Mixing
time complexity of a function to find minimum number of lines to cover all zeros in assignment problem
$begingroup$
I am working on assignment problem by Hungarian method in O(n^3) polynomial time. To draw minimum number of lines to cover all zeros, i have got this function which is working really good. But can anyone please tell me the time complexity of the function. That means if it is O(n^3) or O(n^4) and why? I am really confused.
void step_four()
{
int row;
int col;
bool done;
done = false;
while (!done)
{
find_a_zero(&row,&col);
if (row == -1)
{
done = true;
step = 6;
}
else
{
mask_matrix[row][col] = 2;
if (star_in_row(row))
{
find_star_in_row(row, &col);
RowCover[row] = 1;
ColCover[col] = 0;
}
else
{
done = true;
step = 5;
path_row_0 = row;
path_col_0 = col;
}
}
}
}
void find_a_zero(int *row, int *col)
{
int r=0;
int c;
bool done;
*row = -1;
*col = -1;
done = false;
while (!done)
{ c = 0;
while (true)
{
if (a[r][c] == 0 && RowCover[r] == 0 && ColCover[c] == 0)
{
*row = r;
*col = c;
done = true;
}
c += 1;
if (c >= max || done)
{
break;
}
}
r += 1;
if (r >= max)
done = true;
}
}
computational-complexity
asked Jan 10 at 10:56
AaditiAaditi
12
$endgroup$
add a comment |
$begingroup$
I am working on assignment problem by Hungarian method in O(n^3) polynomial time. To draw minimum number of lines to cover all zeros, i have got this function which is working really good. But can anyone please tell me the time complexity of the function. That means if it is O(n^3) or O(n^4) and why? I am really confused.
void step_four()
{
int row;
int col;
bool done;
done = false;
while (!done)
{
find_a_zero(&row,&col);
if (row == -1)
{
done = true;
step = 6;
}
else
{
mask_matrix[row][col] = 2;
if (star_in_row(row))
{
find_star_in_row(row, &col);
RowCover[row] = 1;
ColCover[col] = 0;
}
else
{
done = true;
step = 5;
path_row_0 = row;
path_col_0 = col;
}
}
}
}
void find_a_zero(int *row, int *col)
{
int r=0;
int c;
bool done;
*row = -1;
*col = -1;
done = false;
while (!done)
{ c = 0;
while (true)
{
if (a[r][c] == 0 && RowCover[r] == 0 && ColCover[c] == 0)
{
*row = r;
*col = c;
done = true;
}
c += 1;
if (c >= max || done)
{
break;
}
}
r += 1;
if (r >= max)
done = true;
}
}
computational-complexity
asked Jan 10 at 10:56
AaditiAaditi
12
$endgroup$
add a comment |
$begingroup$
I am working on assignment problem by Hungarian method in O(n^3) polynomial time. To draw minimum number of lines to cover all zeros, i have got this function which is working really good. But can anyone please tell me the time complexity of the function. That means if it is O(n^3) or O(n^4) and why? I am really confused.
void step_four()
{
int row;
int col;
bool done;
done = false;
while (!done)
{
find_a_zero(&row,&col);
if (row == -1)
{
done = true;
step = 6;
}
else
{
mask_matrix[row][col] = 2;
if (star_in_row(row))
{
find_star_in_row(row, &col);
RowCover[row] = 1;
ColCover[col] = 0;
}
else
{
done = true;
step = 5;
path_row_0 = row;
path_col_0 = col;
}
}
}
}
void find_a_zero(int *row, int *col)
{
int r=0;
int c;
bool done;
*row = -1;
*col = -1;
done = false;
while (!done)
{ c = 0;
while (true)
{
if (a[r][c] == 0 && RowCover[r] == 0 && ColCover[c] == 0)
{
*row = r;
*col = c;
done = true;
}
c += 1;
if (c >= max || done)
{
break;
}
}
r += 1;
if (r >= max)
done = true;
}
}
computational-complexity
asked Jan 10 at 10:56
AaditiAaditi
12
$endgroup$
I am working on assignment problem by Hungarian method in O(n^3) polynomial time. To draw minimum number of lines to cover all zeros, i have got this function which is working really good. But can anyone please tell me the time complexity of the function. That means if it is O(n^3) or O(n^4) and why? I am really confused.
void step_four()
{
int row;
int col;
bool done;
done = false;
while (!done)
{
find_a_zero(&row,&col);
if (row == -1)
{
done = true;
step = 6;
}
else
{
mask_matrix[row][col] = 2;
if (star_in_row(row))
{
find_star_in_row(row, &col);
RowCover[row] = 1;
ColCover[col] = 0;
}
else
{
done = true;
step = 5;
path_row_0 = row;
path_col_0 = col;
}
}
}
}
void find_a_zero(int *row, int *col)
{
int r=0;
int c;
bool done;
*row = -1;
*col = -1;
done = false;
while (!done)
{ c = 0;
while (true)
{
if (a[r][c] == 0 && RowCover[r] == 0 && ColCover[c] == 0)
{
*row = r;
*col = c;
done = true;
}
c += 1;
if (c >= max || done)
{
break;
}
}
r += 1;
if (r >= max)
done = true;
}
}
computational-complexity
computational-complexity
asked Jan 10 at 10:56
AaditiAaditi
12
asked Jan 10 at 10:56
AaditiAaditi
12
edited Jan 10 at 11:12
Aaditi
asked Jan 10 at 10:56
AaditiAaditi
12
asked Jan 10 at 10:56
AaditiAaditi
12
asked Jan 10 at 10:56
AaditiAaditi
12
12
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