I'm sorry to re-start this thread, but the discussion at
http://www.jsoftware.com/help/jforc/loopless_code_i_verbs_have_r.htm#_Toc141157994
is leaving me in the a bit confused at a certain point, so let's take the
discussion there sentence by sentence.
The code I am trying to solve out is 2 4 { 3 1 4 1 5 9 NB. sorry to all
those who wrote in
last thread, but I'm still at it
So here we go:
The formal description that follows is not easy to follow--you might want to
skim over it and read it in detail after you have studied the examples that
follow.
J requires that one of the frames be a prefix of the other (if the frames
are identical, each is a prefix of the other and all the following reduces
to the simple case we have studied). The common frame cf is the part of the
frames that is identical, namely the shorter of the two frames; its length
is designated rcf.
So our first goal is to calcuate rcf
left-noun-rank = 1, right noun rank = 1
left verb rank = 0 , right verb rank = infinite
left rank = 0 , right rank = 1
left shape = 2 right shape = 6
lf (left frame) = 2, rf (right frame) = "" (empty)
cf = ""
rcf = 0
Ok, good we have rcf. Let's continue:
If we look at the cells of the operands relative to this common frame (i.
e. the (-rcf)-cells), we see that for the operand with the shorter frame,
these cells are exactly the rank that will be operated on, while for the
operand with the longer frame, each (-rcf)-cell contains multiple operand
cells.
rcf-cells are 0-cells in this case.
The operand with the shorter frame is the right one. So we should find that
0-cells are exactly the rank (0) that will be operated on. But the right
rank is 1, not 0. This doesnt seem right, but for now, I will continue:
First, the (-rcf)-cells of the two operands are paired one-to-one (because
they have the same frame), leaving each shorter-frame operand cell paired
with a longer-frame (-rcf)-cell.
The 0-cells of the two operands are paired one-to-one? 2 is a 0-cell. 4 is a
0-cell. Same with 3, 1, 4, 1, 5, and 9... resulting in 12 pairings... hmmm,
something is rotten in Denmark...
Then, the longer-frame (-rcf)-cells are broken up into operand cells, with
each operand cell being paired with a copy of the shorter-frame operand cell
that was paired with the (-rcf)-cell
_operand_ cells? rcf is 0, how do you break longer frame 0-cells into
operand cells? How do longer frame operand cells differ from shorter-frame
operand cells.
(an equivalent statement is that the cells of the shorter-frame operand are
replicated to match the surplus frame of the longer-frame operand).
Again, I think I have something switcheroo'ed. I think that 3 1 4 1 5 9 (the
right operand) has the shorter frame, but it looks like somehow I have that
wrong.
This completes the pairing of operand cells, and the operation is then
performed on the paired operand cells, and collected using the longer
frame.
yes, this is what fndisplay would show me... without any intermediate steps
to tutor me.
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