How can I make this more efficient path finding? - java

I am working on a maze solver. It runs very fast on my first 2 mazes, however, my third maze takes forever. I am supposed to be able to do it in under a minute, on reasonable hardware.
The solve method takes an immense amount of time on my high-end gaming rig.
Here is the relevant source code
import java.awt.Point;
import java.io.IOException;
import java.nio.charset.Charset;
import java.nio.file.Files;
import java.nio.file.Paths;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
/**
* Created by jphamlett on 6/16/17.
*/
public class main {
static class fileIO {
public static String readFile(String path, Charset encoding)
throws IOException {
byte[] encoded = Files.readAllBytes(Paths.get(path));
return new String(encoded, encoding);
}
}
static class mazeNode {
private Point point;
private int dist;
public Point getPoint() {
return point;
}
public void setPoint(Point point) {
this.point = point;
}
public int getDist() {
return dist;
}
public void setDist(int dist) {
this.dist = dist;
}
public mazeNode(Point point, int dist) {
setPoint(point);
setDist(dist);
}
}
static class Solver {
private String[] pathGrid;
private int[][] gridLength;
public void setPath(String path) {
try {
this.pathGrid = generatePath(fileIO.readFile(path, Charset.defaultCharset()));
} catch (IOException e) {
e.printStackTrace();
}
}
public Point findA() {
for (int row = 0; row < pathGrid.length; row++) {
int pos = pathGrid[row].indexOf("A");
if (pos != -1) {
return new Point(row, pos);
}
}
return null; // Something went wrong
}
public Point findB() {
for (int row = 0; row < pathGrid.length; row++) {
int pos = pathGrid[row].indexOf("B");
if (pos != -1) {
return new Point(row, pos);
}
}
return null; // Something went wrong
}
public Boolean canMove(char symbol) {
return symbol != '#';
}
public String[] generatePath(String path) {
return path.split("\n");
}
public String[] getPath() {
return this.pathGrid;
}
// Use BFS to solve the maze
public int[][] solve(int[][] gridLength, Point src, Point dest) {
if (src == null || dest == null) {
return null;
}
gridLength[src.x][src.y] = 0; // Distance to self is 0
Boolean visited[][] = new Boolean[gridLength.length][gridLength[0].length]; //Set all booleans to false
for (Boolean[] booleans : visited) {
Arrays.fill(booleans, Boolean.FALSE);
}
//System.out.println("Finished making visited array");
visited[src.x][src.y] = Boolean.TRUE;
Queue<mazeNode> queue = new LinkedList<>();
mazeNode initialNode = new mazeNode(src, 0);
queue.add(initialNode);
while (!queue.isEmpty()) {
mazeNode currentNode = queue.peek();
Point currentPoint = currentNode.getPoint();
//System.out.println("Point: " + currentPoint);
visited[currentPoint.x][currentPoint.y] = Boolean.TRUE;
if (currentPoint.equals(dest)) {
return gridLength;
}
queue.poll();
// Add adjacent valid cells
try {
if (canMove(pathGrid[currentPoint.x].charAt(currentPoint.y - 1)) && !visited[currentPoint.x][currentPoint.y - 1]) {
gridLength[currentPoint.x][currentPoint.y - 1] = currentNode.getDist() + 1;
queue.add(new mazeNode(new Point(currentPoint.x, currentPoint.y - 1), currentNode.getDist() + 1));
}
} catch (IndexOutOfBoundsException e) {
}
try {
if (canMove(pathGrid[currentPoint.x].charAt(currentPoint.y + 1)) && !visited[currentPoint.x][currentPoint.y + 1]) {
gridLength[currentPoint.x][currentPoint.y + 1] = currentNode.getDist() + 1;
queue.add(new mazeNode(new Point(currentPoint.x, currentPoint.y + 1), currentNode.getDist() + 1));
}
} catch (IndexOutOfBoundsException e) {
}
try {
if (canMove(pathGrid[currentPoint.x - 1].charAt(currentPoint.y)) && !visited[currentPoint.x - 1][currentPoint.y]) {
gridLength[currentPoint.x - 1][currentPoint.y] = currentNode.getDist() + 1;
queue.add(new mazeNode(new Point(currentPoint.x - 1, currentPoint.y), currentNode.getDist() + 1));
}
} catch (IndexOutOfBoundsException e) {
}
try {
if (canMove(pathGrid[currentPoint.x + 1].charAt(currentPoint.y)) && !visited[currentPoint.x + 1][currentPoint.y]) {
gridLength[currentPoint.x + 1][currentPoint.y] = currentNode.getDist() + 1;
queue.add(new mazeNode(new Point(currentPoint.x + 1, currentPoint.y), currentNode.getDist() + 1));
}
} catch (IndexOutOfBoundsException e) {
}
}
return null; // Cannot be reached
}
public Solver(String path) {
setPath(path);
}
}
public static void main(String[] args) {
long startTime = System.currentTimeMillis();
Solver solver = new Solver("mazes/maze3.txt");
int[][] path = solver.solve(new int[solver.getPath().length][solver.getPath()[0].length()], solver.findA(), solver.findB());
long endTime = System.currentTimeMillis();
long totalTime = endTime - startTime;
System.out.println(totalTime);
for (int[] i : path) {
for (int j : i) {
System.out.print(j + " ");
}
System.out.println();
}
endTime = System.currentTimeMillis();
totalTime = endTime - startTime;
System.out.println(totalTime);
}
}
Here is maze2.txt
###############B#############################################
##.....########.#......................................#...##
##.###.#........####################################.#.#.#.##
##.###.#.#########..........#########.......########.#.#.#.##
##.#####...........########.#.......#.#####.########.#.#.#.##
##.########################.#.#####.#.#...#.########.#.#.#.##
##............................#####.#.##.##.########.#.#.#.##
##.###.############################.#.##.##.########.#.#.#.##
##.###.##...#...#...#...#...#.......#.##.##.########.#.#.#.##
##.###....#...#...#...#...#...#######.##.##.########.#.#.#.##
##.##################################.##.##.########.#.#.#.##
##.......................................##.########.#.#.#.##
###########################################.########.#.#.#.##
###...............................#########..........#.#.#.##
########################.###########################.#.#.#.##
#........................#...........................#.#.#.##
#.######################.#############################.#.#.##
#.#..........#.........................................#.#.##
#.#.########.#.#########################################.#.##
#.#........#.#.#.........................................#.##
#.##########.#.#.#########################################.##
#............#.#.##........................................##
##############.#.#############################.#####.########
#..............................................#####........#
########################A####################################
I have attached maze3 because the formatting here makes it shift oddly.
https://pastebin.com/c4LhG5hT

Your problem is the visited array.
First, a minor issue: The visited array should not be a Boolean[][]. Just make it a boolean[][], which is automatically initialized to all false values, so that initialization loop can be eliminated too.
Now, the main problem is that visited is not marked true until you actually process that point. This means that the same point is added many times to the queue.
Example maze:
#####################
#...#...#...#...#...#
A.#1..#2..#3..#4..#5B
#...#...#...#...#...#
#####################
In this case, point 1 is added twice to the queue. Each point up to point 2 will also be added twice. Point 2 will be added 4 times, point 3 8 times, point 4 16 times, and point 5 32 times.
