The tricky bit is that ^: isn't a very flexible recursion scheme. The verb it
repeatedly applies effectively has to be monadic. A function on two arguments
can be turned into a function on one argument if we make that argument an
ordered pair. So to make addition monadic, we'll start by splitting a vector
into two pieces:
({. + {:) 1 2
3
We'll designate the head of our two-element vector as the "accumulator" and use
the tail as the TODO list -- the arguments for later. Our monadic adder must
extract the head of the tail and add it to the accumulator (the head).
({. + ({.@{:))
Unfortunately, our accumulator and argument list have different shapes. We'll
have to box them in order to put them into the same vector.
(<20),(<2 3 4)
┌──┬─────┐
│20│2 3 4│
└──┴─────┘
We also have to extract the TODO list, unbox it, then take its head. Then we
can add it to the accumulator.
((>@{.) + ({.@>@{:)) (<20),(<2 3 4)
22
Yay, we have an updated accumulator! We also need to update our TODO list.
That's just beheading the current TODO list.
(}.@>@{:) (<20),(<2 3 4)
3 4
Both the accumulator and the new TODO list will have to get boxed up again and
recombined.
((<@((>@{.) + ({.@>@{:))) , (<@(}.@>@{:))) (<20),(<2 3 4)
┌──┬───┐
│22│3 4│
└──┴───┘
So let's apply *that* big verb three times:
(((<@((>@{.) + ({.@>@{:))) , (<@(}.@>@{:))) ^:3) (<20),(<2 3 4)
┌──┬┐
│29││
└──┴┘
From here, extracting the final value of the accumulator is easy.
None of this relies on anything special about the + verb, so this
transformation should be packageable as an adverb if you're really intent on
using ^: everywhere.
---
Justin Slepak
PhD student, Computer Science dept.
----- Original Message -----
From: Skip Cave <[email protected]>
To: [email protected]
Sent: Tue, 27 Sep 2016 03:35:19 -0400 (EDT)
Subject: [Jprogramming] Learning Recursion
NB. I am attempting to learn recursion using ^:
NB. A simple recursion:
2+^:(3) 2
8
NB. This is equivalent to
2+2+2+2
8
NB. What if I want to change the left argument on each recursion?
NB. Instead of 2 each time, I want to add 2, then 3, then 4.
NB. I know there are much easier ways to do this, but I want to see
NB. if I can avoid do. while. loops if the left argument changes on each
iteration
NB. I tried
(2 3 4)+^:(3) 2
8 11 14
NB. Nope. That didn't work.
NB. What I want to achieve is:
2+3+4+2
11
NB. but do it recursively using ^:
Skip
Skip Cave
Cave Consulting LLC
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