Problems with pathfinding. Weird paths being returned - java

I have been reading some articles to learn A* pathfinding and I was able to make a test program based on it, but the the program I made is giving strange paths which are not a shortest path to the target node.
In the images, the green square unit represents the starting node, the red unit represents target node, blue units are impassable tiles (walls) and the purple units represent the path found from starting node to target node
http://imgur.com/5dJEfYc
http://imgur.com/lHfXEyW
If anybody could find a problem with the pathfinding source code I would be much thankful. I'm burned out from trying to know what caused it to act strange.
Its allowed to cut corners and go diagonal
package com.streak324.pathfinding;
import java.util.Comparator;
import java.util.HashSet;
import java.util.PriorityQueue;
import com.badlogic.gdx.utils.Array;
public class PathFinder {
private boolean foundTarget;
private int width, height;
//list of nodes that leads from starting node to target node is stored here
private Array<PathNode> path;
//all nodes stored in this array
private PathNode[][] nodes;
private PriorityQueue<PathNode> open;
private HashSet<PathNode> closed;
//nodes that must be referenced
private PathNode start, target, current;
//how far the current node can reach for other nodes from its own position
private int range = 1;
public PathFinder(int width, int height, boolean map[][]) {
this.width = width;
this.height = height;
nodes = new PathNode[width][height];
for(int i=0; i<width; i++) {
for(int j=0; j<height; j++) {
nodes[i][j] = new PathNode(i, j);
//if wall tile is spotted, mark the node unwalkable
if(map[i][j] != true) { nodes[i][j].passable = false; }
}
}
open = new PriorityQueue<PathNode>(new CostComparator());
closed = new HashSet<PathNode>();
}
public Array<PathNode> getPath(int sx, int sy ,int tx, int ty) {
path = new Array<PathNode>();
open.clear();
closed.clear();
start = nodes[sx][sy];
start.movementCost = 0;
addToOpenList(start);
target = nodes[tx][ty];
while(foundTarget != true) {
if(open.size() == 0) { return null; }
current = open.poll();
addToClosedList(current);
checkNeighbors(current);
}
traceBack();
return path;
}
// makes its way back adding the parent node until start
private void traceBack() {
while(current != start) {
path.add(current);
current = current.parent;
}
}
//checks for nodes within certain range
private void checkNeighbors(PathNode node) {
//continues loop if i or j goes out of bounds of nodes array
for(int i = node.x - range; i <= (node.x + range); i++) {
if(i >= width || i < 0) { continue; }
for(int j = node.y - range; j <= (node.y + range); j++) {
if( j >= height || j < 0) { continue; }
if((i == node.x && j == node.y) ) { continue; }
PathNode neighbor = nodes[i][j];
identifyNode(neighbor);
}
}
}
//if node is not on open list, add node and calculate it
private void identifyNode(PathNode node) {
if(!node.passable || closed.contains(node) ) return;
if(node == target) {
foundTarget = true;
System.out.println("Target Found: " + node.x + ", " + node.y);
return;
}
else if(!open.contains(node)) {
addToOpenList(node);
calcHeuristic(node);
updateNode(node, current);
}
else {
checkForReparenting(node);
}
}
//is the movement cost less to go from the current node?
private void checkForReparenting(PathNode node) {
float cost = node.movementCost;
float reCost = calcMovementCost(node, current);
if(reCost < cost) {
System.out.println("Reparenting");
updateNode(node, current);
}
}
//updates parent and cost
private void updateNode(PathNode child, PathNode parent) {
child.parent = parent;
child.movementCost = calcMovementCost(child, parent);
child.totalCost = child.movementCost + child.heuristic;
}
private float calcMovementCost(PathNode n1, PathNode n2) {
float dx = n1.x - n2.x;
float dy = n1.y - n2.y;
return (float) Math.sqrt( (dx*dx + dy*dy) ) + n2.movementCost;
}
private float calcHeuristic(PathNode node) {
float dx = node.x - target.x;
float dy = node.y - target.y;
return (float) Math.sqrt( (dx*dx + dy*dy) );
}
private void addToOpenList(PathNode node) {
if(!open.contains(node) && !closed.contains(node)) {
open.add(node);
}
}
private void addToClosedList(PathNode node) {
if(!closed.contains(node)) {
closed.add(node);
}
}
public class PathNode {
public int x, y;
public PathNode parent;
//g, h and f
public float movementCost, heuristic, totalCost;
public boolean passable;
public PathNode(int x, int y) {
this.x = x;
this.y = y;
passable = true;
}
}
private class CostComparator implements Comparator<PathNode> {
#Override
public int compare(PathNode a, PathNode b) {
if(a.totalCost < b.totalCost) return 1;
else return -1;
}
}
}
no comments
http://pastebin.com/rSv7pUrB
I'm guessing something is wrong in the way that the priority queue is ordering the elements or I may have not properly calculated the totalCost, movementCost, and heuristic variables, but I see nothing wrong with it.
Someone that could point me to the right direction of a probable problem or solution is much appreciated

