Sort polygon's points for drawing - algorithm

Question

I have a matrix (0 means nothing, 1 means terrain) that represents a level in my game. The matrix corresponds to a grid that my screen is broken up into, and indicates where my terrain goes.
My terrain is actually composed of 4 points in the corners of each block within the grid. When you have multiple blocks that are connected, I use a merge-cell algorithm that removes the duplicate points and any interior points. The result is that I end up with a list of points representing only the outer edges of the polygon.
To draw this polygon, I need the points to be in some sort of order (either clockwise or counter-clockwise) such that each point is followed by it's neighboring point. Obviously the first and last points need to be neighbors. Since this is all in a grid, I know the exact distance between neighboring points.
The problem is that I am having trouble coming up with an algorithm that allows me to "walk" around the edge of the polygon while putting the points in order. I believe there should be a way to utilize the fact that I have the matrix representing the geometry, meaning there is only 1 possible way to draw the polygon (even if it is concave).
I have tried several approaches using greedy-type algorithms, but can't seem to find a way to know, in every case, which direction I want to travel in. Given that any particular point can have up to 3 neighbors (the fourth isn't included because it is the "starting" point, meaning that I have already sorted it) I need a way of knowing which way to move.
Update
Another approach that I have been trying is to sort the points by their X (with tiebreaker of Y) which gives me the topmost/leftmost edge. It also guarantees that I am starting on an outer edge. However, I'm still struggling to find an algorithm that guarantees that I stay on the outside without crossing over.
Here is an example matrix:
0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 1 1 0 0
Which corresponds to this (black dots represent my points):

Answer 1

First of all please consider that for a general matrix the output can be composed of more than one closed loop; for example boundaries of the matrix
form three distinct loops, one of them placed inside another.
To extract these loops the first step is to build a map of all "walls": you have a vertical wall each time the content of one cell is different from the next cell on the same row; you have instead an horizontal wall when the content is different from the same cell in the next row.
data = [[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ],
[ 0, 1, 1, 1, 1, 0, 0, 0, 0, 0 ],
[ 0, 1, 0, 0, 1, 0, 1, 1, 0, 0 ],
[ 0, 1, 0, 0, 1, 0, 1, 1, 1, 0 ],
[ 0, 1, 1, 1, 1, 0, 0, 1, 1, 0 ],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ]]
rows = len(data)
cols = len(data[0])
walls = [[2*(data[r][c] != data[r][c+1]) + (data[r][c] != data[r+1][c])
for c in range(cols-1)]
for r in range(rows-1)]
In the example above I'm using two bits: 0x01 to mark horizontal walls and 0x02 to mark vertical walls. For a given (r, c) cell the walls are the right and bottom wall of the cell.
