I began developing a "Spaceship vs Comets" style game last week and now I have come to a stop.
The purpose of the game is to shoot the comets before they pass your ship. You make the comets explode by firing at them. Simple idea!
However, sometimes when I play I get the "IndexOutOfBounds" error. It almost always appears when I haven't fired for a while (The size of my shots ArrayList is 0) and when I then fire and it collides it crashes.
So I have some kind of error in my code, but I really can't see it. I now hope that one of you might see why this accurs and save me from further "IndexOutOfBounds" errors! :)
Here is the part of the code that fails, including the functions that I use to move the comets and shots:
GAME CLASS
if(!Game.player.getShots().isEmpty() && !comet.getComets().isEmpty()) { //Om de är tomma så ignorera
for(int x = 0; x < Game.player.getShots().size(); x++) { //Shots X
if(!comet.getComets().isEmpty() && !comet.getComets().isEmpty()) {
for(int y = 0; y < comet.getComets().size(); y++) { //Comets Y
if(comet.getComets().get(y).intersects(Game.player.getShots().get(x)) && !comet.getComets().isEmpty() && !Game.player.getShots().isEmpty()) {
//the for loop above is the line that won't compile sometimes
comet.getComets().remove(y);
Game.player.getShots().remove(x);
score++;
}
}
}
}
}
//Comet spawn timer
comet.addComets();
//Move the comets and shots!
Game.player.moveShots();
comet.moveComets();
repaint();
COMET CLASS
public ArrayList<Rectangle> getComets() {
return comets;
}
public void moveComets() {
if(!comets.isEmpty()) {
for(int x = 0; x < comets.size(); x++) {
comets.get(x).x -= cometSpeed;
}
}
}
PLAYER CLASS (Shots are in this class)
public void fire() {
shots.add(new Rectangle(x + player.width, y + 23, shotWidth,shotHeight));
}
public ArrayList<Rectangle> getShots() {
return shots;
}
public void moveShots() {
if(!shots.isEmpty()) {
for(int x = 0; x < shots.size(); x++) {
shots.get(x).x += fireSpeed;
}
}
}
Keep in mind that comets and shots are both "ArrayList out of the object Rectangle"
I will provide screenshots of error and a picture of the game below!
The error line is marked in the code above, the if statement should block it from crashing (I thought).
Thanks in advance! All help is appreciated! :)
You should change the order at the if statement to avoid it from evaluate one part of it.
You should change your condition to:
if( x < Game.player.getShots().size() && comet.getComets().get(y).intersects(Game.player.getShots().get(x))) {
That's because you're removing a shot and inside the comets for when the shoot is removed at the next comet iteration it will throw IndexOutOfBounds as the array no longer have the shot you're checking at the if, so you'll need to check again for the x existing on shots.
You can also do it at the for, you check for both conditions and you let the intersect as only check at the if.
A better performance if would be:
if(!Game.player.getShots().isEmpty() || !comet.getComets().isEmpty()) {
//if one of them is empty, won't be intersections
for(int x = 0; x < Game.player.getShots().size(); x++) { //Shots X
for(int y = 0; y < comet.getComets().size() && x < Game.player.getShots().size(); y++) {
//Comets Y only if the shoot still available
if(comet.getComets().get(y).intersects(Game.player.getShots().get(x))) {
//the for loop above is the line that won't compile sometimes
comet.getComets().remove(y);
Game.player.getShots().remove(x);
score++;
y = 0; // if you don't set the y = 0 the next shoot (as you removed the x, getShots.get(x) would be the x + 1 shoot) will only evaluate for the comets after y, won't evaluate the firsts comets at the array.
}
}
}
}
Try to adjust here if(!comet.getComets().isEmpty() && !comet.getComets().isEmpty()). You are checking for comet array twice.
Let me know if it works.
Is this code multithreaded?
I am also wondering why you are verifying comet.getComets().isEmpty() so many times.
My guess is that you are manipulating the ArrayList in some other part of your code. Reason I think this is because you are checking the size of the list multiple times and because within the for loop you are only deleting in the end, so that should not be the problem.
