I am trying to find the angle of the outer line of the object in the green region of the image as shown in the image above…
For that, I have scanned the green region and get the points (dark blue points as shown in the image)...
As you can see, the points are not making straight line so I can’t find angle easily.
So I think I have to find a middle way and
that is to find the line so that the distance between each point and line remain as minimum as possible.
So how can I find the line so that each point exposes minimum distance to it……?
Is there any algorithm for this or is there any good way other than this?
The obvious route would be to do a least-squares linear regression through the points.
The standard least squares regression formulae for x on y or y on x assume there is no error in one coordinate and minimize the deviations in the coordinate from the line.
However, it is perfectly possible to set up a least squares calculation such that the value minimized is the sum of squares of the perpendicular distances of the points from the lines. I'm not sure whether I can locate the notebooks where I did the mathematics - it was over twenty years ago - but I did find the code I wrote at the time to implement the algorithm.
With:
n = ∑ 1
sx = ∑ x
sx2 = ∑ x2
sy = ∑ y
sy2 = ∑ y2
sxy = ∑ x·y
You can calculate the variances of x and y and the covariance:
vx = sx2 - ((sx * sx) / n)
vy = sy2 - ((sy * sy) / n)
vxy = sxy - ((sx * sy) / n)
Now, if the covariance is 0, then there is no semblance of a line. Otherwise, the slope and intercept can be found from:
slope = quad((vx - vy) / vxy, vxy)
intcpt = (sy - slope * sx) / n
Where quad() is a function that calculates the root of quadratic equation x2 + b·x - 1 with the same sign as c. In C, that would be:
double quad(double b, double c)
{
double b1;
double q;
b1 = sqrt(b * b + 4.0);
if (c < 0.0)
q = -(b1 + b) / 2;
else
q = (b1 - b) / 2;
return (q);
}
From there, you can find the angle of your line easily enough.
Obviously the line will pass through averaged point (x_average,y_average).
For direction you may use the following algorithm (derived directly from minimizing average square distance between line and points):
dx[i]=x[i]-x_average;
dy[i]=y[i]-y_average;
a=sum(dx[i]^2-dy[i]^2);
b=sum(2*dx[i]*dy[i]);
direction=atan2(b,a);
Usual linear regression will not work here, because it assumes that variables are not symmetric - one depends on other, so if you will swap x and y, you will have another solution.
The hough transform might be also a good option:
http://en.wikipedia.org/wiki/Hough_transform
You might try searching for "total least squares", or "least orthogonal distance" but when I tried that I saw nothing immediately applicable.
Anyway suppose you have points x[],y[], and the line is represented by a*x+b*y+c = 0, where hypot(a,b) = 1. The least orthogonal distance line is the one that minimises Sum{ (a*x[i]+b*y[i]+c)^2}. Some algebra shows that:
c is -(a*X+b*Y) where X is the mean of the x's and Y the mean of the y's.
(a,b) is the eigenvector of C corresponding to it's smaller eigenvalue, where C is the covariance matrix of the x's and y's
Related
Is there any algorithm / method to find the smallest regular hexagon around a set of points (x, y).
And by smallest I mean smallest area.
My current idea was to find the smallest circle enclosing the points, and then create a hexagon from there and check if all the points are inside, but that is starting to sound like a never ending problem.
Requirements
First of all, let's define a hexagon as quadruple [x0, y0, t0, s], where (x0, y0), t0 and s are its center, rotation and side-length respectively.
