Converting switch statements to more elegant solution.
- by masfenix
I have a 9 x 9 matrix. (think of suduko).
4 2 1 6 8 1 8 5 8
3 1 5 8 1 1 7 5 8
1 1 4 0 5 6 7 0 4
6 2 5 5 4 4 8 1 2
6 8 8 2 8 1 6 3 5
8 4 2 6 4 7 4 1 1
1 3 5 3 8 8 5 2 2
2 6 6 0 8 8 8 0 6
8 7 2 3 3 1 1 7 4
now I wanna be able to get a "quadrant". for example (according to my code) the quadrant 2 , 2 returns the following:
5 4 4
2 8 1
6 4 7
If you've noticed, this is the matrix from the very center of the 9 x 9. I've split everything up in to pairs of "3" if you know what i mean. the first "ROW" is from 0 - 3, the second from 3 - 6, the third for 6 - 9.. I hope this makes sense ( I am open to alternate ways to go about this)
anyways, heres my code. I dont really like this way, even though it works. I do want speed though beccause i am making a suduko solver.
//a quadrant returns the mini 3 x 3
//row 1 has three quads,"1", "2", 3"
//row 2 has three quads "1", "2", "3" etc
public int[,] GetQuadrant(int rnum, int qnum) {
int[,] returnMatrix = new int[3, 3];
int colBegin, colEnd, rowBegin, rowEnd, row, column;
//this is so we can keep track of the new matrix
row = 0;
column = 0;
switch (qnum) {
case 1:
colBegin = 0;
colEnd = 3;
break;
case 2:
colBegin = 3;
colEnd = 6;
break;
case 3:
colBegin = 6;
colEnd = 9;
break;
default:
colBegin = 0;
colEnd = 0;
break;
}
switch (rnum) {
case 1:
rowBegin = 0;
rowEnd = 3;
break;
case 2:
rowBegin = 3;
rowEnd = 6;
break;
case 3:
rowBegin = 6;
rowEnd = 9;
break;
default:
rowBegin = 0;
rowEnd = 0;
break;
}
for (int i = rowBegin ; i < rowEnd; i++) {
for (int j = colBegin; j < colEnd; j++) {
returnMatrix[row, column] = _matrix[i, j];
column++;
}
column = 0;
row++;
}
return returnMatrix;
}
