Given a contiguous drawing of arbitrary pixels (e.g. on an HTML5 Canvas) is there any algorithm for finding the axis-aligned bounding box that is more efficient than simply looking at every pixel and recording the min/max x/y values?
Just scanline from top left to right and down to get y top,and similar algorithm with different directions for the rest.
Edit by Phrogz:
Here's a pseudo-code implementation. An included optimization ensures that each scan line does not look at pixels covered by an earlier pass:
function boundingBox()
w = getWidth() # Assuming graphics address goes from [0,w)
h = getHeight() # Assuming graphics address goes from [0,h)
for y=h-1 to 0 by -1 # Iterate from last row upwards
for x=w-1 to 0 by -1 # Iterate across the entire row
if pxAt(x,y) then
maxY=y
break # Break out of both loops
if maxY===undefined then # No pixels, no bounding box
return
for x=w-1 to 0 by -1 # Iterate from last column to first
for y=0 to maxY # Iterate down the column, up to maxY
if pxAt(x,y) then
maxX=x
break # Break out of both loops
for x=0 to maxX # Iterate from first column to maxX
for y=0 to maxY # Iterate down the column, up to maxY
if pxAt(x,y) then
minX=x
break # Break out of both loops
for y=0 to maxY # Iterate down the rows, up to maxY
for x=0 to maxX # Iterate across the row, up to maxX
if pxAt(x,y) then
minY=y
break # Break out of both loops
return minX, minY, maxX, maxY
The result (in practice) performs about the same as the brute-force algorithm for a single pixel, and significantly better as the object gets larger.
Demo: http://phrogz.net/tmp/canvas_bounding_box2.html
For fun, here's a visual representation of how this algorithm works:
It doesn't matter in what order you choose to do the sides, you just have to make sure that you take the previous results into account so that you are not double-scanning the corners.
You might be able to use some kind of binary search, or sample on a coarse grid then a successively finer grid. The correctness of this method depends on if 'holes' are allowed in your drawing.
I dislike the current answer. Here's my code that I plugged into OP website. It's much faster in firefox and chrome.
The idea is check all pixels on x axis to see if there's a hit on the Y axis. If so update Y and increase X so we can scan for max X
function contextBoundingBox(ctx,alphaThreshold){
if (alphaThreshold===undefined) alphaThreshold = 15;
var w=ctx.canvas.width,h=ctx.canvas.height;
var data = ctx.getImageData(0,0,w,h).data;
let minX=w;
let maxX=0
let minY=h
let maxY=0
for(let y=0; y<h; y++)
{
for(let x=0; x<w; x++)
{
if (data[y*w*4 + x*4+3])
{
minX = Math.min(minX, x);
maxX = Math.max(maxX, x);
minY = Math.min(minY, y);
maxY = y;
x=maxX
}
}
}
return {x:minX,y:minY,maxX:maxX,maxY:maxY,w:maxX-minX,h:maxY-minY};
}
Related
I am trying to find the direction of triangles in an image. below is the image:
These triangles are pointing upward/downward/leftward/rightward. This is not the actual image. I have already used canny edge detection to find edges then contours and then the dilated image is shown below.
My logic to find the direction:
The logic I am thinking to use is that among the three corner coordinates If I can identify the base coordinates of the triangle (having the same abscissa or ordinates values coordinates), I can make a base vector. Then angle between unit vectors and base vectors can be used to identify the direction. But this method can only determine if it is up/down or left/right but cannot differentiate between up and down or right and left. I tried to find the corners using cv2.goodFeaturesToTrack but as I know it's giving only the 3 most effective points in the entire image. So I am wondering if there is other way to find the direction of triangles.
