Computing target number from numbers in a set

Question

I\'m working on a homework problem that asks me this:

Tiven a finite set of numbers, and a target number, find if the set can be used to calculate the target number

Answer 1

This isn't meant to be the fastest solution, but rather an instructive one.

• It recursively generates all equations in postfix notation
• It also provides a translation from postfix to infix notation
• There is no actual arithmetic calculation done, so you have to implement that on your own
  • Be careful about division by zero

With 4 operands, 4 possible operators, it generates all 7680 = 5 * 4! * 4^3 possible expressions.

• 5 is Catalan(3). Catalan(N) is the number of ways to paranthesize N+1 operands.
• 4! because the 4 operands are permutable
• 4^3 because the 3 operators each have 4 choice

This definitely does not scale well, as the number of expressions for N operands is [1, 8, 192, 7680, 430080, 30965760, 2724986880, ...].

In general, if you have n+1 operands, and must insert n operators chosen from k possibilities, then there are (2n)!/n! k^n possible equations.

Good luck!

import java.util.*;

public class Expressions {
    static String operators = "+-/*";

    static String translate(String postfix) {
        Stack<String> expr = new Stack<String>();
        Scanner sc = new Scanner(postfix);
        while (sc.hasNext()) {
            String t = sc.next();
            if (operators.indexOf(t) == -1) {
                expr.push(t);
            } else {
                expr.push("(" + expr.pop() + t + expr.pop() + ")");
            }
        }
        return expr.pop();
    }

    static void brute(Integer[] numbers, int stackHeight, String eq) {
        if (stackHeight >= 2) {
            for (char op : operators.toCharArray()) {
                brute(numbers, stackHeight - 1, eq + " " + op);
            }
        }
        boolean allUsedUp = true;
        for (int i = 0; i < numbers.length; i++) {
            if (numbers[i] != null) {
                allUsedUp = false;
                Integer n = numbers[i];
                numbers[i] = null;
                brute(numbers, stackHeight + 1, eq + " " + n);
                numbers[i] = n;
            }
        }
        if (allUsedUp && stackHeight == 1) {
            System.out.println(eq + " === " + translate(eq));
        }
    }
    static void expression(Integer... numbers) {
        brute(numbers, 0, "");
    }

    public static void main(String args[]) {
        expression(1, 2, 3, 4);
    }
}