I tried writing a maze solution method using DFS to find a path through a maze that's generated. But I'm having trouble with it. It doesn't even look like it ever finishes the traversal. Here is some output from a sample run. The maze looks like this:
+ +--+--+--+
| | |
+ +--+ + +
| | |
+--+--+--+ +
| | |
+ + + +--+
| | |
+--+--+--+ +
And my DFS method produces this:
0 1 5 9
In the end, I want to display that same maze but with numbers inside of it's path that represent the order which I have visited and ran through it.
Anyway, here's my code:
public static void depthFirstSearch(){
boolean[] visited = new boolean[totalCells]; // marks which vertices have been visited during the search
Stack<Vertex> st = new Stack<Vertex>();
st.push(graph[0][0]);
while(!st.isEmpty()){
Vertex v = st.pop();
if(!visited[v.label]){
visited[v.label] = true;
System.out.print(v.label + " ");
// auxiliary stack to visit neighbors in the order which they appear
Stack<Vertex> auxStack = new Stack<Vertex>();
for(Vertex w : v.neighbors){
if(!visited[w.label]){
auxStack.push(w);
}
}
while(!auxStack.isEmpty()){
st.push(auxStack.pop());
}
}
}
System.out.println();
}
Here is also the Vertex:
class Vertex{
int label;
int x;
int y;
boolean isVisited = false;
boolean hasNorthWall = true;
boolean hasSouthWall = true;
boolean hasEastWall = true;
boolean hasWestWall = true;
boolean hasAllWalls = true;
ArrayList<Vertex> neighbors = new ArrayList<Vertex>();
public Vertex(int x, int y){
this.x = x;
this.y = y;
}
}
And my constructor:
public Maze(int size)
{
this.SIZE = size;
totalCells = SIZE * SIZE;
cellStack = new Stack<Vertex>();
graph = new Vertex[SIZE][SIZE];
}
Thank you in advance for any help!
EDIT: Adding how neighbors are assigned.
public void assignNeighbors(Vertex v)
{
//This handles the cell north of current cell
if(v.y != 0)
{
v.neighbors.add(graph[v.x][v.y-1]);
}
//This handles the cell south of the current cell
if(v.y != (SIZE-1))
{
v.neighbors.add(graph[v.x][v.y+1]);
}
//This handles the cell left of the current cell
if(v.x != 0)
{
v.neighbors.add(graph[v.x -1][v.y]);
}
//right of the current
if(v.x != SIZE-1)
{
v.neighbors.add(graph[v.x + 1][v.y]);
}
}
EDIT2: Adding in how the label is assigned (it just stores the vertex number)
public void fill()
{
int vertexNumber = 0;
//This loop creates a new vertex
for(int i=0; i < SIZE; i++)
{
for(int j = 0; j < SIZE; j++)
{
Vertex v = new Vertex(j,i);
graph[j][i] = v;
}
}
//adds values to vertex
for(int i = 0; i < SIZE; i++)
{
for(int j = 0; j < SIZE; j++)
{
graph[j][i].label = vertexNumber;
vertexNumber++;
}
}
//This loop assigns the neighbors
for(int i = 0; i < SIZE; i++)
{
for(int j = 0; j < SIZE; j++)
{
assignNeighbors(graph[j][i]);
}
}
mazeGenerator();
}
I'm not quite sure about this part
for(Vertex w : v.neighbors){
if(!visited[w.label]){
auxStack.push(w);
}
}
while(!auxStack.isEmpty()){
st.push(auxStack.pop());
}
If I'm correct you can replace this with just one loop but this shouldn't be the problem. Have you tried debugging it step by step? You know how the maze looks like and thus you know how the alghoritm should behave in every step. I suggest you try it and then maybe update your post and describe at which step your algorithm behaves weird.
I wanted to add this as comment but I lack the reputation.
Related
example - there is an unconnected graph with vertices A - B - C - D and E - F - G. (a hyphen means that they are connected). The code below is using depth-first traversal, I need to modify it to display all connected vertices. eg:
list0: ABCD
list1: EFG
etc...
