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Looking for people who know about semiring geometry to help me with a project

Haskell on Reddit - Thu, 05/01/2014 - 1:48am

I've taken an interest in an area of math known by the name tropical geometry. It's sort of new, been around about ten years (as far as algebraic geometers are concerned).

Anyway, I digress. I've been trying to do research in the area, and it occurred to me that it would really help if I could efficiently perform tropical calculations, and make tropical graphs. So, I've been working on a Haskell library to do so. You can look at what I have so far on GitHub (it's not much).

I'm at the point where I would like to make certain major decisions about the project, but I don't have the foresight nor the expertise to do so, and I'm looking for someone who does.

Any insight, advice, criticism, what-have-you, would be greatly appreciated.

submitted by l1cust
[link] [9 comments]
Categories: Incoming News

LYAH solveRPN code query

Haskell on Reddit - Wed, 04/30/2014 - 10:58pm

Hi I have a query regarding the solveRPN code in The book LYAH , specifically the last line of the complete code which can be found here (, I do understand if read x:[1,2,3] would result in the code being trying to read x as an Int , but why should 'read x:[]' be read as an Int

submitted by nerorevenge
[link] [2 comments]
Categories: Incoming News

Can GHC do purity-to-mutability rewriting?

Haskell on Reddit - Wed, 04/30/2014 - 2:36pm

Let's say you have some record r and you do r { myfield = newvalue }.

With immutability semantic a new value r2 will be created with myfield being set to the newvalue and all other fields being copied over from r; if not shared, r will then get garbage collected at the next occasion (usually very quickly in the next minor GC).

Does there exist some kind of optimisation that recognizes that if r isn't shared (used anywhere else), we could not do the whole allocating and copying and modify myfield in-place?

If yes, how is this optimisation called?

How much of this can GHC already do?

If it cannot do this already, how difficult do you estimate such an analysis to be and what kind of architecture needs to be in place? My first guess is that some kind of liveness analysis should be sufficient.

Note that I'm not talking about the more specific forms of strictness analysis where, say, a tuple (Int, Int) is completely eliminated; I mean precisely the general idea of updating records in-place when there is no other consumer.

