如何降低圈复杂度？

Question

Thanks for reading my question. I am currently taking a Java class on Coursera, and was asked to write a program on minesweeper for the assignment. My code creates the correct result, but my grade was deducted greatly because my code is ”excessively complex, with a cyclomatic complexity of 60“ according to the auto-grader. I understand that there are too many conditionals and loops, but I had a hard time trying to make it more simple.

Here is my code. It takes 3 integer command-line arguments m, n, and k to create an m-by-n grid with k mines in random locations. I use "5" to mark the mines instead of " " because the highest a number in a tile can get is 4 (since a tile has 4 sides). If two mines are located side by side, extra values might be added to its marker of "5". So I make all the values >= 5 become " " when I print them out. Each value is separated by two spaces.

public class Minesweeper {
  public static void main(String[] args) {
    int m = Integer.parseInt(args[0]);
    int n = Integer.parseInt(args[1]);
    int k = Integer.parseInt(args[2]);
    int[][] mine = new int[m][n];
    //put the mines
    for(int z = 0; z < k; z++) {
      int randomX = (int) (Math.random() * m);
      int randomY = (int) (Math.random() * n);
      mine[randomX][randomY] = 5; 
    }

    for(int y = 0; y < n; y++) {
      for(int x = 0; x < m; x++) {
        //first row of the grid
        if(y == 0) {
          //upper left corner
          if(x == 0) {
            if(mine[x + 1][y] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x][y + 1] >= 5) {
              mine[x][y] += 1;
            }
          }
          //upper right corner
          else if(x == m - 1) {
            if(mine[x - 1][y] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x][y + 1] >= 5) {
              mine[x][y] += 1;
            } 
          }
          //mid of first row
          else {
            if(mine[x - 1][y] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x + 1][y] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x][y + 1] >= 5) {
              mine[x][y] += 1;
            } 
          }
        }
        //mid rows
        else if(y > 0 && y < n - 1) {
          //left side
          if(x == 0) {
            if(mine[x][y - 1] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x][y + 1] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x + 1][y] >= 5) {
              mine[x][y] += 1;
            }
          }
          //right side
          else if(x == m - 1) {
            if(mine[x][y - 1] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x][y + 1] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x - 1][y] >= 5) {
              mine[x][y] += 1;
            }
          }
          //mid
          else {
            if(mine[x][y - 1] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x][y + 1] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x - 1][y] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x + 1][y] >= 5) {
              mine[x][y] += 1;
            }
          } 
        }
        //bottom row
        else if(y == n - 1) {
          //bottom left corner
          if(x == 0) {
            if(mine[x + 1][y] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x][y - 1] >= 5) {
              mine[x][y] += 1;
            }
          }
          //bottom right corner
          else if(x == m - 1) {
            if(mine[x - 1][y] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x][y - 1] >= 5) {
              mine[x][y] += 1;
            }
          }
          //middle of the bottom row
          else {
            if(mine[x + 1][y] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x - 1][y] >= 5) {
              mine[x][y] += 1;
            }
            if(mine[x][y - 1] >= 5) {
              mine[x][y] += 1;
            }
          }
        }
      }
    }
  //print out the grid
    for(int y = 0; y < n; y++) {
      for(int x = 0; x < m; x++) {
        //println at the right edge of the grid
        if(x == m - 1) {
          if(mine[x][y] >= 5) {
            System.out.println("*");
          }
          else {
            System.out.println(mine[x][y]);
          }
        }
        //other tiles, no need to switch lines
        else {
          if(mine[x][y] >= 5) {
            System.out.print("*  ");
          }
          else {
            System.out.print(mine[x][y] + "  ");
          }
        }
      } 
    }
  }
}

Thank you for your time, and I'd really appreciate any suggestions.

Answer 1

You can reduce the complexity if you do the left/right/up/down testing in a loop to save a lot of lines of code:

for (int tryX = -1; tryX <= 1; tryX++) {
  for (int tryY = -1; tryY <= 1; tryY++) {
    if(mine[x + tryX][y + tryY] >= 5) {
      mine[x][y] += 1;
    }
  }
}

Since this is taking a lot of code lines, it will reduce the complexity. You should extract code to methods with your IDE (see here for IntelliJ). Good extraction points are loops.

I see two good extraction points:

1. initArrayWithRandomMines()
2. calculateNeighborMines()

Answer 2

I won't share a refactored code, but give you some ideas and examples that you should do. I think it will be more useful for your learning.

You should find the same condition and extract them from several branches to one. I recommend you start from inner conditions and after got to outside.

Let's take an example the branch with y == 0 condition:

if(x == 0) {
  if(mine[x + 1][y] >= 5) {
     mine[x][y] += 1;
  }
  if(mine[x][y + 1] >= 5) {
    mine[x][y] += 1;
  }
}
//upper right corner
else if(x == m - 1) {
  if(mine[x - 1][y] >= 5) {
     mine[x][y] += 1;
  }
  if(mine[x][y + 1] >= 5) {
     mine[x][y] += 1;
  } 
//mid of first row
else {
  if(mine[x - 1][y] >= 5) {
    mine[x][y] += 1;
  }
  if(mine[x + 1][y] >= 5) {
    mine[x][y] += 1;
  }
  if(mine[x][y + 1] >= 5) {
    mine[x][y] += 1;
  } 
}

You can see that you check mine[x + 1][y] >= 5 when x == 0 or in else branch. You can merg two conditions in one and it will looks like x < m-1 and now code will look like:

if(x < m-1 && mine[x + 1][y] >= 5) {
  mine[x][y] += 1;
}

if(x == 0) {
  if(mine[x][y + 1] >= 5) {
    mine[x][y] += 1;
  }
}
//upper right corner
else if(x == m - 1) {
  if(mine[x - 1][y] >= 5) {
     mine[x][y] += 1;
  }
  if(mine[x][y + 1] >= 5) {
     mine[x][y] += 1;
  } 
//mid of first row
else {
  if(mine[x - 1][y] >= 5) {
    mine[x][y] += 1;
  }
  if(mine[x][y + 1] >= 5) {
    mine[x][y] += 1;
  } 
}

Continue with each if statement where do you use x in condition. when you finish with x , do the same actions with conditions with y . Extract the common parts from several branches.

And about part where you print the result. Please think about what you should do. You should print * or mine[x][y] depends on mine[x][y]==5 and print spaces or new_line ( System.out.println() ) depends on x == m - 1 . Now think about how to implement it.