contents of chapter 1

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전자정보공학부

Contents of Chapter 1

• Sections– 1.1 What is an algorithm?– 1.2 Algorithm specification– 1.3 Performance analysis– 1.4 Randomized algorithms– 1.5 References and readings

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1.1 What is an Algorithm?

• The word algorithm– Comes from the name of a Persian author, Abu Ja’far

Mohammed ibn Musa al Khowarizmi (c. 825 A.D.)– Special significance in computer science referring to a

method that can be used by a computer for the solution of a problem

• Definition 1.1 [Algorithm]– A finite set of instructions that, if followed,

accomplishes a particular task– Satisfying the criteria:

• Input • Output• Definiteness• Finiteness• Effectiveness

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1.1 What is an Algorithm?

• Algorithm vs. program– Program: expression of an algorithm in a programming

language– Program may never terminate (e.g, OS)

• Four distinct areas of studying algorithms– How to devise algorithms

• Art which may never be fully automated• Study of design strategies in this book is helpful

– How to validate algorithms• Algorithm validation program verification (two stages)

– How to analyze algorithms• Performance analysis on the best, worst, and average

cases– How to test a program

• Debugging (error correction) • Profiling (performance measurement)

(E. Dijkstra “debugging can only point to the presence of errors, but not to their absence”)

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1.2 Algorithm Specification

• Algorithm vs. program– Use them interchangeably in this textbook

• Algorithm specification– Natural language – Flowchart– Programming language– (Programming language + natural language)

• Example 1.1 [selection sort]– Sort a collection of n1elements of type Type

“From those elements that are currently unsorted, find the smallest and place it next in the sorted list”

Not an algorithm

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Selection Sort Algorithms

• Program 1.1

• Program 1.2

• Type implementation– 1) using typedef, or 2) using template

void SelectionSort(Type a[], int n) // Sort the array a[1:n] into nondecreasing order. {

for (int i=1; i<=n; i++) { int j = i; for (int k=i+1; k<=n; k++)

if (a[k]<a[j]) j=k; Type t = a[i]; a[i] = a[j]; a[j] = t;

} }

for (i=1; i<=n; i++) {

examine a[i] to a[n] and suppose

the smallest element is at a[j];

interchange a[i] and a[j];

}

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• Recursion– Direct recursion: functions calling themselves directly

– Indirect recursion: functions calling indirectly via other functions

1.2.2 Recursive Algorithms

A(…) {

….A(…);

}

A(…) {

….B(…);

}

B(…){

….A(…);

}

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• Example 1.2 [Tower of Hanoi]

• 2 disks– AC; AB; CB;

• 3 disks– AB; AC; BC; AB; CA; CB; AB;

• 4 disks– ?

Tower CTower A Tower B

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• Program 1.3

#include<iostream.h>enum tower { A='A', B='B', C='C'};

void TowersOfHanoi(int n, tower x, tower y, tower z)

// Move the top n disks from tower x to tower y.

{ if (n) {

TowersOfHanoi(n-1, x, z, y);

cout << "move top disk from tower " << char(x)

<< " to top of tower " << char(y) << endl;

TowersOfHanoi(n-1, z, y, x);

}

}

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1.2.2 Recursive Algorithms • 4 disks

– TOH(4,A,B,C)• TOH(3,A,C,B)

– TOH(2,A,B,C)» TOH(1,A,C,B) = AC;» AB;» TOH(1,C,B,A) = CB;

– AC;– TOH(2,B,C,A)

» TOH(1,B,A,C) = BA;» BC;» TOH(1,A,C,B) = AC;

• AB;• TOH(3,C,B,A)

– TOH(2,C,A,B)» TOH(1,C,B,A) = CB;» CA;» TOH(1,B,A,C) = BA;

– CB;– TOH(2,A,B,C)

» TOH(1,A,C,B) = AC;» AB;» TOH(1,C,B,A) = CB;

• Solution=AC;AB;CB;AC;BA;BC;AC;AB;CB;CA;BA;CB;AC;AB;CB

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• Example 1.3: Permutation generator– Eg, permutation of (a,b,c,d)

• a followed by permutations of (b,c,d)• b followed by permutations of (a,c,d)• c followed by permutations of (a,b,d)• d followed by permutations of (a,b,c)

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– Program 1.4

#include<iostream.h>

void Perm(Type a[], int k, int n) {

if (k==n) { // Output permutation. for (int i=1; i<=n; i++) cout << a[i] << " "; cout << endl;

} else // a[k:n] has more than one permutation. // Generate these recursively.

for (int i=k; i<=n; i++) { Type t=a[k]; a[k]=a[i]; a[i]=t; Perm(a, k+1, n); // All permutations of a[k+1:n] t=a[k]; a[k]=a[i]; a[i]=t;