As you can see, that is an exponential number of queue items for each round1 to process, doubling each time multiple paths meet.
Solution: Rename visited to queued, and mark point true at the same time you add it to the queue, thus preventing the addition of the same point multiple times.
Result: Code completes in less then 50 milliseconds for maze 3.
1) By "round" I mean processing of all queued points that is one step further away from start (distance).

Related

How do I fix this Java BubbleSort algorithm?

I'm doing a Java programming assignment which involves bubble sorting a .dat file BetelgeuseNames.dat with strings in it alphabetically. My AP Computer Science A teacher told me my code is correct, but it still gives the wrong output.
There are three classes called BubbleSort, BubbleSortTimer, and StopWatch. The program runs from BubbleSortTimer.
BubbleSort:
import java.util.ArrayList;
import javax.swing.JOptionPane;
import java.io.FileWriter;
import java.io.IOException;
public class BubbleSort {
// Private instance variables:
private ArrayList<String> list;
private int number;
public BubbleSort(ArrayList<String> a_list) {
list = a_list;
}
public void swap(int first, int second) {
String temp1 = list.get(first);
String temp2 = list.get(second);
list.set(first, temp2);
list.set(second, temp1);
}
public int getNumber() {
String numStr;
numStr = JOptionPane.showInputDialog("How many names do you want to sort?");
number = Integer.parseInt(numStr);
return number;
}
public void printSorted() {
try {
FileWriter writer = new FileWriter("sorted.dat");
for (int i = 0; i < number; i++) {
writer.write(list.get(i) + "\n");
}
writer.close();
} catch (IOException exception) {
System.out.println("Error processing file: " + exception);
}
}
public void bubbleSort() {
for (int i = 0; i < number; i++) {
for (int j = 0; j < number - i - 1; j++) {
if (list.get(i).compareTo(list.get(i+1)) > 0) {
swap(i, i + 1);
}
}
}
} // End method
}
BubbleSortTimer:
import java.util.ArrayList;
import java.io.BufferedReader;
import java.io.FileReader;
import javax.swing.JOptionPane;
import java.io.IOException;
public class BubbleSortTimer {
private ArrayList<String> list = new ArrayList<String>();
public void readNames() {
try {
FileReader reader = new FileReader("BetelgeuseNames.dat");
BufferedReader in = new BufferedReader(reader);
boolean done = false;
String name;
while (done == false) {
name = in.readLine();
if (name == null) {
done = true;
} else {
list.add(name);
}
}
reader.close();
} catch (IOException exception) {
System.out.println("Error processing file: " + exception);
}
} // End method
public void runSort() {
readNames();
StopWatch timer = new StopWatch();
BubbleSort sorter = new BubbleSort(list);
int number = sorter.getNumber();
timer.start();
sorter.bubbleSort();
timer.stop();
sorter.printSorted();
String msg = "Number of names sorted: " + number + "\nMilliseconds required to sort: " + timer.getElapsedTime() + "\nOutput file is \"sorted.dat\"";
JOptionPane.showMessageDialog(null, msg);
}
public static void main(String[] args) {
BubbleSortTimer bubble = new BubbleSortTimer();
bubble.runSort();
}
}
StopWatch:
/**
* A stopwatch accumulates time when it is running. You can
* repeatedly start and stop the stopwatch. You can use a
* stopwatch to measure the running time of a program.
* from section 18.2 of Horstmann's CCJ
*/
public class StopWatch {
/**
* Constructs a stopwatch that is in the stopped state
* and has no time accumulated.
*/
public StopWatch() {
reset();
}
/**
* Starts the stopwatch. Times starts accumulating now.
*/
public void start() {
if (isRunning) return;
isRunning = true;
startTime = System.currentTimeMillis();
}
/**
* Stops the stopwatch. Time stops accumulating and is
* added to the elapsed time.
*/
public void stop() {
if (!isRunning) return;
isRunning = false;
long endTime = System.currentTimeMillis();
elapsedTime = elapsedTime + endTime - startTime;
}
/**
* Returns the total elapsed time.
#return the total elapsed time
*/
public long getElapsedTime() {
if (isRunning) {
long endTime = System.currentTimeMillis();
elapsedTime = elapsedTime + endTime - startTime;
startTime = endTime;
}
return elapsedTime;
}
/**
* Stops the watch and resets the elapsed time to 0.
*/
public void reset() {
elapsedTime = 0;
isRunning = false;
}
private long elapsedTime;
private long startTime;
private boolean isRunning;
}
Input:
Moewm
Bmlzvltcso
Aqxjor
Wwgjie
Qqqtpivd
Xgyhaerv
Wqpjwdvxjq
Ecsfnow
Zlptuqxctt
Jhtprwvopk
Expected Output:
Aqxjor
Bmlzvltcso
Ecsfnow
Jhtprwvopk
Moewm
Qqqtpivd
Wqpjwdvxjq
Wwgjie
Xgyhaerv
Zlptuqxctt
Actual Output:
Bmlzvltcso
Aqxjor
Moewm
Qqqtpivd
Wwgjie
Wqpjwdvxjq
Ecsfnow
Xgyhaerv
Jhtprwvopk
Zlptuqxctt
This is how Android did (binary) sorting (edited to fix this situation):
public void binarySort() {
int lo = 0; // sort start
for (int start=lo ; start < number; start++) {
String pivot = list.get(start);
// Set left (and right) to the index where list.get(start) (pivot) belongs
int left = 0;
int right = start;
assert left <= right;
/*
* Invariants:
* pivot >= all in [lo, left].
* pivot < all in [right, start].
*/
while (left < right) {
int mid = (left + right) >>> 1;
if (pivot.compareTo(list.get(mid)) < 0)
right = mid;
else
left = mid + 1;
}
assert left == right;
/*
* The invariants still hold: pivot >= all in [lo, left] and
* pivot < all in [left, start], so pivot belongs at left. Note
* that if there are elements equal to pivot, left points to the
* first slot after them -- that's why this sort is stable.
* Slide elements over to make room for pivot.
*/
int n = start - left; // The number of elements to move
// Switch is just reshifter in default case
switch (n) {
case 2: list.set(left + 2,list.get(left + 1));
case 1: list.set(left + 1,list.get(left));
break;
default:
if(n>0){
list.add(left,list.remove(left+n));
}
}
list.set(left,pivot);
}
}
This is how you can do (bubble) sorting:
public void bubbleSort() {
for (int i = 0; i < number; i++) {
for (int j = i + 1; j < number; j++) {
if (list.get(i).compareTo(list.get(j)) > 0) {
swap(i, j);
}
}
}
}
BUBBLE SORTING V/S BINARY SORTING:
OFF TOPIC: As you can compare above, bubble sorting is easier to code/read/understand and is also faster as compared to binary sorting, because binary sorting (actually) uses array recreation many times which ofcourse takes more time compared to swap.
Because there is a problem with your bubbleSort() method. Please try this way.
public void bubbleSort() {
for (int i = 0; i < number; i++) {
for (int j = 1; j < number - i; j++) {
if (list.get(j - 1).compareTo(list.get(j)) > 0) {
swap(j - 1, j);
}
}
}
}

How can I create a Java algorithm that takes any number of players and pair them up any number of times uniquely?