I don't understand how to implement this.
public class Graph {
private final int MAX_VERTS = 20;
private Vertex vertexList[];
private int matrix[][];
private int countV;
private StackX theStack;
// ------------------------------------------------------------
public Graph() {
vertexList = new Vertex[MAX_VERTS];
matrix = new int[MAX_VERTS][MAX_VERTS];
countV = 0;
for (int x = 0; x < MAX_VERTS; x++)
for (int y = 0; y < MAX_VERTS; y++)
matrix[x][y] = 0;
theStack = new StackX();
}
// -------------------------------------------------------------
public void addVertex(char label) {
vertexList[countV++] = new Vertex(label);
}
// -------------------------------------------------------------
public void addEdge(int x, int y) {
matrix[x][y] = 1;
matrix[y][x] = 1;
}
// -------------------------------------------------------------
public void displayVertex(int v) {
System.out.print(vertexList[v].label);
}
public void dfs() {
vertexList[0].wasVisited = true;
displayVertex(0);
theStack.push(0);
while (!theStack.isEmpty()) {
int v = getUnvisitedVertex(theStack.peek());
if (v == -1)
theStack.pop();
else
{
vertexList[v].wasVisited = true;
displayVertex(v);
theStack.push(v);
}
}
for (int j = 0; j < countV; j++)
vertexList[j].wasVisited = false;
}
// ------------------------------------------------------------
public int getUnvisitedVertex(int vertex) {
for (int j = 0; j < countV; j++)
if (matrix[vertex][j] == 1 && !vertexList[j].wasVisited) {
return j;
}
return -1;
}
}
Your DFS does not need to modified. It needs to be put inside a loop so that each pass will discover one of the lists of connected nodes that your are looking for
LOOP
Select arbitrary vertex
DFS, saving each visited vertex in list.
LOOP over visited vertices
remove from graph
LOOP END
IF all vertices removed
BREAK out of loop
Start new list
LOOP END
Output lists
I have two 2d boolean arrays, the smaller array (shape) is going over the larger array (world).
I am having trouble to find a method to find out when the smaller array can "fit" into the larger one.
When I run the code it either just goes through the larger array, never stopping, or stops after one step (incorrectly).
public void solve() {
ArrayList<Boolean> worldList=new ArrayList<>();
ArrayList<Boolean> shapeList=new ArrayList<>();
for (int i = 0; i < world.length; i++) {
for (int k = 0; k < world[i].length; k++) {
worldList.add(world[i][k]);
display(i, k, Orientation.ROTATE_NONE);
for (int j = 0; j < shape.length; j++) {
for (int l = 0; l < shape[j].length; l++) {
shapeList.add(shape[j][l]);
if(shapeList.equals(worldList)) {
return;
}
}
}
}
}
}
A good place to start with a problem like this is brute force for the simplest case. So, for each index in the world list, just check to see if every following index of world and shapes match.
Notice we only iterate to world.size()-shapes.size(), because naturally if shapes is longer than the portion of world we haven't checked, it won't fit.
import java.util.ArrayList;
public class Test {
ArrayList<Boolean> world = new ArrayList<>();
ArrayList<Boolean> shapes = new ArrayList<>();
public static void main(String[] args) {
new Work();
}
public Test() {
world.add(true);
world.add(false);
world.add(false);
world.add(true);
shapes.add(false);
shapes.add(true);
// Arraylists initialized to these values:
// world: T F F T
// shapes: F T
System.out.println(getFitIndex());
}
/**
* Get the index of the fit, -1 if it won't fit.
* #return
*/
public int getFitIndex() {
for (int w = 0; w <= world.size()-shapes.size(); w++) {
boolean fits = true;
for (int s = 0; s < shapes.size(); s++) {
System.out.println("Compare shapes[" + s + "] and world["+ (w+s) + "]: " +
shapes.get(s).equals(world.get(w+s)));
if (!shapes.get(s).equals(world.get(w+s))) fits = false;
}
System.out.println();
if (fits) return w;
}
return -1;
}
}
When we run this code, we get a value of 2 printed to the console, since shapes does indeed fit inside world, starting at world[2].