Here is my code in python to differentiate between the triangle/square and circle:
#blue_masking
mask_blue=np.copy(img1)
row,columns=mask_blue.shape
for i in range(0,row):
for j in range(0,columns):
if (mask_blue[i][j]==25):
mask_blue[i][j]=255
else:
mask_blue[i][j]=0
blue_edges = cv2.Canny(mask_blue,10,10)
kernel_blue = cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(2,2))
dilated_blue = cv2.dilate(blue_edges, kernel)
blue_contours,hierarchy =
cv2.findContours(dilated_blue,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
for cnt in blue_contours:
area = cv2.contourArea(cnt)
perimeter = cv2.arcLength(cnt,True)
M = cv2.moments(cnt)
cx = int(M['m10']/M['m00'])
cy = int(M['m01']/M['m00'])
if(12<(perimeter*perimeter)/area<14.8):
shape="circle"
elif(14.8<(perimeter*perimeter)/area<18):
shape="squarer"
elif(18<(perimeter*perimeter)/area and area>200):
shape="triangle"
print(shape)
print(area)
print((perimeter*perimeter)/area,"\n")
cv2.imshow('mask_blue',dilated_blue)
cv2.waitKey(0)
cv2.destroyAllWindows()
Source image can be found here: img1
Please help, how can I found the direction of triangles?
Thank you.
Assuming that you only have four cases: [up, down, left, right], this code should work well for you.
The idea is simple:
Get the bounding rectangle for your contour. Use: box = cv2.boundingRect(contour_pnts)
Crop the image using the bounding rectangle.
Reduce the image vertically and horizontally using the Sum option. Now you have the sum of pixels along each axis. The axis with the largest sum determines whether the triangle base is vertical or horizontal.
To identify whether the triangle is pointing left/right or up/down: you need to check whether the bounding rectangle center is before or after the max col/row:
The code (assumes you start from the cropped image):
ver_reduce = cv2.reduce(img, 0, cv2.REDUCE_SUM, None, cv2.CV_32F)
hor_reduce = cv2.reduce(img, 1, cv2.REDUCE_SUM, None, cv2.CV_32F)
#For smoothing the reduced vector, could be removed
ver_reduce = cv2.GaussianBlur(ver_reduce, (3, 1), 0)
hor_reduce = cv2.GaussianBlur(hor_reduce, (1, 3), 0)
_,ver_max, _, ver_col = cv2.minMaxLoc(ver_reduce)
_,hor_max, _, hor_row = cv2.minMaxLoc(hor_reduce)
ver_col = ver_col[0]
hor_row = hor_row[1]
contour_pnts = cv2.findNonZero(img) #in my code I do not have the original contour points
rect_center, size, angle = cv2.minAreaRect(contour_pnts )
print(rect_center)
if ver_max > hor_max:
if rect_center[0] > ver_col:
print ('right')
else:
print ('left')
else:
if rect_center[1] > hor_row:
print ('down')
else:
print ('up')
Photos:
Well, Mark has mentioned a solution that may not be as efficient but perhaps more accurate. I think this one should be equally efficient but perhaps less accurate. But since you already have a code that finds triangles, try adding the following code after you have found triangle contour:
hull = cv2.convexHull(cnt) # convex hull of contour
hull = cv2.approxPolyDP(hull,0.1*cv2.arcLength(hull,True),True)
# You can double check if the contour is a triangle here
# by something like len(hull) == 3
You should get 3 hull points for a triangle, these should be the 3 vertices of your triangles. Given your triangles always 'face' only in 4 directions; Y coordinate of the hull will have close value to the Y coordinate of the centroid for triangle facing left or right and whether it's pointing left or right will depend on whether hull X is less than or greater than centroid X. Similarly use hull and centroid X and Y for triangle pointing up or down.
I have a list of consecutive points and I need to find the coordinates of a polygon some size larger. I can calculate each of the points in the new polygon if it has convex angles, but I'm not sure how to adjust for when the angles are concave.
Concave angles can be treated in exactly the same way as convex ones: For each vertex you generate lines that are parallel to the two original segments but shifted by your offset value. Then the vertex is replaced with the intersection of these two lines.
The difficulty is that the resulting polygon can have intersections if the original one has one or more concave angles. There are different ways to handle these intersections. Generally they can produce inner contours (holes in the polygon) but maybe you are only interested in the outer contour.
