W M
I suppose in my ODE example it's building up expressions somewhere for lazy evaluation,
Exactly right. The trick is spotting which expressions. Until you have some experience of likely causes, rather than guessing I can recommend studying some heap profiles to learn what kind of values are building up - e.g. are they closures for 'integrate' or for '+'? Here are a couple of simple changes that seem to fix your space leak.
iterate2 f x n = if n == 0 then x else iterate2 f (f x) (n - 1)
Try x `seq` iterate2 f (f x) (n - 1) to ensure that the accumulator is holding simple values rather than closures. Also, currently you are building pairs lazily. Making a strict pair type helps: data Pair a = P !a !a
integrate step f xInit xFinal yInit nSteps = let h = (xFinal - xInit) / fromIntegral nSteps next = step f h in iterate2 next (xInit, yInit) nSteps
iterate2 next (P xInit yInit) nSteps
rk4Next f h (x, y) = let a1 = h * f x y a2 = h * f (x + h/2) (y + a1/2) a3 = h * f (x + h/2) (y + a2/2) a4 = h * f (x + h) (y + a3) in (x + h, y + a1/6 + a2/3 + a3/3 + a4/6)
P (x + h) (y + a1/6 + a2/3 + a3/3 + a4/6)
rk4 = integrate rk4Next
stiff x y = -( ((exp (x - 1)) + (x - 1) * y) / (1 - x - 0.0001 * y) )
main = do putStr "Begin\n" putStr (show (rk4 stiff 0 1 (exp (-1)) 1000)) putStr "\n" putStr (show (rk4 stiff 0 1 (exp (-1)) 10000)) putStr "\n" putStr (show (rk4 stiff 0 1 (exp (-1)) 100000)) putStr "\n"