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Discrete Mathematics

Discrete Mathematics. Chapter 7 Advanced Counting Techniques. 大葉大學 資訊工程系 黃鈴玲 (Lingling Huang). Outline. 7.1 Recurrence Relations 7.2 Solving Linear Recurrence Relations 7.4 Generating Functions 7.5 Inclusion-Exclusion 7.6 Applications of Inclusion-Exclusion.

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Discrete Mathematics

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  1. Discrete Mathematics Chapter 7 Advanced Counting Techniques 大葉大學 資訊工程系 黃鈴玲(Lingling Huang)

  2. Outline 7.1 Recurrence Relations 7.2 Solving Linear Recurrence Relations 7.4 Generating Functions 7.5 Inclusion-Exclusion 7.6 Applications of Inclusion-Exclusion

  3. 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}?

  4. 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

  5. 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

  6. 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

  7. … 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 : Let an be the number of bit strings of length n that do not have two consecutive 0s. ∴ 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

  8. … 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

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

  10. 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, not linear) (False , not linear) (False, not homogeneous)

  11. Solving Linear Homogeneous RecurrenceRelations with Constant Coefficients 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)

  12. 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. r1與r2順序可交換，結果會一樣 驗算：a2 = a1 + 2a0 =11a2= 322 -1 =11

  13. 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

  14. 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.

  15. 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

  16. 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 rootsr1, r2,…, rk. Then the solution of an is an = a1r1n +a2r2n + …+akrkn, for n = 0, 1, 2, … where a1, a2,…ak are constants.

  17. 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

  18. 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= 0 hast distinct roots r1, r2, …, rtwith multiplicitiesm1, m2, …, mt respectively, where mi  1,i, and m1+ m2 +…+ mt = k, then where ai,jare constants for 1  it and 0 jmi-1.

  19. 補充說明： 若特徵方程式的root為：1 (2重根), -2 (3重根),3 (無重根) 則上述定理給出的通解為： an= (a1,1+a1,2 n)  1n+ (a2,1+a2,2 n +a2,3 n2)  (-2)n +a3,1  3n (其實變數a的下標可從1開始排起，只要不重複就好) an= (a1+a2 n)  1n+ (a3+a4 n +a5 n2)  (-2)n +a6  3n

  20. 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+2a2+4a3 = -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

  21. Linear Nonhomogeneous RecurrenceRelations with Constant Coefficients Example: an= 3an-1 + 2n A recurrence relation of the forman= c1an-1 + c2an-2 + … + ckan-k+ F(n), wherec1, c2, …, ckare real numbersandF(n) is a function not identically zero dependingonly onn. The recurrence relationan= c1an-1 + c2an-2 + … + ckan-k is called the associated homogeneous recurrence relation.

  22. Example 9: an= an-1 + 2n, associated h.r.r  an= an-1 an= an-1 + an-2 + n2+1, associated h.r.r  an= an-1 + an-2 an= 3an-1 + n3n, associated h.r.r  an= 3an-1 an= an-1 + an-3+ n!, associated h.r.r  an= an-1 + an-3

  23. Theorem 5. If {an (p)} is a particular solution (特解) of an= c1an-1 + c2an-2 + … + ckan-k+ F(n), then every solution is of the form {an (p) + an (h)}, where {an (h)} is a solution of an= c1an-1 + c2an-2 + … + ckan-k Proof. If {an (p)} and {bn} are both solutions of an= c1an-1 + c2an-2 + … + ckan-k+ F(n), then an(p)= c1an-1(p)+ c2an-2 (p)+ … + ckan-k(p)+ F(n), and bn = c1bn-1 + c2bn-2 + … + ckbn-k+ F(n).  an(p) - bn= c1(an-1 - bn-1)+ c2(an-2 - bn-2)+ … + ck(an-k - bn-k)  {an(p) - bn} is a solution of an= c1an-1 + c2an-2 + … + ckan-k  bn = an(p)+ an (h)

