UCI ICS/Math 6D5-Recursion -1 Strong Induction “Normal” Induction “Normal” Induction: If we prove that 1) P(n 0 ) 2) For any k≥n 0, if P(k) then P(k+1) Then P(n) is true for all n≥n 0. “Strong” Induction “Strong” Induction: If we prove that 1) Q(n 0 ) 2) For any k≥n 0, if Q(j) for all j from n 0 to k then Q(k+1) Then Q(k) is true for all k≥n 0. These two are equivalent! Why? Set P(n) to be the predicate “If Q(j)” Why? Set P(n) to be the predicate “If n 0 ≤ j ≤ n then Q(j)” –P(n 0 ) ≡ Q(n 0 ) –P(k) ≡ Q(j) holds for all j between n 0 and k

UCI ICS/Math 6D5-Recursion -2 Proof by Strong Induction: Jigsaw Puzzle Each “step” in assembling a jigsaw puzzle consists of putting together two 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 saw puzzle of n pieces.” Basis Step: P(1) is (trivially) true. Inductive Step: We assume P(j) true for all j≤k and we’ll argue P(k+1): The last step in assembling a puzzle of k+1 pieces is to put together two blocks: one of size j>0 and one of size k+1-j, for some j. Since 0<j,k+1-j≤k, P(j) and P(k+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) + ((k+1-j)-1) = k = (k+1)-1. (This implies P(k+1), and hence ends the inductive part, and thus also the proof.)

UCI ICS/Math 6D5-Recursion -3 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: - First 4 cases: 18=2*7+4, 19=7+3*4, 20=5*4, 21=3*7 - Afterwards, using strong induction: Take any k≥21. By (strong) inductive assumption, and by the 4 cases above, k-3=i*7+j*4 for some i,j. Therefore k+1=i*7+(j+1)*4, which ends the inductive step. (Btw, starting from 18 may seem weird, but you just can’t do it for 17...)

UCI ICS/Math 6D5-Recursion -4 Basis for Induction: Integers are well ordered Induction is based on the fact that the integers are “well ordered”, i.e. 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 smallest element i.e. e in S s.t. for all x in S we have e ≤ x. 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)=T for all n≥0 (thus showing that induction works) : Define (***) S = {n>0 | P(n)=F}. Assume S non-empty. By well-ordering of N, there is a least element e of S. Consider element e–1: Since P(e) = F then by (**) also P(e–1) = F. Note that e–1 cannot be equal to 0 because P(0) = T by (*). But then by (***), e–1 is in S, so e is not the least element of S…  Contradiction! Therefore S must be empty. Therefore P(n)=T for all n ≥ 0.

UCI ICS/Math 6D5-Recursion -5 Well Ordering can be used directly to prove things (i.e. not necessarily via induction)  Z Theorem: If a and b are integers, not both 0, then  s,t  Z (s*a + t*b = gcd( a,b )).  Z Proof: For any a,b define S = {n>0 |  s,t  Z n= s*a + t*b}. By the well-ordering of Z, S has a smallest element, call it d. Choose s and t so that d = s*a + t*b. Claim: d is a common divisor of a and b. Proof that d | a: Writing d = q*a + r where 0≤r<d. If r=0 then d | a. If r>0 then since r = d – q*a = (s*a + t*b) – q*a = (s - q)*a + t*b  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 s*a + t*b = d.

UCI ICS/Math 6D5-Recursion -6 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.

UCI ICS/Math 6D5-Recursion -7 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 (Btw, we can also use F(n) instead of f n to designate n-th Fibonacci number.) The first 18 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: Question: Are they growing exponentially, i.e. like f n ≈ a n for some a?

