Everything goes, everything comes back; eternally rolls the wheel of being. (Friedrich Nietzsche)

Transcription

1 Chapter 6 Linear Recurrences Everything goes, everything comes back; eternally rolls the wheel of being. (Friedrich Nietzsche) This chapter is dedicated to linear recurrences, a special type of equations that defines a sequence, that is a series of terms of the form a 0, a 1, a, a 3,..., a n, a n+1,... recursively, that is such that each term a n is defined as a function of the preceding terms. A recursive linear recurrence must be accompanied by initial conditions, that is information about some of the first terms such as a 0 or a 0, a 1. Example 60. The Fibonacci sequence is defined by f n = f n 1 + f n, f 0 = 0, f 1 = 1. To compute f, we have To compute f 3, we have f = f 1 + f 0 = 1. f 3 = f + f 1 = =. We will see in this chapter two methods to solve linear recurrences involving one or two preceding terms. 145

2 146 CHAPTER 6. LINEAR RECURRENCES Recurrence Relation A recurrence relation is an equation that recursively defines a sequence, i.e., each term of the sequence is defined as a function of the preceding terms A recursive formula must be accompanied by initial conditions (information about the beginning of the sequence). Fibonacci Sequence 0, 1, 1,, 3, 5, 8, 13, 1, 34, 55, Leonardo Pisano Bigolio (c c. 150) Image from wikipedia

3 147 The first method is called backtracking, and consists of taking a linear recurrence defining a n, and replace the terms a n 1, a n,... with the relation that defines a n, but where n is replaced by n 1, n, etc. This is best illustrated on an example. Example 61. Consider the linear recurrence Then and so on and so forth. Therefore a n = a n 1 + 3, a 1 =. a n = a n a n 1 = a n + 3 a n = a n a n 3 = a n = (a n + 3) + 3 = a n + 3 = (a n 3 + 3) + 6 = a n =... = a 1 + 3(n 1). The last equality follows because a generic term is of the form a n i + 3i, therefore when n i = 1, i = n 1. By plugging the initial condition, we conclude a n = + 3(n 1). Once the solution has been found, you may wonder how to check whether this is the right answer. One way to do it is by proving it by induction! Example 6. Let us provide a proof by induction for the above example. Define P (n) = a n = + 3(n 1). Then P (1) = a 1 =, which is the initial condition, is true. Suppose P (k) = a k = + 3(k 1) is true. We want to prove P (k + 1). as desired. a k+1 = a k + 3 = + 3(k 1) + 3 = + 3k

4 148 CHAPTER 6. LINEAR RECURRENCES Solving Recurrence Relation Backtracking is a technique for finding explicit formula for recurrence relation E.g., say a n = a n-1 +3 and a 1 = a n = a n-1 +3 = (a n- +3)+3=a n- +*3 = (a n-3 +3)+*3=a n-3 +3*3 = (a n- +3)+3=a n- +*3 = a 1 +(n-1)*3 a n = +(n-1)*3 Homogeneous Relation Of Degree d A linear homogeneous relation of degree d is of the form Examples The Fibonacci sequence The relation: a n =a n-1 (degree 1) But not the relation: a n =a n-1 +1 The characteristic equation of the above relation is

5 Definition 37. A linear homogeneous relation of degree d is of the form Its characteristic equation is a n = c 1 a n 1 + c a n c d a n d. x d = c 1 x d 1 + c x d c d. 149 The characteristic equation is obtained from a n = c 1 a n 1 + c a n c d a n d by replacing a i by x i : then factor out x n d to get x n c 1 x n 1 c x n... c d x n d = 0, x n d (x d c 1 x d 1 c x d... c d ) = 0. Example 63. The Fibonacci sequence f n = f n 1 + f n is a homogeneous relation. Let us compute its characteristic equation: x n x n 1 x n = 0 x n (x x 1) = 0 therefore x x 1 = 0 is the characteristic equation. Let us focus on quadratic characteristic equations, that is of the form x c 1 x c = 0 which corresponds to linear recurrences of the form a n = c 1 a n 1 + c a n. Suppose that x c 1 x c = 0 has two distinct real roots s 1, s, then Therefore s 1 c 1 s 1 c = 0, s c 1 s c = 0. s n 1 c 1 s n 1 1 c s n = 0, s n c 1 s n 1 c s n = 0 and we have that if s is a solution of x c 1 x c = 0 then s n is a solution of a n = c 1 a n 1 +c a n. This tells us that solutions of a n are composed of s n 1, s n. Note the term composed is used, because if a sequence a n also satisfies the recurrence of a n, then a n + a n satisfies the recurrence of a n as well, as does multiples of a n (see Exercise 53).

