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Advance Data Structure Review of Chapter 2

Advance Data Structure Review of Chapter 2. 張啟中. Review of Chapter 2 Arrays. 1.3 Data Abstraction and Encapsulation 2.2 The Array As An abstract Data Type 2.5 The Representation of Arrays Example 2.3 The Polynomial Abstract Data Type 2.4 The Sparse Matrix Abstract Data Type

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Advance Data Structure Review of Chapter 2

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  1. Advance Data StructureReview of Chapter 2 張啟中

  2. Review of Chapter 2 Arrays • 1.3 Data Abstraction and Encapsulation • 2.2 The Array As An abstract Data Type • 2.5 The Representation of Arrays • Example • 2.3 The Polynomial Abstract Data Type • 2.4 The Sparse Matrix Abstract Data Type • 2.6 The String Abstract Data Type

  3. 定義：Data Type A data type is a collection of objects and a set of operations that act on those objects. 定義：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.

  4. 資料型態 • Data Types • Primitive Data Types • Accumulated Data Types - Array - Structures - Union • Abstract Data Types • Examples

  5. 表一：C++ 的基本資料型態 Example: Data Type of C++

  6. Accumulated Data Type • Array • Struct • Union

  7. 如何理解 Abstract Data Types • ADTs 是一個物件的型態的抽象定義具有一般化的特性，其中也包含這個物件相關的操作。 • Data Structures 課程，旨在瞭解每一個 ADT，要用什麼樣的結構來表達與儲存，而這樣的表達與儲存方式，又對於該物件的操作帶來什麼樣的優缺點。亦即關注在二個重點： • 物件儲存與表達方式（資料結構） • 物件操作方式（演算法）

  8. The Array as an Abstract Data Type • Array • A collection of data of the same type • An array is usually implemented as a consecutive set of memory locations • int list[5], *plist[5] • ADT definition of an Array • More general structure than "a consecutive set of memory locations.“ • An array is a set of pairs, <index, value>, in mathematical, call correspondence or mapping

  9. class GeneralArray { // objects: A set of pairs <index, value> where for each value of index in // IndexSet there is a value of type float. IndexSet is a finite ordered set of one // or more dimensions, for example, {0, …, n - 1} for one dimension, {(0, 0), // (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)} for two dimensions, etc.public: GeneralArray(intj, RangeList list, floatinitValue = defaultValue) ; // The constructor GeneralArray creates a j dimensional array of floats; the // range of the kth dimension is given by the kth element of list. For each // index i in the index set, insert <i, initValue> into the array. floatRetrieve(indexi) ; // if (i is in the index set of the array) return the float associated with i in the // array; else signal an error. voidStore(indexi, float x) ; // if (i is in the index set of the array) delete any pair of the form <i, y> // present in the array and insert the new pair <i, x>; else signal an error. } ; // end of GeneralArray

  10. Array（陣列） • 陣列是用來存放同樣型態的資料 • 陣列的大小必須在程式中預先設定 • 在程式執行中，陣列的大小無法改變 • 陣列中的資料是透過索引（index）來存取 const int ArraySize = 100; int iArray[ArraySize];

  11. 7 51 22 43 9 0 1 2 98 99 一維陣列和記憶體間的對應 Memory m m + 2 m + 4 m + 6 int iArray[100]; m + 198 假定 sizeof(int) = 2

  12. 多維陣列和記憶體間的對應 Memory m m + 22 6 m + 44 5 4 m + 66 3 m + 88 2 m + 110 1 0 m + 132 0 1 2 3 4 5 6 7 8 9 10 int mesh[7][11]; 假定 sizeof(int) = 2

  13. 多維陣列的宣告 typearray_name[arraySize1] ...... [arraySizen]; 【範例】 int mesh[7][11]; float cube[6][8][3]; 6 5 4 5 4 3 3 2 2 1 1 2 0 1 0 0 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 8 9 10

  14. 5 4 3 2 1 2 0 1 0 0 1 2 3 4 5 6 7 float cube[6][8][3]; m m + 48 m + 96 m + 144 m + 192 m + 240 m + 288 假定 sizeof(int) = 2

  15. struct 和記憶體間的對應 m struct studentType { char Name[20]; // 姓名 char Address[30]; // 地址 char PhoneNumber[10]; // 電話 int Age; // 年齡 char Department[4]; // 系別 int Year; // 年級 char Class; // 班級 }; studentType Student1, Student2; m + 20 m + 50 m + 60 m + 62 m + 66 m + 68 假定 sizeof(int) = 2

