ICS141: Discrete Mathematics for Computer Science I

Transcription

1 ICS141: Discrete Mathematics for Computer Science I Dept. Information & Computer Sci., Jan Stelovsky based on slides by Dr. Baek and Dr. Still Originals by Dr. M. P. Frank and Dr. J.L. Gross Provided by McGraw-Hill 13-1

2 Quiz 1. State the 1 st Principle of Mathematical Induction 2. What is the difference between the 1 st, 2 nd and strong principles of Mathematical Induction. (Describe in plain English) 3. What is the big-o complexity of Euclid's Algorithm? 13-2

3 Lecture 21 Chapter 4. Induction and Recursion 4.3 Recursive Definitions and Structural Induction 13-3

4 Recursive Definitions n In induction, we prove all members of an infinite set satisfy some predicate P by: n proving the truth of the predicate for larger members in terms of that of smaller members. n In recursive definitions, we similarly define a function, a predicate, a set, or a more complex structure over an infinite domain (universe of discourse) by: n defining the function, predicate value, set membership, or structure of larger elements in terms of those of smaller ones. 13-4

5 Recursion n Recursion is the general term for the practice of defining an object in terms of itself n or of part of itself. n This may seem circular, but it isn t necessarily. n An inductive proof establishes the truth of P(k+1) recursively in terms of P(k). n There are also recursive algorithms, definitions, functions, sequences, sets, and other structures. 13-5

6 Recursively Defined Functions n Simplest case: One way to define a function f:n S (for any set S) or series a n = f(n) is to: n Define f(0) n For n > 0, define f(n) in terms of f(0),,f(n 1) n Example: Define the series a n = 2 n where n is a nonnegative integer recursively: n a n looks like 2 0, 2 1, 2 2, 2 3, n Let a 0 = 1 n For n > 0, let a n = 2 a n

7 Another Example n Suppose we define f(n) for all n N recursively by: n Let f(0) = 3 n For all n > 0, let f(n) = 2 f(n 1) + 3 n What are the values of the following? n f(1) = 2 f(0) + 3 = = 9 n f(2) = 2 f(1) + 3 = = 21 n f(3) = 2 f(2) + 3 = = 45 n f(4) = 2 f(3) + 3 = =

8 Recursive Definition of Factorial n Give an inductive (recursive) definition of the factorial function, F(n) = n! = 1 i n i = 1 2 n n Basis step: F(1) = 1 n Recursive step: F(n) = n F(n 1) for n > 1 n F(2) = 2 F(1) = 2 1 = 2 n F(3) = 3 F(2) = 3 {2 F(1)} = = 6 n F(4) = 4 F(3) = 4 {3 F(2)} = 4 {3 2 F(1)} = =

9 The Fibonacci Numbers n The Fibonacci numbers f n 0 is a famous series defined by: f 0 = 0, f 1 = 1, f n 2 = f n 1 + f n

10 Inductive Proof about Fibonacci Numbers n Theorem: f n < 2 n. Implicitly for all n N n Proof: By induction n Basis step: f 0 = 0 < 2 0 = 1 f 1 = 1 < 2 1 = 2 Note: use of base cases of recursive definition n Inductive step: Use 2 nd principle of induction (strong induction). Assume 0 i k, f i < 2 i. Then f k+1 = f k + f k 1 is < 2 k + 2 k 1 < 2 k + 2 k = 2 k

11 A Lower Bound on Fibonacci Numbers n Theorem: For all integers n 3, f n > α n 2, where α = ( /2 )/ n Proof. (Using strong induction.) n Let P(n) = (f n > α n 2 ). n Basis step: For n = 3, note that α n 2 = α < 2 = f 3. For n = 4, α n 2 = α 2 = ( /2 + 5)/4 = ( /2 )/ (= α + 1) < 3 = f

12 A Lower Bound on Fibonacci Numbers: Proof Continues n Inductive step: For k 4, assume P(j) for 3 j k, prove P(k+1). n f k+1 = f k + f k 1 > α k 2 + α k 3 (by inductive hypothesis, f k 1 > α k 3 and f k > α k 2 ). n Note that α 2 = α + 1. since ( /2 )/2 = ( /2 )/2 + 1 n Thus, α k 1 = α 2 α k 3 = (α + 1)α k 3 = αα k 3 + α k 3 = α k 2 + α k 3. n So, f k+1 = f k + f k 1 > α k 2 + α k 3 = α k 1. n Thus P(k+1)

13 Recursively Defined Sets n An infinite set S may be defined recursively, by giving: n A small finite set of base elements of S. n A rule for constructing new elements of S from previously-established elements. n Implicitly, S has no other elements but these. base element (basis step) construction rule (recursive step) n Example: Let 3 S, and let x+y S if x,y S. What is S? 13-13

