Data Structures and Algorithm Analysis - Computer Science at ...

Sec. 3.5 Calculating the Running Time for a Program 71When analyzing these two code fragments, we will assume that n isa power of two. The first code fragment has its outer for loop executedlog n + 1 times because on each iteration k is multiplied by two until itreaches n. Because the inner loop always executes n times, the total cost forthe first code fragment can be expressed as ∑ log ni=0n. Note that a variablesubstitution takes place here to create the summation, with k = 2 i . FromEquation 2.3, the solution for this summation is Θ(n log n). In the secondcode fragment, the outer loop is also executed log n + 1 times. The innerloop has cost k, which doubles each time. The summation can be expressedas ∑ log ni=0 2i where n is assumed to be a power of two and again k = 2 i .From Equation 2.8, we know that this summation is simply Θ(n).What about other control statements? While loops are analyzed in a mannersimilar to for loops. The cost of an if statement in the worst case is the greaterof the costs for the then and else clauses. This is also true for the average case,assuming that the size of n does not affect the probability of executing one of theclauses (which is usually, but not necessarily, true). For switch statements, theworst-case cost is that of the most expensive branch. For subroutine calls, simplyadd the cost of executing the subroutine.There are rare situations in which the probability for executing the variousbranches of an if or switch statement are functions of the input size. For example,for input of size n, the then clause of an if statement might be executed withprobability 1/n. An example would be an if statement that executes the thenclause only for the smallest of n values. To perform an average-case analysis forsuch programs, we cannot simply count the cost of the if statement as being thecost of the more expensive branch. In such situations, the technique of amortizedanalysis (see Section 14.3) can come to the rescue.Determining the execution time of a recursive subroutine can be difficult. Therunning time for a recursive subroutine is typically best expressed by a recurrencerelation. For example, the recursive factorial function fact of Section 2.5 callsitself with a value one less than its input value. The result of this recursive call isthen multiplied by the input value, which takes constant time. Thus, the cost ofthe factorial function, if we wish to measure cost in terms of the number of multiplicationoperations, is one more than the number of multiplications made by therecursive call on the smaller input. Because the base case does no multiplications,its cost is zero. Thus, the running time for this function can be expressed asT(n) = T(n − 1) + 1 for n > 1; T (1) = 0.We know from Examples 2.8 and 2.13 that the closed-form solution for this recurrencerelation is Θ(n).