I'm trying to solve a 9x9 sudoku puzzle using Simulated Annealing, but my implementation doesn't seem to be working correctly. It does not even get closer to a lower-cost solution but instead keeps circling around results that cost between 60 and 80.
My cost function returns the sum of three things: Number of repeating digits in each row, column and block (3x3).
And the successor (neighbour) function i implemented changes two randomly selected digits from the 9x9 grid with random values.
And here is my SA function that doesn't work as expected:
public static void simulatedAnnealing() {
Sudoku neighbour; // candidate successor object
final Double temperature = 2.0; // initial temperature
final Double coolingFactor = 0.999; // cooling constant
final int maxIterations = 1000; // number of iterations
for(Double t = temperature; t>1; t*=coolingFactor) {
for(int i = 0; i < maxIterations; i++) {
neighbour = sudoku.generateSuccessor(); // set random neighbour
int delta = neighbour.cost() - sudoku.cost(); // calculate delta
if (delta <= 0) {
sudoku = neighbour; // always accept good step.
} else {
if (Math.exp(-delta / temperature) > Math.random()) { // Simulated annealing
sudoku = neighbour;
}
}
}
System.out.println(sudoku.cost());
if(sudoku.cost() == 0) { break; } // if puzzle is solved
} }
Function for generating successors:
public Sudoku generateSuccessor() {
int[][] newGrid = new int[9][9];
for(int o = 0; o < 9; o ++) { // cloning current grid array
for(int g = 0; g < 9; g ++) {
newGrid[o][g] = grid[o][g];
}
}
Sudoku rndm = new Sudoku(newGrid); // random Sudoku object.
for (int i = 0; i < 2; i++) { // will randomize 2 cells in 9x9 grid.
int rndmCell = rndmValue(); // random digit for randomizing.
int randomRow = rndm(); // random row that will be randomized
int randomCol = rndm(); // random column that will be randomized
// prevent randomizing given cells in sudoku (in problem definition)
boolean shouldContinue = false;
for (Coordinate c : SudokuSolver.concreteCoordinates) {
if (c.row == randomRow && c.col == randomCol) {
shouldContinue = true;
break;
}
}
if (shouldContinue) {
i--;
continue;
}
// prevention end.
rndm.grid[randomRow][randomCol] = rndmCell;
}
return rndm;
}
Cost function:
public int cost() {
if(hasZeros()) { // if grid is not totally filled with numbers don't calculate its cost.
return -1;
}
int cost = 0;
for(int i = 0; i< 9; i++) { // find total collusions in rows&columns.
cost += findNumberOfCollusions(grid[i]); // find collustions at row 'i'.
cost += findNumberOfCollusions(getColumn(grid,i)); // find collustions at column 'i'.
}
for(int r = 0; r < 9; r += 3) { //find total colusions in blocks (3x3).
for(int c = 0; c < 9; c += 3) {
int[] block = new int[9];
int ctr = 0;
for (int i = r; i < r + 3; i++) {
for (int y = c; y < c+ 3; y++) {
block[ctr] = grid[i][y];
ctr++;
}
}
cost += findNumberOfCollusions(block);
}
}
return cost;
}
When i run the program the output is costs between 60 and 80. After that the temperature goes below the limit and the program outputs a solution that costs around that interval. Can anyone tell me what am i doing wrong? Thanks in advance.
I also had a similar problem to the one you describe, my fitness remained stuck (actually though, my problem was with not copying lists in Python). I can't really assure why your code gets stuck, but if I had to guess: the neighbor generation (int rndmCell = rndmValue(); int randomRow = rndm(); int randomCol = rndm();) may be actually doing more harm than good. Imagine that you have a nearly complete sudoku, but out of the blue two of the correct cells that you had now change their value to a complete opposite one, which is not only wrong on the cell itself but also on the row, column and/or 3x3 square. I'm no mathematician, but logic tells me that the more fitting the sudoku is (i.e. the closer its fitness is to 0), the more chances there are to mess up the sudoku by randomly changing cells. This approach may get you stuck on a local minimum easily.
A more "informed" solution for this problem would be to keep one of the three basic restrictions of the sudoku puzzle fixed by, for instance, generating rows that are permutations of the values [1..9], swapping two cells of a random row (thus still fulfilling the restriction), and calculating the fitness only on the columns and on the 3x3 squares. This choice of neighbor generation is usually more effective. If you are interested, this idea comes from the paper Metaheuristics can solve Sudoku puzzles. I can say that this idea helped me a lot and now the algorithm completes sudokus that I provide :)
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I've been trying to dive into java development and have gone for a relatively easy problem of finding prime numbers, however, I keep getting errors and can't see what I've done wrong, any help?
I've been toiling over my computer for an infuriating while and have tried everything, even rewriting the code from beginning
public class HelloWorld{
public static void main(String []args){
int[] check = {2};
//cycle through numbers 1-100
for (int i = 1; i < 100; i++) {
//cycle through numbers to be checked against i
for (int x = 0; x < 101; x++) {
//check if the current itteration of i has no multiples
if (i%check[x] == 0) {
check[i] = i;
} else {
// print any prime numbers
System.out.print(i);
check[i] = i;
}
}
}
}
}
The immediate cause of your error is that you defined the check[] array to have a size of 1, but you are trying to access elements higher than that, which don't exist. However, I don't think that you really need that array here. Consider this version:
for (int i=2; i < 100; i++) {
boolean match = true;
for (int x=2; x <= Math.sqrt(i); x++) {
if (i % x == 0) {
match = false;
break;
}
}
if (match) {
System.out.println("Prime number: " + i);
}
}
Note that the inner loop in x only needs to go as high as the square root of the outer i value. This is because any value greater than sqrt(i) can't possible divide it.
