我正在写一个数独的棋盘游戏生成器。我的代码如下所示:
/**
* Created by szubansky on 3/9/16.
*/
public class SudokuBoard {
static int N = 9;
static int[][] grid = new int[N][N];
static void printGrid()
{
for (int row = 0; row < N; row++)
{
for (int col = 0; col < N; col++) {
System.out.printf("%5d", grid[row][col]);
}
System.out.println("\n");
}
}
private static boolean checkRow(int row, int num)
{
for( int col = 0; col < 9; col++ )
if(grid[row][col] == num)
return false;
return true;
}
private static boolean checkCol(int col, int num)
{
for( int row = 0; row < 9; row++ )
if(grid[row][col] == num)
return false;
return true;
}
private static boolean checkBox(int row, int col, int num)
{
row = (row / 3) * 3;
col = (col / 3) * 3;
for(int r = 0; r < 3; r++)
for(int c = 0; c < 3; c++)
if(grid[row+r][col+c] == num)
return false;
return true;
}
public static boolean fillBoard(int row, int col, int[][] grid)
{
if(row==9) {
row = 0;
if (++col == 9)
return true;
}
if(grid[row][col] != 0)
return fillBoard(row+1, col, grid);
for(int num=1 + (int)(Math.random() * ((9 - 1) + 1)); num<=9; num++)
{
if(checkRow(row,num) && checkCol(col,num) && checkBox(row,col,num)){
grid[row][col] = num;
if(fillBoard(row+1, col, grid))
return true;
}
}
grid[row][col] = 0;
return false;
}
static public void main(String[] args){
fillBoard(0, 0, grid);
printGrid();
}
}
问题在于在fillBoard回溯算法中生成数字
到处都是0
当我将for循环中的num范围更改为10时,它会顺利进行,但我的数字必须低于9
我还可以将回溯fillBoard的开头更改为row==8
和col==8
,并使用随机数正确填充它,使用&#34; 0&#34;留下最后一行和最后一列。
如何从1到9生成随机数并填充我的所有网格?
答案 0 :(得分:1)
试试这个:
public static void main(String[] args) {
int[][] grid = new int[9][9];
randomFillGrid(grid, 1, 10);
for (int[] row : grid) {
System.out.println(Arrays.toString(row));
}
}
static void randomFillGrid(int[][] grid, int randomNumberOrigin, int randomNumberBound) {
PrimitiveIterator.OfInt iterator = ThreadLocalRandom.current()
.ints(randomNumberOrigin, randomNumberBound)
.iterator();
for (int[] row : grid) {
for (int i = 0; i < row.length; i++) {
row[i] = iterator.nextInt();
}
}
}
编辑:
如果你想生成一个数独网格,即
同一个整数可能不会在9x9游戏板的同一行,列或9个3×3子区域中的任何一个中出现两次。
import java.util.Arrays;
import java.util.Random;
import java.util.concurrent.ThreadLocalRandom;
/**
* @author FaNaJ
*/
public class SudokuGenerator {
private static final int N = 9;
private static final int S_N = 3;
public static int[][] generateSudokuGrid() {
int[][] grid = new int[N][N];
int[] row = {1, 2, 3, 4, 5, 6, 7, 8, 9};
for (int y = 0; y < N; y++) {
int attempts = 0;
do {
if (++attempts > 1000000) { // Oops! I know (sometimes :) it's not a good algorithm...
