如何生成一个完整的数独板？算法错误

小编典典

如何生成一个完整的数独板？算法错误

algorithm

我正在尝试生成一个完整的（如每个单元都填充有一个数字）类似Sudoku的板。这是与数独无关的其他事物，因此我对达到具有可以解决的白色方块的数独或与数独无关的数独都不感兴趣。不知道你是否明白我的意思。

我已经在Java中完成了此操作：

private int sudokuNumberSelector(int x, int y, int[][] sudoku) {


    boolean valid = true;
    String validNumbers = new String();
    int[] aValidNumbers;
    int squarexstart = 0;
    int squareystart = 0;
    int b = 0;                          // For random numbers
    Random randnum = new Random();
    randnum.setSeed(new Date().getTime());

    // Check numbers one by one
    for(int n = 1; n < 10; n++) {

        valid = true;

        // Check column
        for(int i = 0; i < 9; i++) {
            if(sudoku[i][y] == n) {
                valid = false;
            }
        }

        // Check file
        for(int j = 0; j < 9; j++) {
            if(sudoku[x][j] == n) {
                valid = false;
            }
        }

        // Check square
        switch (x) {
        case 0: case 1: case 2: squarexstart = 0; break;
        case 3: case 4: case 5: squarexstart = 3; break;
        case 6: case 7: case 8: squarexstart = 6; break;
        }

        switch (y) {
        case 0: case 1: case 2: squareystart = 0; break;
        case 3: case 4: case 5: squareystart = 3; break;
        case 6: case 7: case 8: squareystart = 6; break;
        }

        for(int i = squarexstart; i < (squarexstart + 3); i++ ) {
            for(int j = squareystart; j < (squareystart + 3); j++ ) {
                if(sudoku[i][j] == n) {
                    valid = false;
                }
            }
        }

        // If the number is valid, add it to the String
        if(valid) {
            validNumbers += n;
        }
    }

    if(validNumbers.length() != 0) {

        // String to int[]
        aValidNumbers = fromPuzzleString(validNumbers);

        // By this random number, return the valid number in its position
        Log.d(TAG, "NUMBERS: " + validNumbers.length());

        // Select a random number from the int[]
        b = randnum.nextInt((aValidNumbers.length));


        return aValidNumbers[b];

    } else {
        return 0;
    }
}

从这段代码中调用此方法：

int[][] sudoku = new int[9][9];

for(int i = 0; i < 9; i++) {
            for(int j = 0; j < 9; j++) {
                sudoku[i][j] = sudokuNumberSelector(i, j, sudoku);
            }
        }

但这并不像看起来那么容易！当算法生成了像这样的木板的一部分时，循环就以粗体显示在单元格上：

|||164527389|||
|||983416257|||
|||257938416|||
|||719352648|||
|||3256791**0**0|||
|||000000000|||
|||000000000|||
|||000000000|||
|||000000000|||

没有任何数字可放入此单元格中，因为根据数独规则的所有数字均已在列，行或正方形上！

对我来说这是一场噩梦。有什么办法可以工作吗？如果没有，我想我必须重做所有事情，就像我在做数独游戏一样。

提前致谢。

阅读 301

2020-07-28

共1个答案

小编典典

问题在于，在大多数情况下无法使用随机数生成完整的电路板，如果无法找到下一个单元，则必须使用回溯。我曾经写过一个数独游戏，所以这是生成填充板的代码。

这是Cell类。

 public class SudokuCell implements Serializable {

    private int value;
    private boolean filled;
    private HashSet<Integer> tried;

    public SudokuCell() {
        filled = false;
        tried = new HashSet();
    }

    public boolean isFilled() {
        return filled;
    }

    public int get() {
        return value;
    }

    public void set(final int number) {
        filled = true;
        value = number;
        tried.add(number);
    }

    public void clear() {
        value = 0;
        filled = false;
    }

    public void reset() {
        clear();
        tried.clear();
    }

    public void show() {
        filled = true;
    }

    public void hide() {
        filled = false;
    }

    public boolean isTried(final int number) {
        return tried.contains(number);
    }

    public void tryNumber(final int number) {
        tried.add(number);
    }

    public int numberOfTried() {
        return tried.size();
    }
 }

这是Field类（将所有数据保留在一个对象中非常方便）。

 public class SudokuField implements Serializable {

    private final int blockSize;
    private final int fieldSize;
    private SudokuCell[][] field;

    public SudokuField(final int blocks) {
        blockSize = blocks;
        fieldSize = blockSize * blockSize;
        field = new SudokuCell[fieldSize][fieldSize];
        for (int i = 0; i < fieldSize; ++i) {
            for (int j = 0; j < fieldSize; ++j) {
                field[i][j] = new SudokuCell();
            }
        }
    }

    public int blockSize() {
        return blockSize;
    }

    public int fieldSize() {
        return fieldSize;
    }

    public int variantsPerCell() {
        return fieldSize;
    }

    public int numberOfCells() {
        return fieldSize * fieldSize;
    }

    public void clear(final int row, final int column) {
        field[row - 1][column - 1].clear();
    }

    public void clearAllCells() {
        for (int i = 0; i < fieldSize; ++i) {
            for (int j = 0; j < fieldSize; ++j) {
                field[i][j].clear();
            }
        }
    }

    public void reset(final int row, final int column) {
        field[row - 1][column - 1].reset();
    }

    public void resetAllCells() {
        for (int i = 0; i < fieldSize; ++i) {
            for (int j = 0; j < fieldSize; ++j) {
                field[i][j].reset();
            }
        }
    }

