Hi,
extending Kacper's idea, you could also try:
(X/X)←∈(⍴¨X⊂X)⍴¨⊂1 0
It is a bit longer but may be faster because it avoids \.
/// Jürgen
On 03/20/2014 08:08 PM, Kacper Gutowski wrote:
On 2014-03-21 00:23:41, Elias Mårtenson wrote:
One of my experiments was very short, and tantalisingly close to correct:
≠\V
Unfortunately, it's not quite right. Does anyone have an idea how to coerce
that one into being right?
I couldn't come up with anything better, but if you don't mind having a
separate statement with assignment in place you could do something like:
(V/V)←∊≠\¨V⊂V
-k