In any case you have to find the intersection points first. If you don't find any, you are finished.
Otherwise find a start point of which you can be sure that it is on the outer contour. In many cases you can take the one with smallest X coordinate for that. Then trace the polygon contour until you get to the first intersection. Add the intersection to the polygon. If you are only interested in the outer contour, then skip all following vertices until you get back to the intersection point. Then continue to add the vertexes to the resulting polygon until you get to the next intersection and so on.
If you also need the inner contours (holes) it gets a bit more complicated, but I guess you can figure this out.
I also need to add that you should pe prepared for special cases like (almost) duplicate edges that cause numerical problems. Generally this is not a trivial task, so if possible, try to find a suitable polygon library.
For this problem I found a relatively simple solution for figuring out whether the calculated point was inside or outside the original polygon. Check to see whether the newly formed line intersects the original polygon's line. A formula can be found here http://www.geeksforgeeks.org/orientation-3-ordered-points/.
Suppose your polygon is given in counter-clockwise order. Let P1=(x1,y1), P2=(x2,y2) and P3=(x3,y3) be consecutive vertices. You want to know if the angle at P2 is “concave” i.e. more than 180 degrees. Let V1=(x4,y4)=P2-P1 and V2=(x5,y5)=P3-P2. Compute the “cross product” V1 x V2 = (x4.y5-x5.y4). This is negative iff the angle is concave.
Here is a code in C# that receives a list of Vector2D representing the ordered points of a polygon and returns a list with the angles of each vertex. It first checks if the points are clockwise or counterclockwise, and then it loops through the points calculating the sign of the cross product (z) for each triple of angles, and compare the value of the cross product to the clockwise function result to check if the calculated angle to that point needs to be the calculated angle or adjusted to 360-angle. The IsClockwise function was obtained in this discussion: How to determine if a list of polygon points are in clockwise order?
public bool IsClockwise(List<Vector2> vertices)
{
double sum = 0.0;
for (int i = 0; i < vertices.Count; i++)
{
Vector2 v1 = vertices[i];
Vector2 v2 = vertices[(i + 1) % vertices.Count];
sum += (v2.x - v1.x) * (v2.y + v1.y);
}
return sum > 0.0;
}
List<float> estimatePolygonAngles(List<Vector2> vertices)
{
if (vertices.Count < 3)
return null;
//1. check if the points are clockwise or counterclockwise:
int clockwise = (IsClockwise(vertices) ? 1 : -1);
List<float> angles = new List<float>();
List<float> crossProductsSigns = new List<float>();
Vector2 v1, v2;
//2. calculate the angles between each triple of vertices (first and last angles are computed separetely because index of the array):
v1 = vertices[vertices.Count - 1] - vertices[0];
v2 = vertices[1] - vertices[0];
angles.Add(Vector2.Angle(v1, v2));
crossProductsSigns.Add(Vector3.Cross(v1, v2).z > 0 ? 1 : -1);
for (int i = 1; i < vertices.Count-1; i++)
{
v1 = vertices[i-1] - vertices[i];
v2 = vertices[i+1] - vertices[i];
angles.Add(Vector2.Angle(v1, v2));
crossProductsSigns.Add(Vector3.Cross(v1, v2).z > 0 ? 1 : -1);
}
v1 = vertices[vertices.Count - 2] - vertices[vertices.Count - 1];
v2 = vertices[0] - vertices[vertices.Count - 1];
angles.Add(Vector2.Angle(v1, v2));
crossProductsSigns.Add(Vector3.Cross(v1, v2).z > 0 ? 1 : -1);
//3. for each computed angle, check if the cross product is the same as the as the direction provided by the clockwise function, if dont, the angle must be adjusted to 360-angle
for (int i = 0; i < vertices.Count; i++)
{
if (crossProductsSigns[i] != clockwise)
angles[i] = 360.0f - angles[i];
}
return angles;
}
I am trying to get a y-tick at "zero" for a multi-series d3 plot. My x-axis is a time scale and y-axis is some random data-scale. Here is my plunkr
http://plnkr.co/edit/emOKcxrHP7gU1U1l0Mff?p=preview
If I just add zero to the y-tick values, it does not work (i.e. in the following function if I say var yTickValues=[0] ) and it messes up my plot (draws another x-axis below the existing one)
function getYTickValues(){
var deltaY = Math.round((maxY - minY)/(yTickCount-1));
var yTickValues = [];
for(var i=0;i<yTickCount;i++){
yTickValues.push(((minY + i * deltaY) * 100) / 100);
};
return yTickValues;
}
I am unable to figure out how to fix this so I can always get a y-tick at zero. I would like to not touch my minX, maxX, minY and maxY because the domain range scale will change for the sake of accommodating the zero y-tick.
