On 6/30/07, Tracy Harms <[EMAIL PROTECTED]> wrote:
produces a hook, where +/\ is g and someverb is h in
the monadic template,
g
/ \
y h
|
y
(which may well not be what is happening) then +/\
acts dyadically. I have not comprehended how this
verb works, dyadically. Monadically, I follow how it
gives cummulative sums, or subtotals.
You can use the 'trace' routine recommended by the
parsing and execution page to get definitive answers
on some of these questions.
Anyways, (u v) y is equivalent to y u v y if u and v are
verbs and y is a noun.
So, (+/\ i.) 3 is the same as 3 +/\ i. 3
Note also that verbs of the form u\ use u monadically,
and you can use the < monad to get an idea of what their
arguments look like. In this case;
3 <\ i. 3
+-----+
|0 1 2|
+-----+
And, since +/0 1 2 is 3,
3 +/\ i. 3
3
Not what you wanted (and perhaps even not what you were
asking about), but perhaps something you might have a use
for in a different context.
+/\ i. 5
0 1 3 6 10
+/\~i. 5
0 0 0 0 0 0
0 1 2 3 4 0
1 3 5 7 0 0
3 6 9 0 0 0
6 10 0 0 0 0
Neither of the above are hooks. But ~ is an adverb
which always uses its verb dyadically.
Check out variations like;
<\i.5
<\~i.5
;~i.5
0 <\ i.5
1 <\i.5
2 <\i.5
etc...
(Oh, and pay attention to the ranks of these derived verbs.)
--
Raul
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