I am looking to create an algorithm in Java that can take any number of "players" and group them up a specified number of times each. However, two pairs cannot be the same. So, if we are supplied 9 players (dubbed 0, 1, 2, etc) by the user and each player should be paired 3 times, that means that this algorithm needs to be able to generate a list of pairs where each player is paired 3 times.
So 4 players being paired two times could be: {{0, 1}, {2, 3}, {0, 2}, {1, 3}}.
Obviously, it can be impossible in some scenarios (like 4 players being uniquely paired 20 times), but I have input restrictions to combat that.
{0, 1} and {1, 0} are equal pairs. The order of the numbers does not matter, they are not unique.
The preferable way for input is just given two numbers (number of players, number of pairs per players) and the preferable way for the output to be given is in a two dimensional array of integers (each player being dubbed by an integer), like I gave an example of.
Does anyone have any ideas on how to do this? Pseudo-code, actual code, any ideas are welcome. Thanks!
I think your question is valid and interesting to solve in code.
This is why I coded that whole thing.
There's one downside to my solution, or rather: the problem.
In certain situations, some players can have many matches between them, while others have little. So in the end, some players might not get matched properly.
In this case, you'd need a mathematical trick, or a backtracking algorithm, that steps back on parts of the solution and tries (brute-forces) other combinations. My algorithm has neither, but it indicates Exceptions and validity.
Also check the comments in the code.
package snippet;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map.Entry;
public class BadPairingStuff6 {
static class Player {
public final int mID;
private final BadPairingStuff6 mParentLogic;
public int mMatches;
public Player(final int pID, final BadPairingStuff6 pBadPairingStuff5) {
mID = pID;
mParentLogic = pBadPairingStuff5;
}
#Override public int hashCode() {
return mID;
}
#Override public boolean equals(final Object obj) {
if (this == obj) return true;
if (obj == null) return false;
if (getClass() != obj.getClass()) return false;
final Player other = (Player) obj;
if (mID != other.mID) return false;
return true;
}
#Override public String toString() {
return "Player[" + mID + "]";
}
public void incMatches() {
++mMatches;
}
public int getMatches() {
return mMatches;
}
public boolean canPlayAnotherMatch() {
return getMatches() < mParentLogic.mPairingsAllowed;
}
}
static class Pairing {
public final Player mPlayer1;
public final Player mPlayer2;
public Pairing(final Player pPlayer1, final Player pPlayer2) {
if (pPlayer1.mID < pPlayer2.mID) {
mPlayer1 = pPlayer1;
mPlayer2 = pPlayer2;
} else {
mPlayer1 = pPlayer2;
mPlayer2 = pPlayer1;
}
}
#Override public String toString() {
return mPlayer1 + "+" + mPlayer2;
}
#Override public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + mPlayer1.mID;
result = prime * result + mPlayer2.mID;
return result;
}
#Override public boolean equals(final Object obj) {
if (this == obj) return true;
if (obj == null) return false;
if (getClass() != obj.getClass()) return false;
final Pairing other = (Pairing) obj;
if (mPlayer1 != other.mPlayer1) return false;
if (mPlayer2 != other.mPlayer2) return false;
return true;
}
}
static class PartnerMap {
private final HashMap<Player, ArrayList<Player>> mMap = new HashMap<>();
public PartnerMap(final Iterable<Player> pPlayers) {
for (final Player player : pPlayers) {
final ArrayList<Player> partners = new ArrayList<>();
for (final Player partner : pPlayers) {
if (player != partner) partners.add(partner);
}
mMap.put(player, partners);
}
}
public Player getPartner(final Player pPlayer) {
final ArrayList<Player> possiblePartners = mMap.get(pPlayer);
if (possiblePartners.size() < 1) throw new NotEnoughPartnersException(pPlayer);
return possiblePartners.get((int) (Math.random() * possiblePartners.size()));
}
public void removePartners(final Player pPlayer, final Player pPartner) {
System.out.println("\t\tBadPairingStuff5.PartnerMap.removePartners(" + pPlayer + ", " + pPartner + ")");
System.out.println("\t\t\tRemoving for " + pPlayer);
System.out.println("\t\t\t\tBEFORE: " + toString(mMap.get(pPlayer)));
mMap.get(pPlayer).remove(pPartner);
System.out.println("\t\t\t\tAFTER: " + toString(mMap.get(pPlayer)));
System.out.println("\t\t\tRemoving for " + pPartner);
System.out.println("\t\t\t\tBEFORE: " + toString(mMap.get(pPartner)));
mMap.get(pPartner).remove(pPlayer);
System.out.println("\t\t\t\tAFTER: " + toString(mMap.get(pPartner)));
}
static String toString(final Iterable<Player> pPlayers) {
final StringBuilder sb = new StringBuilder();
sb.append("[");
for (final Player player : pPlayers) {
sb.append(player.mID + " ");
}
sb.append("]");
return sb.toString();
}
public void removePlayerCompletely(final Player pPlayer) {
System.out.println("\t\t\tBadPairingStuff5.PartnerMap.removePlayerCompletely(" + pPlayer + ")");
for (final ArrayList<Player> partnerMap : mMap.values()) {
partnerMap.remove(pPlayer);
}
mMap.get(pPlayer).clear();
}
public void print() {
System.out.println("Partner Map");
for (final Entry<Player, ArrayList<Player>> e : mMap.entrySet()) {
System.out.println("\t" + e.getKey());
for (final Player v : e.getValue()) {
System.out.println("\t\t" + v);
}
}
}
}
public static class NotEnoughPartnersException extends IllegalStateException {
private static final long serialVersionUID = -7249807214069096317L;
private final Player mPlayer;
public NotEnoughPartnersException(final Player pPlayer) {
super("Not enough partners available for " + pPlayer + "!");
mPlayer = pPlayer;
}
public Player getPlayer() {
return mPlayer;
}
}
static class PairingResult {
public final ArrayList<Pairing> mCreatedPairings;
public final ArrayList<Exception> mExceptions;
public PairingResult(final ArrayList<Pairing> pCreatedPairings, final ArrayList<Exception> pExceptions) {
mCreatedPairings = pCreatedPairings;
mExceptions = pExceptions;
}
public boolean isValid() {
return mExceptions.size() < 1;
}
}
public static void main(final String[] args) {
final int players = 10;
final int pairingsAllowed = 4;
final PairingResult result = new BadPairingStuff6(players, pairingsAllowed).createPairings();
System.out.println("All pairings:");
final HashMap<Long, Long> playCounter = new HashMap<>();
for (final Pairing p : result.mCreatedPairings) {
System.out.println("\t" + p);
{
final Long oldCount = playCounter.get(Long.valueOf(p.mPlayer1.mID));
playCounter.put(Long.valueOf(p.mPlayer1.mID), Long.valueOf(oldCount == null ? 1 : (oldCount.longValue() + 1)));
}
{
