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Chapter 7 Advanced Counting Techniques

歐亞書局. Discrete Mathematics and Its Applications Sixth Edition By Kenneth Rosen. Chapter 7 Advanced Counting Techniques. 歐亞書局. 7.1 Recurrence Relations 7.2 Solving Linear Recurrence Relations 7.3 Divide-and-Conquer Algorithms and Recurrence Relations

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Chapter 7 Advanced Counting Techniques

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  1. 歐亞書局 Discrete Mathematics and Its Applications Sixth Edition By Kenneth Rosen Chapter 7Advanced Counting Techniques

  2. 歐亞書局 • 7.1 Recurrence Relations • 7.2 Solving Linear Recurrence Relations • 7.3 Divide-and-Conquer Algorithms and Recurrence Relations • 7.4 Generating Functions • 7.5 Inclusion—Exclusion • 7.6 Applications of Inclusion— Exclusion P. 1

  3. 7.1 Recurrence Relations • Definition 1: A recurrence relation for the sequence {an} is an equation that expresses an in terms of one or more of the previous terms of the sequence, namely, a0, a1, …, an-1, for all integers n with n>=n0, where n0 is a nonnegative integer. • A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation • Ex.1-2

  4. Modeling with Recurrence Relations • Ex.3: Compound interest • Ex.4: Rabbits and the Fibonacci numbers • Ex.5: The Tower of Hanoi • Ex.6 • Ex.7: Codeword enumeration • Ex.8

  5. FIGURE 1 (7.1) 歐亞書局 FIGURE 1 Rabbits on an Island. P. 452

  6. FIGURE 2 (7.1) 歐亞書局 FIGURE 2 The Initial Position in the Tower of Hanoi. P. 452

  7. FIGURE 3 (7.1) 歐亞書局 FIGURE 3 An Intermediate Position in the Tower of Hanoi. P. 453

  8. FIGURE 4 (7.1) 歐亞書局 FIGURE 4 Counting Bit Strings of Length n with No Two Consecutive 0s. P. 455

  9. 7.2 Solving Linear Recurrence Relations • Definition 1: A linearhomogeneous recurrence relation of degree k with constant coefficients:an=c1an-1+c2an-2+…+ckan-k,where c1,c2,…,ck are real numbers, and ck0. • Ex.1 • Pn=(1.11)Pn-1 • fn=fn-1+fn-2 • an=an-5

  10. Solving Linear Homogeneous Recurrence Relations with Constant Coefficients • an=c1an-1+c2an-2+…+ckan-k • an=rn is a solution iff rn=c1rn-1+c2rn-2+…+ckrn-k • rk-c1rk-1-c2rk-2-…-ck-1r-ck=0 • Characteristic equation • Characteristic roots • Theorem 1: Let c1 and c2 are real numbers. Suppose r2-c1r-c2=0 has two distinct roots r1 and r2. Then the sequence {an} is a solution of the recurrence relation an=c1an-1+c2an-2iffan=1r1n+ 2r2n for n=0, 1, 2, …, where 1 and 2 are constants. • Proof.

  11. Ex.3: an=an-1+2an-2, a0=2, a1=7. • Ex.4: Fibonacci numbers. • Theorem 2: Let c1 and c2 are real numbers with c20. Suppose r2-c1r-c2=0 has only one root r0. Then the sequence {an} is a solution of the recurrence relation an=c1an-1+c2an-2iffan=1r0n+ 2nr0n for n=0, 1, 2, …, where 1 and 2 are constants. • Ex. 5: an=6an-1-9an-2, a0=1, a1=6.

  12. Theorem 3: Let c1,c2,…,ck be real numbers. Suppose rk-c1rk-1-…-ck=0 has k distinct roots r1,r2,…, rk. Then the sequence {an} is a solution of the recurrence relation an=c1an-1+c2an-2+…+ckan-kiffan=1r1n+2r2n+…+krkn for n=0, 1, 2, …, where 1,2 …,k are constants. • Ex.6: an=6an-1-11an-2+6an-3, a0=2 , a1=5, a2=15.

