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
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
Interesting!
newfib=: 13 :' }:}.(,[:+/_2&{.) ^: y (0 1)'
newfib 10
1 1 2 3 5 8 13 21 34 55
Linda
-Original Message-
From: Programming On Behalf Of
The3DSquare jOsHUa
Sent: Saturday, July 20, 2019 6:11 PM
To: programm...@jsoftware.com
Subject: [Jprogramming] Question about
I don't think so, but I can offer
4 (,[:+/_2&{.)@] ^:[ 0 1
0 1 1 2 3 5
Henry Rich
On 7/20/2019 6:11 PM, The3DSquare jOsHUa wrote:
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. T
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 ^:.
Than
At long last -
I've only just managed to find some further solutions!
I generate triples of P5s,
such that P5(a) + P5(b) = P5(c) for triple (a,b,c)
and check for which of them P5(d) = P5(a) + P5(c) for an integer d:
|:Ptriples 30
4 7 12 19
7 23 22 22
8 24 25 29
Here are some timings on my la
FWIW, here’s my code for changing back to decimal representation.
z2d =: 3 : 0
if. _1 < <./ y do.
+/ #/ y join Fb
else.
+/ */ y join Fb
end.
)
join =: 3 : 0
join/ y
:
l =. - ({:@$x) >. {:@$y
r =. (l{."1 x),: l{."1 y
if. 2<#$r do.
(l{."1 x), l{."1 y
end.
)
eg
d2z 100
1 0 0 0 0 1
http://www.ramanujanmachine.com
R.E. Boss
> -Oorspronkelijk bericht-
> Van: Programming
> Namens Devon McCormick
> Verzonden: vrijdag 19 juli 2019 16:41
> Aan: J-programming forum
> Onderwerp: [Jprogramming] Program to replicate Ramanujan's methods
>
> Fellow Jedi -
>
> There's a Ne
"In other words, it's not the fibonacci part that's the problem for me,
it's whatever you are using to convert to ascii that's the problem.
(Note also that there's no need to add any 1s into the represented
result - you just terminate the sequence by eliminating what would be
trailing ascii nul c
"You mention the possibility of 1 1 1 appearing in an encode result and that
making decode awkward without a loop. My guess is that that can happen if
you encode multiple numbers into a single result where two adjacent
encodings produce the 1 1 1. Is that right?"
This is correct. Simplest exampl
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