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2016-06-10T00:07:33Z wasamasa: spoiler: for game development you'll want to reuse as many libraries as possible, so unless you're doing text adventures or are jonathan blow, don't invent your own programming language
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2016-06-10T00:10:24Z wasamasa: and if you've been unhappy with the existing implementations targeting the CLR, you'll have even worse tooling (read: none) if you start your own
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2016-06-10T00:12:03Z taylan: noethics: by compiling the expression on the fly
2016-06-10T00:15:34Z noethics: taylan, does that mean that a scheme that conforms to rnrs necessarily has to bundle a compiler with executables it produces?
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2016-06-10T00:20:43Z taylan: noethics: yes
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2016-06-10T00:20:58Z taylan: well, it could statically determine whether they use eval
2016-06-10T00:21:26Z taylan: (and omit the compiler if they don't)
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2016-06-10T00:21:48Z noethics: taylan, i see, thanks
2016-06-10T00:21:59Z taylan: noethics: some implementations have both a compiler and interpreter
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2016-06-10T01:18:33Z makufiru: wasamasa: yeah that's half the reason I'm targeting the CLR so I can use all the monogame libraries too. Just need strong FFI
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2016-06-10T01:39:11Z noethics: makufiru, how about chez?
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2016-06-10T02:38:51Z makufiru @noethics seems interesting in terms of studying the source code, but not having a JIT or the ability to cross-compile with xamarin :/
2016-06-10T02:39:07Z makufiru: i don't know why my message looks like that? lol
2016-06-10T02:40:08Z noethics: makufiru, who needs a JIT when it compiles to native code ._.
2016-06-10T02:40:16Z makufiru: it compiles to C right?
2016-06-10T02:40:21Z noethics: machine code
2016-06-10T02:40:51Z makufiru: oh
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2016-06-10T06:59:40Z webshinra: hello world
2016-06-10T07:00:27Z ecraven: hey webshinra
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2016-06-10T15:24:44Z GreySunshine: Hello all I am Vasanth. I am new to scheme and I need some help debugging some code. Please take a look at this http://pastebin.com/qfrfXtu5
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2016-06-10T15:42:06Z GreySunshine: Well I was able to debug!
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2016-06-10T22:04:32Z n_blownapart: hi this is Binet's formula for calculating fibonacci numbers. I guess the author uses an alias, but what is acc (accumulator?) I don't follow the recursion from line 9-14. please explain - thanks: https://www.refheap.com/120126
2016-06-10T22:05:08Z wasamasa: acc is the traditional name for an accumulator, yes
2016-06-10T22:05:55Z ijp: n_blownapart: if you can understand the usual recursive definition of exponentiation, this one is only slightly different
2016-06-10T22:05:56Z n_blownapart: excellent wasamasa thanks. what about line 14?
2016-06-10T22:06:09Z wasamasa: what about it?
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2016-06-10T22:06:37Z n_blownapart: thanks ijp its at the end why put it there wasamasa ?
2016-06-10T22:06:41Z wasamasa: looks like a fairly standard pattern to recursion
2016-06-10T22:06:52Z ijp: incidentally this is the second time I've seen binet's formula in as many hours
2016-06-10T22:07:13Z wasamasa: well, the thing is that you typically can't use recursion on the function as is to have it tail-recursive
2016-06-10T22:07:14Z n_blownapart: really ijp ? its part of sicp in my case
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2016-06-10T22:07:36Z wasamasa: so you typically write a helper function that takes an accumulator, then invoke it with a neutral starting value
2016-06-10T22:07:47Z wasamasa: in scheme this can be done by nesting a define
2016-06-10T22:08:19Z ijp: you could explicitly pass around the base argument in *^ and pull it out of ^
2016-06-10T22:08:25Z wasamasa: these days you'd probably just use a named let instead
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2016-06-10T22:08:49Z wasamasa: or the appropriate higher-order function
2016-06-10T22:08:56Z ijp: named let isn't always clearer
2016-06-10T22:09:14Z n_blownapart: ok please hold on wasamasa ijp
2016-06-10T22:09:28Z ijp: if you call the recursion multiple times in a form, people expecting a loop get confused
2016-06-10T22:09:47Z n_blownapart: ^ is already defined by scheme, right?
2016-06-10T22:09:56Z wasamasa: nope
2016-06-10T22:09:58Z ijp: we call it expt
2016-06-10T22:10:05Z n_blownapart: ok
2016-06-10T22:10:14Z wasamasa: I have no idea who came up with these silly symbols in your example
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2016-06-10T22:10:59Z ijp: wasamasa: the phi here is the golden ratio
2016-06-10T22:11:25Z n_blownapart: so the acc gets passed around with the recursion ? also what's the difference between a base case and a "neutral " starting point ?
2016-06-10T22:11:26Z wasamasa: I'm speaking of *^
2016-06-10T22:11:30Z ijp: the use of psi is more arbitrary, but psi usually comes with phi for whatever reason
2016-06-10T22:11:49Z ijp: oh right, well sicp wouldn't have
2016-06-10T22:11:57Z n_blownapart: psi makes fairly accurate calculations very accurate.
