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Data structure


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Data structure

  1. 1. Concept of Data Structure Data Structure:A means of storing a collection of data. ORIt is the specification of the elements of the structure, the relationships between them, and the operation that may be performed upon them. 04/09/13 1
  2. 2. Computer Science & Data Structure Computer science is concern with study of Methods for effectively using a computer to solve problems. orIn determining exactly the problem to be solved.This process entails:1. Gaining an understanding of the problem.2. Translating vague descriptions, goals, and contradictory requests, and often unstated desires, into a precisely formulated conceptual Solution04/09/13 2
  3. 3. Computer Science & Data Structure (Continue)3. Implementing the solution with a computer program. This solution typically consists of two parts: algorithms and data structures:Algorithm:An algorithm is a concise specification of a method for solving a problem.Data Structure:A data structure can be viewed as consisting of a setof algorithms for performing operations on the data it stores.04/09/13 3
  4. 4. Computer Science & Data Structure (Continue)Data Structure & AlgorithmAlgorithms are part of what constitutesa data structure. In constructing a solution toa problem, a data structure must be chosenthat allows the data to be operated uponeasily in the manner required by thealgorithm.04/09/13 4
  5. 5.  12 Data Structures vs. Algorithms •Choice of data structures �A primary design consideration at many times �Determines the algorithms to be used �“A well-designed data structure allows a variety of critical operations to be performed using as few resources, both execution time and memory space, as possible.” •The opposite direction sometimes �Certain algorithms require particular data structures �E.g., shortest path algorithm needs priority queues – Data Structures – How do we organize – the data to be – handled?  Algorithms  Describe the process  by which data is  handled.Duality04/09/13 5
  6. 6. 04/09/13 6
  7. 7. Abstract Data Types(ADT)What is Abstraction?To abstract is to ignore some detailsof a thing in favor of others.ADT & Data structureAn Abstract Data Type (ADT) is more a way of looking at a data structure: focusing on what it does and ignoring how it does its job.04/09/13 7
  8. 8. Abstract Data Types(ADT) Why need Abstraction? Abstraction is important in problem solving because it allows problem solvers to focus on essential details while ignoring the inessential, thus simplifying the problem and bringing to attention those aspects of the problem involved in its solution. Once an abstract data type is understood and documented, it serves as a specification that programmers can use to guide their choice of data representation and operation implementation, and as a standard for ensuring program correctness.04/09/13 8
  9. 9. Abstract Data Types(ADT)Abstract data types:Each data structure can be developed around theconcept of an abstract data type that defines both dataorganization and data handling operations. A mathematical entity consisting of a set of values and a collection of operations that manipulate them. For example, the Integer abstract data type consists of a carrier set containing the positive and negative whole numbers and 0, and a collection of operations manipulating these values, such as addition, subtraction, multiplication, equality comparison, and order comparison.04/09/13 9
  10. 10. Abstract Data Types(ADT) Abstract data types are important in computer science because they provide a clear and precise way to specify what data a program must manipulate, and how the program must manipulate its data, without regard to details about how data are represented or how operations are implemented. ADT defines both data organization and data handling operations. The study of data structure is organized around a collection of abstract data types that includes lists, trees, sets, graphs, and dictionaries.04/09/13 10
