虽然循环卡在代码

问题描述：

随机点我需要帮助，我有我的朋友来帮我调试此代码，并且我们发现代码停止在代码上的随机点，有时在第一个循环，有时在最后一个循环等等。代码只是制作一个应用了所有规则的9x9数独板。当我分别测试每个模块时，它工作并打印出了一个出色的数独板，但是这里的问题在于某种程度上被卡住了。帮助将不胜感激。虽然循环卡在代码

int[,] helpblock = new int[10, 10]; 
int[,] helplines = new int[10, 10]; 
int[,] helpcols = new int[10, 10]; 
for (int i = 0; i < 10; i++) { 
    for (int j = 0; j < 10; j++) { 
     helplines[i, j] = 0; 
     helpcols[i, j] = 0; 
     helpblock[i, j] = 0; 
    } 
} 

Random rnf = new Random(); 
int help1; 
for (int i = 0; i < 9; i++) { 
    for (int j = 0; j < 9; j++) { 
     help1 = rnf.Next(1, 10); 
     if (i == 0 || i == 1 || i == 2) { 
      if (j == 0 || j == 1 || j == 2) { 
       if (helplines[i, help1] != 0 || helpblock[0, help1] != 0 || helpcols[j, help1] != 0) { 
        while (helplines[i, help1] != 0 || helpblock[0, help1] != 0 || helpcols[j, help1] != 0) { 
         System.Threading.Thread.Sleep(1); 
         help1 = rnf.Next(1, 10); 
        } 
       } 
       map[i, j] = help1; 
       helplines[i, help1]++; 
       helpcols[j, help1]++; 
       helpblock[0, help1]++; 
      } else if (j == 3 || j == 4 || j == 5) { 
       if (helplines[i, help1] != 0 || helpblock[1, help1] != 0 || helpcols[j, help1] != 0) { 
        while (helplines[i, help1] != 0 || helpblock[1, help1] != 0 || helpcols[j, help1] != 0) { 
         System.Threading.Thread.Sleep(1); 
         help1 = rnf.Next(1, 10); 
        } 
       } 
       map[i, j] = help1; 
       helplines[i, help1]++; 
       helpcols[j, help1]++; 
       helpblock[1, help1]++; 
      } else if (j == 6 || j == 7 || j == 8) { 
       if (helplines[i, help1] != 0 || helpblock[2, help1] != 0 || helpcols[j, help1] != 0) { 
        while (helplines[i, help1] != 0 || helpblock[2, help1] != 0 || helpcols[j, help1] != 0) { 
         System.Threading.Thread.Sleep(1); 
         help1 = rnf.Next(1, 10); 
        } 
       } 
       map[i, j] = help1; 
       helplines[i, help1]++; 
       helpcols[j, help1]++; 
       helpblock[2, help1]++; 
      } 
     } else if (i == 3 || i == 4 || i == 5) { 
      if (j == 0 || j == 1 || j == 2) { 
       if (helplines[i, help1] != 0 || helpblock[3, help1] != 0 || helpcols[j, help1] != 0) { 
        while (helplines[i, help1] != 0 || helpblock[3, help1] != 0 || helpcols[j, help1] != 0) { 
         System.Threading.Thread.Sleep(1); 
         help1 = rnf.Next(1, 10); 
        } 
       } 
       map[i, j] = help1; 
       helplines[i, help1]++; 
       helpcols[j, help1]++; 
       helpblock[3, help1]++; 
      } else if (j == 3 || j == 4 || j == 5) { 
       if (helplines[i, help1] != 0 || helpblock[4, help1] != 0 || helpcols[j, help1] != 0) { 
        while (helplines[i, help1] != 0 || helpblock[4, help1] != 0 || helpcols[j, help1] != 0) { 
         System.Threading.Thread.Sleep(1); 
         help1 = rnf.Next(1, 10); 
        } 
       } 

       map[i, j] = help1; 
       helplines[i, help1]++; 
       helpcols[j, help1]++; 
       helpblock[4, help1]++; 
      } else if (j == 6 || j == 7 || j == 8) { 
       if (helplines[i, help1] != 0 || helpblock[5, help1] != 0 || helpcols[j, help1] != 0) { 
        while (helplines[i, help1] != 0 || helpblock[5, help1] != 0 || helpcols[j, help1] != 0) { 
         System.Threading.Thread.Sleep(1); 
         help1 = rnf.Next(1, 10); 
        } 
       } 

       map[i, j] = help1; 
       helplines[i, help1]++; 
       helpcols[j, help1]++; 
       helpblock[5, help1]++; 