6 150 CHAPTER 6. LINEAR RECURRENCES Theorem If the characteristic equation x -c 1 x-c =0 (of the recurrence relation a n =c 1 a n-1 +c a n- ) has two distinct roots s 1,s, then the explicit formula for the sequence a n is u.s 1n +v.s n a single root s, then the explicit formula for a n is u.s n +v.n.s n where u & v are determined by initial conditions. Example Determine the number of bit strings (i.e., comprising 0/1s) of length n that contains no adjacent 0s. C n = this number of bit strings A binary string with no adjacent 0s is constructed by o Adding 1 to any string w of length n-1 satisfying the condition, or o Adding 10 to any string v of length n- satisfying the condition Thus C n = C n-1 +C n- where C 1 = (0,1), C =3 (01, 10, 11)

7 151 This means that the final solution is really a composition of s n 1, s n, namely a solution for a n = c 1 a n 1 + c a n is given by a n = us n 1 + vs n, where u, v depend on the initial conditions (that is on a 0, a 1 ). Suppose now that x c 1 x c = 0 has one double real root s, that is x c 1 x c = (x s). Then s c 1 s c = 0, and s n is a solution of a n = c 1 a n 1 + c a n as for the case of two distinct roots. We obtained the characteristic equation from x n c 1 x n 1 c x n = x n (x c 1 x c ) = 0. If s is a root of this equation, then s is a root of its derivative: Therefore s satisfies both nx n 1 c 1 (n 1)x n c (n )x n 3 = 0. s n = c 1 s n 1 + c s n, ns n = c 1 (n 1)s n 1 + c (n )s n. (6.1) If we combine s n and ns n, as we did for s n 1 and s n, we get a n = us n + vns n, and a n+1 = us n+1 + v(n + 1)s n+1. We are left to check c 1 a n+1 + c a n+ = c 1 (us n+1 + v(n + 1)s n+1 ) + c (us n+ + v(n + )s n+ ) = u[c 1 s n+1 + c s n+ ] + v[c 1 (n + 1)s n+1 + c (n + )s n+ ] = us n+ + v(n + )s n+ = a n+ using (6.1), which is consistent with our recurrence relation. Example 64. Suppose we want to determine the number of bit strings of length n that contains no adjacent zeroes. We denote this number by C n. We first observe that there are two ways of obtaining such sequences from a smaller sequence. One may take any string: of length n 1 satisfying the condition and add a 1 (one cannot add a 0, since the sequence may finish by 0), of length n satisfying the condition and add 10 (one cannot add 00, or 01 since the string could finish by 0, and 11 is included above).

8 15 CHAPTER 6. LINEAR RECURRENCES Example Now solve C n = C n-1 +C n- where C 1 =, C =3 Characteristic equation: x -x-1=0 Its roots are Recall roots of quadratic eqn. Thus Example Initial conditions give us: Solving, we get

