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I wanted to make an algorithm to create a maze using recursive back-tracker algorithm. I get this problem when running the program:
Exception in thread "main" java.util.NoSuchElementException
at java.base/java.util.Vector.firstElement(Vector.java:481)
at DC_files.Maze.RBT(MazeAlgorithm.java:131)
at DC_files.MazeAlgorithm.main(MazeAlgorithm.java:222)
and honestly, at this point after finding couple of bugs and error i have no idea what might be wrong. The only clue that i have, is that in the switch there is an error, that the stack.firstElement() doesn't update and is always the same and in the cases something is being inserted wrongly.
while(vis < toVis){
Vector<Integer> neighbours = new Vector<>();
int x = stack.firstElement().getKey(); (line 131 with the error.)
int y = stack.firstElement().getValue();
//North neighbour
if(y-1 >= 0 && !matrix[y - 1][x].vis){
neighbours.add(0);
}
entire code
package DC_files;
import javafx.util.Pair;
import java.util.Random;
import java.io.*;
import java.util.*;
class Cell{
/*
each Cell holds information of:
-the state of it's walls,
-the value that the Cell holds(will be used later to customize rooms in the maze),
-and information if the Cell was visited(will be used in the algorithm)
*/
boolean[] walls;
boolean vis;
int value;
public boolean isVis() {
return vis;
}
public boolean[] getWalls() {
return walls;
}
public int getValue() {
return value;
}
/*
by default, each Cell has the value of each wall set to true(they exist/there are no corridors),
vis as visited set to false, because we weren't in the room yet,
value set to zero, for further development phase (0 - not visited, 1 - visited, 2 - ??? (...))
*/
Cell(){
walls = new boolean[]{true, true, true, true}; // {N, S, E, W}
vis = false;
value = 0;
}
}
class Maze{
/*
Maze class was created in order to generate an array of Cells,
that will be later used in the algorithm.
First we create integer ,,size,, to store size of the array, then
we create a matrix, that will hold Cell objects.
In the constructor of the class, we set the required integer s as size,
allocate memory to the matrix, by setting its dimensions to size x size
and we create Cell objects for each spot int the matrix, without any modifications.
*/
int size;
Cell[][] matrix;
Maze(int s){
size = s;
matrix = new Cell[size][size];
for(int y = 0; y < size; y++){
for(int x = 0; x < size; x++){
matrix[y][x] = new Cell();
//System.out.print("1");
}
}
}
void showMaze(){
for(int y = 0; y < size; y++){
for(int x = 0; x < size; x++){
System.out.print(matrix[y][x].getValue() + " ");
}
System.out.println();
}
}
/*
Class ,,MazeAlgorithm'' is responsible for creating connections between cells in matrix.
It uses RBT, which stands for Recursive Back-Tracker, to achieve that goal.
In order to use RBT, we need to give it starting point. After we do that,
it will check, if there is any neighbour, that fulfills the required conditions,
and then goes to that neighbour setting visited to true and value to 1.
After that, it repeats the process, and constantly adds the cells coordinates
on top of the stack, to keep track of the road it went through. If it gets stuck,
it will go back to the previous cell, by removing the top coordinates of the stack,
and going to NOW top coordinates (previous move), checking if that cell has any
neighbours which can fulfill the conditions. If it cannot find any neighbour,
it means the maze was created.
*/
void RBT(int startX, int startY){
//Setting cells to visit to size x size, and the value of visited cells to 0
int toVis = size*size;
int vis = 0;
//Declaring a stack, that will hold the information about our road
Stack<Pair<Integer, Integer>> stack = new Stack<>();
//Setting starting position: pushing it onto the stack,
//setting it's values as visited and 1,
//incrementing visited rooms.
stack.push(new Pair<>(startX, startY));
matrix[startY][startX].vis = true;
matrix[startY][startX].value = 1;
vis += 1;
//we will check, if our current node has any neighbours we can go to,
//saving those neighbours inside a vector
while(vis < toVis){
Vector<Integer> neighbours = new Vector<>();
int x = stack.firstElement().getKey();
int y = stack.firstElement().getValue();
//North neighbour
if(y-1 >= 0 && !matrix[y - 1][x].vis){
neighbours.add(0);
}
//South neighbour
if(y+1 < size && !matrix[y + 1][x].vis){
neighbours.add(1);
}
//East neighbour
if(x+1 < size && !matrix[y][x + 1].vis){
neighbours.add(2);
}
//West neighbour
if(x-1 >= 0 && !matrix[y][x - 1].vis){
neighbours.add(3);
}
//checking, if there are any neighbours we can visit
//if yes, we do our job
//if not, we pop our stack and repeat the process
if(!neighbours.isEmpty()){
Random rand = new Random();
int randDir = neighbours.get(rand.nextInt(neighbours.size()));
switch (randDir) {
//North
case 0 -> {
matrix[y][x].walls[0] = false;
stack.push(new Pair<>(x, y - 1));
matrix[y - 1][x].value = 1;
matrix[y - 1][x].vis = true;
matrix[y - 1][x].walls[1] = false;
}
//South
case 1 -> {
matrix[y][x].walls[1] = false;
stack.push(new Pair<>(x, y + 1));
matrix[y + 1][x].value = 1;
matrix[y + 1][x].vis = true;
matrix[y + 1][x].walls[0] = false;
}
//East
case 2 -> {
matrix[y][x].walls[2] = false;
stack.push(new Pair<>(x + 1, y));
matrix[y][x + 1].value = 1;
matrix[y][x + 1].vis = true;
matrix[y][x + 1].walls[3] = false;
}
//West
case 3 -> {
matrix[y][x].walls[3] = false;
stack.push(new Pair<>(x - 1, y));
matrix[y][x - 1].value = 1;
matrix[y][x - 1].vis = true;
matrix[y][x - 1].walls[2] = false;
}
}
vis += 1;
}else{
stack.pop();
}
}
}
}
public class MazeAlgorithm {
public static void main(String[] args){
Maze m = new Maze(3);
m.RBT(1, 1);
m.showMaze();
}
so i just run the code again, and the error line changed to line 131, i am also watching some tutorial about the Intellij debugger rn.
EDIT: Problem resolved. Had to change lines:
int x = stack.firstElement().getKey();
int y = stack.firstElement().getValue();
into:
int x = stack.peek().getKey();
int y = stack.peek().getValue();
Turns out method .firstElement() doesn't give you first element (the one on top of the stack) but the one that was first inserted (always the same object) so in the end all the operation were performed on one, single element and not multiple ones.
