Induction and Recursion

Odd Powers Are Odd

Fact: If m is odd and n is odd, then nm is odd.

Proposition: for an odd number m, mk is odd for all non-negative integer k.

Proof by induction

Let P(i) be the proposition that mi is odd.

• P(1) is true by definition.

• P(2) is true by P(1) and the fact.

• P(3) is true by P(2) and the fact.

• P(i+1) is true by P(i) and the fact.

• So P(i) is true for all i.

The Induction Rule

0 and (from n to n +1),

proves 0, 1, 2, 3,….

R(0), R(n)R(n+1)

m. R(m)

Like domino effect…

For any n>=0

Very easy to prove

Much easier to prove with R(n) as a

n assumption.

Statements in green form a template for inductive proofs.

Proof: (by induction on n)

The induction hypothesis, P(n), is:

Proof by Induction

Let’s prove:

Proof by Induction

Base Case (n = 0):

11

1

r

r

0 12 0

? 11

11

rr r r

r

Wait: divide by zero bug!

This is only true for r 1

121.::

1(

11)P n r

rr r r

r

nn

12 1

11

nn r

r r rr

Theorem: 1. r

Induction Step: Assume P(n) for some n 0 and prove P(n + 1):

( ) 121.

11

1

rr r rr

r

+1+1

nn

Proof by Induction

Have P (n) by assumption:

So let r be any number 1, then from P (n) we have

12 1

11

rr r r

r

nn

How do we proceed?

Proof by Induction

111

11

nn nrr r r

r

+1 n

adding r n+1 to both sides,

1 1

( ) 1

1 ( 1)

1

1

1

n nr r r

r

r

r

+1n

But since r 1 was arbitrary, we conclude (by UG), that( ) 1

21.1

11

rr r rr

r

+1+1

nn

which is P (n+1). This completes the induction proof.

Summation

Try to prove:

Proving a Property

Base Case (n = 1):

Induction Step: Assume P(i) for some i 1 and prove P(i + 1):

Proving an Inequality

Base Case (n = 3):

Induction Step: Assume P(i) for some i 3 and prove P(i + 1):

Prove P(0).

Then prove P(n+1) assuming all of

P(0), P(1), …, P(n) (instead of just P(n)).

Conclude n.P(n)

Strong Induction

0 1, 1 2, 2 3, …, n-1 n.

So by the time we got to n+1, already know

all of

P(0), P(1), …, P(n)

Strong induction

Ordinary induction

equivalent

Claim: Every integer > 1 is a product of primes.

Prime Products

Proof: (by strong induction)

Base case is easy.

Suppose the claim is true for all 2 <= i < n.

Consider an integer n.

In particular, n is not prime.

So n = k·m for integers k, m where n > k,m >1.

Since k,m smaller than n,

By the induction hypothesis, both k and m are product of primes

k = p1 p2 p94

m = q1 q2 q214

Prime Products

…So

n = k m = p1 p2 p94 q1 q2 q214

is a prime product.

This completes the proof of the induction

step.

Claim: Every integer > 1 is a product of primes.

Available stamps:

5¢ 3¢

Theorem: Can form any amount 8¢

Prove by strong induction on n.

P(n) ::= can form (n +8)¢.

Postage by Strong Induction

What amount can you form?

Postage by Strong Induction

Base case (n = 0):

(0 +8)¢:

Inductive Step: assume (m +8)¢ for 0 m n,

then prove ((n +1) + 8)¢

cases:

n +1= 1, 9¢:

n +1= 2, 10¢:

case n +1 3: let m =n 2.

now n m 0, so by induction hypothesis have:

(n 2)+8

= (n +1)+8+

3

Postage by Strong Induction

We’re done!

In fact, use at most two 5-cent stamps!

Postage by Strong Induction

Given an unlimited supply of 5 cent and 7 cent stamps,

what postages are possible?

Theorem: For all n >= 24,

it is possible to produce n cents of postage from 5¢ and 7¢ stamps.

Recursive Definitions and Structural Induction

• Sometimes it’s easier to define an object in terms of itself. • This process is called Recursion.• Example: the sequence of powers of 2 is given by an = 2n

for n = 0, 1, 2, …. This sequence can also be defined by giving the first term of the sequence, namely, a0 = 1, and a rule finding a term of the sequence from the previous one, namely, an+1 = 2an, for n = 0, 1, 2, ….