return generateSudokuGrid();
}
shuffleArray(row);
} while (!isAllowed(grid, y, row));
System.arraycopy(row, 0, grid[y], 0, N);
}
return grid;
}
static boolean isAllowed(int[][] grid, int y, int[] row) {
// check columns
for (int i = 0; i < y; i++) {
for (int j = 0; j < N; j++) {
if (grid[i][j] == row[j]) {
return false;
}
}
}
// check sub grids
int startY = (y / S_N) * S_N;
for (int x = 0; x < N; x++) {
int startX = (x / S_N) * S_N;
for (int j = startX; j < startX + S_N; j++) {
if (j != x) {
for (int i = startY; i < y; i++) {
if (grid[i][j] == row[x]) {
return false;
}
}
}
}
}
return true;
}
static void shuffleArray(int[] array) {
Random random = ThreadLocalRandom.current();
for (int i = N; i > 1; i--) {
swap(array, i - 1, random.nextInt(i));
}
}
static void swap(int[] array, int i, int j) {
int tmp = array[i];
array[i] = array[j];
array[j] = tmp;
}
public static void main(String[] args) {
int[][] grid = generateSudokuGrid();
for (int[] row : grid) {
System.out.println(Arrays.toString(row));
}
}
}
输出:
[3, 4, 6, 9, 1, 2, 7, 8, 5]
[9, 7, 2, 3, 8, 5, 4, 1, 6]
[5, 8, 1, 6, 7, 4, 3, 2, 9]
[7, 6, 3, 8, 2, 9, 1, 5, 4]
[4, 9, 5, 1, 6, 7, 2, 3, 8]
[2, 1, 8, 4, 5, 3, 6, 9, 7]
[6, 2, 4, 5, 9, 1, 8, 7, 3]
[8, 5, 7, 2, 3, 6, 9, 4, 1]
[1, 3, 9, 7, 4, 8, 5, 6, 2]
答案 1 :(得分:0)
您可以使用Random
类java.util
包来生成随机数。下面是一个生成1到9之间随机数的示例:
Random rn = new Random();
int answer = rn.nextInt(9) + 1;
答案 2 :(得分:0)
在我开始之前,让我这样说,虽然我不会直接告诉你所有错误是什么,但我会帮助您缩小搜索范围并建议进一步调试代码的方法。
现在解决您的问题:您的阵列仅填充零。
只有两行可以填充您的数组。 第一个设置数组中的一个点,使其具有1到9之间的随机int值。如果此随机数未通过一系列测试,则数组中的该点将设置为零。
如果其中一个测试者从未返回true,则数组将仅填充零。
这留下了4个可能被窃听的功能。 -checkRow -checkCol -checkBox -fillBoard
前两个功能相对简单,所以我可以肯定地说这些功能很好。
这导致只有checkBox和fillBoard函数被怀疑是你的bug的原因。
此时,调试就在这里。
这两个函数都包含循环。
查看循环是否正常的一种方法是将一组预期的更改/返回值与实际从程序中获得的值进行比较。
要执行此操作,请使用调试器 (断言语句+ java -ea program.java和/或其他调试器方法)
或者,您可以在循环中添加print语句,以打印各种变量/函数的已获取值。
无论如何,欢迎来到Stacks Overflow。
答案 3 :(得分:0)
你提到你正在使用回溯算法,所以我认为使用回溯算法并试图证明它是什么会很有趣。下面是你和我的代码的混搭。我试图坚持你正在做的事情,只是添加/改变我想要的东西。
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
/**
* SudokuBoard.
*/
public class SudokuBoard {
// Dimension size of everything
static int N = 9;
// Sudoku grid
static int[][] grid = new int[N][N];
// Values that are potentially valid at each position in the sudoku grid
static int[][][] values = new int[N][N][N];
// Current index into the values
static int[][] index = new int[N][N];
/**
* Return a shuffled list of values from 1 - 9 with each value
* appearing only once.
*/
private static List<Integer> getValidValues(){
List<Integer> validValues = new ArrayList<>();
for(int i = 1; i < 10; i++){
validValues.add(i);
}
Collections.shuffle(validValues);
return validValues;
}
/**
* Populate the values array with shuffled values to choose from.
*/
private static void initValues()
{
for(int i = 0; i < values.length; i++){
for(int j = 0; j < values[i].length; j++){
List<Integer> validValues = getValidValues();
for(int k = 0; k < values[j].length; k++){
values[i][j][k] = validValues.get(k);
}
}
}
}
/**
* print the 2D sudoku grid.
*/
public static void printGrid()
{
for (int row = 0; row < N; row++)
{
for (int col = 0; col < N; col++) {
System.out.printf("%5d", grid[row][col]);
}
System.out.println("\n");
}
}
/**
* Check the row for validity.
*/
private static boolean checkRow(int row, int num)
{
for( int col = 0; col < 9; col++ )
if(grid[row][col] == num)
return false;
return true;
}
/**
* Check the col for validity.
*/
private static boolean checkCol(int col, int num)
{
for( int row = 0; row < 9; row++ )
if(grid[row][col] == num)
return false;
return true;
}
/**
* Check the box for validity.
*/
private static boolean checkBox(int row, int col,
int num, boolean testRowCol)
{
int theR = row;
int theC = col;
row = (row / 3) * 3;
col = (col / 3) * 3;
for(int r = 0; r < 3; r++) {
for (int c = 0; c < 3; c++) {
if (testRowCol) {
if (theR == row + r && theC == col + c){
continue;
}
}
if (grid[row + r][col + c] == num) {
return false;
}
}
}
return true;
}
/**
* Build the sudoku board.
