Java 生成一个3x3幻方
我正在使用的3x3魔方中充满了数字1-9。每行、每列和每对角线中的值必须相加到15。我必须遵循以下伪代码:Java 生成一个3x3幻方,java,magic-square,Java,Magic Square,我正在使用的3x3魔方中充满了数字1-9。每行、每列和每对角线中的值必须相加到15。我必须遵循以下伪代码: recursive_funtion(position) { for number from 1 to 9, not used elsewhere already { put this number on this position, make it unavailable if solution valid {
recursive_funtion(position) {
for number from 1 to 9, not used elsewhere already {
put this number on this position, make it unavailable
if solution valid {
if solution complete {
done, show solution
}else{
recursive_function(next_position)
}
}
(make this space blank again, and the number available)
}
}
我让它一直工作,直到它通过第一行,并意识到在左上角点添加1,在中上角点添加2后,无法在该行中添加15。有人能看到我在哪里犯了错误,或者能填写我遗漏的任何逻辑吗?谢谢
public class MagicSquare {
private int[][] magicSquare;
private boolean[] available;
public MagicSquare() {
magicSquare = new int[3][3];
available = new boolean[9];
for (int i = 0; i < available.length; i++) {
available[i] = true;
}
}
public void run() {
System.out.println("Magic Square Puzzle");
solve(0, 0);
}
public boolean solve(int row, int col) {
for (int i = 1; i <= 9; i++) {
if (isAvailable(i)) {
System.out.println("placing " + i + " at (" + row + ", " + col + ")");
magicSquare[row][col] = i;
available[i - 1] = false;
//solution is valid so far
if (isFilledRow(row)) {
if (isValidRow(row) && isValidCol(col)) {
if (solve(row, col)) {
return true;
} else {
magicSquare[row][col] = 0;
magicSquare[row][col - 1] = 0;
col = col - 1;
available[magicSquare[row][col] - 1] = false;
solve(row, col);
}
}
}
if (!isFilledRow(row)) { //check logic here!
if (isValidSolution()) {
System.out.println("You found a correct solution!");
printSolution();
} else {
//new row
if (col == 2 && row != 2) {
row = row + 1;
System.out.println("new row and solve(" + row + ", " + col + ")");
solve(row, 0);
//new col
} else if (col != 2) {
col = col + 1;
System.out.println("new col and solve(" + row + ", " + col + ")");
solve(row, col);
} else if (row == 2 && col == 2) {
//check logic here!
}
}
}
magicSquare[row][col] = 0;
available[i - 1] = true;
}
}
return false;
}
public boolean isAvailable(int value) {
if (available[value - 1] == true) {
System.out.println(value + " is available to be placed");
return true;
} else {
System.out.println(value + " is not available to be placed");
return false;
}
}
public boolean isFilledRow(int row) {
if (magicSquare[row][0] != 0 && magicSquare[row][1] != 0 && magicSquare[row][2] != 0) {
System.out.println("row " + row + " is filled");
return true;
} else {
//System.out.println("row " + row + " is not filled");
return false;
}
}
public boolean isValidRow(int row) {
if (magicSquare[row][0] + magicSquare[row][1] + magicSquare[row][2] == 15) {
System.out.println("row " + row + " adds to 15");
return true;
} else {
System.out.println("row " + row + " does not add to 15");
return false;
}
}
public boolean isValidCol(int col) {
if (magicSquare[0][col] + magicSquare[1][col] + magicSquare[2][col] == 15) {
System.out.println("col " + col + " adds to 15");
return true;
} else {
//System.out.println("col " + col + " does not add to 15");
return false;
}
}
public boolean isValidOnDiagonals() {
if ((magicSquare[0][0] + magicSquare[1][1] + magicSquare[2][2] == 15) && (magicSquare[2][0] + magicSquare[1][1] + magicSquare[0][2] == 15)) {
System.out.println("diagonals add to 15");
return true;
} else {
//System.out.println("diagonals don't add to 15");
return false;
}
}
public boolean isValidSolution() {
for (int i = 0; i < magicSquare.length; i++) {
if (isValidRow(i) && isValidCol(i) && isValidOnDiagonals()) {
System.out.println("solution is valid");
return true;
}
}
//System.out.println("solution is not valid");
return false;
}
public void printSolution() {