  24. Example 10. Find all solutions of the recurrence relation an = 3an-1 + 2n. What is the solution with a1=3? Sol : {associated的部分 an = 3an-1 先解} Characteristic equation: r– 3 = 0  r = 3 an(h)= a 3n. {particular solution} ∵ F(n) = 2n ∴ Letan(p)=cn+d, where c, d R. If an(p)= cn+d is a solution to an = 3an-1 + 2n, then cn+d = 3(c(n-1)+d)+2n =3cn - 3c + 3d+2n  2cn-3c + 2d+2n = (2c+2)n + (2d-3c) = 0 (任何n代入都需為0) ∴2c+2 = 0, and2d-3c= 0  c = -1, d = -3/2 an(p)= -n- 3/2  an=an(h)+an(p) = a 3n-n- 3/2 If a1= a 3-1- 3/2 = 3  a = 11/6  an= (11/6)3n- n- 3/2

  25. Example 11. Find all solutions of the recurrence relation an = 5an-1 -6an-2 + 7n. Sol : {associated的部分 an = 5an-1 -6an-2 先解} Characteristic equation: r2 – 5r + 6 = 0  r1= 3, r2= 2  an(h)= a1 3n + a2 2n. {particular solution} ∵ F(n) = 7n∴ Letan(p) = c7n, where c R. If an(p)= c7nis a solution to an = 5an-1 -6an-2 + 7n, then c7n = 5c7n-1 - 6c7n-2 + 7n  49c = 35c - 6c + 49  c = 49/20 an(p)= (49/20) 7n  an=an(h)+an(p) = a1 3n + a2 2n + (49/20) 7n Exercise : 23

  26. Theorem 6. an= c1an-1 + c2an-2 + … + ckan-k+ F(n), where F(n) = (btnt + bt-1nt-1 +…+ b1n + b0)sn. When s is not a root of the characteristic equationof the associated linear homogeneous recurrence relation, there is a particular solution of the form (ptnt + pt-1nt-1 +…+ p1n + p0)sn. When s is a root of the characteristic equation and its multiplicity is m, there is a particular solution of the form nm(ptnt + pt-1nt-1 +…+ p1n + p0)sn.

  27. Example 12. What form does a particular solution of the linear nonhomogeneous recurrence relationan = 6an-1 - 9an-2 + F(n) have when F(n) =3n, F(n) =n3n,F(n) =n22n , and F(n) = (n2+1)3n. Sol : The associated linear homogeneous recurrence relation is an = 6an-1 - 9an-2. characteristic equation: r2- 6r+ 9 = 0  r = 3 (2重根) F(n) =3n, and 3 is a rootan(p) = p0n23n F(n) =n3n,and 3 is a rootan(p) = n2(p1n+p0)3n F(n) =n22n , and 2 is not a rootan(p) = (p2n2+p1n+p0)2n F(n) = (n2+1)3n , and 3 is a rootan(p) = n2 (p2n2+p1n+p0)3n Exercise : 27

  28. The associated linear homogeneous recurrence relation is an = an-1 . Example 13. Find the solutions of the recurrence relation an = an-1 + n with a1=1. Sol : characteristic eq.: r- 1 = 0  r = 1  an(h) = c(1)n=c F(n) = n = n(1)n, and 1 is a root an(p) = n(p1n+p0)1n= p1n2+p0n 將an(p)代入an = an-1 + n  p1n2+p0n = p1(n-1)2+p0(n-1)+n (2p1-1)n+p0-p1=0  p1= ½, p0= p1= ½ an(p) =(n2+n)/2 an = an(p) + an(h) = (n2+n)/2+c Exercise : 29 a1=1  c=0  an = an(p) + an(h) = (n2+n)/2

  29. ex 40: Solve the simultaneous recurrence relations an = 3an-1 + 2bn-1 bn = an-1 + 2bn-1with a0 = 1 and b0 = 2. Sol : an -bn= 2an-1  bn = an-2an-1  an = 3an-1 + 2bn-1 = 3an-1 + 2an-1- 4an-2  r = 1, 4  an = 5an-1 - 4an-2  r2- 5r+ 4 = 0  an = a1+a24n a0 = a1+a2 = 1  a1 = -1, a2 = 2 a1 = a1+4a2 = 3a0 + 2b0 = 7  an= 24n-1  bn = an-2an-1= 24n-1-4n+2 = 4n+1

  30. 7.4 Generating Functions. Def 1.The generating function for the sequence a0, a1, a2,… of real numbers is the infinite series G(x) = a0 + a1x +… + anxn +… = (若數列{an}是finite，可視為是infinite，但後面的項都等於0)