UCI ICS/Math 6D5-Recursion -8 Fibonacci numbers are related to a Golden Section i.e. the splitting of an interval so that (larger part) / whole = (smaller p. / larger p.) x y Ref:

UCI ICS/Math 6D5-Recursion -9 Fibonacci Growth Theorem: If n≥3 then f n > φ n-2 where φ is the solution to golden ratio, i.e. φ = (1+  5) / 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… Fact you can check : 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

UCI ICS/Math 6D5-Recursion -10 Recursively Defined Sets Always (1) Basis Step and (2) Recursive Step  S ; (2) If x  S and y  S,x+y  S Set S of 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 (1) L( Now recursively define length of a string, L: Σ*→N where (1) L(λ)=0 ; (2) L(wx)=L(w)+1 u   String Catenation (“”) (1) If u  Σ*, then uλ=u (2) If u,w  Σ* and x  Σ, then u(wx)=(uw)x

UCI ICS/Math 6D5-Recursion -11 Recursively Defined Sets Always (1) Basis Step and (2) Recursive Step Well-Formed Boolean Formula. (1)T, F, and s, where s is a propositional variable are all well-formed formula (WFF). (2)If E and F are WFF’s, then (¬E), (E  F), (E  F), (E  F), and (E↔F) are all WFF’s. Well-Formed Arithmetic Expression (1)x is a well-formed arithmetic expression (WFA) if x is either a numeral or a variable. (2)If F and G are WFA’s, then (F+G), (F-G), (F*G), (F/G), and (E  F) are all WFA’s.

UCI ICS/Math 6D5-Recursion -12 Recursively Defined Structures: 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 graph formed by starting with a root, r, which is not in any of these trees and adding an edge from r to each of these roots is also a rooted tree, whose root is r. Basis Step 1 Step 2...

UCI ICS/Math 6D5-Recursion -13 Recursively Defined Structures: Extended Binary Trees Extended Binary Trees (1) The empty set is an Extended Binary Tree (2) If T 1 and T 2 are extended binary trees, then the following tree, denoted T 1T 2, is also an extended binary tree: pick a new root r, and attach T 1 as the left subtree and T 2 as the right subtree. Step 1 Step 2 Step 3...

UCI ICS/Math 6D5-Recursion -14 Full Binary Trees Full Binary Trees (1) A single vertex is a Full Binary Tree (2) If T 1 and T 2 are full binary trees, then the following tree, denoted T 1T 2, is also a full binary tree: pick a new root, r, and attach T 1 as the left subtree and attach T 2 as the right subtree. (Definition change only in the base case, but unlike Extended Binary Trees, Full Binary Tree has exactly 0 or 2 child-nodes) Base Step 1 Step 2

UCI ICS/Math 6D5-Recursion -15 (Recursive) Definitions of Functions on Full Binary Trees Definition of Height, h(T), of a Full Binary Tree T: (1) If T has a single (root) node/vertex then h(T)=0. (2) O/w h(T 1 T 2 )=1+max(h(T 1 ),h(T 2 )) The Number of vertices, n(T), of a Full Binary Tree, T, is given by (1) If T has a single (root) node/vertex, n(T)=1. (2) n(T 1 T 2 )=1+n(T 1 )+n(T 2 )

UCI ICS/Math 6D5-Recursion -16 Structural Induction If a set is recursively defined, to show a predicate true for all elements in the set one needs to: (1) Show the predicate true on all base cases; (2) Show that if the predicate is true for each of the elements used to construct a new element, then the same predicate is also true for that new element. E.g. to prove P(T) holds for all T ∈ FBT, these steps are: (1) showing P(vertex) holds, because base case of FBT definition is just a vertex (2) Showing that if P(T 1 ) and P(T 2 ) then P(T 1 T 2 ), because the recursive clause of FBT definition defines a new FBT as T 1 T 2. Example: We’ll show that if T is a full binary tree, then n(T) ≤ 2 h(T)+1 -1 (proof on next slide)

UCI ICS/Math 6D5-Recursion -17 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 therefore n(T)=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) Inductive 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)

UCI ICS/Math 6D5-Recursion -18 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)