} }

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Programming Report

• Section 1.2 Exercises (pp.9-11)– 세 문제 이상을 선택하여 프로그래밍 (6, 7, 8, 11 중의 하나는 반드시

포함할 것 )– 제출일 : 2 주 후 ( 늦으면 하루 당 5% 씩 감점 )– 제출 방법

• report (hwp 파일 )• source code (p*.cpp 파일 )• outputs (p*.txt 파일 ) * 는 문제 번호• 모든 파일을 하나의 파일로 압축하여 TA 에게 email 로 제출 • 압축 파일명은 r1_ 학번 .zip

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1.3 Performance Analysis

• Criteria for making evaluative judgments about software– Correctness– Documentation– Modularity– Readability– Etc

• Criteria for performance evaluation of algorithms– Definition 2.1:

• Space complexity: amount of memory the algorithm needs to run to completion

• Time complexity: amount of computer time the algorithm needs to run to completion

• Two kinds of performance evaluation– Performance analysis: a priori estimate – Performance measurement: a posteriori testing

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1.3.2 Time Complexity

• One method– tp(n) = caADD(n) + csSUB(n) + cmMULT(n) …

– n: instance characteristic

– Impossible to estimate tp(n)

• Another method (counting the number of program steps)– Program step: syntactically or semantically meaningful

segment of a program that has an execution time being independent of the instance characteristics

– Eg, return(a+b+b*c+(a+b-c)/(a+b)+4.0);

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– Program 1.8 (counting the steps of the function Sum)

– Each invocation of Sum executes a total of 2n+3 steps


float Sum(float a[], int n) { float s = 0.0;

count++; // count is global for (int i=1; i<=n; i++) {

count++; // For `for' s += a[i]; count++; // For assignment

} count++; // For last time of `for' count++; // For the return return s;

}

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– Program 1.10 (counting the steps of the function RSum)

– tRSum(n) = 2 if n=0,

2+tRSum(n-1) if n>0

– tRSum(n) = 2+tRSum(n-1)

= 2+2+tRSum(n-2) …. = 2n+2


float RSum(float a[], int n) { count++; // For the if conditional

if (n <= 0) { count++; // For the return return 0.0;

} else { count ++; // For the addition, function // invocation and return return (RSum(a, n-1) + a[n]);

} }

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• Sum is linear time algorithm– 2n+2– If n is doubled, the run time also doubles approximately.– “The time complexity of Sum is linear in the input size

n.”

• Definition 1.3 [input size]– One of the instance characteristics – Number of elements needed to describe the instance– Example 1.9 (matrix addition)

• Program 1.11

• Input size = 2mn+2• Step count = 2mn+2m+1

void Add(Type a[][SIZE], Type b[][SIZE], Type c[][SIZE], int m, int n)

{ for (int i=1; i<=m; i++)

for (int j=1; j<=n; j++)

c[i][j] = a[i][j] + b[i][j];

}

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• Summary of time complexity– Sum, Rsum, Add is very simple examples– Not solely dependent on input sizes, but dependent on

actual values of inputs • Eg, sequential searching algorithm

– Best case: 1 comparison (query is the same as the first element)

– Worst case: n+1 comparisons (query is the same as the last element)

– Average case: n/2 – Which one is faster between n2+2n and 100n ?

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1.3.3 Asymptotic Notation

• Definition 1.4 [Big “oh”]– f(n)=O(g(n)) iff there exist positive constants c and n0

such that f(n)c*g(n) for all n, nn0.

– Example 1.11• 3n+2 = O(n) (because 3n+24n for all n2)• 3n+3 = O(n) ?• 100n+6 = O(n) ?• 10n2+4n+2 = O(n2) ?• 1000n2+100n-6 = O(n2) ?• 6*2n+n2 = O(2n) ?

• 3n+3 = O(n2) ?• 10n2+4n+2 = O(n) ?

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• Comparison of speed– O(1): constant– O(log n)– O(n): linear– O(n log n)– O(n2): quadratic– O(n3): cubic– O(2n): exponential

• In f(n)=O(g(n)), – g(n) is an upper bound of the value of f(n) for all n,

nn0.

– So, 3n+3=O(n2), 3n+3=O(2n) are correct– But, we say 3n+3=O(n) (almost never say 3n+3=O(n2),

3n+3=O(2n))

• Theorem 1.2– If f(n)=amnm +…+ a1n + a0, then f(n)=O(nm)

faster

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• Definition 1.5 (Omega)– f(n)=(g(n)) iff there exist positive constants c and n0

such that f(n)c*g(n) for all n, nn0.

– Example 1.12• 3n+2 = (n) (because 3n+23n for all n1)• 3n+3 = (n) ?• 100n+6 = (n) ?• 10n2+4n+2 = (n2) ?• 1000n2+100n-6 = (n2) ?• 6*2n+n2= (2n) ?• 3n+3 = (n2) ?• 10n2+4n+2 = (n) ?

