COMP2230 Tutorial 2

Mathematical Induction

A tool for proving the truth of a statement for some integer “n”.“n” is usually a +ve integer from 1 to infinity.Why using M.I.?

Usual Procedure• Prove the statement is true for n=k0 (e.g. n=1, n=10,…)• Assume it is true for n• Prove it is true for (n+1)

e.g. Prove 1+2+3+…..+ n =n(n+1)/2……………..eqn.1

n=1,

l.h.s of eqn 1 = 1,

r.h.s of eqn 1 = 1(1+1)/2=1. (i.e. eqn. 1 is true for n=1)

Mathenatical Induction II

Assume eqn. 1. is true for n

i.e. 1+2+…..+n = n(n+1)/2

For (n+1),

l.h.s. = 1+2+….+n +(n+1) = n(n+1)/2 +(n+1) =(n+1)(n+2)/2

r.h.s = (n+1)(n+2)/2=l.h.s.

Mathematical Induction III

How does M.I. work?

We have proved: 1. The statement/equation is true for n=1

2. If it is true for n, then it is true for (n+1)

The Equation 1+2+3+…….+n = n(n+1)/2

As The Equation is true for n=1, The Equation is true for n=2

As The Equation is true for n=2, The Equation is true for n=3

…………

The Equation is true for any +ve integer n.

Useful assumptions

•For simplifying analysis

•Do not affect the asymptotic results

•E.g (ceiling, flooring) Why ceiling and flooring?

ceiling(2.3)=3, flooring(2.3)=2, round(2.3)=2.

ceiling(2.7)=3, flooring(2.7)=2, round(2.7)=3.

• For asymptotic analysis, T(1) can be regarded as the smallest T(n).

•If there is a recurrence T(n)=2T(n/4)+n, then T(1), T(2), T(3) can be treated as the same constant.

Computing Complexity

1.Direct Counting. (for simple algorithms, non-recurrent algorithms)

You can refer to Tutorial 1.

2.Subsitution method (for recurrent algorithms).

Guess the form of the solution

Use M.I. to prove your guess.

3. Recursion tree

4. Master Theorems

If both algorithms can be used to solve a problem, which one is more efficient, in time?

16

Another Example

T(n)=Θ(1) =1for 1≤n ≤ 3T(n)=3T(n/4)+Θ (n2), for n4

We make a simplifying assumption: n is a power of 4.

Without expansion: T(n)First expansion

Cn2

T(n/4)

T(n/4) T(n/4)Second expansion C(n/4)2

C(n/4)2 C(n/4)2

…………………………………..T(n/16)……………………………….C(n/42)2

L0: Cn2

L1: 3C(n/4)2

L2: 32C(n/42)2

L3: 33C(n/43)2

…..

Lk: 3kC(n/4k)2

If (n/4k) = 1, k=log4n

Total:k

k

i

iCn 3)16/3(1

0

2

nk

i

iCn 4log1

0

2 3)16/3(

3log1

0

2 4)16/3( nCnk

i

i

Ans: O(n2) , Can it be Θ(n2)?

……. (what should we do to do to get the answer below)

Exercise

Use substitution method, andMaster method to verify the

above result.

note

Big-O concepts can’t be used within M.I. process.

A physical function, (e.g. cn2) which is a member of Big-O should be defined before M.I.

At the end of M.I. if it comes up with

T(n) ≤ guess(n) – (some +ve lower order terms), just ignore the lower order terms.

If there is a +ve lower order term left during M.I., add a lower order term to the physical function, guess(n).