recurrence relations

Recurrence Relations

Usually we'll deal with work functions that tell us something about how much work T(n) an algorithm A does depending on the size n of the problem. Typically T(n) will denote the amount of work done by A, in the worst case, on a problem of size n or less. In addition, we'll assume T(n) > 0 for all n > 0. Suppose n = 2^m Suppose also, for some positive constant c and integer q > 2 (1) T(n) < qT(n/2) + cn when n > 1 and (2) T(1) = c (An example of this would be where T(n) < 3T(n/2) + 8.75n for n > 1 and T(1) = 8.75) By line (1), T(2) < qT(1) + 2c, which by line (2) is cq+2c = c(q+2) By line (1), T(4) < qT(2) + 4c < q(c(q+2)) + 4c = c(q²+2q+4) By line (1), T(8) < qT(4) + 8c < q(c(q²+2q+4)) + 8c = c(q³+2q²+4q+8) So we see there is a pattern: T(2^m) < c(q^m + 2q^m-1 + ... + 2^m-1q + 2^m) That is the same as: T(2^m) < c2^m((q/2)^m + (q/2)^m-1 + ... + (q/2) + 1) = c2^m((q/2)^m+1-1)/((q/2)-1) (By the way, because q!=2, (q/2)-1 is not zero) Now, c2^m((q/2)^m+1-1)/((q/2)-1) is Θ(nr^log(n)), where r = q/2. Also ... this may seem strange, but since log(r^log(n)) = log(n)log(r) = log (n^log(r)), we see that r^log(n) = n^log(r) = n^log(q)-1 Therefore T(n) is O(n^log(q)) Since q > 2, O(n^log(q)) is "polynomial time". So, for example, when q=4, that means log(q)=2 and T(n) is O(n²). And if q=3, then log(q) is less than 1.59, so in that case T(n) is O(n^1.59) Note that the calculation above is valid if q=1. When q=1 we find that T(2^m) < c(q^m + 2q^m-1 + ... + 2^m-1q + 2^m) = c(1 + 2 + 4 + ... + 2^m-1 + 2^m) = c(2^m+1-1) = 2c(2^m-(1/2)) = 2c(n-(1/2)) So it's clear that T(n) is O(n) when q=1. When q=2, we can carry the derivation above to the point where we get: T(2^m) < c2^m((q/2)^m + (q/2)^m-1 + ... + (q/2) + 1) and at this point it becomes clear that T(2^m) < c2^m(1^m + 1^m-1 + ... + 1¹ + 1⁰) and this just says that T(2^m) < c2^m(1^m + 1^m-1 + ... + 1¹ + 1⁰) = c(m+1)2^m Since n=2^m, this last expression is clearly O(nlog(n)). Suppose we have a recurrence where n = 2^m and for some positive constant c (3) T(n) < 2T(n/2) + cn² when n > 1 and (4) T(1) = c In this case we have T(2) < 2T(1) + c2² < c(2+2²) = c(2+4) T(4) < 2T(2) + c4² < 2c(2+2²) + c4² = c(4+8+16) T(8) < 2T(4) + c8² < 2c(4+8+16) + c64 = c(8+16+32+64) The pattern developing is: T(2^m) < c(2^m + 2^m+1 + ...+ 2^m+m) m (1 + 2 + 4 + 8 + ...+ 2^m) = c2^m(2^m+1-1) This is clearly O(n²).