1. CHAPTER 1 BASIC CONCEPT
All the programs in this file are selected from
Ellis 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
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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
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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
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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.
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6. 在已排序的資料中找值18 Binary Search Sequential Search 第3次找到 第7次找到 3 4 6 8 9 15
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7. Translating a Problem into an Algorithm
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]; }
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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; }
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9. Data Type
Data Type A 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 (equality/inequality), assigns, functions
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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
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11. Abstract Data Type 不管如何(How)作，只管作了什麼(What) Abstraction is …
Abstract Data Type An 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
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12. Abstract data type model
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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
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14. Structure 1. 1:Abstract data type Natural_Number (p
*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 ( ) ::= 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 Nat_No Subtract(x,y) ::= if (x<y) return else return x-y end Natural_Number
Creator
Observer
Transformer
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15. Play with ADT Add(Zero(),y) ::= return y
Add(Successor(x),y) ::= return Successor(Add(x,y))
Equal(Zero(),Zero()) ::= return TRUE
Equal(Successor(x),Zero()) ::= return FALSE
Equal(Successor(x), Successor(y)) ::= return Equal(x, y)
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16. Play with ADT (cont’d) 0Zero() 1Successor(Zero())
2Successor(Successor(Zero()))
1 + 2 = ? …
1 + 2 == 3 ?
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17. Recursive Algorithms Direct recursion Indirect 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.
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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 algorithm
1. 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!!
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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
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
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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; }
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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 searchnum
int 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.
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);
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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
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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).
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.
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26. 執行 Hanoi(1,z,y,x) 列印 Move disk 1 from z to x.
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).
列印 Move disk 2 from x to z.
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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?
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28. Performance Analysis(Cont.)
Practice Practice Practice
Evaluate a program
Meet 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
Think
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29. I: instance (input, output)
S(P)= c+ Sp(I)
P: a program
I: instance (input, output)
The space needed is the sum of
Fixed space and Variable space
Fixed space: c
Includes the instructions, variables, and constants
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)
CHAPTER 1

30. Recall: pass the address of the first element of the array &
*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) ; } *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 the
first element of the array &
pass by value
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31. Program 1. 11: Recursive function for summing a list of numbers (p
*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)
Ssum(I)=Ssum(n)=6n *
CHAPTER 1

32. Time Complexity Total time How to evaluate?
T(P)= compile time + run (or execution) time
May run it many times without recompilation. Run time
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
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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. Iterative summing of a list of numbers
*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
CHAPTER 1

35. Program 1. 13: Simplified version of Program 1. 12 (p
*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. Recursive summing of a list of numbers
*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
CHAPTER 1

37. Matrix addition
*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]; }
CHAPTER 1

38. Program 1. 16: Matrix addition with count statements (p
*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
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39. Suggestion: Interchange the loops when rows >> cols
*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)
Tabular Method
Iterative function to sum a list of numbers
*Figure 1.2: Step count table for Program 1.10 (p.26)
steps/execution
CHAPTER 1

41. *Figure 1.3: Step count table for recursive summing function (p.27)
Recursive Function to sum of a list of numbers
*Figure 1.3: Step count table for recursive summing function (p.27)
CHAPTER 1

42. *Figure 1.4: Step count table for matrix addition (p.27)
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. n=O(n2)=O(n2.5)= O(n3)= O(2n)
Asymptotic Notation O
Definition
f(n)= O(g(n)) iff there exist positive constants c and n0 such 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 such 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 Examples Remarks Theorem
f(n)= (g(n)) iff there exist positive constants c1, c2, and n0 such 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. Performance Measurement (machine dependent)
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 time that has elapsed since the program began running.
Method 2 uses time to time events. This function returns the time, measured in seconds, 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
CHAPTER 1
CLOCKS_PER_SEC