*/
public static boolean fillBoard(int row, int col)
{
// if we are back at the beginning then success!
// but just for sanity we will check that its right.
if(row != 0 && col != 0){
if(row % 9 == 0 && col % 9 == 0){
return checkBoard();
}
}
// don't go out of range in the grid.
int r = row % 9;
int c = col % 9;
// get the index in the values array that we care about
int indexIntoValues = index[r][c];
// if the index is out of range then we have domain wipe out!
// lets reset the index and try to back up a step. Backtrack!
if(indexIntoValues > 8){
index[r][c] = 0;
// there are a few cases to cover
// if we are at the beginning and the index is out
// of range then failure. We should never get here.
if(row == 0 && col == 0) {
return false;
}
grid[r][c] = 0;
// if the row is at 0 then back the row up to row - 1 and
// the col - 1
if(r == 0 && c > 0) {
return fillBoard(row - 1, col - 1);
}
// if the row is greater than 0 then just back it up by 1
if(r > 0){
return fillBoard(row - 1, col);
}
}
index[r][c] += 1;
// get the value that we care about
int gridValue = values[r][c][indexIntoValues];
// is this value valid
if(checkRow(r,gridValue) && checkCol(c,gridValue) &&
checkBox(r,c,gridValue,false)){
// assign it and move on to the next one
grid[r][c] = gridValue;
return fillBoard(row+1, r == 8 ? col + 1 : col);
}
// the value is not valid so recurse and try the next value
return fillBoard(row, col);
}
/**
* This is a sanity check that the board is correct.
* Only run it after a solution is returned.
*/
private static boolean checkBoard(){
//the grid is N X N so just use N as the condition.
//check rows are ok.
for(int i = 0; i < N; i++){
for(int j = 0; j < N; j++){
//for each of the elements in the row compare against
//every other one except its self.
int toTest = grid[i][j];
//check that the digits in the elements are in the valid range.
if(toTest > 9 || toTest < 1)
return false;
for(int k = 0; k < N; k++){
//don't test me against myself
if(k == j)
continue;
//if i am equal to another in the row there is an error.
if(toTest == grid[i][k])
return false;
}
}
}
//check the cols are ok.
for(int i = 0; i < N; i++){
for(int j = 0; j < N; j++){
//flip i and j to go for cols.
int toTest = grid[j][i];
for(int k = 0; k < N; k++){
if(k == j)
continue;
if(toTest == grid[k][i])
return false;
}
}
}
//check blocks are ok
for(int i = 0; i < N; i++){
for(int j = 0; j < N; j++){
int toTest = grid[i][j];
if(!checkBox(i, j, toTest, true))
return false;
}
}
return true;
}
static public void main(String[] args){
initValues();
if(fillBoard(0, 0))
printGrid();
else
System.out.println("Something is broken");
}
}
所以我添加的主要内容是两个多维数组,一个是值列表,另一个是该值列表的当前索引。因此values
数组会保留每个单元格可以拥有的所有有效值的列表。我拖拽了这些值,给你随意的感觉。出于任何性能原因,我们并不真正关心价值订购,但我们希望它是&#34;随机&#34;为了从运行中获得不同的板。接下来,index
数组会跟踪您在values
数组中的位置。在index
数组中的索引超出范围的任何时候,这意味着没有任何值可以工作。这是您需要将索引重置为零并返回到您来自的单元格以在那里尝试新值的点。这是回溯步骤。我一如既往地按顺序移动阵列,但没有必要。我认为你实际上可以转到有效值最小的变量,所以在这种情况下最高的索引,搜索应该更快,但是谁现在关心,它是一个小搜索,深度是81,你赢了!我还包括董事会有效的健全检查。
这是一个输出:
1 3 2 5 7 6 4 9 8
7 5 8 1 9 4 3 2 6
6 4 9 8 2 3 1 7 5
8 6 3 2 5 9 7 1 4
5 7 4 6 3 1 9 8 2
9 2 1 4 8 7 5 6 3
4 9 7 3 6 8 2 5 1
3 8 5 7 1 2 6 4 9
2 1 6 9 4 5 8 3 7
我希望这会以某种方式帮助你,这是一个有趣的。