Any help is appreciated.
Change the y domain to start at 0:
y.domain([0, maxY]);
and then also including 0 in the yTickValues array as you suggest above:
var yTickValues = [0];
The data values still remain between minY and maxY, but the y-axis runs to 0. I think that's what the question was getting at?
I also made a couple of changes to the getYTickValues() function to evenly space the rest of the y tick values. See http://plnkr.co/edit/q6XnujIyB8JdzN8AA88j?p=preview
So, I've been struggling with a frankly now infuriating problem all day today.
Given a set of verticies of a triangle on a plane (just 3 points, 6 free parameters), I need to calculate the area of intersection of this triangle with the unit square defined by {0,0} and {1,1}. (I choose this because any square in 2D can be transformed to this, and the same transformation can move the 3 vertices).
So, now the problem is simplified down to only 6 parameters, 3 points... which I think is short enough that I'd be willing to code up the full solution / find the full solution.
( I would like this to run on a GPU for literally more than 2 million triangles every <0.5 seconds, if possible. as for the need for simplification / no data structures / libraries)
In terms of my attempt at the solution, I've... got a list of ways I've come up with, none of which seem fast or ... specific to the nice case (too general).
Option 1: Find the enclosed polygon, it can be anything from a triangle up to a 6-gon. Do this by use of some intersection of convex polygon in O(n) time algorithms that I found. Then I would sort these intersection points (new vertices, up to 7 of them O(n log n) ), in either CW or CCw order, so that I can run a simple area algorithm on the points (based on Greens function) (O(n) again). This is the fastest i can come with for an arbitrary convex n-gon intersecting with another m-gon. However... my problem is definitely not that complex, its a special case, so it should have a better solution...
Option 2:
Since I know its a triangle and unit square, i can simply find the list of intersection points in a more brute force way (rather than using some algorithm that is ... frankly a little frustrating to implement, as listed above)
There are only 19 points to check. 4 points are corners of square inside of triangle. 3 points are triangle inside square. And then for each line of the triangle, each will intersect 4 lines from the square (eg. y=0, y=1, x=0, x=1 lines). that is another 12 points. so, 12+3+4 = 19 points to check.
Once I have the, at most 6, at fewest 3, points that do this intersection, i can then follow up with one of two methods that I can think of.
2a: Sort them by increasing x value, and simply decompose the shape into its sub triangle / 4-gon shapes, each with an easy formula based on the limiting top and bottom lines. sum up the areas.
or 2b: Again sort the intersection points in some cyclic way, and then calculate the area based on greens function.
Unfortunately, this still ends up being just as complex as far as I can tell. I can start breaking up all the cases a little more, for finding the intersection points, since i know its just 0s and 1s for the square, which makes the math drop out some terms.. but it's not necessarily simple.
Option 3: Start separating the problem based on various conditions. Eg. 0, 1, 2, or 3 points of triangle inside square. And then for each case, run through all possible number of intersections, and then for each of those cases of polygon shapes, write down the area solution uniquely.
Option 4: some formula with heaviside step functions. This is the one I want the most probably, I suspect it'll be a little... big, but maybe I'm optimistic that it is possible, and that it would be the fastest computationally run time once I have the formula.