There are several issues with your code:
You never really use the heuristic. The following statement (the only call to calcHeuristic) just "throws the result away".
calcHeuristic(node);
That alone can't be the error here, since it's a valid admissible heuristic to guess the distance to the target to be 0. However the algorithm degenerates that way (to what I think is the Dijkstra algorithm).
You never update the position of the node in the priority queue. That means a node with updated totalDistance will never move up in the proirity queue, even if it's totalCost becomes less than the totalCost of another node. You have to remove the node and add it again to do that with a PriorityQueue:
open.remove(node);
// ... update totalDistance
open.add(node);
You terminate too early for general A* (however that wouldn't be an issue here, since totalDistance is equal to the real distance, for expanded neighbors of the target IF you use the heuristic; here the distance real distance is different by either sqrt(2) or 1). In general the distance heuristic for the last step can be arbitrary bad (and here it's bad, see (1.)) and you can only be sure you found the real solution, if you run the algorithm to the point where you would expand the target node.

Estimated Streak324:
Altough your implementation of A* is now working properly, I recommend you to do a quick search in Internet for java search libraries. Your code will look much more simple, scalable and modular, and the implementations are very efficient and well tested. This will be your code using Hipster:
//HERE YOU DEFINE THE SEARCH PROBLEM
// The maze is a 2D map, where each tile defined by 2D coordinates x and y
// can be empty or occupied by an obstacle. We have to define de transition
// function that tells the algorithm which are the available movements from
// a concrete tile point.
SearchProblem p = ProblemBuilder.create()
.initialState(origin)
.defineProblemWithoutActions()
.useTransitionFunction(new StateTransitionFunction<Point>() {
#Override
public Iterable<Point> successorsOf(Point state) {
// The transition function returns a collection of transitions.
// A transition is basically a class Transition with two attributes:
// source point (from) and destination point (to). Our source point
// is the current point argument. We have to compute which are the
// available movements (destination points) from the current point.
// Class Maze has a helper method that tell us the empty points
// (where we can move) available:
//TODO: FILL WITH YOUR CODE GENERATING THE NEIGHBORS, FILTERING
//THOSE WHICH ARE NOT ACCESIBLE DUE TO OBSTACLES IN YOUR MAP
return [...]
}
})
.useCostFunction(new CostFunction<Void, Point, Double>() {
// We know now how to move (transitions) from each tile. We need to define the cost
// of each movement. A diagonal movement (for example, from (0,0) to (1,1)) is longer
// than a top/down/left/right movement. Although this is straightforward, if you don't
// know why, read this http://www.policyalmanac.org/games/aStarTutorial.htm.
// For this purpose, we define a CostFunction that computes the cost of each movement.
// The CostFunction is an interface with two generic types: S - the state, and T - the cost
// type. We use Points as states in this problem, and for example doubles to compute the distances:
#Override
public Double evaluate(Transition<Void, Point> transition) {
Point source = transition.getFromState();
Point destination = transition.getState();
// The distance from the source to de destination is the euclidean
// distance between these two points http://en.wikipedia.org/wiki/Euclidean_distance
return source.distance(destination);
}
})
.build();
//HERE YOU INSTANTIATE THE ALGORITHM AND EXECUTE THE SEARCH
//MazeSearch.printSearch(Hipster.createAStar(p).iterator(), maze);
System.out.println(Hipster.createAStar(p).search(goal));
As you can see, you only need to define the components to be used in the search problem, and then execute the algorithm. The library will do the rest of the operations for you.
Also, the library is open-source and licsensed Apache2. You have access to several examples that will help you to start working with the library.
In your case, as you are using a custom 2D grid, the only thing you need to adapt is the transition function which checks your grid to filter those neighbors not accesible due to obstacles.
The inmediate benefit of using this implementation is, apart of the scalability and modularity of the code, avoid instantiating all the nodes in the path, as the library will do it dynamically for you, reducing memory and increasing performance (specially in cases of huge grids).
I hope my answer helps,