9 153 Therefore the linear recurrence involved is C n = C n 1 + C n. The characteristic equation is obtained from x n x n 1 x n = x n (x x 1) = 0, it is x x 1 = 0. To find its roots, we compute The solution is then 1 ± 1 4( 1) C n = u( = 1 ± 5. ) n + v( 1 5 ) n. We are left to find u, v based on the initial conditions. They are C 1 = (the strings are 0 and 1), while C = 3 (the strings are 11,01,10). This gives us two equations for two unknowns: ( 1 + ) ( 5 1 ) 5 u + v = u ( which can be simplified to u + v 3 u + v ) + v ( u v ) = ) = u v ) = 3. Set a = (u + v)/, b = 5(u v)/. We need to solve a + b =, 3a + b = 3 b = 3, a = 1. Thus v = , u = 1 5 =

10 154 CHAPTER 6. LINEAR RECURRENCES Exercises for Chapter 6 Exercise 51. Consider the linear recurrence a n = a n 1 a n with initial conditions a 1 = 3, a 0 = 0. Solve it using the backtracking method. Solve it using the characteristic equation. Exercise 5. What is the solution of the recurrence relation with a 0 = and a 1 = 7? a n = a n 1 + a n Exercise 53. Let a n = c 1 a n 1 +c a n +...+c k a n k be a linear homogeneous recurrence. Assume both sequences a n, a n satisfy this linear homogeneous recurrence. Show that a n + a n and αa n also satisfy it, for α some constant.

11 155 Examples for Chapter 6 Linear Recurrence relations are often useful to analyze algorithms, such as divide-and-conquer algorithms. We will illustrate this using the game of Hanoi tower. In this game, the goal is to move n disks ranked from the largest at the bottom to the smallest on top from one post to another. The only permitted action is to remove the top disk from a post and drop it onto another post. The rule is that a larger disk can never lie above a smaller disk on any post. When n = 3 disks, the Hanoi tower game can be solved in 7 steps. But say one would like to know how many steps it would take to solve it for n = 50 disks, how could this be figured out? The method is to derive a linear recurrence relation, and then to solve it. To find a linear recurrence relation, notice that to solve the Hanoi tower game for n = 3 disks, the following steps are done: 1. Solve a Hanoi tower game for n = disks,. Move the largest disk, 3. Solve another Hanoi tower game for n = disks.. In fact, this is true in general, which yields the linear recurrence T n = T n 1 + 1, where T n denotes the number of steps for n disks. We solve this linear recurrence using backtracking. Note that Then T n 1 = T n + 1, T n = T n 3 + 1, T n 3 = T n 4 + 1,... T n = T n = (T n + 1) = 4T n + 3 = 4(T n 3 + 1) + 3 = 8T n = 8(T n 4 + 1) + 7 = 16T n =...

12 156 CHAPTER 6. LINEAR RECURRENCES Hanoi Tower Goal: move all n disks in the same order, but on a different post. Only permitted action: remove the top disk from a post and drop it onto another post. Rule: a larger disk can never lie above a smaller disk on any post. Hanoi Tower (n=3)

13 157 Find a Recurrence T n = minimum number of steps needed to move an n-disk tower from one post to another T n = T n 1 +1 T n 1 1 T n 1 T 1 = 1 T n = T n 1 +1 Backtracking

14 158 CHAPTER 6. LINEAR RECURRENCES Backtracking T 1 = 1 T n = T n 1 +1 T n = T n = T n = 4T n + 3 = 4 T n = 8T n T n = n 1 Induction P(n)=``T n = n 1 Basis step: P(1)=``T 1 = 1 Inductive step: suppose P(n) is true. To show, P(n+1). T n+1 = T n + 1 = n = n+1 1

15 159 We notice that a general term is of the form i T n i + ( i 1), therefore when n i = 1, i = n 1, and we get with T 1 = 1. Thus finally T n = n 1 T 1 + ( n 1 1) T n = n 1. Once a solution has been found by backtracking, it is advised to confirm that the solution is sound, by performing a proof by induction. Here P (n) = T n = n 1. The basis step is P (1) = T 1 = 1 1 = 1 which is true. The induction step is to assume that P (k) = T k = k 1 is true. We need to prove P (k + 1). But as desired! T k+1 = T k + 1 = k + 1 = k+1.

16 160 CHAPTER 6. LINEAR RECURRENCES