14 Example cont. n Let 3 S, and let x+y S if x,y S. What is S? n 3 S (basis step) n 6 (= 3 + 3) is in S (first application of recursive step) n 9 (= 3 + 6) and 12 (= 6 + 6) are in S (second application of the recursive step) n 15 (= or 6 + 9), 18 (= or 9 + 9), 21 (= ), 24 (= ) are in S (third application of the recursive step) n so on n Therefore, S = {3, 6, 9, 12, 15, 18, 21, 24, } = set of all positive multiples of

15 The Set of All Strings n n Given an alphabet Σ, the set Σ * of all strings over Σ can be recursively defined by: n Basis step: λ Σ * (λ : empty string) n Recursive step: (w Σ * x Σ) wx Σ * Example: If Σ = {0, 1} then n λ Σ* (basis step) n 0 and 1 are in Σ* (first application of recursive step) n 00, 01, 10, and 11 are in Σ* (second application of the recursive step) n so on n Therefore, Σ* consists of all finite strings of 0 s and 1 s together with the empty string 13-15

16 String: Example n Show that if Σ = {a, b} then aab is in Σ*. Proof: We construct it with a finite number of applications of the basis and recursive steps in the definition of Σ*: 1. λ Σ* by the basis step. 2. By step 1, the recursive step in the definition of Σ* and the fact that a Σ, we can conclude that λa = a Σ*

17 Proof cont. 3. Since a Σ* from step 2, and a Σ, applying the recursive step again we conclude that aa Σ*. 4. Since aa Σ* from step 3 and b Σ, applying the recursive step again we conclude that aab Σ*. n Since we have shown aab Σ* with a finite number of applications of the basis and recursive steps in the definition we have finished the proof

18 Rooted Trees n Trees will be covered in more depth in chapter 10. n n n Briefly, a tree is a graph in which there is exactly one undirected path between each pair of nodes. An undirected graph can be represented as a set of unordered pairs (called arcs) of objects called nodes. Definition of the set of rooted trees: n Basis step: Any single node r is a rooted tree. n Recursive step: If T 1,, T n are disjoint rooted trees with respective roots r 1,, r n, and r is a node not in any of the T i s, then another rooted tree is {(r, r 1 ),, (r, r n )} T 1 T n

19 Illustrating Rooted Tree Definition n How rooted trees can be combined to form a new rooted tree r T 1 r 1 T 2 r 2 T n r n 13-19

20 Building Up Rooted Trees 13-20

21 Extended Binary Trees n A special case of rooted trees. n Recursive definition of extended binary trees: n Basis step: The empty set is an extended binary tree. n Recursive step: If T 1, T 2 are disjoint extended binary trees, then e 1 e 2 T 1 T 2 is an extended binary tree, where e 1 = if T 1 =, and e 1 = {(r, r 1 )} if T 1 and has root r 1, and similarly for e 2. (T 1 is the left subtree and T 2 is the right subtree.) 13-21

22 Building Up Extended Binary Trees 13-22

23 Lamé s Theorem n Theorem: a,b N, a b>0, and let n be the number of steps Euclid s algorithm needs to compute gcd(a,b). Then n 5k, where k = log 10 b +1 is the number of decimal digits in b. n Thus, Euclid s algorithm is linear-time in the number of digits in b. (or, Euclid s algorithm is O(log a)) n Proof: n Uses the Fibonacci sequence! (See next!) 13-23

24 Proof of Lamé s Theorem n Consider the sequence of division-algorithm equations used in Euclid s alg.: r 0 = r 1 q 1 + r 2 with 0 r 2 < r 1 r 1 = r 2 q 2 + r 3 with 0 r 3 < r 2 Where a = r 0, b = r 1, and gcd(a,b)=r n. r n 2 = r n 1 q n 1 + r n with 0 r n < r n 1 r n 1 = r n q n + r n+1 with r n+1 = 0 (terminate) n The number of divisions (iterations) is n. Continued on next slide 13-24

25 Lamé Proof cont. n Since r 0 r 1 > r 2 > > r n, each quotient q i r i 1 /r i 1. n Since r n 1 = r n q n and r n 1 > r n, q n 2. n So we have the following relations between r and f: r n 1 = f 2 r n 1 2r n 2f 2 = f 3 r n 2 r n 1 + r n f 2 + f 3 = f 4 r 2 r 3 + r 4 f n 1 + f n 2 = f n b = r 1 r 2 + r 3 f n + f n 1 = f n+1. n Thus, if n > 2 divisions are used, then b f n+1 > α n 1. n Thus, log 10 b > log 10 (α n 1 ) = (n 1)log 10 α (n 1)0.208 > (n 1)/5. n If b has k decimal digits, then log 10 b < k, so n 1 < 5k, so n 5k