Related

Depth First Search recursive calls cause StackOverflow error

I mam using an implementation of the Depth First Search Algorithm in order to solve a maze. I do not wish for the shortest path to be found but for the fastest way to find a valid path. This is the method i use to solve mazes. Mazes are 2d int arrays where 0 is an open square, 1 is a wall, 2 is a visited square and 9 is the destination.
public class DepthAlgorithm {
/**
* This method returns true when a path is found. It will also fill up the
* list with the path used. It will start from (xn,yn,.....,x2,y2,x1,y2)
* because it will use recursion.
* #param maze 2d array of maze
* #param x the x of the starting position
* #param y the y of the starting position
* #param path a List of the path
* #return True if a path is found
*/
public static boolean searchPath(int [][] maze, int x, int y, List<Integer> path){
// Check if the destination (9) is reached
if (maze[y][x] == 9) {
path.add(x);
path.add(y);
return true;
}
// When the current position is not visited (0) we shall make it visited (2)
if (maze[y][x] == 0) {
maze[y][x] = 2;
//Here we visit all neighbour squares recursively and if a path is found
// we shall fill the path list with the current position.
int dx = -1; // Start and search from starting
int dy = 0; // position (x-1,y)
if (searchPath(maze, x + dx, y + dy, path)) {
path.add(x);
path.add(y);
return true;
}
dx = 1; // Start and search from starting
dy = 0; // position (x+1,y)
if (searchPath(maze, x + dx, y + dy, path)) {
path.add(x);
path.add(y);
return true;
}
dx = 0; // Start and search from starting
dy = -1; // position (x,y-1)
if (searchPath(maze, x + dx, y + dy, path)) {
path.add(x);
path.add(y);
return true;
}
dx = 0; // Start and search from starting
dy = 1; // position (x,y+1)
if (searchPath(maze, x + dx, y + dy, path)) {
path.add(x);
path.add(y);
return true;
}
}
return false;
}
}
My algorithm works fine for small maze sizes. When i want to solve a 50 * 50 maze it's quite quick. When i move to 500 * 500 a Stack Overflow error shows up. I can understand that it shows up because of the many recursive calls of the function but i do not know how to fix it. Is there another way so I can still use the Depth First Search and store my path but without the Stack Overflow? Or are there any changes that can be done in my code so it can be fixed?
public class MazeRunner {
// Maze is a 2d array and it has to be filled with walls peripherally
// with walls so this algorithm can work. Our starting position in this
// will be (1,1) and our destination will be flagged with a 9 which in
// this occasion is (11,8).
private int[][] maze ;
private final List<Integer> path = new ArrayList<>();
