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CSE 221/ICT221 Analysis and Design of Algorithms

CSE 221/ICT221 Analysis and Design of Algorithms. Lecture 2: Definition of algorithm and Mathematical Background for algorithm analysis. Asst.Proof.Dr.Surasak Mungsing E-mail: surasak.mu@spu.ac.th. Topics. Definition of algorithm Mathematical background Performance of algorithm Recursion.

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CSE 221/ICT221 Analysis and Design of Algorithms

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  1. CSE 221/ICT221 Analysis and Design of Algorithms Lecture 2:Definition of algorithm and Mathematical Backgroundfor algorithm analysis Asst.Proof.Dr.SurasakMungsing E-mail: surasak.mu@spu.ac.th

  2. Topics • Definition of algorithm • Mathematical background • Performance of algorithm • Recursion

  3. Definition • Algorithm(in mathematic and computer Science) is • an effective method expressed as a finite of well-defined for calculating a function. Theyare used for calculation, data processing, and automated reasoning. [from Wikipedia] • a step-by-step problem-solving procedure, especially an established, recursive computational procedure for solving a problem in a finite number of steps. [from the free online dictionary] • a step-by-step problem-solving procedure, especially an established, recursive computational procedure for solving a problem in a finite number of steps. [from answer.com ] • Algorithms that require very long time to solve problem, in practice, may be useless • Algorithms that require extraordinary large memory for processing may not be practical for running on most of current computers

  4. Selection Problem Problem: Select the kth largest integer from the given set of N integers Solutions: at least two algorithms that can solve this problem Algorithm 1Read N integers and put them in an array, then sort them with a simple sorting algorithm, such as Bubble Sort, and return the value at the index kth of the array Algorithm 2 Read only the first k integer into an array, then sort them in ascending. Next read in the rest of integer one-by -one and compare the value with the kth one in the array, if it is less than the kth value then do nothing, else replace the value in the arayy at the right position. Repeat the process until no integer left.

  5. Mathematical Background • Exponential • Logarithms • Series • Proof

  6. Exponential XAXB=XA+B XA/XB = XA-B (XA)B=XAB XN + XN = ? 2N + 2N = ?

  7. Logarithms DefinitionXA = B ifflogxB = A Theorem 1logBA = logCB/logCA ; A, B, C > 0, A  1 Theorem 2log(AB) = log A + log B; A, B > 0 Remark logarithm written as log (without base number mean log base2 or log2 ค่าที่ควรจำไว้อ้างอิง log1 = 0 because 20 = 1 log2 = 1 because 21 = 2 log1,024 = 10 because 210 = 1,024 log 1,048,576 = 20 because 220 = 1,048,576

  8. Series N AN+1 -1 If A=2 ; Ai  = A-1 i=0 N N If 0<A<1 ; N 2N+1 -1 Ai = 1   1 Ai 2i ≤  i=0 i=0 1- A 1- A i=0 If N=∞; approaches

  9. Series (cont.) Nk+1 N N N ik i  i2  = |k + 1 |    = i=1 i=1 i=1 N(N+1)(2N+1) N(N+1) 2 6  N2 N3 2 3 k ≠ -1

  10. Proof • Proof by Induction • Proof by Contradiction • Proof by Counterexample

  11. Proof by Induction Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers (non-negative integers). It is done by proving that the first statement in the infinite sequence of statements is true, and then proving that if any one statement in the infinite sequence of statements is true, then so is the next one. [http://en.wikipedia.org/wiki/Mathematical_induction ]

  12. Proof by Induction The simplest and most common form of mathematical induction proves that a statement involving a natural number n holds for all values of n. The proof consists of two steps: 1. The basis (base case): showing that the statement holds when n is equal to the lowest value that n is given in the question. Usually, n = 0 or n = 2. The inductive step: showing that if the statement holds for some n, then the statement also holds when n + 1 is substituted for n. The assumption in the inductive step that the statement holds for some n is called the induction hypothesis (or inductive hypothesis)

  13. Example : Proof by Induction Prove that Fi < (5/3)i for i  1 1.Provebase case;F1 = 1 < 5/3,F2 = 1 <25/9,F3=2 <125/27 2. Show inductive hypothesis i = 1, 2, …, k 2.Show that the statement is true for k+1 Fk+1 < (5/3)k+1 from definition; Fk+1= Fk+Fk-1 Fk+1 < (5/3)k+ (5/3)k-1 <(3/5) (5/3)k+1 +(3/5)2 (5/3)k+1 < (3/5) (5/3)k+1 +(9/25) (5/3)k+1 Fk+1 <(3/5 +9/25) (5/3)k+1 < (24/25) (5/3)k+1 < (5/3)k+1

  14. Example:Proof by Induction Prove that 1+2+3+…+n = n(n+1)/2 1. Basis: Show that the statement holds for n = 1 ifn=1 then 1*(1+1)/2 = 1 (prove base case) 2. Inductive step: show that if P(n) holds, then also P(n+1) holds. Assume P(n) holds (for some unspecified value of n). It must then be shown that P(n + 1) holds, that is: 1+2+3+…+n+(n+1)= n(n+1)/2 +(n+1) = (n2+n+2n+2)/2 = (n2+3n+2)/2 = (n+1)(n+2)/2 hereby showing that indeed P(n + 1) holds.

