CHAPTER 1 1

CHAPTER 1CHAPTER 1BASIC CONCEPT

All the programs in this file are selected fromEllis Horowitz, Sartaj Sahni, and Susan Anderson-Freed“Fundamentals of Data Structures in C”,Computer Science Press, 1992.

CHAPTER 1 2

Data Structure

What is data structure?Ans. Data structure 探討一群相關資料的資料表示方

法與資料運作方法目的 : 使用最有效率的方式 , 對一群相關資料進行處理如何設計 :

1. 找出對該資料的各種運算2. 考慮最適當的 Data Structure, 使得各種運算的 效率最佳3. 設計一個完整的 Algorithm

CHAPTER 1 3

How to create programs

Requirements: What inputs, functions, and outputs

Analysis: bottom-up vs. top-down

Design: data objects and operations

Refinement and Coding Verification

Program Proving Testing Debugging

CHAPTER 1 4

Algorithm Definition

An algorithm is a finite set of instructions that accomplishes a particular task.

Criteria Input: zero more quantities are externally supplied. Output: at least one quantity is produced. definiteness: clear and unambiguous finiteness: terminate after a finite number of steps effectiveness: instruction is basic enough to be

carried out

CHAPTER 1 5

Algorithm

One distinguishes between an algorithm and a program, the latter of which does not have to satisfy the fourth condition (Finiteness).

For example, we can think of an operating system that continues in a wait loop until more jobs are entered.

CHAPTER 1 6

在已排序的資料中找值 18

Sequential Search

第 7 次找到

Binary Search

第 3 次找到

3

4

6

8

9

15

18

3

4

6

8

9

15

18

CHAPTER 1 7

Problem Devise a program that sorts a set of n>= 1 integers

Solution I From those integers that are currently unsorted, find

the smallest and place it next in the sorted list Solution II

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

Examine list[i] to list[n-1] and suppose that the smallest integer is list[min];

Interchange list[i] and list[min];

}

Translating a Problem into an Algorithm

CHAPTER 1 8

Translating a Problem into an Algorithm Solution III

#define swap(x,y,t) ((t)= (x), (x)= (y), (y)= (t))

void sort(int list[], int n){ int i, j, min, temp; for (i= 0; i< n-1; i++){ min= i; for (j= i+1; j< n; j++){ if (list[j]< list[min]) min= j; } swap(list[i], list[min], temp);}

上面 swap 意思即類似void swap(int *x, int *y){ int temp= *x; *x= *y; *y= temp;}

CHAPTER 1 9

Data Type

Data TypeA data type is a collection of objects and a set of operations that act on those objects.

Example of "int" Objects: 0, +1, -1, ..., Int_Max, Int_Min Operations: arithmetic(+, -, *, /, and %), testing (equa

lity/inequality), assigns, functions

CHAPTER 1 10

Data encapsulation and Data abstraction

Data encapsulation or Information hiding 是把資料物件的內部程式碼內容隱藏不被外部的使用者知道目的 :1. 讓往後可能的修改能夠局限在此程式單元之中2. 讓程式單元不需隨著 ADT 內部的表示方式的改

變而修改外部的使用方式3. 可提高程式的可重用度 , 可讀性 , 方便除錯…

Data abstraction 是將一個資料物件的 specification 和 implementation 分離目的 : 只描述主要特徵而隱藏大部分的次要特徵 , 可以將複 雜的物件加以簡化

CHAPTER 1 11

Abstract Data Type

Abstract Data TypeAn abstract data type (ADT) is a data type that is organized in such a way that the specification of the objects and the operations on the objects is separated from the representation of the objects and the implementation of the operations.

ADT Operations — only the operation name and its parameters are visible to the user — through interface

Why abstract data type ? implementation-independent

不管如何 (How)作，只管作了什麼 (What) Abstraction is …

Generalization of operations with unspecified implementation

CHAPTER 1 12

Abstract data type model

CHAPTER 1 13

Classifying the Functions of a Data Type

Creator/constructor Create a new instance of the designated type

Transformers Also create an instance of the designated type by

using one or more other instances Observers/reporters

Provide information about an instance of the type, but they do not change the instance