  13. Theorem 4: Let c1,c2,…,ck be real numbers. Suppose rk-c1rk-1-…-ck=0 has t distinct roots r1,r2,…, rt with multiplicities m1,m2,…, mt such that mi>=1 and m1+m2+…+mt=k.Then the sequence {an} is a solution of the recurrence relation an=c1an-1+c2an-2+…+ckan-kiffan=(1,0+1,1n+…+1,m1-1)r1n +(2,0+2,1n+…+2,m2-1)r2n +…+(t,0+t,1n+…+t,mt-1)rtn for n=0, 1, 2, …, where i,i are constants. • Ex.7 • Ex.8: an=-3an-1-3an-2-an-3, a0=1 , a1=-2, a2=-1.

  14. Linear Nonhomogeneous Recurrence Relations with Constant Coefficients • Nonhomogeneous: • an=c1an-1+c2an-2+…+ckan-k+F(n) • Associated homogeneous recurrence relation: an=c1an-1+c2an-2+…+ckan-k • Ex.9: an=an-1+2n, an=an-1+an-2+n2+n+1, an=3an-1+n3n, an=an-1+an-2+an-3+n!

  15. Theorem 5: If {an(p)} is a particular solution of the nonhomogeneous recurrence relation with constant coefficients 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 the associated homogeneous recurrence relation an=c1an-1+c2an-2+…+ckan-k. • Proof. • Ex.10. an=3an-1+2n, a1=3. • Ex.11: an=5an-1-6an-2+7n.

  16. Theorem 6: Suppose {an} satisfies the nonhomogeneous recurrence relation with constant coefficients an=c1an-1+c2an-2+…+ckan-k +F(n), and F(n)=(btnt+bt-1nt-1+…+b1n+b0)sn. When s is not a root of the associate 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. • Ex.12: an=6an-1-9an-2+F(n), F(n)=3n, n3n, n22n, (n2+1)3n. • Ex.13: an=an-1+n.

  17. 7.3 Divide-and-Conquer Algorithms and Recurrence Relations • Divide-and-conquer algorithms • Divide: a problem of size n into a subproblems, each of size n/b • Conquer: combine the solutions of subproblems • Divide-and-conquer recurrence relation • f(n)=af(n/b)+g(n) • Ex.1: binary search • Ex.2: finding the maximum and minimum of a sequence • Ex.3: merge sort • Ex.4: fast multiplication of integers • Ex.5: fast matrix multiplication

  18. n=bk • f(n)=akf(1)+j=0..k-1ajg(n/bj) • Theorem 1: Let f be an increasing function that satisfies the recurrence relationf(n)=af(n/b)+c. Thenf(n) is O(nlogba) if a>1, O(logn) if a=1.When n=bk, f(n)=C1nlogba+C2, whereC1=f(1)+c/(a-1), C2=-c/(a-1) • Proof. • Ex. 6: f(n)=5f(n/2)+3, f(1)=7. Find f(2k). • Ex.7: binary search • Ex.8: locate the maximum and minimum elements in a seqeunce.

  19. Theorem 2 (Master Theorem): Let f be an increasing function that satisfies the recurrence relationf(n)=af(n/b)+cnd. When n=bk, Thenf(n) is O(nd) if a<bd, O(ndlogn) if a=bd, O(nlogba) if a>bd. • Ex.9: merge sort • Ex.10: fast multiplication algorithm • Ex.11: fast matrix multiplication • Ex.12: The Closest-Pair Problem

  20. FIGURE 1 (7.3) 歐亞書局 FIGURE 1 The Recursive Step of the Algorithm for Solving the Closest-Pair Problem. P. 480

  21. FIGURE 2 (7.3) 歐亞書局 FIGURE 2 Showing That There Are at Most Seven Other Points to Consider for Each Point in the Strip. P. 481

  22. 7.4 Generating Functions • To represent sequences efficiently by coding the terms of a sequence as coefficients of powers of a variable x in a formal power series • To solve recurrence relations • To prove combinatorial identities • To study properties of sequences

  23. Definition 1: The generating function for the sequence a0, a1, …, ak, … of real numbers is the infinite seriesG(x) = a0+a1x+…+akxk+…=k=0..akxk. • Ordinary generating functions • Ex.1 • Ex.2 • Ex.3

  24. Useful Facts about Power Series • Ex.4: f(x)=1/(1-x) • Ex.5: f(x)=1/(1-ax) • Theorem 1: Let f(x)=k=0..akxk and g(x)=k=0..bkxk. Then f(x)+g(x)=k=0..(ak+bk)xk and f(x) g(x)=k=0..(j=0..kajbk-j)xk and. • It’s valid only for power series that converge in an interval. • Ex.6: f(x)=1/(1-x)2.

