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How can I make this more efficient path finding?

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).

for Loop repeating itself once, even when told not to

I can't for the life of me figure out why this string comparison loop will loop again after even when finding that no values in the array = that value.
Query p = new Query(section).addSort("Login", Query.SortDirection.DESCENDING) ;
PreparedQuery qp = datastore.prepare(p);
int listSize = 0;
for(Entity amount : qp.asIterable()){
listSize++;
}
for (Entity result2 : qp.asIterable()) {
if (breakOff==true) {
break;
} else {
System.out.println("1stloop size is " + size + result2.getProperty("Login"));
MyBean temp2 = new MyBean();
temp2.setData((String) result2.getProperty("Login"));
Query q = new Query("sent").addSort("Login", Query.SortDirection.DESCENDING);
PreparedQuery pq = datastore.prepare(q);
int i = 0;
for (Entity result : pq.asIterable()) {
i++;
System.out.println("2ndloop is "+ result.getProperty("Login"));
MyBean temp = new MyBean();
temp.setData((String) result.getProperty("Login"));
if (temp.getData().equals(temp2.getData())) {
System.out.println("broken");
break;
} else if (i >= size) {
sentNum.setData(temp2.getData());
Entity logins = new Entity("sent");
logins.setProperty("Login", temp2.getData());
datastore.put(logins);
System.out.println("sizematch wrote " + temp2.getData());
breakOff = true;
} else {
System.out.println("increase" + i);
}
}
}
}
As you can see, I call a breakOff==true so the loop wouldn't go through it again, but it always does. Once exactly.

Well, look at the place you check if you need to break, it is in the start of the loop. That means that the program will break in the next enter to the loop, so it will run one more time exactly.
To fix it, just change the statement location to the end of the second loop.
here is your fixed code: (by the way it is a mess, try to organize it a bit)
Query p = new Query(section).addSort("Login", Query.SortDirection.DESCENDING) ;
PreparedQuery qp = datastore.prepare(p);
int listSize = 0;
for(Entity amount : qp.asIterable()){
listSize++;
}
for (Entity result2 : qp.asIterable()) {
System.out.println("1stloop size is " + size + result2.getProperty("Login"));
MyBean temp2 = new MyBean();
temp2.setData((String) result2.getProperty("Login"));
Query q = new Query("sent").addSort("Login", Query.SortDirection.DESCENDING);
PreparedQuery pq = datastore.prepare(q);
int i = 0;
for (Entity result : pq.asIterable()) {
i++;
System.out.println("2ndloop is "+ result.getProperty("Login"));
MyBean temp = new MyBean();
temp.setData((String) result.getProperty("Login"));
if (temp.getData().equals(temp2.getData())) {
System.out.println("broken");
break;
} else if (i >= size) {
sentNum.setData(temp2.getData());
Entity logins = new Entity("sent");
logins.setProperty("Login", temp2.getData());
datastore.put(logins);
System.out.println("sizematch wrote " + temp2.getData());
breakOff = true;
break;
} else {
System.out.println("increase" + i);
}
}
if (breakOff)
break;
}
notice that there is no reason to check if(breakOff==true) as you can also check if(breakOff) and get the same answer. The reason is that breakOff==true will give you true if and only if breakOff is true, but if(breakOff) will be true if and only if breakOff is true, so they are same.

