Integer exponentiation in OCaml

Question

Is there a function for integer exponentiation in OCaml? ** is only for floats. Although it seems to be mostly accurate, isn't there a possibility of precision errors, something like 2. ** 3. = 8. returning false sometimes? Is there a library function for integer exponentiation? I could write my own, but efficiency concerns come into that, and also I'd be surprised if there isn't such a function already.

Answer 1

Not in the standard library. But you can easily write one yourself (using exponentiation by squaring to be fast), or reuse an extended library that provides this. In Batteries, it is Int.pow.

Below is a proposed implementation:

let rec pow a = function
  | 0 -> 1
  | 1 -> a
  | n -> 
    let b = pow a (n / 2) in
    b * b * (if n mod 2 = 0 then 1 else a)

If there is a risk of overflow because you're manipulating very big numbers, you should probably use a big-integer library such as Zarith, which provides all sorts of exponentiation functions.

(You may need the "modular exponentiation", computing (a^n) mod p; this can be done in a way that avoids overflows by applying the mod in the intermediary computations, for example in the function pow above.)

Answer 2

Regarding the floating-point part of your question: OCaml calls the underlying system's pow() function. Floating-point exponentiation is a difficult function to implement, but it only needs to be faithful (that is, accurate to one Unit in the Last Place) to make 2. ** 3. = 8. evaluate to true, because 8.0 is the only float within one ULP of the mathematically correct result 8.

All math libraries should(*) be faithful, so you should not have to worry about this particular example. But not all of them actually are, so you are right to worry.

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A better reason to worry would be, if you are using 63-bit integers or wider, that the arguments or the result of the exponentiation cannot be represented exactly as OCaml floats (actually IEEE 754 double-precision numbers that cannot represent 9_007_199_254_740_993 or 2⁵³ + 1). In this case, floating-point exponentiation is a bad substitute for integer exponentiation, not because of a weakness in a particular implementation, but because it is not designed to represent exactly integers that big.

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(*) Another fun reference to read on this subject is “A Logarithm Too Clever by Half” by William Kahan.

Answer 3

A simpler formulation of the solution above:

let pow =
  let rec pow' a x n =
    if n = 0 then a else pow' (a * (if n mod 2 = 0 then 1 else x)) (x * x) (n / 2) in
  pow' 1

pow' a x n computes a times x to the n.

Answer 4

Here's another implementation which uses exponentiation by squaring (like the one provided by @gasche), but this one is tail-recursive

let is_even n = 
  n mod 2 = 0

(* https://en.wikipedia.org/wiki/Exponentiation_by_squaring *)
let pow base exponent =
  if exponent < 0 then invalid_arg "exponent can not be negative" else
  let rec aux accumulator base = function
    | 0 -> accumulator
    | 1 -> base * accumulator
    | e when is_even e -> aux accumulator (base * base) (e / 2)
    | e -> aux (base * accumulator) (base * base) ((e - 1) / 2) in
  aux 1 base exponent