Data Structures and Algorithms

Transcription

1 Data Structures and Algorithms Prof. Ajit A. Diwan Prof. Ganesh Ramakrishnan Prof. Deepak B. Phatak Department of Computer Science and Engineering Session: Infix to Postfix (Program) Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 1

2 Algorithm Input: Infix Expression Output: Postfix Expression 1. Validate infix expression 1.1 If not valid, display appropriate error message, else proceed for conversion 2. Traverse the infix expression (character by character) 2.1 If character is operand, then append to the string postfix 2.2 else, do the following: If operator is (, push it to the operator stack and continue with point 2. Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 2

3 Algorithm If stack is empty and operator is not ), then push the operator on the stack If stack is not empty, then compare the current operator with the one which is present on top of the stack. (a) If it has higher precedence: push the operator on the stack (b) If it has lower precedence then, i. Append the top element of the stack to string postfix ii. Pop the top operator from the stack iii. Continue i and ii till ( is encountered or the stack iv. becomes empty. Finally, push the current operator, but not ) on the stack 3. Append the remaining operators from the stack to the string postfix by using top followed by pop, till stack becomes empty. Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 3

4 : Functions Name Parameter Returns Purpose checkparentheses() Infix expression (character array) Called from validate(). Returns true or false isvalid() Current character Called from validate(). Returns true or false validate() Infix expression (character array) Returns true or false to the main() function isoperand() Current character Called from infix2postfix(). Returns true or false To check whether open and close braces match To check whether characters in expression are valid To check whether the expression is valid To check whether the current character is an operand i.e. a to z or A to Z Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 4

5 : Functions Name Parameter Returns Purpose prioritynumber() checkpriority() infix2postfix() character i.e. Operator Current operator, and top operator on stack Infix expression, Postfix expression. (Both are character arrays) Called from checkpriority(). Returns an integer. Called from infix2postfix(). Returns true or false Returns the postfix string to the main() function To determine the priority of the operator To determine the priority of the operator Converts a valid infix expression to postfix expression Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 5

6 1. Traverse the entire infix expression i. Count the number of ( i. Count the number of ) a. If count of ) is more than (, then return false 2. If the total count of ( and ) matches, return true 3. Else, return false bool checkparentheses(char infix[]) { int open=0, close=0, i=0; while (infix[i]!= '\0') { if(infix[i] == '(') open++; if(infix[i] == ')') { close++; if (close>open) return false; i++; if(open==close) return true; else return false; // End of function Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 6

7 bool isvalid(char currentchar) { return ((currentchar>=65 && currentchar<=90) (currentchar>=97 && currentchar<=122) (currentchar>=40 && currentchar<=43) currentchar==45 currentchar==47 currentchar==94); // end of fucntion Check whether the character is either of the following A to Z or a to z or (, ), *, + or - or / or ^ Return true if the above condition is satisfied, else return false Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 7

8 1. Traverse the string i. If character is not an operand or operator, then return false ii. Else, point to the next character 2. Examine whether checkparentheses() i. If false was returned, return false 3. Return true. This means that the expression passed all tests and it is valid. bool validate(char infix[]) { int i=0; while (infix[i]!= '\0') { if (! isvalid(infix[i]) ) return false; else i++; if (checkparentheses(infix) == false) return false; return true; // end of fucntion Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 8

9 int prioritynumber(char Operator) { if (Operator=='^') return 0; if (Operator=='*' Operator=='/') return 1; if (Operator=='+' Operator=='-') return 2; if (Operator=='(') return 3; // End of function Return value 0, 1, 2, or 3 depending on the operator bool checkpriority(char currentoperator, char topoperator) { if ( prioritynumber(currentoperator) < prioritynumber(topoperator)) return true; else return false; // End of function Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 9

10 bool isoperand(char currentchar) { return ( (currentchar>=65 && currentchar<=90) (currentchar>=97 && currentchar<=122) ); // End of function Check whether the character is either of the following A to Z or a to z or Return true if the above condition is satisfied, else return false Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 10

11 void infix2postfix(char infix[], char answer[]) { int i=0, j=0; char currentchar; char postfix[50]=""; stack<char> operatorstack; //Operator stack while(infix[i]!='\0') { //Traverse the infix expression currentchar = infix[i]; Code for 2.1 Code for 2.2 i++; //Point to next character of infix expression //End of while Code for 3 strcpy(answer,postfix); //End of function Note: 2.1, 2.2, and 3 are point number given in the Algorithm Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 11

12 2.1 If character is operand, then append to the string postfix // Code for 2.1 if ( isoperand(currentchar) ) { postfix[j]=currentchar; j++; Note: 2.1, is point number given in the Algorithm Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 12

13 2.2 else, do the following (It is an operator): If operator is (, push it to the operator stack and continue with point 2. // Code for 2.2 else { //It is operator Code for Code for Code for if (currentchar=='(') { operatorstack.push(currentchar); i++; continue; Note: 2.2, 2.2.1, 2.2.2, and are point number given in the Algorithm Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 13

14 2.2.2 If stack is empty and operator is not ), then push the operator on the stack // Code for 2.2 else { //It is operator Code for Code for Code for if(operatorstack.empty() && currentchar!=')') { operatorstack.push(currentchar); Note: 2.2, 2.2.1, 2.2.2, and are point number given in the Algorithm Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 14

15 2.2.3 If stack is not empty, then compare the current operator with the one which is present on top of the stack. (a) If it has higher precedence: push the operator on the stack (b) If it has less precedence then, i. Append the top element of the stack to string postfix ii. Pop the top operator from the stack iii. Continue i and ii till ( is encountered or the stack becomes empty. iv. Finally, push the current operator, but not ) on the stack Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 15

16 // Code for 2.2 else { //It is operator Code for Code for Code for else { if (checkpriority(currentchar,operatorstack.top())) operatorstack.push(currentchar); else { while (!operatorstack.empty()) { if (operatorstack.top()=='(') { operatorstack.pop(); break; postfix[j]=operatorstack.top(); j++; operatorstack.pop(); //End of while if(currentchar!=')') operatorstack.push(currentchar); //End of else //End of else a b Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 16

17 3. Append the remaining operators from the stack to the string postfix by using top followed by pop, till stack becomes empty. // Code for 3 //Transfer remaining items in stack while(!operatorstack.empty()) { postfix[j]=operatorstack.top(); operatorstack.pop(); j++; Note: 3, is point number given in the Algorithm Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 17

18 int main() { char infix[50], postfix[50]; cout << "Enter an infix expression without space. Use parentheses to indicate priority. Only +, -, *, /, ^ symbols are allowed\n"; cin >> infix; if (validate(infix) == false) { cout << "The expression contains a number, or an unwanted symbol, or the round braces do not match\n"; return -1; cout << "Infix expression : " << infix << endl; infix2postfix(infix,postfix); cout << "Postfix expression : " << postfix << endl; return 0; Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 18

19 References Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 19

20 Thank you Ajit A. Diwan, Ganesh Ramakrishnan, and Deepak B. Phatak, 20