  16. Self-Referential Structures One or more of its components is a pointer to itself. typedef struct list { char data; list *link; } list item1, item2, item3; item1.data=‘a’; item2.data=‘b’; item3.data=‘c’; item1.link=item2.link=item3.link=NULL; Construct a list with three nodes item1.link=&item2; item2.link=&item3; malloc: obtain a node a b c

  17. union { field1 declaration; field2 declaration; } variable_name; typedef union { field1 declaration; field2 declaration; } type_name; union 的宣告 宣告變數 宣告型態 【範例】 union { char charValue; int intValue; float floatValue; } dataCell; typedef union { char charValue; int intValue; float floatValue; } dataCellType;

  18. union 和記憶體間的對應 union { char charValue; /* 1 byte */ int intValue; /* 2 byte */ float floatValue; /* 4 byte */ } dataCell; m m + 4 typedef struct { char opcode; union { int intValue; char strValue[256]; } data; } instruction; m m+1 m+5 m + 257

  19. Example • Ordered List • Polynomial ADT • Sparse Matrix ADT • String ADT

  20. Ordered List Examples ordered (linear) list: (item1, item2, item3, …, itemn) • (MONDAY, TUEDSAY, WEDNESDAY, THURSDAY, FRIDAY, SATURDAYY, SUNDAY) • (2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, ACE) • (1941, 1942, 1943, 1944, 1945) • (a1, a2, a3, …, an-1, an) • ()

  21. Operations on Ordered List • Find the length, n , of the list. • Read the items from left to right (or right to left). • Retrieve the i’th element. • Store a new value into the i’th position. • Insert a new element at the position i , causing elements numbered i, i+1, …, n to become numberedi+1, i+2, …, n+1 • Delete the element at position i , causing elements numbered i+1, …, n to become numbered i, i+1, …, n-1 array (sequential mapping)? (1)~(4) O (5)~(6) X

  22. Addition: Multiplication: Polynomial（多項式）

  23. classPolynomial { // objects:p(x) = a0xe0 + … + anxen; a set of ordered pairs of <ei, ai>, where // ai∈Coefficient and ei ∈ Exponent // We assume that Exponent consists of integer ≥０ public: Polynomial() ; // return the polynomial p(x) = 0 intoperator!() ; // if *this is the zero polynomial, return 1; else return 0; CoefficientCoef(Exponente) ; // return the coefficient of e in *this ExponentLeadExp() ; // return the largest exponent in *this Polynomial Add(Polynomialpoly) ; // return the sum of the polynomials *this and poly PolynomialMult(Polynomial poly); // return the product of the polynomials *this and poly floatEval(float f) ; //Evaluate the polynomial *this at f and return the result. }; // end of Polynomial

  24. Polynomial: Representation 1 • Representation by Degrees private intdegree; // degree MaxDegree floatcoef [MaxDegree + 1]; • Example: Let A(x)=Σaixi, then a.degree= n, a.coef[i]=an-i, ,0<= i<= n

  25. n degree coef 0 1 2 n-1 n Polynomial: Representation 1

  26. Polynomial: Representation 2 • Representation by Degrees, but dynamic allocation space private: int degree; //degree <= MaxDegree float *coef; polynomial::polynomial(int d) { degree= d, coef=new float[degree+1]; }

  27. Polynomial: Representation 3 • Representation by Terms. Class polynomial; Class term { Friend polynomial; private: int exp float coef; }; • In polynomial class • In Implement file of polynomial class private: static term termArray[MaxTerms]; // MaxTerms is a constant. // termArray[MaxTerms] shared by all polynomial objects. static int free; int start, finish; term Polynomial::termArray[MaxTerms]; int polynomial::free = 0;

  28. Polynomial: Representation 3-1 A(X)=2X1000+1 B(X)=X4+10X3+3X2+1 A.start A.finish B.start B. finish free coef exp 0 1 2 3 4 5 6 • In general, A.Finish = A.Start + n –1. • For zero polynomial, A.Finish = A.Start – 1

  29. Polynomial: Representation 3-2 k numTerm ? coef 0 1 2 k-1 ? expon 0 1 2 k-1

  30. The Representations of Polynomials Compare

  31. 操作分析 • 多項式相加 • As Representation 3，we take O(m+n) at time complexity，if A(x) has m terms, B(x) has n terms. See Book pp.83-84 • 多項式相乘

  32. Sparse Matrix col1 col2 col3 col4 col5 col6 row0 row1 row2 row3 row4 row5 5*3 6*6 (a) (b) *Figure 2.3:Two matrices 15/15 8/36 sparse matrix data structure?