  25. Definition 2: Let u be a real number and k a nonnegative integer. Then the extended binomial coefficient (u,k) is defined by (u,k)=u(u-1)…(u-k+1)/k! if k>0, or 1 if k=0. • Ex.7: (-2, 3), (1/2, 3) • Ex.8: (-n, r)=(-1)rC(n+r-1,r) • Theorem 2: (The Extended Binomial Theorem) Let x be a real number with |x|<1 and let u be a real number. Then(1+x)u= k=0..(u,k)xk. • Ex.9: (1+x)-n, (1-x)-n.

  26. TABLE 1 (7.4) 歐亞書局 P. 489

  27. Counting Problems and Generating Functions • Counting the r-combinations from n elements when repetition is allowed • e1+e2+…+en=C • Ex.10: e1+e2+e3=17, e1:2-5, e2:3-6, e3:4-7. • Ex.11 • Ex.12 • Ex.13: k-combinations from n elements. • Ex.14: r-combinations from n elements when repetition is allowed. • Ex.15: select r objects of n different kinds if we must select at least one object of each kind.

  28. Using Generating Functions to Solve Recurrence Relations • Ex.16: ak=3ak-1, a0=2. • Ex.17: an=8an-1+10n-1, a1=9.

  29. Proving Identities via Generating Functions • Ex.18

  30. 7.5 Inclusion-Exclusion • The principle of inclusion-exclusion • |AB|=|A|+|B|-|AB| • Ex.1: (Fig.1) • Ex.2: (Fig.2) • Ex.3

  31. FIGURE 1 (7.5) 歐亞書局 FIGURE 1 The Set of Students in a Discrete Mathematics Class. P. 501

  32. FIGURE 2 (7.5) 歐亞書局 FIGURE 2 The Set of Positive Integers Not Exceeding 1000 Divisible by Either 7 or 11. P. 501

  33. For three sets: (Fig. 3) • |AB C|=|A|+|B|+|C|-|AB|-|BC|-|CA|+|AB C| • Ex. 4 (Fig. 4)

  34. FIGURE 3 (7.5) 歐亞書局 FIGURE 3 Finding a Formula for the Number of Elements in the Union of Three Sets. P. 502

  35. FIGURE 4 (7.5) 歐亞書局 FIGURE 4 The Set of Students Who Have Taken Courses in Spanish, French, and Russian. P. 503

  36. Theorem 1: (The Principle of Inclusion-Exclusion) Let A1, A2, …, An be finite sets. Then|A1A2  … An|=i=1..n|Ai|-i,j=1..n|AiAj|+i,j,k=1..n|AiAj Ak|-…+(-1)n+1|A1A2  …An|. • Ex.5

  37. 7.6 Applications of Inclusion-Exclusion • An alternative form of inclusion-exclusion • To solve: the number of elements in a set that have none of n properties P1, P2, …, Pn. • Let Ai be the subset that have property Pi. • The number of elements with all the properties Pi1, Pi2, …, Pik: N(Pi1, Pi2, Pik). • N(P1’,P2’,…,Pn’)=N-|A1A2  … An|=N-i=1..nN(Pi)+i,j=1..nN(PiPj)-i,j,k=1..nN(PiPjPk)+…+(-1)nN(P1P2…Pn). • Ex.1

  38. The Sieve of Eratosthenes • To find all primes not exceeding a specified positive integer • For example, the primes not exceeding 100 • P1: divisible by 2 • P2: divisible by 3 • P3: divisible by 5 • P4: divisible by 7

  39. TABLE 1 (7.6) 歐亞書局 P. 508

  40. The Number of Onto Functions • Ex.2 • Theorem 1: Let m and n be positive integers with m>=n. Then, there arenm-C(n,1)(n-1)m+C(n,2)(n-2)m+…+(-1)n-1C(n,n-1)1monto functions from a set with m elements to a set with n elements. • =n!S(m,n), where S(m,n) is a Stirling number of the second kind • Ex.3

  41. Derangement • Derangement: a permutation of n objects that leave no objects in their original positions • Ex.4: The Hatcheck problem (later) • Ex.5 • Let Dn denote the number of derangements of n objects, D3=2 • Theorem 2: The number of derangements of a set with n elements is Dn=n![1-1/1!+1/2!-1/3!+…+(-1)n1/n!] • Proof.

  42. TABLE 2 (7.6) 歐亞書局 P. 512

  43. Thanks for Your Attention!

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