
derived from upper case "A" which IS the adjacency matrix, which is of
course essential. "revise" need not be provided with a as well as A,
and can make do with the 2-column matrix of node-pai
On Sun, Nov 4, 2012 at 6:29 PM, Michal D. wrote:
>> for all w in Dy there exists a v in Dx such that the relationship v
>> compare w is true.
>>
>> Does this sound right?
>>
> Yes, that sounds right so long as compare implies looking up the values in
> the correct constraint table. Your formulatio
Haha, hello Mike D!
I figured you were talking about "a". I think I followed the 2-column
matrix of node-pairs conversation for the most part, except for RE Boss's
contribution which I didn't appreciate until just now. The -: initially
threw me off so I thought the whole expression was replaced
On Sun, Nov 4, 2012 at 9:46 PM, Michal D. wrote:
> to J, it would be nice to use more primitive operations. Why the dislike
> for "a"? I'm guessing it's a stab at removing some of the code?
Passing the same data, multiple ways, to J function is something of a
"bad code smell".
Sometimes it's
>
>
> I had somehow gotten a different idea of consistency:
>
for all w in Dy for all v in Dx the relationship v compare w is true
>
Ah, ok, that's a little bit too strong (it might remove values that can
participate in a solution).
I should have said: for all w in Dy, there exists v in Dx su
Henry, what do you make of
i. :. (i. :. +) b. _1
i. :.+ :.i.
?
Kip Murray
Sent from my iPad
On Nov 4, 2012, at 5:45 PM, Henry Rich wrote:
> i. :. (i. :. +) ^:_1 ^:_1 ]5
> 0 1 2 3 4
>
> The obverse of i. :. (i. :. +) should be (i. :. +), and
> the obverse of that should be + . I think
Yes, that seems wrong. It should be
i. :.+
shouldn't it?
Henry Rich
On 11/4/2012 10:15 PM, km wrote:
Henry, what do you make of
i. :. (i. :. +) b. _1
i. :.+ :.i.
?
Kip Murray
Sent from my iPad
On Nov 4, 2012, at 5:45 PM, Henry Rich wrote:
i. :. (i. :. +) ^:_1 ^:_1 ]5
0 1 2 3
My inclination is that it is correct in enforcing that the obverse of
the obverse is the original function. I'm open to good reasons why this
isn't always the case, but I think if you need to break this rule, then
what you're looking for probably isn't the obverse. Using &. in this
case will just c
On Sun, Nov 4, 2012 at 10:00 PM, Michal D. wrote:
>> A tricky part seems to be that the upper triangle of A corresponds to
>> the forward direction in C (the bit in D that selects a row in A is
>> also selecting a row in C) while the lower triangle of A corresponds
>> to the reverse direction in C
> >> variables. In other words, if we specify the constraint x < y it
> >> looks like either x1 < x2 or x2 < x1 is sufficient to satisfy arc
> >> consistency. In other words, i think we should always use the
> >> symmetric closure of the constraint.
> >
> >> Does this sound valid to you?
> >
> >
It appears the obverse of u :. v is v :. u and not simply v . This explains
the behavior Henry saw.
Kip Murray
Sent from my iPad
On Nov 4, 2012, at 9:33 PM, Marshall Lochbaum wrote:
> My inclination is that it is correct in enforcing that the obverse of
> the obverse is the original functio
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