2016-06-10T22:12:15Z ijp: I have no idea what you could possibly mean
2016-06-10T22:12:48Z n_blownapart: phi works alone, the addition of psi makes the code dead accurate
2016-06-10T22:13:56Z ijp: I'd have assumed it was for alphabetic reasons, if χ wasn't in the way. I presume it is for euphony
2016-06-10T22:14:14Z n_blownapart: I'm a beginner I don't follow that ijp
2016-06-10T22:14:34Z ijp: euphony (n) the quality of being pleasing to the ear
2016-06-10T22:15:30Z n_blownapart: nice ... could you guys walk me through the recursive part? the "neutral case" is confusing, I'm used to reading base cases at the top.
2016-06-10T22:16:13Z n_blownapart: btw these are the best solutions for sicp I've found so far, so I am subject to them for now
2016-06-10T22:16:19Z wasamasa: the inner function almost reads like plain english
2016-06-10T22:16:30Z wasamasa: if the argument is zero, return what's been accumulated
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2016-06-10T22:16:37Z n_blownapart: ok
2016-06-10T22:16:41Z n_blownapart: ..
2016-06-10T22:17:06Z wasamasa: otherwise repeat with the argument decremented by one and the accumulator times the base
2016-06-10T22:17:47Z ijp: working through an example looking at the arguments to the recurrence is probably more helpful than a prose explanation
2016-06-10T22:18:07Z n_blownapart: 9-14 can pretty much function without the phi / psi part , correct?
2016-06-10T22:18:09Z wasamasa: so, assuming that the power (= argument) is a positive number, it will eventually reach zero and the accumulator will be multiplied every time with the base
2016-06-10T22:18:31Z ijp: 9 - 14 doesn't depend on them at all, only fib~ does
2016-06-10T22:18:35Z wasamasa: now all that's needed is to kick it off with suitable starting values
2016-06-10T22:18:57Z wasamasa: for multiplication that happens to be 1
2016-06-10T22:19:24Z n_blownapart: ok thanks kindly. one thing...
2016-06-10T22:19:52Z n_blownapart: something general about binet's formula ....
2016-06-10T22:19:54Z wasamasa: I'd instrument the definition with some print/newline things
2016-06-10T22:20:28Z n_blownapart: like display ? I sort of can use it , wasamasa
2016-06-10T22:20:28Z ijp: it's not a practical algorithm if that's what you are going to ask
2016-06-10T22:21:05Z n_blownapart: ijp the gist of the exercise was to do a formal proof. this is just the code for binet's
2016-06-10T22:21:18Z n_blownapart: but still...
2016-06-10T22:21:21Z wasamasa: just to understand that the recursion happening here is of loop-like nature
2016-06-10T22:21:29Z n_blownapart: cool ok
2016-06-10T22:21:56Z n_blownapart: this is my first exposure to programming except a brief foray with ruby
2016-06-10T22:21:56Z wasamasa: display prints strings with quotes, so not quite like C's printf
2016-06-10T22:22:21Z n_blownapart: C ? no clue
2016-06-10T22:22:34Z n_blownapart: tho I see
2016-06-10T22:22:40Z wasamasa: it's the same in ruby
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2016-06-10T22:23:25Z ijp: afaict "binet's formula" mostly gets used as a 2nd example of how to use generating functions
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2016-06-10T22:24:00Z n_blownapart: ok that 's a lot to digest. here's the thing :
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2016-06-10T22:24:10Z wasamasa: and as a lazy way of generating small fibonacci numbers
2016-06-10T22:24:33Z n_blownapart: wasamasa: you mean not valid for larger nums . ? haven't tried it.
2016-06-10T22:24:43Z wasamasa: n_blownapart: IIRC that was the downside of that method
2016-06-10T22:24:44Z ijp: it's valid, it just won't work on a computer
2016-06-10T22:24:58Z ijp: but this is all background
2016-06-10T22:25:18Z n_blownapart: the first example is the real slow one. you mean that one? I'll post it if you want to see it.
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2016-06-10T22:25:56Z n_blownapart: psi is - 0.618 ....
2016-06-10T22:26:20Z n_blownapart: phi is 1.618
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2016-06-10T22:27:30Z n_blownapart:  at first I thought phi minus psi gave some sort of identity. but its negative to the addition would not create an identity. what is going on with the math here?
2016-06-10T22:27:44Z n_blownapart: so*
2016-06-10T22:27:52Z ijp: wasamasa: you could adapt it by having arithmetic in Q(√5), like jordigh posted in emacs earlier, but it'd probably be about the same efficiency as the matrix exponentiation method
2016-06-10T22:29:31Z n_blownapart: oh I see ..
2016-06-10T22:29:36Z ijp: i.e. [[0,1],[1,1]]^n * [[0],[1]]
2016-06-10T22:30:26Z ijp: spoilers for another sicp exercise btw
2016-06-10T22:31:10Z ijp: n_blownapart: are you familiar with proof by induction?