final Long oldCount = playCounter.get(Long.valueOf(p.mPlayer2.mID));
playCounter.put(Long.valueOf(p.mPlayer2.mID), Long.valueOf(oldCount == null ? 1 : (oldCount.longValue() + 1)));
}
}
System.out.println("Pairings per Player: ");
for (final Entry<Long, Long> e : playCounter.entrySet()) {
System.out.println("\t" + e.getKey() + " -> " + e.getValue());
}
System.out.println("Exceptions:");
System.out.flush();
sleep();
for (final Exception e : result.mExceptions) {
e.printStackTrace();
}
System.err.flush();
sleep();
System.out.println("Valid result: " + result.isValid());
System.out.println("All done.");
}
/*
* OBJECT
*/
final int mPairingsAllowed;
private final ArrayList<Player> mPlayers = new ArrayList<>();
public BadPairingStuff6(final int pPlayersCount, final int pPairingsAllowed) {
mPairingsAllowed = pPairingsAllowed;
// create players
for (int i = 0; i < pPlayersCount; i++) {
mPlayers.add(new Player(i, this));
}
}
public PairingResult createPairings() {
final ArrayList<Pairing> createdPairings = new ArrayList<>();
final ArrayList<Exception> exceptions = new ArrayList<>();
final PartnerMap possiblePairings = new PartnerMap(mPlayers);
final HashSet<Player> playersToHandle = new HashSet<>(mPlayers);
while (!playersToHandle.isEmpty()) {
final ArrayList<Player> removePlayersPerRound = new ArrayList<>();
for (final Player player : playersToHandle) {
if (!player.canPlayAnotherMatch()) {
possiblePairings.removePlayerCompletely(player);
removePlayersPerRound.add(player);
continue;
}
try {
System.out.println("Creating matches for " + player + " (" + player.getMatches() + ")");
final Player partner = possiblePairings.getPartner(player);
if (!partner.canPlayAnotherMatch()) continue;
final Pairing newPairing = new Pairing(player, partner);
if (createdPairings.contains(newPairing)) System.out.println("WARNING! Double hit for " + newPairing);
createdPairings.add(newPairing);
possiblePairings.removePartners(player, partner);
player.incMatches();
partner.incMatches();
System.out.println("\tMatched with " + partner);
if (!partner.canPlayAnotherMatch()) {
possiblePairings.removePlayerCompletely(partner);
removePlayersPerRound.add(partner);
}
} catch (final NotEnoughPartnersException e) {
// the flushes and sleeps are only a cheap shot to keep System.out and System.err outputs in somewhat chronological order.
// this is for proof/debug/answer only, and should NOT be used in production!
System.out.flush();
sleep();
e.printStackTrace();
// throw e; // if you want to abort early
removePlayersPerRound.add(e.getPlayer());
exceptions.add(e);
System.err.flush();
sleep();
}
}
playersToHandle.removeAll(removePlayersPerRound);
}
possiblePairings.print();
return new PairingResult(createdPairings, exceptions);
}
// the sleeps are only a cheap shot to keep System.out and System.err outputs in somewhat chronological order.
// this is for proof/answer only, and should NOT be used in production
static void sleep(final long pMilliSec) {
try {
Thread.sleep(pMilliSec);
} catch (final InterruptedException e1) { /* */ }
}
static void sleep() {
sleep(100);
}
}
I use lots of inner static classes. This is for demonstration purposes only.
If you want to actually use those classes, put each of them into its separate file (remove the "static class", and add a "public class" where it's missing).
Also note that this complexity is needed for random assignments. If the algorithm could always churn out the same combinations, it would be about 1/10th of the code.

Towers of Hanoi with Stacks Java

I am working on a Towers of Hanoi problem in Java. I chose to use Stacks as the pegs and have everything working except for the move method. I have the specification and a JUnit test class and am currently passing 6 of the 7 tests but am failing on the move test. The specs are as follows:
Here is my Towers class:
package edu.metrostate.ics240.p2.towers;
import java.util.Stack;
public class Towers {
private static final int DEFAULT_SIZE = 5;
private static final int MAX_SIZE = 64;
private static final int MIN_PEG = 1;
private static final int MAX_PEG = 3;
private static Stack<Integer>[] tower = new Stack[4];
private int numOfRings;
public Towers(int n) {
if (n < 1 || n > MAX_SIZE)
throw new IllegalArgumentException(
String.format("Number of rings (%s) cannot be less than 1 or exceed 64 ", n));
numOfRings = n;
tower[1] = new Stack<Integer>();
tower[2] = new Stack<Integer>();
tower[3] = new Stack<Integer>();
for (int i = 1; i <= numOfRings; i++)
tower[1].push(i);
}
public Towers() {
numOfRings = DEFAULT_SIZE;
tower[1] = new Stack<Integer>();
tower[2] = new Stack<Integer>();
tower[3] = new Stack<Integer>();
for (int i = 1; i <= numOfRings; i++)
tower[1].push(i);
}
private static void pegCheck(int pegNumber){
if (pegNumber < MIN_PEG || pegNumber > MAX_PEG)
throw new IllegalArgumentException(
String.format("Peg number (%s) cannot be less than 1 or exceed 3 ", pegNumber));
}
public int getRingCount(int pegNumber) {
pegCheck(pegNumber);
switch (pegNumber) {
case 1:
if (tower[1].isEmpty())
return 0;
else
return tower[1].size();
case 2:
if (tower[2].isEmpty())
return 0;
else
return tower[2].size();
case 3:
if (tower[3].isEmpty())
return 0;
else
return tower[3].size();
default:
return 0;
}
}
public int getTopDiameter(int pegNumber) {
pegCheck(pegNumber);
switch (pegNumber) {
case 1:
if(getRingCount(1) > 0){
return tower[1].get(tower[1].peek() - tower[1].size());
}else
return 0;
case 2:
if(getRingCount(2) > 0){
return tower[2].get(tower[2].peek() - tower[2].size());
}else
return 0;
case 3:
if(getRingCount(3) > 0){
return tower[3].get(tower[3].peek() - tower[3].size());
}else
return 0;
default:
return 0;
}
}
public boolean move(int startPeg, int endPeg) {
pegCheck(startPeg);
pegCheck(endPeg);
Stack<Integer> startTower = tower[startPeg];
Stack<Integer> endTower = tower[endPeg];
if (getRingCount(startPeg) > 0 && endPeg != startPeg && getRingCount(endPeg) > 0 && getTopDiameter(startPeg) < getTopDiameter(endPeg)) {
int topRing = startTower.pop();
endTower.push(topRing);
return true;
}else
return false;
}
}
and finally the JUnit test(s):
import static org.junit.Assert.*;
import org.junit.Test;
import edu.metrostate.ics240.p2.towers.*;
import java.util.Random;
public class TowersTest {
private static final int MAX_NUM_RINGS = 64;
private static final long SEED = 20170604001L;
private static final Random RAND = new Random(SEED);
#Test
public void testDefaultConstruction() {
Towers t = new Towers();
assertEquals(5, t.getRingCount(1));
assertEquals(0, t.getRingCount(2));
assertEquals(0, t.getRingCount(3));
assertEquals(1, t.getTopDiameter(1));
assertEquals(0, t.getTopDiameter(2));
assertEquals(0, t.getTopDiameter(3));
}
#Test
public void testConstruction() {