CHAPTER 1 14

*Structure 1.1:Abstract data type Natural_Number (p.17)structure Natural_Number is (denoted by Nat_No) objects: an ordered subrange of the integers starting at zero and ending at the maximum integer (INT_MAX) on the computer functions: for all x, y Nat_Number; TRUE, FALSE Boolean and where +, -, <, and == are the usual integer operations. Nat_No Zero ( ) ::= 0 Boolean Is_Zero(x) ::= if (x) return FALSE else return TRUE Nat_No Add(x, y) ::= if ((x+y) <= INT_MAX) return x+y else return INT_MAX Boolean Equal(x,y) ::= if (x== y) return TRUE else return FALSE Nat_No Successor(x) ::= if (x == INT_MAX) return x else return x+1 Nat_No Subtract(x,y) ::= if (x<y) return 0 else return x-y end Natural_Number

Creator

Observer

Transformer

CHAPTER 1 15

Play with ADT

Add(Zero(),y) ::= return yAdd(Successor(x),y) ::=

return Successor(Add(x,y))

Equal(Zero(),Zero()) ::= return TRUEEqual(Successor(x),Zero()) ::=

return FALSE Equal(Successor(x), Successor(y)) ::=

return Equal(x, y)

CHAPTER 1 16

Play with ADT (cont’d)

0Zero()1Successor(Zero())2Successor(Successor(Zero()))1 + 2 = ?

…

1 + 2 == 3 ?

CHAPTER 1 17

Recursive Algorithms

Direct recursion Functions call themselves

Indirect recursion Functions call other functions that invoke the

calling function again

Any function that we can write using assignment, if-else, and while statements can be written recursively.

CHAPTER 1 18

Recursive algorithm

How to determine that express an algorithm recursively ? The problem itself is defined recursively Statements: if-else and while can be written recursively

How to design recursive algorithm1. 找出遞迴關係 找出遞迴關係才能夠對問題進行切割

2. 找出終止條件 找出終止條件避免無窮遞迴下去 非常重

要 !!

CHAPTER 1 19

遞迴和迴圈的比較

遞迴的優點 : 簡潔易懂 缺點 : 執行效率差因為執行時需要對

遞迴副程式進行呼叫 , 由於呼叫遞 迴副程式需要處理額外的工作

迴圈的優點 : 執行效率佳 缺點 : 有的問題不易用迴圈來表示 ,

即使寫得出來也低階難懂。 如河內塔問題

哪些工作 ?Trace 一個

CHAPTER 1 20

Binary Search

int binsearch(int list[], int searchnum, int left, int right){// search list[0]<= list[1]<=...<=list[n-1] for searchnum int middle; while (left<= right){ middle= (left+ right)/2; switch(compare(list[middle], searchnum)){ case -1: left= middle+ 1;

break; case 0: return middle; case 1: right= middle- 1; } } return -1;}

int compare(int x, int y){ if (x< y) return -1; else if (x== y) return 0; else return 1;}

int compare(int x, int y){ if (x< y) return -1; else if (x== y) return 0; else return 1;}

CHAPTER 1 21

Recursive Implementation of Binary Search

int binsearch(int list[], int searchnum, int left, int right){// search list[0]<= list[1]<=...<=list[n-1] for searchnumint middle; if (left<= right){ middle= (left+ right)/2; switch(compare(list[middle], searchnum)){ case -1:return binsearch(list, searchnum, middle+1, right); case 0: return middle; case 1: return binsearch(list, searchnum, left, middle- 1); } } return -1;}

CHAPTER 1 22

河內塔問題 有三根木樁 , 在其中一根木樁上放上 N 個鐵盤 , 規則是 :

1. 一次移動一個鐵盤 .2. 小的鐵盤永遠在大鐵盤的上方 .

請問將 N 個鐵盤全部移動到另一鐵盤需要多少次 ??? 設三個木樁為 x , y , z, 數量為 n, 函式取名為為 Hanoi (n , x , y ,

z)1. 當 N 為 1 時表示只要再移動一次就完成工作 . 也就是只要把 一個鐵盤從 x 移到 z 就完成工作 .2. 當 N 不為 1 時作 3 個動作 . ㄅ . 執行 Hanoi(n-1,x,z,y); ㄆ . 將一個鐵盤從 x 移到 z; ㄇ . 執行 Hanoi(n-1,y,x,z);