I'm currently trying to implement the A* (A star) - Algorithm. I already got it working when I don't have walls and when it is only needed to go besides the wall. My problem now is that when I'm placing the start inside of my walls, my algorithm is calculating infinitely as I think it doesn't go backward. Can you guys please help me?
Here is the code from the node class:
int[][] mMap; // if there is a wall => 1
int[][] mAStarField;
ArrayList<AStarNode> mAStarPath;
public class AStarNode implements Comparable<AStarNode>{
public int x;
public int y;
public float c;
public AStarNode p;
public AStarNode(int x, int y, float c, AStarNode p) {
this.x = x; //X pos
this.y = y; //Y pos
this.c = c; //Cost to get to the node
this.p = p; //Parent of the node
}
//override the compareTo method
public int compareTo(AStarNode node)
{
if (c == node.c)
return 0;
else if (c > node.c)
return 1;
else
return -1;
}
}
public class Node {
public int x;
public int y;
public int z;
public int w;
public Node(int x, int y, int z, int w) {
this.x = x;
this.y = y;
this.z = z;
this.w = w;
}
}
This is the code from my A* algorithm:
//Pathfinding with A*
//return path length
int updateAStar() {
//Needed for drawing:
//Array containing the distance to the start node (filled with max int at start)
mAStarField = new int[mMap.length][mMap[0].length];
//List containing the found path
mAStarPath = new ArrayList<AStarNode>();
for (int j = 0; j < mMap.length; j++) {
for (int k = 0; k < mMap[0].length; k++) {
mAStarField[j][k]=Integer.MAX_VALUE;
}
}
//AStarNode(x,y,c,w)
//x X pos
//y Y pos
//c Cost to get to the node
//p Parent of the node
//List can be sorted expensive but simple by c value with
//Collections.sort(openList);
ArrayList<AStarNode> openList = new ArrayList<AStarNode>();
ArrayList<AStarNode> closedList = new ArrayList<AStarNode>();
int dist = abs(mStartNode[0]-mEndNode[0])+abs(mStartNode[1]-mEndNode[1]);
//If there is any target, that isn't on my field, add start node to list
if (dist>0 && mMap[mStartNode[0]][mStartNode[1]] != 1 && mMap[mEndNode[0]][mEndNode[1]]!=1) {
openList.add(new AStarNode(mStartNode[0], mStartNode[1], 0, null));
mAStarField[mStartNode[0]][mStartNode[1]] = 0;
}
// my code begins here (only everything from here on can be edited!)
while(!openList.isEmpty())
{
Collections.sort(openList);
AStarNode current = openList.get(0);
if(current.x == mEndNode[0] && current.y == mEndNode[1])
{
return 1;
}
openList.remove(0);
closedList.add(current);
ArrayList<AStarNode> neighbors = new ArrayList<AStarNode>();
neighbors.add(new AStarNode(current.x - 1, current.y, current.c + 1, current));
neighbors.add(new AStarNode(current.x + 1, current.y, current.c + 1, current));
neighbors.add(new AStarNode(current.x, current.y - 1, current.c + 1, current));
neighbors.add(new AStarNode(current.x, current.y + 1, current.c + 1, current));
for(AStarNode n : neighbors)
{
if(n.x >= 0 && n.y >= 0 && n.x < mMap.length && n.y < mMap.length && mMap[n.x][n.y] != 1){
float cost = estimateDistanceEnd(n.x, n.y);
n.c = cost;
if(closedList.contains(n) && cost >= n.c) continue;
if(!openList.contains(n) || cost < n.c)
{
n.p = current;
if(!openList.contains(n)){
mAStarField[n.x][n.y] = (int) n.c;
openList.add(n);
Collections.sort(openList);
}
}
}
}
}
return -1;
}
int estimateDistanceEnd(int x, int y){
return abs(x-mEndNode[0])+abs(y-mEndNode[1]);
}
int estimateDistanceStart(AStarNode a){
return abs(a.x-mStartNode[0])+abs(a.y-mStartNode[1]);
}
int estimateDistance(AStarNode a, AStarNode b){
return abs(a.x-b.x)+abs(a.y-b.y);
}
A picture of my current path solving result
Important: I'm only allowed to change code inside of the area I marked.
Thank you!
I can't run the code, but I have a little idea about the the source of the issue. See, the A* algorithm doesn't "walk back" while looking for the best path. It solves the pathfinding issue by calculating the less costly way to get to the end for literally every node that it evaluates. It starts by calculating the itineraries which are the most straightforward, then if it didn't work it'll enlarge it's options until it either, uh, finds a way - or run out of options.
The principle of the closed list to avoid evaluating a node twice. As you guessed, the problem here is that you are creating new nodes for neighbors at every iteration of the pathfinding algorithm, thus making it harder for the closed list to be used correctly.
A complex object like a custom class can be compared through 3 means: it either is the same object (it refer to the same pointer (it's the same instance, it's at the same place in the computer's memory)), or the values are all the same whatever it's pointer is pointing, or you can define a rule to compare them. These methods would be: comparing by reference, comparing by value and operator overloading - although that last one isn't possible in java, but you can write a method to do the same.
When doing closedList.contains(n), you are comparing by reference (which is the default for this kind of operation). Since all the nodes have been created on the fly, even if their coordinates are the same they all have different address in memory, and this is why this condition will never be met.
Assuming that you cannot mess with your tutor's code, you can still fix this. You almost got it right the first time! There are many ways to to fix this, in fact, and as I miss some of the context I'll be rather plain in my suggested approach: you'll write a method to fetch a specific node from a list (like the operator overloading I spoke about, but with the bare minimum effort) and we'll work by reference from this point onward.
First, create a master list of all your AStarNode (if you don't already have one, if you do then use that one instead):
// my code begins here (only everything from here on can be edited!)