18

Recursively Defined Functions

• Use two steps to define a function with the set of nonnegative integers as its domain:BASIS STEP: Specify that value of the function at zero. RECURSIVE STEP: Give a rule for finding its value at an integer from its values at smaller integers.

• Such a definition is called a recursive or inductive definition.• Examples:

– Suppose that f is defined recursively byf(0) = 3,f(n+1) = 2f(n) + 3Find f(1), f(2), f(3), and f(4).Solution: f(1) = 2f(0) + 3 = 2*3 + 3 = 9f(2) = 2f(1) + 3 = 2*9 + 3 = 21f(3) = 2f(2) + 3 = 2*21 + 3 = 45f(4) = 2f(3) + 3 = 2*45 + 3 = 93

19

Recursive Examples

– Give an inductive definition of the factorial function F(n) = n!.Solution: Basis Step: F(0) = 1Inductive Step: F(n+1) = (n+1)F(n)

E.g. Find F(5). F(5) = 5F(4) = 5*4F(3) = 5*4*3F(2) = 5 * 4 * 3 * 2F(1)

= 5 * 4 * 3 * 2 * 1F(0) = 5 * 4 * 3 * 2 * 1 * 1 = 120

– Give a recursive definition of .

Solution: Basis Step:

Inductive Step:20

n

kka

0

0

0

0

aak

k

n

knk

n

kk aaa

01

1

0

)(

• Example: Find the Fibonacci numbers f2, f3, f4, f5, and f6. Solution:f2 = f1 + f0 = 1 + 0 = 1,f3 = f2 + f1 = 1 + 1 = 2,f4 = f3 + f2 = 2 + 1 = 3,f5 = f4 + f3 = 3 + 2 = 5, f6 = f5 + f4 + 5 + 3 = 8.

• Example: Find the Fibonacci numbers f7. Solution ?

21

DEFINITION 1

The Fibonacci numbers, f0, f1, f2, …, are defined by the equations f0= 0, f1 =

1, and fn = fn-1 + fn-2

for n = 2, 3, 4, ….

Fibonacci Number

• The basis step of the recursive definition of strings says that the empty string belongs to ∑*. The recursive step states that new strings are produced by adding a symbol from ∑ to the end of strings in ∑*. At each application of the recursive step, strings containing one additional symbol are generated.

• Example: If ∑ = {0,1}, the strings found to be in ∑*, the set of all bit strings are λ, specified to be in ∑* in the basis step, 0 and 1 formed during the first application of the recursive step, 00, 01, 10, and 11 formed during the second application of the recursive step, and so on.

22

DEFINITION 2The set ∑* of strings over the alphabet ∑ can be defined recursively

by BASIS STEP: λ ∑* (where λ is the empty string containing no

symbols).RECURSIVE STEP: If w ∑* and x ∑ , then wx ∑*.

Recursively Defined Sets and Structures

We can define the set of well-formed formulae for compound statement forms involving T, F, propositional variables, and operators from the set {¬, Λ, V, →, ↔}. BASIS STEP: T, F, and s, where s is a propositional variable, are well-formed formulae.RECURSIVE STEP: If E and F are well-formed formulae, then (¬E), (E Λ F), (E V F), (E → F), and (E ↔ F) are well-formed formulae.

By the basis step we know that T, F, p, and q are well-formed formulae, where p and q are propositional variables. From an initial application of the recursive step, we know that (p V q), (p → F), (F → q), and (q Λ F) are well-formed formulae. A second application of the recursive step shows that ((p V q) → (q Λ F)), (q V (p V q)), and ((p → F) → T) are well-formed formulae.

23

Well-Formed Formulae for Compound Statement Forms.

24

DEFINITION 4The set of rooted trees, where a rooted tree consists of a set of

vertices containing a distinguished vertex called the root, and edges connecting these vertices, can be defined recursively by these steps:

BASIS STEP: A single vertex r is a rooted tree.RECURSIVE STEP: Suppose that T1, T2, …, Tn are disjoint rooted

trees with roots r1, r2, …, rn, respectively. Then the graph formed by starting with a root r, which is not in any of the rooted trees, T1, T2, …, Tn, and adding an edge from r to each of the vertices r1, r2, .., rn, is also a rooted tree.

