1 / 36

Discrete Mathematics

Discrete Mathematics. Chapter 6 Advanced Counting Techniques. 7.1 Recurrence Relations( 遞迴關係 ). Example 1. Let { a n } be a sequence that satisfies the recurrence relation a n = a n - 1 - a n - 2 for n =2,3,…, and suppose that a 0 =3 ,and a 1 =5 .

lizethk
Download Presentation

Discrete Mathematics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Discrete Mathematics Chapter 6 Advanced Counting Techniques

  2. 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 a0=3,and a1=5. Here a0=3 and a1=5 are the initial conditions. By the recurrence relation, a2 = a1-a0 = 2 a3 = a2-a1 = -3 a4 = a3-a2 = -5 : Q1: Applications ? Q2: Are there better ways for computing the terms of{an}?

  3. Example 3.Compound Interest (複利) Suppose that a person deposits(存款) $10000 ina saving account at a bank yielding 11% per year with interest compounded annually. How much will be in the account after 30 years ? ※Modeling with Recurrence Relations We can use recurrence relations to model (describe) a wide variety of problems. Sol : Let Pn denote the amount in the account after n years. Pn=Pn-1 + 0.11Pn-1=1.11 Pn-1, ∴ P30=1.11 P29=(1.11)2P28=…=(1.11)30P0=228922.97 P0=10000

  4. H4moves peg 2 peg 3 peg 1 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}. Sol :Hn=2Hn-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) 目標 : n 個disk都從 peg 1 移到 peg 2 H1=1

  5. 上例中 Hn=2Hn-1+1, H1=1 ∴Hn=2Hn-1+1 =2(2Hn-2+1)+1 =22Hn-2+2+1 =22(2Hn-3+1)+2+1 =23Hn-3+(22+2+1) : =2n-1H1+(2n-2+2n-3+…+1) =2n-1+2n-2+…+1 = =2n-1

  6. 2 n-2 n-1 n n-3 1 an-1種 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 0s. How many such bit strings are there of length 5 ? Sol : ∴ an = an-1+an-2, n  3 a1=2 (string : 0,1) a2=3 (string : 01,10,11) 1 an-2種 0 1 ∴ a3=a2+a1=5, a4=8, a5=13

  7. 2 3 n-1 n 1 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 : ∴ an = 9an-1 + (10n-1-an-1) = 8an-1 + 10n-1 , n2 a1 = 9 an-1種 1~9 10n-1 - an-1種 0

  8. 求an通解 : Exercise : 3,23,25,27,29,41 (41推廣成n)

  9. 7.2 Solving Recurrence Relations Def 1. A linearhomogeneous recurrence relationof degree k (i.e., k terms) with constant coefficients is a recurrence relation of the form where ciR and ck≠0 an = c1an-1+c2an-2+…+ckan-k Example 1 and 2. fn = fn-1 + fn-2 an = an-5 an= an-1 + an-22 an= nan-1 Hn= 2Hn-1 + 1 (True, deg=2) (True, deg=5) (False, 不是linear) (False , 不是linear, not constant coeff. ) (False, 不是homogeneous)

  10. Theorem 1. Let an = c1an-1+ c2an-2 be a recurrence relation with c1,c2R. If r2 - c1r - c2= 0 (稱為characteristic equation)has two distinct roots r1 and r2. Then the solution of an is an = a1r1n + a2r2n, for n=0,1,2,…, where a1 , a2 are constants. (a1, a2可利用 a0, a1算出)

  11. Example 3. What’s the solution of the recurrence relation an = an-1 + 2an-2 with a0=2 anda1=7 ? Sol : The characteristic equation is r2 – r - 2=0. Its two roots are r1= 2 and r2 = -1. Hence an=a12n +a2 (-1)n . ∵a0 = a1+a2 = 2, a1=2a1-a2=7 ∴a1 = 3, a2 = -1 an = 32n - (-1)n. 驗算:a2 = a1 + 2a0 =11a2= 322 -1 =11

  12. Example 4. Find an explicit formula for the Fibonacci numbers. Sol : fn = fn-1 + fn-2 , n  2, f0=0 , f1=1. The characteristic equation is r2 - r- 1=0. Its two roots are , . So we have

  13. Thm 2. Let an = c1an-1+c2an-2 be a recurrence relation with c1,c2R. If r2 -c1r-c2= 0 has only one root r0. Then the solution of an is an = a1  r0n + a2 n r0n for n=0,1,2,…, where a1 and a2 are constants.