int numRings = RAND.nextInt(MAX_NUM_RINGS);
Towers t = new Towers(numRings);
assertEquals(numRings, t.getRingCount(1));
assertEquals(0, t.getRingCount(2));
assertEquals(0, t.getRingCount(3));
assertEquals(1, t.getTopDiameter(1));
assertEquals(0, t.getTopDiameter(2));
assertEquals(0, t.getTopDiameter(3));
}
#Test
public void testMove() {
int numRings = RAND.nextInt(64);
Towers t = new Towers(numRings);
assertTrue(t.move(1, 2));
assertEquals(numRings - 1, t.getRingCount(1));
assertEquals(1, t.getRingCount(2));
assertEquals(0, t.getRingCount(3));
assertEquals(2, t.getTopDiameter(1));
assertEquals(1, t.getTopDiameter(2));
assertEquals(0, t.getTopDiameter(3));
assertTrue(t.move(1, 3));
assertEquals(numRings - 2, t.getRingCount(1));
assertEquals(1, t.getRingCount(2));
assertEquals(1, t.getRingCount(3));
assertEquals(3, t.getTopDiameter(1));
assertEquals(1, t.getTopDiameter(2));
assertEquals(2, t.getTopDiameter(3));
}
#Test
public void testInvalidConstructor(){
Towers t = null;
try {
t = new Towers(0); // illegal value
fail("Expected exception");
} catch (IllegalArgumentException iae) {
// expected
}
try {
t = new Towers(MAX_NUM_RINGS + 1); // illegal value
fail("Expected exception");
} catch (IllegalArgumentException iae) {
// expected
}
}
#Test
public void testPreconditionGetRingCount() {
Towers t = new Towers();
try {
t.getRingCount(0);
fail("Exception expected");
} catch (IllegalArgumentException iae) {
// expected
}
try {
t.getRingCount(4);
fail("Exception expected");
} catch (IllegalArgumentException iae) {
// expected
}
}
#Test
public void testPreconditionTopRing() {
Towers t = new Towers();
try {
t.getTopDiameter(0);
fail("Exception expected");
} catch (IllegalArgumentException iae) {
// expected
}
try {
t.getTopDiameter(4);
fail("Exception expected");
} catch (IllegalArgumentException iae) {
// expected
}
}
#Test
public void testIllegalMoves(){
Towers t = new Towers();
t.move(1, 2);
t.move(1, 3);
assertFalse(t.move(1, 1)); // can't move to itself
assertFalse(t.move(1, 2)); // moving larger ring to smaller
assertFalse(t.move(1, 3)); // moving larger ring to smaller
assertFalse(t.move(3, 2));
}
}
I think I know where my issue lies. The precondition of getTopDiameter() returns the top ring size if getRingCount(pegNum) > 0 but returns 0 if the stack is empty or there are no rings on the peg. Since tower[1] is the only peg that gets initialized with rings and the other two do not, getTopDiameter() returns 0 since there are no rings currently on tower[2] and tower[3]. In the move() method one of the preconditions requires that getTopdiameter(startPeg) be less than getTopDiamater(endPeg) but if the endPeg was initialized with 0 rings and is therefore empty, getTopDiamater(endPeg) returns 0 which is obviously not less than 1 in this case. I just cannot figure this bit out. Any help is greatly appreciated, thank you in advance!
UPDATE
revised code that is passing all test cases:
package edu.metrostate.ics240.p2.towers;
import java.util.Stack;
public class Towers {
private static final int DEFAULT_SIZE = 5;
private static final int MAX_SIZE = 64;
private static final int MIN_PEG = 1;
private static final int MAX_PEG = 3;
#SuppressWarnings("unchecked")
private static Stack<Integer>[] tower = new Stack[4];
private int numOfRings;
public Towers(int n) {
if (n < 1 || n > MAX_SIZE)
throw new IllegalArgumentException(
String.format("Number of rings (%s) cannot be less than 1 or exceed 64 ", n));
numOfRings = n;
tower[1] = new Stack<Integer>();
tower[2] = new Stack<Integer>();
tower[3] = new Stack<Integer>();
for (int i = numOfRings; i >= 1; i--)
tower[1].push(i);
}
public Towers() {
numOfRings = DEFAULT_SIZE;
tower[1] = new Stack<Integer>();
tower[2] = new Stack<Integer>();
tower[3] = new Stack<Integer>();
for (int i = numOfRings; i >= 1; i--)
tower[1].push(i);
}
private static void pegCheck(int pegNumber) {
if (pegNumber < MIN_PEG || pegNumber > MAX_PEG)
throw new IllegalArgumentException(
String.format("Peg number (%s) cannot be less than 1 or exceed 3 ", pegNumber));
}
public int getRingCount(int pegNumber) {
pegCheck(pegNumber);
if (tower[pegNumber].isEmpty()) {
return 0;
} else
return tower[pegNumber].size();
}
public int getTopDiameter(int pegNumber) {
pegCheck(pegNumber);
if (getRingCount(pegNumber) > 0) {
return tower[pegNumber].get(tower[pegNumber].size() - 1);
}
return 0;
}
public boolean move(int startPeg, int endPeg) {
pegCheck(startPeg);
pegCheck(endPeg);
if (endPeg != startPeg) {
if (getRingCount(startPeg) > 0) {
if (getRingCount(endPeg) == 0 || getTopDiameter(startPeg) < getTopDiameter(endPeg)) {
int topRing = tower[startPeg].pop();
tower[endPeg].push(topRing);
return true;
}
}
}
return false;
}
}
You say:
In the move() method one of the preconditions requires that getTopdiameter(startPeg) be less than getTopDiamater(endPeg) but if the endPeg was initialized with 0 rings and is therefore empty, getTopDiamater(endPeg) returns 0 which is obviously not less than 1 in this case
But if you read the preconditions in the image you provide - it is stating that getTopdiameter(startPeg) be less than getTopDiamater(endPeg) if endPeg has at least one ring so to write this as conditions you need
getRingCount(endPeg) > 0 && getTopdiameter(startPeg) < getTopDiamater(endPeg))
-- Edit --
You need to separate the conditions into different if statements (or have and or condition also) to handle the case when the towers have no pegs - currently with your conditions as is, it fails on the first move as the condition getRingCount(endPeg) > 0 will be false. If getRingCount == 0 then you can just do the move without needing to check if the diameters are compatible. For readability I would suggest you separate your conditions out initially - you can always combine them as required later - something like this pseudo code
if not same peg
if start peg has rings
if end peg is empty or (end peg has rings and diameters are compatible)
do move and return true
return false

Numbering A Crossword Java ACM Graphics

The problem asks for an acm graphics program that reads a txt file like this:
R
FUN
SALES
RECEIPT
MERE#FARM
DOVE###RAIL
MORE#####DRAW
HARD###TIED
LION#SAND
EVENING
EVADE
ARE
D
and makes a crossword puzzle, with blank squares on letters, black squares on '#', and nothing on empty spaces. The problem also asks that "if the square is at the beginning of a word running across, down, or both, the square should contain a number that is assigned sequentially through the puzzle."
I have the square drawing working, but I'm stuck on drawing the numbers correctly. There is something wrong with how I'm detecting null space and black squares. Can someone tell me what I'm doing wrong, please?
