1. UCI ICS/Math 6A, Summer 20075-Recursion -1 stasio@ics.uci.edu, after franklin@uci.edu Strong Induction “Normal” Induction “Normal” Induction: If we prove that 1) P(n 0 ) is true for some n 0 (typically 0 or 1), and 2) If P(k) is true for any k≥n 0, then P(k+1) is also true. Then P(n) is true for all n≥n 0. “Strong” Induction “Strong” Induction: If we prove that 1) Q(n 0 ) is true for some n 0 (typically 0 or 1), and 2) If Q(j) is true for all j from n 0 to k (for any k≥n 0 ), then Q(k+1) is also true. Then Q(k) is true for all k≥n 0. These 2 forms are equivalent: Let P(n) = “ These 2 forms are equivalent: Let P(n) = “n 0 ≤j≤n  Q(j)” Ref: http://en.wikipedia.org/wiki/Strong_inductionhttp://en.wikipedia.org/wiki/Strong_induction

2. UCI ICS/Math 6A, Summer 20075-Recursion -2 stasio@ics.uci.edu, after franklin@uci.edu Proof by Strong Induction: Jigsaw Puzzle Each “step” in assembling a jigsaw puzzle consists of putting together 2 already assembled blocks of pieces where each single piece is considered a block itself. P(n) = “It takes exactly n-1 steps to assemble a jigsaw puzzle of n pieces.” Basis Step: P(1) is (trivially) true. Inductive Step: We assume P(k) true for k≤n and we’ll argue P(n+1): The last step in assembling a puzzle with n+1 steps is to put together 2 blocks: one of size j>0 and one of size n+1-j. By since 0<j,n+1-j≤n, P(j) and P(n+1-j) are both assumed true. And so, the total number of steps to assemble a puzzle with n+1 pieces is 1 + (j-1) + ((n+1-j)-1) = n = (n+1)-1. (this implies P(n+1), and hence ends the inductive part, and thus also the whole proof)

3. UCI ICS/Math 6A, Summer 20075-Recursion -3 stasio@ics.uci.edu, after franklin@uci.edu More Examples of Theorems with easy Proofs using Strong Induction Thm1: The second player always wins the following game: –Starting with 2 piles each containing the same number of matches, players alternately remove any non-zero number of matches from one of the piles. –The winner is the person who removes the last match. Thm2: Every n>1 can be written as the product of primes. Thm3: Every postage amount of at least 18 cents can be formed using just 4-cent and 7-cent stamps.

4. UCI ICS/Math 6A, Summer 20075-Recursion -4 stasio@ics.uci.edu, after franklin@uci.edu Basis for Induction: Integers are well ordered Induction is based on the fact that the integers are “well ordered” in the following sense: Any non-empty set of integers which is bounded below i.e. there exists b (not necessarily in S) s.t. b ≤ x for all x in S contains a least element i.e. element e in S s.t. for all x in S we have e ≤ x. Fact: The integers are well ordered. Why does well-ordering imply induction? Assume P(0) and P(k)→P(k+1) for all k≥0. We’ll show that P(n) is true for all n≥0 (thus showing that induction works) : Define S = {n>0 | P(n) is not true}. Assume S non-empty. By well-ordering of N, there is a least element e of S. Let k=e-1≥0. If (not P(e)) then (not P(e-1)). We conclude that (not P(e-1)). e-1 cannot be equal to 0 because P(0) is true. But then e-1 must be in S, and so e is not the least element of S! => Contradiction => Therefore S must b empty. Therefore P(n) is true for all n>0. Since P(0) is also true, P(n) is true for all n≥0.

5. UCI ICS/Math 6A, Summer 20075-Recursion -5 stasio@ics.uci.edu, after franklin@uci.edu Well Ordering can be used directly to prove things (i.e. not necessarily via induction)  Z Thrm: If a and b are integers, not both 0, then  s,t  Z (sa + tb = gcd(a,b)).  Z Proof: For any a,b define S = {n>0 |  s,t  Z n= sa + tb}. By the well-ordering of Z, S has a smallest element, call it d. Choose s and t so that d = sa + tb. Claim: d is a common divisor of a and b. Proof that d|a: Writing d=qa+r where 0≤r<d. If r=0 then d|a. If r>0 then since r=d-qa=(sa+tb)-qa=(s-q)a+tb  we would have r  S. But since r<d it would mean that d is not the smallest element of S => contradiction => and therefore r=0 (and hence d|a). Similarly, d|b. d is the greatest common divisor since any common divisor of a and b must also divide sa+tb=d.