  33. 若以二維陣列的表示法： #define MAX_ROW 100 #define MAX_COL 100 typedef int matrix[MAX_ROW][MAX_COL]; matrix m1, m2; 來儲存稀疏矩陣，那麼，矩陣中含有許多的 0。有沒有必要 儲存這些 0 呢？有沒有比較節省記憶體空間的另類表示法？

  34. Abstract Data Type Sparse Matrix classSparseMatrix { //objects: a set of triples, <row, column, value>, where row and column are integers and // form a unique combination, and value comes from the set item. public: SparseMatrix(int MaxRow, int MaxCol); //create a SparseMatrix that can hold up to MaxItems= MaxRow*MaxCol and whose //maximum row size is MaxRow and whose maximum column size is MaxCol SparseMatrix Transpose(); // return the matrix produced by interchanging the row and column value of every triple. SparseMatrix Add(SparseMatrix b); //if the dimensions of a(*this) and b are the same, return the matrix produced by adding //corresponding items, namely those with identical row and column values. else return //error. SparseMatrix Multiply(a, b); //if number of columns in a equals number of rows in b return the matrix d produced by //multiplying a by b according to the formula: d[i][j]= Sum(a[i][k](b[k][j]), //where d(i, j) is the (i, j)th element, k=0 ~ ((columns of a) –1) else return error. };

  35. Representation of Sparse Matrix classSparseMatrix; class MatrixTerm { friend classSparseMatrix private: int col, row, value; }; • In class SparseMatrix private: int col, row,Terms; MatrixTerm smArray[MaxTerms]; // Note: triples are ordered by row and within rows by columns

  36. col 0 1 2 3 4 5 term row 0 15 0 0 22 0 -15 0 11 3 0 0 0 0 0 0 -6 0 0 0 0 0 0 0 0 91 0 0 0 0 0 0 0 28 0 0 0 row 1 row 2 row 3 row 4 row 5 row col value a[0] 0 0 15 a[1] 0 3 22 a[2] 0 5 -15 a[3] 1 1 11 a[4] 1 2 3 a[5] 2 3 -6 a[6] 4 0 91 a[7] 5 2 28

  37. Sparse Matrix 運算分析 • 轉置 (transpose) • 相加 • 相乘

  38. Abstract Data Type String Class String { //objects: a finite set of zero or more characters. public: String (char *init, int m); //Constructor that initializes *this to string init of length is m. intoperator==(string t); // if the string represented by *this equal string t return 1(true) else return 0(false) int operator!(); // if *this is empty then return 1(TRUE); else return 0 (FALSE). intLength(); //return the number of characters in *this. String Concat(String t); //return a string whose elements are those of *this followed by those of t. String Substr(int i, int j); //return the string containing j characters of *this at positions i, i+1, ..., i+j-1, //if these are valid positions of *this; else return empty string. intFind(String pat); //return an index i such that pat matches the substring of *this that begins at //position i. Return –1 if pat is either empty or not a substring of *this. };

  39. Pattern Matching • Given two strings, string and pat, where pat is a pattern to be searched for in string • Two methods • a simple algorithm • O(S*P)  O(n2) • optimal algorithm (by Knuth-Morris-Pratt) • Linear complexity • O(S+P)  O(n)

  40. a a a a b b pattern pattern a b a b b a a a a a no match a b a b b a b a a a match A simple algorithm O(n*m)

  41. The Failure Function • Definition • If p= p0p1. . .pn-1 is a pattern, then its failure function, f is defined as: f(j) = largest i< j such that p0p1. . .pi= pj-ipj-i+1. . .pj , i>= 0 = -1, otherwise • Example j 0 1 2 3 4 5 6 7 8 9 pat a b c a b c a c a b f -1 -1 -1 0 1 2 3 -1 0 1

  42. Rule for Optimal Pattern Matching • If a partial match is found such that si-j. . .si-1= p0p1. . .pj-1 and si<>pj then matching may be resumed by comparing si and pf(j-1)+1 if j<>0. If j= 0, continue by comparing si+1 and p0 string a b c a ? ? . . . ? pat a b c a b c a c a b f -1 -1 -1 0 1 2 3 -1 0 1 Continue here

  43. Example of Optimal Pattern Matching j 0 1 2 3 4 5 6 7 8 9 pat a b c a b c a c a b f -1 -1 -1 0 1 2 3 -1 0 1 str c b a b c a a b c a b c a b c a c a b

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