Next, we need to find whether an arbitrary point is inside the hexagon. The following functions do this:
function getHexAlpha(t, hex)
t = t - hex.t0;
t = t - 2*pi * floor(t / (2*pi));
return pi/2 - abs(rem(t, pi/3) - (pi/6));
end
function getHexRadious( P, hex )
x = P.x - hex.x0;
y = P.y - hex.y0;
t = atan2(y, x);
return hex.s * cos(pi/6) / sin(getHexAlpha(t, hex));
end
function isInHex(P, hex)
r = getHexRadious(P, hex);
d = sqrt((P.x - hex.x0)^2 + (P.y - hex.y0)^2);
return r >= d;
end
Long story short, the getHexRadious function formulates the hexagon in polar form and returns distance from center of hexagon to its boundary at each angle. Read this post for more details about getHexRadious and getHexRadious functions. This is how these work for a set of random points and an arbitrary hexagon:
The Algorithm
I suggest a two-stepped algorithm:
1- Guess an initial hexagon that covers most of points :)
2- Tune s to cover all points
Chapter 1: (2) Following Tarantino in Kill Bill Vol.1
For now, let's assume that our arbitrary hexagon is a good guess. Following functions keep x0, y0, t0 and tune s to cover all points:
function getHexSide( P, hex )
x = P.x - hex.x0;
y = P.y - hex.y0;
r = sqrt(x^2 + y^2);
t = atan2(y, x);
return r / (cos(pi/6) / sin(getHexAlpha(t, hex)));
end
function findMinSide( P[], hex )
for all P[i] in P
S[i] = getHexSide(P, hex);
end
return max(S[]);
end
The getHexSide function is reverse of getHexRadious. It returns the minimum required side-length for a hexagon with x0, y0, t0 to cover point P. This is the outcome for previous test case:
Chapter 2: (1)
As a guess, we can find two points furthest away from each other and fit one of hexagon diameters' on them:
function guessHex( P[] )
D[,] = pairwiseDistance(P[]);
[i, j] = indexOf(max(max(D[,])));
[~, j] = max(D(i, :));
hex.x0 = (P[i].x + P[j].x) / 2;
hex.y0 = (P[i].y + P[j].y) / 2;
hex.s = D[i, j]/2;
hex.t0 = atan2(P.y(i)-hex.y0, P.x(i)-hex.x0);
return hex;
end
Although this method can find a relatively small polygon, but as a greedy approach, it never guarantees to find the optimum solutions.
Chapter 3: A Better Guess
Well, this problem is definitely an optimization problem with its objective being to minimize area of hexagon (or s variable). I don't know if it has an analytical solution, and SO is not the right place to discuss it. But any optimization algorithm can be used to provide a better initial guess. I used GA to solve this with findMinSide as its cost function. In fact GA generates many guesses about x0, y0, and t0 and the best one will be selected. It finds better results but is more time consuming. Still no guarantee to find the optimum!
Optimization of Optimization
When it comes to optimization algorithms, performance is always an issue. Keep in mind that hexagon only needs to enclose the convex-hall of points. If you are dealing with large sets of points, it's better to find the convex-hall and get rid of the rest of the points.
For a given line of an equation y = mx + c where m is the gradient and c is the y-intercept. How would I determine the "edge-points" on a graph?
To clarify what I mean by "edge-points", I've added an example below.
The edge-points are circled in red. To determine the edge-points here it would simply be (0, c) & (maximum x-value, m * maximum x-value + c). However, the problem arises when I consider lines with a different m value. For example:
I can't apply the same logic here, instead the edge-points would be ((maximum y-value - c) * m, maximum y-value) & (-c * m, 0), which was derived from the equation y = mx + c. So my question is how would I determine these 2 edge-points given any m or c? Is there a certain pattern I'm not seeing here?
You need to solve 4 simple linear equations (not equation system!):
y = m * 0 + c = c
y = m * maxX + c
0 = m * x + c
maxY = m * x + c
and get points of intersections with axes and with max lines. Then filter out points with negative coordinates and too large ones, because you want only the 1st quadrant
The first one equation is already solved y=c
The second gives point of intersection with right vertical line
The third gives point of intersection with OX axis
The fourth gives point of intersection with top horizontal line
Example:
maxX = 5
maxY = 5
line y = 2 *x - 1
x0, y0 = 0, -1
x1, y1 = 5, 9
x2, y2 = 1/2, 0
x3, y3 = 2, 5
First pair contains negative coordinate y=-1
Second pair contains y=9 > maxY
Third and fourth ones fulfill your constraints.
So this line gives segment (1/2, 0)-(2, 5) (like near vertical segment at your second picture)
This algo might be considered as simple kind of line clippping by rectangle
For the line to be in the given rectangle there is a constraint given by the x values and a constraint given by the y values.