For instance if you run this method at two threads at the same time, the ArrayList can be checked at one point, but can reduce after checking the size. then when for instance the size was 10, but became 9, but you still try to delete x with value 10, you will get the out of bounds error.
I'm having hard time coding this ... I want to find a certain combination of numbers in a 2D grid. This combination is already known and is stored in an ArrayList. The catch is that the combination can be completed from ANY of the 8 cardinal directions (stored as ints from 0 to 7) like a word search puzzle. I'm trying to find that direction but honestly I'm not sure where to go from here ...
[...] //This is only a part of the bigger code I'm working on
for (i = 0; i < grid.length; ++i){
for (j = 0; j < grid[i].length; ++j){
if (grid[i][j] == digits.get(0){ //Here I find the 1st digit of my number
}
}
}
Keep in mind that the number entered can be of ANY length and that the direction is only returned IF I get the whole number. I'm so lost literally any advice is appreciated, thanks.
for (i = 0; i < grid.length; ++i){
for (j = 0; j < grid[i].length; ++j){
int gridItemIndex = digits.indexOf( grid[i][j] )
{
if( gridItemIndex != -1 )
{
// now you are sure grid[i][j] IS IN LIST
// gridItemIndex tells THE INDEX NUMBER OF grid[i][j] in ArrayList(digits)
}
}
}
I hope this gets to close to what you are trying to say??
int gridItemIndex = digits.indexOf( grid[i][j] )
{
if( gridItemIndex != -1 )
{
// now you are sure grid[i][j] IS IN LIST
// gridItemIndex tells THE INDEX NUMBER OF grid[i][j] in ArrayList(digits)
}
}
Deals with the case where you are trying to locate the existence and index number of grid[i][j] in digits ArrayList, And please don't -vefy my answer before not even discussing why and the end result of your expected combination is a bit blurry still, would be glad to help
I think the problem here is that you don't see the bigger picture. The below is your starting point, cleaned up a bit.
Now what I would do is write a method that checks the table in every direction for the length of your combination. Checking for north is basically doing grid[startingX][startingY--]. If you apply that logic to every direction and account for 'out of bounds'-errors, you will solve it in no time.
//2D array
int[][] grid = new int[25][25];
//Your combination
List<Integer> combination = new ArrayList<>();
//Loop over the 2D array to find your starting point
for (int i = 0; i<25; i++) {
for (int j = 0; j<25; j++) {
if (grid[i][j] == combination.get(0)) { //If we find the first element of your combination
//Check in all cardinal directions
checkDirections(i, j);
}
}
}
To this question:
The superqueen is a chess piece that can move like a queen, but also like a knight. What is the maximal number of superqueens on an 8X8 chessboard such that no one can capture an other?
I want to write a brute force algorithm to find the maximum. Here's what I wrote:
public class Main {
public static boolean chess[][];
public static void main(String[] args) throws java.lang.Exception {
chess = new boolean[8][8];
chess[0][0] = true;
for (int i = 0; i < 8; i++) {
for (int j = 0; j < 8; j++) {
/*Loop to check various possibilities*/
if (!checkrow(i) && !checkcolumn(j) && !checkdiagonals(i, j) && !checkknight(i, j)) {
if (i != 0 || j != 0) {
chess[i][j] = true;
}
}
}
}/*printing the array*/
for (int i = 0; i < 8; i++) {
for (int j = 0; j < 8; j++) {
System.out.print(((chess[i][j]) ? "T" : "x") + "|");
}
System.out.println();
}
}
/*All working fine here*/
public static boolean checkrow(int a) {
for (int i = 0; i < 8; i++) {
if (chess[a][i]) {
return true;
}
}
return false;
}
/*All working fine here*/
public static boolean checkcolumn(int a) {
for (int i = 0; i < 8; i++) {
if (chess[i][a]) {
return true;
}
}
return false;
}
/*All working fine here*/
public static boolean checkdiagonals(int pi, int pj) {
int i = pi - Math.min(pi, pj);
int j = pj - Math.min(pi, pj);
for (int k = i, l = j; k < 8 && l < 8; k++, l++) {
if (chess[k][l]) {
return true;
}
}
int i_2 = pi - Math.min(pi, pj);
int j_2 = pj + Math.min(pi, pj);
for (int k = i_2, l = j_2; k < 8 && l > 1; k++, l--) {
if (chess[k][l]) {
return true;
}
}
return false;
}
/*Not All working fine here try commenting out this method above so that that it doesn't run during the check*/
public static boolean checkknight(int pi, int pj) {
for (int i = -1; i <= 1; i++) {
for (int j = -1; j <= 1; j++) {
if (0 <= pi + 2 * i && pi + 2 * i <= 8 && 0 <= pj + j && pj + j <= 8) {
if (chess[pi + 2 * i][pj + j]) {
return true;
}
}
if (0 <= pi + i && pi + i <= 8 && 0 <= pj + 2 * j && pj + 2 * j <= 8) {
if (chess[pi + i][pj + 2 * i]) {
return true;
}
}
}
}
return false;
}
}
I have two questions:
My algorithm for checkknight looks for all knight positions, is it wrong? or there is some coding error.Everything is working fine when I comment out it and I get a nice solution.
