Monday, January 26, 2015



TopCoder Problem Statement - Pool
Single Round Match 163 Round 1 - Division I, Level Two
Single Round Match 163 Round 1 - Division II, Level Three


Determine the fewest number of swaps necessary to re-order a set of billiard balls into a legal racking.

Java Source:
import java.util.PriorityQueue;

public class Pool {

    private enum BallType {SOLID, STRIPE, EIGHT}

    private static BallType[][] legalRacks = {{
            BallType.SOLID, BallType.SOLID,
            BallType.STRIPE, BallType.EIGHT, BallType.STRIPE,
            BallType.SOLID, BallType.STRIPE, BallType.SOLID, BallType.STRIPE,
            BallType.STRIPE, BallType.SOLID, BallType.STRIPE, BallType.SOLID,
    }, {
            BallType.STRIPE, BallType.STRIPE,
            BallType.SOLID, BallType.EIGHT, BallType.SOLID,
            BallType.STRIPE, BallType.SOLID, BallType.STRIPE, BallType.SOLID,
            BallType.SOLID, BallType.STRIPE, BallType.SOLID, BallType.STRIPE,

    * Represents a configuration of balls, and how many moves it took to
    * get there.  Includes some methods for swapping balls, and determining
    * how far away it is from a legal rack.
    private class SearchNode implements Comparable {

        private BallType[] rack;
        private final int moves;
        private int numOutOfPlace = -1;

        SearchNode(BallType[] rack, int moves) {

            // Make a defensive copy of the rack that was passed in.
            this.rack = new BallType[rack.length];
            System.arraycopy(rack, 0, this.rack, 0, rack.length);

            this.moves = moves;

        * Returns the number of balls that are out of place.  Since there
        * are more than one legal racks, it checks them all, and returns
        * the minimum.  Note, this calculation is only done once.
        int getNumOutOfPlace() {

            if (numOutOfPlace >= 0)  {
                return numOutOfPlace;

            numOutOfPlace = Integer.MAX_VALUE;

            // Check each legal configuration.
            for (int i = 0; i < legalRacks.length; i++) {

                int n = 0;

                for (int b = 0; b < rack.length; b++) {
                    if (rack[b] != legalRacks[i][b]) {

                numOutOfPlace = Math.min(numOutOfPlace, n);

            return numOutOfPlace;

        * A heuristic representing how ideal this state is.  It is better
        * to have less balls out of place in a lower number of moves.
        int getPriority() {
            return getNumOutOfPlace() + moves;

        * Returns a new SearchNode with one pair of balls swapped.  Number
        * of moves is incremented by one.  If the 8 ball is out of position,
        * it will be swapped into the correct place first.
        SearchNode swapBalls(BallType[] legalRack) {

            int ball1;
            int ball2;

            SearchNode newNode = new SearchNode(rack, moves + 1);

            // If the 8 ball is out of place, fix that first.
            if (rack[4] != BallType.EIGHT) {
                ball1 = 4;
                ball2 = 0;
                while (rack[ball2] != BallType.EIGHT) ball2++;  // Find the 8.
            } else {

                * ball1 is the first ball that is out of place - regardless
                * of it's type (solid / stripe).
                * ball2 is the first ball out of place of the type that
                * should be at the location of ball1.
                ball1 = getFirstOutOfPlace(legalRack, null);
                ball2 = getFirstOutOfPlace(legalRack, legalRack[ball1]);

            newNode.rack[ball1] = rack[ball2];
            newNode.rack[ball2] = rack[ball1];

            return newNode;

        * Find the first ball that does not match the corresponding legal
        * rack.  If type is null, it can be either stripe or solid.  If
        * type is given, then the out of place ball must be of that type.
        private int getFirstOutOfPlace(BallType[] legalRack, BallType type) {

            for (int i = 0; i < rack.length; i++) {
                if ((type == null) || (rack[i] == type)) {
                    if (rack[i] != legalRack[i]) {
                        return i;

            return -1;

        // Sort SearchNodes based on their priority.
        public int compareTo(SearchNode o) {
            return, o.getPriority());


    public int rackMoves(int[] triangle) {

        * Using a PriorityQueue to process the possible solutions will allow
        * us to focus on the most promising paths first.  That is those closest
        * to a legal rack in the fewest number of moves.
        PriorityQueue possibleSolutions = new PriorityQueue<>();

        * Convert the given input array into an array of BallType objects.
        * These will be easier to work with then constantly having to check
        * ranges to determine if the ball is a stripe or solid.
        BallType[] rack = new BallType[triangle.length];
        for (int i = 0; i < triangle.length; i++) {
            if (triangle[i] < 8) rack[i] = BallType.SOLID;
            else if (triangle[i] > 8) rack[i] = BallType.STRIPE;
            else rack[i] = BallType.EIGHT;

        // Put our starting position into the PriorityQueue
        SearchNode n = new SearchNode(rack, 0);

        // The fewest moves seen so far that leads to a legal rack.
        int minNumberOfMoves = Integer.MAX_VALUE;

        while (true) {

            if (possibleSolutions.isEmpty()) {
                return minNumberOfMoves;

            // Get the most promising node.
            SearchNode s = possibleSolutions.poll();

            * If there are no balls out of place, check to see if this
            * is the best solution found so far.
            if (s.getNumOutOfPlace() == 0) {
                minNumberOfMoves = Math.min(minNumberOfMoves, s.moves);

            } else {

                * Make sure the move that we're about to make keeps us below
                * the number of moves in the best solution.  Otherwise, there's
                * no point in trying it.
                if ((s.moves + 1) < minNumberOfMoves) {
                    for (int i = 0; i < legalRacks.length; i++) {

                        * Add to possible solutions swappings that will
                        * get us closer to each of the legal racks.




The first thing I did was to create an enumerated type to represent the three types of billiard balls: Solid, Stripe, and 8-ball. It seemed easier to work with just these three and avoid having to deal with the individual values of the balls. With that, I created an array of legal rack configurations. Note that the legal racks are inverses of each other. Looking back, I probably could have made better use of that fact.

Next, I define an inner class called SearchNode. This holds a rack configuration, the number of moves required to reach this state, and some code to count the number of balls that are out of place, and to swap balls - leading to a new configuration. The method getPriority() gives us an idea of how good the current state is. The fewer balls out of place, and the less moves needed to reach this state, the better.

The swapBalls() method first looks to see if the 8 ball is in position 4. If it's not, then we'll swap the ball in position 4 with the 8 ball. Otherwise, we look for the first ball that is out of position (i.e. a Stripe where a Solid belongs). Then we look for an opposing ball to swap with that.

Finally, the SearchNode class provides a compareTo() method. The algorithm uses a priority queue to examine the SearchNode with the lowest priority. The compareTo() method allows SearchNodes to be sorted.

The heart of the algorithm is in the while loop of the rackMoves() method. It's based on the A* algorithm. We start by placing the initial state into the priority queue. Then we remove the lowest priority state from the queue and add two new states representing steps toward one of the two legal end states.

Once we've found a state that is legal, we update minNumberOfMoves. Any state that cannot reach the end state in less moves will not be added to the priority queue, and that line of progression will terminate.

Once the queue is empty, we'll have found our minimum possible number of moves.

This solution is probably overkill given the problem. With only 15 balls, and noting that each swap (not including the 8 ball) brings two more balls into alignment, it shouldn't be hard to write a brute-force approach. I like this solution becase it is more general. It can be expanded to take any starting condition and find the best sequence to get to any ending state. The same algorithm can be used to solve a rubik's cude, those sliding tile games, or a host of other problems.

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