Consider a dynamic partitioning scheme.show that on average the memory contains half as many holes as segments
Let s and h denote the average number of segments and holes, respectively. The probability that a given segment is followed by a hole in memory (and not by another segment) is 0.5, because deletions and creations are equally probable in equilibrium. So with s segments in memory, the average number of holes must be s/2. It is intuitively reasonable that the number of holes must be less than the number of segments because neighboring segments can be combined into a single hole on deletion.