Discrete Mathematics Chapter 6 Advanced Counting Techniques 7

Discrete Mathematics Chapter 6 Advanced Counting Techniques

7. 1 Recurrence Relations(遞迴關係 ) Example 1. Let {an} be a sequence that satisfies the recurrence relation an=an-1 -an-2 for n=2, 3, …, and suppose that a 0=3, and a 1=5. Here a 0=3 and a 1=5 are the initial conditions. By the recurrence relation, a 2 = a 1 -a 0 = 2 a 3 = a 2 -a 1 = -3 a 4 = a 3 -a 2 = -5 : Q 1: Applications ? Q 2: Are there better ways for computing the terms of {an}? n

※Modeling with Recurrence Relations We can use recurrence relations to model (describe) a wide variety of problems. Example 3. Compound Interest (複利) Suppose that a person deposits(存款) $10000 in a saving account at a bank yielding 11% per year with interest compounded annually. How much will be in the account after 30 years ? Sol : Let Pn denote the amount in the account after n years. P 0=10000 Pn=Pn-1 + 0. 11 Pn-1=1. 11 Pn-1, ∴ P 30=1. 11 P 29=(1. 11)2 P 28=…=(1. 11)30 P 0 =228922. 97

Example 5. (The Tower of Hanoi) The rules of the puzzle allow disks to be moved one at a time from one peg to another as long as a disk is never placed on top of a smaller disk. Let Hn denote the number of moves needed to solve the Tower of Hanoi problem with n disks. Set up a recurrence relation for the sequence {Hn}. 目標 : n 個disk都從 peg 1 移到 peg 2 H 4 moves peg 1 Sol : peg 2 peg 3 Hn=2 Hn-1+1, H 1=1 ( n-1個 disk 先從peg 1→peg 3, 第 n 個 disk 從 peg 1→peg 2, n-1個 disk 再從 peg 3→peg 2)

上例中 Hn=2 Hn-1+1, H 1=1 ∴Hn=2 Hn-1+1 =2(2 Hn-2+1)+1 =22 Hn-2+2+1 =22(2 Hn-3+1)+2+1 =23 Hn-3+(22+2+1) : =2 n-1 H 1+(2 n-2+2 n-3+…+1) =2 n-1+2 n-2+…+1 = =2 n-1

Example 6. Find a recurrence relation and give initial conditions for the number of bit strings of length n that do not have two consecutive 0 s. How many such bit strings are there of length 5 ? Sol : 2 … n-3 n-2 n-1 1 n ∴ an = an-1+an-2, n 3 1 an-1 種 a 1=2 (string : 0, 1) an-2 種 a 2=3 (string : 01, 10, 11) 1 0 ∴ a 3=a 2+a 1=5, a 4=8, a 5=13

Example 7. (Codeword enumeration) A computer system considers a string of decimal digits a valid codeword if it contains an even number of 0 digits. Let an be the number of valid n-digit codewords. Find a recurrence relation for an. Sol : 1 2 3 … n-1 an-1 種 10 n-1 - an-1 種 n 1~9 0 ∴ an = 9 an-1 + (10 n-1 -an-1) = 8 an-1 + 10 n-1 , n 2 a 1 = 9

求an通解 : Exercise : 3, 25, 27, 29, 41 (41推廣成n)

7. 2 Solving Recurrence Relations Def 1. A linear homogeneous recurrence relation of degree k (i. e. , k terms) with constant coefficients is a recurrence relation of the form an = c 1 an-1+c 2 an-2+…+ckan-k where ci R and ck≠ 0 Example 1 and 2. fn = fn-1 + fn-2 an = an-5 an = an-1 + an-22 an = nan-1 Hn = 2 Hn-1 + 1 (True, deg=2) (True, deg=5) (False, 不是linear) (False , 不是linear, not constant coeff. ) (False, 不是homogeneous)