CHAPTER 1 23

CHAPTER 1 24

河內塔程式碼

void Hanoi(int n, char x, char y, char z){

if (n > 1) {

Hanoi(n-1,x,z,y);printf("Move disk %d from %c to %c.\n",n,x,z);Hanoi(n-1,y,x,z);

} else{

printf("Move disk %d from %c to %c.\n",n,x,z); }

}

CHAPTER 1 25

Hanoi (n , x , y , z) n=4 的結果執行 Hanoi(3,x,z,y). 執行 Hanoi(2,x,y,z). 執行 Hanoi(1,x,z,y). 列印 Move disk 1 from x to y. 列印 Move disk 2 from x to z. 執行 Hanoi(1,y,x,z). 列印 Move disk 1 from y to z. 列印 Move disk 3 from x to y. 執行 Hanoi(2,z,x,y). 執行 Hanoi(1,z,y,x). 列印 Move disk 1 from z to x. 列印 Move disk 2 from z to y.

CHAPTER 1 26

執行 Hanoi(1,x,z,y). 列印 Move disk 1 from x to y. 列印 Move disk 4 from x to z. 執行 Hanoi(3,y,x,z). 執行 Hanoi(2,y,z,x). 執行 Hanoi(1,y,x,z). 列印 Move disk 1 from y to z. 列印 Move disk 2 from y to x. 執行 Hanoi(1,z,y,x) 列印 Move disk 1 from z to x. 列印 Move disk 3 from y to z. 執行 Hanoi(2,x,y,z). 執行 Hanoi(1,x,z,y). 列印 Move disk 1 from x to y. 列印 Move disk 2 from x to z. 執行 Hanoi(1,y,x,z). 列印 Move disk 1 from y to z.

CHAPTER 1 27

Performance Analysis (machine independent)

Space and Time Does the program efficiently use primary and secondary

storage? Is the program's running time acceptable for the task?

Evaluate a program generally Does the program meet the original specifications of the

task? Does it work correctly? Does the program contain documentation that show how

to use it and how it works? Does the program effectively use functions to create

logical units? Is the program's code readable?

CHAPTER 1 28

Performance Analysis(Cont.)

Evaluate a programMeet specifications, Work correctly, Good user-interface, Well-documentation,Readable, Effectively use functions, Running time acceptable, Efficiently use space/storage

How to achieve them? Good programming style, experience, and practice Discuss and think

Practice Practice Prac

tice

ThinkThinkThink

CHAPTER 1 29

S(P)= c+ Sp(I)

The space needed is the sum of Fixed space and Variable space

Fixed space: c Includes the instructions, variables, and constant

s Independent of the number and size of I/O

Variable space: Sp(I) Includes dynamic allocation, functions' recursion

Total space of any program S(P)= c+ Sp(Instance)

P: a programI: instance

(input, output)

CHAPTER 1 30

*Program 1.9: Simple arithmetic function (p.19)float abc(float a, float b, float c){ return a + b + b * c + (a + b - c) / (a + b) + 4.00; }

*Program 1.10: Iterative function for summing a list of numbers (p.20)float sum(float list[ ], int n){ float tempsum = 0; int i; for (i = 0; i<n; i++) tempsum += list [i]; return tempsum;}

Sabc(I) = 0

Ssum(I) = 0

Recall: pass the address of thefirst element of the array &pass by value

CHAPTER 1 31

*Program 1.11: Recursive function for summing a list of numbers (p.20)float rsum(float list[ ], int n){ if (n) return rsum(list, n-1) + list[n-1]; return 0; }

*Figure 1.1: Space needed for one recursive call of Program 1.11 (p.21) ( 以 16bit x86 small/near 為例 )

Type Name Number of bytes parameter: float parameter: integer return address:(used internally)

list [ ] n

2 2 2(unless a far address)

TOTAL per recursive call 6

Ssum(I)=Ssum(n)=6n

*

CHAPTER 1 32

Time Complexity

Total time T(P)= compile time + run (or execution) time May run it many times without recompilation. Run tim

e How to evaluate?

