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2015-12-16T15:28:18Z LeoNerd: Hmm?
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2015-12-16T15:34:16Z lupine: i've been programming professionally for a decade and still don't know a lisp. so, yeah :D
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2015-12-16T16:20:30Z jackdaniel: lisp is a state of mind, not the programming language ;]
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2015-12-16T16:41:45Z oleo: jaaa, i enjoy the borg collective too from time to time......
2015-12-16T16:41:49Z oleo: lol
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2015-12-16T16:45:53Z cousteau: how is a vector different from a list?  is it more like a C array?
2015-12-16T16:46:16Z LeoNerd: Yes
2015-12-16T16:46:30Z LeoNerd: A vector has O(1) access time for any element; but also like a C array it is a fixed length once constructed
2015-12-16T16:46:54Z LeoNerd: You can mutate the individual slots in a vector, but the length of it remains constant
2015-12-16T16:47:09Z cousteau: I see
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2015-12-16T16:47:38Z cousteau: so it's an array with no fixed type (I guess scheme is kinda flexible with types in general)
2015-12-16T16:47:49Z LeoNerd: Indeed
2015-12-16T16:50:46Z cousteau: ...damn, I'm having trouble getting all this list and pair thing.  I know it should be trivial but for some reason I fail to comprehend it.
2015-12-16T16:51:34Z cousteau: is a dotted pair conceptually different from a vector of length 2?
2015-12-16T16:51:39Z LeoNerd: A pair is a blob that has two little things inside it
2015-12-16T16:51:45Z LeoNerd: Not really... it could be implemented in a similar way
2015-12-16T16:52:00Z LeoNerd: The language would have to know which it was though, because  (vector? (cons 1 2))  has to be false
2015-12-16T16:52:03Z davexunit: cousteau: they are fundamentally different structures, yes.
2015-12-16T16:52:11Z LeoNerd: But conceptually, a cons feels quite similar to a two-element vector
2015-12-16T16:52:16Z davexunit: sure
2015-12-16T16:52:58Z davexunit: but the similarities no longer hold when you change N
2015-12-16T16:53:38Z cousteau: ...so maybe like an int[2] vs a struct {int a; int d;}?  (although those 2 are conceptually equivalent, even if they have different properties and methods)
2015-12-16T16:54:12Z davexunit: sure something like that
2015-12-16T16:54:23Z cousteau: davexunit, well yeah, but maybe you could say a pair is a subset of a vector where N=2
2015-12-16T16:54:43Z cousteau: (you could also say that an object in C is an array of 1 element... actually in some contexts it is)
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2015-12-16T17:07:19Z davexunit: yes that works
2015-12-16T17:07:37Z davexunit: in practice, it would be silly to use a vector where a pair would do
2015-12-16T17:07:52Z cousteau: ok, makes sense
2015-12-16T17:08:24Z davexunit: vectors are an inherently imperative data structure
2015-12-16T17:08:36Z davexunit: whereas pairs form persistent structures
2015-12-16T17:08:39Z davexunit: like lists
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2015-12-16T17:13:45Z cousteau: hm, this tutorial seems reluctant to (define (foo x y z) ...) and instead insists in (define foo (lambda (x y z) ...))
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2015-12-16T17:13:47Z cojy: vector has to store a length so they aren't equivalent even when n=2
2015-12-16T17:15:12Z davexunit: it all depends on what we define sameness to be here
2015-12-16T17:15:21Z davexunit: we've been talking more on a conceptual level
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2015-12-16T17:15:38Z davexunit: you *could* use a vector of length 2 in situations where you would use a pair
2015-12-16T17:15:49Z f-a: http://paste.lisp.org/display/219851 I have got this `factorise` function, which chokes on non-small numbers. What's the trick to speed it up?
2015-12-16T17:18:24Z f-a: or if that's too difficult to read, what's a good, efficient (possibly clearly writen)factorisation in scheme?