6. UCI ICS/Math 6A, Summer 20075-Recursion -6 stasio@ics.uci.edu, after franklin@uci.edu Recursive (Inductive) Definitions A function f:N→R is defined “recursively” by specifying (1) f(0), its value at 0, and (2) f(n), for n>0, in terms of f(1),….,f(n-1), i.e. in terms of f(k)’s for k<n. Examples: –f(n)=n! can be specified as f(0)=1 and, for n>0, f(n)=n*f(n-1). –For any a, f a (n)=a n, can be specified as f a (0)=1 and, for n>0, f a (n)=a*f a (n-1) Note: In many cases, we specify f(k) explicitly not only for f(0) but also for f(1), f(2), …, f(m) for some m>0, and then use a recursive formula to define f(n) for all n>m.

7. UCI ICS/Math 6A, Summer 20075-Recursion -7 stasio@ics.uci.edu, after franklin@uci.edu Fibonacci Numbers are Recursively Defined The Fibonacci numbers f 0,f 1,f 2,…,f n,…, are defined by (1) f 0 =0, f 1 =1 (2) f n =f n-1 +f n-2, for n≥2 Subscripts can be a real pain and can interfere with understanding. It is often convenient to use F(n) instead of f n The Fibonacci numbers are f 0 =0, f 1 =1, f 2 =1, f 3 =2, f 4 =3, f 5 =5, f 6 =8, f 7 =13, f 8 =21, f 9 =34, f 10 =55, f 10 =89, f 11 =144, f 12 =233, f 13 =377, f 14 =610, f 15 =987, f 11 =1597, f 12 =2584, f 13 =4181, f 14 =6765, f 15 =10946, f 16 =17711, f 17 =28657, f 18 =46368,... Ref: http://en.wikipedia.org/wiki/Fibonacci_numberhttp://en.wikipedia.org/wiki/Fibonacci_number

8. UCI ICS/Math 6A, Summer 20075-Recursion -8 stasio@ics.uci.edu, after franklin@uci.edu Fibonacci numbers are related to a Golden Section x y Ref: http://en.wikipedia.org/wiki/Golden_ratiohttp://en.wikipedia.org/wiki/Golden_ratio

9. UCI ICS/Math 6A, Summer 20075-Recursion -9 stasio@ics.uci.edu, after franklin@uci.edu Fibonacci Growth Theorem: Let φ=(1+  5)/2and n≥3 then f n > φ n-2 Proof by Strong Induction: Basis Step (for n=3 and 4): f 3 = 2 > φ =1.618… f 4 = 3 > φ 2 = (1+2  5+5)/4 = (3+  5)/2 = φ+1 = 2.168… (Note: Along the way, we showed that φ 2 = φ+1) Inductive Step: Assume f k >φ k-2 for all k s.t. 3≤k≤n. f n+1 = f n +f n-1 > φ n-2 +φ n-3 = φ n-3 (φ+1) = φ n-3 φ 2 = φ (n+1)-2

10. UCI ICS/Math 6A, Summer 20075-Recursion -10 stasio@ics.uci.edu, after franklin@uci.edu Recursively Defined Sets Always (1) “Basis Step” and (2) “Recursive Step”  S. (2) If x  S and y  S,x+y  S. Multiples of 3 (1) 3  S. (2) If x  S and y  S, then x+y  S. . (2) If w  and x ,wx . Strings Σ* over an alphabet Σ. Let λ be the empty string. (1) λ  Σ*. (2) If w  Σ* and x  Σ, then wx  Σ*. Examples: Examples: Σ={0,1}; Σ={0,1,2,3,4,5,6,7,8,9}; Σ={a,b,c,d,e,…,x,y,z} Now →N by (1) L( Now recursively define (“length”) L: Σ*→N by (1) L(λ)=0; (2)L(wx)=L(x)+1 w   String Catenation (“”) (1) If w  Σ*, then wλ=w (2) If u,w  Σ* and x  Σ, then u(wx)=(uw)x

11. UCI ICS/Math 6A, Summer 20075-Recursion -11 stasio@ics.uci.edu, after franklin@uci.edu Recursively Defined Sets Always (1) “Basis Step” and (2) “Recursive Step” Well-Formed Boolean Formulae. (1)T, F, and s, where s is a propositional variable are all well-formed Boolean formulae (WFBF). (2)If E and F are WFBF’s, then (¬E), (E  F), (E  F), (E  F), and (E↔F) are all WFBF’s. Well-Formed Arithmetic Expressions (1)x is a well-formed arithmetic expression (WFAE) if x is a numeral or a variable. (2)If F and G are WFAE’s, then (F+G), (F-G), (F*G), (F/G), and (E  F) are all WFAE’s.