--- Overall, I know that it can be solved using some high level library (clipper for instance). I also realize that writing general solutions isn't so hard when using data structures of various kinds (linked list, followed by sorting it). And all those cases would be okay, if I just needed to do this a few times. But, since I need to run it as an image processing step, on the order of >9 * 1024*1024 times per image, and I'm taking images at .. lets say 1 fps (technically I will want to push this speed up as fast as possible, but lower bound is 1 second to calculate 9 million of these triangle intersection area problems). This might not be possible on a CPU, which is fine, I'll probably end up implementing it in Cuda anyways, but I do want to push the limit of speed on this problem.
Edit: So, I ended up going with Option 2b. Since there are only 19 intersections possible, of which at most 6 will define the shape, I first find those 3 to 6 verticies. Then i sort them in a cyclic (CCW) order. And then I find the area by calculating the area of that polygon.
Here is my test code I wrote to do that (it's for Igor, but should be readable as pseudocode) Unfortunately it's a little long winded, but.. I think other than my crappy sorting algorithm (shouldn't be more than 20 swaps though, so not so much overhead for writing better sorting)... other than that sorting, I don't think I can make it any faster. Though, I am open to any suggestions or oversights I might have had in chosing this option.
function calculateAreaUnitSquare(xPos, yPos)
wave xPos
wave yPos
// First, make array of destination. Only 7 possible results at most for this geometry.
Make/o/N=(7) outputVertexX = NaN
Make/o/N=(7) outputVertexY = NaN
variable pointsfound = 0
// Check 4 corners of square
// Do this by checking each corner against the parameterized plane described by basis vectors p2-p0 and p1-p0.
// (eg. project onto point - p0 onto p2-p0 and onto p1-p0. Using appropriate parameterization scaling (not unit).
// Once we have the parameterizations, then it's possible to check if it is inside the triangle, by checking that u and v are bounded by u>0, v>0 1-u-v > 0
variable denom = yPos[0]*xPos[1]-xPos[0]*yPos[1]-yPos[0]*xPos[2]+yPos[1]*xPos[2]+xPos[0]*yPos[2]-xPos[1]*yPos[2]
//variable u00 = yPos[0]*xPos[1]-xPos[0]*yPos[1]-yPos[0]*Xx+yPos[1]*Xx+xPos[0]*Yx-xPos[1]*Yx
//variable v00 = -yPos[2]*Xx+yPos[0]*(Xx-xPos[2])+xPos[0]*(yPos[2]-Yx)+yPos[2]*Yx
variable u00 = (yPos[0]*xPos[1]-xPos[0]*yPos[1])/denom
variable v00 = (yPos[0]*(-xPos[2])+xPos[0]*(yPos[2]))/denom
variable u01 =(yPos[0]*xPos[1]-xPos[0]*yPos[1]+xPos[0]-xPos[1])/denom
variable v01 =(yPos[0]*(-xPos[2])+xPos[0]*(yPos[2]-1)+xPos[2])/denom
variable u11 = (yPos[0]*xPos[1]-xPos[0]*yPos[1]-yPos[0]+yPos[1]+xPos[0]-xPos[1])/denom
variable v11 = (-yPos[2]+yPos[0]*(1-xPos[2])+xPos[0]*(yPos[2]-1)+xPos[2])/denom
variable u10 = (yPos[0]*xPos[1]-xPos[0]*yPos[1]-yPos[0]+yPos[1])/denom
variable v10 = (-yPos[2]+yPos[0]*(1-xPos[2])+xPos[0]*(yPos[2]))/denom
if(u00 >= 0 && v00 >=0 && (1-u00-v00) >=0)
outputVertexX[pointsfound] = 0
outputVertexY[pointsfound] = 0
pointsfound+=1
endif
if(u01 >= 0 && v01 >=0 && (1-u01-v01) >=0)
outputVertexX[pointsfound] = 0
outputVertexY[pointsfound] = 1
pointsfound+=1
endif
if(u10 >= 0 && v10 >=0 && (1-u10-v10) >=0)
outputVertexX[pointsfound] = 1
outputVertexY[pointsfound] = 0
pointsfound+=1
endif
if(u11 >= 0 && v11 >=0 && (1-u11-v11) >=0)
outputVertexX[pointsfound] = 1
outputVertexY[pointsfound] = 1
pointsfound+=1
endif
// Check 3 points for triangle. This is easy, just see if its bounded in the unit square. if it is, add it.