You can find the row and column of fitting like this
public void fit() {
int h = world.length - shape.length;
int w = world[0].length - shape[0].length;
for (int i = 0; i <= h; i++) {
for (int k = 0; k <= w; k++) {
boolean found = true;
for (int j = 0; j < shape.length && found; j++) {
for (int l = 0; l < shape[j].length && found; l++) {
if (shape[j][l] != world[i + j][k + l])
found = false;
}
}
if (found) {
//Your shape list fit the world list at starting index (i, k)
//You can for example save the i, k variable in instance variable
//Or return then as an object for further use
return;
}
}
}
at the Moment i do a 2D strategy game using pathfinding to navigate my Units over the (still small) tilemap. The tiles are 32x32 and the map is 50x100 big so ist very small :) . It works all so far but i have the more laggs the more Units i create. Till 30 Units it works as it should but more makes my Programm lagg very strong.
So i use an ArrayList for my openSet and (after doing some googling) i know thats bad. So i need to Keep my openList sorted by using TreeSet, but by using TreeSet its necessary to Override compareTo(). Im not fit enough with comparisions like this.
What must i compare exactly, the f value or the Signum? I dont know that and i Need some help.
Here is the A* Algorithm:
public static List<Tile> findPath(int startx,int starty,int endx,int endy){
for(int i = 0; i < width; i++){
for(int j = 0;j < height;j++){
tiles[i][j] = new Tile(i,j,size,size,obstacles[i][j],false);
}
}
for(int i = 0; i < width; i++){
for(int j = 0;j < height;j++){
tiles[i][j].addNeighbours(tiles,width,height);
}
}
List<Tile> openList = new ArrayList<Tile>(); // Here i want a TreeSet
HashSet<Tile> closedList = new HashSet<Tile>();
List<Tile> path = null;
Tile start = tiles[startx][starty];
Tile end = tiles[endx][endy];
Tile closest = start;
closest.h = heuristic(closest,end);
openList.add(start);
while(!openList.isEmpty()) {
int winner = 0;
for (int i = 0; i < openList.size(); i++) {
if (openList.get(i).f < openList.get(winner).f) {
winner = i;
}
}
Tile current = openList.get(winner);
openList.remove(current);
if (current == end) {
path = new ArrayList<Tile>();
Tile tmp = current;
path.add(tmp);
while (tmp.previous != null) {
path.add(tmp);
tmp = tmp.previous;
}
return path;
}
closedList.add(current);
List<Tile> neighbours = current.neighbours;
for (int i = 0; i < neighbours.size(); i++) {
Tile neighbour = neighbours.get(i);
int cost = current.g + heuristic(current,neighbour);
if (openList.contains(neighbour) && cost < neighbour.g) {
openList.remove(neighbour);
}
if (closedList.contains(neighbour) && cost < neighbour.g) {
closedList.remove(neighbour);
}
int newcost = heuristic(neighbour, end);
if (!openList.contains(neighbour) && !closedList.contains(neighbour) && !neighbour.obstacle) {
neighbour.h = newcost;
if (neighbour.h < closest.h) {
closest = neighbour;
}
}
if (!openList.contains(neighbour) && !closedList.contains(neighbour) && !neighbour.obstacle) {
neighbour.g = cost;
openList.add(neighbour);
neighbour.f = neighbour.g + neighbour.h;
neighbour.previous = current;
}
}
}
Tile tmp = closest;
path = new ArrayList<Tile>();
path.add(tmp);
while (tmp.previous != null) {
path.add(tmp);
tmp = tmp.previous;
}
return path;
}
public static int heuristic(Tile A,Tile B) {
int dx = Math.abs(A.x - B.x);
int dy = Math.abs(A.y - B.y);
return 1 * (dx + dy) + (1 - 2 * 1) * Math.min(dx,dy);
}
And i have another Problem. I load the whole entire map inclusive ist obstacle during calling the finPath-Method, but i didnt find another solution, where i can load it only once. And i really tried a lot believe me... .