public long startTime,stopTime;
public MazeRunner(int [][] maze){
this.maze = maze;
}
public void runDFSAlgorithm(int startingX,int startingY){
startTime = System.nanoTime();
DepthAlgorithm.searchPath(maze,startingX,startingY,path);
stopTime = System.nanoTime();
printPath();
System.out.println("Time for Depth First Algorithm: "+((double) (stopTime-startTime) / 1_000_000)+" milliseconds");
}
public void printPath(){
ListIterator li = path.listIterator(path.size());
int lengthOfPath = (path.size()/2-1);
int fromX,fromY,bool = 0,toX = 0,toY = 0,i = 2;
while(li.hasPrevious()){
if (bool == 0) {
fromX = (int) li.previous();
fromY = (int) li.previous();
toX = (int) li.previous();
toY = (int) li.previous();
System.out.println("From ("+fromY+", "+fromX+") to ("+toY+", "+toX+")");
bool++;
continue;
}
if (bool == 1){
fromX = toX;
fromY = toY;
toX = (int) li.previous();
toY = (int) li.previous();
System.out.println("From ("+fromY+", "+fromX+") to ("+toY+", "+toX+")");
i++;
}
}
System.out.println("\nLength of path is : "+lengthOfPath);
}
public static void main(String[] args){
int [][] maze = {{1,1,1,1,1,1,1,1,1,1,1,1,1},
{1,0,1,0,1,0,1,0,0,0,0,0,1},
{1,0,1,0,0,0,1,0,1,1,1,0,1},
{1,0,0,0,1,1,1,0,0,0,0,0,1},
{1,0,1,0,0,0,0,0,1,1,1,0,1},
{1,0,1,0,1,1,1,0,1,0,0,0,1},
{1,0,1,0,1,0,0,0,1,1,1,0,1},
{1,0,1,0,1,1,1,0,1,0,1,0,1},
{1,0,0,0,0,0,0,0,0,0,1,9,1},
{1,1,1,1,1,1,1,1,1,1,1,1,1}};
MazeRunner p = new MazeRunner(maze);
p.runDFSAlgorithm(startingPoint[0],startingPoint[1]);
}
}
And this is the class i use for my testings. It for sure works for this example but for greater arrays it doesn't. Any suggestions would be appreciated. When i run my program on big arrays i get this following error:
Generally speaking there are only two possibilities that could cause a stackoverflow exception
1.memory of stack is not enough
2.there exist a dead loop which in using recursive means it does't exist an end condition.
Since you algo works for small size maze. It’s possible the reason one. As you know the rule of recursive is first in last out, in JVM all the data of unexecuted functions will be stored in stack which owns a much smaller memory than heap.
You should never use a recursive algorithm for real work unless you know for certain that the depth of the stack will be limited to a reasonable number. Usually that means O(log N), or O(log^2 N) at most.
DFS for a 500x500 maze could put around 250000 function calls on the stack, which is way too many.
You can use DFS if you really want to, but you should maintain your own stack in a separate data structure. BFS would be better though, unless there's some reason you really don't want the shortest path.

