October 3, 2001 CSE 373, Autumn 2001 1

Mathematical Background

• Exponents

XA XB = XA+B

XA / XB = XA-B

(XA)B = XAB

XN+XN = 2XN

2N+2N = 2N+1

• Logarithms

Definition: XA = B if and only if logX B = A.

October 3, 2001 CSE 373, Autumn 2001 2

Logarithms, continued

• log AB = log A + log B

Proof: Let X = log A, Y = log B, and Z = log AB. Then 2X = A, 2Y = B, and 2Z = AB.

So, 2X 2Y = AB = 2Z.

Therefore, X + Y = Z.

• 1,0,,log

loglog ACBA

A

BB

C

CA

October 3, 2001 CSE 373, Autumn 2001 3

Logarithms, Series

• log A/B = log A – log B

• log (AB) = B log A

Series

•

• binary representation of numbers

•

122 1

0

N

N

i

i

2

)1(

1

NNi

N

i

October 3, 2001 CSE 373, Autumn 2001 4

Boolean Logic

• Let P and Q be statements.

• “not P” is true if P is false.

• “P and Q” is true if both P and Q are true.

• “P or Q” is true if one of or both P or Q are true.

• “P implies Q” is true if P is false or Q is true (or both).

P

QP

QP

QPQP

October 3, 2001 CSE 373, Autumn 2001 5

Proof by Induction

1. Formulate a statement P(n) that depends on n.

2. Prove P(1). (Basis Step)

3. Prove that for any n 1, if P(m) for all 1 m n, then P(n+1). (Induction Step)

(The red statement is called the induction hypothesis.)

October 3, 2001 CSE 373, Autumn 2001 6

Proof by induction

• Why does it work? We don’t actually prove, for example, P(503).

P(1) P(2) P(3) P(4) P(503)

October 3, 2001 CSE 373, Autumn 2001 7

Proof that

1. P(N) is

2. P(1):

3. Suppose P(m) is true for all

1 m N, N 1.

2

)1(

1

NNi

N

i

2

)1(

1

NNi

N

i

2

)11(11

1

1

i

i

October 3, 2001 CSE 373, Autumn 2001 8

P(N) implies P(N+1)

2

)2)(1(2

)1)(2(2

)1(

2

)1(2

.)hypind.(2

)1()1(

)1(1

1

1

NN

NN

NNN

NNN

iNiN

i

N

i

October 3, 2001 CSE 373, Autumn 2001 9

log N < N

• For all integers N > 0,

log N < N.

• P(N) is log N < N.

• P(1): log 1 = 0 < 1.

• Suppose P(m) is true for all

1 m N, N 1.

log (N+1) log (N+N)

= log 2N

= (log N) + 1

< N + 1 (ind. hyp.)

October 3, 2001 CSE 373, Autumn 2001 10

Proof by Contradiction

• Assume the theorem is false and show that that assumption leads to a known falsehood. This shows that the original assumption was false.

• The only way the whole line of reasoning is true is if is false.

FalseP

P

October 3, 2001 CSE 373, Autumn 2001 11

Proof by Counterexample

• To show a general statement to be false, you need only exhibit one specific instance of the statement to be false.

• “For all N > 1, N > N2 ” is false.

For N=5, 5 < 52 .

October 3, 2001 CSE 373, Autumn 2001 12

Recursion

• A recursive function is one that calls itself.

• n! = n (n-1) (n-2) … 1

int factorial (const int n)

{

if (n == 1)

return 1;

else

return (n * factorial(n-1));

}

October 3, 2001 CSE 373, Autumn 2001 13

Anatomy of a Recursive Function

• Base case: does not use recursion.

if (n == 1)

return 1;

• Making progress: the recursive call needs to get “closer” to a base case.

else

return (n * factorial(n-1));

October 3, 2001 CSE 373, Autumn 2001 14

Recursion and Induction

• Prove that factorial(n) returns n!• P(n) is • P(1): factorial(1) follows the if

part of the if statement and returns 1 = 1! So, P(1) is true.

• Suppose P(m) is true for all 1 m n, n 1. Then factorial(n+1) follows the else part of the if statement and returns (n+1)*factorial(n).So, factorial(n+1) returns (n+1)*n! = (n+1)! P(n+1) is true.

October 3, 2001 CSE 373, Autumn 2001 15

Iterative factorial

int factorial_iterative(const int n){

int i;int product = 1;

for (i = 1; i <= n; i++)product *= i;

return product;}

• How do we prove that this function computes factorial correctly?

n

i

in1

!

October 3, 2001 CSE 373, Autumn 2001 16

Loop invariants

• Loop invariant: a statement that is true after each iteration of a loop.

• After the kth iteration of the loop (before i++), i=k and

product = k!

• If the loop invariant is true, then after the nth iteration, the loop will end with i=n and the function will return product = n!

• Exercise: Prove that factorial_iterative(n) returns n!