  15. Proof by contradiction • In logic, proof by contradiction is a form of proof that establishes the truth or validity of a proposition by showing that the proposition being false would imply a contradiction. Since by the law of bivalence a proposition must be either true or false, and its falsity has been shown impossible, the proposition must be true. • The idea: • Assume that a given proposition is untrue. • Based on that assumption reach two conclusions that contradict each other.

  16. Prove that there is no largest prime number(this is the idea of Euclid's original proof) Prime numbers are integers with no exact integer divisors except 1 and themselves. 1. To prove: "There is no largest prime number" by contradiction.   2. Assume: There is a largest prime number, call it p. 3. Consider the number N  that is one larger than the product of all of the primes smaller than or equal to p.  N=1*2*3*5*7*11...*p + 1.  Is it prime?  4. N is at least as big as p+1 and so is larger than p and so, by Step 2, cannot be prime. On the other hand,  N has no prime factors between 1 and p because they would all leave a remainder of 1.  It has no prime factors larger than p because Step 2 says that there are no primes larger than p.  So N has no prime factors and therefore must itself be prime We have reached a contradiction (N is not prime by Step 4, and N is prime by Step 5) and therefore our original assumption that there is a largest prime must be false

  17. Proof by Counterexample In logic, and especially in its applications to mathematics and philosophy, a counterexample is an exception to a proposed general rule. In mathematics, counterexamples are often used to prove the boundaries of possible theorems. In philosophy, counterexamples are usually used to argue that a certain philosophical position is wrong by showing that it does not apply in certain cases.

  18. Example: Proof by Counterexample Prove that the statement Fk k2 is not true To prove , select k =11 and show that F11=144 > 112 that is the statement does not hold ( Fibonacci numbers, Fi: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, … )

  19. Recurrence Relation Some Examples

  20. The Pizza Slicing Problem • How many pieces of pizza can you get with N straight cuts ?

  21. The Pizza Slicing Problem (cont.) • ... N cutsBut ... who said you should cut • 2N slices through the center every time?

  22. A Better Slicing Method ... • When cutting, intersect all previous cuts and avoid previous intersection points!

  23. So ... How Many Pieces? • The N-th cut creates N new pieces. • Hence, the total number of pieces given by N cuts, denoted P(N), is given by the following two rules: • P(1) =2 • P(N) =P(N-1) + N • Recursive definition of P(N)!

  24. Recurrence Relations • The pizza-cutting problem is an example of recurrence relation, where a function f (N) is recursively defined. (Base Case)f (1) = 2 (Recursive Case) f (N) = f (N - 1)+ N for N > 2 • The standard method for solving recurrence relations, called“unfolding”, makes repeated substitutionsapplying the recursive rule until the base case is reached.

  25. Recurrence Relations (cont.) f (N) = f (N - 1)+ N f (N) = f (N - 2)+ (N - 1)+ N f (N) = f (N - 3) + (N - 2)+ (N - 1)+ N ... f (N) = f (N - i) + (N - i -1)+ … + (N - 1)+ N • The base case is reached when i =N - 1 f (N) = 2 + 2 + 3 + … + (N - 2) + (N -1) + N

  26. Towers of Hanoi • Goal: transfer all N disks from peg A to peg C • Rules: • move one disk at a time • never place larger disk above smaller one

  27. Recursive Solution • Recursive solution: • transfer N -1 disks from A to B • move largest disk from A to C • transfer N -1 disks from B to C • Total number of moves: T(N) =2 T(N -1) +1

  28. Solution of the Recurrence forTowers of Hanoi • Recurrence relation: T(N) =2 T(N-1) +1 T(1) =1 • Solution by unfolding: T(N) = 2 ( 2 T(N-2) +1 ) + 1 = = 4T(N -2) + 2 + 1 = = 4 (2T(N -3)+1 ) + 2 +1 = = 8T(N -3) + 4 + 2 +1 = … =2i T(N - i) + 2i-1 + 2i-2 + ... + 21 + 20

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