+ - * / … ( 最細的估計 ) Use the system clock 直接用量的比較省事 Number of steps performed ( 中等的估計 )

machine-independent Instance 及所費時間函數的趨勢 ( 粗略的估計 , p.39)

Definition of a program step A program step is a syntactically or semantically meaningful

program segment whose execution time is independent of the instance characteristics

CHAPTER 1 33

Methods to compute the step count

Introduce variable count into programs Tabular method

Determine the total number of steps contributed by each statement steps_per_statement * frequency_for_that_statement s/e, steps/execution

add up the contribution of all statements

CHAPTER 1 34

*Program 1.12: Program 1.10 with count statements (p.23)

float sum(float list[ ], int n){ float tempsum = 0; count++; /* for assignment */ int i; for (i = 0; i < n; i++) { count++; /*for the for loop */ tempsum += list[i]; count++; /* for assignment */ } count++; /* last execution of for */

count++; /* for return */ return tempsum; } 2n + 3 steps

Iterative summing of a list of numbers

CHAPTER 1 35

*Program 1.13: Simplified version of Program 1.12 (p.23)

float sum(float list[ ], int n){ float tempsum = 0; int i; for (i = 0; i < n; i++) count += 2; count += 3; return 0;}

2n + 3 steps

CHAPTER 1 36

*Program 1.14: Program 1.11 with count statements added (p.24)

float rsum(float list[ ], int n){

count++; /*for if conditional */if (n) {

count++; /* for return and rsum invocation */ return rsum(list, n-1) + list[n-1]; } count++; return list[0];} 2n+2 steps

Recursive summing of a list of numbers

CHAPTER 1 37

*Program 1.15: Matrix addition (p.25)

void add( int a[ ] [MAX_SIZE], int b[ ] [MAX_SIZE], int c [ ] [MAX_SIZE], int rows, int cols){ int i, j; for (i = 0; i < rows; i++) for (j= 0; j < cols; j++) c[i][j] = a[i][j] +b[i][j]; }

Matrix addition

CHAPTER 1 38

*Program 1.16: Matrix addition with count statements (p.25)

void add(int a[ ][MAX_SIZE], int b[ ][MAX_SIZE], int c[ ][MAX_SIZE], int row, int cols ){ int i, j; for (i = 0; i < rows; i++){ count++; /* for i for loop */ for (j = 0; j < cols; j++) { count++; /* for j for loop */ c[i][j] = a[i][j] + b[i][j]; count++; /* for assignment statement */ } count++; /* last time of j for loop */ } count++; /* last time of i for loop */}

2rows * cols + 2 rows + 1

CHAPTER 1 39

*Program 1.17: Simplification of Program 1.16 (p.26)

void add(int a[ ][MAX_SIZE], int b [ ][MAX_SIZE], int c[ ][MAX_SIZE], int rows, int cols){ int i, j; for( i = 0; i < rows; i++) { for (j = 0; j < cols; j++) count += 2; count += 2; } count++; } 2rows cols + 2rows +1

Suggestion: Interchange the loops when rows >> cols

CHAPTER 1 40

*Figure 1.2: Step count table for Program 1.10 (p.26)

Statement s/e Frequency Total steps

float sum(float list[ ], int n){ float tempsum = 0; int i; for(i=0; i <n; i++)

tempsum += list[i]; return tempsum;}

0 0 00 0 01 1 10 0 01 n+1 n+11 n n1 1 10 0 0

Total 2n+3

Tabular Method

steps/execution

Iterative function to sum a list of numbers

CHAPTER 1 41

*Figure 1.3: Step count table for recursive summing function (p.27)

Statement s/e Frequency Total steps

float rsum(float list[ ], int n){ if (n) return rsum(list, n-1)+list[n-1]; return list[0];}

0 0 00 0 01 n+1 n+11 n n1 1 10 0 0

Total 2n+2

Recursive Function to sum of a list of numbers

CHAPTER 1 42

*Figure 1.4: Step count table for matrix addition (p.27)

Statement s/e Frequency Total steps

Void add (int a[ ][MAX_SIZE]‧ ‧ ‧ ){ int i, j; for (i = 0; i < row; i++) for (j=0; j< cols; j++) c[i][j] = a[i][j] + b[i][j];}

0 0 00 0 00 0 01 rows+1 rows+11 rows‧ (cols+1) rows‧ cols+rows1 rows‧ cols rows‧ cols0 0 0

Total 2rows‧ cols+2rows+1

Matrix Addition

CHAPTER 1 43

Asymptotic Notation(O, , )