Neo4j - Unable to rollback transaction in creating relationships with java

I am trying to create some nodes in Neo4j through a Maven Java Application and create relationships between those nodes. Exactly i want to create 16807 nodes and 17210368 relationships. I read a file and get the row variable which has the number of nodes i must create and i also have a list which has 34420736 elements (=17210368*2). I want to create a relationship from node[element 0 of list] to node[element 1 from list], from node[element 2 of list] to node[element 3 from list] etc. Also the max element of list is 16807. I create an ArrayList<Node> as to create the nodes dynamically cause i want the programm to run with different files(and with different row values).
Here is my code:
GraphDatabaseFactory dbFactory = new GraphDatabaseFactory();
GraphDatabaseService graphDb = dbFactory.newEmbeddedDatabase("C:\Users\....\default.graphdb");
Transaction tx = graphDb.beginTx();
try {
final RelationshipType type2 = DynamicRelationshipType.withName("KNOW");
ArrayList<Node> nodelist = new ArrayList<Node>();
for (int k = 0; k < row; k++) { //row=16807
nodelist.add(graphDb.createNode());
nodelist.get(k).setProperty("Name", "ListNode " + k);
}
int count=0;
for (int j = 0; j < list.size() ; j++) { //list.size()=34420736
nodelist.get(list.get(count)).createRelationshipTo(nodelist.get(list.get(count+1)), type2);
count=count+2;
}
tx.success();
}
finally {
tx.close();
}
graphDb.shutdown();
If i run the code without trying to create relationships it creates the nodes and run correctly. When i add the for loop that creates the realtionships it throws me the following error:
Exception in thread "main" org.neo4j.graphdb.TransactionFailureException: Unable to rollback transaction
at org.neo4j.kernel.TopLevelTransaction.close(TopLevelTransaction.java:131)
at com.mycompany.traverse_test.traverse_main.main(traverse_main.java:138)
Caused by: java.lang.IllegalStateException: No RelationshipState for added relationship!
at org.neo4j.kernel.api.txstate.RelationshipChangeVisitorAdapter$1.visit(RelationshipChangeVisitorAdapter.java:132)
at org.neo4j.kernel.api.txstate.RelationshipChangeVisitorAdapter.visitAddedRelationship(RelationshipChangeVisitorAdapter.java:83)
at org.neo4j.kernel.api.txstate.RelationshipChangeVisitorAdapter.visitAdded(RelationshipChangeVisitorAdapter.java:106)
at org.neo4j.kernel.api.txstate.RelationshipChangeVisitorAdapter.visitAdded(RelationshipChangeVisitorAdapter.java:47)
at org.neo4j.kernel.impl.util.diffsets.DiffSets.accept(DiffSets.java:76)
at org.neo4j.kernel.impl.api.state.TxState.accept(TxState.java:156)
at org.neo4j.kernel.impl.api.KernelTransactionImplementation.rollback(KernelTransactionImplementation.java:542)
at org.neo4j.kernel.impl.api.KernelTransactionImplementation.close(KernelTransactionImplementation.java:404)
at org.neo4j.kernel.TopLevelTransaction.close(TopLevelTransaction.java:112)
... 1 more
Any ideas??

Neo4j is trying to roll back your transaction due to a bug in your code. The fact that it is failing to roll back may be a bug in neo4j, but that is really not your main problem.
Looking at your code, it looks like you are iterating through your list too many times. That is, the code in the list loop is using up 2 list elements at a time, so you should only be looping list.size()/2 times.
Here is code that should fix that bug, and it also makes a few other improvements.
GraphDatabaseFactory dbFactory = new GraphDatabaseFactory();
GraphDatabaseService graphDb = dbFactory.newEmbeddedDatabase("C:\Users\....\default.graphdb");
Transaction tx = graphDb.beginTx();
try {
final RelationshipType type2 = DynamicRelationshipType.withName("KNOW");
ArrayList<Node> nodelist = new ArrayList<Node>();
for (int k = 0; k < row; k++) { //row=16807
Node node = graphDb.createNode();
node.setProperty("Name", "ListNode " + k);
nodelist.add(node);
}
for (int j = 0; j < list.size() ; j += 2) { //list.size()=34420736
nodelist.get(list.get(j)).createRelationshipTo(
nodelist.get(list.get(j+1)), type2);
}
tx.success();
} catch(Throwable e) {
e.printStackTrace();
// You may want to re-throw the exception, rather than just eating it here...
} finally {
tx.close();
}
graphDb.shutdown();
[EDITED]
However, the above code can still run out of memory, since it is trying to create so many resources (16K nodes and 17M relationships) in a single transaction.
The following example code does the work in multiple transactions (one for creating the nodes and node list, and multiple transactions for the relationships).
NUM_RELS_PER_CHUNK specifies the maximum number of relationships to be created in each transaction. The createRelEndpointList() method must be modified to fill in the list of relationship endpoint (node) indices (each index is the 0-origin position of a node in nodeList).
public class MyCode {
private static final int NODE_COUNT = 16807;
private static final int NUM_RELS_PER_CHUNK = 1000000;
public static void main(String[] args) {
doIt();
}
private static void doIt() {
GraphDatabaseFactory dbFactory = new GraphDatabaseFactory();
GraphDatabaseService graphDb = dbFactory.newEmbeddedDatabase(new File("C:\\Users\\....\\default.graphdb"));
try {
RelationshipType type = DynamicRelationshipType.withName("KNOW");
List<Node> nodeList = createNodes(graphDb, NODE_COUNT);
List<Integer> list = createRelEndpointList();
final int numRels = list.size() / 2;
final int numChunks = (numRels + NUM_RELS_PER_CHUNK - 1)/NUM_RELS_PER_CHUNK;
int startRelIndex = 0, endRelIndexPlus1;
for (int i = numChunks; --i >= 0 && startRelIndex < numRels; ) {
endRelIndexPlus1 = (i > 0) ? startRelIndex + NUM_RELS_PER_CHUNK : numRels;
createRelationships(graphDb, nodeList, list, startRelIndex, endRelIndexPlus1, type);
startRelIndex = endRelIndexPlus1;
}
} finally {
graphDb.shutdown();
}
}
private static List<Node> createNodes(GraphDatabaseService graphDb, int rowCount) {
ArrayList<Node> nodeList = new ArrayList<Node>(rowCount);
Transaction tx = graphDb.beginTx();
try {
final StringBuilder sb = new StringBuilder("ListNode ");
final int initLength = sb.length();
for (int k = 0; k < rowCount; k++) {
Node node = graphDb.createNode();
sb.setLength(initLength);
sb.append(k);
node.setProperty("Name", sb.toString());
nodeList.add(node);
}
tx.success();
System.out.println("Created nodes.");
} catch(Exception e) {
e.printStackTrace();
tx.failure();
return null;
} finally {
tx.close();
}
return nodeList;
}
private static List<Integer> createRelEndpointList() {
final List<Integer> list = new ArrayList<Integer>();
// Fill
// list
// ...
return list;
}
private static void createRelationships(GraphDatabaseService graphDb, List<Node> nodeList, List<Integer> list, int startRelIndex, int endRelIndexPlus1, RelationshipType type) {
Transaction tx = graphDb.beginTx();
try {
final int endPlus2 = endRelIndexPlus1 * 2;
for (int j = startRelIndex * 2; j < endPlus2; ) {
Node from = nodeList.get(list.get(j++));
Node to = nodeList.get(list.get(j++));
from.createRelationshipTo(to, type);
}
tx.success();
System.out.println("Created rels. Start: " + startRelIndex + ", count: " + (endRelIndexPlus1 - startRelIndex));
} catch(Exception e) {
e.printStackTrace();
tx.failure();
// You may want to re-throw the exception, rather than just eating it here...
} finally {
tx.close();
}
}
}

