Binary search of an ordered list

• Given: x, and a₁ < ... < a_n
  a₀ defined to be -∞
  a_n+1 defined to be +∞
• Return: i such that x = a_i
  0 if no such i exists

BINARY SEARCH
  i := 1
  j := n
  while (i<j) do
        m := [ (i+j)/2 ]
        if (x > am) then
           i := m+1
        else
           j := m
  if (x ≠ ai) then
     i := 0
  return i

Worst case analysis

Average case analysis

Let n = 2^k + r,     0 ≤ r < 2^k     (r < n/2)

We always do k iterations;
we sometimes (2r/n of the time) do an extra one.

A = k + 2r/n,     where k = [lg n]

CASE 2:  x is not in the list (n+1 intervals equally likely)

Each interval goes with the succeeding list member.
(The last interval goes with the last list member.)

The last list member never requires an extra iteration.

A = k + 2r/(n+1),     where k = [lg n]