Theorem 1. Let an = c 1 an-1+ c 2 an-2 be a recurrence relation with c 1, c 2 R. If r 2 - c 1 r - c 2= 0 (稱為characteristic equation) has two distinct roots r 1 and r 2. Then the solution of an is an = a 1 r 1 n + a 2 r 2 n , for n=0, 1, 2, …, where a 1 , a 2 are constants. (a 1 , a 2可利用 a 0, a 1算出)

Example 3. What’s the solution of the recurrence relation an = an-1 + 2 an-2 with a 0=2 and a 1=7 ? Sol : The characteristic equation is r 2 – r - 2=0. Its two roots are r 1= 2 and r 2 = -1. Hence an=a 1 2 n +a 2 (-1)n. ∵a 0 = a 1+a 2 = 2, a 1=2 a 1 -a 2=7 ∴a 1 = 3, a 2 = -1 驗算：a 2 = a 1 + 2 a 0 =11 a 2= 3 22 - 1 =11 an = 3 2 n - (-1)n.

Example 4. Find an explicit formula for the Fibonacci numbers. Sol : fn = fn-1 + fn-2 , n 2, f 0=0 , f 1=1. The characteristic equation is r 2 - r - 1=0. Its two roots are So we have , .

Thm 2. Let an = c 1 an-1+c 2 an-2 be a recurrence relation with c 1, c 2 R. If r 2 - c 1 r - c 2= 0 has only one root r 0. Then the solution of an is an = a 1 r 0 n + a 2 n r 0 n for n=0, 1, 2, …, where a 1 and a 2 are constants.

Example 5. What’s the solution of an= 6 an-1 - 9 an-2 with a 0=1 and a 1=6 ? Sol : The root of r 2 - 6 r + 9 = 0 is r 0 = 3. Hence an = a 1．3 n +a 2．n．3 n. ∵a 0 = a 1 = 1 a 1 = 3 a 1 + 3 a 2 = 6 ∴ a 1 = 1 and a 2 = 1 驗算：a 2 = 6 a 1 - 9 a 0 =27 n n an = 3 + n．3 a 2= 32 +2 32 =27

Thm 3. Let an = c 1 an-1 + c 2 an-2 + … + ckan-k be a recurrence relation with c 1, c 2, …, ck R. If rk - c 1 rk-1 - c 2 rk-2 -…- ck = 0 has k distinct roots r 1, r 2, …, rk. Then the solution of an is an = a 1 r 1 n +a 2 r 2 n + …+akrkn, for n = 0, 1, 2, … where a 1, a 2, …ak are constants.

Example 6 (k = 3) Find the solution of an = 6 an-1 - 11 an-2 + 6 an-3 with initial conditions a 0=2, a 1=5 and a 2=15. Sol : The roots of r 3 - 6 r 2 + 11 r – 6 = 0 are r 1 = 1, r 2 = 2, and r 3 = 3 ∴an = a 1 1 n + a 2 2 n + a 3 3 n ∵a 0 = a 1 + a 2 + a 3 = 2 a 1 = 1, a 2 = -1, a 1 = a 1 + 2 a 2 + 3 a 3 = 5 a 3 = 2 a 2 = a 1 + 4 a 2 + 9 a 3 = 15 0 =47 ∴an = 1 - 2 n + 2 3 n 驗算：a 3 = 6 a 2 -311 a 1+ 6 a 3 a 3= 1 - 2 + 2 3 =47

Thm 4. Let an = c 1 an-1 + c 2 an-2 + … + ckan-k be a recurrence relation with c 1, c 2, …, ck R. If rk - c 1 rk-1 - c 2 rk-2 - … - ck = 0 has t distinct roots r 1, r 2, …, rt with multiplicities m 1, m 2, …, mt respectively, where mi 1, i, and m 1+ m 2 +…+ mt = k, then (接下一頁)

where ai, j are constants. (1 i t , 0 j mi-1)