Secondly it'll result only in one solution.For other solutions I have to offset(or change position) of other pieces bit by bit after each mega-loop of this, I am confused about implementing it. My instincts guide me that I need to change whole of the code. Is there a modification or a way to do it?
Additional Thoughts: I think we would add to a counter each time we place a piece and add to a long array and output the maximum and array after storing the relevant data.
Code Location: You may view/edit/fork/download it at http://ideone.com/gChD8a
This a rough brute-force method starting from the opposite direction, i.e. from the solved eight-queens puzzle. This will allow us to find a bunch of viable solutions.
The brute-force technique for going from a single superqueen to potentially 8 seems to be especially complex due to the knight's traversal. Based on the runs, about 60% of the viable paths for normal queens are invalid with superqueens. So if we were to instead brute force normal queens, and then work backwards, that is potential time saved for finding a solution, and we can better determine the run-time. Because we know normal queens is easier.
We start off with the 12 fundamental solutions, we would then use these as inputs. Solving normal queens is outside this, but the wiki page has a fantastic article describing everything.
In my case, I stored them as Strings representing the coordinate of the queen (the rows are indices).
So: "17468253" = A1, B7, C4, D6, E8, F2, G5, H3
By brute-forcing the opposite direction from solved queens, we only have to test at most 12 x 8! possible solutions. Because order doesn't matter, additional optimization could occur by eliminating duplicate boards and solutions for processing.
First up, checkKnight, which appears to be your source of confusion. Using absolute values, you can reasonably determine whether or not a piece is within knight's range by checking whether the X offset is 2 and Y offset is 1, or vice versa. You've made a complex checkKnight function to check each individual location and whether or not a piece is on the border. Working the other way by hitscanning each queen to each other queen is logically simpler and less of a nightmare to debug.
Queen class
public class Queen {
int i, j;
public Queen(int i, int j) {
this.i = i;
this.j = j;
}
public boolean checkKnight(Queen queen) { // if any queen meets another
// queen at 2 and 1 offset, we
// eliminate it.
return (Math.abs(i - queen.i) == 2 && Math.abs(j - queen.j) == 1)
|| (Math.abs(i - queen.i) == 1 && Math.abs(j - queen.j) == 2);
}
}
This board has been modified since I originally posted. It takes a String input and converts it to a full chessboard. It has some minor work towards the potential any-size board, but right now it handles child board creation. When a child board is created, the queens are passed by reference rather than making a whole new set of queens. A total of 96 queens are stored in memory, 1 for each one on the original 12-board solution. Not perfectly optimized, but better than 96 -> 672 -> 4032 -> ...