variable i = 0
for(i=0; i<3; i+=1)
if(xPos[i] >= 0 && xPos[i] <= 1 )
if(yPos[i] >=0 && yPos[i] <=1)
if(!((xPos[i] == 0 || xPos[i] == 1) && (yPos[i] == 0 || yPos[i] == 1) ))
outputVertexX[pointsfound] = xPos[i]
outputVertexY[pointsfound] = yPos[i]
pointsfound+=1
endif
endif
endif
endfor
// Check intersections.
// Procedure is: loop over 3 lines of triangle.
// For each line
// Check if vertical
// If not vertical, find y intercept with x=0 and x=1 lines.
// if y intercept is between 0 and 1, then add the point
// Check if horizontal
// if not horizontal, find x intercept with y=0 and y=1 lines
// if x intercept is between 0 and 1, then add the point
for(i=0; i<3; i+=1)
variable iN = mod(i+1,3)
if(xPos[i] != xPos[iN])
variable tx0 = xPos[i]/(xPos[i] - xPos[iN])
variable tx1 = (xPos[i]-1)/(xPos[i] - xPos[iN])
if(tx0 >0 && tx0 < 1)
variable yInt = (yPos[iN]-yPos[i])*tx0+yPos[i]
if(yInt > 0 && yInt <1)
outputVertexX[pointsfound] = 0
outputVertexY[pointsfound] = yInt
pointsfound+=1
endif
endif
if(tx1 >0 && tx1 < 1)
yInt = (yPos[iN]-yPos[i])*tx1+yPos[i]
if(yInt > 0 && yInt <1)
outputVertexX[pointsfound] = 1
outputVertexY[pointsfound] = yInt
pointsfound+=1
endif
endif
endif
if(yPos[i] != yPos[iN])
variable ty0 = yPos[i]/(yPos[i] - yPos[iN])
variable ty1 = (yPos[i]-1)/(yPos[i] - yPos[iN])
if(ty0 >0 && ty0 < 1)
variable xInt = (xPos[iN]-xPos[i])*ty0+xPos[i]
if(xInt > 0 && xInt <1)
outputVertexX[pointsfound] = xInt
outputVertexY[pointsfound] = 0
pointsfound+=1
endif
endif
if(ty1 >0 && ty1 < 1)
xInt = (xPos[iN]-xPos[i])*ty1+xPos[i]
if(xInt > 0 && xInt <1)
outputVertexX[pointsfound] = xInt
outputVertexY[pointsfound] = 1
pointsfound+=1
endif
endif
endif
endfor
// Now we have all 6 verticies that we need. Next step: find the lowest y point of the verticies
// if there are multiple with same low y point, find lowest X of these.
// swap this vertex to be first vertex.
variable lowY = 1
variable lowX = 1
variable m = 0;
for (i=0; i<pointsfound ; i+=1)
if (outputVertexY[i] < lowY)
m=i
lowY = outputVertexY[i]
lowX = outputVertexX[i]
elseif(outputVertexY[i] == lowY)
if(outputVertexX[i] < lowX)
m=i
lowY = outputVertexY[i]
lowX = outputVertexX[i]
endif
endif
endfor
outputVertexX[m] = outputVertexX[0]
outputVertexY[m] = outputVertexY[0]
outputVertexX[0] = lowX
outputVertexY[0] = lowY
// now we have the bottom left corner point, (bottom prefered).
