# Appendix B: Solving Recurrence Equations—With Applications to Analysis of Recursive Algorithms The analysis of recursive algorithms is not as straight forward as it is for iterative algorithms. Ordinarily, however, it is not difficult to represent the time complexity of a recursive algorithm by a recurrence equation. The recurrence equation must then be solved to determine the time complexity. We discuss techniques for solving such equations and for using the solutions in the analysis of recursive algorithms. ## B.1 Solving Recurrences Using Induction Mathematical induction is reviewed in [Appendix A](LiB0093.html#1281). Here we show how it can be used to analyze some recursive algorithms. We consider first a recursive algorithm that computes *n*!. Algorithm B.1: Factorial****Problem: Determine *n*! = *n* (*n* − 1) (*n* − 2) … (3) (2) (1) when *n* ≥ 1. 0! = 1 Inputs: a nonnegative integer *n*. Outputs: *n*!. ``` int fact (int n) { if (n == 0) return 1; else return n * fact (n - 1); } ``` **** To gain insight into the efficiency of this algorithm, let's determine how many times this function does the multiplication instruction for each value of *n*. For a given *n*, the number of multiplications done is the number done when *fact* (*n* − 1) is computed plus the one multiplication done when *n* is multiplied by *fact* (*n* − 1). If we represent the number of multiplications done for a given value of *n* by *tn*, we have established that  An equation such as this is called a ***recurrence equation*** because the value of the function at *n* is given in terms of the value of the function at a smaller value of *n*. A recurrence by itself does not represent a unique function. We must also have a starting point, which is called an ***initial condition.*** In this algorithm, no multiplications are done when *n* = 0. Therefore, the initial condition is  We can compute *tn* for larger values of *n* as follows:  Continuing in this manner gives us more and more values of *tn*, but it does not enable us to compute *tn*, for an arbitrary *n* without starting at 0. We need an explicit expression for *tn*. Such an expression is called a ***solution*** to the recurrence equation. Recall that it is not possible to find a solution using induction. Induction can only verify that a candidate solution is correct. (Constructive induction, which is discussed in [Section 8.5.4](LiB0072.html#848), can help us discover a solution.) We can obtain a candidate solution to this recurrence by inspecting the first few values. An inspection of the values just computed indicates that  is the solution. Now that we have a candidate solution, we can use induction to try to prove that it is correct. Induction base: For *n* = 0,  Induction hypothesis: Assume, for an arbitrary positive integer *n*, that  Induction step: We need to show that  If we insert *n* + 1 in the recurrence, we get  This completes the induction proof that our candidate solution *tn* is correct. Notice that we highlight the terms that are equal by the induction hypothesis. We often do this in induction proofs to show where the induction hypothesis is being applied. There are two steps in the analysis of a recursive algorithm. The first step is determining the recurrence; the second step is solving it. Our purpose here is to show how to solve recurrences. Determining the recurrences for the recursive algorithms in this text is done when we discuss the algorithms. Therefore, in the remainder of this appendix we do not discuss algorithms; rather, we simply take the recurrences as given. We now present more examples of solving recurrences using induction. Example B.1****Consider the recurrence  The first few values are  It appears that  We use induction to prove that this is correct. Induction base: For *n* = 1,  Induction hypothesis: Assume, for an arbitrary *n* > 0 and *n* a power of 2, that  Induction step: Because the recurrence is only for power of 2, the next value to consider after *n* is 2*n*. Therefore, we need to show that  If we insert 2*n* in the recurrence, we get  **** Example B.2****Consider the recurrence  The first few values are  It appears that  We use induction to prove that this is correct. Induction base: For *n* = 1,  Induction hypothesis: Assume, for an arbitrary *n* > 0 and *n* a power of 2, that  Induction step: We need to show that  If we insert 2*n* in the recurrence, we get  This completes the induction proof. Finally, because  the solution to this recurrence is usually given as  **** Example B.3****Consider the recurrence  The first few values are  There is no obvious candidate solution suggested by these values. As mentioned earlier, induction can only verify that a solution is correct. Because we have no candidate solution, we cannot use induction to solve this recurrence. However, it can be solved using the technique discussed in the next section. ****