我在学年的最后一个项目(我作为CS学生的第一年)的代码中找不到错误.在执行骑士巡回赛问题时,我一直坚持递归.这是有问题的文件：

https://github.com/sheagunther/tsp161knightstour/blob/master/KnightsTourRecursiveSearch.java

具体来说,我的问题是这部分代码(从第265行开始)：

else{

for(int i = 0; i < numberOfPossibleNextMoves; i++){

Cell nextCellToMoveTo = candidateNextMoves.get(i);

int currentRowStorage = currentRow;

int currentColumnStorage = currentColumn;

currentRow = nextCellToMoveTo.row;

currentColumn = nextCellToMoveTo.column;

listOfMoves.add(nextCellToMoveTo);

chessBoard[currentRow][currentColumn] = 1;

currentMoveNumber++;

boolean tourFound = findTour();

if(tourFound){

return true;

}

else{ // Undo the last move just made

backtrackCount++;

chessBoard[currentRow][currentColumn] = -1;

currentRow = currentRowStorage;

currentColumn = currentColumnStorage;

listOfMoves.remove(nextCellToMoveTo);

currentMoveNumber--;

}

}

return false;

这是findTour()的结尾.这是程序的一部分,用于测试当前正方形(也称为单元格)的所有可能移动,如果可以从新移动到正方形完成巡回,则返回true.如果游览不能从广场完成,它会进入其他地方{并撤消移动.这就是我认为的问题所在.

现在,如上面的代码设置,程序陷入无限递归循环.

记下else {声明的这一部分：

chessBoard[currentRow][currentColumn] = -1;

currentRow = currentRowStorage;

currentColumn = currentColumnStorage;

这部分代码将chessBoard中的方块更改为-1,这意味着它是未访问的(1 =已访问).如上所述,新移动的currentRow和currentColumn用于将方块设置为未访问的.然后使用currentRowStorage和currentColumnStorage将这些值重置为先前的跳转值.

如果我将代码更改为

currentRow = currentRowStorage;

currentColumn = currentColumnStorage;

chessBoard[currentRow][currentColumn] = -1;

它成功地找到了一个不正确的巡回演出,其中最后1/3左右的动作只是在几个方格之间来回跳跃.这是预期的,因为它没有正确处理重置过程.

我怀疑我的问题是由于我在宣布变量的地方.这是我的第一个复杂的递归问题,我不确定我是否正确处理currentRow / Column和currentRow / ColumnStorage之间的切换.我应该在当地或多或少地宣布它们吗？

以下是要求的相关部分：

If the tour is not completed, then findTour determines the (possibly

empty) list of vacant cells that are reachable from the knight’s

current cell, and stores this list in a locally declared list

variable, candidateNextMoves. It is critical that this list variable

be declared local to the method. If this list is empty, then there is

no way to extend the current partial tour, so findTour should return

false. If the list is not empty, then findTour tries extending the

tour by each move on the list as follows. It iterates over the list,

and for each cell of the list it makes the next move to that cell,

updating all of L (the list of moves in the tour), B (the 2D array of

the state of the board (visited, not visted)), currRow, and currCol to

reflect this move. It then recursively calls itself, assigning the

result of the call to a locally declared boolean variable, which you

might name “success”. If success is assigned true, findTour returns

true. If success is false, findTour undoes the move it just made, or

“backtracks”, and tries the next move of candidateNextMoves. You will

maintain a static int variable, backtrackCount, which is initialized

to 0, and incremented for each undo of a move.

一个注意事项 – 我调用了我的布尔值’tourFound’而不是’success’.