// calculate the cos(theta) of unit x hat vector to the other verticies
make/o/N=(pointsfound) angles = (p!=0)?( (outputVertexX[p]-lowX) / sqrt( (outputVertexX[p]-lowX)^2+(outputVertexY[p]-lowY)^2) ) : 0
// Now sort the remaining verticies based on this angle offset. This will orient the points for a convex polygon in its maximal size / ccw orientation
// (This sort is crappy, but there will be in theory, at most 25 swaps. Which in the grand sceme of operations, isn't so bad.
variable j
for(i=1; i<pointsfound; i+=1)
for(j=i+1; j<pointsfound; j+=1)
if( angles[j] > angles[i] )
variable tempX = outputVertexX[j]
variable tempY = outputVertexY[j]
outputVertexX[j] = outputVertexX[i]
outputVertexY[j] =outputVertexY[i]
outputVertexX[i] = tempX
outputVertexY[i] = tempY
variable tempA = angles[j]
angles[j] = angles[i]
angles[i] = tempA
endif
endfor
endfor
// Now the list is ordered!
// now calculate the area given a list of CCW oriented points on a convex polygon.
// has a simple and easy math formula : http://www.mathwords.com/a/area_convex_polygon.htm
variable totA = 0
for(i = 0; i<pointsfound; i+=1)
totA += outputVertexX[i]*outputVertexY[mod(i+1,pointsfound)] - outputVertexY[i]*outputVertexX[mod(i+1,pointsfound)]
endfor
totA /= 2
return totA
end
I think the Cohen-Sutherland line-clipping algorithm is your friend here.
First off check the bounding box of the triangle against the square to catch the trivial cases (triangle inside square, triangle outside square).
Next check for the case where the square lies completely within the triangle.
Next consider your triangle vertices A, B and C in clockwise order. Clip the line segments AB, BC and CA against the square. They will either be altered such that they lie within the square or are found to lie outside, in which case they can be ignored.
You now have an ordered list of up to three line segments that define the some of the edges intersection polygon. It is easy to work out how to traverse from one edge to the next to find the other edges of the intersection polygon. Consider the endpoint of one line segment (e) against the start of the next (s)
If e is coincident with s, as would be the case when a triangle vertex lies within the square, then no traversal is required.
If e and s differ, then we need to traverse clockwise around the boundary of the square.
Note that this traversal will be in clockwise order, so there is no need to compute the vertices of the intersection shape, sort them into order and then compute the area. The area can be computed as you go without having to store the vertices.
Consider the following examples:
In the first case:
We clip the lines AB, BC and CA against the square, producing the line segments ab>ba and ca>ac
ab>ba forms the first edge of the intersection polygon
To traverse from ba to ca: ba lies on y=1, while ca does not, so the next edge is ca>(1,1)
(1,1) and ca both lie on x=1, so the next edge is (1,1)>ca
The next edge is a line segment we already have, ca>ac
ac and ab are coincident, so no traversal is needed (you might be as well just computing the area for a degenerate edge and avoiding the branch in these cases)
In the second case, clipping the triangle edges against the square gives us ab>ba, bc>cb and ca>ac. Traversal between these segments is trivial as the start and end points lie on the same square edges.
In the third case the traversal from ba to ca goes through two square vertices, but it is still a simple matter of comparing the square edges on which they lie:
ba lies on y=1, ca does not, so next vertex is (1,1)
(1,1) lies on x=1, ca does not, so next vertex is (1,0)
(1,0) lies on y=0, as does ca, so next vertex is ca.
Given the large number of triangles I would recommend scanline algorithm: sort all the points 1st by X and 2nd by Y, then proceed in X direction with a "scan line" that keeps a heap of Y-sorted intersections of all lines with that line. This approach has been widely used for Boolean operations on large collections of polygons: operations such as AND, OR, XOR, INSIDE, OUTSIDE, etc. all take O(n*log(n)).
It should be fairly straightforward to augment Boolean AND operation, implemented with the scanline algorithm to find the areas you need. The complexity will remain O(n*log(n)) on the number of triangles. The algorithm would also apply to intersections with arbitrary collections of arbitrary polygons, in case you would need to extend to that.
On the 2nd thought, if you don't need anything other than the triangle areas, you could do that in O(n), and scanline may be an overkill.
