Your diagram isn't right. Rather than a column y{UV, you need a column
y, the first entry in which would be 3.
The first application of foo is
3 foo 0 0 0
which gives a result of shape 3 3 .
The next application of foo is
2 foo (3x3 array)
giving shape 3 3 3, etc.
The y in foo is NOT the items of the y input to Fold, but rather the
initialization/feedback value.
The x in foo is the items of the y input to Fold.
Henry Rich
On 9/7/2021 12:04 PM, R.E. Boss wrote:
In my example
UV=:0,(,-@|.)=i.3
foo=: {{ x+ y{UV }}
and with the fold-diagram as guide, I thought the working of 0 0 0 ]F:. foo 3
2 _3 would be
y{UV x= 0 0 0
v v
0 0 1 ---> 0 0 1
v v
0 1 0 ---> 0 1 1
v v
0 0 -1 --> 0 1 0
but it is not. The output produced is completely incomprehensible to me, with a
shape of 3 3 3 3 3.
With some reverse engineering I found out that swapping the y and x gives what
I want, so
0 0 0 ]F:. (foo~) 3 2 _3
0 0 1
0 1 1
0 1 0
Memo to self: swap arguments (of inner verb) before fold.
R.E. Boss
-----Original Message-----
From: Programming <[email protected]> On Behalf Of Henry
Rich
Sent: dinsdag 7 september 2021 16:44
To: [email protected]
Subject: Re: [Jprogramming] Fold experience
Fold principles:
1. The items of y [and x] are applied to the x argument of v. This is
analogousto u/ y and u/\. y. The underlying J principle is that when a verb
operates multiple times, it operates repeatedly on its y argument. Thus the
result of each application of v is the y argument to the next v.
2. The same verb could be used for Fold Forward or Fold Reverse, with the only
difference being the order in which items are processed.
3. The optional x argument should be thought of as an additional item of y, or
if you like an initial value for the iteration. If given, it is the first y
argument to v.
Henry Rich
On 9/7/2021 7:15 AM, R.E. Boss wrote:
OK, I thought that x was always on the left of a verb, and y on the right. Did
not expect the diagram would have priority over that J axiom.
Never too old to learn, although it will be quite confusing. Now I have normal,
old-fashioned verbs, and verbs which will be folded.
Apart from that, why does swapping the parameters (with ~) not work?
Thanks.
R.E. Boss
-----Original Message-----
From: Programming <[email protected]> On Behalf
Of Raul Miller
Sent: dinsdag 7 september 2021 11:28
To: Programming forum <[email protected]>
Subject: Re: [Jprogramming] Fold experience
The diagram shows the y argument to the fold on the left and the x argument to
the fold on the right. So that cannot be the issue you are talking about.
The diagram shows y0 being used first, y1 second and so on, which matches what
we saw happening in action. So that cannot be the issue you are talking about.
But that leaves me confused (an all-too-common event) -- what is the behavior
which you are talking about which does not conform to that diagram?
Thanks,
--
Raul
On Tue, Sep 7, 2021 at 5:19 AM R.E. Boss <[email protected]> wrote:
The main point I was trying to make was that F:. showed behaviour not conform
the diagram on https://code.jsoftware.com/wiki/Vocabulary/fcap.
The role of x and y seem to be interchanged.
However, then u F:.v~ should work, which does not.
The advantage of fold is just that x can differ essentially from an item of y,
contrary to /\. , where all items are equal, and you are forced to use boxes.
R.E.Boss
-----Original Message-----
From: Programming <[email protected]> On
Behalf Of Raul Miller
Sent: dinsdag 7 september 2021 10:16
To: Programming forum <[email protected]>
Subject: Re: [Jprogramming] Fold experience
Swapping the argument does not appear to influence the shape of the result.
The results themselves are different,
UV=:0,(,-@|.)=i.3
foo=: {{ x+ y{UV }}
oof=: {{ y+ x{UV }}
0 0 0 (]F:.foo -: ]F:.foo~) 3 2 _3
0
0 0 0 (]F:.oof -: ]F:.oof~) 3 2 _3
0
That said, focusing on the variant which you described as providing the desired
behavior:
0 0 0 ]F:.oof 3 2 _3
0 0 1
0 1 1
0 1 0
This corresponds to:
3 oof 0 0 0
0 0 1
2 oof 3 oof 0 0 0
0 1 1
_3 oof 2 oof 3 oof 0 0 0
0 1 0
Which, in turn, somewhat matches the value selection mechanism of J's scan:
<\ 3 2 _3
+-+---+------+
|3|3 2|3 2 _3|
+-+---+------+
The difference between fold and scan being that fold reuses the result of the
previous evaluation while scan does not.
I hope this makes sense,
--
Raul
On Tue, Sep 7, 2021 at 2:09 AM R.E. Boss <[email protected]> wrote:
[UV=:0,(,-@|.)=i.3
0 0 0
1 0 0
0 1 0
0 0 1
0 0 _1
0 _1 0
_1 0 0
foo=: {{ x+ y{UV }} NB. constructed conform the diagram on
https://code.jsoftware.com/wiki/Vocabulary/fcap
$0 0 0 (]F:. foo) 3 2 _3
3 3 3 3 3
$0 0 0 (]F:. foo)~ 3 2 _3
3 3 3 3 3
foo=: {{ y+ x{UV }} NB. y and x are swapped, counterintuitive
$0 0 0 (]F:. foo) 3 2 _3
3 3
0 0 0 (]F:. foo) 3 2 _3 NB. this is the desired behaviour
0 0 1
0 1 1
0 1 0
$0 0 0 (]F:. foo)~ 3 2 _3
3 3
Also remarkable is that swapping the arguments seems not to influence the
result.
Any clarification?
R.E. Boss
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