• In f(n)=(g(n)), – g(n) is a lower bound of the value of f(n) for all n, nn0.

– So, 3n+3= (1) is correct– But, we say 3n+3= (n) (almost never say 3n+3= (1))

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• Definition 1.6 (Theta)– f(n)=(g(n)) iff there exist positive constants c1, c2, and

n0 such that c1*g(n) f(n)c2*g(n) for all n, nn0.

– Example 1.13• 3n+2 = (n)• 3n+3 = (n) ?• 100n+6 = (n) ?• 10n2+4n+2 = (n2) ?• 1000n2+100n-6 = (n2) ?• 6*2n+n2= (2n) ?• 3n+3 = (n2) ?• 10n2+4n+2 = (n) ?

• In f(n)=(g(n)), – g(n) is both an upper and lower bound of the value of

f(n) for all n, nn0.

– So, 3n+3=O(1) is not correct

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• Example 1.17 [Magic square]– Figure 1.2

11182529

101219213

46132022

23571416

17241815

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– Program 1.15

#include<iostream.h>#include<iomanip.h> void Magic(int n) // Create a magic square of size n, n being odd.

{ int square[SIZE][SIZE];

if (!(n%2)) {cout << "n is even" << endl; return;}

else {

for (int i=0; i<n; i++) // Initialize square to zero.

for (int j=0; j<n; j++) square[i][j] = 0;

square[0][(n-1)/2]=1; // Middle of first row

// (i, j) is the current position.

int j = (n-1)/2, k, l;

for (int key = 2; key <= n*n; key++) {// Move up and left. The next two if statements

// may be replaced by the % operator if -1%n is

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– Time complexity: (n2)

// implemented to have value n-1.

k = (i) ? (i-1) : (n-1);

l = (j) ? (j-1) : (n-1);

if (square[k][l]) i = (i+1)%n;

else { // square[k][l] is empty.

i = k; j = l; }

square[i][j] = key; }

// Output the magic square.

for (i=0; i<n; i++)

for (j=0; j<n; j++) cout << setw(5) << square[i][j];

cout << endl; }

} }

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• Example 1.18 [Computing xn]– Program 1.16

– Time complexity: (log n)

Type Exponentiate(Type x, int n) // Return xn for an integer n>=0 {

int m=n; Type power=1, z=x; while (m>0) {

while (!(m%2)) { m/=2; z*=z;

} m--; power*=z;

} return power;

}

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1.3.4 Practical Complexities

• When program P has (n) and Q has (n2), P is faster than Q for sufficiently large n.

• Table 1.7

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• Figure 1.3

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• Table 1.8

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1.3.5 Performance Measurement

• Time functions in C++#include <time.h>…

T(){

clock_t start_time, end_time; …..

start_time = clock(); …… /* program segment to measure the time

*/ end_time = clock(); printf("time elapsed=%f\n",

(float)(end_time-start_time)/CLOCKS_PER_SEC); ….}

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• To time a short event,– Repeat it several times and divide the total time for the

event by the number of repetitions.

• Program 1.19

void TimeSearch()

{ // Repetition factors int r[21] = {0, 200000, 200000, 150000, 100000, 100000, 100000, 50000, 50000, 50000, 50000, 50000, 50000, 50000, 50000, 50000, 50000, 25000, 25000, 25000, 25000}; int a[1001], n[21]; for (int j=1; j<=1000; j++) a[j]=j; for (j=1; j<=10; j++) {

n[j]= 10*(j-1); n[j+10]=100*j; } cout << " n t1 t" << endl << endl; cout << setiosflags (ios::fixed); cout << setprecision(3);

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for (j=1; j<=20; j++) {

int h = GetTime();

for (int i=1; i<=r[j]; i++)

int k = SeqSearch(a, 0, n[j]);

int h1 = GetTime();

int t1 = h1 - h;

float t = t1; t /= r[j];

cout << setw(5) << n[j] << << setw(5) << t1 << << setw(8)

<< t << endl;

}

cout << "Times are in milliseconds" << endl;

}

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• Improper timing– Programs 1.20

float t = 0;

for (int i=1; i <= r[j]; i++) {

int h = GetTime();


int h1 = GetTime();

t += h1 - h;

}

t /= r[j];

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• Improper timing– Programs 1.21

int t = 0;

while (t < DESIRED_TIME) { int h = GetTime(); int k = SeqSearch(a, 0, n[j]) int h1 = GetTime(); t += h1 - h;

}

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• Proper timing– Program 1.22

int h = GetTime(), t = 0;

while (t < DESIRED_TIME) {


int h1 = GetTime();

t = h1 - h;

}

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