Board class
public class Board {
static int boardSize = 8;
ArrayList<Queen> queens = new ArrayList<Queen>();
public Board(String s) {
for (int i = 0; i < s.length(); i++) {
queens.add(new Queen(i, s.charAt(i) - 49)); // you could implement
// base 16 here, for
// example, for a 15x15
// board
}
}
public Board(Board b) { // duplicates the board, but keeps references to
// queens to conserve memory, only 96 total queens
// in existence through search!
for (Queen q : b.queens) {
queens.add(q);
}
}
public boolean checkForImpact() {
for (int i = 0; i < queens.size(); i++) {
for (int j = i + 1; j < queens.size(); j++) {
if (queens.get(i).checkKnight(queens.get(j))) { // just check
// for any
// queens
// intersecting,
// one hit is
// enough
return true;
}
}
}
return false;
}
public ArrayList<Board> getChildBoards() { // create child boards with a
// single queen removed
ArrayList<Board> boards = new ArrayList<Board>();
for (int i = 0; i < queens.size(); i++) {
boards.add(new Board(this));
}
int i = 0;
for (Board b : boards) {
b.queens.remove(i);
i++;
}
return boards;
}
public String drawBoard() {
String s = "";
char[][] printableBoard = new char[boardSize][boardSize];
for (int i = 0; i < 8; i++) {
for (int j = 0; j < 8; j++) {
printableBoard[i][j] = '_';
}
}
for (Queen q : queens) {
printableBoard[q.i][q.j] = 'Q';
}
s += " A B C D E F G H\n";
for (int i = 0; i < 8; i++) {
s += (8 - i) + "|";
for (int j = 0; j < boardSize; j++) {
s += printableBoard[i][j];
s += "|";
}
s += "\n";
}
return s;
}
}
Test class
import java.util.ArrayList;
public class Test {
static String[] boards = { "24683175", "17468253", "17582463", "41582736",
"51842736", "31758246", "51468273", "71386425", "51863724",
"57142863", "63184275", "53172864" }; // all 12 solutions for the 8
// queens problem
static ArrayList<Board> boardObjects = new ArrayList<Board>();
public static void main(String[] args) {
for (String queens : boards) { // create starter boards
boardObjects.add(new Board(queens));
}
int i;
ArrayList<Board> foundBoards = null;
for (i = 8; i > 0; i--) {
ArrayList<Board> newBoards = new ArrayList<Board>();
foundBoards = new ArrayList<Board>();
for (Board b : boardObjects) {
if (b.checkForImpact()) { // if any queen intercepts we get
// children
ArrayList<Board> boardsToBeAdded = b.getChildBoards(); // pass
// all
// permutations
// of
// queens
// once
// removed
for (Board bo : boardsToBeAdded) {
newBoards.add(bo); // add it in to the next list
}
} else {
foundBoards.add(b); // if we have no impact, we have a
// solution
}
}
if (!foundBoards.isEmpty())
break;
boardObjects.clear();
boardObjects = newBoards;
}
System.out.println("The maximum number of super-queens is: " + i);
ArrayList<String> winningCombinations = new ArrayList<String>();
for (Board board : foundBoards) {
String createdBoard = board.drawBoard();
boolean found = false;
for (String storedBoard : winningCombinations) {
if (storedBoard.equals(createdBoard))
found = true;
}
if (!found)
winningCombinations.add(createdBoard);
}
for (String board : winningCombinations) {
System.out.println(board);
}
}
}
The end output is:
The maximum number of super-queens is: 6
A B C D E F G H
8|Q|_|_|_|_|_|_|_|
7|_|_|_|_|_|_|Q|_|
6|_|_|_|Q|_|_|_|_|
5|_|_|_|_|_|_|_|_|
4|_|_|_|_|_|_|_|Q|
3|_|Q|_|_|_|_|_|_|
2|_|_|_|_|Q|_|_|_|
1|_|_|_|_|_|_|_|_|
A B C D E F G H
8|Q|_|_|_|_|_|_|_|
7|_|_|_|_|_|_|_|_|
6|_|_|_|_|Q|_|_|_|
5|_|_|_|_|_|_|_|Q|
4|_|Q|_|_|_|_|_|_|
3|_|_|_|_|_|_|_|_|
2|_|_|_|_|_|Q|_|_|
1|_|_|Q|_|_|_|_|_|
A B C D E F G H
8|_|_|_|_|Q|_|_|_|
7|Q|_|_|_|_|_|_|_|
6|_|_|_|_|_|_|_|Q|
5|_|_|_|Q|_|_|_|_|
4|_|_|_|_|_|_|_|_|
3|_|_|_|_|_|_|_|_|
2|_|_|Q|_|_|_|_|_|
1|_|_|_|_|_|Q|_|_|
A B C D E F G H
8|_|_|_|_|Q|_|_|_|
7|Q|_|_|_|_|_|_|_|
6|_|_|_|_|_|_|_|Q|
5|_|_|_|Q|_|_|_|_|
4|_|_|_|_|_|_|_|_|
3|_|_|_|_|_|_|Q|_|
2|_|_|Q|_|_|_|_|_|
1|_|_|_|_|_|_|_|_|
A B C D E F G H
8|_|_|_|_|Q|_|_|_|
7|Q|_|_|_|_|_|_|_|
