How to do exponentiation in clojure?

问题

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

回答1:

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)

回答2:

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'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:

(use 'clojure.math.numeric-tower)  ; as of Clojure 1.3
;; (use 'clojure.contrib.math)     ; before Clojure 1.3

(expt 2 200)
=> 1606938044258990275541962092341162602522202993782792835301376

回答3:

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

(Math/pow base exponent)

(.pow (bigint base) exponent)

回答4:

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

回答5:

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

回答6:

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.

回答7:

I think this would work too:

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

回答8:

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)))

回答9:

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))

回答10:

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)))))))

回答11:

A simple one-liner using reduce:

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

回答12:

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.

回答13:

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

来源：https://stackoverflow.com/questions/5057047/how-to-do-exponentiation-in-clojure

标签

clojure

exponentiation