    public boolean isFilled(final int row, final int column) {
        return field[row - 1][column - 1].isFilled();
    }

    public boolean allCellsFilled() {
        for (int i = 0; i < fieldSize; ++i) {
            for (int j = 0; j < fieldSize; ++j) {
                if (!field[i][j].isFilled()) {
                    return false;
                }
            }
        }
        return true;
    }

    public int numberOfFilledCells() {
        int filled = 0;
        for (int i = 1; i <= fieldSize; ++i) {
            for (int j = 1; j <= fieldSize; ++j) {
                if (isFilled(i, j)) {
                    ++filled;
                }
            }
        }
        return filled;
    }

    public int numberOfHiddenCells() {
        return numberOfCells() - numberOfFilledCells();
    }

    public int get(final int row, final int column) {
        return field[row - 1][column - 1].get();
    }

    public void set(final int number, final int row, final int column) {
        field[row - 1][column - 1].set(number);
    }

    public void hide(final int row, final int column) {
        field[row - 1][column - 1].hide();
    }

    public void show(final int row, final int column) {
        field[row - 1][column - 1].show();
    }

    public void tryNumber(final int number, final int row, final int column) {
        field[row - 1][column - 1].tryNumber(number);
    }

    public boolean numberHasBeenTried(final int number, final int row, final int column) {
        return field[row - 1][column - 1].isTried(number);
    }

    public int numberOfTriedNumbers(final int row, final int column) {
        return field[row - 1][column - 1].numberOfTried();
    }

    public boolean checkNumberBox(final int number, final int row, final int column) {
        int r = row, c = column;
        if (r % blockSize == 0) {
            r -= blockSize - 1;
        } else {
            r = (r / blockSize) * blockSize + 1;
        }
        if (c % blockSize == 0) {
            c -= blockSize - 1;
        } else {
            c = (c / blockSize) * blockSize + 1;
        }
        for (int i = r; i < r + blockSize; ++i) {
            for (int j = c; j < c + blockSize; ++j) {
                if (field[i - 1][j - 1].isFilled() && (field[i - 1][j - 1].get() == number)) {
                    return false;
                }
            }
        }
        return true;
    }

    public boolean checkNumberRow(final int number, final int row) {
        for (int i = 0; i < fieldSize; ++i) {
            if (field[row - 1][i].isFilled() && field[row - 1][i].get() == number) {
                return false;
            }
        }
        return true;
    }

    public boolean checkNumberColumn(final int number, final int column) {
        for (int i = 0; i < fieldSize; ++i) {
            if (field[i][column - 1].isFilled() && field[i][column - 1].get() == number) {
                return false;
            }
        }
        return true;
    }

    public boolean checkNumberField(final int number, final int row, final int column) {
        return (checkNumberBox(number, row, column)
                && checkNumberRow(number, row)
                && checkNumberColumn(number, column));
    }

    public int numberOfPossibleVariants(final int row, final int column) {
        int result = 0;
        for (int i = 1; i <= fieldSize; ++i) {
            if (checkNumberField(i, row, column)) {
                ++result;
            }
        }
        return result;
    }

    public boolean isCorrect() {
        for (int i = 0; i < fieldSize; ++i) {
            for (int j = 0; j < fieldSize; ++j) {
                if (field[i][j].isFilled()) {
                    int value = field[i][j].get();
                    field[i][j].hide();
                    boolean correct = checkNumberField(value, i + 1, j + 1);
                    field[i][j].show();
                    if (!correct) {
                        return false;
                    }
                }
            }
        }
        return true;
    }

    public Index nextCell(final int row, final int column) {
        int r = row, c = column;
        if (c < fieldSize) {
            ++c;
        } else {
            c = 1;
            ++r;
        }
        return new Index(r, c);
    }

    public Index cellWithMinVariants() {
        int r = 1, c = 1, min = 9;
        for (int i = 1; i <= fieldSize; ++i) {
            for (int j = 1; j <= fieldSize; ++j) {
                if (!field[i - 1][j - 1].isFilled()) {
                    if (numberOfPossibleVariants(i, j) < min) {
                        min = numberOfPossibleVariants(i, j);
                        r = i;
                        c = j;
                    }
                }
            }
        }
        return new Index(r, c);
    }

    public int getRandomIndex() {
        return (int) (Math.random() * 10) % fieldSize + 1;
    }
}

最后是充满游戏板的功能

private void generateFullField(final int row, final int column) {
    if (!field.isFilled(field.fieldSize(), field.fieldSize())) {
        while (field.numberOfTriedNumbers(row, column) < field.variantsPerCell()) {
            int candidate = 0;
            do {
                candidate = field.getRandomIndex();
            } while (field.numberHasBeenTried(candidate, row, column));
            if (field.checkNumberField(candidate, row, column)) {
                field.set(candidate, row, column);
                Index nextCell = field.nextCell(row, column);
                if (nextCell.i <= field.fieldSize()
                        && nextCell.j <= field.fieldSize()) {
                    generateFullField(nextCell.i, nextCell.j);
                }
            } else {
                field.tryNumber(candidate, row, column);
            }
        }
        if (!field.isFilled(field.fieldSize(), field.fieldSize())) {
            field.reset(row, column);
        }
    }
}

关键是在继续操作之前，您要保存已经为每个单元格尝试过的数字。如果要走到尽头，您只需要为上一个单元格尝试另一个数字即可。如果不可能，请擦除该单元格并向后退一个单元格。迟早您会完成它。（精算只需要很少的时间）。

2020-07-28