So here my two Questions:
What must i exactly compare within the compareTo Method to make it work?
Where can i load my TiledMap once, so A* havent got to update it during it is called?
I am doing a game called 1010! Probably some of you have heard of it. Bascially I encouter some trouble when writing the Algorithm for clearance.
The rule is such that if any row or any column is occupied, then clear row and column respectively.
The scoring is such that each move gains a+10*b points. a is the number of square in the input piece p and b is the total number of row&column cleared.
To start, I create a two dimensional Array board[10][10], poulate each elements in the board[][] with an empty square.
In the class of Square, it has public void method of unset()-> "empty the square" & boolean status() -> "judge if square is empty"In the class of piece, it has int numofSquare -> "return the number of square in each piece for score calculation"
In particular, I don't know how to write it if both row and column are occupied as they are inter-cross each other in an two dimensional array.
It fail the test under some condition, in which some of the squares are not cleared but they should have been cleared and I am pretty sure is the logic problem.
My thinking is that:
Loop through squares in first row and first column, record the number of square that are occupied (using c and r); if both are 10, clear row&column, otherwise clear row or column or do nothing.
reset the c &r to 0, loop through square in the second row, second column…
update score.
Basically the hard part is that if I seperate clear column and clear row algorithm ,I will either judge row or column first then clear them . However, as every column contains at least one square belong to the row, and every row contains at least one square belong to the column, there will be mistake when both row and column are full.
Thanks for help.
import java.util.ArrayList;
public class GameState{
public static final int noOfSquares = 10;
// the extent of the board in both directions
public static final int noOfBoxes = 3;
// the number of boxes in the game
private Square[][] board; // the current state of the board
private Box[] boxes; // the current state of the boxes
private int score; // the current score
// initialise the instance variables for board
// all squares and all boxes are initially empty
public GameState()
{
getboard();
score = 0;
board = new Square[10][10];
for(int i =0;i<board.length;i++){
for(int j =0;j<board[i].length;j++){
board[i][j] = new Square();
}
}
boxes = new Box[3];
for(int k =0;k<boxes.length;k++){
boxes[k] = new Box();
}
}
// return the current state of the board
public Square[][] getBoard()
{
return board;
}
// return the current score
public int getScore()
{
return score;
}
// place p on the board with its (notional) top-left corner at Square x,y
// clear columns and rows as appropriate
int r =0;
int c = 0;
int rowandcolumn = 0;
for (int row=0;row<10;row++){
for (int column=0;column<10;column++) {
if (board[row][column].status() == true){
c = c + 1;
if( c == 10 ) {
rowandcolumn = rowandcolumn + 1;
for(int z=0;z<10;z++){
board[row][z].unset(); //Clear column
}
}
}
if (board[column][row].status() == true){
r = r + 1;
if( r == 10) {
rowandcolumn = rowandcolumn + 1;
for(int q=0;q<10;q++){
board[q][row].unset(); //Clear row
}
}
}
}
r=0; //reset
c=0;
}
score = score + p.numberofBox()+10*rowandcolumn;
}
how about this
void Background::liquidate(int &score){
