Divide and conquer

1. Divide the problem into smaller parts
2. Solve the subproblems
3. Combine the solutons of the subproblems to produce the solution of the original problem

MaxMin

MAXMIN( S )    Assumes that n = 2m 
    if |S| = 2 then
       let S = {a,b}
       return < MAX(a,b), MIN(a,b) >     needs only 1 comparison
    else
       divide S into S1 and S2, each with ½ the elements
       < max1, min1 > := MAXMIN( S1 )
       < max2, min2 > := MAXMIN( S2 )
       return < MAX(max1,max2), MIN(min1,min2) >

Comparisons are made only by the MAX or MIN functions.
Let T(n) = the number of comparisons needed by MAXMIN when |S| = n.
T(2) = 1, since MIN(a,b) = a+b-MAX(a,b).
For n > 2, T(n) = 2T(n/2) + 2.

Solve recurrence:

We expect that T(n) will be linear in n, so assume that T(n) = an + b.

T(2) = 1 → 2a + b = 1
T(n) = 2T(n/2) + 2 → an + b = 2(an/2 + b) + 2
This second equation gives:   b = 2b + 2, and therefore b = -2.
Plugging that into the first equation gives:   2a - 2 = 1, and therefore a = 3/2.

Thus, T(n) = (3/2)n - 2.