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2015-12-16T17:21:28Z wasamasa: cousteau: what if I told you that in elisp, a cons is indeed a two-element C array
2015-12-16T17:24:16Z cousteau: then I'd be very happy :)
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2015-12-16T17:25:04Z wasamasa: it's an implementation detail though
2015-12-16T17:25:12Z cousteau: yeah, got that :)
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2015-12-16T17:27:18Z wasamasa: f-a: memoization would be such a trick
2015-12-16T17:27:30Z wasamasa: f-a: keep in mind though that it only works for pure functions
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2015-12-16T17:28:40Z f-a: thanks wasamasa
2015-12-16T17:28:52Z wasamasa: f-a: the prime candidate (sorry for the pun) would be the prime number generation
2015-12-16T17:29:39Z wasamasa: after all, once you know the nth prime number, you don't need to calculate it again
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2015-12-16T17:33:32Z f-a: indeed I see what you say and it should work perfectly, I'll modify the code and report back if I have further problems
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2015-12-16T23:07:37Z f-a: I am a bit puzzled by memoization
2015-12-16T23:07:48Z f-a: https://codeimmersion.i3ci.hampshire.edu/2009/10/09/memoization/ look at the `memoize` function
2015-12-16T23:08:30Z f-a: inside it there is a   (set! memo ..etcetc..
2015-12-16T23:10:07Z f-a: but memo is defined inside `memoize` and inside a `let`. Wouldn't it, err, die (get garbage collected?) once the function is over?
2015-12-16T23:11:37Z f-a: so say I call (memoize 8) at the beginning of my program, and them same call much later on, where is the memoised result of that call exactly stored?
2015-12-16T23:14:36Z f-a: I would have tohught of first creating a global variable (a dictionary) and then populating it
2015-12-16T23:16:15Z pjb: f-a: the function memoize returns a function (lambda args …).  This function has a free variable: memo.  This free variable is found in the lexical environment around this (lambda args …) function.  Therefore scheme will ENCLOSE the free variable memo with the procedure (lambda args …) so that the free variable is bound at all time, when the procedure is called.
2015-12-16T23:16:22Z pjb: f-a: in short, you have a closure.
2015-12-16T23:17:02Z pjb: A closure encloses a lexical environment with a procedure having free variables, such that those free variables are bound in the enclosed lexical environment.
2015-12-16T23:17:28Z pjb: Then the closure can be called at anytime, and the lexical environment will be present.
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2015-12-16T23:18:12Z pjb: Notice that f, parameter of memoize is also enclosed, as well as assoc cadr apply cons list (since they are all free variables).
2015-12-16T23:22:19Z f-a: thanks pjb
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2015-12-16T23:23:22Z pjb: f-a: notice that when implementing closures, Sussman realized that it was exactly the same code as to implement actors (objects).
2015-12-16T23:23:48Z pjb: f-a: closures and objects are equivalent, enclosing both data and code in a single entity.
2015-12-16T23:24:14Z f-a: pjb: no idea what actors or objects are though (the only language I know is haskell) :P
2015-12-16T23:24:50Z f-a: but I think I get the point
2015-12-16T23:24:51Z pjb: ok. From haskell, you already have closures, just like in maths, without realizing it.  Let's say that closures just implement the normal math expectations.
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2015-12-16T23:25:24Z f-a: pjb: I thought I know what closures where, but I never saw something like this function
2015-12-16T23:25:36Z f-a: of course, there is no set! in haskell
2015-12-16T23:26:00Z pjb: But even without sets, you can have closures, just by having embedded let in haskell.
2015-12-16T23:26:02Z lupine: I was wondering about the whole mutable thing
2015-12-16T23:26:12Z lupine: isn't that normally a bit of a no-no in functional langs?
2015-12-16T23:26:25Z pjb: Mutation is secondary here.
2015-12-16T23:26:35Z lupine: sure, I'm at a tangent
2015-12-16T23:26:41Z pjb: Well, scheme and lisp is not "purely functional".
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2015-12-16T23:28:00Z f-a: pjb: what I meant was that f x = (\y -> x + y) <-- this I see and understand easily (should be a closure, no?)
2015-12-16T23:28:23Z pjb: definitely, since x is a free variable of the lambda form.
2015-12-16T23:28:50Z f-a: while with scheme memoisation there was a step more in my eyes, the dictionary/hash stored somewhere
2015-12-16T23:28:54Z pjb: f 42 will return a closure with an environment where x is bound to 42.
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2015-12-16T23:31:17Z pjb: f-a: granted, the difference is that now you have mutation with set!.  But the variable x in your haskell function, or the variables f and memo in memoize are the same.
2015-12-16T23:31:48Z pjb: f-a: compare f and memo: f is not mutated, but memo is mutated. Both are enclosed in the closure.
2015-12-16T23:33:05Z pjb: (define s (memoize sin)) (define c (memoize cos)) (s 4) (c 4) (s 4) will show that you have two variables f, one in s and one in c, since s compute sin while c computes cos.
2015-12-16T23:33:32Z pjb: Each closure has its own f enclosed.
2015-12-16T23:33:39Z f-a: yeah
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