for (int i = 0; i < magicSquare.length; i++) {
for (int j = 0; j < magicSquare[i].length; j++) {
System.out.print(magicSquare[i][j] + " ");
}
System.out.println();
}
}
}
公共类MagicSquare{
私人int[][]magicSquare;
私有布尔[]可用;
公共魔法广场(){
magicSquare=新整数[3][3];
可用=新布尔值[9];
for(int i=0;i
isValidSolution()
可以返回true
,即使平方无效
- 您的
solve()
方法的逻辑有点太复杂了。我试图简化它
- 有这么多的
System.out.println
语句实际上对算法的执行时间有很大的影响。所以我注释掉了其中的大部分
我相信你可以不断改进,目前,它在找到第一个解决方案后不会停止。它实际上会打印出所有的解决方案
希望您能利用此功能更接近最终/完善的算法:
public class MagicSquare {
private int[][] magicSquare;
private boolean[] available;
public MagicSquare() {
magicSquare = new int[3][3];
available = new boolean[9];
for (int i = 0; i < available.length; i++) {
available[i] = true;
}
}
public void run() {
System.out.println("Magic Square Puzzle");
solve(0, 0);
}
public void solve(int row, int col) {
for (int i = 1; i <= 9; i++) {
if (isAvailable(i)) {
//System.out.println("placing " + i + " at (" + row + ", " + col + ")");
magicSquare[row][col] = i;
available[i - 1] = false;
//solution is valid so far
if (isFilled()) {
if (isValidSolution()) {
System.out.println("You found a correct solution!");
printSolution();
}
} else {
if (col != 2) {
//System.out.println("new col and solve(" + row + ", " + col + ")");
solve(row, col + 1);
} else if (row != 2) {
//System.out.println("new row and solve(" + row + ", " + col + ")");
solve(row + 1, 0);
//new col
}
}
magicSquare[row][col] = 0;
available[i - 1] = true;
}
}
}
public boolean isAvailable(int value) {
if (available[value - 1] == true) {
//System.out.println(value + " is available to be placed");
return true;
} else {
//System.out.println(value + " is not available to be placed");
return false;
}
}
public boolean isValidRow(int row) {
if (magicSquare[row][0] + magicSquare[row][1] + magicSquare[row][2] == 15) {
//System.out.println("row " + row + " adds to 15");
return true;
} else {
//System.out.println("row " + row + " does not add to 15");
return false;
}
}
public boolean isValidCol(int col) {
if (magicSquare[0][col] + magicSquare[1][col] + magicSquare[2][col] == 15) {
//System.out.println("col " + col + " adds to 15");
return true;
} else {
//System.out.println("col " + col + " does not add to 15");
return false;
}
}
public boolean isValidOnDiagonals() {
if ((magicSquare[0][0] + magicSquare[1][1] + magicSquare[2][2] == 15) && (magicSquare[2][0] + magicSquare[1][1] + magicSquare[0][2] == 15)) {
//System.out.println("diagonals add to 15");
return true;
} else {
//System.out.println("diagonals don't add to 15");
return false;
}
}
public boolean isValidSolution() {
for (int i = 0; i < magicSquare.length; i++) {
if (!isValidRow(i) || !isValidCol(i)) {
//System.out.println("solution is valid");
return false;
}
}
//System.out.println("solution is not valid");
return isValidOnDiagonals();
}
public boolean isFilled() {
for (int i = 0; i < available.length; i++) {
if (available[i]) {
return false;
}
}
return true;
}
public void printSolution() {
for (int i = 0; i < magicSquare.length; i++) {
for (int j = 0; j < magicSquare[i].length; j++) {
System.out.print(magicSquare[i][j] + " ");
}
System.out.println();
}
}
}
公共类MagicSquare{
私人int[][]magicSquare;
私有布尔[]可用;
公共魔法广场(){
magicSquare=新整数[3][3];
可用=新布尔值[9];
for(int i=0;i 对于(int i=1;i对于给定的填充正方形,您必须测试a/它是否填充任何行/列,如果是,它是否符合约束条件;如果使用该值,您能够完成问题,即,您的递归是否产生有效的解决方案。如果不是,您需要清空正方形并尝试下一个值。如果您尝试了所有值,则返回错误。基本上,在解决问题之前,您需要能够回溯您的步骤。例如,您可能结束的一个点是[[1,5,9],[4,8,3],[0,0,0]]
。此时,您需要回溯3个步骤才能到达[[1,5,9],[6,0,0]]
这将为您提供solution@njzk2我编辑了我的solve方法,试图按照上面的步骤a进行操作。我在哪里实现回溯?你能给我一个回溯可能是什么样子的例子吗?通常它涉及到让你的solve
方法在到达死胡同时返回布尔值,false(当您尝试了单元格的所有可用值时)或当问题解决时为真。要使用它,您必须做的是,当您有一个不矛盾的值时(无效的填充行或列)也不是解决方案,测试调用递归函数的结果。如果为true,则返回true,如果为false,则尝试使用另一个值。在第一个示例中,一旦您有[1,2,9]
,则返回false,因为您测试了[3,4,5,6,7,8,9]
,这意味着对于调用方法,2无效,请尝试3。