It is because your array has length 1
and you are trying to access out of bound indexes. In case you are lopping 0 to 101 You can initialize your array like this int [] check =new int [101]
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 am trying to solve the Knights tour problem on a 4x4 board with backtracking and recursion in java, and on the output I get this step sequence:
1 13 16 15
10 7 4 14
5 2 11 8
12 9 6 3
in the right upper corner, the 14, 15 and 16 neighbour with each other, which is impossible, because the knight moves on the chessboard into an L-shape. I would be thankful if someone could help me solve this.
the code:
public class KnightsTour {
private static int board[][] = new int[4][4];
private static int stepCounter = 1;
public Test() {
initBoard(board);
tour(0,0);
printSol(board);
}
public static void printSol(int[][] a) {
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < a[i].length; j++) {
if(a[i][j]>9){
System.out.print(a[i][j] + " ");
}else{
System.out.print(a[i][j] + " ");
}
}
System.out.println();
}
System.out.println();
}
public static void initBoard(int[][] a) {
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < a[i].length; j++) {
a[i][j] = -1;
}
}
}
public void tour(int x, int y) {
if (((x < 0) || (x >= 4) || (y < 0) || (y >= 4)) || (board[x][y] != -1)) {
return;
}else{
board[x][y] = stepCounter++;
tour(x+2, y+1);
tour(x+1, y-2);
tour(x+1, y+2);
tour(x-1, y+2);
tour(x-2, y-1);
tour(x-2, y+1);
tour(x-1, y-2);
tour(x+2, y-1);
}
}
public static void main(String[] args){
new KnightsTour();
}
}
You need to make the function return a boolean so it can tell the calling function whether or not it succeeded. Otherwise you just carry on until you've tried every possible combination, even after you've found a solution.
Then, at each call of the function, you need to check the return value and return true if it succeeded.
Then you also obviously need to return true when done.
I suggest something like:
if (stepCounter == 1 + board.length * board[0].length)
return true;
Right after board[x][y] = stepCounter++;.
You need to revert any changes made at the end of the function call, i.e. stepCounter needs to decrease and board[x][y] needs to be set to -1.
After you've successfully made these changes, you should actually see a result of all -1's, because it's not possible on a 4x4 board, but changing it to 8x8 should succeed.
Note that I didn't use 17 above - it's good practice to not use hard-coded values (in, for example, x >= 4). Use either the size of the array, or final values, instead.
Your tour() function appears to set the value of a square to the step-counter of the order it's visited for the first time. But you're trying multiple options. Your first tour dead ends when you get to 12 - one of the subsequent attempts touches each of the 13-16 squares.
You need to store the state of the current tour, rather than the order in which you visited. E.g. if you backtrack, the current square is no longer part of the tour. If you find a tour, you should stop, because you're done.
I've a problem with my sudoku solving method. The program works like this; the board is empty when started, the users adds a couple of numbers to the board and then by hitting a Solve-button the program tries to solve it. Everything works fine besides if I put the same number in the same row. So if the user adds 1,1,0,0 ... 0. In the puzzle it can't solve it because its two 1's next to each other and will just go on forever trying to find a sulotion even though its an unsolvable puzzle. However if they were all 0's(empty) it would solve it right away, same as if Id put 1 and 2 in the top left corner. If I'd just put some random numbers in it will detect it as unsolvable (or will solve it if it's a valid puzzle)
I'm thinking around the lines of saying, when theNumber == (row, col) equals thenNumber == (row+1, col), it should return false because it's a duplicated number.
This is the code I tried to add in the solve method, obviously without success.
if ((puzzle.getNum(row, col) == a) == (puzzle.getNum(row + 1, col) == a)) {
return false;
}
Help is greatly appreciated
Validate the puzzle like this:
Create a boolean array of 9 elements.
Loop through every row, column and 9x9 box.
If you read a number, set the corresponding value in the array to true.
If it is already true throw an error (impossible puzzle).
After reading a row, column or 9x9 box reset the boolean array.
Then, if the validation succeeded call the solving method.
EDIT: Source code
public boolean checkPuzzle() {
boolean[] nums = new boolean[9];
for (int row = 0; row < panel.puzzleSize; row++) {
for (int cell = 0; cell < panel.puzzleSize; cell++) {
if (nums[puzzle[row][cell]]) return false;
nums[puzzle[row][cell]] = true;
}
nums = new boolean[9];
}
for (int col = 0; col < panel.puzzleSize; col++) {
for (int cell = 0; cell < panel.puzzleSize; cell++) {
if (nums[puzzle[cell][col]]) return false;
nums[puzzle[cell][col]] = true;
}
nums = new boolean[9];
}
for (int square = 0; square < panel.puzzleSize; square++) {
int squareCol = panel.squareSize * (square % panel.squareSize);
int squareRow = panel.squareSize * Math.floor(square / panel.squareSize);
for (int cell = 0; cell < panel.puzzleSize; cell++) {
int col = cell % panel.squareSize;
int row = Math.floor(cell / panel.squareSize);
if (nums[puzzle[squareCol + col][squareRow + row]]) return false;
nums[puzzle[squareCol + col][squareRow + row]] = true;
}
nums = new boolean[9];
}
return true;
}
Didn't have too much time to test out, but it might work (?). The row/col variable namings might be incorrect, because I didn't have time to find that in your code, but it shouldn't matter for it to work or not.