The key is that the operation (x u^:v y) is a dyad, so both arguments go
to both u and v. The function of ] [ is to DISCARD an unwanted argument.
x stuff@] y is the same as stuff y
x stuff@[ y is the same as stuff x
Thus
x stuff@] ^: [ y is
(x&(stuff@])) ^: (x [ y) y is
(stuff ^: x) y
The tacit language is intoxicatingly powerful. If you just want to get
a result, you can do the same thing with
fib =: 3 : '(,+/@(_2 _1&{)) ^: y 0 1'
fib 4
0 1 1 2 3 5
Henry Rich
On 7/20/2019 10:40 PM, The3DSquare jOsHUa wrote:
Thanks Henry and Linda
I came up with this:
fib =: (,+/@(_2 _1&{))@]^:[&0 1
fib 8
0 1 1 2 3 5 8 13 21 34
...which works quite well.
With some rearranging, I got the two inputs on the right hand side to
switch after looking at Henry's suggestion.
Also, would you mind explaining the *]^:[* in your code, Henry? The
left/right brackets confuse me and I don't quite
understand how it moves one of the right hand side argument to the left
hand side of the conjunction.
Thanks again!
On Sat, Jul 20, 2019 at 7:08 PM Linda Alvord <[email protected]>
wrote:
Interesting!
newfib=: 13 :' }:}.(,[:+/_2&{.) ^: y (0 1)'
newfib 10
1 1 2 3 5 8 13 21 34 55
Linda
-----Original Message-----
From: Programming <[email protected]> On Behalf Of
The3DSquare jOsHUa
Sent: Saturday, July 20, 2019 6:11 PM
To: [email protected]
Subject: [Jprogramming] Question about ^:
Hello, I'm new to this forum. However, I have a quick question. The
conjugation ^: is represented as u ^: n y. Is there any way to make it like
u ^: y n?
Ie. Turning this:
(,[:+/_2&{.) ^: 4 (0 1)
into this:
(,[:+/_2&{.) ^: (0 1) 4
So essentially, swapping the two right-hand inputs for ^:.
Thanks!
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