Handout 7: Recurrences (Cont d)

Transcription

1 ENGG 2440B: Discrete Mathematics for Engineers Handout 7: Recurrences (Cont d) First Term Instructor: Anthony Man Cho So October 8, 2018 In the last handout, we studied techniques for solving linear homogeneous recurrences. We now build upon those techniques to solve a more general class of recurrences, namely, the class of linear inhomogeneous recurrences. 1 Introduction Consider the following recurrence: T (n) = a 1 T (n 1) + a 2 T (n 2) + + a d T (n d) + F (n). (1) At first sight, this is similar to a linear homogeneous recurrence. However, there is an extra term F (n) at the end, which is unrelated to T (n) and requires some new machinery to handle. Before we discuss the techniques for solving (1), let us see how such recurrence arises. Example 1 Consider 3 pegs, where n circular disks of increasing size are placed in one peg with the largest disk at the bottom. These disks are to be transferred one at a time onto another of the pegs, with the requirements that (i) at no time is one allowed to place a larger disk on top of a smaller one and (ii) at the end one must have all the disks in one peg that is different from the initial one. How many moves are needed to complete the task? This is the famous Tower of Hanoi puzzle. Let T (n) be the number of moves needed to transfer the n disks. Then, it is easy to convince youself that T (1) = 1 and T (2) = 3. Now, observe that to transfer n disks to another peg, we must first transfer the top n 1 disks to a peg, then transfer the largest disk to the vacant peg, and finally transfer the n 1 disks to the peg that contains the largest disk. The above process can be described by the recurrence T (n) = T (n 1) T (n 1) = 2T (n 1) + 1 for n 1 (2) with the initial condition T (1) = 1. The recurrence (2) is a special case of (1), with d = 1, a 1 = 2, and F (n) = 1. 2 Solving Linear Inhomogeneous Recurrences The technique for solving the linear inhomogeneous recurrence (1) largely relies on that for solving the associated linear homogeneous recurrence Indeed, we have the following result: T (n) = a 1 T (n 1) + a 2 T (n 2) + + a d T (n d). (3) Theorem 1 Let T p (n) be a particular solution to the linear inhomogeneous recurrence (1) and T h (n) be a homogeneous solution to the associated linear homogeneous recurrence (3). Then, every solution to (1) is of the form T 0 (n) = T h (n) + T p (n). 1

2 The proof of Theorem 1 is rather straightforward. Proof If T h (n) solves (3) and T p (n) solves (1), then by definition, T h (n) = a 1 T h (n 1) + a 2 T h (n 2) + + a d T h (n d), T p (n) = a 1 T p (n 1) + a 2 T p (n 2) + + a d T p (n d) + F (n). Upon adding the above two equations, we obtain (T h (n) + T p (n)) = a 1 (T h (n 1) + T p (n 1)) + a 2 (T h (n 2) + T p (n 2)) + + a d (T h (n d) + T p (n d)) +F (n), which implies that T 0 (n) = T h (n) + T p (n) also solves (1). Since we already know how to find solutions to the linear homogeneous recurrence (3), to apply Theorem 1, it remains to investigate how we can come up with a particular solution to the linear inhomogeneous recurrence (1). In many cases, the particular solution will be of the same function class as the inhomogeneous term F (n). For instance, if F (n) is a polynomial in n of degree t, then T p (n) will also be a polynomial in n whose degree is at least t. Let us illustrate this by continuing Example 1. Example 1 (cont d) Since F (n) = 1, which is a polynomial of degree 0, we can try the particular solution T p (n) = x, where x is a constant to be determined. Note that T p (n) = x is a polynomial of degree 0. Now, if T p (n) solves the recurrence (2), then we must have x = 2x + 1, which means that x = 1. Hence, T p (n) = 1 is a particular solution to (2). Now, by Theorem 1, in order to solve (2), we also need the solution to its associated linear homogeneous recurrence T (n) = 2T (n 1). (4) The characteristic equation associated with (4) is x = 2. Hence, for any real number θ, is a solution to (4). By Theorem 1, T h (n) = θ2 n T 0 (n) = T h (n) + T p (n) = θ2 n 1 is a solution to (2). It remains to impose the initial condition to determine θ. Since T (1) = 1, we have 1 = 2θ 1, which gives θ = 1. It follows that T (n) = 2 n 1 is the solution to the Tower of Hanoi recurrence (2). We shall further elaborate on how to find a particular solution to a given linear inhomogeneous recurrence in Section 3. Now, let us first summarize the procedure we developed so far for solving the linear inhomogeneous recurrence (1): Procedure for Solving Linear Inhomogeneous Recurrence 2