      } 
     } else if (i == 6 || i == 7 || i == 8) { 
      if (j == 0 || j == 1 || j == 2) { 
       if (helplines[i, help1] != 0 || helpblock[6, help1] != 0 || helpcols[j, help1] != 0) { 
        while (helplines[i, help1] != 0 || helpblock[6, help1] != 0 || helpcols[j, help1] != 0) { 
         System.Threading.Thread.Sleep(1); 
         help1 = rnf.Next(1, 10); 
        } 
       } 

       map[i, j] = help1; 
       helplines[i, help1]++; 
       helpcols[j, help1]++; 
       helpblock[6, help1]++; 
      } else if (j == 3 || j == 4 || j == 5) { 
       if (helplines[i, help1] != 0 || helpblock[7, help1] != 0 || helpcols[j, help1] != 0) { 
        while (helplines[i, help1] != 0 || helpblock[7, help1] != 0 || helpcols[j, help1] != 0) { 
         System.Threading.Thread.Sleep(1); 
         help1 = rnf.Next(1, 10); 
        } 
       } 

       map[i, j] = help1; 
       helplines[i, help1]++; 
       helpcols[j, help1]++; 
       helpblock[7, help1]++; 
      } else if (j == 6 || j == 7 || j == 8) { 
       if (helplines[i, help1] != 0 || helpblock[8, help1] != 0 || helpcols[j, help1] != 0) { 
        while (helplines[i, help1] != 0 || helpblock[8, help1] != 0 || helpcols[j, help1] != 0) { 
         System.Threading.Thread.Sleep(1); 
         help1 = rnf.Next(1, 10); 
        } 
       } 

       map[i, j] = help1; 
       helplines[i, help1]++; 
       helpcols[j, help1]++; 
       helpblock[8, help1]++; 
      } 
     } 
    } 
} 

PrintSudoku(map);

在C＃ – EliKatz

请注明您所使用的语言进行编程。 – Rajesh

了解如何使用调试器。它会准确显示发生了什么，特别是当它“有时”发生时。还有，为什么只使用9x9时分配10x10数组？ –

答

这不是一个真正的答案，但希望它可以帮助您调试程序。里面你for循环您有if报表一大堆，看起来是这样的：

if (i == 0 || i == 1 || i == 2) { 
    if (j == 0 || j == 1 || j == 2) {

它看起来像你使用这个逻辑，以确定哪些阻碍你的，因为这是一个数组，你正在使用第一个索引的硬编码值。

不要这样做，请考虑使用基于当前行和列值的公式计算块。

网格目前看起来是这样的：

| 0 1 2 | 3 4 5 | 6 7 8 | 
--+-------+-------+-------- 
0 |  |  |  | 
1 | [0] | [1] | [2] | 
2 |  |  |  | 
--+-------+-------+-------- 
3 |  |  |  | 
4 | [3] | [4] | [5] | 
5 |  |  |  | 
--+-------+-------+-------- 
6 |  |  |  | 
7 | [6] | [7] | [8] | 
8 |  |  |  | 
---------------------------

什么，你可能会注意到的是，如果你把行号和3做一个整数除法，你会得到Block行数（0，1 ，或2）。将此数字乘以3会得到每行中的第一个块（0,3和6）。现在，如果我们将当前列和整数除以3，我们将得到当前块列（0,1或2）。添加“块行”到“块列”，那么将会给我们的块索引：

int block = ((row/3) * 3) + (column/3);

（请注意，它看起来像你可以简化这个，因为行被分成三部分，然后乘以三，但请记住，整数除法舍去，所以2/3 == 0，而不是.6667）

通过这种方式计算块，您可以大幅减少你必须走过的行数。

此外，使用一些更有意义的变量名将有助于提高可读性。

我将你的代码简化为我认为完全一样的东西。也许这会帮助你分析：

int[,] blocks = new int[10, 10]; 
int[,] rows = new int[10, 10]; 
int[,] columns = new int[10, 10]; 

for (int i = 0; i < 10; i++) 
{ 
    for (int j = 0; j < 10; j++) 
    { 
     rows[i, j] = 0; 
     columns[i, j] = 0; 
     blocks[i, j] = 0; 
    } 
} 

Random rnd = new Random(); 
int randomNumber; 

for (int row = 0; row < 9; row++) 
{ 
    for (int column = 0; column < 9; column++) 
    { 
     int block = ((row/3) * 3) + (column/3); 
     randomNumber = rnd.Next(1, 10);      

     while (rows[row, randomNumber] != 0 || 
      blocks[block, randomNumber] != 0 || 
      columns[column, randomNumber] != 0) 
     { 
      System.Threading.Thread.Sleep(1); 
      randomNumber = rnd.Next(1, 10); 
     } 

     map[row, column] = randomNumber; 
     rows[row, randomNumber]++; 
     columns[column, randomNumber]++; 
     blocks[block, randomNumber]++; 
    } 
}