submitted by nh2_
[link] [22 comments]
Categories: Incoming News

Douglas M. Auclair (geophf): 'Z' is for the ζ-calculus

Planet Haskell - Wed, 04/30/2014 - 11:32am

'Z' is for ζ-calculus  (pronounced 'zeta-calculus')
So, 'Z' is the last letter of the alphabet, unless you're Greek, then ζ is the sixth letter. The last letter of the Greek alphabet is Ω.
But we already did Ωm-Ωm-Ωm. Ram-Ram-Sita-Ram-Sita-Ram-Ram, so we're good there.
As α denotes the beginning of things, Ω (which on a Mac, you get by selecting option-Z. Fitting, that) denotes the end of things. That's why everybody back then got it when Jesus said: "I am the α and the Ω."
He was the package deal.
But look at you all, doing your A-to-Z challenge faithfully. Did you survive? Are you shell-shocked?
To put this into perspective, my mom was a journalist for local news for our local newspaper. She did the A-to-Z challenge, every day (except Sundays), every month, year-after-year.
So you've completed your A-to-Z challenge. Congratulations! Now, lather, rinse, repeat, every month, for the rest of your life, right?
And you thought one month was hard?
Okay, last entry for my A-to-Z guide book for surveyists, explorers and tourists of mathematics.
Okay, so the lambda calculus is the commingling of the κ-calculus (the contextual calculus) and the ζ-calculus (the control calculus), which we talk about in this post (disclaimer/citation: I get most of my information on the ζ-calculus from Hasegawa's seminal 1995 paper on the κζ-calculi). A good deal of attention, when it has been given at all, has been focused principally on the κ-calculus...
'And why not!' says most. In the κ-calculus you get things done. There's no application, but through composition of (first-order) functions and first-order tuple-types you have numbers, counting, addition, and all the rest come out of that.
With the control calculus you make functions that take functions as arguments.
τ : 1 | ((—) ↠ (—))
So reduction (to your answer) is of the form of the continuation-passing form, for a function in the ζ-calculus f : a → b is of the ζ-term form of:
ζ f : a → (c ↠ b)
Which is to say, the zeta-term, f, take a functionally-typed argument, a, and returns the continuation (c ↠ b).
Now, we got the κ-calculus to work by lifting function arguments to tupled-typed terms, and we saw that with the implementation of addition (here). Now, instead of lifting unit types to tuples, in the ζ-calculus we pass a function on x to the zeta-term by using functional composition, i.e.:
pass(c) o (ζ x . f) ~> f[c/x]
To be able to construct a function in the ζ-calculus we have the function-constructor, code:
given f : c → d and x is the 'constifying function' : 1 → cwe have f o x : 1 → d
from that we have code:
code(f) ≡ ζ x . (f o x) : 1 → (c ↠ d)
Boom! code, then, is a function that creates a function from nothing (the '1' type).
But, okay, what can you do with the ζ-calculus?
Well, since we don't have unit cartesian types, like we have in the κ-calculus, then the answer to that is, well, ... nothing, really.
I mean, it's possible to have function represent unit types then start composing them to build an algebra, but this approach is rather unwieldy. The ζ-calculus exists to show that the λ-calculus is perfectly decomposable into the κ-calculus and the ζ-calculus, and so control is usually mapped out in the ζ-calculus (although some directed flow is possible in the κ-calculus, itself alone, as we've seen) (link).
For example, to create the numbers zero and one are possible in the ζ-calculus, but when we try to represent two (that is, functionally, double application), we get this:
x : 1 → cz : 1 → (c ↠ c)  pass(x) : (c ↠ c) → c  pass(x) o z : 1 → c  -- so we've got our constant continuation (step 1) pass(pass(x) o z) : c ↠ (c → c) -- now we've got a continuation-creator pass(pass(x) o z) o z) : 1 → c ζx (pass (pass(x) o z) o z) : 1 → (c ↠ c) -- continuation-creator for the right-hand sidedupl ≡ ζz . ζx (pass (pass(x) o z) o z))) : 1 → ((c ↠ c) ↠ (c ↠ c))
And there we have it, the dupl(icate) function, or the number two, in the ζ-calculus.
Well, you did say you wanted to see it. I did, too. It's a morbid fascination of mine, watching this train wreck transpire.
The Takeaway here
But that's a reveal. Math isn't necessarily easy (shocker!) nor is it necessarily neat. Here we see in the ζ-calculus that it's difficult to express things (like, for example, the number two, or any unit types).
So, if you've got a correspondence (a provable one), all you have to do is to change categories to one where you can express what you need to simply and cleanly.
The application works here, and now, too.
I'm doing things slow and stupid, because I'm doing them the way I've always done them, the way I see how they're done.
Change categories. Somebody is doing what I'm doing, but doing it well. Somebody isn't doing what I'm doing, because they've moved onto better things — more sublime problems — ... I can do the same thing. All I have to do is change my perspective, how I see things, and then, seeing them anew, I can do what I need to do, what I want to do, simply and cleanly, then translate back down to the way I always do things, ...
Or hey, just move on, after having moved up.
It's called a 'lifting function.' It lifts an object from one category to another one, and in being lifted, the object is changed, and what it can do (its morphisms) are changed. And you can change it back 'down' (the co-ajoint function) or you can stay in the newly-lifted category, or you can lift again.
Math is neat that way, once you see that as a possibility.
Life can be like that, too.
Life is neat that way, once you see that as a possibility.
'Z' is for the ζ-calculus.
And with 'Z' we're all done doing our A-to-Z petit survey of mathematics and logic. I hope you had fun, for I had a blast writing these posts.
Thank you for reading them.
Categories: Offsite Blogs

Hylomorphisms and Treesort

Haskell on Reddit - Wed, 04/30/2014 - 9:30am
Categories: Incoming News

The beauty of Haskell

Haskell on Reddit - Wed, 04/30/2014 - 4:38am
Categories: Incoming News

ELI5: How do you write complete complex applications with Haskell?

Haskell on Reddit - Wed, 04/30/2014 - 3:38am

Hi folks,

Every time I look at functional programming languages (I've been a software developer for about two decades now but have never made more intense contact to functional languages) I wonder how you actually write applications with them? All examples that I have seen yet just deal with specific problems like sorting, searching, data aggregation and mathematical problems. But I haven't seen any real complete application yet that has a UI and all that fancy stuff an application consists of.

So are functional languages like Haskell only suitable for solving specific problems and algorithms and are other paradigms necessary to build the actual application around the functional algorithm? Or am I getting something completely wrong here?

Can someone please explain this to me? I'll be very grateful!

Cheers, Hendrik

submitted by SeveQStorm
[link] [9 comments]
Categories: Incoming News

The Identity monad trick

Haskell on Reddit - Wed, 04/30/2014 - 2:27am
Categories: Incoming News - Wed, 04/30/2014 - 1:46am
Categories: Offsite Blogs

Question on list comprehension

Haskell on Reddit - Wed, 04/30/2014 - 12:08am

I'm reading Hitchhiker's guide to Haskell, and chanced upon this list comprehension

let solve x = [ y | x <- [0..], y<-[0..], y == x * x ]

I tried running this on ghci and it doesn't evaluated, neither does

take 10 $ solve 16

My understanding is that the

y <- [0..]

does not ever end, hence the expression isn't evaluated

my second confusion is that below the list comprehension the author mentions

Yes, I know that finding a square root does not require list traversals,

which seem to suggest that the list comprehension above is used to find square root, but to me it looks like it wants to find the squares instead of the square root. Is that right?

submitted by ngzhian
[link] [8 comments]
Categories: Incoming News