  11. 11. Abstract Data Types(ADT) ADT is not a part of a program ,since a program written in a programming language requires the definition of data structure, not just the operations on the data structure. ADT is a useful tool for specifying the logical properties of a data type. ADT is not concerned with the time/space efficiency of data type.04/09/13 11
  12. 12. Specification of ADT ADT consist of 2 parts.1. Value Definition.2. Operator Definition.04/09/13 12
  13. 13. Specification of ADT1. Value Definition.It defines the collection of values for theADT and consist of two parts.a) Definition Clauseb) Condition Clause e.g. For ADT RATIONALValue definition: it states that RATIONAL valueconsist of two integers ,the second of which does not equalto zero.The keyword abstract typedef introduce a value definition.04/09/13 13
  14. 14. Specification of ADT Condition Clause Used to specify any conditions on the newly defined type. The keyword condition is used to specify any conditions on the newly defined type.For e.g. for RATIONAL ADTCondition Clause: denominator may not be zero.Note :The definition clause is required but thecondition clause may not be necessary for everyADT.04/09/13 14
  15. 15. Specification of ADT2. Operator Definition. It defines the operations that are to be performed on a data set. Each operator is defined as an abstract function with three parts: Header. Optional preconditions . Post conditions.04/09/13 15
  16. 16. Specification of ADT2. Operator Definition.For e.g. :Operator definition for RATIONAL ADTincludes the operations of creation ,addition ,multiplication and equality. Now for multiplication operation the operator definition for RATIONAL ADT is. Abstract RATIONAL mult(a,b)……/*header*/ RATIONAL a, b; ………. /*header*/ mul[0]=a[0]*b[0]; ………………….. /*postcond*/ mul[1]=a[1]*b[1];……………………/postcond*/04/09/13 16
  17. 17. Complete example of RATIONL ADT /* value definition*/ Abstract typedef< int , int >RATIONAL; Condition RATIONAL[1]!=0; /* Operator definition */ Abstract equal(a,b) RATIONAL a ,b; Postcondition equal==(a[0]*b[1]==b[0]*b[1]); Like this we can declare the other operations.04/09/13 17
  18. 18. Classification of Data StructureLinear and Non-Linear: In linear data structure, the data items are arranged in linear sequence. e.g. Array, Stack, Queue, Linked List. In non-linear data items are not in sequence. e.g. Tree , graph ,Heap04/09/13 18
  19. 19. Classification of Data Structure (Continue)Homogenous and Non-homogenous. In homogenous data structure ,all the elements are of same type e.g. Array. In non-homogenous data structure, the elements may or may not be of same type e.g. Record.04/09/13 19
  20. 20. Classification of Data Structure (Continue)Primitive and Non-Primitive:Primitive data structures constitute thenumber and the characters which are builtprograms. examples int. reals,pointerdata,charaters data, logical data.04/09/13 20
  21. 21. What is STACK or Pushdown ListA ordered collection of items into which new items may be inserted and from which items may be deleted at one end, called the top of stack.04/09/13 21
  22. 22. Operations on STACK PUSH:When item is added in stack POP:When item is removed from stack EMPTY:Stack containing no items. StackTop:Determine what the top item on a stack without removing it.this is a combination of push & pop04/09/13 22
  23. 23. Push & Pop in STACK PUSH Operation PUSH(STACK ,TOP,MAX,ITEM) STACK--Name given to Array. TOP----It contains the location of top elements. MAX---It is maximum size of STACK. ITEM--->Value that is to be store in stack if top = MAX then stackfull top = top+1 stack(top) = item Return04/09/13 23
  24. 24. Push & Pop in STACKPOP OperationPOP(STACK ,TOP,ITEM)if top = 0 then Underflow item = stack(top) top = top-1 Return04/09/13 24
  25. 25. Applications of STACKExpression EvaluationParenthesis CheckerRecursion04/09/13 25
  26. 26. Infix To Postfix Conversion of an Arithmetic ExpressionOperator preecedence^ ---------Higher precedence*,/--------Next Precedence.+,_-------Least precedence.04/09/13 26