玩这个当然，以后，我感兴趣的问题。下面是我想出了解决这个问题，这是一个很值得穷举法：

开始用数字1-9可用于每个行，列，并阻止
从可用列表数字为一个单元格，随机分配其中一个，一次一个单元格
从该行，列和块的可用数字中删除该数字
如果我们遇到没有有效数字的情况可用于单元格，重置所有内容并重试。

请注意，这可能会运行，永远不会完成，但在我的测试中，它通常会在少于400次尝试（从用户的角度来看相对较快）中找到解决方案。

我做的不同之处在于，我不是使用数组来存储行/列/块信息，而是决定使用存储行/列/块索引作为关键字的Dictionary<int, List<int>>，以及每个关键字的值是该索引的可用数字的List。

然后在我们的循环中，我只将当前行，列和块的剩余可用数字交叉到List。然后我在该列表中选择一个随机索引并将该值分配给单元格（并将其从该行，列和块中的可用项目中移除）。

我还添加了一些日志记录来找出在找到解决方案之前失败了多少次。

下面的代码，这也打印在最后的解决方案：

static void Main() 
{ 
    int[,] map = new int[9, 9]; 
    var blocks = new Dictionary<int, List<int>>(); 
    var rows = new Dictionary<int, List<int>>(); 
    var columns = new Dictionary<int, List<int>>(); 

    // initialize lists of available numbers 
    for (int i = 0; i < 9; i++) 
    { 
     blocks.Add(i, new List<int> { 1, 2, 3, 4, 5, 6, 7, 8, 9 }); 
     rows.Add(i, new List<int> { 1, 2, 3, 4, 5, 6, 7, 8, 9 }); 
     columns.Add(i, new List<int> { 1, 2, 3, 4, 5, 6, 7, 8, 9 }); 
    } 

    Random rnd = new Random(); 
    int failCount = 0; 

    // For each row 
    for (int row = 0; row < 9; row++) 
    { 
     // For each column in this row 
     for (int column = 0; column < 9; column++) 
     { 
      // Calculate block based on current row and column 
      int block = ((row/3) * 3) + (column/3); 

      // Get set of available numbers for this cell by getting the intersection 
      // of available numbers for the row, column, and block 
      var availableItems = 
       rows[row].Intersect(
        columns[column].Intersect(
         blocks[block])).ToList(); 

      // If we reach a point where there are no availableItems, then this is 
      // not a valid Sudoku pattern. Reset everything and try again 
      if (availableItems.Count == 0) 
      { 
       failCount++; 
       Console.WriteLine($"Failed {failCount} times. Trying again..."); 

       // Reset to the first cell 
       row = 0; 
       column = 0; 
       block = 0; 

       // Initialize data 
       for (int i = 0; i < 9; i++) 
       { 
        blocks[i] = new List<int> { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; 
        rows[i] = new List<int> { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; 
        columns[i] = new List<int> { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; 
       } 
       availableItems = new List<int> { 1, 2, 3, 4, 5, 6, 7, 8, 9 }; 
      } 

      // Grab a number from the available numbers by choosing a random index 
      var randomNumber = availableItems[rnd.Next(0, availableItems.Count)]; 

      // Update our map with this item 
      map[row, column] = randomNumber; 

      // Remove this item from our lists 
      rows[row].Remove(randomNumber); 
      columns[column].Remove(randomNumber); 
      blocks[block].Remove(randomNumber); 
     } 
    } 

    // Print our Sudoku map: 
    Console.WriteLine(new string('-', 25)); 

    for(int i = 0; i < 9; i++) 
    { 
     Console.Write("|"); 

     for (int j = 0; j < 9; j++) 
     { 
      Console.Write($" {map[i, j]}"); 
      if ((j + 1) % 3 == 0) Console.Write(" |"); 
     } 

     Console.WriteLine(); 

     if ((i + 1) % 3 == 0) Console.WriteLine(new string('-', 25)); 
    } 

    Console.Write("\nDone!\nPress any key to exit..."); 
    Console.ReadKey(); 
}

输出继电器：

谢谢你，帮助我 – EliKatz

答

这些循环看起来致命的：

while (helplines[i, help1] != 0 || helpblock[5, help1] != 0 || helpcols[j, help1] != 0) { 
    System.Threading.Thread.Sleep(1); 
    help1 = rnf.Next(1, 10); 
}

如果help1新值不会改变while条件false，它会永远循环下去。

我会试着进一步研究一下 – EliKatz

此外，如果没有任何数组项都等于零，这也将永远循环。 –

@RufusL它比这更糟糕。它将永远循环，如果*任何*的项目是非零（无论多少为零） – Bohemian

虽然循环卡在代码

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