Recursive Definitions

1
Recursive Definitions

• Rosen, 3.3

2
Recursive (or inductive) Definitions

• Sometimes easier to define an object in terms of
  itself.
• This process is called recursion.
• Sequences
• s0,s1,s2, defined by s0 a and sn 2sn-1
  b for constants a and b and n? Z
• Sets
• 3 ? S and xy ? S if x?S and y?S
• Functions
• Example f(n) 2n, f(n) 2f(n-1) and f(0) 1

3
Recursively Defined Functions

• To define a function with the set of nonnegative
  integers as its domain
• Specify the value of the function at zero (or
  sometimes, it first k terms).
• Give a rule for finding its value as an integer
  from its values at smaller integers.

4
Examples of Recursively Defined Functions

• Factorial Function n!
• n! n(n-1)(n-2).(1)
• f(0) 1, f(n) n(f(n-1))
• an
• f(0) 1, f(n1) ana
• Fibonacci Numbers
• f00, f11, f n1 fn f n-1
• 0,1,1,2,3,5,8,13,...

5
Prove that the nth term in the Fibonacci sequence
is when n?2

Induction Proof Basic Step Let n 2, then f2
1 10 Inductive Step Consider k?2
and assume that the expression is true for 2 ? n
? k. We must show that the expression is true
for n k1, I.e., that fk1 fk1 fk fk-1
by definition by the inductive
hyposthesis
6
Fibonacci Proof (cont.)
Since f2 is true and fn is true for 2 ? n ? k ?
fk1 is true, then fn is true for all positive
integers n?2.
7
Recursively Defined Set Example

• Let the set S be defined recursively as 3 ? S and
  xy ? S if x?S and y?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.

8
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(n1)
• Let p(n) ? S. Then p(n1) 3(n1) 3n 3.
  Since 3n?S and 3?S, then 3n3?S.
• Therefore every element in A must be in S.

9
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 z xy ? S and x,y are in A. 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 xy 3j3k 3(jk) which is divisible by 3
  so z ?A
• Therefore every element in S must be in A

10
Prove fn gt ?n-2 where ? (1?5)/2 whenever n?3

• Inductive Proof
• Basis Step First we will show fn gt ?n-2 for f3
  and f4.
• ?3-2 ?1 (1?5)/2 lt (1?9)/2 4/2 2 f3
• ?4-2 ?2 (1?5)/22 (1 2?5 5)/4 (6
  2?5)/4 3/2 ?5/2 lt 3/2 ?9/2 3 f4

11
fn gt ?n-2 (cont.)

• Inductive Step
• Assume that fk gt ?k-2 is true for all nonnegative
  integers 3 ? k ? n when ngt4 . We must show that
  fn1gt ?n-1.
• Lemma ?2 (1?5)/22 (12?5 5)/4 3/2
  (?5)/2 1 (1?5)/2 1?

12
fn gt ?n-2 (cont.)

• ?n-1 ?2?n-3 (?1)(?n-3) ?n-2?n-3
• By our inductive hypothesis
• fn-1 gt ?n-3 and fn gt ?n-2
• fn1 fn fn-1 gt ?n-2?n-3 ?n-1
• Since f3 and f3 are true and fk is true for 3 ?
  k ? n ? fn1 is true, then fn is true for all
  positive integers n?3

13
Find a closed form solution to the recursive
equation T(n) T(n-1) c1. T(0) c0

• 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) c0
• T(1) T(0) c1 c0 c1
• T(2) T(1) c1 c0 2c1
• T(3) T(2) c1 c0 3c1
• T(4) T(3) c1 c0 4c1
• Guess that T(n) c0 nc1

14
Induction proof

• Basis Step for n 0, T(0) c0 (0)c1 c0
• Inductive Step Assume that T(n) c0 nc1, we
  must show that T(n1) c0 (n1)c1.
• T(n1) T(n) c1 c0 nc1 c1 c0 (n1)c1.

15
Find a closed form solution to T(1) c0, T(n)
2T(n-1)c1

• T(1) c0
• T(2) 2T(1) c1 2c0 c1
• T(3) 2T(2)c1 2(2c0c1)c1 4c03c1
• T(4) 2T(3)c1 2(4c03c1)c1 8c07c1
• T(5) 2T(4)c1 2(8c07c1)c1 16c015c1
• Guess that T(n) 2n-1c0 (2n-1-1)c1

16
Prove that T(1) c0, T(n) 2T(n-1)c1has closed
form solution T(n) 2n-1c0 (2n-1-1)c1

• Basis Step T(1) 21-1c0 (21-1-1)c1 c0
• Induction Step Assume that T(n) 2n-1c0
  (2n-1-1)c1. We must show that T(n1)
  2(n1)-1c0 (2(n1)-1-1)c1 2nc0 (2n-1)c1.
• T(n1) 2T(n) c1 22n-1c0 (2n-1-1)c1 c1
  2nc0 (2n-2)c1 c1 2nc0 2nc1 - c1 2nc0
  (2n-1)c1.