Data Structure Lecture 3

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Linear Data Structure:
This is about stack.

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Data Structure Lecture 3

  1. 1. Linear Data Structures Stack
  2. 2. Topics Covered in This Lecture• Applications of stack – Reversing the string – Balancing symbols – Postfix expression evaluation – Translating infix expression to postfix expression
  3. 3. Applications of Stack• Reversing the string – push each character on to a stack as it is read. – When the line is finished, we then pop characters off the stack, and they will come off in the reverse order.
  4. 4. Applications of StackReversing the stringvoid ReverseRead(void){ //using static Stack class of Lecture#3 Stack<char> stack;//The Stack ‘stack’ is created char item; cin>>item; while (!stack.isFull()&&item!=$)//type in $ from keyboard to stop input { stack.push(item); // push each character onto the stack cin>>item; } while(!stack.isEmpty() ) { item=stack.pop ();//Pop an element from stack cout<<item; }}
  5. 5. Applications of Stack• Balancing Symbols Compilers use a program that checks whether every symbol (like braces, parenthesis, brackets, etc) in a program is balanced. The simple algorithm for this purpose is:1. Make an empty stack.2. Read the characters until end of file.3. If the character is an open any thing, push it onto the stack.4. If it is a close any thing, then if the stack is empty report an error. Otherwise Pop the Stack. If the popped symbol is not the corresponding opening symbol, then report an error.5. At the end of the file, if the stack is not empty report an error.
  6. 6. Polish Notationa–b/c+d*ePrecedence?1. b/c2. d*e3. a – a1 /a1 = b/c /4. t2 + a2 / t2 = a – b/c / /a2 = d*e/
  7. 7. Infix, Suffix, PrefixInfix = a * b + c((a*b) +c)Priority:1. a * b2. a1 + c / a1 = a * b /Prefix =* a b , +a1 c+*abcSuffix =ab* , a1c+ab*c+
  8. 8. Infix, Suffix, Prefixinfix = a- b * c / d + e / fsuffix =a – bc* / d + e / f a – bc*d/ + e / f a – bc*d/ + ef/ abc*d/- + ef/ abc*d/-ef/+prefix =a - *bc / d + e / f a - /*bcd + e / f a - /*bcd + /ef -a/*bcd + /ef +-a/*bcd/ef
  9. 9. Infix, Suffix, PrefixInfix: a+b*c–d/f+eSuffix: abc*+df/-e+Prefix: +-+a*bc/dfe
  10. 10. Applications of Stack Postfix Expression Evaluation – When a number is seen, it is pushed onto the stack – When an operator is seen, then pop two elements from stack and push the result onto the stack. Now we evaluate the following postfix expression. 6523+8*+3+* 31. The first four are placed on the stack. The resulting stack is 2 5 6 stack
  11. 11. Applications of Stack• evaluating the following postfix expression. 3 6523+8*+3+* 2 5 6 stack 2. Next a + is read, so 3 and 2 are popped from the stack and their sum 5 is pushed. 5 5 6 stack
  12. 12. Applications of Stack• evaluating the following postfix expression. 6523+8*+3+* 5 5 6 stack 83. Next 8 is read and pushed. 5 5 6 stack
  13. 13. Applications of Stack• evaluating the following postfix expression. 8 6523+8*+3+* 5 5 6 stack4. Next a * is seen so 8 and 5 are popped as 8 * 5 = 40 is pushed 40 5 6 stack
  14. 14. Applications of Stack• evaluating the following postfix expression. 6523+8*+3+* 40 5 6 stack5. Next a + is read so 40 and 5 are popped and 40 + 5 = 45 is pushed. 45 6 stack
  15. 15. Applications of Stack• evaluating the following postfix expression. 6523+8*+3+* 45 6 stack6. Now 3 is pushed 3 45 6 stack
  16. 16. Applications of Stack• evaluating the following postfix expression. 6523+8*+3+* 3 45 6 stack7. Next symbol is + so pops 3 and 45 and pushes 45 + 3 = 48, so push 48 in stack. 48 6 stack
  17. 17. Applications of Stack• evaluating the following postfix expression. 6523+8*+3+* 48 6 stack7. Finally a * is seen and 48 and 6 are popped, the result 6 * 48 = 288 is pushed.8. As there is no input, so pop the stack and we get the result. 288 stack
  18. 18. Applications of Stack• Translating infix expressions to postfix expression – When an operand is read, it is immediately placed onto the output. – When an operator or left parenthesis comes then save it in the stack initially stack is empty. – If we see a right parenthesis, then we pop the stack, writing symbols until we encounter a (corresponding) left parenthesis, which is popped but not output. – If we see any other symbol (‘+’, ‘*’, ‘(‘, etc) then we pop entries form the stack until we find an entry of lower priority. One exception is that we never remove a ‘(‘ from the stack except when processing a ‘)’. – When the popping is done, we push the operand onto the stack. – Finally, if we read the end of input, we pop the stack until it is empty, writing symbols onto the output.
  19. 19. Applications of Stack• Translating infix expressions to postfix expressionConvert the following infix expression to postfix expression.a+b*c+(d*e+f)*g1. First the symbol a is read, so it is passed through to the output a2. Then + is read and pushed onto the stack. output + stack3. Next b is read and passed through to the output. ab output *4. Next a * is read. The top entry on the operator stack has lower + precedence than *, so nothing is output and * is put on the . stack
  20. 20. Applications of StackConverting the following infix expression to postfix expression.a+b*c+(d*e+f)*g5. Next, c is read and output. abc output6. The next symbol is a +. Checking the stack, we find that priority of stack top symbol * is higher than + . So we pop a * and place it on the output, Pop the other +, which is not of lower but equal priority, and then push +. * abc*+ + + output stack stack
  21. 21. Applications of StackConverting the following infix expression to postfix expression.a+b*c+(d*e+f)*g 7. The next symbol read is an ‘(‘, which, being of highest precedence, is placed on the stack. ( + stack 8. Then d is read and output. abc*+d output
  22. 22. Applications of StackConverting the following infix expression to postfix expression.a+b*c+(d*e+f)*g9. We continue by reading a *. Since open parenthesis do not get removed except when a closed parenthesis is being processed, there is no output and we push * in stack * ( + stack10. Next, e is read and output. abc*+de output
  23. 23. Applications of StackConverting the following infix expression to postfix expression.a+b*c+(d*e+f)*g11. The next symbol read is a +, since priority of stack top value is higher so we pop * and push +. + abc*+de* ( output + stack12. Now we read f and output f. abc*+de*f output
  24. 24. Applications of StackConverting the following infix expression to postfix expression.a+b*c+(d*e+f)*g 13. Now we read a ‘)’, so the stack is emptied back to the ‘(‘, we output a +. abc*+de*f+ + output stack 14. We read a * next; it is pushed onto the stack. * + stack 15. Now, g is read and output. abc*+de*f+g output
  25. 25. Applications of StackConverting the following infix expression to postfix expression. *a+b*c+(d*e+f)*g + stack16. The input is now empty, so pop output symbols from the stack until it is empty. abc*+de*f+g*+ output stack
  26. 26. Assignment 1 Group Size: 2, Due Date : Feb 27, 2012• Implement Applications of stack 1. Balancing of symbols 2. Decimal to Binary Conversion 3. Infix to Postfix conversion 4. Postfix expression evaluation 5. Infix to Prefix conversion 6. Prefix expression evaluation1. Each application will be implemented as a separate function. Function will use stack, which is already implemented in a header file and is available for reuse.2. Difference between CGPAs of both group members should not be more than ONE

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