ArrayList<AStarNode> nodesList = new ArrayList<AStarNode>();
for (int j = 0; j < mapWidth; j++) {
for (int k = 0; k < mapHeight; k++) {
nodesList.add(new AStarNode(j, k)); // I gimmicked the constructor for my own confort, you'll have to tweak this line so it fits in your code
}
}
Then, write yourself a method which will return a node from an array based on given xy coordinates:
AStarNode GetAStarNodeByPosition(int x, int y, ArrayList<AStarNode> list) {
for (AStarNode m : list) {
if (m.x == x && m.y == y) {
return m;
}
}
return null;
}
Now, you can use these to compare all your nodes by reference. So, now instead of instantiating new nodes all the time, you'll always fetch them from the master list, by reference:
ArrayList<AStarNode> neighbors = new ArrayList<AStarNode>();
neighbors.add(GetAStarNodeByPosition(current.x - 1, current.y, nodesList));
neighbors.add(GetAStarNodeByPosition(current.x + 1, current.y, nodesList));
neighbors.add(GetAStarNodeByPosition(current.x, current.y - 1, nodesList));
neighbors.add(GetAStarNodeByPosition(current.x, current.y + 1, nodesList));
Also, don't forget to fix this line:
//openList.add(new AStarNode(mStartNode[0], mStartNode[1], 0, null));
openList.add(GetAStarNodeByPosition(mStartNode[0], mStartNode[1], nodesList));
Lastly, always remember to test for null if you know that you may have some in your arrays. In this case, the GetAStarNodeByPosition method can return a null if you're too close to the boundaries of the maze. You can either modify the way you add to the neighbors list so there will be no null in there, or you can check for null on this line:
if(n != null && n.x >= 0 && n.y >= 0 && n.x < mMap.length && n.y < mMap.length && mMap[n.x][n.y] != 1){
Honestly I would prevent the inclusion of null in the array at all, that's much safer if you modify the code further later.
Now all your nodes will relate and you'll be able to overcome obstacles which needs your algorithm to search in a more clever way than a straight line.
Have fun!
So I'm making a program that simulates Life-like cellular automata, but I'm having some trouble with the method used to count a cell's live neighbors. The problem is that I want to be able to change how the grid wraps around -- that is, whether it wraps around from left to right (i.e., cylindrical), from top to bottom and left to right (i.e., toroidal), or not at all (i.e., flat) -- and I can't figure out how to make my method account for that. Here's what I have so far:
public int getLiveNeighbors(int row, int col)
{
int count = 0;
// "topology" is an int that represents wraparound:
// 0 = flat; 1 = cylindrical; 2 = toroidal
int top = topology != 2 ? row - 1 : (row + ROWS - 1) % ROWS;
int bottom = topology != 2 ? row + 1 : (row + 1) % ROWS;
int left = topology != 0 ? (col + COLS - 1) % COLS : col - 1;
int right = topology != 0 ? (col + 1) % COLS : col + 1;
for (int r = top; r < bottom + 1; r++)
for (int c = left; c < right + 1; c++)
if (!(r == row && c == col) && getCell(r, c).equals(LIVE))
count++;
}
The key, I think, is the if-statement in the for-loop -- there has to be some way to check whether r and c are within the bounds of the grid, while keeping in mind that the definition of "bounds" will vary depending on whether/how the grid wraps around. In the past I've gotten around this by having three different sets (one for each wraparound setting) of eight different if-statements to individually check each of the eight cells comprising the original cell's neighborhood; as you can imagine, it was not very pretty, but at least it worked.
I'm not so great at explaining my own code, so I hope that wasn't too confusing -- I'm feeling a little loopy myself (ha). If anyone has any questions, feel free to ask!
You probably already have a class like Board with a method like getCell(x, y) (at least a method of this kind is present in your code).
I'd just make this method lenient in a sense that it would accept negative x and y or x and y greater or equal to COLS and ROWS. Thus you could just iterate over col - 1 to col + 1 and row - 1 to row + 1 (minus col and row) and not care that these coordinates go "over the board". It's the task of the Board to do coordinate lookups correctly.
What makes your code harder is also that you handle different topologies in one place. It's quite hard to follow.
You could make it simpler by implementing different subclasses of Board like CylindricalBoard, ToroidalBoard and FlatBoard. Each of the subclasses would implement getCell differently, but in the context of the subclass it will be clearly understandable.
You're looking for the Strategy Pattern:
There are common situations when classes differ only in their behavior. For this cases is a good idea to isolate the algorithms in separate classes in order to have the ability to select different algorithms at runtime.
In this case you'd want something like this (abbreviated for clarity):
class Point {
int x;
int y;
}
interface WrapStrategy {
Point moveUp(Point p);
Point moveDown(Point p);
Point moveLeft(Point p);
Point moveRight(Point p);
}
class CylinderWrapping implements WrapStrategy {
int height;
int circumference;
Point moveUp(Point p) {
if (p.y <= 0)
return null; // cannot move up
return new Point(p.x, p.y - 1);
}
Point moveDown(Point p) {
if (p.y >= height - 1)
return null; // cannot move down
return new Point(p.x, p.y + 1);
}
Point moveLeft(Point p) {
if (p.x <= 0)
return new Point(circumference - 1, p.y);
return new Point(p.x - 1, p.y);
}
Point moveRight(Point p) {
if (p.x >= circumference - 1)
return new Point(0, p.y);
return new Point(p.x + 1, p.y);
}
}
Try this:
import java.awt.Point;
public class Neighbours {
public static void main(String[] args) {
Neighbours inst=new Neighbours();
int r=3;//<ROWS
int c=3;//<COLS
for(int i :new int[]{0,1,2}){
inst.type=i;
System.out.format("There are %d neighbours of point (%d,%d), topography type %d\n", inst.countLiveNeighbours(r, c), c, r,i);
}
}
int ROWS=4;
int COLS=4;
int type=0;//0=flat, 1=cylinder, 2=toroid
/**
* Is x,y a neighbour of r,c?
* #return coordinates of neighbour or null
*/
Point neighbour(int x, int y, int r, int c){
if((x==c)&&(y==r))
return null;
switch (type){
/*this is wrong for the reasons explained below
case 0: return ((x<COLS)&&(y<ROWS)) ? new Point (x,y) : null;
case 1: return y<ROWS ? new Point(x%COLS,y) : null;
case 2: return new Point(x%COLS,y%ROWS);
*/
//replacement statements produce the correct behaviour
case 0: return ((x<COLS)&&(x>-1)&&(y<ROWS)&&(y>-1)) ? new Point (x,y) : null;
case 1: return ((y<ROWS)&&(y>-1)) ? new Point(Math.floorMod(x,COLS),y) : null;
case 2: return new Point(Math.floorMod(x,COLS),Math.floorMod(y,ROWS));
}
return null;
}
int countLiveNeighbours(int r, int c){
int result=0;
for(int x=c-1; x<c+2; x++)
for(int y=r-1; y<r+2; y++){
Point p=neighbour(x,y,r,c);
if(live(p)){
System.out.format("\tpoint (%d,%d)\n",(int)p.getX(),(int)p.getY());
result++;
}
}
return result;
}
boolean live(Point p){
boolean result=true;
if(p==null)
return false;
//perform tests for liveness here and set result
return result;
}
}
There's an error in the logic of what I've build at the moment.
