[erlang-questions] RV: Joe (final)
Ivan Carmenates García
Wed Mar 6 00:30:27 CET 2013
De: Ivan Carmenates García [mailto:co7eb@REDACTED]
Enviado el: martes, 05 de marzo de 2013 18:28
Para: 'Danil Zagoskin'
Asunto: RE: [erlang-questions] Joe (final)
Hi, Danil, Joe, Richard,
Well I think I finally did understand. The most tricky part was the swapping of the values.
Here’s my understanding…
First I came up with an idea I tried this (it was before reading the Richard’s email lol)
[[H,T] || H<-[1,2,3], T<-[1,2,3]]
and I could see that list of comprehension behave like nested for loops in imperative programming because it generate pair of numbers like this
So the most outer loop (the first generator) iterates only when the most inner loop (lasted generator) does the complete cycle. Now what list of comprehension does is that combines each item of the outer loop (first generator) with all the items of the second generator (inner loop) that’s why the head remains in many steps, like Joe says that I can stick for example the number 1 in each permutation of the rest of the list, and so on with the rest of the items in the list.
perms() -> [];
perms(L) -> [[H|T] || H <- L, T <- perms(L--[H])].
[ 1 | [ 2 | [ 3 | [ 4 | [] ] ] ] ] here the first recursive journey ends by returning an empty list within a list []. So the value [] is yielded
Now the rolling back stands in the call to perms([3,4]) where the [[3,4]] list is emitted, [ H|T || H<-[3,4], T<-perms() (this yield [] ) ] (this yield [[3,4]])
So now the first generator has to advance to the second value in the list [3,4], because no more recursive call are made until now, and the second generator
only has one value (4) to yield, so the next value to yield for the first generator in the list [3,4] is 4, so H = 4, T = perms() (which yield []) leading to the
list L’ = [[4, 3]] here is where the swap is made, that’s the part that I couldn’t understand before. So because there is no other value to retrieve in H from the
list [3,4] the evaluation of the comprehension list the and call to the function perms([3,4]) has ended yielding the list [[3,4], [4,3]] and rolling back to the call
of perms([2,3,4]) and so on…
I’m alright? At least a fifty percent?
Danil’s last debug info really helped me to understand the last part.
So Thanks all of you, for your help and time, uff it was really hard but finally I could archive knew knowledge that I hope will never forget.
… well that is if my understanding was at least fifty percent right! Lol.
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