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 Incoming links: Coupled binary exponentiation

Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Imperative programming / 1 / c / Solution 2026-10-31

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First rearrange the ratios so that the intermediate powers remain moderate:

e_{c} (n) = (n + 1) (1 + \frac{1}{n})^{n} - n (1 + \frac{1}{n - 1})^{n - 1} .

(1)

This avoids separately forming enormous powers such as

n^{n}

. Perform the two instances of exponentiation by squaring in one while loop, using coupled binary exponentiation:

def ec(n: Int): Double = {
  require(n > 1)
  val dn = n.toDouble
  var x = 1.0 + 1.0 / dn
  var y = 1.0 + 1.0 / (dn - 1.0)
  var k = n
  var l = n - 1
  var p = 1.0
  var q = 1.0
  while (k > 0 || l > 0) {
    if (k > 0) {
      if ((k & 1) != 0) p *= x
      k /= 2
      if (k > 0) x *= x
    }
    if (l > 0) {
      if ((l & 1) != 0) q *= y
      l /= 2
      if (l > 0) y *= y
    }
  }
  (dn + 1.0) * p - dn * q
}

With

A = 1 + 1/ n

and

B = 1 + 1/ (n - 1)

, the loop invariant is

k, l \geq 0

p x^{k} = A^{n}

, and

q y^{l} = B^{n - 1}

. Each active branch preserves its own power identity by the parity argument in part (a)(ii), and an inactive branch changes nothing. The loop variant is the mathematical integer

k + l

, which need not be computed in a fixed-width variable: it decreases whenever either exponent is positive. At exit

p = A^{n}

q = B^{n - 1}

, proving the returned expression is $e_{c} (n)$ . There are

O (lo g n)

iterations, at most four loop multiplications each, and two final multiplications. Converting n before adding one avoids integer overflow at Int.MaxValue.

The rearrangement avoids floating-point overflow in the original powers, but subtraction still combines two quantities of order

n

to obtain one of order one. Catastrophic cancellation can amplify their accumulated roundoff error. The mathematical

O (n^{- 2})

approximation error therefore does not imply that the Double implementation improves indefinitely as n grows.

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