What should be happening is that my code should display a grid of 0's and 1's.
Like so:
001001
101101
010110
110010
001101
So what has to happen here is that:
For each row there can't be more than 2 numbers of the same type consecutively
the numbers are picked randomly
for each column there can't be more than 2 numbers of the same type consecutively
there can be a maximum of 3 of each type of number going by column or row
edit: to further clarify
ok so I have a row like this:
0 1 0 1 1 0
- As you can see there will always be 3 x 1, and 3 x 0
- the order of numbers is picked randomly (so it might go 0 1, or 1 1, or 0 0 to start etc)
- there can never be more than 2 numbers of the same type consecutively, for instance if it's 001100, you can see that there were 2 0's, then it had to display a 1, but then there were 2 1's, so it had to display an 0. So 011100 couldn't happen (3 1's consecutively) or 000101 (3 0's consecutively)
Based upon this, but for now not essential, the same no 2 numbers consecutively must apply in columns (so in my successful example it goes 001001 across, there are at most 2 0's consecutively. But looking down you get 010101 (that is to say, once again, no more than 2 consecutively)
So my code is as follows:
import java.util.Random;
public class Main {
public static void main(String[] args) {
int l = 6;
int w = 6;
Random rd = new Random();
// Create a grid that is 6 x 6
int[][] grid = new int[l][w];
// for each row
for (int i = 0; i < l; i++) {
int zCount = 0;
int oCount = 0;
int current;
int lastA = 2;
int lastB = 2;
// for each item in the row
for (int j = 0; j < w; j++) {
// set the current item to either 0 or 1
current = rd.nextInt(2);
// make sure there aren't already (e.g. 3 items out of 6)
// items in the row
if (j % 2 == 1) {
// hold every second element
lastA = current;
} else {
// hold every first element
lastB = current;
}
if (current == 1) {
if (oCount != 3) {
if (lastA != lastB) {
// if the two previous items aren't the same
grid[i][j] = current;
// add to the counter
oCount++;
}
}
}
if (current == 0) {
if (zCount != 3) {
if (lastA != lastB) {
// if the two previous items aren't the same
grid[i][j] = current;
// add to the counter
zCount++;
}
}
}
System.out.print(grid[i][j]);
}
System.out.println(" ");
}
}
}
The problem is it generates as follows:
010010
100001
100010
000010
100001
001000
So obviously it doesn't conform to the first, third or fourth points.
I have absolutely no idea why! Except for the columns (third point) which I haven't initialised.
Can anybody work out what the logical failure is in my code?
Thanks for your help!
Here is my procedural solution which tries to keep the amount of required code as small as possible. It is capable of computing 2D-Arrays with arbitrary rows and columns like [6, 6] or [4, 7] or [3, 8] for example. The complexity of the algorithm is O(n) with n = rows * columns.
The program computes an arbitrary 2D-Array (grid) populated with either a 0 or 1. The grid guarantees the following characteristics, formulated mathematically:
∀ r,c ∈ Integer | 0 ≤ r < grid.rows, 0 ≤ c < grid.columns :
r - 2 ≥ 0 ⇒ cardinality( distinct( grid[r][c], grid[r-1][c], grid[r-2][c] )) = 2
r + 2 < grid.rows ⇒ cardinality( distinct( grid[r][c], grid[r+1][c], grid[r+2][c] )) = 2
c - 2 ≥ 0 ⇒ cardinality( distinct( grid[r][c], grid[r][c-1], grid[r][c-2] )) = 2
c + 2 < grid.columns ⇒ cardinality( distinct( grid[r][c], grid[r][c+1], grid[r][c+2] )) = 2
or in other words:
the grid does neither contain a row nor a column which has three or more consecutive 0's or 1's.
Below the Java code I will explain how the algorithm works and why it is designed as it is:
public static void main(String[] args) {
int[][] grid = anyGrid(8, 13);
}
private static int[][] anyGrid(int rows, int cols) {
int[][] grid = new int[rows][cols];
int row = 0;
for (int col = 0; col - row < cols; col++) {
for (int r = row; r >= 0 && col - r < cols;) {
setBit(grid, r, col - r--);
}
if (row < rows - 1) row++;
}
return grid;
}
private static void setBit(int[][] grid, int row, int col) {
int vInd = calcVerticalIndicator(grid, row, col);
int hInd = calcHorizontalIndicator(grid, row, col);
if (isPartiallyRestricted(vInd, hInd)) {
grid[row][col] = flip(vInd);
} else if (isFullyRestricted(vInd, hInd)) {
grid[row][col] = vInd;
grid[row - 1][col] = flip(vInd);
} else {
grid[row][col] = Math.abs(vInd) <= 1
? flip(vInd)
: Math.abs(hInd) <= 1 ? flip(hInd) : anyBit();
}
}
private static boolean isPartiallyRestricted(int vInd, int hInd) {
return vInd == hInd;
}
private static boolean isFullyRestricted(int vInd, int hInd) {
return vInd + hInd == 1;
}
private static int calcVerticalIndicator(int[][] grid, int row, int col) {
return calcIndicator(grid, row - 1, col, row - 2, col, 2);
}
private static int calcHorizontalIndicator(int[][] grid, int row, int col) {
return calcIndicator(grid, row, col - 1, row, col - 2, 4);
}
private static int calcIndicator(int[][] grid, int row1, int col1, int row2, int col2, int unrestricted) {
try {
return grid[row1][col1] * grid[row2][col2] + (grid[row1][col1] - grid[row2][col2]) * unrestricted;
} catch (IndexOutOfBoundsException e) {
return unrestricted;
}
}
private static int anyBit() {
return (int) (Math.random() * 2);
}
private static int flip(int bit) {
return bit == 0 ? 1 : 0;
}
The challenge we face is not to ensure that there are no three consecutive 0's or 1's in a row only or in a column only. The challenge is to ensure that no three consecutive 0's or 1's are neither in a row nor in a column by providing an efficient algorithm.