  31. Example 1. Find the generating functions for the sequences {ak} with (1) ak= 3 (2) ak = k+1 (3) ak = 2k Sol : (1) G(x) = (2) G(x) = (3) G(x) =

  32. a0 a1 a2 a3 a4 a5 a6及之後都=0 Example 2. What is the generating function for the sequence 1,1,1,1,1,1 ? Sol : G(x) = = a0+ a1x + a2x2 + a3x3 +… (expansion，展開式) (closed form) Exercise : 2

  33. 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下面的二項式定理)

  34. Useful Facts About Power Series Example 4. The function f (x) = is the generating function of the sequence 1, 1, 1, …, because = 1 + x + x2 + …= when |x| < 1. Example 5. The function f (x) = is the generating function of the sequence 1, a, a2, …, because = 1 + ax + a2x2 + …= when |ax| < 1 for a≠0. Exercise : 5(a)(b), 11(a)

  35. Theorem 1. Let f(x) = and g(x) = . Thenf(x) + g(x) = . f(x) g(x) = (a0+a1x+a2x2 +…)(b0+b1x +b2x2+…) = (a0b0)+(a0b1+a1b0) x+(a0b2+a1b1+a2b0) x2+… =

  36. Example 6. Let f (x) = . Use Example 4 to find the coefficients a0, a1, a2,… in the expansionf (x)= . Sol : = 1 + x + x2 + … = = ak= k+1

  37. Def 2. Let uR and kN. Then the extended binomial coefficient is defined by Example 7. Find and Sol :

  38. Example 8 When the top parameter is a negative integer, the extended binomial coefficient can be expressed in terms of an ordinary binomial coefficient.

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

  40. (跳過) Example 9. Find the generating functions for (1+x)-n and (1-x)-n where nZ+ Sol : By the Extended Binomial Theorem, By replacing x by –x we have Exercise : 11(b)(d)

  41. Counting Problems and Generating Functions Generating functions can be used to count the numberof combinations of various types. Example 10. Find the number of solutions of e1 + e2 + e3 = 17, where e1, e2, e3are integers with 2 e1  5, 3 e2  6, and 4 e3  7. Sol : The number of solutions with the indicated constraints is the coefficient of x17 in the expansionof (x2 + x3 + x4 + x5)(x3 + x4 + x5 + x6)(x4 + x5 + x6 + x7) (即相當於找 e1, e2, e3 使 xe1 xe2 xe3 = x17) (e1, e2, e3)=(4, 6, 7), (5, 5, 7), (5, 6, 6) 共3種

  42. Example 11. In how many different ways can eight identical cookies be distributed among three distinct children if each child receives at least two cookies and no more than four cookies? Sol : The number of solutions is the coefficient of x8 in the expansion of (x2 + x3 + x4)3 (c1, c2, c3) = (2, 2, 4), (2, 3, 3), (2, 4, 2), (3, 2, 3), (3, 3, 2), (4, 2, 2) 共6種 Exercise: 23

  43. ※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節Thm 1公式) r – 3 = 0 r = 3 an = a  3n ∵ a0 = 2 = a ∴ an = 2  3n

  44. Let be the generating function for {ak}. First note that ak = 3ak-1   G(x) -a0 = 3x G(x) ∵a0 = 2 G(x) - 3x  G(x) = G(x)(1-3x) = 2 從k = 1開始，以避免ak-1變成a-1 ∴ ak = 2  3k Exercise : 5,7,11,33

  45. Example 17 Solving ak= 8ak-1+10k-1 for k =1,2,3,… and initial condition a1 = 9. (Sec. 7.1 Example 7) Sol : Let a0 = 1 (為計算方便而假設). Let be the generating function for {ak}.

  46. ∴ ak = (10k+ 8k)/2 Exercise: 33

  47. 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

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

  49. 7.6 Applications of Inclusion and Exclusion Example 1. How many solutions does x1 + x2 + x3 = 11have,where x1, x2, x3are nonnegative integers withx1  3, x2  4, andx3  6? Sol : Let a solution have property P1 if x1  4, property P2 if x2  5, and property P3 if x3  7. N(P1’P2’P3’) = N - N(P1) - N(P2)- N(P3) + N(P1P2) + N(P2P3) + N(P1P3) - N(P1P2P3)

  50. The Number of Onto Functions 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 ? The numberof 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) =

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