Terrence,
From this and earlier msg, I think that you cannot terminate frame/cell
analysis until 0-cell. The frame/cell analysis should terminate once it meets
the verb's rank. The rank of verb cannot be modified and does not depend on
rank of its argument.
eg. if rank of verb is (lu ru ) 0 _ then you only have to break the left
argument into 0-cells. if rank of verb is _ _ then you don't need to analyze the
frame/cell of left and right. Just feed them to the black (white) box.
May be try this. Just forget frame/cell and "J for C" , experiment with
sentences for different rank on the J session, such as
'abc' , 'defg'
abcdefg
'abc' ,"0 1 'defg'
adefg
bdefg
cdefg
'abc' ,"1 0 'defg'
abcd
abce
abcf
abcg
(2 3$'abc') , (3 4$'defg')
abc
abc
defg
defg
defg
(2 3$'abc') ,"1 (3 4$'defg')
|length error
| (2 3$'abc') ,"1(3 4$'defg')
until you can know intuitively how it works, and then re-read documentation for
explanation why it works in that manner.
Henry Rich wrote:
It IS a property of J. In your example, both frames are empty.
There is no frame that is longer than the other, so there's
nothing to repeat.
It's like
'abc' , 'defg'
The rank of , is infinite, so there is no frame. You have to go to
the definition of , to see what it does with its operands.
Henry Rich
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Terrence Brannon
Sent: Sunday, April 22, 2007 5:38 PM
To: General forum
Subject: Re: [Jgeneral] long-frame rationing when the shapes
are not equal
Well in that case, Henry, "J for C" needs to state that long-frame
rationing is not a property of J, but of particular verbs of j... I
had put it in my dyad processing worksheet as an always, when it is
not.
The whole dyad section of "J for C" had me thinking it was a
property of J.
For example, take:
http://www.jsoftware.com/help/jforc/loopless_code_i_verbs_have
_r.htm#_Toc141157994
but also the case here:
'abc' ,"_ 0 'defg'
On 4/22/07, Henry Rich <[EMAIL PROTECTED]> wrote:
You have it right: there's only 1 cell for each operand, so
they're paired. Now you have to consult the Dictionary so
see what the -. verb will do with its operands. You see
that it removes items of x that appear in y. So it removes
the 2 and the 4s.
Henry Rich
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Terrence Brannon
Sent: Sunday, April 22, 2007 5:19 PM
To: General forum
Subject: [Jgeneral] long-frame rationing when the shapes
are not equal
"long-frame rationing" is my term for what happens when
the noun with
the larger frame has to broken up into n-cells to pair up with the
noun with the smaller frame.
It is demonstrated here:
http://www.jsoftware.com/help/jforc/loopless_code_i_verbs_have
_r.htm#_Toc141157994
For the example:
100 200 + i. 2 3
we see long-frame rationing occur when Henry says <QUOTE>The
longer-frame operand (the right one) is broken up into operand
0-cells, each being paired with a copy of the
shorter-frame operand
cell. Each paired _1-cell becomes a row of paired operand
cells:</QUOTE>
so in my solution of 1 2 3 4 5 4 3 -. 2 4
I am at the long-frame rationing step where the frames
are the same
length, but I cannot pair them 1 for 1 because their shape is
different:
Write out the entire expression: 1 2 3 4 5 4 3 -. 2 4
Expand the verb rank(*1): -. b. 0
0 _ _
Rank Calculation:
----------------
left-noun-rank: 1 right-noun-rank: 1
left-verb-rank: _ right-verb-rank: _
lesser left rank, lr = 1 lesser right rank, rr = 1
The verb will be applied to 1-cells of the left operand
and to 1-cells of the right operand.
left-noun-shape: 7 right-noun-shape: 2
Frame Calulation(*2):
--------------------
left-frame: lf = empty
right-frame: rf = empty
common frame: cf = empty
length of this common frame: rcf = 0
-rcf = (_0)
Operand Pairing:
---------------
1. Pair by common frame consisting of (-rcf)-cells:
[1 2 3 4 5 4 3][2 4]
2. Pair by long-frame rationing. If the frames are the
same length,
then you simply pair them 1 by 1.
*** HELP NEEDED HERE ***
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--
regards,
bill
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