Clojure Koan Factorial Function Implementations

Question

I'm learning clojure and going through the clojure-koan exercises. One of the exercises is about implementing a factorial function. While I was playing around I noticed:

My recursive implementation seems to work for large numbers:

(defn factorial-1 [n]
   (loop [n n f 1]
      (if (= n 1)
          f
          (recur (dec n) (* f n)))))

Calling (factorial-1 1000N) in a REPL yields a number: 402387260077093773...

However when I try the following lazy sequence approach:

(defn factorial-2 [n]
   (reduce * 1 (range 1 (inc n))))

Calling (factorial-2 1000N) in a REPL yields an error:

ArithmeticException integer overflow  clojure.lang.Numbers.throwIntOverflow (Numbers.java:1388)

Why does the seemingly lazy sequence approach result in an integer overflow error?

amalloy · Accepted Answer · 2015-11-09 23:17:37Z

8

You never actually use any bigints in your multiplications, despite passing 1000N, because you only use that number to determine the end of your computation. You start multiplications at 1, and multiply by 1, then 2, and so on. If you modify your definition of factorial-2 to use a bigint, you get the expected behavior:

(defn factorial-2 [n]
  (reduce * 1N (range 1 (inc n))))

answered Nov 9, 2015 at 23:17

amalloy

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1 Comment

indraniel Over a year ago

Thanks for additional insight! I didn't realize the type of initial value to reduce changes the behavior .

ntalbs · Accepted Answer · 2015-11-11 05:16:25Z

5

Or you can simply use *' function.

(defn factorial-3 [n]
  (reduce *' (range 1 (inc n))))

*' function supports arbitrary precision. It will return Long if the number is in the range of Long. If the range exceeds the Long range, it will return BigInt.

edited Nov 11, 2015 at 5:16

answered Nov 10, 2015 at 1:53

ntalbs

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1 Comment

indraniel Over a year ago

Interesting. I didn't know about the *' function. I assume by the presence of this function in the clojure.core namespace that this considered more idiomatic clojure.

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