For simplicity I'm also assuming that the interesting areas are not touching the limits of the matrix; this can be solved by either adding extra rows and cols of zeros or by wrapping matrix access in a function that returns 0 for out-of-matrix virtual elements.
To build the list of boundaries you need to simply start from any point on a wall and move following walls, removing the walls from the map as you process them. When you cannot move any more a cycle has been completed (you're guaranteed to complete cycles because in a graph built in this way from a matrix of inside/outside flags the degree is guaranteed to be even in all vertices).
Filling all those cycles simultaneously using odd-even filling rules is also guaranteed to reproduce the original matrix.
In the code following I'm using r and c as row/col index and i and j instead to represent points on the boundary... for example for cell (r=3, c=2) the schema is:
where the red wall corresponds to bit 0x02 and the green wall to bit 0x01. The walls matrix has one row and one column less than the original data matrix because it's assumed that no walls can be present on last row or column.
result = []
for r in range(rows-1):
for c in range(cols-1):
if walls[r][c] & 1:
i, j = r+1, c
cycle = [(i, j)]
while True:
if i < rows-1 and walls[i][j-1] & 2:
ii, jj = i+1, j
walls[i][j-1] -= 2
elif i > 0 and walls[i-1][j-1] & 2:
ii, jj = i-1, j
walls[i-1][j-1] -= 2
elif j < cols-1 and walls[i-1][j] & 1:
ii, jj = i, j+1
walls[i-1][j] -= 1
elif j > 0 and walls[i-1][j-1] & 1:
ii, jj = i, j-1
walls[i-1][j-1] -= 1
else:
break
i, j = ii, jj
cycle.append((ii, jj))
result.append(cycle)
Basically the code starts from a point on a boundary and the checks if it can move on a wall going up, down, left or right. When it cannot move any more a cycle has been completed and can be added to the final result.
The complexity of the algorithm is O(rows*cols), i.e. it's proportional to the input size and it's optimal (in big-O sense) because you cannot compute the result without at least reading the input. This is easy to see because the body of the while cannot be entered more times than the total number of walls in the map (at each iteration a wall is removed).
Edit
The algorithm can be modified to generate as output only simple cycles (i.e. paths in which each vertex is visited only once).
result = []
index = [[-1] * cols for x in range(rows)]
for r in range(rows-1):
for c in range(cols-1):
if walls[r][c] & 1:
i, j = r+1, c
cycle = [(i, j)]
index[i][j] = 0
while True:
if i > 0 and walls[i-1][j-1] & 2:
ii, jj = i-1, j
walls[i-1][j-1] -= 2
elif j > 0 and walls[i-1][j-1] & 1:
ii, jj = i, j-1
walls[i-1][j-1] -= 1
elif i < rows-1 and walls[i][j-1] & 2:
ii, jj = i+1, j
walls[i][j-1] -= 2
elif j < cols-1 and walls[i-1][j] & 1:
ii, jj = i, j+1
walls[i-1][j] -= 1
else:
break
i, j = ii, jj
cycle.append((ii, jj))
ix = index[i][j]
if ix >= 0:
# closed a loop
result.append(cycle[ix:])
for i_, j_ in cycle[ix:]:
index[i_][j_] = -1
cycle = cycle[:ix+1]
index[i][j] = len(cycle)-1
This is implemented by adding to the output a separate cycle once the same vertex is met twice in the processing (the index table stores for a given i,j point the 0-based index in the current cycle being built).

