(declare fibo-m)

(defn fibo [n]
  (if (<= n 2)
    1
    (+ (fibo-m (- n 1)) (fibo-m (- n 2)))))

(def fibo-m (memoize fibo))

;; ---------------------------------------------------------
;; the (fibo-m 35) should NOT be slow, it has alreay been
;; calculated by the previous call to (fibo-m 36)
;; what would be the idiomatic clojure way to apply memoize
;; to fibo such as the recursives call inside fibo body get
;; memoized as-well ?
;; ---------------------------------------------------------

;; for comparaison i put the working equivalent in python:
def memo(f, cache={}):
    def caching_func(n):
        if n not in cache:
            cache[n] = f(n)
        return cache[n]
    return caching_func

@memo
def fibo(n):
    """
    i know it's not an efficient way to do fibo
    it's just made for example purpose
    """
    if n <= 2:
        return 1
    return fibo(n - 1) + fibo(n - 2)

######################
import time
def time_it(f, *args):
    start = time.clock()
    print f(*args)
    return (time.clock() - start)

print time_it(fibo, 500)
# Elapsed time: 0.004
#= 139423224561697880139724382870407283950070256587697307264108962948325571622863290691557658876222521294125

print time_it(fibo, 499)
# Elapsed time: 0.0
#= 86168291600238450732788312165664788095941068326060883324529903470149056115823592713458328176574447204501