The tricky situation we may run into looks like this:
Let's consider the situation where all the cells at the top and to the left of the cell outlined in blue are already populated and do not violate the rules define above.
picture a) we want to populate the cell having a blue outline. The two cells at it's top are populated with two 0's while the cells at it's left are populated with two 1's. Which value should we choose? Due to symmetry it doesn't matter if we choose a 0 or a 1. Hence, let's go with a 0.
picture b) populating the cell outlined in blue with a 0 violates one rule defined above: the grid does not contain a column with three or more consecutive 0's or 1's. Hence we have to change the value of one of the two cells above of the blue cell.
picture c) say we change the value of the cell which is immediately above the blue cell, from 0 to 1. This could result in the violation of some rules, caused by the already populated cells to the left of the modified cell.
picture d) but a violation would mean that both cells to the left must have a value of 1.
picture e) this would imply that both cells to their top must have a value of 0 which is a contradiction to a situation we assumed. Therefore, changing the cell immediately at the top of the cell outlined in blue will not cause any violation of the rules.
To address the precondition, that no cells to the right of the modified cell are already populated, the algorithm populates the grid in a diagonal way. The population of cells occur in the order as shown below:
The final thing I like to explain is how the algorithm decides which values are available to choose from for each cell. For each cell it inspects the two top-most and two left-most cells and calculates an indication value. This value is used to determine the possible values for a cell by using arithmetic calculation as follows:
if the two cells inspected are both populated with 0's return an indicator value of 0.
if the two cells inspected are both populated with 1's return an indicator value of 1.
I have selected those two values because they communicate the fact, that this values are not permitted, in an intuitive way.
Then I selected a function to communicate if both, the column cells and the row cells, restrict the cell to populate by the same value. This is the case if both indicator values are equal. Keep this characteristic in mind, because we have to find values for the situation when no restriction applies from the column cells or the row cells.
If both indicators restrict the value to populate the cell with by a different value, the sum of them is 1. This is the second characteristic we have to keep in mind when searching for proper indicator values when no restriction applies.
The last thing the algorithm has to achieve is to find proper values when no restriction applies without compromising the unique indicators defined above.
Preserving the indication when the cell is restricted by the same value can be achieved by selecting values for the row and column indicators which are different from 0 and 1 and different from each other.
Preserving the indication when the cell is restricted by both values can be achieved by selecting values being greater than 1 and having a delta to each other of at least 2.
The algorithm does indicate no restriction for a row by the values 2 and -2 and for a column by the values 4 and -4. This values do not conflict with the operations used to identify the other two cases.
I hope this documentation helps to understand the whole program and how it does solve the problem statement. I am glad to hear your comments.
Many of the solutions given are extremely long and complicated. Here's a solution with very minimal code (Ideone Example here):
int row, col, n = 8;
int[][] grid = new int[n][n], cCount = new int[n][2], rCount = new int[n][2];
Deque<Entry<Integer,Integer>> freeInd = new ArrayDeque<Entry<Integer,Integer>>();
Random rand=new Random();
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
// Calcualte constraints: row, col = {-1, 0, 1}, -1 => no constraint.
row = j > 1 && grid[i][j-2] == grid[i][j-1] ? (grid[i][j-1] == 0 ? 1:0):
(rCount[i][0] >= n/2 ? 1: // too many 0's
(rCount[i][1] >= n/2 ? 0:-1)); // too many 1's
col = i > 1 && grid[i-2][j] == grid[i-1][j] ? (grid[i-1][j] == 0 ? 1:0):
(cCount[j][0] >= n/2 ? 1: // too many 0's
(cCount[j][1] >= n/2 ? 0:-1)); // too many 1's
grid[i][j] = row == -1 && col == -1 ? rand.nextInt(2):(row > -1 ? row:col);
// Handle Constraints
if( row == -1 && col == -1){ // no constraint
freeInd.push(new SimpleEntry<Integer,Integer>(i, j)); // add to free indices
} else if( (row > -1 && col > -1 && row != col) // constraint conflict
|| (row > -1 && rCount[i][row] >= n/2) // count conflict
|| (col > -1 && cCount[j][col] >= n/2)){ // count conflict
Entry<Integer, Integer> last = freeInd.pop(); // grab last free index
while(i > last.getKey() || j > last.getValue()){
j = (j-1+ n)%n; // step indices back
i = (j == n-1) ? i-1:i;
rCount[i][grid[i][j]]--; // reduce counters
cCount[j][grid[i][j]]--;
}
grid[i][j] = grid[i][j] == 0 ? 1:0; // flip value
}
rCount[i][grid[i][j]]++; // increment counters
cCount[j][grid[i][j]]++;
}
}
The idea here is that you walk along each row of the matrix adding 0's and 1's abiding by the following rules:
If the current index is unconstrained (i.e. it can be 0 or 1) we choose a value randomly.
If the current index is constrained we force it to have the constrained value.
If there are multiple constraints that do not agree, we revert back to the last unconstrained index (freeInd) by first incrementally stepping backwards along the rows of the matrix, decrementing the count for the given value (0 or 1). E.g. this is done for rows with rCount[i][grid[i][j]]--. When the unconstrained vertex is finally reached, flip it's value.
Finally, increment the count of the value (0 or 1) for the current row and column. E.g. this is done for rows with rCount[i][grid[i][j]]++
The 1st problem which i found in your solution is it's initializing the value of counter value (ocount and zcount) as zero and the only way grid(array) is assigned a value is when if it's greater than three, and the way i see if i am not mistaken the value of counter is incremented in the loop in which they are checked to be greater than 3, and that condition can never be reached .