The x constraint trivially leads to an interval in which the x values must be.
The y constraint also gives you such an interval for the x values but only after some easy calculation.
Now determine the intersection of the two intervals (which may also be empty).
Heading
I need to find the indices of the polygon nearest to a point
So in this case the ouput would be 4 and 0. Such that if the red point is added I know to where to place the vertex in the array. Does anyone know where to start?
(Sorry if the title is misleading, I wasnt sure how to phrase it properly)
In this case the ouput would be 0 and 1, rather than the closest 4.
Point P lies on the segment AB, if two simple conditions are met together:
AP x PB = 0 //cross product, vectors are collinear or anticollinear, P lies on AB line
AP . PB > 0 //scalar product, exclude anticollinear case to ensure that P is inside the segment
So you can check all sequential vertice pairs (pseudocode):
if (P.X-V[i].X)*(V[i+1].Y-P.Y)-(P.Y-V[i].Y)*(V[i+1].X-P.X)=0 then
//with some tolerance if point coordinates are float
if (P.X-V[i].X)*(V[i+1].X-P.X)+(P.Y-V[i].Y)*(V[i+1].Y-P.Y)>0
then P belongs to (i,i+1) segment
This is fast direct (brute-force) method.
Special data structures exist in computer geometry to quickly select candidate segments - for example, r-tree. But these complicated methods will gain for long (many-point) polylines and for case where the same polygon is used many times (so pre-treatment is negligible)
I'll assume that the new point is to be added to an edge. So you are given the coordinates of a point a = (x, y) and you want to find the indices of the edge on which it lies. Let's call the vertices of that edge b, c. Observe that the area of the triangle abc is zero.
So iterate over all edges and choose the one that minimizes area of triangle abc where a is your point and bc is current edge.
a = input point
min_area = +infinity
closest_edge = none
n = number of vertices in polygon
for(int i = 1; i <= n; i++)
{ b = poly[ i - 1 ];
c = poly[ i % n ];
if(area(a, b, c) < min_area)
{ min_area = area(a, b, c);
closest_edge = bc
}
}
You can calculate area using:
/* Computes area x 2 */
int area(a, b, c)
{ int ans = 0;
ans = (a.x*b.y + b.x*x.y + c.x*a.y) - (a.y*b.x + b.y*c.x + c.y*a.x);
return ABS(ans);
}
I think you would be better off trying to compare the distance from the actual point to a comparable point on the line. The closest comparable point would be the one that forms a perpendicular line like this. a is your point in question and b is the comparable point on the line line between the two vertices that you will check distance to.
However there's another method which I think might be more optimal for this case (as it seems most of your test points lie pretty close to the desired line already). Instead of find the perpendicular line point we can simply check the point on the line that has the same X value like this. b in this case is a lot easier to calculate:
X = a.X - 0.X;
Slope = (1.Y - 0.Y) / (1.X - 0.X);
b.X = 0.X + X;
b.Y = 0.Y + (X * Slope);
And the distance is simply the difference in Y values between a and b:
distance = abs(a.Y - b.Y);
One thing to keep in mind is that this method will become more inaccurate as the slope increases as well as become infinite when the slope is undefined. I would suggest flipping it when the slope > 1 and checking for a b that lies at the same y rather than x. That would look like this:
Y = a.Y - 0.Y;
Inverse_Slope = (1.X - 0.X) / (1.Y - 0.Y);
b.Y = 0.Y + Y;
b.X = 0.Y + (Y * Inverse_Slope);
distance = abs(a.X - b.X);
Note: You should also check whether b.X is between 0.X and 1.X and b.Y is between 0.Y and 1.Y in the second case. That way we are not checking against points that dont lie on the line segment.
I admit I don't know the perfect terminology when it comes to this kind of thing so it might be a little confusing, but hope this helps!
Rather than checking if the point is close to an edge with a prescribed tolerance, as MBo suggested, you can fin the edge with the shortest distance to the point. The distance must be computed with respect to the line segment, not the whole line.
How do you compute this distance ? Let P be the point and Q, R two edge endpoints.
