Comments on: I think my brain just exploded

By: kshitij

kshitij — Tue, 26 Feb 2008 10:47:17 +0000

for parselmouths:

filter(lambda (x, y, z): x*x + y*y == z*z, [(a, b, c) for a in range (1, n+1) for b in range (a, n+1) for c in range(b, n+1)])

obviously, the haskell and python versions are idea-wise isomorphic.

and here’s a kind of O(n^2) version . the range for the multiplying factor (mu) is off the top of my head, and is there to take the output from the p-q stuff (eg (3,4,5)) and get the derived triplets (eg (6,8,10) and (12,16,20)). i say “kind of” O(n^2) because the generation of unique triplets is O(n^2) at minimum, and multiplying each of the integers in the triplet to get all the derived triplets is an additional operation. i don’t have the upper bound for the multiplying factor, though if it can be proved that such a bound exists and is independent of n, the algorithm would become strictly — as opposed to “kind of” — O(n^2)

filter(lambda (a, b, c): max(a, b, c) <= n, [(p*q*mu, ((p**2 - q**2)/2)*mu, ((p**2 + q**2)/2)*mu) for q in range(1,n,2) for p in range(q+2,n,2) for mu in range(1, 5)])

By: DavidLG

DavidLG — Mon, 31 Dec 2007 10:44:37 +0000

Replying to the original post, you can do the same thing in Scala:

def f(n:Int) = for(x <- 1 to n; y <- (x+1) to n; z <- (y+1) to n; if x*x + y*y == z*z) yield (x,y,z);

The Scala compiler infers the return type of the function, Seq[(Int,Int,Int)], and compiles the code into statically-typed JVM bytecodes.

By: Brian

Brian — Mon, 31 Dec 2007 01:26:23 +0000

Spruce Moose: actually, I’m not working on Project Euler. This was something different.

By: Brian

Brian — Sun, 30 Dec 2007 23:17:05 +0000

Programming is a bad choice for people who are bad in math stuff, IMHO.

By: Spruce Moose

Spruce Moose — Sun, 30 Dec 2007 22:43:12 +0000

Oh man - spoilers for Problem 9 of Project Euler or what. (-: I’m working on that one at the moment, but using Erlang.

By: she

she — Sun, 30 Dec 2007 06:55:24 +0000

Hmm… unfortunately i am not good at math, but i hear more praise about haskell than ocaml. The question still remains however, aren’t both bad choice for people who are bad in math stuff?

By: Justin Dubs

Justin Dubs — Sun, 30 Dec 2007 03:24:18 +0000

Here is a slight tweak of your haskell code to make it generate ALL of the pythagorean triples, in order of hypotenuse length:

pythagoreanTriples :: [(Int, Int, Int)]
pythagoreanTriples = [(a, b, c) | c <- [1..], b <- [1..c-1], a <- [1..b-1], a^2 + b^2 == c^2]

Prelude> take 5 pythagoreanTriples
[(3,4,5),(6,8,10),(5,12,13),(9,12,15),(8,15,17)]

By: Justin Dubs

Justin Dubs — Sun, 30 Dec 2007 03:20:29 +0000

Here’s a slight tweak of your haskell code to generate ALL of the pythagorean triples, in order of hypotenuse length:

pythagoreanTriples :: [(Int, Int, Int)]
pythagoreanTriples = [(a, b, c) | c <- [1..], b <- [1..c-1], a take 5 pythagoreanTriples
[(3,4,5),(6,8,10),(5,12,13),(9,12,15),(8,15,17)]

By: ChriS

ChriS — Sun, 30 Dec 2007 03:10:15 +0000

In passing, let me mention that the monadic version which looks quite similar too (even more so with the do notation not shown here):
let f n = range 1 n >>= (fun x -> range (x+1) n >>= (fun y -> range (y+1) n >>= (fun z -> if x * x + y * y = z * z then return(x,y,z) else [] )))
(the >>= and return are the standard monadic operators on lists — see http://enfranchisedmind.com/blog/2007/08/06/a-monad-tutorial-for-ocaml/).

By: James

James — Sun, 30 Dec 2007 03:04:03 +0000

Funny, I just did this today for Project Euler (I’d already solved the problem using smarter methods than this, but I was curious as to how to do this with list comprehensions). While it was too slow for that problem, it’s great how a one-liner like this can do so much.