To solve this problem use the algo of backtracking by assigning the new value to a different value if the calue
A working code in jsFiddle (for 6x6 grids):
$(function(){
function print(str){
$("body").append(str + "<br/>");
}
function toBin(num, length){
if(!length){
length = 3;
}
var str = num.toString(2);
while(str.length < length){
str = 0 + str;
}
return str;
}
var wrongAnds = [
parseInt('000000111', 2),
parseInt('000111000', 2),
parseInt('111000000', 2),
parseInt('100100100', 2),
parseInt('010010010', 2),
parseInt('001001001', 2),
];
var wrongOrs = [
parseInt('111111000', 2),
parseInt('111000111', 2),
parseInt('000111111', 2),
parseInt('011011011', 2),
parseInt('101101101', 2),
parseInt('110110110', 2),
];
function test(mask){
for (var i = 0; i < 6; i++) {
if((wrongAnds[i] & mask) == wrongAnds[i]){
return false;
}
if((wrongOrs[i] | mask) == wrongOrs[i]){
return false;
}
}
return true;
}
var threeGrid = [];
var toRight = [];
var toBottom = [];
for(var mask = 1<<9-1; mask >= 0; mask--){
if(test(mask)){
threeGrid.push(mask);
}
}
function numberOfSetBits(i)
{
i = i - ((i >> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
}
function getCol(grid, col){
var ret = 0;
for(var i=0; i<3; i++){
ret += (grid & (1 << (i*3+col))) >> (i*2+col);
}
return ret;
}
var wrongAnds6 = [
parseInt('011100', 2),
parseInt('001110', 2)
];
var wrongOrs6 = [
parseInt('100011', 2),
parseInt('110001', 2)
];
for(var i = 0; i < threeGrid.length; i++){
for(var j = 0; j < threeGrid.length; j++){
var grid1 = threeGrid[i];
var grid2 = threeGrid[j];
var toRightOk = true;
var toBottomOk = true;
var printit = (i==0);
for(var k=0;k<3;k++){
var row = ((grid1 & wrongAnds[k]) << 3 >> (k*3)) + ((grid2 & wrongAnds[k]) >> (k*3));
var col = ((getCol(grid1, k)) << 3) + ((getCol(grid2, k)));
if(numberOfSetBits(row) != 3
|| ((wrongAnds6[0] & row) == wrongAnds6[0])
|| ((wrongAnds6[1] & row) == wrongAnds6[1])
|| ((wrongOrs6[0] | row) == wrongOrs6[0])
|| ((wrongOrs6[1] | row) == wrongOrs6[1])
) {
toRightOk = false;
}
if(numberOfSetBits(col) != 3
|| ((wrongAnds6[0] & col) == wrongAnds6[0])
|| ((wrongAnds6[1] & col) == wrongAnds6[1])
|| ((wrongOrs6[0] | col) == wrongOrs6[0])
|| ((wrongOrs6[1] | col) == wrongOrs6[1])
) {
toBottomOk = false;
}
}
if(toRightOk){
if(!toRight[grid1]){
toRight[grid1] = [];
}
toRight[grid1].push(grid2);
}
if(toBottomOk){
if(!toBottom[grid1]){
toBottom[grid1] = [];
}
toBottom[grid1].push(grid2);
}
}
}
function intersect(arr1, arr2){
var results = [];
for (var i = 0; i < arr1.length; i++) {
if (arr2.indexOf(arr1[i]) !== -1) {
results.push(arr1[i]);
}
}
return results;
}
var found = false;
while(!found){
var grid1 = threeGrid[0];
var grid1 = threeGrid[Math.floor(Math.random()*threeGrid.length)];
var grid2 = toRight[grid1][Math.floor(Math.random()*toRight[grid1].length)];
var grid3 = toBottom[grid1][Math.floor(Math.random()*toBottom[grid1].length)];
var arr4 = intersect(toBottom[grid2], toRight[grid3]);
if(arr4.length > 0){
var grid4 = arr4[Math.floor(Math.random()*arr4.length)];
found = true;
}
}
function gridToStrings(grid){
var rowS = [];
for(var i=0; i<3; i++){
rowS.push(toBin(((grid & wrongAnds[i]) >> (i*3))));
}
return rowS;
}
var grid1S = gridToStrings(grid1);
var grid2S = gridToStrings(grid2);
var grid3S = gridToStrings(grid3);
var grid4S = gridToStrings(grid4);
print(grid1S[0] + grid2S[0]);
print(grid1S[1] + grid2S[1]);
print(grid1S[2] + grid2S[2]);
print(grid3S[0] + grid4S[0]);
print(grid3S[1] + grid4S[1]);
print(grid3S[2] + grid4S[2]);
});
Theory:
Find all possible 3x3 grids
Find all possible left-to-right and top-to-bottom pairings
get 4 random grids to form the 6x6 grid
Implementation:
Represent 3x3 grids as 9bit integers. A 3x3 grid is wrong if there are 3 1s or 3 0s in it. This can be easily filtered with a couple bitwise operations.
Test the Cartesian product of these 3x3 grids (Compare every grid with every grid). Check if there are exactly 3 0s and 3 1s in all rows and columns (put the second grid right to the first grid to check 3 rows, and put it below the first grid to check 3 columns), and that there are no consecutive 3 0s or 1s.
get the top-left, top-right and bottom-right grids. Check if there is an available 4th grid that can go below the top-right grid and right to the bottom-left grid. If there is none, restart step 4, otherwise pick one.
A couple outputs:
011010
100101
001011
110100
101100
010011
110010
101100
010011
001101
100110
011001
001101
110010
010011
101100
110100
001011
Edit:
there is only 1120 solutions to this problem (jsFiddle). There are 2^36 ways to fill a 6x6 grid with 0s and 1s. If you used brute force (get a random 6x6 grid, then check if its right), that would mean an average ~61356676 (6.1*10^7) executions to find a correct solution. Even thought your method is somewhat faster (it can fail sooner if its not the last digit thats wrong), it might still be slow.
I think there are two problems with your code:
If oCount or zCount have become 3 there are no more assignments grid[i][j]=current if the random value is not acceptable. You get zeroes at these positions (to which the grid was initialized).
Near the right bottom there might not be any more valid solutions. You would have to undo previous assignments, i.e. you would need to do some kind of backtracking.