int arr_flag[2][10]; //0 is row,1 is column。
for (int i = 0; i < 2; i++)
{
for (int j = 0; j < 10; j++)
{
arr_flag[i][j] = 1;
}
}
//column
for (int i = 0; i < 10; i++)
{
for (int j = 0; j < 10; j++)
{
if (arr[i][j].type == 0)
{
arr_flag[0][i] = 0;
break;
}
}
}
//row
for (int i = 0; i < 10; i++)
{
for (int j = 0; j < 10; j++)
{
if (arr[j][i].type == 0)
{
arr_flag[1][i] = 0;
break;
}
}
}
//clear column
for (int i = 0; i < 10; i++)
{
if (arr_flag[0][i] == 1)
{
for (int j = 0; j < 10; j++)
{
arr[i][j].Clear();
}
}
}
//clear row
for (int i = 0; i < 10; i++)
{
if (arr_flag[1][i] == 1)
{
for (int j = 0; j < 10; j++)
{
arr[j][i].Clear();
}
}
}
}
I tried to write somme code for the idea I posted
// place p on the board with its (notional) top-left corner at Square x,y
// clear columns and rows as appropriate
int r =0;
int c = 0;
int rowandcolumn = 0;
int row=FindFirstRow();
int column=FindFirstColumn();
if(row!=-1 && column!=-1)
{
rowandcolumn++;
//actions here: row found and column found
//clear row and column
clearRow(row);
clearColumn(column);
}
else if(row!=-1)
{
//only row is found
//clear row
clearRow(row);
}
else if(column!=-1)
{
//only column is found
//clear column
clearColumn(column);
}
else
{
//nothing is found
}
public void clearRow(int row)
{
for(int i=0; i<10;i++)
{
board[row][i].unset();
}
}
public void clearColumn(int column)
{
for(int i=0; i<10;i++)
{
board[i][column].unset();
}
}
//this method returns the first matching row index. If nothing is found it returns -1;
public int FindFirstRow()
{
for (int row=0;row<10;row++)
{
int r=0;
for (int column=0;column<10;column++)
{
if (board[row][column].status() == true)
{
r = r + 1;
if( r == 10)
{
//row found
return row;
}
}
}
r=0; //reset
}
//nothing found
return -1;
}
//this method returns the first matching column index. If nothing is found it returns -1;
public int FindFirstColumn()
{
for (int column=0;column<10;column++)
{
int c=0;
for (int row=0;row<10;row++)
{
if (board[row][column].status() == true)
{
c = c + 1;
if( c == 10 )
{
//matching column found
return column;
}
}
}
c=0; //reset
}
//nothing found
return -1;
}
In my last question seen here: Sudoku - Region testing I asked how to check the 3x3 regions and someone was able to give me a satisfactory answer (although it involved a LOT of tinkering to get it working how I wanted to, since they didn't mention what the class table_t was.)
I finished the project and was able to create a sudoku generator, but it feels like it's contrived. And I feel like I've somehow overcomplicated things by taking a very brute-force approach to generating the puzzles.
Essentially my goal is to create a 9x9 grid with 9- 3x3 regions. Each row / col / region must use the numbers 1-9 only once.
The way that I went about solving this was by using a 2-dimensional array to place numbers at random, 3 rows at a time. Once the 3 rows were done it would check the 3 rows, and 3 regions and each vertical col up to the 3rd position. As it iterated through it would do the same until the array was filled, but due to the fact that I was filling with rand, and checking each row / column / region multiple times it felt very inefficient.
Is there an "easier" way to go about doing this with any type of data construct aside from a 2d array? Is there an easier way to check each 3x3 region that might coincide with checking either vert or horizontal better? From a standpoint of computation I can't see too many ways to do it more efficiently without swelling the size of the code dramatically.
