Recursive Definitions

Rosen, 3.4

Recursive

(or inductive)

Definitions

• Sometimes easier to define an object in terms of itself.

• This process is called recursion.

– Sequences • {s 0 ,s 1 ,s 2 , …} defined by s 0 constants a and b and n  = a and s n Z+ – Sets • 3  S and x+y  S if x  S and y  S = 2s n-1 + b for – Functions • Example: f(n) = 2 n , f(n) = 2f(n-1) and f(0) = 1

Recursively Defined Functions

To define a function with the set of nonnegative integers as its domain 1. Specify the value of the function at zero (or sometimes, it first k terms).

2. Give a rule for finding its value at an integer from its values at smaller integers.

Examples of Recursively Defined Functions

• Factorial Function n!

  n! = n(n-1)(n-2)….(1) f(0) = 1, f(n) = n(f(n-1)) • a n  f(0) = 1, f(n+1) = f(n)*a • Fibonacci Numbers  f 0 =0, f 1 =1, f n+1 = f n  {0,1,1,2,3,5,8,13,...} + f n-1

Prove that the nth term in the Fibonacci sequence is 1  n  2  i  0 f i when n  2 Induction Proof: Basic Step: Let n = 2, then f 2 true for 2  n  n = k+1, i.e., that f k+1  (1  = 1 =  i  0 f i ) 1  i 2  2   0 f i =1+0 Inductive Step: Consider k  2 and assume that the expression is k. We must show that the expression is true for k 1 f k+1 = f k  +f k-1 (1  by definition k  2  i  0 f i )  f k  1 by the inductive hypothesis

Fibonacci Proof

(cont.)  (1  k  2  i  0 f i )  f k  1  (1  k  1  i  0 f i ) Since f 2 is true and [f true, then f n n is true for 2  n  k  f k+1 ] is is true for all positive integers n  2.

Recursively Defined Set Example

Let the set S be defined recursively as 3  and x+y  S if x  S and y  S.

S Prove that S is equal to the set of positive integers divisible by 3 (set A).

Proof:

To show that S = A, we must show that A  S and that S  A.

Recursive Set Example

(cont.) First we will show that A  S. To do this we must show that every positive integer divisible by 3 is in S. This is the same as saying that 3n is in S for n  Z+. We will do an inductive proof.

Let p(n) be the statement that 3n  S for n a positive integer.

Basis Step: p(1) = 3(1) = 3 Inductive Step: We will show that p(n)  p(n+1) Let p(n)  S. Then p(n+1) = 3(n+1) = 3n + 3. Since 3n  S and 3  S, then 3n+3  S.

Therefore every element in A must be in S.

Recursive Set Example

(cont.) Now we will show that S  A. To do this we must show that every element in S is divisible by 3, i.e., that it can be represented as 3(j) where j is a positive integer.

Basis Step: 3  A since 3 = 3(1) Inductive Step: Assume x,y are in A (where x,y  S). x and y are divisible by 3 since they are both in A. Therefore x = 3j and y = 3k for j,k  Z+. Then z = x+y = 3j+3k = 3(j+k), which is divisible by 3, so z  A (and by definition, z  S).

Therefore every element in S must be in A

Prove f

n

>

 n-2

where



whenever n



3 = (1+



5)/2

Inductive Proof

Basis Step: First we will show f n >  n-2 for f 3 and  f 4 .

3-2 =  1 = (1+  5)/2 < (1+  9)/2 = 4/2 = 2 = f 3  4-2 =  2 = [(1+  5)/2] 2 = (1+ 2  5 + 5)/4 = (6+ 2  5)/4 = 3/2 +  5/2 < 3/2 +  9/2 = 3 = f 4

f

n

>

 n-2 (cont.) Inductive Step: Assume that f k >  k-2 is true for all nonnegative integers 3  k  We must show that f n+1 >  n-1 .

n when n>4 . Lemma:  2 = [(1+  5)/2] 2 = (1+2  5 +5)/4 = 3/2 + (  5)/2 = 1 + (1+  5)/2 = 1+ 

f

n

>

 n-2 (cont.)  n-1 =  2  n-3 = (  +1)(  n-3 ) =  n-2 +  n-3 By our inductive hypothesis f f n-1 n+1 > = f  n n-3 + f and f n n-1 >  n-2 >  n-2 +  n-3 =  n-1 Since f 3 and f 3 true, then f n are true and f k is true for 3  k  is true for all positive integers n  3 n  f n+1 is

Find a closed form solution to the recursive equation: T(n) = T(n-1) + c 1 . T(0) = c 0

One approach is to write down several terms, guess what the equation is, and then prove that your guess is correct (or not) using an induction proof.

T(0) = c 0 T(1) = T(0) + c 1 = c 0 T(2) = T(1) + c 1 = c 0 + c 1 + 2c 1 T(3) = T(2) + c 1 = c 0 T(4) = T(3) + c 1 = c 0 + 3c 1 + 4c 1

Guess that T(n) = c 0 + nc 1

Induction proof

Basis Step: for n = 0, T(0) = c 0 + (0)c 1 = c 0 Inductive Step: Assume that T(n) = c 0 we must show that T(n+1) = c 0 + nc + (n+1)c 1 1 .

, T(n+1) = T(n) + c 1 (n+1)c 1 .

= c 0 + nc 1 + c 1 = c 0 +

Find a closed form solution to T(1) = c

0

, T(n) = 2T(n-1)+c

1 T(1) = c 0 T(2) = 2T(1) + c 1 = 2c 0 + c 1 T(3) = 2T(2)+c 1 = 2(2c 0 +c 1 )+c 1 = 4c 0 +3c 1 T(4) = 2T(3)+c 1 = 2(4c 0 +3c 1 )+c 1 = 8c 0 +7c 1 T(5) = 2T(4)+c 1 = 2(8c 0 +7c 1 )+c 1 = 16c 0 +15c 1

Guess that T(n) = 2 n-1 c 0 + (2 n-1 -1)c 1

Prove that T(1) = c 0 , T(n) = 2T(n-1)+c 1 has closed form solution T(n) = 2 n-1 c 0 + (2 n-1 -1)c 1 Basis Step: T(1) = 2 1-1 c 0 + (2 1-1 -1)c 1 = c 0 Induction Step: Assume that T(n) = 2 n-1 c 0 (2 n-1 -1)c 1 . We must show that T(n+1) = + 2 (n+1)-1 c 0 + (2 (n+1)-1 -1)c 1 = 2 n c 0 + (2 n -1)c 1 .

T(n+1) = 2T(n) + c 1 c 1 = 2 n c 0 2 n c 0 + (2 n + (2 -1)c n 1 .

-2)c 1 = 2[2 n-1 c 0 + c 1 + (2 n-1 -1)c 1 ]+ = 2 n c 0 + 2 n c 1 - c 1 =