A* Pathfinding problems Processing(Java)

I'm quite new to programming though following a bunch of tutorials I've ended up with this code to deal with the pathfinding of a small game I'm trying to make.
If works for small and straight paths but not for complex routes (it freezes and closedSet.size() gets larger than 70000 in a grid that is only 54 * 46).
Note that wall is true depending on the height of the colliding tiles, so it may be true coming from a point but false coming from another. Is that the problem?
import java.util.*;
int heuristic(int x,int y,int x_,int y_){
int dstX = abs(x - x_);
int dstY = abs(y - y_);
if(dstX > dstY){
return 14*dstY + 10*(dstX - dstY);
}else{
return 14*dstX + 10*(dstY - dstX);
}
}
boolean wall(int x, int y, int x_, int y_){
Tile tileS = getTile(x, y);
Tile tileCurr = getTile(x_, y_);
if(abs(tileS.altitude - tileCurr.altitude) > 1 || tileS.altitude < 1){
return true;
}else{
return false;
}
}
ArrayList<PVector> findPath(int startx, int starty, int endx, int endy){
Queue<Spot> openSet = new PriorityQueue<Spot>(fComparator);
ArrayList<Spot> closedSet = new ArrayList<Spot>();
Spot start = new Spot(startx, starty);
Spot end = new Spot(endx, endy);
Spot current = start;
openSet.add(start);
while(!openSet.isEmpty()){
current = openSet.poll();
closedSet.add(current);
println(closedSet.size());
if (current.x == end.x && current.y == end.y) {
break;
}
ArrayList<Spot> successors = new ArrayList<Spot>();
for(int i = 0; i < collidingTiles.size(); i++){
JSONObject difference = collidingTiles.getJSONObject(i);
/*JSONArray such as
[{x: -1, y: -1},{x: 0, y: -1},...](not including {x:0, y:0})
*/
int x_ = difference.getInt("x");
int y_ = difference.getInt("y");
int x = x_ + current.x;
int y = y_ + current.y;
if(x >= 0 && x <= map.columns && y >= 0 && y <= map.rows){
Spot s = new Spot(x, y);
successors.add(s);
}
}
for(Spot s: successors){
if (!closedSet.contains(s) && !wall(s.x, s.y, current.x, current.y)) {
int tempG = current.g + heuristic(s.x, s.y, current.x, current.y);
if(tempG < s.g || !openSet.contains(s)){
s.g = tempG;
s.h = heuristic(s.x, s.y, end.x, end.y);
s.f = s.g + s.h;
s.parent = current;
if (!openSet.contains(s)) {
openSet.add(s);
}
}
}
}
successors.clear();
}
ArrayList<PVector> path = new ArrayList<PVector>();
Spot temp = current;
PVector tile = new PVector(temp.x + 0.5, temp.y + 0.5);
path.add(tile);
while (temp.parent != null) {
tile = new PVector(temp.parent.x + 0.5, temp.parent.y + 0.5);
path.add(0, tile);
temp = temp.parent;
}
return path;
}
class Spot{
int x, y;
int f, g, h = 0;
Spot parent;
Spot(int x_, int y_){
x = x_;
y = y_;
}
}
Comparator<Spot> fComparator = new Comparator<Spot>() {
#Override
int compare(Spot s1, Spot s2) {
return s1.f - s2.f;
}
};
Any recommendations or minor changes are also appreciated.
closedSet.size() gets larger than 70000 in a grid that is only 54 * 46
Your code does implement some logic that says
"if a node is closed, don't process it again", and
"if the node is already in the open set, compare G scores"
But in both cases it does not work, because Spot does not implement equals, and therefore contains is comparing for reference equality and it will always be false. So, implement Spot.equals. Specifically, make it compare only x and y, because f/g/h/parent are allowed to be different for nodes that are considered equal for this purpose.
Even when it works, using contains on an ArrayList and a PriorityQueue is not so good for performance. For the closed list, it is easy to use a HashSet (of course, also implement Spot.hashCode, in some way that depends only on x and y). For the open list, getting rid of slow contains is more work. One trick you can use is manually maintaining a binary heap, and additionally have a HashMap which maps an x,y pair to the index in the heap where the corresponding node is located. The reason for manually maintaining a heap is that the HashMap must be updated whenever nodes are moved in the queue, and the normal PriorityQueue does not have such functionality.
The way that finding successors works also worries me from a performance perspective, but I cannot see the details.
Note that wall is true depending on the height of the colliding tiles, so it may be true coming from a point but false coming from another. Is that the problem?
That's fine, A* can tolerate a spot being reachable from one side but not an other. What it cannot natively take into account is if the direction a spot was reached from affects which successors that node has, but that does not happen here.