I built a sudoku game a while ago and used the dancing links algorithm by Donald Knuth to generate the puzzles. I found these sites very helpful in learning and implementing the algorithm
http://en.wikipedia.org/wiki/Dancing_Links
http://cgi.cse.unsw.edu.au/~xche635/dlx_sodoku/
http://garethrees.org/2007/06/10/zendoku-generation/
import java.util.Random;
import java.util.Scanner;
public class sudoku {
/**
* #antony
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int p = 1;
Random r = new Random();
int i1=r.nextInt(8);
int firstval = i1;
while (p == 1) {
int x = firstval, v = 1;
int a[][] = new int[9][9];
int b[][] = new int[9][9];
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if ((x + j + v) <= 9)
a[i][j] = j + x + v;
else
a[i][j] = j + x + v - 9;
if (a[i][j] == 10)
a[i][j] = 1;
// System.out.print(a[i][j]+" ");
}
x += 3;
if (x >= 9)
x = x - 9;
// System.out.println();
if (i == 2) {
v = 2;
x = firstval;
}
if (i == 5) {
v = 3;
x = firstval;
}
}
int eorh;
Scanner in = new Scanner(System.in);
System.out
.println("hey lets play a game of sudoku:take down the question and replace the 0's with your digits and complete the game by re entering your answer");
System.out.println("enter your option 1.hard 2.easy");
eorh = in.nextInt();
switch (eorh) {
case 1:
b[0][0] = a[0][0];
b[8][8] = a[8][8];
b[0][3] = a[0][3];
b[0][4] = a[0][4];
b[1][2] = a[1][2];
b[1][3] = a[1][3];
b[1][6] = a[1][6];
b[1][7] = a[1][7];
b[2][0] = a[2][0];
b[2][4] = a[2][4];
b[2][8] = a[2][8];
b[3][2] = a[3][2];
b[3][8] = a[3][8];
b[4][2] = a[4][2];
b[4][3] = a[4][3];
b[4][5] = a[4][5];
b[4][6] = a[4][6];
b[5][0] = a[5][0];
b[5][6] = a[5][6];
b[6][0] = a[6][0];
b[6][4] = a[6][4];
b[6][8] = a[6][8];
b[7][1] = a[7][1];
b[7][2] = a[7][2];
b[7][5] = a[7][5];
b[7][6] = a[7][6];
b[8][4] = a[8][4];
b[8][5] = a[8][5];
b[0][0] = a[0][0];
b[8][8] = a[8][8];
break;
case 2:
b[0][3] = a[0][3];
b[0][4] = a[0][4];
b[1][2] = a[1][2];
b[1][3] = a[1][3];
b[1][6] = a[1][6];
b[1][7] = a[1][7];
b[1][8] = a[1][8];
b[2][0] = a[2][0];
b[2][4] = a[2][4];
b[2][8] = a[2][8];
b[3][2] = a[3][2];
b[3][5] = a[3][5];
b[3][8] = a[3][8];
b[4][0] = a[4][0];
b[4][2] = a[4][2];
b[4][3] = a[4][3];
b[4][4] = a[4][4];
b[4][5] = a[4][5];
b[4][6] = a[4][6];
b[5][0] = a[5][0];
b[5][1] = a[5][1];
b[5][4] = a[5][4];
b[5][6] = a[5][6];
b[6][0] = a[6][0];
b[6][4] = a[6][4];
b[6][6] = a[6][6];
b[6][8] = a[6][8];
b[7][0] = a[7][0];
b[7][1] = a[7][1];
b[7][2] = a[7][2];
b[7][5] = a[7][5];
b[7][6] = a[7][6];
b[8][2] = a[8][2];
b[8][4] = a[8][4];
b[8][5] = a[8][5];
break;
default:
System.out.println("entered option is incorrect");
break;
}
for (int y = 0; y < 9; y++) {
for (int z = 0; z < 9; z++) {
System.out.print(b[y][z] + " ");
}
System.out.println("");
}
System.out.println("enter your answer");
int c[][] = new int[9][9];
for (int y = 0; y < 9; y++) {
for (int z = 0; z < 9; z++) {
c[y][z] = in.nextInt();
}
}
for (int y = 0; y < 9; y++) {
for (int z = 0; z < 9; z++)
System.out.print(c[y][z] + " ");
System.out.println();
}
int q = 0;
for (int y = 0; y < 9; y++) {
for (int z = 0; z < 9; z++)
if (a[y][z] == c[y][z])
continue;
else {
q++;
break;
}
}
if (q == 0)
System.out
.println("the answer you have entered is correct well done");
else
System.out.println("oh wrong answer better luck next time");
System.out
.println("do you want to play a different game of sudoku(1/0)");
p = in.nextInt();
firstval=r.nextInt(8);
/*if (firstval > 8)
firstval -= 9;*/
}
}
}
I think you can use a 1D array, in much the same way a 1D array can model a binary tree. For example, to look at the value below a number, add 9 to the index.