I came to this question late, but I think I've come up with a more fully flushed out solution along the lines of ryanm's answer. I'll give an outline of for others trying to do this problem at least somewhat efficiently.
First you have two trivial cases to check:
1) Triangle lies entirely within the square
2) Square lies entirely within the triangle (Just check if all corners are inside the triangle)
If neither is true, then things get interesting.
First, use either the Cohen-Sutherland or Liang-Barsky algorithm to clip each edge of the triangle to the square. (The linked article contains a nice bit of code that you can essentially just copy-paste if you're using C).
Given a triangle edge, these algorithms will output either a clipped edge or a flag denoting that the edge lies entirely outside the square. If all edges lie outsize the square, then the triangle and the square are disjoint.
Otherwise, we know that the endpoints of the clipped edges constitute at least some of the vertices of the polygon representing the intersection.
We can avoid a tedious case-wise treatment by making a simple observation. All other vertices of the intersection polygon, if any, will be corners of the square that lie inside the triangle.
Simply put, the vertices of the intersection polygon will be the (unique) endpoints of the clipped triangle edges in addition to the corners of the square inside the triangle.
We'll assume that we want to order these vertices in a counter-clockwise fashion. Since the intersection polygon will always be convex, we can compute its centroid (the mean over all vertex positions) which will lie inside the polygon.
Then to each vertex, we can assign an angle using the atan2 function where the inputs are the y- and x- coordinates of the vector obtained by subtracting the centroid from the position of the vertex (i.e. the vector from the centroid to the vertex).
Finally, the vertices can be sorted in ascending order based on the values of the assigned angles, which constitutes a counter-clockwise ordering. Successive pairs of vertices correspond to the polygon edges.
I have a system that requires moving an image on the screen. I am currently using a png and just placing it at the desired screen coordinates.
Because of a combination of the screen resolution and the required frame rate, some frames are identical because the image has not yet moved a full pixel. Unfortunately, the resolution of the screen is not negotiable.
I have a general understanding of how sub-pixel rendering works to smooth out edges but I have been unable to find a resource (if it exists) as to how I can use shading to translate an image by less than a single pixel.
Ideally, this would be usable with any image but if it was only possible with a simple shape like a circle or a ring, that would also be acceptable.
Sub-pixel interpolation is relatively simple. Typically you apply what amounts to an all-pass filter with a constant phase shift, where the phase shift corresponds to the required sub-pixel image shift. Depending on the required image quality you might use e.g. a 5 point Lanczos or other windowed sinc function and then apply this in one or both axes depending on whether you want an X shift or a Y shift or both.
E.g. for a 0.5 pixel shift the coefficients might be [ 0.06645, 0.18965, 0.27713, 0.27713, 0.18965 ]. (Note that the coefficients are normalised, i.e. their sum is equal to 1.0.)
To generate a horizontal shift you would convolve these coefficients with the pixels from x - 2 to x + 2, e.g.
const float kCoeffs[5] = { 0.06645f, 0.18965f, 0.27713f, 0.27713f, 0.18965f };
for (y = 0; y < height; ++y) // for each row
for (x = 2; x < width - 2; ++x) // for each col (apart from 2 pixel border)
{
float p = 0.0f; // convolve pixel with Lanczos coeffs
for (dx = -2; dx <= 2; ++dx)
p += in[y][x + dx] * kCoeffs[dx + 2];
out[y][x] = p; // store interpolated pixel
}
Conceptually, the operation is very simple. First you scale up the image (using any method of interpolation, as you like), then you translate the result, and finally you subsample down to the original image size.
The scale factor depends on the precision of sub-pixel translation you want to do. If you want to translate by 0.5 degrees, you need scale up the original image by a factor of 2 then you translate the resulting image by 1 pixel; if you want to translate by 0.25 degrees, you need to scale up by a factor of 4, and so on.
Note that this implementation is not efficient because when you scale up you end up calculating pixel values that you won't actually use because they're just dropped when you subsample back to the original image size. The implementation in Paul's answer is more efficient.