  14. Example 5. What’s the solution of an= 6an-1 - 9an-2with a0=1 and a1=6 ? Sol : The root of r2 - 6r + 9 = 0 is r0 = 3. Hence an = a1.3n+a2.n.3n . ∵a0 = a1 = 1 a1 = 3a1 + 3a2 = 6 ∴ a1 = 1 and a2 = 1 an = 3n+ n.3n 驗算:a2 = 6a1 - 9a0 =27a2= 32 +2 32 =27

  15. Thm 3. Let an= c1an-1 + c2an-2 + … + ckan-k be a recurrence relation with c1, c2, …, ck R. If rk -c1rk-1 -c2rk-2 -…-ck= 0 has k distinct roots r1, r2,…, rk. Then the solution of an is an = a1r1n +a2r2n + …+akrkn, for n = 0, 1, 2, … where a1, a2,…ak are constants.

  16. Example 6 (k = 3) Find the solution of an = 6an-1 - 11an-2 + 6an-3 with initial conditions a0=2, a1=5 and a2=15 . Sol : The roots of r3 - 6r2 + 11r – 6 = 0 are r1 = 1, r2 = 2, and r3 = 3 ∴an = a1  1n + a2 2n + a3 3n ∵a0 = a1 + a2 + a3 = 2 a1 = a1 + 2a2 + 3a3 = 5 a2 = a1 + 4a2 + 9a3 = 15 ∴an = 1 - 2n+ 2  3n a1 = 1, a2 = -1, a3 = 2 驗算:a3 = 6a2 - 11a1+6a0 =47a3= 1 - 23 + 2  33=47

  17. Thm 4. Let an= c1an-1 + c2an-2 + … + ckan-k be a recurrence relation with c1, c2, …, ck R. If rk - c1rk-1 - c2rk-2 - … - ck= 0has t distinct roots r1, r2, …, rt with multiplicities m1, m2, …, mt respectively, where mi  1,i, and m1+ m2 +…+ mt = k, then (接下一頁)

  18. where ai,jare constants. (1  i t , 0 jmi-1)

  19. Example 8.Find the solution to the recurrence relation an = -3an-1 - 3an-2 -an-3with initial conditionsa0 = 1, a1 = -2 and a2 = -1. Sol : r3 + 3r2 + 3r + 1 = 0 has a single root r0 = -1 of multiplicity three. ∴ an= (a1+a2n+a3n2) r0n= (a1+a2n+a3n2)(-1)n ∵ a0 = a1 = 1 a1 = (a1+a2+a3)  (-1) = -2 a2 = (a1+a2+a3) = -1 ∴a1 = 1, a2 = 3, a3 = -2 an = (1+3n-2n2)  (-1)n 驗算:a3 = - 3a2 - 3a1-a0 =8a3= (1+33-232)(-1)3=8 Exercise : 3,13,15,19

  20. 7.4 Generating Functions. Def 1. The generating function for the sequence {an}is the infinite power series. G(x) = a0 + a1x +… + anxn +… = (若 {an}是finite,可視為是infinite,但後面的term都等於0)

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

  22. Example 3. Let mZ+ and ,for k = 0, 1, …, m. What is the generating function for the sequence a0, a1,…, am ? Sol : G(x) = a0 + a1x + a2x2 + … + amxm = (1+x)m(by 二項式定理)

  23. Example 5. The function f (x) = is the generating function of the sequence 1, a, a2, …, since = 1 + ax + a2x2 + …= when |ax| < 1 for a≠0

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

  25. Example 7. Find and Sol :

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

  27. 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

  28. ※Using Generating Functions to solve Recurrence Relations. Example 16. Solving the recurrence relation ak = 3ak-1 for k=1,2,3,… and initial condition a0 = 2. Sol : 另法: (by 7.2公式 ) r – 3 = 0 r = 3 an = a  3n ∵ a0 = 2 = a ∴ an = 2  3n

  29. Let be the generating function for {ak}. First note that ak xk= 3ak-1 xk  • G(x) -a0 = 3x G(x) • ∵a0 = 2 G(x) - 3x  G(x) = G(x)(1-3x) = 2 ∴ ak = 2  3k Exercise : 5,7,11,33

  30. A -|AB|-|AC|-|BC|後 +|ABC|後 |A|+|B|+|C| 時 各部分被計算的次數 C B 7.5 Inclusion-Exclusion 排容原理 A,B,C,D : sets 1 2 1 2 1 3 1 2 0 1 2 1 1

  31. Theorem 1. A1, A2, …, An : sets Exercise : 17

  32. 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)= {a, b, c} f (2)= ︰ ︰ f (6)= 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) =

  33. Thm 1.|A| = m , |B| = n There are onto functions f : A → B. pf : A = {a1, a2, …, am}. B = {b1, b2, …, bn} f (a1)= f (a2)= ︰ ︰ f (am)= b1, b2, …, bn

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

  35. Example 5. derangements of 12345 : 21453, 23451, 34512, … Def. Let Dn denote the number of derangements of n objects. D4 = (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) =

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

More Related