Each pair returns a value:
'0#n' returns a 0-length value;
'1#n' returns a 1-length value.
From this, what would you suppose is the length of
'2#n'?
On 4/18/07, Terrence Brannon <[EMAIL PROTECTED]> wrote:
1 0 1 0 0 0 # 3 1 4 1 5 9
# b. 0
_ 1 _
Rank Calculation:
----------------
left-noun-rank: 1 right-noun-rank: 1
left-verb-rank: 1 right-verb-rank: _
lesser left rank, lr = 1 lesser right rank, rr = 1
The verb will be applied to one-cells of the left operand
and to one-cells of the right operand.
left-noun-shape: 6 right-noun-shape: 6
Frame Calulation(**):
--------------------
left-frame: lf = "" right-frame: rf = ""
common frame: cf = ""
length of this common frame: rcf = 0
-rcf = _0
Operand Pairing:
---------------
1. Pair by common frame consisting of (-rcf)-cells (****):
a _0 cell is the entire datum
<box><box1>1 0 1 0 0 0</box1><box2>3 1 4 1 5 9</box2></box>
2. Pair by long-frame rationing:
The frames are the same length, so we just go pair-wise
1 # 3
0 # 1
1 # 4
0 # 1
0 # 5
0 # 9
but it is not clear what is being returned in the '0 # n' cases and how
the isolated pair-wise actions, some of which return a value and some
of which do not, are accumulated into a return value. Please advise on
this.
I'm thinking it's something like in Lisp with append:
(append (# 1 3) (# 0 1) (# 1 4) (# 0 1) (# 0 5) (# 0 9) )
(append 3 nil 4 nil nil nil )
(append 3 4)
'(3 4)
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Devon McCormick, CFA
^me^ at acm.
org is my
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