I would recommend starting with a valid solution and transforming this solution step by step according to random values for grid positions - but only if this is possible without breaking validity. If have prepared an example implementation:
public static void main(String[] args) {
int l = 6, w = 6;
Grid g = new Grid(l, w);
Random rd = new Random();
// initialize with checkerboard pattern (which is a valid solution)
for (int y = 0; y < l; y++) for (int x = 0; x < w; x++) g.arr[y][x] = (x ^ y) & 1;
// construct a valid grid by transformation of grids while preserving validity
for (int y = 0; y < l; y++) for (int x = 0; x < w; x++) {
int v = rd.nextInt(2), v2 = v ^ 1;
if (g.arr[y][x] == v) continue;
// try to modify current grid by exchanging values: 01/10=>10/01 or 10/01=>01/10
// (keep parts of the grid which have already been adapted to random values)
rotating: for (int y2 = y + 1; y2 < l; y2++) for (int x2 = x; x2 < w; x2++) {
if (g.arr[y2][x] == v && g.arr[y][x2] == v && g.arr[y2][x2] == v2) {
g.rotate(x, y, x2, y2);
// keep result if grid is still valid, undo otherwise
if (g.rotatedOk(x, y, x2, y2)) break rotating;
g.rotate(x, y, x2, y2);
}
}
}
g.printOn(System.out);
}
public static class Grid {
int l, w;
int[][] arr;
Grid(int l, int w) {
this.arr = new int[this.l = l][this.w = w];
}
void rotate(int x, int y, int x2, int y2) {
int v;
v = arr[y][x]; arr[y][x] = arr[y2][x]; arr[y2][x] = v;
v = arr[y][x2]; arr[y][x2] = arr[y2][x2]; arr[y2][x2] = v;
}
boolean rotatedOk(int x, int y, int x2, int y2) { // check after rotation
return okAt(x, y) && okAt(x2, y) && okAt(x, y2) && okAt(x2, y2);
}
private boolean okAt(int x, int y) { // check single position in grid
int v = arr[y][x];
if (count(x, y, -1, 0, v) + count(x, y, 1, 0, v) > 1) return false;
if (count(x, y, 0, -1, v) + count(x, y, 0, 1, v) > 1) return false;
return true;
}
private int count(int x, int y, int dx, int dy, int v) {
for (int n = 0; ; n++) {
x += dx; y += dy;
if (x < 0 || x >= w || y < 0 || y >= l || arr[y][x] != v) return n;
}
}
void printOn(PrintStream s) {
for (int y = 0; y < l; y++) { for (int x = 0; x < w; x++) s.print(arr[y][x]); s.println(); }
}
}
The problem with your approach is that you need a mechanism that handles when a new value can't be used because it follows two similar values, but the other value can't be used because it is under two other values. For example, say your grid has got this far:
101010
011010
00?
You would then need to slowly roll back positions and try different values.
The following code solves that problem using recursion:
import java.util.Random;
public class Main {
final int height = 6;
final int width = 6;
int[][] grid;
Random rd = new Random();
public static void main(final String[] args) {
Main main = new Main();
main.process();
}
private void process() {
// Create a grid that is 6 x 6
grid = new int[height][width];
for(int x = 0; x < width; x++) {
for(int y = 0; y < height; y++) {
grid[x][y] = -1;
}
}
recurseFillMatrix(0, 0);
}
private boolean recurseFillMatrix(final int x, final int y) {
// first, try putting a random number in the cell
int attempt = 1;
grid[x][y] = Math.abs(rd.nextInt()%2);
do {
if(isGridValid()) {
if(x == (width - 1) && y == (height - 1)) {
printGrid();
return true;
}
boolean problemSolved;
if(x == (width - 1)) {
problemSolved = recurseFillMatrix(0, y + 1);
} else {
problemSolved = recurseFillMatrix(x + 1, y);
}
if(problemSolved) {
return true;
}
}
attempt++;
grid[x][y] = 1 - grid[x][y];
} while(attempt <= 2);
grid[x][y] = -1;
return false;
}
private boolean isGridValid() {
for(int y = 0; y < height; y++) {
for(int x = 0; x < width; x++) {
// if the current item is -1, then we are finished
if(grid[x][y] == -1) {
return true;
}
// if we are after the second column
if(x > 1) {
if(grid[x-2][y] == grid[x-1][y] && grid[x-1][y] == grid[x][y]) {
return false;
}
}
// if we are after the second row
if(y > 1) {
if(grid[x][y-2] == grid[x][y-1] && grid[x][y-1] == grid[x][y]) {
return false;
}
}
// total the values in this column
int total = 0;
for(int i = 0; i <= y; i++) {
total += grid[x][i];
}
if(y == (height - 1)) {
if(total != 3) {
return false;
}
} else {
if(total > 3) {
return false;
}
}
// total the values in this row
total = 0;
for(int i = 0; i <= x; i++) {
total += grid[i][y];
}
if(x == (width - 1)) {
if(total != 3) {
return false;
}
} else {
if(total > 3) {
return false;
}
}
}
}
return true;
}
private void printGrid() {
for(int y = 0; y < height; y++) {
for(int x = 0; x < width; x++) {
System.out.print(grid[x][y]);
}
System.out.println("");
}
}
}
The isGridValid() method uses your defined rules to check if the grid (as it is filled so far) complies with the rules. At the first sign that it does not, it returns false.
If I have to change your solution to achieve the result, here is what it should look like..
Take the incrementors for oCount and zCount in a separate if-else
Take the assignment to grid(i,j) outside the loop
Your if-else block is not taking into account every condition possible, like
What about when last 2 items are same
What about when the zCount or oCount has reached 3
Taking into account these consideration, this code works fine.
import java.util.Random;
public class Main {
public static void main(String[] args) {
int l = 6;
int w = 6;
Random rd = new Random();
// Create a grid that is 6 x 6
int[][] grid = new int[l][w];
// for each row
for (int i = 0; i < l; i++) {
int zCount = 0;
int oCount = 0;
int current;
int lastA = 2;
int lastB = 2;
// for each item in the row
for (int j = 0; j < w; j++) {
// set the current item to either 0 or 1
current = rd.nextInt(2);
// make sure there aren't already (e.g. 3 items out of 6)
// items in the row
if (current == 1) {
if (oCount != 3) {
if (lastA == lastB) {
current = lastA == 1 ? 0 : 1;
}
} else {
current = current == 1 ? 0 : 1;
}
} else if (current == 0) {
if (zCount != 3) {
if (lastA == lastB) {
current = lastA == 1 ? 0 : 1;
}
} else {
current = current == 1 ? 0 : 1;
}
}
grid[i][j] = current;
if (current == 1) {
oCount++;
} else {
zCount++;
}
if (j % 2 == 1) {
// hold every second element
lastA = current;
} else {
// hold every first element
lastB = current;
}
System.out.print(grid[i][j]);
}
System.out.println(" ");
}
}
}
Again, This solution takes care of row conditions only. You would need to do similar checks for columns as well, to achieve the full result
HTH
here I tested you problem and seems that it is what you need.