6|_|_|_|_|_|_|_|Q|
5|_|_|_|_|_|_|_|_|
4|_|_|Q|_|_|_|_|_|
3|_|_|_|_|_|_|Q|_|
2|_|_|_|_|_|_|_|_|
1|_|_|_|Q|_|_|_|_|
I've removed the duplicates and made a nice board printing method. don't remember the exact math, but this highlights 40 possible locations. There are others, just by looking, but we've found a fair chunk of them already! From here, we can gently shift individual queens around. From a cursory look, each board has a single piece that can be moved to 3 additional spaces, so now we know there are probably about 160 solutions.
Conclusions
With this application, the run-time on my machine was less than a second, meaning that if we attached this to a standard queens application, the additional knight's brute-forcing would have no impact on that process and have almost the same run-time. In addition, because only 6-piece puzzles are possible, we know that your eventual application run will finish its finding at the 6th piece being placed, as no more solutions are possible, since there are no viable 7-piece and 8-piece solutions.
In other words, finding the maximum super-queen layout is likely actually shorter than the maximum queen layout due to the additional restrictions!
Trying to brute-force such a question is a good way to get a feel for it. So I won't suggest looking up pre-cooked solutions at first.
One little remark though: I don't see the reason for the condition if (i != 0 || j != 0) { that you have there. You are working on Java arrays. Instead of 1 through 8, they go 0 through 7, but the 0 is the first column, you should not eliminate it, otherwise it's only a 7x7 board.
First, let me address your technical question: how to calculate the knight positions.
Take a sheet of quad paper, put a queen somewhere not less than two squares away from the edge. Then mark the end positions of a knight-move from it.
You'll end up with just 8 squares that need to be considered. There is no point in doing a 3x3 loop to find them. A better idea would be to prepare a static array with the relative coordinates of the knight moves - an array of 8 pairs of numbers - and loop on that. So you have only an 8-step loop. In each step of the loop, check for bounds (0 ≤ X + Xoffset < 8, 0 ≤ Y + Yoffset < 8 ), and you have the knight coordinates.
Second, there is no point checking the part of the board that's ahead of you. Since you have not covered the next row and those below it, there is no point in looking for queens there. The implications of this:
You'll never put another queen in the same row where you have just marked a queen position (because you threaten it horizontally). This means that if you mark a queen, you should use continue to break out of the inner loop to the next row.
You don't need checkrow(). When you start a row, there is no queen ahead of you. And if you followed the above bullet point, there is no queen on your back, either.
When you use checkcolumn, you start at row 0, but you can finish at the row before the one you are on (i-1). There are still no queens in the rows below you! The same is true for the diagonal checks.
Earlier I said that you need to prepare and check 8 knight positions. But now you know there is no queen at the knight positions ahead of you. So you only need to prepare an array with four knight positions - the ones above your position.
But most importantly... once you have finished and you have your queens in positions and print the board: you have a single solved board. You have proved that this number of queens is possible. But is it the highest number possible? You have not checked what happens if you don't put a queen on the first square of the first row, but on the second. Perhaps this will allow you to put in an extra queen later. And what about the queen in the second row? Maybe if you moved that, you would be able to put a queen somewhere below where you couldn't before?
