Hello Clojure experts,
I am trying to do some timing tests in Clojure 1.3 and I thought I'd ask a question based on an existing piece of code that solves a differential equation that is adapted from this blog post.
Code follows:
;; the differential equation is
;; dy/dt = f(y,t) = t - y
(defn f [t y] (- t y))
;; solve using Euler's method
(defn solveEuler [t0 y0 h iter]
(if (> iter 0)
(let [t1 (+ t0 h)
y1 (+ y0 (* h (f t0 y0)))]
(recur t1 y1 h (dec iter)))
[t0 y0 h iter]))
(defn multipleSolveEuler []
(let [steps '(1 10 100 1000 10000 100000)
results (map #(second (solveEuler 0.0 0.0 (/ 1.0 %) %)) steps)
errors (map #(- (Math/exp -1) %) results)]
(partition 3 (interleave steps results errors))))
(def *cpuspeed* 2.0)
(defmacro cyclesperit [expr its]
`(let [start# (. System (nanoTime))
ret# ( ~@expr (/ 1.0 ~its) ~its )
finish# (. System (nanoTime))]
(int (/ (* *cpuspeed* (- finish# start#)) ~its))))
(defn solveEuler-2 [t0 y0 h its]
(let [zero (int 0)]
(loop [t0 (double t0), y0 (double y0), h (double h), its (int its)]
(if (> its zero)
(let [t1 (+ t0 h)
y1 (+ y0 (* h (- t0 y0)))]
(recur t1 y1 h (dec its)))
[t0 y0 h its]))))
So when I say
(time solveItEuler-2 0.0 1.0 (/ 1.0 1000000000) 1000000000))
I get a time 6004.184 msecs on a 6 month old Macbook pro. I issue the command again, and I get around the same time. But when I run it 3 other times I get times in the range of 3500 msec. I have noticed this before for other code snippets, and was wondering why this is so. I suppose I would expect a roughly similar execution time across successive runs.
Is it my lack of understanding how "time" works, something I am missing, or some kind of caching happening under the hood?
Thanks.