Thread-safe circular buffer in Java

Consider a few web server instances running in parallel. Each server holds a reference to a single shared "Status keeper", whose role is keeping the last N requests from all servers.
For example (N=3):
Server a: "Request id = ABCD" Status keeper=["ABCD"]
Server b: "Request id = XYZZ" Status keeper=["ABCD", "XYZZ"]
Server c: "Request id = 1234" Status keeper=["ABCD", "XYZZ", "1234"]
Server b: "Request id = FOO" Status keeper=["XYZZ", "1234", "FOO"]
Server a: "Request id = BAR" Status keeper=["1234", "FOO", "BAR"]
At any point in time, the "Status keeper" might be called from a monitoring application that reads these last N requests for an SLA report.
What's the best way to implement this producer-consumer scenario in Java, giving the web servers higher priority than the SLA report?
CircularFifoBuffer seems to be the appropriate data structure to hold the requests, but I'm not sure what's the optimal way to implement efficient concurrency.

Buffer fifo = BufferUtils.synchronizedBuffer(new CircularFifoBuffer());

Here's a lock-free ring buffer implementation. It implements a fixed-size buffer - there is no FIFO functionality. I would suggest you store a Collection of requests for each server instead. That way your report can do the filtering rather than getting your data structure to filter.
/**
* Container
* ---------
*
* A lock-free container that offers a close-to O(1) add/remove performance.
*
*/
public class Container<T> implements Iterable<T> {
// The capacity of the container.
final int capacity;
// The list.
AtomicReference<Node<T>> head = new AtomicReference<Node<T>>();
// TESTING {
AtomicLong totalAdded = new AtomicLong(0);
AtomicLong totalFreed = new AtomicLong(0);
AtomicLong totalSkipped = new AtomicLong(0);
private void resetStats() {
totalAdded.set(0);
totalFreed.set(0);
totalSkipped.set(0);
}
// TESTING }
// Constructor
public Container(int capacity) {
this.capacity = capacity;
// Construct the list.
Node<T> h = new Node<T>();
Node<T> it = h;
// One created, now add (capacity - 1) more
for (int i = 0; i < capacity - 1; i++) {
// Add it.
it.next = new Node<T>();
// Step on to it.
it = it.next;
}
// Make it a ring.
it.next = h;
// Install it.
head.set(h);
}
// Empty ... NOT thread safe.
public void clear() {
Node<T> it = head.get();
for (int i = 0; i < capacity; i++) {
// Trash the element
it.element = null;
// Mark it free.
it.free.set(true);
it = it.next;
}
// Clear stats.
resetStats();
}
// Add a new one.
public Node<T> add(T element) {
// Get a free node and attach the element.
totalAdded.incrementAndGet();
return getFree().attach(element);
}
// Find the next free element and mark it not free.
private Node<T> getFree() {
Node<T> freeNode = head.get();
int skipped = 0;
// Stop when we hit the end of the list
// ... or we successfully transit a node from free to not-free.
while (skipped < capacity && !freeNode.free.compareAndSet(true, false)) {
skipped += 1;
freeNode = freeNode.next;
}
// Keep count of skipped.
totalSkipped.addAndGet(skipped);
if (skipped < capacity) {
// Put the head as next.
// Doesn't matter if it fails. That would just mean someone else was doing the same.
head.set(freeNode.next);
} else {
// We hit the end! No more free nodes.
throw new IllegalStateException("Capacity exhausted.");
}
return freeNode;
}
// Mark it free.
public void remove(Node<T> it, T element) {
totalFreed.incrementAndGet();
// Remove the element first.
it.detach(element);
// Mark it as free.
if (!it.free.compareAndSet(false, true)) {
throw new IllegalStateException("Freeing a freed node.");
}
}
// The Node class. It is static so needs the <T> repeated.
public static class Node<T> {
// The element in the node.
private T element;
// Are we free?
private AtomicBoolean free = new AtomicBoolean(true);
// The next reference in whatever list I am in.
private Node<T> next;
// Construct a node of the list
private Node() {
// Start empty.
element = null;
}
// Attach the element.
public Node<T> attach(T element) {
// Sanity check.
if (this.element == null) {
this.element = element;
} else {
throw new IllegalArgumentException("There is already an element attached.");
}
// Useful for chaining.
return this;
}
// Detach the element.
public Node<T> detach(T element) {
// Sanity check.
if (this.element == element) {
this.element = null;
} else {
throw new IllegalArgumentException("Removal of wrong element.");
}
// Useful for chaining.
return this;
}
public T get () {
return element;
}
#Override
public String toString() {
return element != null ? element.toString() : "null";
}
}
// Provides an iterator across all items in the container.
public Iterator<T> iterator() {
return new UsedNodesIterator<T>(this);
}
// Iterates across used nodes.
private static class UsedNodesIterator<T> implements Iterator<T> {
// Where next to look for the next used node.
Node<T> it;
int limit = 0;
T next = null;
public UsedNodesIterator(Container<T> c) {
// Snapshot the head node at this time.
it = c.head.get();
limit = c.capacity;
}
public boolean hasNext() {
// Made into a `while` loop to fix issue reported by #Nim in code review
while (next == null && limit > 0) {
// Scan to the next non-free node.
while (limit > 0 && it.free.get() == true) {
it = it.next;
// Step down 1.