Answer 2

This seems like it would work to me:
For every filled square, check which of its neighbours are filled. For those that aren't, add the appropriate edges to a list of edges. Generate those edges as directed, either clockwise or anticlockwise as you prefer.
To construct a full path, start by pulling any edge from the set and add it to the path. It has an order so look at the second vertex. Find the edge in the set with the first vertex that is equal to that second vertex. Pull that edge from the set and add it to the path. Continue until the path is closed.
Repeat to generate a list of paths. A simple polygon should end up as one path. A complex polygon — one with holes in the middle in this case — will be several.

Answer 3

I guess there are different ways to do this, I suppose there is quite simple one for case when diagonal connected cells counted as different contours:
You just need too keep cell and corner direction. For example you started from upper right corner of some earth cell (it supposed that either upper or right cell, or both are nothing if it is bourder) and want to go clockwise.
If cell to the right is earth, than you change current cell to it and change corner to upper left (it is the same point). Then you go to next iteration.
In other case, if you started from upper right corner of some earth cell and want to go clockwise. If cell to the right is NOT earth than you don't change current cell and change corner to bottom right, (it's next point)
So you also have symmetrical situation for other three possible corners, and you can go to next iteration until returning to start point.
So here is pseudo-code I wrote, it uses the same indexing as picture uses, and supposes that all cells along borders are free, otherwise you will need to check if index id not out of range.
I will also need additional array with almost the same dimensions as matrix to mark processed contours, it need to be 1 cell wider than matrix cause I'm going to mark vertical lines, and each vertical line is supposed to have coordinates of cell to the right of it. Note that there are only 2 cases midst 8 dwscribed above when you need to mark vertical line.
int mark[,] = new int[height,width+1]
start_i = i = 0;
start_j = j = 0;
direction = start_direction = top_left;
index = 0;
//outer cycle through different contours
while(true)
{
++index;
//scanning for contours through all the matrix
//continue from the same place, we stopped last time
for(/*i = i*/; i < n; i++)
{
for(/*j = j*/; j < n; j++)
{
//if we found earth
if(m[i,j] == 1)
{
//check if previous cell is nothing
//check if line between this and previous contour doesn't already added
if(m[i,j - 1] == 0 && mark[i,j] == 0)
{
direction = bottom_left;
break;
}
//the same for next cell
if(m[i,j + 1] == 0 && mark[i,j+1] == 0)
{
direction = top_right;
break;
}
}
}
//break if we found contour
if(i != start_i || j != start_j)
break;
}
//stop if we didn't find any contour
if(i == start_i && j == start_j)
{
break;
}
polygon = new polygon;
start_i = i;
start_j = j;
start_direction = direction;
//now the main part of algorithm described above
do
{
if(direction == top_left)
{
if(n(i-1,j) == 1)
{
direction = bottom_left;
position = (i-1,j)
}
else
{
direction = top_right;
polygon.Add(i,j+1);
}
}
if(direction == top_right;)
{
if(n[i,j + 1] == 1)
{
direction = top_left;
position = (i,j + 1)
}
else
{
direction = bottom_right;
mark[i, j + 1] = index;//don't forget to mark edges!
polygon.Add(i+1,j+1);
}
}
if(direction == bottom_right;
{
if(n[i+1,j] == 1)
{
direction = top_right;
position = (i+1,j)
}
else
{
direction = bottom_left;
polygon.Add(i+1,j);
}
}
if(direction == bottom_left)
{
if(n[i,j - 1] == 1)
{
direction = bottom_right;
position = [i,j - 1]
}
else
{
direction = top_left;
mark[i, j] = index;//don't forget to mark edges!
polygon.Add(i,j);
}
}
//and we can stop as we reached the starting state
}while(i != start_i || j != start_j || direction != start_direction);
//stop if it was last cell
if(i == n-1 && j == n- 1)
{
break;
}
}
Also you may need to know which contour is inside which, and you mat need a stack to keep what contours you are inside while you are scanning, so every time you are crossing the existing contour you need to add it to the stack or remove if it is already at the top of the stack.
It will cause the next changes in code:
...
//continue from the same place, we stopped last time
for(/*i = i*/; i < n; i++)
{
for(/*j = j*/; j < n; j++)
{
if(mark[i,j] != 0)
{
if(stack.top() == mark [i,j])
{
stack.pop();
}
else
{
stack.push(mark [i,j]);
}
}
//if we found earth
if(m[i,j] == 1)
{
...

Answer 4

If your matrix can contain random patterns, the answer is far more complicated than it seems.
For one thing they may be an arbitrary number of distinct polygons, and each of them might be hollow.
Besides, finding the contour of a region (even with no holes) is of little help for drawing the surface. Your GPU will eventually need triangles, which means you will need to decompose your polygons into rectangles.
Finding an optimal decomposition of a hollow bunch of squares (i.e. the smallest set of rectangles that will cover them all) is a well studied NP-complete problem with no known solution.
There exist algorithms to find an optimal decomposition of such shapes with no holes, but they are very complex.
A greedy algorithm is much easier to implement and usually yields acceptable results.
So I would do a greedy search on your matrix, collecting rectangles until all "1" values have been visited. Turning these rectangles into coordinates should be easy enough, since you know exactly where the top left and bottom right corners are.
The greedy scan will look like:
while your matrix is not empty
move to first "1" value. This is the rectangle top left corner
from this corner, extend the rectangle in x and y to maximize its surface
store the rectangle in a list and clear all corresponding "1" values