I used a functional approach using Guava, it is quite simple, readable and has a short code.
#Test
public void test_permutations()
{
List<Integer> binary = Lists.newArrayList(1,0,1,0,1,0); // Domain list
Set<String> flattenSet = Sets.newHashSet(); // Store non-repetitive values
// Create list of possible values
Collection<List<Integer>> permutations = Collections2.permutations(binary);
for (List<Integer> permutation : permutations)
{
String joinString = StringUtils.join(permutation, "");
flattenSet.add(joinString);
}
// Create predicate to filter positive values
Predicate<String> predicate = new Predicate<String>() {
public boolean apply(String input) {
// Discard wrong values
if (input.contains("000") || input.contains("111")) {
return false;
} else {
return true;
}
}
};
// Use predicate to filter values
Collection<String> filteredList = Collections2.filter(flattenSet, predicate);
// Display result
for (String result : filteredList) {
System.out.println(result);
}
}
It is simple, I've commented the code to be clear but you can debug it to understand step by step.
The generated output is:
010011
110010
010101
010110
100110
101001
011010
110100
001011
001101
011001
101010
101100
100101
Hope to help
I think that its a mistake to think of generating it one element at at time. Instead imagine that I generate the entire set of permissible rows {001100,101010,....etc} There are only 6!/(3!3!)=20 ways to arrange three ones and three and some of them will be excluded. Now I am going to generate a game tree by saying that a move is selecting a valid row for the next row. If I discover at some point that there are no more valid moves then i will back track and try a different move.
To generate a move I randomly select a row, if its a valid move, I try to select another move, if that is impossible I backtrack, effectively doing a (random) depth first search of the game tree.
public class gametree {
public static ImmutableList<Row> allValidRows = // create a list of all valid rows.
public static List<Rows> getValidMoves(Move parent){ //Backtracks up the
//tree to the root to find the current state of the board, and returns
//which ever of allValidRows are valid given the game board.
}
public class Move {
public final Move parent;
public List<Rows> validMoves;
public final Row thisMove;
public int depth=0;
Move(Move parent, Row thisMove){
this.thisMove = thisMove;
this.parent = parent;
this.validMoves = getValidMoves(parent);
Move hold=parent;
while(hold!=null){
depth++; hold = parent.parent;
}
}
}
void run {
//pick first move
Move Root = new Move(null, Collections.Shuffle(allValidRows).get(0));
Move FinalMove = search(Root);
//Something to print out the answer here
}
public Move search(Move move){
if(depth==5){ return Move} //If I get to row six I win.
else if(move.validMoves.isEmpty()) { //If there are no valid moves,
//then this move wasnt valid, to strip it from the parent's
//possible moves and try again
move.parent.validMoves.remove(move.thisMove);
search(move.parent);
} else { //pick a random valid move and create a nextMove
Move nextMove = new Move(move, Collection.Shuffle(move.getValidMoves).get(0))
search(nextMove);
}
}
The worst case for this algorithm is that there is only one victory state and it has to try every possible state, but in practice this game does not seem very restrictive so it will probably not take long at all.
This code is strictly illustrative.
So I need to take a 2D array do calculations to each elements and transfer that into another 2D array while using the values to the "left" "right" "up" and "down" of the current element. If the current element is on the edge (x = 0, y = 0, x = array.length , y = array.length) I will get an array out of bounds error. I want to create a for loop that deals with each of those cases but I don't know how to do it. A sample of my code is
private void buildE(int[][] array, int y, int x)
{
int up = array[y - 1][x];
int down = array[y + 1][x];
int left = array[y][x - 1];
int right = array[y][x + 1];
if(up == 0){
buildETopRow(array);
}
E will be my new array. This method does not work because y does not equal 0, it just doesn't exist but I can't set ints to null either. In the case of an out of bounds error I need the element (up, down, left, or right) that is out of bounds to equal the current element. Is there a way I can still use a for loop for this or do I need to do something else?
If I read this correctly you want to effectively treat the difference of an element on the edge with an element off the edge as 0. If that's true I would write four methods right(), left(), up() and down(), with down() shown below as an example:
/*
* Return the difference between an element an the element below it
*/
public void down(int x, int y) {
if (y == array.length - 1) {
\\ on the bottom edge
return 0;
}
return array[y][x] - array[y + 1][x];
}
And inside your loop you'd calculate:
up(x,y) + down(x,y) + right(x,y) + left(x,y)
or whatever calculation it is you need to sum up.
The easiest way it to surround your array with a border region. So that your x dimension is really width+2.
import java.util.*;
import java.lang.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
int realWidth = 10;
int realHeight = 10;
int[][] in = new int[(realWidth+2)][(realHeight+2)];
int[][] out = new int[(realWidth+2)][(realHeight+2)];
for (int j = 1;j<realHeight+1;j++)
{
for (int i = 1;i<realWidth+1;i++)
{
int top = in[j-1][i];
int bottom = in[j+1][i];
int left= in[j][i-1];
int right = in[j][i+1];
out[j][i] = operation(top,bottom,left,right);
}
}
}
public static int operation (int top,int bottom,int left,int right)
{
return top+bottom+left+right;
}
}
I'm not totally sure what your question is, but (1) the usual structure for traversing a 2D array is to use nested for loops (one inside the other), and (2) when you want wrap-around counters (e.g. 2, 3, 0, 1, 2, ...) use the remainder operator %.
int numRows = theArray.length;
int numCols = theArray[0].length;
for (int i = 0; i < numRows; i++) {
for (int j = 0; j < numCols; j++) {
int right = theArray[(j+1) % numCols];
int down = theArray[(i+1) % numRows];
int left = theArray[(j+numCols-1) % numCols];
int up = theArray[(i+numRows-1) % numCols];
/* right, down, left, and up will be the elements to the right, down,
left, and up of the current element. Npw that you have them, you can
process them however you like and put them in the other array. */
}
}
What the remainder operator A%B does is sets A back to zero once it gets as large as B. Since B is the size of your array, that's exactly when it is too large and will cause an IndexOutOfBounds error. Note: That's not how % works but it's an ok way to think of what it does. To find out more about it you can google it, I found an ok explanation here.