Exact step count Compare the time complexity of two programs that

computing the same function Difficult task of most of programs

Asymptotic notation Big “oh”

upper bound(current trend) Omega

lower bound Theta

upper and lower bound

CHAPTER 1 44

Asymptotic Notation O

Definition f(n)= O(g(n)) iff there exist positive constants c and n0 suc

h that f(n)<= cg(n) for all n, n>= n0

Examples 3n+ 2= O(n) as 3n+ 2<= 4n for all n>= 2 10n2+ 4n+ 2= O(n2) as 10n2+ 4n+ 2<= 11n2 for n>= 5 3n+2<> O(1), 10n2+ 4n+ 2<> O(n)

Remarks g(n) is upper bound, the least? ( 估 algorithm, 作答時… )

n=O(n2)=O(n2.5)= O(n3)= O(2n) O(1): constant, O(n): linear, O(n2): quadratic, O(n3): cubic,

and O(2n): exponential ( 各舉一例 )

CHAPTER 1 45

Asymptotic Notation

Definition f(n)= (g(n)) iff there exist positive constants c and n0 s

uch that f(n)>= cg(n) for all n, n>= n0

Examples 3n+ 2= (n) as 3n+ 2>= 3n for n>= 1 10n2+ 4n+ 2= (n2) as 10n2+4n+ 2>= n2 for n>= 1 6*2n+ n2= (2n) as 6*2n+ n2 >= 2n for n>= 1

Remarks lower bound, the largest ? (used for problem)

3n+3= (1), 10n2+4n+2= (n); 6*2n+ n2= (n100) Theorem

If f(n)= amnm+ ...+ a1n+ a0 and am> 0, then f(n)= (nm)

CHAPTER 1 46

Asymptotic Notation

Definition f(n)= (g(n)) iff there exist positive constants c1, c2, and n0 s

uch that c1g(n)<= f(n) <= c2g(n) for all n, n>= n0

Examples 3n+2=(n) as 3n+2>=3n for n>1 and 3n+2<=4n for all n>= 2 10n2+ 4n+ 2= (n2); 6*2n+n2= (2n)

Remarks Both an upper and lower bound 3n+2!=(1); 10n2+4n+ 2!= (n)

Theorem If f(n)= amnm+ ... +a1n+ a0 and am> 0, then f(n)= (nm)

CHAPTER 1 47

Example of Time Complexity Analysis

Statement Asymptotic complexity

void add(int a[][Max.......) 0{ 0 int i, j; 0 for(i= 0; i< rows; i++) (rows) for(j=0; j< cols; j++) (rows*cols) c[i][j]= a[i][j]+ b[i][j]; (rows*cols)} 0

Total (rows*cols)

CHAPTER 1 48

Example of Time Complexity Analysis(Cont.)

int binsearch(int list[], int left, int right){ int middle; while (left<= right){ middle= (left+ right)/2; switch(compare(list[middle], searchnum)){ case -1: left= middle+ 1;

break; case 0: return middle; case 1: right= middle- 1; } } return -1;}

worst case (log n)best case (1)

CHAPTER 1 49

*Figure 1.9:Times on a 1 billion instruction per second computer(p.40)

CHAPTER 1 50

*Figure 1.8:Plot of function values(p.39)

nlogn

n

logn

CHAPTER 1 51

注意 : 一般常以 average case 和 worst case 來討論 ( 真的要怎麼作 ?)

Estimate average case : 用隨機的測試資料 , 取結果的平均時間 Estimate worst case : 用隨機的測試資料 , 取結果的最大時間

There are actually two different methods for timing events in C. Figure 1.10 shows the major differences between these two methods. Method 1 uses clock to time events. This function gives the amount of processor tim

e that has elapsed since the program began running. Method 2 uses time to time events. This function returns the time, measured in seco

nds, as the built-in type time_t. The exact syntax of the timing functions varies from computer to computer and also

depends on the operating system and compiler in use.Method 1 Method 2

Start timing start=clock(); start=time(NULL);

Stop timing stop=clock(); stop=time(NULL);

Type returned clock_t time_t

Result in seconds

duration=((double)(stop-start)/CLK_TCK;

duration=(double)difftime(stop, start);

Figure 1.10

Performance Measurement (Performance Measurement (machine machine dependent)dependent)

CLOCKS_PER_SEC