I just made this up, but could something like this work?
private boolean makePuzzle(int [] puzzle, int i)
{
for (int x = 0; x< 10 ; x++)
{
if (//x satisfies all three conditions for the current square i)
{
puzzle[i]=x;
if (i==80) return true //terminal condition, x fits in the last square
else
if makePuzzle(puzzle, i++);//find the next x
return true;
}// even though x fit in this square, an x couldn't be
// found for some future square, try again with a new x
}
return false; //no value for x fit in the current square
}
public static void main(String[] args )
{
int[] puzzle = new int[80];
makePuzzle(puzzle,0);
// print out puzzle here
}
Edit: its been a while since I've used arrays in Java, sorry if I screwed up any syntax. Please consider it pseudo code :)
Here is the code as described below in my comment.
public class Sudoku
{
public int[] puzzle = new int[81];
private void makePuzzle(int[] puzzle, int i)
{
for (int x = 1; x< 10 ; x++)
{
puzzle[i]=x;
if(checkConstraints(puzzle))
{
if (i==80)//terminal condition
{
System.out.println(this);//print out the completed puzzle
puzzle[i]=0;
return;
}
else
makePuzzle(puzzle,i+1);//find a number for the next square
}
puzzle[i]=0;//this try didn't work, delete the evidence
}
}
private boolean checkConstraints(int[] puzzle)
{
int test;
//test that rows have unique values
for (int column=0; column<9; column++)
{
for (int row=0; row<9; row++)
{
test=puzzle[row+column*9];
for (int j=0;j<9;j++)
{
if(test!=0&& row!=j&&test==puzzle[j+column*9])
return false;
}
}
}
//test that columns have unique values
for (int column=0; column<9; column++)
{
for(int row=0; row<9; row++)
{
test=puzzle[column+row*9];
for (int j=0;j<9;j++)
{
if(test!=0&&row!=j&&test==puzzle[column+j*9])
return false;
}
}
}
//implement region test here
int[][] regions = new int[9][9];
int[] regionIndex ={0,3,6,27,30,33,54,57,60};
for (int region=0; region<9;region++) //for each region
{
int j =0;
for (int k=regionIndex[region];k<regionIndex[region]+27; k=(k%3==2?k+7:k+1))
{
regions[region][j]=puzzle[k];
j++;
}
}
for (int i=0;i<9;i++)//region counter
{
for (int j=0;j<9;j++)
{
for (int k=0;k<9;k++)
{
if (regions[i][j]!=0&&j!=k&®ions[i][j]==regions[i][k])
return false;
}
}
}
return true;
}
public String toString()
{
String string= "";
for (int i=0; i <9;i++)
{
for (int j = 0; j<9;j++)
{
string = string+puzzle[i*9+j];
}
string =string +"\n";
}
return string;
}
public static void main(String[] args)
{
Sudoku sudoku=new Sudoku();
sudoku.makePuzzle(sudoku.puzzle, 0);
}
}
Try this code:
package com;
public class Suduku{
public static void main(String[] args ){
int k=0;
int fillCount =1;
int subGrid=1;
int N=3;
int[][] a=new int[N*N][N*N];
for (int i=0;i<N*N;i++){
if(k==N){
k=1;
subGrid++;
fillCount=subGrid;
}else{
k++;
if(i!=0)
fillCount=fillCount+N;
}
for(int j=0;j<N*N;j++){
if(fillCount==N*N){
a[i][j]=fillCount;
fillCount=1;
System.out.print(" "+a[i][j]);
}else{
a[i][j]=fillCount++;
System.out.print(" "+a[i][j]);
}
}
System.out.println();
}
}
}