Android Game Logic - Obsticles

I have been working on android apps for a long time but now I have decided to create a game aside with my pre-calculus final. I have completed the whole code and it works perfectly except one tiny issue. First of the game is based on flying pig(my classmate's face) avoiding top and bottom osticle. I developed an algorithm so that the obsticles are evenly spaced and based on random selection placed either as the top or bottom of the screen but never both at the same time!. The algorithm that needs improvement is in the 3rd code segment!
This is the screenshot of the problem: screenshot here
(My account is new so stackoverflow wont let me to share photos directly)
This is the class that calls updates for all dynamic objects (ship = pig, bacon = bonus item, BM = BarrierManager class's update() which updates the obsticles)
//this will update the resources
void Update(float dt) {
ship.update(dt);
//bumbing
if (!ship.death) {
background.update(dt);
**BM.update(dt);**
for (int i = 0; i % 5 == 0; i++) {
bacon.update(dt, BM.position);
}
}
ArrayList<Point> bacon_point = new ArrayList<Point>(bacon.getArray());
if (ship.bump(bacon_point.get(0), bacon_point.get(1), bacon_point.get(2), bacon_point.get(3))) {
bacon.setX(-200);
bacon.setY(-200);
Message msg = BM.game_panel.game.handler.obtainMessage();
msg.what = 0;
BM.game_panel.game.handler.sendMessage(msg);
}
for (int i = 0; i < BM.TopWalls.size(); i++) {
ArrayList<Point> temp = new ArrayList<Point>(BM.TopWalls.get(i).getArray());
//after we have accest the TopWalls arraylist we can call bump that check TopWalls object's points position with the pig's points
ArrayList<Point> temp2 = new ArrayList<Point>(BM.BottomWalls.get(i).getArray());
//after we have accest the TopWalls arraylist we can call bump that check BottomWalls object's points position with the pig's points
if ((ship.bump(temp.get(0), temp.get(1), temp.get(2), temp.get(3))) || (ship.bump(temp2.get(0), temp2.get(1), temp2.get(2), temp2.get(3))) || ship.death) {
ship.death = true;
game.onStop();
while(f==0) {
MediaPlayer mp = MediaPlayer.create(game, R.raw.explosion_fart);
mp.start();
f++;
}
f++;
Message msg = BM.game_panel.game.handler.obtainMessage();
msg.what = 1;
BM.game_panel.game.handler.sendMessage(msg);
i = BM.TopWalls.size()-1;
if(f == 8){
thread.setRunning(false);
}
}
}
}
In the BarrierManager I have created this update method which takes float dt = MainTheards general time for the game.
TopWalls is ArrayList this ArrayList is composed of top obsticles. Bottom walls is the same but BottomWalls.
//zreb decides if it should create TopWalls or BottomWalls object. This parameter is later transfered to the Barier.update method where I work with it
public void update(float dt){
for (int i=0;i<Num+1; i++) {
int zreb = new Random().nextInt(2);
//{if} to make the spacing bigger
if (i % 5 == 0){
**TopWalls.get(i).update(dt, true, zreb);
BottomWalls.get(i).update(dt, false, zreb);**
if(zreb == 0){
position.set(TopWalls.get(i).getX(), TopWalls.get(i).getY());
}
else{
position.set(BottomWalls.get(i).getX(),BottomWalls.get(i).getY());
}
}
}
}
Now this algoritm in the Barrier.class is where the magic went wrong. This update method takes the float dt mentioned earlier, a boolean variable for determining if the obsticle we work with at that instance is the Top or Bottom, and the zreb random int that determines if the top or bottom obsticle is going to be shown.
//problem! needs to be discussed
public void update(float dt, boolean b, int zreb) {
//checking if the barrier is still there
if (x<-bitmap.getWidth()){
//'b'is passed from the Barriermanager - 'update' method, determining if we have to use monkey-true or farmer-false
int raz = 200;
int vyska = BM.dpos;
//'zreb' helps me to randomly determine if monkey or ballard is going to appear
//here it determines if we are going to have Top or Bottom obsticle in the game
if(zreb == 1) {
vyska = BM.dpos - raz;
}
else {
vyska = BM.dpos + raz;
}
//koniec tohto trienedia
if (b)
{
//setting the y value for the top wall
y = vyska - BM.dl/2 - bitmap.getHeight()/2;
}
else{
//setting the y value for bottom wall
y = vyska + BM.dl/2 + bitmap.getHeight()/2;
}
//setting x-value
x = (int) (x +bitmap.getWidth()*(BM.TopWalls.size()-1));
}
x = (int) (x - BM.game_panel.ShipSpeed*dt);
}
Do you have any idea why this "one-or-the-other" condition is not working properly?
This would help me lot because this error made me deactivate the app from the store.