2016-06-10T22:31:37Z n_blownapart: It's my first experience : http://www.billthelizard.com/2009/12/sicp-exercise-113-fibonacci-and-golden.html
2016-06-10T22:31:38Z rudybot: http://teensy.info/DYJpjQ4BVL
2016-06-10T22:31:54Z n_blownapart: I like spoilers ijp !
2016-06-10T22:32:34Z n_blownapart: hold on I'm getting bogged down. it turns out it is an identity so I have to look closer at the math ..
2016-06-10T22:32:43Z zv: it is described in detail in sicp
2016-06-10T22:32:53Z zv: not just of the fibonacci sequence, but it's generalization as well.
2016-06-10T22:33:08Z n_blownapart: zv really ? that book seems to skip around. that is good news
2016-06-10T22:33:10Z zv: (that is to say, f(n-x)
2016-06-10T22:33:37Z zv: that is to say f(n)+f(n-1)...f(n-x)
2016-06-10T22:33:49Z n_blownapart: yeah the post above has the full proof, but seeing it in code is confusing. my math is weak but I generally follow it.
2016-06-10T22:34:46Z n_blownapart: (f n) + (f (- n 1) = (f (+ n 1)  << that?
2016-06-10T22:35:09Z galex|713: you forgot parenthesis
2016-06-10T22:35:17Z n_blownapart: correction (fib n)
2016-06-10T22:35:25Z n_blownapart: word galex|713
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2016-06-10T22:37:44Z n_blownapart: ((- phi^n psi^n) / sqrt 5 ) is 1 / sqrt 5 , is that right?
2016-06-10T22:38:57Z ijp: that's fib n
2016-06-10T22:42:18Z n_blownapart: that is fascinating. what I cannot grasp is that it resolves to an identity in the numerator. btw , are there more identities than 1 in mathematics?
2016-06-10T22:42:55Z n_blownapart: the definition of an identity is always just 1 , right ?
2016-06-10T22:43:13Z ijp: sure, 0 is the identity for +, "" is the identity for string concatenation, ...
2016-06-10T22:43:25Z ijp: the identity function is the identity for function composition
2016-06-10T22:43:49Z n_blownapart: ijp please explain that if you have time.
2016-06-10T22:44:20Z n_blownapart: or is it a joke. I'm a bit saturated.
2016-06-10T22:44:23Z ijp: there isn't much to say
2016-06-10T22:44:25Z wasamasa: it's basic algebra
2016-06-10T22:44:35Z ijp: (+ 0 x) = x = (+ x 0)
2016-06-10T22:44:49Z ijp: (string-append "" x) = x = (string-append x "")
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2016-06-10T22:46:48Z n_blownapart: so phi^n / sqrt 5 is less accurate for our purposes than the binet scenario ? ( 1 / sqrt 5 )
2016-06-10T22:51:10Z ijp: no, the question is really saying that the (+ psiⁿ / √5) stays within 1/2 in either direction
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2016-06-10T22:52:18Z ijp: so since that term is there to make it an integer, you can instead just round to the nearest integer
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2016-06-10T22:52:49Z n_blownapart: sorry which term is referred?
2016-06-10T22:52:53Z n_blownapart: to
2016-06-10T22:54:23Z ijp: the + ψⁿ/√5
2016-06-10T22:55:06Z ijp: (aside for other people: here is an even more opaque and amusing way to generate the fibonacci numbers https://paulhankin.github.io/Fibonacci/)
2016-06-10T22:55:46Z n_blownapart: you mean - ψⁿ/√5 ? ijp
2016-06-10T22:56:00Z ijp: er, yes
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2016-06-10T22:56:46Z n_blownapart: ok . btw are you using shortcuts for those characters or cutting and pasting?
2016-06-10T22:57:01Z ijp: emacs input method
2016-06-10T22:57:06Z n_blownapart: sweet
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2016-06-10T23:00:02Z n_blownapart: the question is asking to be within 1/2 in either direction , but the author of the solution decided to add the psi business. I haven't read ahead in sicp obviously. I mean, staying within 1/2 , it must be gearing up to give the binet solution or similar, correct ?
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2016-06-10T23:00:25Z n_blownapart: author on 'bill the lizard ' blog meaning..
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2016-06-10T23:05:59Z n_blownapart: cool paulhankin gives the same method 'closed form'
2016-06-10T23:06:05Z ijp: you just need to glue together everything that's been covered so far. The binet formula tells us that fib(n) = (φⁿ-ψⁿ)/√5 = φⁿ/√5 - ψⁿ/√5. |ψⁿ/√5| < 1/2. So fib(n) - 1/2 < φⁿ/√5 < fib(n)+1/2. fib(n) is an integer, so fib(n) is the only integer in that range, i.e. the one closest to φⁿ/√5
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2016-06-10T23:12:32Z n_blownapart: very kind of you, thanks ijp , wasamasa . I need to work on it. more drama later !
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