12. UCI ICS/Math 6A, Summer 20075-Recursion -12 stasio@ics.uci.edu, after franklin@uci.edu Recursively Defined Structures Example 1: Rooted Trees Rooted Trees (1) A single vertex, r, is a rooted tree with root r. (2) Suppose that T 1,T 2,…,T n are rooted trees with roots r 1,r 2,…,r n respectively. Then the following graph is a rooted tree with root r: Take a free node r and add an edge from r to each of root r 1,…,r n. Basis Step 1 Step 2...

13. UCI ICS/Math 6A, Summer 20075-Recursion -13 stasio@ics.uci.edu, after franklin@uci.edu Recursively Defined Structures Example 2: Extended Binary Trees Extended Binary Trees (1) The empty set is an Extended Binary Tree (EBT) (2) If T 1 and T 2 are EBT’s, then the following tree, denoted T 1T 2, is also an EBT: Pick a new root, r, and attach T 1 as r’s left subtree and attach T 2 as r’s right subtree. - “Extended” means that even an empty set is considered a tree. - “Binary” means that every node has at most 2 children. Basis Step 1 Step 2 Step 3 One element: An Empty Set!...

14. UCI ICS/Math 6A, Summer 20075-Recursion -14 stasio@ics.uci.edu, after franklin@uci.edu Recursively Defined Structures Example 3: Full Binary Trees Full Binary Trees (1) A single vertex is an Full Binary Tree (FBT) (2) If T 1 and T 2 are FBT’s, then the following tree, denoted T 1T 2, is also an FBT: Pick a new root, r, and attach T 1 as r’s left subtree and attach T 2 as r’s right subtree. - Unlike “Extended” BT’s we don’t count an empty set in. - In “Full” BTs each node has exactly 0 or 2 children Base Step 1 Step 2 Step 3...

15. UCI ICS/Math 6A, Summer 20075-Recursion -15 stasio@ics.uci.edu, after franklin@uci.edu Recursive Definitions of Functions on recursively defined objects (e.g. Full Binary Trees) Trees were defined recursively. Natural definition of a function on trees will be recursive as well! Examples: The Height function, h(T): (1) If T has a single (root) node/vertex, h(T)=0. (2) o/w, i.e. if T=T 1 T 2,, then h(T) = h(T 1 T 2 ) = 1+max(h(T 1 ),h(T 2 )) The Number of vertices (function), n(T): (1) If T has a single (root) node/vertex, n(T)=1. (2) o/w, i.e. if T=T 1 T 2,, then n(T) = n(T 1 T 2 ) = 1+n(T 1 )+n(T 2 )

16. UCI ICS/Math 6A, Summer 20075-Recursion -16 stasio@ics.uci.edu, after franklin@uci.edu Structural Induction If a set is recursively defined, to show a result true for all elements in the set: (1) Show the result is true for all elements the basis step includes. (2) Show that if the result is true for each of the elements used to construct a new element, it is also true for that new element. Thrm: If T is a full binary tree, then n(T)≤2 h(T)+1 -1 (proof on next slide) Ref: http://en.wikipedia.org/wiki/Structural_inductionhttp://en.wikipedia.org/wiki/Structural_induction

17. UCI ICS/Math 6A, Summer 20075-Recursion -17 stasio@ics.uci.edu, after franklin@uci.edu Structural Induction Example Thrm: If T is a full binary tree, then n(T)≤2 h(T)+1 -1 Proof: Basis Step: If T is just the root vertex, n(T)=1, h(T)=0 and n(T)=1≤1=2 0+1 -1 Inductive Step: When T= T 1T 2, we compute n(T)=1+n(T 1 )+n(T 2 ) Definition of n(T) ≤ 1+(2 h(T 1 )+1 -1)+(2 h(T 2 )+1 -1) Induction hypothesis ≤ 2·max(2 h(T 1 )+1,2 h(T 2 )+1 )-1 Arithmetic = 2·2 1+max(h(T 1 ),h(T 2 )) -1 Arithmetic = 2·2 h(T) -1 Definition of h(T)

18. UCI ICS/Math 6A, Summer 20075-Recursion -18 stasio@ics.uci.edu, after franklin@uci.edu Recursive Algorithms An algorithm is “recursive” if it solves a problem by reducing it to an instance of the same problem with smaller input. Examples: 1.procedure factorial (n: nonnegative integer) if n=0 then factorial(n) := 1 else factorial(n) := n ·factorial(n-1) 2.procedure power(a: nonzero real, n: nonnegative integer) if n=0 then power(a,n) := 1 else power(a,n) := a·power(a,n-1) 3.procedure gcd(a,b: nonnegative integers with a<b) if a=0 then gcd(a,b) := b else gcd(a,b) := gcd(b mod a, a) 4.procedure fibonacci (n: nonnegative integer) if n≤1 then fibonacci(n) := 1 else fibonacci(n) := fibonacci(n-1) + fibonacci(n-2)