limit -= 1;
}
if (limit != 0) {
next = it.element;
}
}
return next != null;
}
public T next() {
T n = null;
if ( hasNext () ) {
// Give it to them.
n = next;
next = null;
// Step forward.
it = it.next;
limit -= 1;
} else {
// Not there!!
throw new NoSuchElementException ();
}
return n;
}
public void remove() {
throw new UnsupportedOperationException("Not supported.");
}
}
#Override
public String toString() {
StringBuilder s = new StringBuilder();
Separator comma = new Separator(",");
// Keep counts too.
int usedCount = 0;
int freeCount = 0;
// I will iterate the list myself as I want to count free nodes too.
Node<T> it = head.get();
int count = 0;
s.append("[");
// Scan to the end.
while (count < capacity) {
// Is it in-use?
if (it.free.get() == false) {
// Grab its element.
T e = it.element;
// Is it null?
if (e != null) {
// Good element.
s.append(comma.sep()).append(e.toString());
// Count them.
usedCount += 1;
} else {
// Probably became free while I was traversing.
// Because the element is detached before the entry is marked free.
freeCount += 1;
}
} else {
// Free one.
freeCount += 1;
}
// Next
it = it.next;
count += 1;
}
// Decorate with counts "]used+free".
s.append("]").append(usedCount).append("+").append(freeCount);
if (usedCount + freeCount != capacity) {
// Perhaps something was added/freed while we were iterating.
s.append("?");
}
return s.toString();
}
}
Note that this is close to O1 put and get. A Separator just emits "" first time around and then its parameter from then on.
Edit: Added test methods.
// ***** Following only needed for testing. *****
private static boolean Debug = false;
private final static String logName = "Container.log";
private final static NamedFileOutput log = new NamedFileOutput("C:\\Junk\\");
private static synchronized void log(boolean toStdoutToo, String s) {
if (Debug) {
if (toStdoutToo) {
System.out.println(s);
}
log(s);
}
}
private static synchronized void log(String s) {
if (Debug) {
try {
log.writeLn(logName, s);
} catch (IOException ex) {
ex.printStackTrace();
}
}
}
static volatile boolean testing = true;
// Tester object to exercise the container.
static class Tester<T> implements Runnable {
// My name.
T me;
// The container I am testing.
Container<T> c;
public Tester(Container<T> container, T name) {
c = container;
me = name;
}
private void pause() {
try {
Thread.sleep(0);
} catch (InterruptedException ex) {
testing = false;
}
}
public void run() {
// Spin on add/remove until stopped.
while (testing) {
// Add it.
Node<T> n = c.add(me);
log("Added " + me + ": " + c.toString());
pause();
// Remove it.
c.remove(n, me);
log("Removed " + me + ": " + c.toString());
pause();
}
}
}
static final String[] strings = {
"One", "Two", "Three", "Four", "Five",
"Six", "Seven", "Eight", "Nine", "Ten"
};
static final int TEST_THREADS = Math.min(10, strings.length);
public static void main(String[] args) throws InterruptedException {
Debug = true;
log.delete(logName);
Container<String> c = new Container<String>(10);
// Simple add/remove
log(true, "Simple test");
Node<String> it = c.add(strings[0]);
log("Added " + c.toString());
c.remove(it, strings[0]);
log("Removed " + c.toString());
// Capacity test.
log(true, "Capacity test");
ArrayList<Node<String>> nodes = new ArrayList<Node<String>>(strings.length);
// Fill it.
for (int i = 0; i < strings.length; i++) {
nodes.add(i, c.add(strings[i]));
log("Added " + strings[i] + " " + c.toString());
}
// Add one more.
try {
c.add("Wafer thin mint!");
} catch (IllegalStateException ise) {
log("Full!");
}
c.clear();
log("Empty: " + c.toString());
// Iterate test.
log(true, "Iterator test");
for (int i = 0; i < strings.length; i++) {
nodes.add(i, c.add(strings[i]));
}
StringBuilder all = new StringBuilder ();
Separator sep = new Separator(",");
for (String s : c) {
all.append(sep.sep()).append(s);
}
log("All: "+all);
for (int i = 0; i < strings.length; i++) {
c.remove(nodes.get(i), strings[i]);
}
sep.reset();
all.setLength(0);
for (String s : c) {
all.append(sep.sep()).append(s);
}
log("None: " + all.toString());
// Multiple add/remove
log(true, "Multi test");
for (int i = 0; i < strings.length; i++) {
nodes.add(i, c.add(strings[i]));
log("Added " + strings[i] + " " + c.toString());
}
log("Filled " + c.toString());
for (int i = 0; i < strings.length - 1; i++) {
c.remove(nodes.get(i), strings[i]);
log("Removed " + strings[i] + " " + c.toString());
}
c.remove(nodes.get(strings.length - 1), strings[strings.length - 1]);
log("Empty " + c.toString());
// Multi-threaded add/remove
log(true, "Threads test");
c.clear();
for (int i = 0; i < TEST_THREADS; i++) {
Thread t = new Thread(new Tester<String>(c, strings[i]));
t.setName("Tester " + strings[i]);
log("Starting " + t.getName());
t.start();
}
// Wait for 10 seconds.
long stop = System.currentTimeMillis() + 10 * 1000;
while (System.currentTimeMillis() < stop) {
Thread.sleep(100);
}
// Stop the testers.
testing = false;
// Wait some more.
Thread.sleep(1 * 100);
// Get stats.
double added = c.totalAdded.doubleValue();
double skipped = c.totalSkipped.doubleValue();
//double freed = c.freed.doubleValue();
log(true, "Stats: added=" + c.totalAdded + ",freed=" + c.totalFreed + ",skipped=" + c.totalSkipped + ",O(" + ((added + skipped) / added) + ")");
}

