clojure Fibonacci analytical column a solution of

fib problem recently made a sequence of http://www.4clojure.com/problem/26 4Clojure

They used the


#(take % (map first (iterate (fn [[a b]] [b (+ a b)]) [1 1])))

After seeing a finished answer it is used in this way


#((apply comp (repeat (- % 2) (fn [x] (conj x (+ (peek x) (peek (pop x))))))) [1 1])

Read a long time came to understand how it is down here recorded

Here the code apart to similar


(def compute (fn [x] (conj x (+ (peek x) (peek (pop x))))))
(def fn-seq (repeat (- n 2) compute)
(def fib (apply comp fn-seq))
(fib [1 1])

First of all


(fn [x] (conj x (+ (peek x) (peek (pop x)))))

This end view of a single function is better understood, that the tail of the vector adding two records, x is added to the


(repeat (- % 2) (fn [x] (conj x (+ (peek x) (peek (pop x))))))

Repeatedly generating the (- 2%) that function defined later times, the same returns a seq, fn


(apply comp fn-seq)

Is to fn-seq as an argument, call the comp, to generate new fn, fn this role is kept calling the fn compute defined above

Finally, call fn, passing parameters [11]
When n = 3 when
Will (compute [1 1]) => [1 1 2]
when n = 4
(compute (compute [1 1])) => [1 1 2 3]

But this solution still has a bug
n <= 1 when the return or [11]
Anti-repeat back empty seq ()

Looked comp source code, will return identity parameter is null
This return directly [11]

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