The language is Java 1. Convert infix to postfix:Infix:a * b +(cd)Postfix:2. Draw the sequence of stack configurations in the evaluation of the followingpostfix expression:3 5 5 *3. Convert infix (4 + 3)+ 4 7 + / 6 ** 5(73) / 2 to postfix.4. Evaluate the value of the postfix from question #3.Hint:Convert infix to postfix: store operators +, -, *, / and openparentheses to a stack.Evaluate the value of postfix: store numbers to a stack.

Note:-As per policy We have entitled to answer the first question Step 1:- Infix Expression is define as operator is available in between the operands. For example C * D C and D are operands and * is the operator Post-fix Expression is define as operator is available in after the operands. For Example AB+ Program Approach:- 1.import the java.util.stack2.Use switch case to work on the operator of the infix expression 3.Declare the user-defined function whose name is static String infixToPostfix(String exp)4.Use the stack Step 2:- Program:- import java.util.Stack; public class Test1 { // A utility function to return // precedence of a given operator // Higher returned value means // higher precedence static int Prec(char ch) { switch (ch) { case '+': case '-': return 1; case '*': case '/': return 2; case '^': return 3; } return -1; } // The main method that converts // given infix expression // to postfix expression. static String infixToPostfix(String exp) { // initializing empty String for result String result = new String(""); // initializing empty stack Stack<Character> stack = new Stack<>(); for (int i = 0; i<exp.length(); ++i) { char c = exp.charAt(i); // If the scanned character is an // operand, add it to output. if (Character.isLetterOrDigit(c)) result += c; // If the scanned character is an '(', // push it to the stack. else if (c == '(') stack.push(c); // If the scanned character is an ')', // pop and output from the stack // until an '(' is encountered. else if (c == ')') { while (!stack.isEmpty() && stack.peek() != '(') result += stack.pop(); stack.pop(); } else // an operator is encountered { while (!stack.isEmpty() && Prec(c) <= Prec(stack.peek())){ result += stack.pop(); } stack.push(c); } } // pop all the operators from the stack while (!stack.isEmpty()){ if(stack.peek() == '(') return "Invalid Expression"; result += stack.pop(); } return result; } // Driver method public static void main(String[] args) { String exp = "a*b+(c-d)"; System.out.println(infixToPostfix(exp)); //Call the function } } Explanation:- i)Stack is LIFO based.LIFO stands for Last In Fast Out. The stack is the linear data structure. The data structure is defined as a way to arrange the element either in ascending or descending order.ii)push() function is used to add the element into the stackiii)pop() function is used to delete the element from the stackStep 3:- Output:-

Answered: 1. Convert infix to postfix: Infix: * b…

Engineering

Computer Science

1. Convert infix to postfix: Infix: * b + (c - d) Postfix: a

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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