Let t be in range [0,1], you need to minimize
D²(P, QR) = D²(P, Q + t QR) = (PQ + t QR)² = PQ² + 2 t PQ.QR + t² QR².
The minimum is achieved when the derivative cancels, i.e. t = - PQ.QR / QR². If this quantity exceeds the range [0,1], just clamp it to 0 or 1.
To summarize,
if t <= 0, D² = PQ²
if t >= 1, D² = PR²
otherwise, D² = PQ² - t² QR²
Loop through all the vertices, calculate the distance of that vertex to the point, find the minimum.
double min_dist = Double.MAX_VALUE;
int min_index=-1;
for(int i=0;i<num_vertices;++i) {
double d = dist(vertices[i],point);
if(d<min_dist) {
min_dist = d;
min_index = i;
}
}
I'm working on a project in Matlab and need to find the area between two lines inside of the square [-1,+1]x[-1,+1] intersecting in a point (xIntersection,yIntersection). So the idea is to subtract the two lines and integrate between [-1, xIntersection] and [xIntersection, +1], sum the results and if it's negative, change its sign.
For details on how I find the intersection of the two lines check this link.
I'm using Matlab's function integral(), here a snippet of my code:
xIntersection = ((x_1 * y_2 - y_1 * x_2) * (x_3 - x_4) - (x_1 - x_2) * (x_3 * y_4 - y_3 * x_4) ) / ((x_1 - x_2) * (y_3 - y_4) - (y_1 - y_2) * (x_3 - x_4));
d = #(x) g(x) - f(x);
result = integral(d, -1, xIntersection) - int( d, xIntersection, 1)
if(result < 0),
result = result * -1;
end
Note that I defined previously in the code g(x) and f(x) but haven't reported it in the snippet.
The problem is that I soon realized that the lines could intersect either inside or outside of the square, furthermore they could intersect the square on any of its sides and the number of possible combinations grows up very quickly.
I.e.:
These are just 4 cases, but considering that f(+1), f(-1), g(+1), g(-1) could be inside the interval [-1,+1], above it or under it and that the intersection could be inside or outside of the square the total number is 3*3*3*3*2 = 162.
Obviously in each case the explicit function to integrate in order to get the area between the two lines is different, but I can't possibly think of writing a switch case for each one.
Any ideas?
I think my answer to your previous question still applies, for the most part.
If you want to compute the area of the region bounded by the smaller angle of the two lines and the boundaries of the square, then you can forget about the intersection, and forget about all the different cases.
You can just use the fact that
the area of the square S is equal to 4
the value of this integral
A = quadgk(#(x) ...
abs( max(min(line1(x),+1),-1) - max(min(line2(x),+1),-1) ), -1, +1);
gives you the area between the lines (sometimes the large angle, sometimes the small angle)
the value of of min(A, S-A) is the correct answer (always the small angle).
Assuming “between the lines” means “inside the smaller angle formed by the lines”:
With the lines l and h, S := [-1,+1]x[-1,+1] and B as the border of S.
Transform l to the form l_1 + t*l_2 with l_1 and l_2 beeing vectors. Do the same for h.
If the intersection is not inside S, find the 4 intersections of l and h with B. Sort them so you get a convex quadrilateral. Calculate its area.
Else:
Find the intersection point p and find the intersection angle α between l_2 and h_2. and then check:
If α in [90°,180°] or α in [270°,360°], swap l and h.
If α > 180°, set l_2 = −l_2
Set l_1 := h_1 := p. Do once for positive t and negative t
(forwards and backwards along l_2 and h_2 from p):
Find intersections s_l and s_h of l and h with B.
If on same border of B: compute area of triangle s_l, s_h, p
If on adjacent borders of B: find the corner c of B in between the hit borders and once again sort the four points s_l, s_h, p and c so you get an convex quadrilateral and calculate it’s area.
If on opposite borders, find the matching side of B (you can do this by looking at the direction of s_l-p). Using its two corner points c_1 and c_2, you now have 5 points that form a polygon. Sort them so the polygon is convex and calculate its area.
