How to do exponentiation in clojure?

How can I do exponentiation in clojure? For now I'm only needing integer exponentiation, but the question goes for fractions too.

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You can use java's Math.pow or BigInteger.pow methods:

(Math/pow base exponent)


(.pow (bigdec base) exponent)

If you really need a function and not a method you can simply wrap it:

 (defn pow [b e] (Math/pow b e))

And in this function you can cast it to int or similar. Functions are often more useful that methods because you can pass them as parameters to another functions - in this case map comes to my mind.

If you really need to avoid Java interop, you can write your own power function. For example, this is a simple function:

 (defn pow [n p] (let [result (apply * (take (abs p) (cycle [n])))]
(if (neg? p) (/ 1 result) result)))

That calculates power for integer exponent (i.e. no roots).

Also, if you are dealing with large numbers, you may want to use BigInteger instead of int.

And if you are dealing with very large numbers, you may want to express them as lists of digits, and write your own arithmetic functions to stream over them as they calculate the result and output the result to some other stream.

How about clojure.contrib.genric.math-functions

There is a pow function in the clojure.contrib.generic.math-functions library. It is just a macro to Math.pow and is more of a "clojureish" way of calling the Java math function.

http://clojure.github.com/clojure-contrib/generic.math-functions-api.html#clojure.contrib.generic.math-functions/pow

When this question was originally asked, clojure.contrib.math/expt was the official library function to do this. Since then, it has moved to clojure.math.numeric-tower

classic recursion (watch this, it blows stack)

(defn exp [x n]
(if (zero? n) 1
(* x (exp x (dec n)))))

tail recursion

(defn exp [x n]
(loop [acc 1 n n]
(if (zero? n) acc
(recur (* x acc) (dec n)))))

functional

(defn exp [x n]
(reduce * (repeat n x)))

sneaky (also blows stack, but not so easily)

(defn exp-s [x n]
(let [square (fn[x] (* x x))]
(cond (zero? n) 1
(even? n) (square (exp-s x (/ n 2)))
:else (* x (exp-s x (dec n))))))

library

(require 'clojure.contrib.math)

Clojure has a power function that works well: I'd recommend using this rather than going via Java interop since it handles all the Clojure arbitrary-precision number types correctly. It is in namespace clojure.math.numeric-tower.

It's called expt for exponentiation rather than power or pow which maybe explains why it's a bit hard to find ... anyway here's a small example (note that use works but better use require):

(require '[clojure.math.numeric-tower :as math :refer [expt]])  ; as of Clojure 1.3
;; (use 'clojure.contrib.math)     ; before Clojure 1.3
(expt 2 200)
=> 1606938044258990275541962092341162602522202993782792835301376

Reminder about package installation

You must first install the Java package org.clojure.math.numeric-tower to make the Clojure namespace clojure.math.numeric-tower accessible!

On the command line:

$ lein new my-example-project
$ cd lein new my-example-project

Then edit project.clj and add [org.clojure/math.numeric-tower "0.0.4"] to the dependencies vector.

Start a lein REPL (not a clojure REPL)

$ lein repl

Now:

(require '[clojure.math.numeric-tower :as math])
(math/expt 4 2)
;=> 16

or

(require '[clojure.math.numeric-tower :as math :refer [expt]])
(expt 4 2)
;=> 16

I think this would work too:

(defn expt [x pow] (apply * (repeat pow x)))

Try

(defn pow [x n]
(loop [x x n n r 1]
(cond
(= n 0) r
(even? n) (recur (* x x) (/ n 2) r)
:else (recur x (dec n) (* r x)))))

for a tail-recursive O(log n) solution, if you want to implement it yourself (only supports positive integers). Obviously, the better solution is to use the library functions that others have pointed out.

user=> (.pow (BigInteger. "2") 10)
1024
user=> (.pow (BigInteger. "2") 100)
1267650600228229401496703205376

SICP inspired full iterative fast version of 'sneaky' implementation above.

(defn fast-expt-iter [b n]
(let [inner (fn [a b n]
(cond
(= n 0) a
(even? n) (recur a (* b b) (/ n 2))
:else (recur (* a b) b (- n 1))))
]
(inner 1 b n)))

Use clojure.math.numeric-tower, formerly clojure.contrib.math.


API Documentation


(ns user
(:require [clojure.math.numeric-tower :as m]))


(defn- sqr
"Uses the numeric tower expt to square a number"
[x]
(m/expt x 2))

Implementation of "sneaky" method with tail recursion and supporting negative exponent:

(defn exp
"exponent of x^n (int n only), with tail recursion and O(logn)"
[x n]
(if (< n 0)
(/ 1 (exp x (- n)))
(loop [acc 1
base x
pow n]
(if (= pow 0)
acc
(if (even? pow)
(recur acc (* base base) (/ pow 2))
(recur  (* acc base) base (dec pow)))))))

A simple one-liner using reduce:

(defn pow [a b] (reduce * 1 (repeat b a)))

I personally use:

(defn pow [x n] (reduce *' (repeat n x)))

Notice the apostrophe (') after the asterisk.

Works well for all sizes of integers.

Note: This might be a little slow for some implementations. (time (pow 2 200000)) took 1.2 seconds to resolve on my system.

Since Clojure 1.11, clojure.math/pow ships with the standard library, and works both for Clojure and ClojureScript.