Here is the code:
import acm.program.*;
import java.io.*;
import java.util.*;
import acm.graphics.*;
import java.awt.*;
public class Crossword extends GraphicsProgram {
public void run() {
String fileName = "crosswordfile.txt";
makeCrosswordPuzzle(fileName);
}
private static final int sqCon = 15; // constant for square x and y dimensions
private int y = 0;
public void makeCrosswordPuzzle(String fileName) {
BufferedReader rd;
int y = 0; // y value for the square being added during that loop. increments by sqCon after every line
int wordNumber = 1; // variable for numbers added to certain boxes. increments every time the program adds a number
try {
rd = new BufferedReader(new FileReader(fileName));
String line = rd.readLine(); //reads one line of the text document at a time and makes it a string
while (line != null) {
int x = 0;
for (int i = 0; i < line.length(); i++) {
char lineChar = line.charAt(i);// the character being examined for each loop
GRect whiteSq = new GRect(sqCon,sqCon); //GRect for blank squares
GRect blackSq = new GRect(sqCon,sqCon);//GRect for black squares
blackSq.setFilled(true);
blackSq.setFillColor(Color.BLACK);
if (lineChar == '#'){
add (blackSq,x,y);
}
if (Character.isLetter(lineChar)) {
add (whiteSq, x, y);
// if the element above or to the left of the current focus is null or blackSq, place the number and then increment wordNumber
GObject above = getElementAt(x+sqCon/2,y-sqCon/2);
GObject left = getElementAt(x-sqCon/2, y+sqCon/2);
GLabel wordNumberLabel = new GLabel(Integer.toString(wordNumber));
if (above == null || left == null || above == blackSq || left == blackSq) {
add(wordNumberLabel,x,y+sqCon);
wordNumber++;
}
}
x += sqCon;
}
line = rd.readLine();
y += sqCon;
}
rd.close();
}
catch (IOException e) {
throw new ErrorException(e);
}
}
}
Edited to add:
I copied your code over to my Eclipse and ran it. Here's the result.
You did fine on the upper half, but you missed the down numbers on the lower half.
Here's the same code, reformatted so it's easier to read.
import java.awt.Color;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import acm.graphics.GLabel;
import acm.graphics.GObject;
import acm.graphics.GRect;
import acm.program.GraphicsProgram;
import acm.util.ErrorException;
public class Crossword extends GraphicsProgram {
private static final long serialVersionUID = -7971434624427958742L;
public void run() {
// String fileName = "crosswordfile.txt";
String fileName = "C:/Eclipse/eclipse-4.2-work/com.ggl.testing/crosswordfile.txt";
makeCrosswordPuzzle(fileName);
}
private static final int sqCon = 15; // constant for square x and y
// dimensions
private int y = 0;
public void makeCrosswordPuzzle(String fileName) {
BufferedReader rd;
int y = 0; // y value for the square being added during that loop.
// increments by sqCon after every line
int wordNumber = 1; // variable for numbers added to certain boxes.
// increments every time the program adds a number
try {
rd = new BufferedReader(new FileReader(fileName));
String line = rd.readLine(); // reads one line of the text document
// at a time and makes it a string
while (line != null) {
int x = 0;
for (int i = 0; i < line.length(); i++) {
char lineChar = line.charAt(i);// the character being
// examined for each loop
GRect whiteSq = new GRect(sqCon, sqCon); // GRect for blank
// squares
GRect blackSq = new GRect(sqCon, sqCon);// GRect for black
// squares
blackSq.setFilled(true);
blackSq.setFillColor(Color.BLACK);
if (lineChar == '#') {
add(blackSq, x, y);
}
if (Character.isLetter(lineChar)) {
add(whiteSq, x, y);
// if the element above or to the left of the current
// focus is null or blackSq, place the number and then
// increment wordNumber
GObject above = getElementAt(x + sqCon / 2, y - sqCon
/ 2);
GObject left = getElementAt(x - sqCon / 2, y + sqCon
/ 2);
GLabel wordNumberLabel = new GLabel(
Integer.toString(wordNumber));
if (above == null || left == null || above == blackSq
|| left == blackSq) {
add(wordNumberLabel, x, y + sqCon);
wordNumber++;
}
}
x += sqCon;
}
line = rd.readLine();
y += sqCon;
}
rd.close();
} catch (IOException e) {
throw new ErrorException(e);
}
}
}
I followed the advice of my own comment. I created the crossword puzzle answer, numbered the crossword puzzle answer, and finally drew the crossword puzzle answer.
Here's the applet result:
I kept a List of crossword puzzle cells. That way, I could determine the length and the width of the puzzle by the number of characters on a row and the number of rows of the input text file. I didn't have to hard code the dimensions.
For each crossword cell, I kept track of whether or not it was a letter, and whether or not it was a dark space.
When determining where to put the numbers, I followed 2 rules.
An across number is placed where the cell left of the cell is empty or dark, and there are three or more letters across.
A down number is placed where the cell above the cell is empty or dark, there are three or more letters down, and there is no across number.
You can see in the code that I had to do some debug printing to get the crossword puzzle clue numbering correct. I broke the process into many methods to keep each method as simple as possible.
Finally, I drew the crossword puzzle answer from the information in the List.
