How do I optimize casting number to boolean, the most efficient way, Optimizing Tic Tac Toe

Question

I am new to java and I am trying to optimize my code as far as possible.

I want to know what is the most efficient way for the machine to convert a number, lets say int or char, to a boolean.

I am implementing TicTacToe in java. I am using bitmaps to represent the state of the game. I am aware of the unreadability, but I don't want to optimize that. Code efficient < machine efficient.

import java.util.Scanner;

public class TicTacToe {
    boolean swap = false;
    char player1 = 0b000000000;
    char player2 = 0b000000000;
    static char[] winconditions = new char[]{0b111000000, 0b000111000, 0b000000111, 0b100100100, 0b010010010, 0b001001001, 0b100010001, 0b001010100};
    Scanner scanner = new Scanner(System.in);

    boolean evaluateWinner() {
        char current = swap ? player2 : player1;
        for (char condition : winconditions)
            if ((condition == (current & condition))) return true;
        return false;
    }

    int getInput() {
        int field = player1 | player2;
        while (true) {
            System.out.println("Give Input: 0 1 ... 8 ");
            int out = 1 << scanner.nextInt();
            if ((out & field) == 0) return out;
        }
    }

    public void start() {
        do {
            draw();
            if (swap) player2 |= getInput();
            else player1 |= getInput();
            swap = !swap;
        } while (evaluateWinner());
    }

    void draw() {
        for (int i = 0; i < 9; i++) {
            boolean isP1 = (player1 & 1 << i) != 0;
            boolean isAny = (player2 & 1 << i) != 0 || isP1;
            String out = isAny ? (is1 ? "X" : "O") : String.valueOf(i);
            if (i % 3 == 2) out += "|\n";
            System.out.print("|" + out);
        }
    }
}

Call var game = new TicTacToe(); game.start(); to run the game.

In TicTacToe.draw I need to determine if a specific bit is set to 1 or 0. With a few bitwise operations I am getting a number. In C I wouldn't have to convert the number to boolean. But in Java I have to. What is the most efficient way to do that?

1. n != 0

boolean isP1 = (player1 & 1 << i) != 0;
boolean isAny = (player2 & 1 << i) != 0 || isP1;

2. (boolean) n

boolean isP1 = (boolean) (player1 & 1 << i) ;
boolean isAny = ((boolean)(player2 & 1 << i))|| isP1;

Is it the same or are there more efficient ways to do convert numbers to a boolean?

---

If you have more ideas how to optimize the code in cpu runtime and in memory usage. Let me know. I don't think there is much more I can do. But proof me wrong if you want.

Kind reagards : )

---

Answer 1

As long as your code lacks correctness, i.e. the condition of while(evaluateWinner()) is wrong and there’s no check for a draw, you should not worry about performance.

Letting aside that in this simple program, performance doesn’t matter at all, you are focusing on the wrong things. You are performing string concatenation in a loop and calling print nine times in a row. The form of the boolean expression is entirely irrelevant. Even performing modulo 3 is more expensive here (unless the JVM’s optimizer replaces it).

If you want to continue, as a fun exercise, consider the following points:

• Reduce the number of I/O operations (i.e. one print instead of nine)
• Prefer local variables
• Use the smaller integer types (byte, short, char) only for arrays. For most operations, especially local variables and arithmetic, they may counter the intended effect: Java actually uses int, but has to perform additional truncation
• If you need an immutable char[], use String. The encapsulation has no noticeable performance penalty but enables lots of potential optimizations.
• If your replace player1 & 1 << i with player1 >> i & 1 you get zero or one, then, instead of testing against zero and conditionally using 'X' or 'O', you can use it as an index into "OX", avoiding the conditional.

E.g.

public class TicTacToe {
    static final String WIN_CONDITIONS="\u01c0\070\007\u0124\222\111\u0111\124";

    final Scanner scanner = new Scanner(System.in);
    int player1 = 0b000_000_000, player2 = 0b000_000_000;

    boolean evaluateWinner(int current) {
        for(int ix = 0; ix < WIN_CONDITIONS.length(); ix++) {
            char condition = WIN_CONDITIONS.charAt(ix);
            if((condition == (current & condition))) return true;
        }
        return false;
    }

    int getInput() {
        int field = player1 | player2;
        for(;;) {
            System.out.println("Give Input: 0 1 ... 8 ");
            int out = 1 << scanner.nextInt();
            if((out & field) == 0) return out;
            System.out.println("already occupied");
        }
    }

    public void start() {
      for(boolean swap = false; ; swap = !swap) {
            draw();
            int p = swap? player2: player1;
            p |= getInput();
            if(evaluateWinner(p)) {
                System.out.println(swap? "O wins": "X wins");
                break;
            }
            if(swap) player2 = p; else player1 = p;
            if((player1 | player2) == 0x1FF) {
                System.out.println("Draw");
                break;
            }
        }
    }

    void draw() {
        char[] chars = "|0|1|2|\n|3|4|5|\n|6|7|8|\n".toCharArray();
        int p1 = player1, field = p1 | player2;
        for(int i = 0, j = 1; i < 9; j += 2)
            for(int c = 0; c < 3; c++, i++, j+=2)
              if((field & 1 << i) != 0) chars[j] = "OX".charAt(p1 >> i & 1);
        System.out.print(chars);
    }
}

As said, you won’t notice actual performance differences with such a simple program, but it’s a fun exercise providing some ideas.