问题
I'm working on a program to generate random grids, latin squares, and sudoku. I am working on the latin squares and pretty much have it all working except I am in a continuous loop. If I break them up they work fine. There is probably something small I am doing wrong and I can't find it. Can you spot what's wrong?
EDIT: For those that don't know what Latin Square is (if someone doesn't know) its usually a 9x9 grid that doesn't have repeats in either rows nor columns.
UPDATE: I found a problem with the notSame equaling true right before if(notSame) statement. It was always equaling true so wouldn't finish checking the row. Now when I run it is no longer in continuous loop but instead the rows have no repeats but columns still do.
UPDATE #2: I now redid a lot of the coding for the columns. My professor asked me to change a few things but it still puts me in to a continuous loop.
int row = 0, col = 0, count = 0;
bool notSame = true;
// setting up rows and columns
for (row = 0; row < grid.GetLength(0); row++)
{
for (col = 0; col < grid.GetLength(1); col++)
{
grid[row, col] = rnd.Next(1, 10);
//for loop to check rows for repeats
for (int c = 0; c < col; c++)
{
// if there is repeat go back a column and set bool = false
if (grid[row, col] == grid[row, c])
{
col--;
count++;
notSame = false;
break;
}
//notSame = true;
}
// if bool = true loop to check columns for repeats
if (notSame)
{
for (int r = 0; r < row; r++)
{
// if repeat then go back row
if (grid[row, col] == grid[r, col])
{
notSame = false;
count++;
break;
}
}
if (notSame == false && count <= 50)
{
row--;
//break;
}
else if (notSame == false && count > 50)
{
count = 0;
col = 0;
row = 0;
break;
}
}
}
}
I am using a 2D array called grid.
回答1:
I do not know where your coding error is. But your algorithm is not very efficient.
Both Latin squares and sudokus are actually special cases of the "graph colouring" problem. That is, given a bunch of "nodes" that are arbitrarily "connected" together, find a way to colour each node so that no two nodes that are connected have the same colour.
This problem is in general quite difficult to solve quickly, but for the specific cases of sudokus and Latin squares it is pretty straightforward and can be done easily in C#. You create a "graph" that has 81 nodes, and each node is "connected" to the other nodes in its row and column. The "colours" are the numbers 1 through 9.
In my five-part series of articles I walk you through how to create an efficient graph colouring algorithm that can solve sudokus. It would not be difficult to adapt the algorithm to instead generate sudokus.
http://blogs.msdn.com/b/ericlippert/archive/tags/graph+colouring/
回答2:
My problem was an extra count in the check for rows. The count would always go over 50 and therefore would cause an infinite loop. To clarify:
grid[row, col] = rnd.Next(1, 10);
//for loop to check rows for repeats
for (int c = 0; c < col; c++)
{
// if there is repeat go back a column and set bool = false
if (grid[row, col] == grid[row, c])
{
col--;
count++;
notSame = false;
break;
}
//notSame = true;
}
The count++ would increment and would at times end up > = 50 which would then kick in this code:
else if (notSame == false && count > 50)
{
count = 0;
col = 0;
row = 0;
break;
}
Which then cause everything to be set back to 0 and would restart. Therefore, it caused an infinite loop. Thanks all for the help!
回答3:
I think about the simplest way to generate a latin square is to start with a know latin square. Say:
1 2 3 4
2 3 4 1
3 4 1 2
4 1 2 3
Which is easy to generate for any size by cyclic permutations of rows, then simply assign each value in the grid (1,2,3,4) to another value, randomly generated. For example, you could simply shuffle a list of the numbers 1,2,3,4 and get, let's say 2,4,3,1. Now just replace the 1's in your square with the first entry in you shuffled list, the 2's with your second, and so on to give you:
2 4 3 1
4 3 1 2
3 1 2 4
1 2 4 3
Now you can also shuffle the order of rows (and/or columns) as well if you wish and it should still be valid.
Edit: actually, thinking about this, it's probably easiest to just start with the first square and then shuffle the columns and rows. No need to do the substitution part.
回答4:
doing an explicit decrements of the variable that you are iterating with, is a no no. that's probably your problem. This sounds like a good place to use backtracking to avoid it :)
EDIT: I see saw many problems I don't know where to start. This algorithm will never give you what you need. First of all it can lead to a deadlock problem when you start filling it, and there is no way you can add a number to a particular row/column.. imagine that you have 12345 on row 5, and then you have on column 6 numbers 6 7 8 9.. well, you can't add a number to row 5, column 6 ;) see the problem there?? besides that, your code has several problems: getting the iteration variables change while iterating is a big problem and should be avoided.
once notSame = false; then it remain that way for the rest of your execution.
columns go vertical, and rows horizontal, so this (1,2) is row 1 column 2.. you are checking rows in the first phase.. and columns on the second..
// if bool = true loop to check columns for repeats
if (notSame)
{
for (int r = 0; r < row; r++)
{
// if repeat then genereate new random and go back row
if (grid[row, col] == grid[r , col])
{
grid[row, col] = rnd.Next(1, 10);
that has a problem on itself.. if you change the number there, you should have been checked all the same that before!
tell you teacher to come here and read this ;).. I don't know how else to help you, this algorithm is totally wrong, and need a total refactor (and yes, you can do it using iteration, but not for, you need to use while and flags).
来源：https://stackoverflow.com/questions/9913139/generating-random-latin-square-continuous-loop