Maybe you want to look at Disruptor - Concurrent Programming Framework.
Find a paper describing the alternatives, design and also a performance comparement to java.util.concurrent.ArrayBlockingQueue here: pdf
Consider to read the first three articles from BlogsAndArticles
If the library is too much, stick to java.util.concurrent.ArrayBlockingQueue

I would have a look at ArrayDeque, or for a more concurrent implementation have a look at the Disruptor library which is one of the most sophisticated/complex ring buffer in Java.
An alternative is to use an unbounded queue which is more concurrent as the producer never needs to wait for the consumer. Java Chronicle
Unless your needs justify the complexity, an ArrayDeque may be all you need.

Also have a look at java.util.concurrent.
Blocking queues will block until there is something to consume or (optionally) space to produce:
http://docs.oracle.com/javase/1.5.0/docs/api/java/util/concurrent/BlockingQueue.html
Concurrent linked queue is non-blocking and uses a slick algorithm that allows a producer and consumer to be active concurrently:
http://docs.oracle.com/javase/1.5.0/docs/api/java/util/concurrent/ConcurrentLinkedQueue.html

Hazelcast's Queue offers almost everything you ask for, but doesn't support circularity. But from your description I am not sure if you actually need it.

If it were me, I would use the CircularFIFOBuffer as you indicated, and synchronize around the buffer when writing (add). When the monitoring application wants to read the buffer, synchronize on the buffer, and then copy or clone it to use for reporting.
This suggestion is predicated on the assumption that latency is minimal to copy/clone the buffer to a new object. If there are large number of elements, and copying time is slow, then this is not a good idea.
Pseudo-Code example:
public void writeRequest(String requestID) {
synchronized(buffer) {
buffer.add(requestID);
}
}
public Collection<String> getRequests() {
synchronized(buffer) {
return buffer.clone();
}
}

Since you specifically ask to give writers (that is web servers) higher priority than the reader (that is monitoring), I would suggest the following design.
Web servers add request information to a concurrent queue which is read by a dedicated thread, which adds requests to a thread-local (therefore non-synchronized) queue that overwrites the oldest element, like EvictingQueue or CircularFifoQueue.
This same thread checks a flag which indicates if a report has been requested after every request processed, and if positive, produces a report from all elements present in the thread-local queue.

how to implement multithreaded Breadth-first search in java?

I've done the breadth-first search in a normal way.
now I'm trying to do it in a multithreaded way.
i have one queue which is shared between the threads.
i use synchronize(LockObject) when i remove a node from the queue ( FIFI queue )
so what I'm trying to do is that.
when i thread finds a solution all the other threads will stop immediately.

i assume you are traversing a tree for your BFS.
create a thread pool.
for each unexplored children in the node, retrieve a thread from the thread pool (perhaps using a Semaphore). mark the child node as 'explored' and explore the node's children in a BFS manner. when you have found a solution or done exploring all the nodes, release the semaphore.
^ i've never done this before so i might have missed out something.

Assuming you want to do this iteratively (see note at the bottom why there may be better closed solutions), this is not a great problem for exercising multi threading. The problem is that multithreading is great if you don't depend on previous results, but here you want the minimum amount of coins.
As you point out, a breadth first solution guarantees that once you reach the desired amount, you won't have any further solutions with less coins in a single threaded environment. However, in a multithreaded environment, once you start calculating a solution, you cannot guarantee that it will finish before some other solution. Let's imagine for the value 21: it can be a 20c coin and a 1c or four 5c coins and a 1c; if both are calculating simultaneously, you cannot guarantee that the first (and correct) solution will finish first. In practice, it is unlikely the situation will happen, but when you work with multithreading you want the solution to work in theory, because multithreads always fail in the demonstration, no matter if they should not have failed until the death heat of the universe.