This is still quite a few cases, but I don’t think there’s a way around that. You can simplify this somewhat by also using the polygon formula for the triangle and the quadrilateral.
How to sort the points so you get a convex polygon/quadrilateral: select any one of them as p_1 and then sort the rest of the points according to angle to p_1.
If you define an intersection function and a polygon_area that takes a list of points, sorts them and returns the area, this algorithm should be rather simple to implement.
edit: Image to help explain the comment:
Hey guys thanks for your answers, I also thought of an empirical method to find the area between the lines and wanted to share it for the sake of discussion and completeness.
If you take a large number of random points within the square [-1,+1]x[-1,+1] you can measure the area as the fraction of the points which fall in the area between the two lines.
Here's a little snippet and two images to show the different accuracy of empirical result obtained with different number of points.
minX = -1;
maxX = +1;
errors = 0;
size = 10000;
for j=1:size,
%random point in [-1,+1]
T(j,:) = minX + (maxX - minX).*rand(2,1);
%equation of the two lines is used to compute the y-value
y1 = ( ( B(2) - A(2) ) / ( B(1) - A(1) ) ) * (T(j,1) - A(1)) + A(2);
y2 = (- W(1) / W(2)) * T(j,1) -tresh / W(2);
if(T(j,2) < y1),
%point is under line one
Y1 = -1;
else
%point is above line one
Y1 = +1;
end
if(T(j,2) < y2),
%point is under line two
Y2 = -1;
else
%point is above line two
Y2 = +1;
end
if(Y1 * Y2 < 0),
errors = errors + 1;
scatter(T(j,1),T(j,2),'fill','r')
else
scatter(T(j,1),T(j,2),'fill','g')
end
end
area = (errors / size) / 4;
And here are two images, it sure takes longer than the solution posted by #Rody but as you can see you can make it accurate.
Number of points = 2000
Number of points = 10000
Actually this is a classic problem as SO user Victor put it (in another SO question regarding which tasks to ask during an interview).
I couldn't do it in an hour (sigh) so what is the algorithm that calculates the number of integer points within a triangle?
EDIT: Assume that the vertices are at integer coordinates. (otherwise it becomes a problem of finding all points within the triangle and then subtracting all the floating points to be left with only the integer points; a less elegant problem).
Assuming the vertices are at integer coordinates, you can get the answer by constructing a rectangle around the triangle as explained in Kyle Schultz's An Investigation of Pick's Theorem.
For a j x k rectangle, the number of interior points is
I = (j – 1)(k – 1).
For the 5 x 3 rectangle below, there are 8 interior points.
(source: uga.edu)
For triangles with a vertical leg (j) and a horizontal leg (k) the number of interior points is given by
I = ((j – 1)(k – 1) - h) / 2
where h is the number of points interior to the rectangle that are coincident to the hypotenuse of the triangles (not the length).
(source: uga.edu)
For triangles with a vertical side or a horizontal side, the number of interior points (I) is given by
(source: uga.edu)
where j, k, h1, h2, and b are marked in the following diagram
(source: uga.edu)
Finally, the case of triangles with no vertical or horizontal sides can be split into two sub-cases, one where the area surrounding the triangle forms three triangles, and one where the surrounding area forms three triangles and a rectangle (see the diagrams below).
The number of interior points (I) in the first sub-case is given by
(source: uga.edu)
where all the variables are marked in the following diagram
(source: uga.edu)
The number of interior points (I) in the second sub-case is given by
(source: uga.edu)
where all the variables are marked in the following diagram
(source: uga.edu)
Pick's theorem (http://en.wikipedia.org/wiki/Pick%27s_theorem) states that the surface of a simple polygon placed on integer points is given by:
A = i + b/2 - 1
Here A is the surface of the triangle, i is the number of interior points and b is the number of boundary points. The number of boundary points b can be calculated easily by summing the greatest common divisor of the slopes of each line:
b = gcd(abs(p0x - p1x), abs(p0y - p1y))
+ gcd(abs(p1x - p2x), abs(p1y - p2y))
+ gcd(abs(p2x - p0x), abs(p2y - p0y))
The surface can also be calculated. For a formula which calculates the surface see https://stackoverflow.com/a/14382692/2491535 . Combining these known values i can be calculated by:
i = A + 1 - b/2
My knee-jerk reaction would be to brute-force it:
Find the maximum and minimum extent of the triangle in the x and y directions.