3 1. Solve the associated linear homogeneous recurrence (3) using the techniques introduced in Handout 6 to get the homogeneous solution T h (n). Note that T h (n) takes the form T h (n) = θ 1 T 1 (n) + θ 2 T 2 (n) + + θ d T d (n), where T 1 (n),..., T d (n) are the functions determined by the roots of the characteristic equation associated with (3) and θ 1,..., θ d are any real numbers. 2. Find a particular solution T p (n) to the linear inhomogeneous recurrence (1) by examining the function class of F (n). 3. Form the candidate solution T 0 (n) = T h (n)+t p (n) and use the initial conditions on T (1),..., T (d) to fix the parameters θ 1,..., θ d in T h (n). 3 Finding a Particular Solution As mentioned earlier, the particular solution is often of the same function class as the inhomogeneous term F (n) in (1). More precisely, we have the following result, which informally states that if F (n) takes the form F (n) = p(n)s n, where p(n) is a polynomial in n and s is a constant, then there is a particular solution T p (n) to (1) that takes the form T p (n) = q(n)s n, where q(n) is another polynomial in n. Theorem 2 Suppose that the inhomogeneous term F (n) in (1) takes the form F (n) = (b 0 + b 1 n + b 2 n b t n t )s n (5) for some real numbers b 0, b 1,..., b t and s. If s is not a root of the characteristic equation associated with the linear homogeneous recurrence (3), then there is a particular solution T p (n) to (1) of the form T p (n) = (x 0 + x 1 n + x 2 n x t n t )s n, for some real numbers x 0, x 1,..., x t. On the other hand, if s is a root of the characteristic equation associated with the linear homogeneous recurrence (3) of multiplicity m 1, then there is a particular solution T p (n) to (1) of the form T p (n) = n m (x 0 + x 1 n + x 2 n x t n t )s n, for some real numbers x 0, x 1,..., x t. In the context of Example 1, we have F (n) = 1, which means that b 0 = 1, b 1 = b 2 = = b t = 0, and s = 1 in (5). Since s = 1 is not a root of the characteristic equation (4), by Theorem 2, there is a particular solution of the form T p (n) = x 0. This is consistent with our choice in Example 1. Let us present some more examples to illustrate the procedure for solving linear inhomogeneous recurrences. Example 2 Consider the recurrence { T (n) = 5T (n 1) 6T (n 2) + 7 n for n 2, T (0) = 0, T (1) = 1. (6) 3

4 The associated linear homogeneous recurrence is whose characteristic equation is given by T (n) = 5T (n 1) 6T (n 2), The roots are r 1 = 3 and r 2 = 2. Hence, for any real numbers θ 1, θ 2, x 2 = 5x 6. (7) T h (n) = θ 1 3 n + θ 2 2 n is a homogeneous solution. Now, let us find a particular solution to the recurrence (6). Note that F (n) = 7 n and s = 7 is not a root of the characteristic equation (7). Hence, by Theorem 2, the particular solution T p (n) should take the form T p (n) = x 0 7 n for some constant x 0. Substituting this into (6) yields x 0 7 n = 5x 0 7 n 1 6x 0 7 n n. Upon simplifying and solving for x 0, we get x 0 = Thus, our candidate solution to the recurrence (6) is T 0 (n) = T h (n) + T p (n) = θ 1 3 n + θ 2 2 n n. Lastly, using our initial conditions, we get 0 = T 0 (0) = θ 1 + θ , 1 = T 0 (1) = 3θ 1 + 2θ This gives θ 1 = and θ 2 = Hence, the solution to the recurrence (6) is given by T (n) = n n n. Example 3 Consider the recurrence { T (n) = 6T (n 1) 9T (n 2) + 3 n for n 2, T (0) = 0, T (1) = 1 2. (8) The associated linear homogeneous recurrence is whose characteristic equation is given by T (n) = 6T (n 1) 9T (n 2), x 2 = 6x 9. (9) 4

5 The root r = 3 is repeated and has multiplicity 2. Hence, by Theorem 2 of Handout 6, for any real numbers θ 1, θ 2, T h (n) = θ 1 3 n + θ 2 n3 n is a homogeneous solution. To find a particular solution to (8), we note that F (n) = 3 n and s = 3 is a root of the characteristic equation (9) of multiplicity 2. Hence, by Theorem 2, the particular solution T p (n) should take the form T p (n) = x 0 n 2 3 n for some constant x 0. Substituting this into (8) yields x 0 n 2 3 n = 6x 0 (n 1) 2 3 n 1 9x 0 (n 2) 2 3 n n. Upon simplifying and solving for x 0, we get x 0 = 1 2. Thus, our candidate solution to the recurrence (8) is given by Lastly, using the initial conditions, we have T 0 (n) = T h (n) + T p (n) = θ 1 3 n + θ 2 n3 n n2 3 n. 0 = T 0 (0) = θ 1, 1 2 = T 0(1) = 3θ 1 + 3θ , which gives θ 1 = 0 and θ 2 = 1 3. It follows that the solution to the recurrence (8) is given by T (n) = 1 3 n3n n2 3 n. 5