Now you have 2 possible solutions: one is to introduce choke points at the beginning of each level; you don't start that level until the previous level is finished. The other is once you reach a solution continue doing all the calculations with a lower level than the current result (which means you cannot purge the others). Probably with all the synchronization needed you get better performance by going single threaded, but let's go on.
For the first solution, the natural form is to iterate increasing the level. You can use the solution provided by happymeal, with a Semaphore. An alternative is to use the new classes provided by java.
CoinSet getCoinSet(int desiredAmount) throws InterruptedException {
// Use whatever number of threads you prefer or another member of Executors.
final ExecutorService executor = Executors.newFixedThreadPool(10);
ResultContainer container = new ResultContainer();
container.getNext().add(new Producer(desiredAmount, new CoinSet(), container));
while (container.getResult() == null) {
executor.invokeAll(container.setNext(new Vector<Producer>()));
}
return container.getResult();
}
public class Producer implements Callable<CoinSet> {
private final int desiredAmount;
private final CoinSet data;
private final ResultContainer container;
public Producer(int desiredAmount, CoinSet data, ResultContainer container) {
this.desiredAmount = desiredAmount;
this.data = data;
this.container = container;
}
public CoinSet call() {
if (data.getSum() == desiredAmount) {
container.setResult(data);
return data;
} else {
Collection<CoinSet> nextSets = data.addCoins();
for (CoinSet nextSet : nextSets) {
container.getNext().add(new Producer(desiredAmount, nextSet, container));
}
return null;
}
}
}
// Probably it is better to split this class, but you then need to pass too many parameters
// The only really needed part is to create a wrapper around getNext, since invokeAll is
// undefined if you modify the list of tasks.
public class ResultContainer {
// I use Vector because it is synchronized.
private Vector<Producer> next = new Vector<Producer>();
private CoinSet result = null;
// Note I return the existing value.
public Vector<Producer> setNext(Vector<Producer> newValue) {
Vector<Producer> current = next;
next = newValue;
return current;
}
public Vector<Producer> getNext() {
return next;
}
public synchronized void setResult(CoinSet newValue) {
result = newValue;
}
public synchronized CoinSet getResult() {
return result;
}
}
This still has the problem that existing tasks are executed; however, it is simple to fix that; pass the thread executor into each Producer (or the container). Then, when you find a result, call executor.shutdownNow. The threads that are executing won't be interrupted, but the operation in each thread is trivial so it will finish fast; the runnables that have not started won't start.
The second option means you have to let all the current tasks finish, unless you keep track of how many tasks you have run at each level. You no longer need to keep track of the levels, though, and you don't need the while cycle. Instead, you just call
executor.submit(new Producer(new CoinSet(), desiredAmount, container)).get();
And then, the call method is pretty similar (assume you have executor in the Producer):
public CoinSet call() {
if (container.getResult() != null && data.getCount() < container.getResult().getCount()) {
if (data.getSum() == desiredAmount)) {
container.setResult(data);
return data;
} else {
Collection<CoinSet> nextSets = data.addCoins();
for (CoinSet nextSet : nextSets) {
executor.submit(new Producer(desiredAmount, nextSet, container));
}
return null;
}
}
}
and you also have to modify container.setResult, since you cannot depend that between the if and setting the value it has not been set by some other threads (threads are really annoying, aren't they?)
public synchronized void setResult(CoinSet newValue) {
if (newValue.getCount() < result.getCount()) {
result = newValue;
}
}
In all previous answers, CoinSet.getSum() returns the sum of the coins in the set, CoinSet.getCount() returns the number of coins, and CoinSet.addCoins() returns a Collection of CoinSet in which each element is the current CoinSet plus one coin of each possible different value
Note: For the problem of the coins with the values 1, 5, 10 and 20, the simplest solution is take the amount and divide it by the largest coin. Then take the modulus of that and use the next largest value and so on. That is the minimum amount of coins you are going to need. This rule applies (AFAICT) when the following property if true: if for all consecutive pairs of coin values (i.e. in this case, 1-5, 5-10, 10-20) you can reach any int multiple of the lower element in the pair with with a smaller number of coins using the larger element and whatever coins are necessary. You only need to prove it to the min common multiple of both elements in the pair (after that it repeats itself)

