Linear Recurrence Relations

Transcription

1 Linear Recurrence Relations

2 Linear Homogeneous Recurrence Relations The Towers of Hanoi According to legend, there is a temple in Hanoi with three posts and 64 gold disks of different sizes. Each disk has a hole through the center so that it fits on a post. In the misty past, all the disks were on the first post, with the largest on the bottom and the smallest on top. Monks in the temple have labored through the years since to move all the disks to one of the other two posts according to the following rules: The only permitted action is removing the top disk from one post and dropping it onto another post. A larger disk can never lie above a smaller disk on any post. E. Lehman, T. Leighton and A. Meyer Mathematics for Computer Science

3 Linear Homogeneous Recurrence Relations The Towers of Hanoi 7 step solution for n = 3: E. Lehman, T. Leighton and A. Meyer Mathematics for Computer Science

4 Linear Homogeneous Recurrence Relations The Towers of Hanoi recursive solution: E. Lehman, T. Leighton and A. Meyer Mathematics for Computer Science

5 The Towers of Hanoi T 0 = 1 T 1 = 1 T 2 = 3 T 3 = 7... T n = 2T n-1 + 1, for n 2

6 Linear Homogeneous Recurrence Definition Relations A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n-1 + c 2 a n c k a n-k Where c 1, c 2,...,c k are real numbers, c k 0

7 Fibonacci numbers F n = F n F n 2, for n 2 (1) F 0 = 1 F 1 = 1 Since F n grows exponentially, we will assume that F n is proportional to x n for some base x. Let F n = x n x n = x n 1 + x n 2 x 2 = x + 1 x 2 - x - 1 = 0 x 1 = x 2 = Do they satisfy (1)?

8 Linear Homogeneous Recurrence Theorem Relations Let c 1 and c 2 be real numbers. Suppose that r 2 -c 1 r-c 2 =0 has two distinct roots r 1 and r 2. Then the sequence {a n } is a solution of the recurrence relation a n = c 1 a n-1 + c 2 a n-2 if and only if a n = α 1 r 1 n + α 2 r 2 n for n=0, 1, 2,...,where α 1 and α 2 are constants.

9 Example Linear Homogeneous Recurrence Relations

10 Linear Homogeneous Recurrence Relations Example Solve the recurrence equation a n = 5a n-1-6a n-2 for a 0 = 1, a 1 = 4 Solution: The characteristic equation and its roots: r 2-5r + 6 = 0, r 1 =2, r 2 =3 General solution: a n = 1 2 n n

11 General solution: a n = 1 2 n n System of linear equations for initial conditions and its solution: a 0 = a 1 = = = -1 4 = = 2 Final answer: a n = -1 2 n n

12 Linear Homogeneous Recurrence Example Solve the recurrence relation a n = 4 a n-1-4 a n-2, with a 0 = 1 and a 1 = 3? x 2-4x + 4 = 0 x 1,2 = 2?

13 Linear Homogeneous Recurrence Theorem Relations Let c 1 and c 2 be real numbers with c 2 0. Suppose that r 2 - c 1 r - c 2 = 0 has only one root r 0. A sequence {a n } is a solution of the recurrence relation a n = c 1 a n-1 + c 2 a n-2 if and only if a n = α 1 r 0n + α 2 nr 0 n for n=0, 1, 2,...,where α 1 and α 2 are constants.

14 Linear Homogeneous Recurrence Example Solve the recurrence relation a n = 4 a n-1-4 a n-2, with a 0 = 1 and a 1 = 3? x 2-4x + 4 = 0 x 1,2 = 2 a n = α 1 2 n + α 2 n2 n a n = 2 n n2n

15 Linear Homogeneous Recurrence Example Relations Find the solution to the recurrence relation a n = -3a n-1-3a n-2 - a n-3 With the initial conditions a 0 = 1 and a 1 = -2, and a 2 = -1. x 3 + 3x 2 + 3x + 1 = 0 x 1, 2, 3 = -1 a n = α 1 r 0n + α 2 nr 0 n + α 3 n 2 r 0 n

16 Linear Homogeneous Recurrence Word problem Relations Find a recurrence relation for the number of ternary strings that do not contain two consecutive 0s.