Position and orientation of robot in a grid

I want to make a 10x10 grid and put the robot in position (10,1) (bottom left). I want this robot to be able to move forward, turn left/right and to pick up/put objects in a grid. When put in any position, there should be number in a grid which shows how many objects is put in this position, just like this:
..........
...1......
..2.......
....3.....
..........
..........
......9...
.....4....
.........1
..........
We will not see the robot in a grid. I have two classes. Class Robot:
public class Robot {
private Area area;
private Robot rob;
public Robot(Area area){
this.area = area;
rob = new Robot(area);
}
public void Right(){
}
public void Left(){
}
public void Forward(){
}
public void Put(){
}
public void PickUp(){
}
public (?) getPosition(){ // should return robot's position
}
}
Class Area:
private int numberOfObjects;
private Robot robot;
private static final int X = 10;
private static final int Y = 10;
private Object [][] area; // grid
public Area(){ // defines a grid and robot
area = new Area[X][Y];
for(int a=0;a<X;a++){
for(int b=0;b<Y;b++)
area[a][b]=".";
}
numberOfObjects = 0; // grid is initially empty
Area ar = new Area();
robot = new Robot(ar);
}
public void Put(int x,int y){ // put the object to position (x,y)
area[x][y]=numberOfObjects++;
}
public void PickUp(int x,int y){ // pick up the object in position (x,y)
if(area[x][y]!=null){
area[x][y]=numberOfObjects--;
}
}
public void PrintAGrid(){
for(int r=0;r<X;r++){
for(int c=0;c<Y;c++)
System.out.print(area[r][c]+" ");
System.out.println();
}
System.out.println();
}
}
How can I put the robot in position (10,1)? How can I declare and set its orientation (i.e. on the right)? I guess it will be easy to write other methods, so I do not focus on it.
There are several issues with your code.
Why do you have an instance of Robot inside the class Robot? You have not used that instance at all!
private Object [][] area; should be int[][] area. You always save int in this, right?
If I understand your requirements correctly, Your implementation of pick and put is not correct.
Here is a help how you can solve the problems. I had to think several times if Robot should be in Grid or it should be the other way. I ended up with Grid in Robot.
May be Grid could be a singleton.
Here is our Grid
public class Grid {
private int[][] numberOfObjects = new int[10][10];
public void put(int x, int y) {
numberOfObjects[y][x]++;
}
public void pick(int x, int y) {
numberOfObjects[y][x]--;
}
}
You can replace parameters int x, int y with a Point.
And here is the robot
public class Robot {
private static final int NORTH = 0, EAST = 1, SOUTH = 2, WEST = 3;
private int direction;
private int x, y;
private Grid grid;
public Robot(Grid grid) {
this.x = 0;
this.y = 0;
this.grid = grid;
direction = NORTH;
}
public void right() {
direction++;
if (direction == 4) {
direction = 0;
}
}
public void left() {
direction--;
if (direction == -1) {
direction = 3;
}
}
public void forward() {
if (direction == NORTH) {
y--;
} else if (direction == SOUTH) {
y++;
} else if (direction == EAST) {
x++;
} else if (direction == WEST) {
x--;
}
}
public void put() {
grid.put(x, y);
}
public void pick() {
grid.pick(x, y);
}
}
You need to represent the curent location with a variable and initialize it to the 10 1 postion, though your array goes 0-9 and 0-9 so this may be 9,0. to store this position maybe try a Point object that contains a Point x,y.
If someone is interested in a JavaScript version, you can have a look at this repo right here. In general:
The robot must have a facing direction (left, up, right, down).
The are three possible commands: left, right, move.
With that being said, the algorithm is quite straightforward:
totalScore = 0
Foreach i in input
computeCurrentDirection()
if input != MOVE: continue
totalScore += i
return totalScore
There are several sweet-tricks that someone might do to optimize the functions. Take a look at switchDirection.
const directionArray = [Directions.RIGHT, Directions.DOWN, Directions.LEFT, Directions.UP];
const switchDirection = (currDirection, command) => {
if (command === Commands.MOVE) {
return currDirection
}
const currDirectionIndex = directionArray.indexOf(currDirection);
if (command === Commands.RIGHT) {
return directionArray[(currDirectionIndex + 1) % 4];
}
return directionArray[((currDirectionIndex - 1) + 4) % 4];
}
Instead of an exhaustive approach, someone might use an array to help compute the upcoming direction of the robot. This significantly reduces the amount of needed code.
Note this implementation can be easily expanded to accommodate any new requirements needed for project expansion. When faced with such questions, try to architect your codebase in a testable and expandable way, because it's usually the case where reviewers are interested in your coding organizational skills, rather than whether you are able to solve the problem or not.

