truth table in 2D arry

Question

Below, you'll see how I am building my truth table.

    //Tab 2D represents truth table
    //tt [nbr of combinaisons] [nbr of variables + S]
    boolean tt [][] = new boolean [nbrCombinaisons][nbrVariables+1];
    for (int j = 0; j < nbrVariables; j++) {    
        for (int i = 0; i < nbrCombinaisons; i++) { 
            tt[i][j] = true;
        }
    }
    //Display truth tab in console
    for (boolean[] row : tt) { System.out.println(Arrays.toString(row));}

    }
}

Do you know how can I store my array to have somethin like this :

False False  **False**
False True   **False**
True  False  **False**
True  True   **False**

** S ** will be stored after.

tks

I'm not sure I understand what you're asking for. The output you've displayed looks like a 4x3 2d array (4 rows, 3 columns) - where's the problem? — Thomas, Feb 27 '19 at 12:19
Ok, from any boolean expression, I would like to create his truth table. That's why I am using a 2D array. If in my expression I have only 2 varables (a - b), so I have 4 combinaisons (4 rows and 3 columns[a, b and my output]) . But i don't know how store all combinaisons in my table... About my output, for the moment, i don't care, I will do it after. I hope be more understood. Thx — , Feb 27 '19 at 12:25
still not clear, you are already storing all combinations in a `boolean[][]` with all columns for the variables and the output. Is your question how to fill the matrix, that is, how to generate all the combinations? — user85421, Feb 27 '19 at 12:32
maybe you should edit the question to include that (generate the matrix). Simple solution `tt[0][0] = false; tt[0][1] = false; tt[1][0]= ...` or using loops and some bit logic (hint: `for (var i=0; i<4; i++) System.out.println(((i&2)!=0) + " " + ((i&1)!=0));` — user85421, Feb 27 '19 at 12:49

score 0 · Accepted Answer · answered Feb 27 '19 at 15:50

If you interpret false as 0 and true as 1 all possible combinations can be generated using the binary numbers from 0 to number of all combinations minus 1. If you have x variables the number of possible combinations is 2^x. For example

for 2 variables count of combinations is 2^2 = 4 and the binary numbers from 0 to 4-1 are

00
01
10
11

for 3 variables count of combinations is 2^3 = 8 and the binary numbers from 0 to 8-1 are

000
001
010
011
100
101
110
111

Using above insights your code could be something like:

public static void main(String[]args) {        
    int nbrVariables = 2;
    int nbrCombinaisons = (int) Math.pow(2, nbrVariables);

    boolean tt [][] = new boolean [nbrCombinaisons][nbrVariables+1];        
    for (int j = 0; j < nbrCombinaisons; j++) {             
        String    tempStr  = String.format("%"+nbrVariables+"s", Integer.toBinaryString(j)).replace(" ", "0");
        boolean[] tempBool = new boolean[tempStr.length()+1];
        boolean total = tempStr.charAt(0)=='1';
        for(int i=0; i<tempStr.length(); i++){
            tempBool[i]= tempStr.charAt(i)=='1';
            if(i>0){
                total = total && tempBool[i];  //table for logical AND change operator to || for OR or ^ for XOR
            }
        }
        tempBool[tempStr.length()] = total;
        tt[j] = tempBool;
    }

    for (boolean[] row : tt) {            
        for (boolean c : row) { 
            System.out.print(c + "\t");        
        }
        System.out.println();
    }      
}

truth table in 2D arry

1 Answers1