Loop over all combinations of integer points within those extents.
For each set of points, use one of the standard tests (Same side or Barycentric techniques, for example) to see if the point lies within the triangle. Since this sort of computation is a component of algorithms for detecting intersections between rays/line segments and triangles, you can also check this link for more info.
This is called the "Point in the Triangle" test.
Here is an article with several solutions to this problem: Point in the Triangle Test.
A common way to check if a point is in a triangle is to find the vectors connecting the point to each of the triangle's three vertices and sum the angles between those vectors. If the sum of the angles is 2*pi (360-degrees) then the point is inside the triangle, otherwise it is not.
Ok I will propose one algorithm, it won't be brilliant, but it will work.
First, we will need a point in triangle test. I propose to use the "Barycentric Technique" as explained in this excellent post:
http://www.blackpawn.com/texts/pointinpoly/default.html
Now to the algorithm:
let (x1,y1) (x2,y2) (x3,y3) be the triangle vertices
let ymin = floor(min(y1,y2,y3)) ymax = ceiling(max(y1,y2,y3)) xmin = floor(min(x1,x2,x3)) ymax = ceiling(max(x1,x2,3))
iterating from xmin to xmax and ymin to ymax you can enumerate all the integer points in the rectangular region that contains the triangle
using the point in triangle test you can test for each point in the enumeration to see if it's on the triangle.
It's simple, I think it can be programmed in less than half hour.
I only have half an answer for a non-brute-force method. If the vertices were integer, you could reduce it to figuring out how to find how many integer points the edges intersect. With that number and the area of the triangle (Heron's formula), you can use Pick's theorem to find the number of interior integer points.
Edit: for the other half, finding the integer points that intersect the edge, I suspect that it's the greatest common denominator between the x and y difference between the points minus one, or if the distance minus one if one of the x or y differences is zero.
Here's another method, not necessarily the best, but sure to impress any interviewer.
First, call the point with the lowest X co-ord 'L', the point with the highest X co-ord 'R', and the remaining point 'M' (Left, Right, and Middle).
Then, set up two instances of Bresenham's line algorithm. Parameterize one instance to draw from L to R, and the second to draw from L to M. Run the algorithms simultaneously for X = X[L] to X[M]. But instead of drawing any lines or turning on any pixels, count the pixels between the lines.
After stepping from X[L] to X[M], change the parameters of the second Bresenham to draw from M to R, then continue to run the algorithms simultaneously for X = X[M] to X[R].
This is very similar to the solution proposed by Erwin Smout 7 hours ago, but using Bresenham instead of a line-slope formula.
I think that in order to count the columns of pixels, you will need to determine whether M lies above or below the line LR, and of course special cases will arise when two points have the same X or Y co-ordinate. But by the time this comes up, your interviewer will be suitably awed and you can move on to the next question.
Quick n'dirty pseudocode:
-- Declare triangle
p1 2DPoint = (x1, y1);
p2 2DPoint = (x2, y2);
p3 2DPoint = (x3, y3);
triangle [2DPoint] := [p1, p2, p3];
-- Bounding box
xmin float = min(triangle[][0]);
xmax float = max(triangle[][0]);
ymin float = min(triangle[][1]);
ymax float = max(triangle[][1]);
result [[float]];
-- Points in bounding box might be inside the triangle
for x in xmin .. xmax {
for y in ymin .. ymax {
if a line starting in (x, y) and going in any direction crosses one, and only one, of the lines between the points in the triangle, or hits exactly one of the corners of the triangle {
result[result.count] = (x, y);
}
}
}
I have this idea -
Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of the triangle. Let 'count' be the number of integer points forming the triangle.
If we need the points on the triangle edges then using Euclidean Distance formula http://en.wikipedia.org/wiki/Euclidean_distance, the length of all three sides can be ascertained.
The sum of length of all three sides - 3, would give that count.