Rooted Tree

25

26

DEFINITION 5The set of extended binary trees can be defined recursively by

these steps:

BASIS STEP: The empty set is an extended binary tree.RECURSIVE STEP: IfT1 and T2 are disjoint extended binary trees,

there is an extended binary tree, denoted by T1 ·T2 , consisting of a root r together with edges connecting the root to each of the roots of the left subtree T1 and the right subtree T2 when these trees are nonempty.

Extended Binary Tree

27

28

DEFINITION 6The set of full binary trees can be defined recursively by these steps:

BASIS STEP: There is a full binary tree consisting only of a single vertex r.RECURSIVE STEP: IfT1 and T2 are disjoint full binary trees, there is full

binary tree, denoted by T1 ·T2 , consisting of a root r together with

edges connecting the root to each of the roots of the left subtree T1 and the

right subtree T2.

Full Binary Tree

29

• Example: Give a recursive algorithm for computer n!, when n is a nonnegative integer.Basis Step: F(0) = 1Inductive Step: F(n+1) = (n+1)F(n)E.g. Find F(5).

F(5) = 5F(4) = 5*4F(3) = 5*4*3F(2) = 5 * 4 * 3 * 2F(1) = 5 * 4 * 3 * 2 * 1F(0) = 5 * 4 * 3 * 2 * 1 * 1 = 120

30

DEFINITION 1An algorithm is called recursive if it solves a problem by reducing it

to an instance of the same problem with smaller input.

Recursive Algorithms

ALGORITHM 1 A Recursive Algorithm for Computing n!.procedure factorial(n: nonnegative integer)if n = 0 then factorial(n):=1else factorial(n):=n*factorial(n-1)

• Example: Give a recursive algorithm for computer an, where a is a nonzero number and n is a nonnegative integer

31

Recursive Algorithms

ALGORITHM 2 A Recursive Algorithm for Computing an.procedure power(a: nonzero real number, n: nonnegative

integer)if n = 0 then power(a,n):=1else power(a,n):=a*power(a, n-1)

• Example: Express the linear search algorithm as a recursive procedure.

32

Linear Search

ALGORITHM 5 A Recursive Linear Search Algorithm.procedure search(i,j,x: i,j,x integers, 1 <=i<=n, 1<=j<=n)if ai = x then location := i

else if i = j thenlocation : = 0

else search(i+1,j, x)

• Example: Construct a recursive version of a binary search algorithm

33

Binary Search

ALGORITHM 6 A Recursive Binary Search Algorithm.procedure binarySearch(i,j,x: i,j,x integers, 1 <= i <= n, 1<= j

<= n)m := if x = am then location := m

else if (x < am and i < m ) then

binarySearch(x, i, m-1)else if (x > am and j > m ) then

binarySearch(x, m+1, j) else location := 0

2/)( ji

• Sometimes we start with the value of the computation at one or more integers, the base cases, and successively apply the recursive definition to find the values of the function at successive larger integers. Such a procedure is called iterative.

• How many additions are performed to find fibonacci(n)?

34

ALGORITHM 7 A Recursive Algorithm for Fibonacci Numbers.procedure fibonacci(n: nonnegative integer)if n = 0 then fibonacci(0) := 0else if (n = 1) then fibonacci(1) := 1else fibonacci(n) := fibonacci(n-1) + fibonacci(n-2)

Recursion and Iteration

35

f4

f3 f2

f2f1 f1

f0

f1f0

fn+1 -1 addition to find fn

36

ALGORITHM 8 An Iterative Algorithm for Computing Fibonacci Numbers.

procedure iterativeFibonacci(n: nonnegative integer)if n = 0 then y:= 0else

beginx := 0y := 1for i := 1 to n -1begin z := x + y x : = y y : = zend

end{y is the nth Fibonacci number}

n – 1 additions are used for an iterative algorithm.

• Recursive algorithm may require far more computation than an iterative one.

• Sometimes it’s preferable to use a recursive procedure even if it is less efficient.

• Sometimes an iterative approach is preferable.

37

Tradeoff