Tag Archives: java

What the $heck?

Did you know the dollar sign ($) is a valid character in a Java variable name? Imagine my surprise when a student showed me the method he wrote. It’s supposed to calculate the amount of tip to leave at a restaurant table.

/**
 * Given the dollar amount of the bill, return 15% tip.
 */
public double tip(double $)
{
    return $ * 0.15;
}

I was so sure this was a bug in the IDE that I submitted a bug report. Then I read the Java spec on identifier names.

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Computational Games, part 3

If you’ve been following along, you should have a working Java program that builds the entire game tree for the “1, 2, … 10” two-player game. We are now going to add some code to compute each node’s win/loss status.

(This is the third part of a series on computational games. Read part 1 and part 2 before tackling the code here.)

Start by refactoring the class to add a winLoss attribute. Its values will be W (win), L (loss), or ? (don’t know yet). Also modify the constructor and toString methods accordingly. The changes are highlighted in green.

public class OneTwoTen
{
    int player;
    int pile;
    OneTwoTen add1;
    OneTwoTen add2;
    char winLoss;

    public OneTwoTen(int player, int pile)
    {
        this.player = player;
        this.pile = pile;
        this.add1 = null;
        this.add2 = null;
        this.winLoss = '?';
    }

    public String toString()
    {
        return "(" + player + ", " + pile + ", " + winLoss + ")";
    }

All the other methods remain as they are.  Don’t you love nicely modularized code that “just works” when you make changes?

We can’t know whether a particular node is a win or a loss until we know what its children are. So we’ll do this from the bottom up as described in part 1 of this series.

The leaves are handled specially: they are automatically losses. (By the way, this is only true for games where the goal is to be first to finish. In other games, such as “Even or Odd,” the goal is to have an even number of tokens regardless of who took the last one. Therefore, the leaves are not strictly losses.)

For all other nodes, we call doWinLoss on the children, then figure out if there are any losses among them. If there are, we are a win. Otherwise, a loss.

public void doWinLoss()
{
    if (add1 == null && add2 == null) winLoss = 'L';
    else
    {
        if (add1 != null) add1.doWinLoss();
        if (add2 != null) add2.doWinLoss();

        if ( ((add1 != null) && (add1.winLoss == 'L'))  ||
             ((add2 != null) && (add2.winLoss == 'L')) )
             winLoss = 'W';
        else
             winLoss = 'L';
    }
}

That’s all there is to it! Run it like this:

OneTwoTen root = new OneTwoTen(1,0);
root.buildTree();
root.doWinLoss();
root.printTree();

You should see that the root (1,0) node is labeled with a W to indicate it is a winning position — the first player is guaranteed a way to win. All other nodes will be labeled according to the rules discussed in part 1.

Want to try this on your own? Modify the program to implement the Even or Odd game. Here are the rules:

  • Start with 15 tokens in the pile. Each player maintains their own piles, initially with zero tokens in them.
  • At each turn, a player can take one, two, or three tokens from the central pile and adds them to their pile.
  • The game is over when the central pile is empty.
  • The winner is the player with an even number in their pile, regardless of who took the last tokens or who has more.

Some key points to consider when writing the program:

  • Each node will have three children because you can take 1, 2, or 3.
  • The win/loss status of the leaves depend on who has an even number and are therefore not automatically losses.
  • The in-order traversal of the tree won’t be pretty. You might be better off with a pre-order traversal: display the node first, then call printTree on each of the children.

When you’re done, you’ll be able to answer the central questions: will the first player win or lose the game? What moves should each player make?

In later posts, we’ll tackle some harder problems:

  • “Loopy” games, in which a move may take you back to a state you’ve already seen. Checkers is a well-known example of a loopy game — once a piece is kinged, it can move back and forth on the board.
  • You may have noticed a lot of duplication among the nodes. A total of 232 states are in the tree, but there are only twenty-one distinct states. How can we avoid computing and storing the duplicates? This will help us deal with games that have a lot of states.

Along the way, you’ll learn some more advanced data structures, such as HashMaps, and Java techniques for using them effectively.

Computational Games, part 2

Let’s see about coding up the “1, 2, … 10” game so the computer can arrive at the same conclusion we did: that player 1 is going to win the game.