To find the number of points inside the triangle we need to use a triangle fill algorithm and instead of doing the actual rendering i.e. executing drawpixel(x,y), just go through the loops and keep updating the count as we loop though.
A triangle fill algorithm from
Fundamentals of Computer Graphics by
Peter Shirley,Michael Ashikhmin
should help. Its referred here http://www.gidforums.com/t-20838.html
cheers
I'd go like this :
Take the uppermost point of the triangle (the one with the highest Y coordinate). There are two "slopes" starting at that point. It's not the general solution, but for easy visualisation, think of one of both "going to the left" (decreasing x coordinates) and the other one "going to the right".
From those two slopes and any given Y coordinate less than the highest point, you should be able to compute the number of integer points that appear within the bounds set by the slopes. Iterating over decreasing Y coordinates, add all those number of points together.
Stop when your decreasing Y coordinates reach the second-highest point of the triangle.
You have now counted all points "above the second-highest point", and you are now left with the problem of "counting all the points within some (much smaller !!!) triangle, of which you know that its upper side parallels the X-axis.
Repeat the same procedure, but now with taking the "leftmost point" instead of the "uppermost", and with proceedding "by increasing x", instead of by "decreasing y".
After that, you are left with the problem of counting all the integer points within a, once again much smaller, triangle, of which you know that its upper side parallels the X-axis, and its left side parallels the Y-axis.
Keep repeating (recurring), until you count no points in the triangle you're left with.
(Have I now made your homework for you ?)
(wierd) pseudo-code for a bit-better-than-brute-force (it should have O(n))
i hope you understand what i mean
n=0
p1,p2,p3 = order points by xcoordinate(p1,p2,p3)
for int i between p1.x and p2.x do
a = (intersection point of the line p1-p2 and the line with x==i).y
b = (intersection point of the line p1-p3 and the line with x==i).y
n += number of integers between floats (a, b)
end
for i between p2.x+1 and p3.x do
a = (intersection point of the line p2-p3 and the line with x==i).y
b = (intersection point of the line p1-p3 and the line with x==i).y
n += number of integers between floats (a, b)
end
this algorithm is rather easy to extend for vertices of type float (only needs some round at the "for i.." part, with a special case for p2.x being integer (there, rounded down=rounded up))
and there are some opportunities for optimization in a real implementation
Here is a Python implementation of #Prabhala's solution:
from collections import namedtuple
from fractions import gcd
def get_points(vertices):
Point = namedtuple('Point', 'x,y')
vertices = [Point(x, y) for x, y in vertices]
a, b, c = vertices
triangle_area = abs((a.x - b.x) * (a.y + b.y) + (b.x - c.x) * (b.y + c.y) + (c.x - a.x) * (c.y + a.y))
triangle_area /= 2
triangle_area += 1
interior = abs(gcd(a.x - b.x, a.y - b.y)) + abs(gcd(b.x - c.x, b.y - c.y)) + abs(gcd(c.x - a.x, c.y - a.y))
interior /= 2
return triangle_area - interior
Usage:
print(get_points([(-1, -1), (1, 0), (0, 1)])) # 1
print(get_points([[2, 3], [6, 9], [10, 160]])) # 289
I found a quite useful link which clearly explains the solution to this problem. I am weak in coordinate geometry so I used this solution and coded it in Java which works (at least for the test cases I tried..)
Link
public int points(int[][] vertices){
int interiorPoints = 0;
double triangleArea = 0;
int x1 = vertices[0][0], x2 = vertices[1][0], x3 = vertices[2][0];
int y1 = vertices[0][1], y2 = vertices[1][1], y3 = vertices[2][1];
triangleArea = Math.abs(((x1-x2)*(y1+y2))
+ ((x2-x3)*(y2+y3))
+ ((x3-x1)*(y3+y1)));
triangleArea /=2;
triangleArea++;
interiorPoints = Math.abs(gcd(x1-x2,y1-y2))
+ Math.abs(gcd(x2-x3, y2-y3))
+ Math.abs(gcd(x3-x1, y3-y1));
interiorPoints /=2;
return (int)(triangleArea - interiorPoints);
}