Here's the code:
import java.awt.Color;
import java.awt.Point;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import acm.graphics.GLabel;
import acm.graphics.GRect;
import acm.program.GraphicsProgram;
import acm.util.ErrorException;
public class Crossword extends GraphicsProgram {
private static final boolean DEBUG = false;
private static final long serialVersionUID = -7971434624427958742L;
private List<CrosswordCell> crosswordCellList;
#Override
public void run() {
this.crosswordCellList = new ArrayList<CrosswordCell>();
// String fileName = "crosswordfile.txt";
String fileName = "C:/Eclipse/eclipse-4.2-work/" +
"com.ggl.testing/crosswordfile.txt";
try {
readCrosswordAnswer(fileName);
if (DEBUG) printCrosswordAnswer();
numberCrosswordCells();
if (DEBUG) printCrosswordAnswer();
drawCrosswordAnswer();
} catch (FileNotFoundException e) {
throw new ErrorException(e);
} catch (IOException e) {
throw new ErrorException(e);
}
}
private void readCrosswordAnswer(String fileName)
throws FileNotFoundException, IOException {
BufferedReader reader =
new BufferedReader(new FileReader(fileName));
String line = "";
int row = 0;
while ((line = reader.readLine()) != null) {
for (int column = 0; column < line.length(); column++) {
CrosswordCell cell = new CrosswordCell(column, row);
char lineChar = line.charAt(column);
if (lineChar == '#') {
cell.setDarkCell(true);
} else if (Character.isLetter(lineChar)) {
cell.setLetter(true);
}
crosswordCellList.add(cell);
}
row++;
}
reader.close();
}
public void printCrosswordAnswer() {
for (CrosswordCell cell : crosswordCellList) {
System.out.println(cell);
}
}
private void numberCrosswordCells() {
int clueNumber = 1;
for (CrosswordCell cell : crosswordCellList) {
if (cell.isLetter()) {
clueNumber = testCell(cell, clueNumber);
}
}
}
private int testCell(CrosswordCell cell, int clueNumber) {
Point p = cell.getLocation();
CrosswordCell leftCell = getLeftCell(p.x, p.y);
List<CrosswordCell> acrossList = getRightCells(p.x, p.y);
if (DEBUG) {
System.out.print(p);
System.out.println(", " + leftCell + " " +
acrossList.size());
}
if ((leftCell == null) && (acrossList.size() >= 3)) {
cell.setClueNumber(clueNumber++);
} else {
CrosswordCell aboveCell = getAboveCell(p.x, p.y);
List<CrosswordCell> downList = getBelowCells(p.x, p.y);
if (DEBUG) {
System.out.print(p);
System.out.println(", " + aboveCell + " " +
downList.size());
}
if ((aboveCell == null) && (downList.size() >= 3)) {
cell.setClueNumber(clueNumber++);
}
}
return clueNumber;
}
private CrosswordCell getAboveCell(int x, int y) {
int yy = y - 1;
return getCell(x, yy);
}
private CrosswordCell getLeftCell(int x, int y) {
int xx = x - 1;
return getCell(xx, y);
}
private List<CrosswordCell> getBelowCells(int x, int y) {
List<CrosswordCell> list = new ArrayList<CrosswordCell>();
for (int i = y; i < (y + 3); i++) {
CrosswordCell cell = getCell(x, i);
if (cell != null) {
list.add(cell);
}
}
return list;
}
private List<CrosswordCell> getRightCells(int x, int y) {
List<CrosswordCell> list = new ArrayList<CrosswordCell>();
for (int i = x; i < (x + 3); i++) {
CrosswordCell cell = getCell(i, y);
if (cell != null) {
list.add(cell);
}
}
return list;
}
private CrosswordCell getCell(int x, int y) {
for (CrosswordCell cell : crosswordCellList) {
Point p = cell.getLocation();
if ((p.x == x) && (p.y == y)) {
if (cell.isDarkCell()) {
return null;
} else if (cell.isLetter()){
return cell;
} else {
return null;
}
}
}
return null;
}
private void drawCrosswordAnswer() {
int sqCon = 32;
for (CrosswordCell cell : crosswordCellList) {
Point p = cell.getLocation();
if (cell.isDarkCell()) {
drawDarkCell(p, sqCon);
} else if (cell.isLetter()) {
drawLetterCell(cell, p, sqCon);
}
}
}
private void drawDarkCell(Point p, int sqCon) {
GRect blackSq = new GRect(sqCon, sqCon);
blackSq.setFilled(true);
blackSq.setFillColor(Color.BLACK);
add(blackSq, p.x * sqCon, p.y * sqCon);
}
private void drawLetterCell(CrosswordCell cell, Point p, int sqCon) {
GRect whiteSq = new GRect(sqCon, sqCon);
add(whiteSq, p.x * sqCon, p.y * sqCon);
if (cell.getClueNumber() > 0) {
String label = Integer.toString(cell.getClueNumber());
GLabel wordNumberLabel = new GLabel(label);
add(wordNumberLabel, p.x * sqCon + 2, p.y * sqCon + 14);
}
}
class CrosswordCell {
private boolean darkCell;
private boolean isLetter;
private int clueNumber;
private Point location;
public CrosswordCell(int x, int y) {
this.location = new Point(x, y);
this.clueNumber = 0;
this.darkCell = false;
this.isLetter = false;
}
public boolean isDarkCell() {
return darkCell;
}
public void setDarkCell(boolean darkCell) {
this.darkCell = darkCell;
}
public boolean isLetter() {
return isLetter;
}
public void setLetter(boolean isLetter) {
this.isLetter = isLetter;
}
public int getClueNumber() {
return clueNumber;
}
public void setClueNumber(int clueNumber) {
this.clueNumber = clueNumber;
}
public Point getLocation() {
return location;
}
#Override
public String toString() {
StringBuilder builder = new StringBuilder();
builder.append("CrosswordCell [location=");
builder.append(location);
builder.append(", clueNumber=");
builder.append(clueNumber);
builder.append(", darkCell=");
builder.append(darkCell);
builder.append(", isLetter=");
builder.append(isLetter);
builder.append("]");
return builder.toString();
}
}
}

I am trying to solve '15 puzzle', but I get 'OutOfMemoryError' [closed]

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 12 years ago.
Is there a way that I can optimize this code as to not run out of memory?
import java.util.HashMap;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.Stack;
public class TilePuzzle {
private final static byte ROWS = 4;
private final static byte COLUMNS = 4;
private static String SOLUTION = "123456789ABCDEF0";
private static byte RADIX = 16;
private char[][] board = new char[ROWS][COLUMNS];
private byte x; // Row of the space ('0')
private byte y; // Column of the space ('0') private String representation;
private boolean change = false; // Has the board changed after the last call to toString?
private TilePuzzle() {
this(SOLUTION);
int times = 1000;
Random rnd = new Random();
while(times-- > 0) {
try {
move((byte)rnd.nextInt(4));
}
catch(RuntimeException e) {
}
}
this.representation = asString();
}
public TilePuzzle(String representation) {
this.representation = representation;
final byte SIZE = (byte)SOLUTION.length();
if (representation.length() != SIZE) {
throw new IllegalArgumentException("The board must have " + SIZE + "numbers.");
}
boolean[] used = new boolean[SIZE];
byte idx = 0;
for (byte i = 0; i < ROWS; ++i) {
for (byte j = 0; j < COLUMNS; ++j) {
char digit = representation.charAt(idx++);
byte number = (byte)Character.digit(digit, RADIX);
if (number < 0 || number >= SIZE) {
throw new IllegalArgumentException("The character " + digit + " is not valid.");
} else if(used[number]) {
throw new IllegalArgumentException("The character " + digit + " is repeated.");
}
used[number] = true;
board[i][j] = digit;
if (digit == '0') {
x = i;
y = j;
}
}
}
}
/**
* Swap position of the space ('0') with the number that's up to it.
*/
public void moveUp() {
try {
move((byte)(x - 1), y);
} catch(IllegalArgumentException e) {
throw new RuntimeException("Move prohibited " + e.getMessage());
}
}
/**
* Swap position of the space ('0') with the number that's down to it.
*/
public void moveDown() {
try {
move((byte)(x + 1), y);
} catch(IllegalArgumentException e) {
throw new RuntimeException("Move prohibited " + e.getMessage());
}
}
/**
* Swap position of the space ('0') with the number that's left to it.
*/
public void moveLeft() {
try {
move(x, (byte)(y - 1));
} catch(IllegalArgumentException e) {
throw new RuntimeException("Move prohibited " + e.getMessage());
}
}
/**
* Swap position of the space ('0') with the number that's right to it.
