Yes, it does indeed:
u=. +
((u^:((# - 1:))@:{.) -: u/) (i.1)
1
Also,
erase'u'
1
((u^:((# - 1:))@:{.) -: u/) (i.1)
1
I have to digest it a little bit more but so far everything seems to be
falling into place; at least, your interpretation is consistent.
Thanks
On Thu, Sep 11, 2014 at 7:24 PM, Raul Miller <rauldmil...@gmail.com> wrote:
> Remember that the definition of u/ was:
>
> "u/y applies the dyad u between the items of y".
>
> The items of i.1 are {.i.1
>
> And:
> ({.i.1) -: +/i.1
> 1
>
> Does that make sense? (And, yes, I intentionally mangled english
> grammar in my third sentence of this message.)
>
> Thanks,
>
> --
> Raul
>
>
> On Thu, Sep 11, 2014 at 7:18 PM, Jose Mario Quintana
> <jose.mario.quint...@gmail.com> wrote:
> > Now, that is interesting. Thus, it seems, the unofficial documentation
> > about / is not quite correct (depending on what the meaning of "is" is.).
> >
> > I have another question: if "u is applied (#y)-1 times." How come,
> >
> > (-: u^:0)i.1
> > 1
> >
> > ?
> >
> >
> >
> > On Thu, Sep 11, 2014 at 6:36 PM, Raul Miller <rauldmil...@gmail.com>
> wrote:
> >
> >> y always remains unchanged. Note also that the result is distinct from
> y:
> >>
> >> (-: +/) i.1
> >> 0
> >>
> >> Thanks,
> >>
> >> --
> >> Raul
> >>
> >> On Thu, Sep 11, 2014 at 3:13 PM, Jose Mario Quintana
> >> <jose.mario.quint...@gmail.com> wrote:
> >> > My only guess is:
> >> > "m/y inserts successive verbs from the gerund m between items of y"
> >> >
> >> > So, if there is no "between items of y" inserts nothing and y remains
> >> > unchanged; but, it seems to me that the Dictionary could be more
> >> assertive
> >> > in this instance.
> >> >
> >> >
> >> > On Thu, Sep 11, 2014 at 2:13 PM, Dan Bron <j...@bron.us> wrote:
> >> >
> >> >> Pepe wrote:
> >> >> > My only question is: Does the Dictionary support this behavior?
> >> >>
> >> >> > Raul responded:
> >> >> > Yes, it does.
> >> >>
> >> >> I replied:
> >> >> > I am intrigued. Can you elaborate?
> >> >>
> >> >> Thomas followed-up:
> >> >> > I assumed that by not mentioning it, the implementation
> >> >> > is free to do what it chooses. It could be anything!
> >> >>
> >> >> That's what I think too. The behavior is, in the strictest literal
> >> sense,
> >> >> undefined. But Raul differs. I'm interested in his rationale (which,
> >> >> historically, has been both solid and instructive).
> >> >>
> >> >> -Dan
> >> >>
> >> >>
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