1 / 16

COMP2230 Tutorial 2

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=k 0 (e.g. n=1, n=10,…) Assume it is true for n

violet-mcintosh
Download Presentation

COMP2230 Tutorial 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. COMP2230 Tutorial 2

  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)

  3. 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.

  4. 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.

  5. 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.

  6. 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. • Recursion tree • Master Theorems

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

  8. 16

  9. Another Example • T(n)=Θ(1) =1for 1≤n ≤ 3 • T(n)=3T(n/4)+Θ (n2), for n4 • We make a simplifying assumption: n is a power of 4.

  10. 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

  11. 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:

  12. ……. (what should we do to do to get the answer below) Ans: O(n2) , Can it be Θ(n2)?

  13. Exercise • Use substitution method, and • Master method to verify the above result.

  14. 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).

More Related