I gather from your comment on happymeal's anwer that you are trying to find how to reach a specific amount of money by adding coins of 1c, 5c, 10c and 20c.
Since each coin denomination divides the denomination of the next bigger coin, this can be solved in constant time as follows:
int[] coinCount(int amount) {
int[] coinValue = {20, 10, 5, 1};
int[] coinCount = new int[coinValue.length];
for (int i = 0; i < coinValue.length; i++) {
coinCount[i] = amount / coinValue[i];
amount -= coinCount[i] * coinValue[i];
}
return coinCount;
}
Take home message: Try to optimize your algorithm before resorting to multithreading, because algorithmic improvements can yield much greater improvements.

I've successfully implemented it.
what i did is that i took all the nodes in the first level, let's say 4 nodes.
then i had 2 threads. each one takes 2 nodes and generate their children. whenever a node finds a solution it has to report the level that it found the solution in and limit the searching level so other threads don't exceed the level.
only the reporting method should be synchronized.
i did the code for the coins change problem. this is my code for others to use
Main Class (CoinsProblemBFS.java)
package coinsproblembfs;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Scanner;
/**
*
* #author Kassem M. Bagher
*/
public class CoinsProblemBFS
{
private static List<Item> MoneyList = new ArrayList<Item>();
private static Queue<Item> q = new LinkedList<Item>();
private static LinkedList<Item> tmpQ;
public static Object lockLevelLimitation = new Object();
public static int searchLevelLimit = 1000;
public static Item lastFoundNode = null;
private static int numberOfThreads = 2;
private static void InitializeQueu(Item Root)
{
for (int x = 0; x < MoneyList.size(); x++)
{
Item t = new Item();
t.value = MoneyList.get(x).value;
t.Totalvalue = MoneyList.get(x).Totalvalue;
t.Title = MoneyList.get(x).Title;
t.parent = Root;
t.level = 1;
q.add(t);
}
}
private static int[] calculateQueueLimit(int numberOfItems, int numberOfThreads)
{
int total = 0;
int[] queueLimit = new int[numberOfThreads];
for (int x = 0; x < numberOfItems; x++)
{
if (total < numberOfItems)
{
queueLimit[x % numberOfThreads] += 1;
total++;
}
else
{
break;
}
}
return queueLimit;
}
private static void initializeMoneyList(int numberOfItems, Item Root)
{
for (int x = 0; x < numberOfItems; x++)
{
Scanner input = new Scanner(System.in);
Item t = new Item();
System.out.print("Enter the Title and Value for item " + (x + 1) + ": ");
String tmp = input.nextLine();
t.Title = tmp.split(" ")[0];
t.value = Double.parseDouble(tmp.split(" ")[1]);
t.Totalvalue = t.value;
t.parent = Root;
MoneyList.add(t);
}
}
private static void printPath(Item item)
{
System.out.println("\nSolution Found in Thread:" + item.winnerThreadName + "\nExecution Time: " + item.searchTime + " ms, " + (item.searchTime / 1000) + " s");
while (item != null)
{
for (Item listItem : MoneyList)
{
if (listItem.Title.equals(item.Title))
{
listItem.counter++;
}
}
item = item.parent;
}
for (Item listItem : MoneyList)
{
System.out.println(listItem.Title + " x " + listItem.counter);
}
}
public static void main(String[] args) throws InterruptedException
{
Item Root = new Item();
Root.Title = "Root Node";
Scanner input = new Scanner(System.in);
System.out.print("Number of Items: ");
int numberOfItems = input.nextInt();
input.nextLine();
initializeMoneyList(numberOfItems, Root);
System.out.print("Enter the Amount of Money: ");
double searchValue = input.nextDouble();