  27. 27. Infix To Postfix Conversion of an Arithmetic Expression The rules to remember.1. Parenthesize the expression starting fro left to right.2. During Parenthesizing the expression the operands associated with operator having higher precedence are first parenthesized.04/09/13 27
  28. 28. Infix To Postfix Conversion of an Arithmetic Expression3. The sub-expression which has been converted into postfix is to be treated as single operand.4. Once the expression is converted to postfix form remove the parenthesis.Example: A+B*C A+[(B+C)+(D+E)*F]/G Answer: ABC+DE+F*+G/+04/09/13 28
  29. 29. Algorithm for Conversing Infix Expression to Postfix Form Postfix(Q,P) Q—>Given Infix expression. P->Equivalent Postfix expression.1. Push “(“ onto STACK and add “)” to the end of Q.2. Scan Q from left to right and repeat step 3 to 6 for each element of Q until the STACK is Empty.3. If an operand is encountered ,add it to P.4. If a left parenthesis is encountered ,push it onto STACK.04/09/13 29
  30. 30. Algorithm for Conversing Infix Expression to Postfix Form5. If an operator ® is encountered then:(a) ADD ® to STACK[End of IF](b) Repeatedly pop from STACK and add P each operator which has same precedence as or higher precedence than ®.04/09/13 30
  31. 31. Algorithm for Conversing Infix Expression to Postfix Form6. If a right parenthesis is encountered then:(a) Repeatedly pop from STACK and add to P each operator until a left parenthesis is encountered.(b) Remove the left parenthesis. do not add it of P][End of If][End of step 2 loop]7. Exit.04/09/13 31
  32. 32. Converting Infix Expression to Prefix Expression.1. Reverse the input string.2. Examine the next element in the input.3. If it is operand ,add it to the output string.4. If it is closing parenthesis , push it on STACK.04/09/13 32
  33. 33. Converting Infix Expression to Prefix Expression.5. If it is operator , then: i) if STACK is empty, push –operation on STACK. ii) if the top of the stack is closing parenthesis push operator on STACK. iii) if it has same or higher priority then the top of STACK, push operator on S. iv) else pop the operator from the STACK and add it to output string, repeat S.04/09/13 33
  34. 34. Converting Infix Expression to Prefix Expression.6. If it is a opening parenthesis , pop operator from STACK and add them to S until a closing parenthesis is encountered. POP and discard the closing parenthesis.7. If there is more input go to step 2.8. If there is more input , unstack the remaining operators and add them.9. Reverse the output string.04/09/13 34
  35. 35. Algorithm to Evaluate a Postfix ExpressionThis algorithm finds the value of anarithmetic expressions P written inpostfix notation. The algorithm uses aSTACK to hold operands , evaluate P.1. Add right parenthesis “)” at the end of P.[This acts as sentinel]2. Scan P from left to right and repeat step 3 & 4 for each element of P until the sentinel “)” is encountered.04/09/13 35
  36. 36. Algorithm to Evaluate a Postfix Expression3. If an operand is encountered , put it on STACK.4. If an operator ® is encountered , then: A) Remove the two top elements of STACK, Where A is the top element and B is the next-to-top element. B)Evaluate B ®A. C)Place the result of (b) back o STACK. [End of IF] [End of step 2 loop]6. Set value equal to the top element on STACK.7. Exit.04/09/13 36
  37. 37. Parenthesis Checker Read exp Create empty Stack For each character C in exp If(current character is left symbol) then Push the character C onto Stack. Elseif(current character C is a right Symbol)then If(stack is empty)then Print:”Error:No matching open symbol” Exit04/09/13 37
  38. 38. Parenthesis Checker Else POP a symbol S from the Stack If (S doesnot correspond to C)then Print:”Error:Incorrect nesting of symbols” Exit Endif Endif Endif End for04/09/13 38
  39. 39. Parenthesis Checker If(stack is not empty)then Print:”Error:Missing closing symbols(s)” Else Print:”Input expression is ok” Endif End04/09/13 39