17 Linear Homogeneous Recurrence Word problem Relations

18 Linear Homogeneous Recurrence Word problem Relations

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20 Linear Homogeneous Recurrence Word problem Relations

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23 Linear Non-Homogeneous Recurrence Relations with Constant Coefficients Linear homogeneous recurrence relations of degree k with constant coefficients: a n = 5a n-1-6a n-2 a n = 4 a n-1-4 a n-2, a n = -3a n-1-3a n-2 - a n-3 Linear non-homogeneous recurrence relation of degree k with constant coefficients: a n = 5a n-1-6a n n a n = 4 a n-1-4 a n-2 + n a n = -3a n-1-3a n-2 - a n n (n+1)

24 Linear Non-Homogeneous Recurrence Relations with Constant Coefficients A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n-1 + c 2 a n c k a n-k where c 1, c 2,...,c k are real numbers, c k 0 A linear non-homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form a n = c 1 a n-1 + c 2 a n c k a n-k + g(n) where c 1, c 2,...,c k are real numbers, c k 0, and g(n) a function not identically zero depending only on n.

25 Linear Non-Homogeneous Recurrence Relations with Constant Coefficients If non-homogeneous recurrence is f n = c 1 f n-1 + c 2 f n c k f n-k + g(n) the associate homogeneous recurrence is f nc = c 1 f n-1 + c 2 f n c k f n-k Let f n be the general solution of f nc, and f n a particular ( arbitrary) solution of f n. Then f n = f n + f n

26 Linear Non-Homogeneous Recurrence Relations Theorem (particular solution) Suppose {f n } satisfies the linear nonhomogeneous recurrence relation f n = c 1 f n-1 + c 2 f n c k f n-k + g(n), and g(n) = (b t n t + b t-1 n t b 1 n + b 0 ) s n where b 0,b 1,..,b n,s R. If s is not a root of f nc, there is a particular solution of the form f n = (p t n t + p t-1 n t-1 + +p 1 n + p 0 ) s n If s is a root of f nc, and its multiplicity is m, there is a particular solution of the form f n = n m (p t n t + p t-1 n t-1 + +p 1 n + p 0 ) s n

27 Linear Non-Homogeneous Recurrence Relations What form does a particular solution of the linear nonhomogeneous recurrence relation f n = 6f n-1-9f n-2 + g(n) g(n) = 5 have when g(n) = 5n + 1 g(n) = 5n g(n) = 5n 2 + n + 1 g(n) = n2 n g(n) = 2 n (5n 2 + n + 1) g(n) = n3 n g(n) = 3 n (5n 2 + n + 1)

28 Linear Non-Homogeneous Recurrence Relations f n = 6f n-1-9f n-2 + g(n) f nc = 1 3 n + 2 n3 n f n = (p t n t + p t-1 n t-1 + +p 1 n + p 0 ) s n g(n) = 5 f n = p 0 g(n) = 5n + 1 f n = p 1 n + p 0 g(n) = 5n f n = p 2 n 2 +p 1 n + p 0 g(n) = 5n 2 + n + 1 f n = p 2 n 2 +p 1 n + p 0 g(n) = n2 n f n = (p 1 n + p 0 ) 2 n g(n) = 2 n (5n 2 + n + 1) f n = (p 2 n 2 +p 1 n + p 0 ) 2 n

29 Linear Non-Homogeneous Recurrence Relations f n = 6f n-1-9f n-2 + g(n) f nc = 1 3 n + 2 n3 n f n = (p t n t + p t-1 n t-1 + +p 1 n + p 0 ) s n g(n) = n2 n g(n) = 2 n (5n 2 + n + 1) g(n) = n3 n g(n) = 3 n (5n 2 + n + 1) f n = (p 1 n + p 0 ) 2 n f n = (p 2 n 2 +p 1 n + p 0 ) 2 n f n = n 2 (p 1 n + p 0 ) 3 n f n = n 2 (p 2 n 2 + p 1 n + p 0 ) 3 n

30 Linear Non-Homogeneous Recurrence Relations f n = (p t n t + p t-1 n t-1 + +p 1 n + p 0 ) s n g(n) = 5 f n = p 0 g(n) = 5n + 1 f n = p 1 n + p 0 What if s = 1? Solve f n = f n-1 + n with f 1 = 1; f n = n(p 1 n + p 0 ) and f n = n(n + 1)/2

31 Solve Example

32 Example (cont.)

33 Linear Non-Homogeneous Recurrence Relations Word Problems Example

34 Linear Non-Homogeneous Recurrence Relations Word Problems Example Recurrence

35 Recurrence Word Problems

36 Recurrence Word Problems

37 Extra Credit, Max number of points 5(HW) DUE Tuesday, February 10, 11AM, in LATEX NO HELP FROM ANYBODY! from quiz 3, sample 3

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