CHAPTER 1 BASIC CONCEPT

CHAPTER 1BASIC CONCEPT All the programs in this file are selected from Ellis Horowitz, Sartaj Sahni, and Susan Anderson-Freed “Fundamentals of Data Structures in C”, 1/e, Computer Science Press, 1992, 2/e,Silicon press, 2008. CHAPTER 1

Pointers • * • dereferencing operator, indirection operator • & • address operator • int i, *pi; • pi = &i; • *pi = 10; or i = 10; CHAPTER 1

Dynamic Memory Allocation • int i, *pi; • float f, *pf; • pi = (int *) malloc (sizeof(int)); • pf = (float *) malloc (sizeof(float)); • *pi = 1024; • *pf = 3.14; • printf (“Integer = %d; float = %f\n”, *pi, *pf); • free (pi); • free (pf); CHAPTER 1

Dynamic Memory Allocation • #define MALLOC(p, s) \ • if (!((p) = malloc(s))) { \ • fprintf(stderr, “Insufficient memory”); \ • exit (EXIT_FAILURE); \ • } • …… • int *pi; • …… • MALLOC(pi, sizeof(int)); • *pi = 1024; CHAPTER 1

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

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

Algorithm • DefinitionAn 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

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

Sequential Search 第7次找到 Binary Search 第3次找到在已排序的資料中找值18 CHAPTER 1

Translating a Problem into an Algorithm • 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]; } CHAPTER 1

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

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 (equality/inequality), assigns, functions CHAPTER 1

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

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

Abstract data type model CHAPTER 1

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

*ADT 1.1:Abstract data type Natural_Number (p.20)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 computerfunctions: for all x, y  Nat_Number; TRUE, FALSE  Boolean and where +, -, <, and == are the usual integer operations.Nat_No Zero ( ) ::= 0Boolean Is_Zero(x) ::= if (x) returnFALSEelse returnTRUENat_No Add(x, y) ::= if ((x+y) <= INT_MAX) return x+y else returnINT_MAXBoolean Equal(x,y) ::= if (x== y) returnTRUEelse returnFALSENat_No Successor(x) ::= if (x == INT_MAX) return xelse return x+1Nat_No Subtract(x,y) ::= if (x<y) return 0else return x-yend Natural_Number Creator Observer Transformer CHAPTER 1

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) CHAPTER 1

Play with ADT (cont’d) • 0Zero() • 1Successor(Zero()) • 2Successor(Successor(Zero())) • 1 + 2 = ? • … • 1 + 2 == 3 ? CHAPTER 1

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

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!! CHAPTER 1

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

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; } CHAPTER 1

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

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

CHAPTER 1

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

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

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

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 workcorrectly? • 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

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 Think Think CHAPTER 1

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

*Program 1.10: Simple arithmetic function (p.23)float abc(float a, float b, float c){ return a + b + b * c + (a + b - c) / (a + b) + 4.00; }*Program 1.11: Iterative function for summing a list of numbers (p.24)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 CHAPTER 1

*Program 1.12: Recursive function for summing a list of numbers (p.24)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.25) (以16bit x86 small/near為例) Ssum(I)=Ssum(n)=6n * CHAPTER 1

Time Complexity • Total time • 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.33) • 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

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

*Program 1.13: Program 1.11 with count statements (p.26)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; } Iterative summing of a list of numbers 2n + 3 steps CHAPTER 1

*Program 1.14: Simplified version of Program 1.13 (p.27)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

*Program 1.15: Program 1.12 with count statements added (p.27)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];} Recursive summing of a list of numbers 2n+2 steps CHAPTER 1

*Program 1.16: Matrix addition (p.28)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

*Program 1.17: Matrix addition with count statements (p.29)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

*Program 1.18: Simplification of Program 1.17 (p.29)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

*Figure 1.2: Step count table for Program 1.11 (p.30) Tabular Method Iterative function to sum a list of numbers steps/execution CHAPTER 1

*Figure 1.3: Step count table for recursive summing function (p.31) Recursive Function to sum of a list of numbers CHAPTER 1

*Figure 1.4: Step count table for matrix addition (p.31) Matrix Addition CHAPTER 1

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

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

Asymptotic Notation  • Definition • f(n)= (g(n)) iff there exist positive constants c and n0such 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

Asymptotic Notation  • Definition • 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

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

CHAPTER 1 BASIC CONCEPT

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