(This is a knee-jerk response: I haven't thought about this much yet. I
reserve the right to change my mind about any of the following as I consider
it more.)
Yuvaraj Athur Raghuvir wrote:
> (c) F a dyad that takes scalar inputs
> (6 1 5 2 3 4) -: \:; L F"0 0/
First, (c) and "0 0 are redundant. Second I assume from ; that the output
is boxed, so that what you really have is F=:G&.> and you don't need (c)
or "0 0. Third, do you know anything about the (shape of) the output of F
? If you intentionally used ; instead of > , I think that's a big hint, but
I'm not sure how to use it quite yet.
> (a) W where 2 = $W = w1, w2
> (b) L where 3 = $ L = l1 , l2 , l3
> (6 1 5 2 3 4)
Since F is rank 0, then W must be a prefix of L, hence W -: 2 {. L . And
since #6 1 5 2 3 4 is 6, then w1=l1=6 .
BTW, there are an infinite number of possible Fs, and if \:W or /:L is
already 6 1 5 2 3 4 then F could be [ or ] . If you post more specifics
about the problem, we could help you narrow it down more. Meanwhile, I'll
give it more thought.
-Dan
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