Example 8. Find the solution to the recurrence relation an = -3 an-1 - 3 an-2 - an-3 with initial conditions a 0 = 1, a 1 = -2 and a 2 = -1. Sol : r 3 + 3 r 2 + 3 r + 1 = 0 has a single root r 0 = -1 of multiplicity three. ∴ an = (a 1+a 2 n+a 3 n 2) r 0 n = (a 1+a 2 n+a 3 n 2)(-1)n ∵ a 0 = a 1 = 1 a 1 = (a 1+a 2+a 3) (-1) = -2 a 2 = (a 1+a 2+a 3) = -1 驗算：a 3 = - 3 a 2 - 3 a 1 - a 0 =8 a 3= (1+3 3 -2 32) (-1)3 =8 ∴a 1 = 1, a 2 = 3, a 3 = -2 an = (1+3 n-2 n 2) (-1)n Exercise : 3, 15, 19

7. 4 Generating Functions. Def 1. The generating function for the sequence {an} is the infinite power series. G(x) = a 0 + a 1 x +… + anxn +… = (若 {an} 是finite，可視為是infinite，但後面的term 都等於 0)

Example 2. What is the generating function for the sequence 1, 1, 1, 1 ? Sol : (expansion，展開式) (closed form)

Example 3. Let m Z+ and , for k = 0, 1, …, m. What is the generating function for the sequence a 0, a 1, …, am ? Sol : G(x) = a 0 + a 1 x + a 2 x 2 + … + amxm = (1+x)m (by 二項式定理)

Example 5. The function f (x) = is the generating function of the sequence 1, a, a 2, …, since = 1 + ax + a 2 x 2 + …= when |ax| < 1 for a≠ 0

Def 2. Let u R and k Z+∪{0}. Then the extended binomial coefficient is defined by

Example 7. Find Sol : and

Thm 2. (The Extended Binomial Theorem) Let x R with |x|<1 and let u R, then

Example 9. Find the generating functions for (1+x)-n and (1 -x)-n where n Z+ Sol : By the Extended Binomial Theorem, Def 2. By replacing x by –x we have

※Using Generating Functions to solve Recurrence Relations. Example 16. Solving the recurrence relation ak = 3 ak-1 for k=1, 2, 3, … and initial condition a 0 = 2. Sol : 另法： (by 7. 2公式 ) r – 3 = 0 r = 3 an = a 3 n ∵ a 0 = 2 = a ∴ an = 2 3 n

Let generating function for {ak}. First note that ak xk = 3 ak-1 xk be the - a 0 = 3 x G(x) ∵a 0 = 2 k G(x) - 3 x G(x) = G(x)(1 -3 x) = 2 ∴a =2 3 G(x) k Exercise : 5, 7, 11, 33

7. 5 Inclusion-Exclusion 排容原理 A, B, C, D : sets A |A|+|B|+|C| 時 各部分被計算的次數 1 2 1 B 1 2 3 1 0 2 1 21 1 C -|A B|-|A C|-|B C|後 +|A B C|後

Theorem 1. A 1, A 2, …, An : sets Exercise : 17

7. 6 Applications of Inclusion and Exclusion Example 2. How many onto functions are there form set A={1, 2, 3, 4, 5, 6} to set B={a, b, c} ? Sol : f : A → B f (1)= f (2)= ︰ ︰ f (6)= {a, b, c} 不同的填法造出不同的函數 如何使a, b, c都出現 ? ＃ of onto functions = (所有函數個數) - (a, b, c中有一個沒被對應) + (a, b, c中二個沒被對應) - (a, b, c都沒被對應) =

Thm 1. |A| = m , |B| = n There are onto functions f : A → B. pf : A = {a 1, a 2, …, am}. B = {b 1, b 2, …, bn} f (a 1)= b 1, b 2, …, bn f (a 2)= ︰ f (am)= ︰

※Derangements 亂序 Def. A derangement is a permutation of objects that leaves no object in its original position.

Example 5. derangements of 12345 : 21453, 23451, 34512, … Def. Let Dn denote the number of derangements of n objects. D 4 = (所有4個元素的permutation數) - (4個元素有一個在原位置的permutation數) + (4元素中有二個在原位置的個數) - (4個元素中有三個在原位置的個數) + (4元素都在原位置的個數) =

Theorem 2. (亂序公式) Exercise : 8 參考： 12, 13