Question: In the language Lisp, each of the four basic arithmetic operators appears before an arbitrary number of operands, which are separated by spaces. The resulting
In the language Lisp, each of the four basic arithmetic operators appears before an arbitrary number of operands, which are separated by spaces. The resulting expression is enclosed in parentheses.
The operators behave as follows:
(+ a b c ...) returns the sum of all the operands, and (+) returns 0.
(- a b c ...) returns a - b - c - ..., and (- a) returns -a. The minus operator must have at least one operand.
(* a b c ...) returns the product of all the operands, and (*) returns 1.
(/ a b c ...) returns a / b / c / ..., and (/ a) returns 1 / a. The divide operator must have at least one operand.
You can form larger arithmetic expressions by combining these basic expressions using a fully parenthesized prefix notation.
For example, the following is a valid Lisp expression: (+ (- 6) (* 2 3 4) (/ (+ 3) (*) (- 2 3 1)))
This expression is evaluated successively as follows: (+ (- 6) (* 2 3 4) (/ 3 1 -2)) (+ -6 24 -1.5) 16.5
Design and implement an algorithm that uses any stack implementation (e.g. java.util.Stack ) to evaluate a legal Lisp expression composed of the four basic operators and integer values. Write a program that reads such expressions and demonstrates your algorithm.
Precondition :You can assume opening and closing parenthesis are all balanced in any given expression. Also you can assume all numbers are integer.
Demonstrate your implementation with the following expressions:
Expression (+ (- 6) (* 2 3 4) (/ (+ 3) (*) (- 2 3 1))) evaluates to 16.5
Expression (+ (- 632) (* 21 3 4) (/ (+ 32) (*) (- 21 3 1))) evaluates to -378.11764705882354
Expression (+ (/ 2) (* 2) (/ (+ 1) (+) (- 2 1 ))) evaluates to Infinity
Hint: You may use more than one stack if needed.
Make it simple as possible.
Codes I have so far:
import java.util.*; public class LispExpression { private static double evaluateLisp(String lisp) { StackInterface exprStack = new ArrayStack<>(); StackInterface currentOpStack = new ArrayStack<>(); int characterCount = lisp.length(); double result = 0; for (int index = 0; index < characterCount; index++) { switch () { case "*": result = currentOpStack.pop(); while (!currentOpStack.isEmpty()) { result *= currentOpStack.pop(); } break; case "/": result = currentOpStack.pop(); while (!currentOpStack.isEmpty()) { result /= currentOpStack.pop(); } break; case "+": result = currentOpStack.pop(); while (!currentOpStack.isEmpty()) { result += currentOpStack.pop(); } break; case "-": result = currentOpStack.pop(); while (!currentOpStack.isEmpty()) { result -= currentOpStack.pop(); } break; default: result = currentOpStack.pop(); break; } exprStack.push(result); } return result; } public static void main(String args[]) { System.out.println(evaluateLisp("(+(- 6)(* 2 3 4)(/(+ 3)(*)(- 2 3 1)))") + " "); System.out.println(evaluateLisp("(+(- 632)(* 21 3 4)(/ (+ 32)(*)(- 21 3 1)))") + " "); System.out.println(evaluateLisp("(+(/ 2)(* 2)(/(+ 1)(+)(- 2 1 )))") + " "); } } Step by Step Solution
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To evaluate a Lispstyle expression using stacks you need to account for multiple operations occurring concurrently given the prefix notation Heres a plan and implementation to enhance the provided cod... View full answer
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