int searchLimit = (int) Math.ceil((searchValue / MoneyList.get(MoneyList.size() - 1).value));
System.out.print("Number of Threads (Muste be less than the number of items): ");
numberOfThreads = input.nextInt();
if (numberOfThreads > numberOfItems)
{
System.exit(1);
}
InitializeQueu(Root);
int[] queueLimit = calculateQueueLimit(numberOfItems, numberOfThreads);
List<Thread> threadList = new ArrayList<Thread>();
for (int x = 0; x < numberOfThreads; x++)
{
tmpQ = new LinkedList<Item>();
for (int y = 0; y < queueLimit[x]; y++)
{
tmpQ.add(q.remove());
}
BFS tmpThreadObject = new BFS(MoneyList, searchValue, tmpQ);
Thread t = new Thread(tmpThreadObject);
t.setName((x + 1) + "");
threadList.add(t);
}
for (Thread t : threadList)
{
t.start();
}
boolean finish = false;
while (!finish)
{
Thread.sleep(250);
for (Thread t : threadList)
{
if (t.isAlive())
{
finish = false;
break;
}
else
{
finish = true;
}
}
}
printPath(lastFoundNode);
}
}
Item Class (Item.java)
package coinsproblembfs;
/**
*
* #author Kassem
*/
public class Item
{
String Title = "";
double value = 0;
int level = 0;
double Totalvalue = 0;
int counter = 0;
Item parent = null;
long searchTime = 0;
String winnerThreadName="";
}
Threads Class (BFS.java)
package coinsproblembfs;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
/**
*
* #author Kassem M. Bagher
*/
public class BFS implements Runnable
{
private LinkedList<Item> q;
private List<Item> MoneyList;
private double searchValue = 0;
private long start = 0, end = 0;
public BFS(List<Item> monyList, double searchValue, LinkedList<Item> queue)
{
q = new LinkedList<Item>();
MoneyList = new ArrayList<Item>();
this.searchValue = searchValue;
for (int x = 0; x < queue.size(); x++)
{
q.addLast(queue.get(x));
}
for (int x = 0; x < monyList.size(); x++)
{
MoneyList.add(monyList.get(x));
}
}
private synchronized void printPath(Item item)
{
while (item != null)
{
for (Item listItem : MoneyList)
{
if (listItem.Title.equals(item.Title))
{
listItem.counter++;
}
}
item = item.parent;
}
for (Item listItem : MoneyList)
{
System.out.println(listItem.Title + " x " + listItem.counter);
}
}
private void addChildren(Item node, LinkedList<Item> q, boolean initialized)
{
for (int x = 0; x < MoneyList.size(); x++)
{
Item t = new Item();
t.value = MoneyList.get(x).value;
if (initialized)
{
t.Totalvalue = 0;
t.level = 0;
}
else
{
t.parent = node;
t.Totalvalue = MoneyList.get(x).Totalvalue;
if (t.parent == null)
{
t.level = 0;
}
else
{
t.level = t.parent.level + 1;
}
}
t.Title = MoneyList.get(x).Title;
q.addLast(t);
}
}
#Override
public void run()
{
start = System.currentTimeMillis();
try
{
while (!q.isEmpty())
{
Item node = null;
node = (Item) q.removeFirst();
node.Totalvalue = node.value + node.parent.Totalvalue;
if (node.level < CoinsProblemBFS.searchLevelLimit)
{
if (node.Totalvalue == searchValue)
{
synchronized (CoinsProblemBFS.lockLevelLimitation)
{
CoinsProblemBFS.searchLevelLimit = node.level;
CoinsProblemBFS.lastFoundNode = node;
end = System.currentTimeMillis();
CoinsProblemBFS.lastFoundNode.searchTime = (end - start);
CoinsProblemBFS.lastFoundNode.winnerThreadName=Thread.currentThread().getName();
}
}
else
{
if (node.level + 1 < CoinsProblemBFS.searchLevelLimit)
{
addChildren(node, q, false);
}
}
}
}
} catch (Exception e)
{
e.printStackTrace();
}
}
}
Sample Input:
Number of Items: 4
Enter the Title and Value for item 1: one 1
Enter the Title and Value for item 2: five 5
Enter the Title and Value for item 3: ten 10
Enter the Title and Value for item 4: twenty 20
Enter the Amount of Money: 150
Number of Threads (Muste be less than the number of items): 2