[erlang-questions] How to understanding recursive inside comprehension lists?

[I Gusti Ngurah Oka Prinarjaya]

Hi,

Wow.. after reading the blog's introduction, i realize, it turns out a very
special beautiful function and it turns out R. Virding is a fans of this
looks-simple-and-beautiful function hahaha. Wow it's amazing and looks
great for me.

Thank you




Pada tanggal Jum, 23 Agu 2019 pukul 21.39 Eckard Brauer <
eckard.brauer@REDACTED> menulis:

> Nice explanation, thank you!
>
> For me, it helps to think of mathematical induction:
>
> First trying to get a model of the base case(s) - occasionally more
> than only one exist.
>
> After that I (try to) think of the step case - how to evolve from base
> or any given case to the next (more complex) one.
>
> The brilliance of the example is to show multiple conditions - one for
> head, one (dependent) for tail - are given the probably elegantest
> possible way.
>
> Eckard
>
> Am Fri, 23 Aug 2019 11:09:42 -0300
> schrieb Brujo Benavides <elbrujohalcon@REDACTED>:
>
> > Since I needed a bit more space to write an answer �� I replied on my
> > blog
> > <
> https://medium.com/erlang-battleground/how-to-comprehend-comprehensions-c924f92a97e1?sk=b668da926f05b7293f34dafa48fa5654
> >
> > �� Cheers :)
> >
> > Brujo Benavides <http://about.me/elbrujohalcon>
> >
> >
> >
> > > On 23 Aug 2019, at 10:47, Jesper Louis Andersen
> > > <jesper.louis.andersen@REDACTED> wrote:
> > >
> > > On Fri, Aug 23, 2019 at 2:03 PM I Gusti Ngurah Oka Prinarjaya
> > > <okaprinarjaya@REDACTED <mailto:okaprinarjaya@REDACTED>> wrote:
> > > Hi,
> > >
> > > Now I read Joe's book titled Programming Erlang 2nd Edition. I
> > > practice some functions such as for/3, quicksort/1, pythag/1, and
> > > perms/1, and perms/1 is the function that hard to understand.
> > >
> > > I understand comprehension lists, I fully understand for/3, I fully
> > > understand quicksort/1, pythag/1. But it's really hard for me to
> > > understand perms/1. Please teach me how to read and understand this
> > > perms/1 function.
> > >
> > > perms([]) -> [[]];
> > > perms(List) -> [ [H|T] || H <- List, T <- perms(List--[H]) ].
> > >
> > >
> > > Note you have two generators in the comprehension. So for each
> > > generated H, it generates all the possible T's. Also note that the
> > > T's depend on the H. It is akin to having a for-loop within a
> > > for-loop:
> > >
> > > for H := range(List) {
> > >   for T := perms(List -- [H]) {
> > >      Res := append(Res, [H|T])
> > >   }
> > > }
> > >
> > > Now, to understand why this works, the argument is based on an
> > > induction principle:
> > >
> > > Observe that to generate a permutation, you first select something
> > > which goes first, H, and then you need to generate a permutation
> > > out of the remaining elements, List -- [H]. Suppose we fix H to
> > > some value in the list. Then surely, we can generate all the
> > > permutations where H goes first, by generating perms(List - [H])
> > > and then attaching H to all the results:
> > >
> > > perms(List) ->
> > >   H = pick_among(List),
> > >   [ [H|T] || T <- perms(List -- [H]) ].
> > >
> > > But now note that to generate all permutations, any element could
> > > have been picked by pick_among/1. So we need to loop over all
> > > elements in the list one at a time, and consider what happens if
> > > that element goes first. This is what the original code is doing.
> > >
> > > Alternative wordings by Sverker and Fred :)
> > >
> > > _______________________________________________
> > > erlang-questions mailing list
> > > erlang-questions@REDACTED
> > > http://erlang.org/mailman/listinfo/erlang-questions
> >
>
>
>
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