Generating truth tables in Java

Question

I'm trying to print some truth tables as part of a school assignment. How can I generate a dynamic size truth table in Java?

So that printTruthTable(1) prints:

0
1

printTruthTable(3) prints:

0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

And so on. I have been trying to implement it using recursion, but I just can't get it right.

Answer 1

here's my take on your problem, all written nice and tight in a small class, just copy/paste

notice how I used modulo2 (the % sign) to get 0's and 1's from the loop indices

public class TruthTable {
    private static void printTruthTable(int n) {
        int rows = (int) Math.pow(2,n);

        for (int i=0; i<rows; i++) {
            for (int j=n-1; j>=0; j--) {
                System.out.print((i/(int) Math.pow(2, j))%2 + " ");
            }
            System.out.println();
        }
    }
    public static void main(String[] args) {
        printTruthTable(3); //enter any natural int
    }
}

Answer 2

This is not a truth table - rather, it's a table of binary numbers. You can use Java's Integer.toBinaryString method to generate the zeros and ones that you need; inserting spaces should be trivial.

int n = 3;
for (int i = 0 ; i != (1<<n) ; i++) {
    String s = Integer.toBinaryString(i);
    while (s.length() != 3) {
        s = '0'+s;
    }
    System.out.println(s);
}

Answer 3

The magic of recursion:

public static void main(String args[]) {
    int size = 3;
    generateTable(0, size, new int[size]);
}

private static void generateTable(int index, int size, int[] current) {
    if(index == size) { // generated a full "solution"
        for(int i = 0; i < size; i++) {
            System.out.print(current[i] + " ");
        }
        System.out.println();
    } else {
        for(int i = 0; i < 2; i++) {
            current[index] = i;
            generateTable(index + 1, size, current);
        }
    }
}