*/
public void moveRight() {
try {
move(x, (byte)(y + 1));
} catch(IllegalArgumentException e) {
throw new RuntimeException("Move prohibited " + e.getMessage());
}
}
private void move(byte movement) {
switch(movement) {
case 0: moveUp(); break;
case 1: moveRight(); break;
case 2: moveDown(); break;
case 3: moveLeft(); break;
}
}
private boolean areValidCoordinates(byte x, byte y) {
return (x >= 0 && x < ROWS && y >= 0 && y < COLUMNS);
}
private void move(byte nx, byte ny) {
if (!areValidCoordinates(nx, ny)) {
throw new IllegalArgumentException("(" + nx + ", " + ny + ")");
}
board[x][y] = board[nx][ny];
board[nx][ny] = '0';
x = nx;
y = ny;
change = true;
}
public String printableString() {
StringBuilder sb = new StringBuilder();
for (byte i = 0; i < ROWS; ++i) {
for (byte j = 0; j < COLUMNS; ++j) {
sb.append(board[i][j] + " ");
}
sb.append("\r\n");
}
return sb.toString();
}
private String asString() {
StringBuilder sb = new StringBuilder();
for (byte i = 0; i < ROWS; ++i) {
for (byte j = 0; j < COLUMNS; ++j) {
sb.append(board[i][j]);
}
}
return sb.toString();
}
public String toString() {
if (change) {
representation = asString();
}
return representation;
}
private static byte[] whereShouldItBe(char digit) {
byte idx = (byte)SOLUTION.indexOf(digit);
return new byte[] { (byte)(idx / ROWS), (byte)(idx % ROWS) };
}
private static byte manhattanDistance(byte x, byte y, byte x2, byte y2) {
byte dx = (byte)Math.abs(x - x2);
byte dy = (byte)Math.abs(y - y2);
return (byte)(dx + dy);
}
private byte heuristic() {
byte total = 0;
for (byte i = 0; i < ROWS; ++i) {
for (byte j = 0; j < COLUMNS; ++j) {
char digit = board[i][j];
byte[] coordenates = whereShouldItBe(digit);
byte distance = manhattanDistance(i, j, coordenates[0], coordenates[1]);
total += distance;
}
}
return total;
}
private class Node implements Comparable<Node> {
private String puzzle;
private byte moves; // Number of moves from original configuration
private byte value; // The value of the heuristic for this configuration.
public Node(String puzzle, byte moves, byte value) {
this.puzzle = puzzle;
this.moves = moves;
this.value = value;
}
#Override
public int compareTo(Node o) {
return (value + moves) - (o.value + o.moves);
}
}
private void print(Map<String, String> antecessor) {
Stack toPrint = new Stack();
toPrint.add(SOLUTION);
String before = antecessor.get(SOLUTION);
while (!before.equals("")) {
toPrint.add(before);
before = antecessor.get(before);
}
while (!toPrint.isEmpty()) {
System.out.println(new TilePuzzle(toPrint.pop()).printableString());
}
}
private byte solve() {
if(toString().equals(SOLUTION)) {
return 0;
}
PriorityQueue<Node> toProcess = new PriorityQueue();
Node initial = new Node(toString(), (byte)0, heuristic());
toProcess.add(initial);
Map<String, String> antecessor = new HashMap<String, String>();
antecessor.put(toString(), "");
while(!toProcess.isEmpty()) {
Node actual = toProcess.poll();
for (byte i = 0; i < 4; ++i) {
TilePuzzle t = new TilePuzzle(actual.puzzle);
try {
t.move(i);
} catch(RuntimeException e) {
continue;
}
if (t.toString().equals(SOLUTION)) {
antecessor.put(SOLUTION, actual.puzzle);
print(antecessor);
return (byte)(actual.moves + 1);
} else if (!antecessor.containsKey(t.toString())) {
byte v = t.heuristic();
Node neighbor = new Node(t.toString(), (byte)(actual.moves + 1), v);
toProcess.add(neighbor);
antecessor.put(t.toString(), actual.puzzle);
}
}
}
return -1;
}
public static void main(String... args) {
TilePuzzle puzzle = new TilePuzzle();
System.out.println(puzzle.solve());
}
}
The problem
The root cause is the tons of String objects you are creating and storing in the toProcess Queue and the antecessor Map. Why are you doing that?
Look at your algorithm. See if you really need to store >2 million nodes and 5 million strings in each.
The investigation
This was hard to spot because the program is complex. Actually, I didn't even try to understand all of the code. Instead, I used VisualVM – a Java profiler, sampler, and CPU/memory usage monitor.
I launched it:
And took a look at the memory usage. The first thing I noticed was the (obvious) fact that you're creating tons of objects.
This is an screenshot of the app:
As you can see, the amount of memory used is tremendous. In as few as 40 seconds, 2 GB were consumed and the entire heap was filled.
A dead end
I initially thought the problem had something to do with the Node class, because even though it implements Comparable, it doesn't implement equals. So I provided the method:
public boolean equals( Object o ) {
if( o instanceof Node ) {
Node other = ( Node ) o;
return this.value == other.value && this.moves == other.moves;
}
return false;
}
But that was not the problem.
The actual problem turned out to be the one stated at the top.
The workaround
As previously stated, the real solution is to rethink your algorithm. Whatever else can be done, in the meantime, will only delay the problem.
But workarounds can be useful. One is to reuse the strings you're generating. You're very intensively using the TilePuzzle.toString() method; this ends up creating duplicate strings quite often.
Since you're generating string permutations, you may create many 12345ABCD strings in matter of seconds. If they are the same string, there is no point in creating millions of instances with the same value.
The String.intern() method allows strings to be reused. The doc says:
Returns a canonical representation for the string object.
A pool of strings, initially empty, is maintained privately by the class String.
When the intern method is invoked, if the pool already contains a string equal to this String object as determined by the equals() method, then the string from the pool is returned. Otherwise, this String object is added to the pool and a reference to this String object is returned.
For a regular application, using String.intern() could be a bad idea because it doesn't let instances be reclaimed by the GC. But in this case, since you're holding the references in your Map and Queue anyway, it makes sense.
So making this change:
public String toString() {
if (change) {
representation = asString();
}
return representation.intern(); // <-- Use intern
}
Pretty much solves the memory problem.
This is a screenshot after the change:
Now, the heap usage doesn't reach 100 MB even after a couple of minutes.
Extra remarks
Remark #1
You're using an exception to validate if the movement is valid or not, which is okay; but when you catch them, you're just ignoring them:
try {
t.move(i);
} catch(RuntimeException e) {
continue;
}
If you're not using them anyway, you can save a lot of computation by not creating the exceptions in the first place. Otherwise you're creating millions of unused exceptions.
Make this change:
if (!areValidCoordinates(nx, ny)) {
// REMOVE THIS LINE:
// throw new IllegalArgumentException("(" + nx + ", " + ny + ")");
// ADD THIS LINE:
return;
}
And use validation instead:
// REMOVE THESE LINES:
// try {
// t.move(i);
// } catch(RuntimeException e) {
// continue;
// }
// ADD THESE LINES:
if(t.isValidMovement(i)){
t.move(i);
} else {
continue;
}
Remark #2
You're creating a new Random object for every new TilePuzzle instance. It would be better if you used just one for the whole program. After all, you are only using a single thread.
Remark #3
The workaround solved the heap memory problem, but created another one involving PermGen. I simply increased the PermGen size, like this:
java -Xmx1g -Xms1g -XX:MaxPermSize=1g TilePuzzle
Remark #4
The output was sometimes 49 and sometimes 50. The matrices were printed like:
1 2 3 4
5 6 7 8
9 A B C
D E 0 F
1 2 3 4
5 6 7 8
9 A B C
D E F 0
... 50 times

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