  40. 40. Example of STACK Many computer algorithms work best with stacks --- stacks are used for remembering partially completed tasks, and undoing (backtracking from) an action. An example of 1. is described in the next section, where sub expressions of an arithmetic expression are remembered for later computation. Another example is presented in the next lecture, where we see how the Java Virtual Machine uses a stack to remember all of a programs methods that have been called but are not yet finished. An example of 2. is the ``undo button on most text editors, which lets a person undo a typing error, or the ``back button on a web browser, which lets a user backtrack to a previous web page. Another example is a searching algorithm, which searches a maze and keeps a history of its moves in a stack. If the algorithm makes a false (bad) move, the move can be undone by retrieving the previous position from the stack.04/09/13 40
  41. 41. STACK Implementation Static and dynamic data structures A stack can be stored in: • a static data structure OR • a dynamic data structure Static data structures These define collections of data which are fixed in size when the program is compiled. An array is a static data structure. Dynamic data structures These define collections of data which are variable in size and structure. They are created as the program executes, and grow and shrink to accommodate the data being stored.04/09/13 41
  42. 42. TRANSPOSE OF SPARSE MATRIX #include<stdio.h> #include<conio.h> // transpose for the sparse matrix void main() { //clrscr(); int a[10][10],b[10][10]; int m,n,p,q,t,col; int i,j; printf("enter the no of row and columns :n"); scanf("%d %d",&m,&n);04/09/13 42
  43. 43. TRANSPOSE OF SPARSE MATRIX // assigning the value of matrix for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("a[%d][%d]= ",i,j); scanf("%d",&a[i][j]); } }04/09/13 43
  44. 44. TRANSPOSE OF SPARSE MATRIX printf("nn"); //displaying the matrix printf("nnThe matrix is :nn"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("%d",a[i][j]); } printf("n"); }04/09/13 44
  45. 45. TRANSPOSE OF SPARSE MATRIX t=0; printf("nnthe non zero value matrix are :nn"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) {// accepting only non zero value if(a[i][j]!=0) {04/09/13 45
  46. 46. TRANSPOSE OF SPARSE MATRIX t=t+1; b[t][1]=i; b[t][2]=j; b[t][3]=a[i][j]; } } } b[0][0]=m; b[0][1]=n; b[0][2]=t; // displaying the matrix of non-zero value printf("n"); For(i=0;i<=t;i++) printf(“%d “,a[i][0],a[i][1],a[i][2]);04/09/13 46
  47. 47. TRANSPOSE OF SPARSE MATRIX // transpose of the matrix printf("nnthe transpose of the matrix :n "); if(t>0) { for(i=1;i<=n;i++) { for(j=1;j<=t;j++) { if(b[j][2]==i) { a[q][1]=b[j][2]; a[q][2]=b[j][1]; a[q][3]=b[j][3]; q=q+1; } }04/09/13 } } 47
  48. 48. TRANSPOSE OF SPARSE MATRIX for(i=0;i<=t;i++) { printf("a[%d %d %d %dn",i,a[i][1],a[i][2],a[i][3]); } getch(); }04/09/13 48
  49. 49. Tower of Hanoi(An application of Recursion & Stack)This problem is not specified in terms of recursion but the recursive technique can be used to produce a logical solution . Problem Definition: There are three pegs A,B,& C exist, any number of disks (n) of different diameters are placed on peg A so that a larger disk is always below a smaller disk.04/09/13 49
  50. 50. Tower of Hanoi(An application of Recursion & Stack) The object is to move n disks to peg C ,using peg B as auxiliary. Only the top disk on any peg may be moved to any other peg, and a larger disk may never rest on a smaller one. Note:in general it requires f(n)=2n-1 moves 04/09/13 50
  51. 51. Recursive solution to tower of Hanoi problem To move n disks from A to C using B as auxiliary:1. If n==1 move the single disk from A to C and stop04/09/13 51
  52. 52. Recursive solution to tower of Hanoi problem2. Move the top n-1 disks from A to B , using C as auxiliary.3. Move the remaining disks from A to C4. Move the n-1 disks from B to C,using A as auxiliary.04/09/13 52
  53. 53. Recursive Implementation of Tower of Hanoi.main(){int n;scanf (“%d”, &n);towers (n, ’A’, ’C’ ,’B’);}04/09/13 53
  54. 54. towers(int n, char frompeg, char topeg,char auxpeg){If (n==1){printf (“n %s %c %s %c”, “move disk 1 from peg “,frompeg, “to peg”, topeg);return;}04/09/13 54
  55. 55. towers(n-1,frompeg,auxpeg,topeg)printf (“n %s %d %s %c %s %c”, “move disk”, n, “from peg”, frompeg, “to peg”, topeg);tower(n-1,auxpeg,topeg,frompeg);} 04/09/13 55