Java Tetris rotation

I know this has been asked a lot but I'd like to know how to rotate a Tetris piece?
I already made a long and bad solution (~170 lines of code) but there should be easier way to do it.
My tetris pieces consist of 4 blocks which all know their location (row and column) in the matrix. Matrix itself is char-typed, so 4 blocks are all letters. It looks like this for example:
......
..T...
.TTT..
......
I tried to simulate my matrix as coordinate system by counting the middle row and column and using it as an origo and then tried to apply this simple algorithm I found:
90 degree rotation (x,y) = (-y,x)
It appears that it works only if my piece is in the center of matrix. I have no idea what I should do, I've been thinking this all day. Here's my method:
public void rotatePiece(ArrayList<Block> random) {
int distance = 0; // how far is the origo
for (int i=0; i < 4; ++i)
board[random.get(i).getRow()][random.get(i).getColumn()] = '.'; // erases the current location of the piece
for (int i=0; i < 4; ++i) {
distance = Math.abs(random.get(i).getColumn()-middleColumn);
if (random.get(i).getColumn() < middleColumn)
random.get(i).setColumn(random.get(i).getColumn()+(distance*2)); // "throws" the location of the block to the other side of the origo
else
random.get(i).setColumn(random.get(i).getColumn()-(distance*2));
int help = random.get(i).getColumn();
random.get(i).setColumn(random.get(i).getRow()); // (x, y) = (-y, x)
random.get(i).setRow(help);
}
for (int i=0; i < 4; ++i)
board[random.get(i).getRow()][random.get(i).getColumn()] = random.get(0).getStyle(); // saves the new location of the piece in the matrix
I would recommend defining four states for each block-group.
enum ROTATION {
UP, DOWN, LEFT, RIGHT;
ROTATION rotateLeft() {
switch(this) {
case UP: return LEFT;
case LEFT: return DOWN;
case DOWN: return RIGHT;
case RIGHT: return UP;
}
return null; // wont happen;
}
ROTATION rotateRight() {
ROTATION r = this;
// wow I'm lazy, but I've actually seen this in production code!
return r.rotateLeft().rotateLeft().rotateLeft();
}
}
abstract class Brick {
Point centerPos;
ROTATION rot;
abstract List<Point> pointsOccupied();
}
class TBrick extends Brick {
List<Point> pointsOccupied() {
int x = centerPos.x();
int y = centerPos.y();
List<Point> points = new LinkedList<Point>();
switch(rot) {
case UP: points.add(new Point(x-1,y);
points.add(new Point(x,y);
points.add(new Point(x+1,y);
points.add(new Point(x, y+1);
break;
case Down: points.add(new Point(x-1,y);
points.add(new Point(x,y);
points.add(new Point(x+1,y);
points.add(new Point(x, y-1);
break;
// finish the cases
}
}
}
You can use a rotation matrix.
You will need to set the origin of your rotation appropriately, which may mean translating the location of the piece with respect to the playing field (such that the origin is in the centre, for example), applying the rotation matrix and then translating it back to its correct location on the playing field coordinates.
The easiest and computational fastest way to do this, would to use precompute them.
That means a tetris piece will look like
class TetrisBlock {
String position[4];
int curPos = 0;
void rotateLeft() {
curPos++;
if (curPos > 3)
curPos = 0;
}
....
}
And then you could define something like
class TetrisTBlock extends TetrisBlock {
...
// in constructor
position={"....\n.T..\nTTT.\n....",
".T..\nTT..\n.T..\n.....",
// I guess you get the idea
...
You do this for every type of block and then you can also add members for adding/removing them from the board.
If you optimize you would go away from the chars....
I think the best way is to hard-code it. Take into consideration that each figure is different and each of the figure's rotation phase is also different. And for each rotation phase - determine which parts of the grid you need to be free (avoid collision).
For a visual representation check this

Categories