So, now you have to actually do the same thing over again, changing one decision every time and working from there. In effect, you have many potential boards. Why? Because there may be more than one valid position on each row where you put that row's queen. So you have decided to put it in the first valid position. But what if you decide to put it in the second valid position? Or leave that row empty? Each such decision is followed by another set of decisions on the next row.
The different boards created by different decisions form a decision tree. The problem for you to consider, therefore, is how to work such a tree out. How to write your decision trail and then backtrack, change, fill another board and count the queens at each level. People here suggested recursion, because it lends itself nicely to such problems. Or you can keep a stack of decisions if you want. You can eliminate some of the potential boards based on symmetries.
I suggest you first make sure you understand your single board well, and then consider how to represent your decision tree and how to traverse it.
There are several questions here.
The first is: how many knight-queens can be placed on an nxn chessboard? Since a k-piece solution can trivially be reduced to a k-1 piece solution, it makes sense to start from the upper bound. That is, look for an n-piece solution, if that fails look for an n-1 piece solution, and so forth.
The second question is: how should I look for a k-piece solution? There are two classic strategies: depth-first and breadth-first. In the former, you consider one vertex of the search tree at a time and use backtracking on failure. In the latter, you consider one complete level of the search tree at a time.
Something that can make a great deal of difference to your search is to account for symmetry (in this case, rotations and reflections).
The third (implicit) question is: what is a good representation here? If your chess-boards are less than 8x8 in size then a 64-bit bit-pattern will do very nicely!
In practical terms, try to separate the three levels of your problem as far as you can. If you don't, you'll find that a choice in one level will severely limit your options at another level.
I've made a simple game, where you are the box and you are shooting little boxes.
Now to check if the shoot tocuhed the monster, i check if its x is highter or equals to the monster's x, and then i check if its less or equals to the monsters x, and then the same for y.
But the thing is, I have two for loops for this, therefore when I shoot and it touched the monster, the monster will not always get deleted cause the for loop did not reach that monster yet, and im sure it's bad for performance.
IS there a faster way to do this? why is it happening?
private void checkKilled() {
for (int i = 0; i < this.shoots.size(); i++) {
for (int j = 0; j < this.monsters.size(); j++) {
Monster m = this.monsters.get(j);
Shoot s = this.shoots.get(i);
if (s.getX() >= m.getX() && s.getX() <= m.getX() + 15
&& s.getY() >= m.getY() && s.getY() <= m.getY() + 15) {
m.isDead = true;
s.isExploded = true;
}
}
}
}
Not fully understanding the problem, but for performance you can break from inner loop if condition is true; it will then go to next shoot.
...
s.isExploded = true;
break;
}
I'm working on this bacteria life game thing I have to make.
Basically I have a 2d string array let's say 20 by 20.
What would be the best way to check all 8 spots around a certain index. Each index is suppose to represent a bacteria. For each bacteria(index) I have to check to see if any of the 8 spots around this index has another bacteria in it, if the index has a bacteria in it, it's represented simply by a "*", asterik.
What would be the best way to go about checking all 8 spots around each index, because based on what is in the indices around a certain index I have to make certain changes etc.
The only idea I have come up with is having a bunch of if statements to check all 8 spots, I was wondering if there is a better way to do this
ex:
row 1 - www , row 2 = wOw , row 3 - www ,
if I am at the O index, what would be the best way to check all the index spots around it for a certain string.
Sorry, I am not very good at explaining my problems, bad english :o.
thanks for any of the help.
so you have something like this
char[][] table = new char[20][20]
for(int i = 0; i < 20; i++) {
for(int j = 0; j < 20; j++) {
int surroundingBacteria = 0;
for(int x = max(i-1,0); x < min(20,i+1); x++) {
for(int y = max(i-1,0); y < min(20,i+1); y++) {
if(table[x][y] == '*') surroundingBacteria++;
}
}
switch(surroundingBacteria) {
// put your case logic here
}
}
}
Here is how I've accomplished this in the past:
for(int x = -1; x<=1; x++){
if ( i+x < xLength && i+x >= 0){
for(int y = -1; y<=1; y++){
if(j+y < yLength && j+y >= 0){
//logic goes here
}
}
}
}