Read Computational Games, Part 1 for the background material.

Each node in the tree is represented with:

  • The player number (1 or 2)
  • The number of tokens remaining in the pile
  • Pointers to the child nodes

(By the way, the content of this blog is aimed at university/college freshman and sophomores. The code is straightforward with not a lot of fancy syntax until it’s needed.)

Let’s define a Java class to represent a OneTwoTen node. We’ll also provide a convenient constructor for setting the attribute values and a toString for displaying them.

public class OneTwoTen
{
    int player;
    int pile;
    OneTwoTen add1;
    OneTwoTen add2;

    public OneTwoTen(int player, int pile)
    {
        this.player = player;
        this.pile = pile;
        this.add1 = null;
        this.add2 = null;
    }

    public String toString()
    {
        return "(" + this.player + ", " + this.pile + ")";
    }
}

Here’s how to define a few nodes:

OneTwoTen n1 = new OneTwoTen(1, 0);    // The initial state
OneTwoTen n2 = new OneTwoTen(2, 10);   // One of the leaf nodes
OneTwoTen n3 = new OneTwoTen(1, 6);    // A mid-game node

(I use BlueJ for a lot of my coding. Its Code Pad feature lets me simply type in statements like the ones above. With BlueJ, I can quickly test code without laboriously defining a main method. If you’re not using BlueJ, you’ll need to put the statements into a main.)

We can inspect the nodes, again in the CodePad:

> n1.player
1 (int)
> n1.pile
0 (int)
> n2.player
2 (int)
> n2.pile
10 (int)
> n1.toString()
“(1, 0)” (String)
> n3.toString()
“(1, 6)” (String)

Next, define a method called spawn that creates a node’s children. For a node where the pile is 10, there’s nothing to do. If the pile is 9, only one child is created. In all other cases, both children are spawned.

The otherPlayer method is a helper. It simply returns the number of the other player.

public void spawn()
{
    if (pile == 10) return;   // Nothing to do for the leaves
    else if (pile == 9)
    {
        // Only one child here
        add1 = new OneTwoTen(otherPlayer(), pile + 1);
    }
    else
    {
        add1 = new OneTwoTen(otherPlayer(), pile + 1);
        add2 = new OneTwoTen(otherPlayer(), pile + 2);
    }
}

public int otherPlayer()
{
    if (player == 1) return 2;
    else return 1;
}

Try it out:

> OneTwoTen n1 = new OneTwoTen(1, 0);
> n1.spawn();
> n1.add1.toString()
“(2, 1)” (String)
> n1.add2.toString()
“(2, 2)” (String)

Looks good! To build the whole tree, we’ll spawn a node’s children, then call buildTree on each of the children, telling each one to build their own trees.

public void buildTree()
{
    spawn();
    if (add1 != null) add1.buildTree();
    if (add2 != null) add2.buildTree();
}

Did it work? Let’s see…

> OneTwoTen n1 = new OneTwoTen(1, 0);
> n1.buildTree();
> n1.add1.add1.toString()
“(1, 2)” (String)
> n1.add1.add2.toString()
“(1, 3)” (String)
> n1.add1.add1.add1.toString()
“(2, 3)” (String)
> n1.add1.add1.add2.toString()
“(2, 4)” (String)

Sure looks like it! It would be nice to see the whole tree, though. The next method below prints the tree, but it’s sideways. You’ll need to turn your head 90 degrees to the left to see it.

The first printTree method takes an integer argument that tells it how far to indent the node. It calls printTree on the take2 node (so it appears on the right side of the sideways tree), then prints itself, then calls printTree on the take1 node. This is called an in-order traversal.

The second no-argument method just calls the first one with an indent of 0.

public void printTree(int indent)
{        
    if (add2 != null) add2.printTree(indent + 1);

    for (int i = 0; i < indent; i++) System.out.print("   ");
    System.out.println(toString());

    if (add1 != null) add1.printTree(indent + 1);
}

public void printTree()
{
    printTree(0);
}

I’ll leave it to you to call printTree after building the tree and seeing the fruits of your (well, my) labor.

In the next post, I’ll show you how to assign the win-loss label to each node.