If I have two points say A(x,y) & B(p,q), how can I find the coordinates of the third point on both sides of AB?
I have the formula for one side, but that cannot give the other one..
Need the formula for coordinates of third point on each sides.
Since you have the result on one side then it's easy to find the symmetrical point.
Let's say your result for the third point is C(r, s). You need to find D(t, w), the symmetrical point of C, with respect to the segment AB.
For this we consider the middle of AB: M(u, v) = (A(x, y) + B(p, q)) / 2;
We have the following equation: M(u, v) = (D(t, w) + C(r, s))/2.
We get D(t, w) = 2 * M(u, v) - C(r, s).
We further get: D(t, w) = A(x, y) + B(p, q) - C(r, s).
A possible solution, based on a rotation matrix R is:
A = [0 0];
B = [0 1];
AB = B-A;
theta = deg2rad(60);
R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
C = A + AB*R';
X = [A;B;C; A];
plot(X(:, 1), X(:, 2));
axis equal
Related
I am testing MATLAB capabilities in solving equations for a project that I intend to do, so I gave it a test run with something simple, but the results that it gives me are incorrect. I tried to solve two non-linear equations with two unknowns, one of the solutions is correct the other is not.
syms theta d x y
eq1 = d * cos(theta) == x;
eq2 = d * sin(theta) == y;
sol = solve(eq1, eq2, theta, d)
sol.theta
sol.d
The solutions for d are correct, but for theta I get:
-2*atan((x - (x^2 + y^2)^(1/2))/y)
-2*atan((x + (x^2 + y^2)^(1/2))/y)
And the correct answer for theta is simply atan(y/x)
Then when I evaluate these solutions with x = 1, y = 0, I get:
eval(sol.d)
eval(sol.theta)
d = 1, -1
theta = NaN, -3.1416
Solutions for d are correct, but theta in that scenario should be 0.
What am I doing wrong?
EDIT: solving it by hand it looks like this: Divide the y equation by the x equation
y/x = (d * sin(theta)) / (d * cos(theta))
y/x = sin(theta)/cos(theta)
y/x = tan(theta)
theta = atan(y/x)
Even if matlab solves it in some other way and gets a different expression, it should still yield the same final result when I use numbers and it PARTIALLY does.
For x = 1 and y = 0, theta should be 0, => this doesnt work, it gives NaN (explanation bellow)
for x = 1 and y = 1, theta should be 45 degrees => this works
for x = 0 and y = 1 theta should be 90 degrees => this works
And I just checked it again with the 45 and 90 degree values for x and y and it works, but for x = 1 and y = 0 it still gives NaN as one of the answers and that is because it gets a 0/0 from the way it is expressing it
-2*atan((x - (x^2 + y^2)^(1/2))/y)
-2*(1 - (1^2 + 0^2))^(1/2)/0
-2*(1 - 1)^(1/2)/0
0/0
but if its in the form of atan(y/x) the result is
theta = atan(0/1)
theta = atan(0)
theta = 0
Did you mean to solve this:
syms a b theta d real
eq1 = a==d * cos(theta) ;
eq2 = b==d * sin(theta) ;
[sol] = solve([eq1 eq2],[d theta] ,'IgnoreAnalyticConstraints', true,'Real',true,'ReturnConditions',true);
When solving the equations with symbolic x and y, the solver will find a solution with a certain condition, which can be obtained using the argument 'ReturnCondition':
syms x y theta d real
eq1 = d*cos(theta) == x;
eq2 = d*sin(theta) == y;
sol = solve([eq1; eq2],[d theta],'ReturnConditions',true);
This gives the following result for sol
>> sol.d
(x^2 + y^2)^(1/2)
-(x^2 + y^2)^(1/2)
>> sol.theta
2*pi*k - 2*atan((x - (x^2 + y^2)^(1/2))/y)
2*pi*k - 2*atan((x + (x^2 + y^2)^(1/2))/y)
>> sol.parameters
k
>> sol.conditions
y ~= 0 & in(k, 'integer')
y ~= 0 & in(k, 'integer')
As you can see, y = 0 does not fulfill this general solution given by the solver, resulting in your problem for y = 0. You can find solutions for y = 0 by either making y numeric instead of symbolic, or by adding an assumption:
syms x y theta d real
assume(y==0)
sol = solve([eq1; eq2],[d theta],'ReturnConditions',true);
I guess its easier to just set y=0 numeric, for this one condition, since there are already 4 possible solutions and conditions for the three lines above.
I have game object that moves along a path on a 2D plane, between 2 points (x1,y1) and (x2,y2). Occasionally it gets moved off the path and needs to be put back on it. When this happens I'll know the x-coordinate, but need to calculate the y-coordinate along the path given the x-coordinate.
Here's an illustration of what I mean:
You have a line segment, i.e., the set of all convex combinations of the given endpoints. You would like to find the coefficients that yield the convex combination (x3,y3), where y3 is unknown.
t (x1,y1) + (1-t) (x2,y2) = (x3,y3)
Since x3 is known, we obtain
t = (x3 - x2) / (x1 - x2)
and, thus,
y3 = ((x3-x2) y1 + (x1-x3) y2) / (x1 - x2)
The general equation of a line in 2D is a.x + b.y + c = 0 where the vector U = (-b, a) is a direction vector of the line.
Because (x1, y1) and (x2, y2) are on the line, you know that:
a.x1 + b.y1 + c = 0
a.x2 + b.y2 + c = 0
(x2-x1, y2-y1) is a direction vector of the line so:
-b = x2-x1 and a = y2-y1
So one equation of your line ax + by + c = 0 with :
a = y2-y1
b = x1-x2
c = -a.x1 - b.y1 = x1(y1-y2) + y1(x2-x1)
Knowing a, b, c and x3, you can easily find y3:
y3 = -(c + a.x3) / b
Pay attention nevertheless to the case where b = 0 (case of a vertical line)
How to find an point where line1 and lin2 intersect, if both lines are defined by x,y,alpha where x,y are coordinates of a point on a line and alpha is the angle between the line and x=const?
I tried applying sine theorem but it yields two answers (triangles can be built on both sides of a line). I can check which point forms correct slope with one of the points, but that is ugly.
I can switch to y=ax+b representation, but then I have special cases that I have to worry about. Vertical and horizontal lines should be differently to avoid division by zero in 1/sin(alpha) and 1/cos(alpha) cases.
I am not looking for an implementation in a certain language, just a formula.
These questions are not relevant because they deal with finite line segments, NOT lines.
Given two points and two vectors, find point of intersection
How do you detect where two line segments intersect?
Suppose line 1 is defined by [x1, y1] and alpha1 and line 2 by [x2, y2] and alpha2.
Suppose k1 = tan(alpha1) and k2 = tan(alpha2).
Then the formula for the x-coordinate of the intersection is
x = (y2 - y1 + k1 * x1 - k2 * x2) / (k1 - k2)
Note: Function tan is undefined for angles pi / 2 + k * pi (where k is an arbitrary integer), so:
if k1 is undefined, then x = x1 and y = y2 + k2 * (x1 - x2)
if k2 is undefined, then x = x2 and y = y1 + k1 * (x2 - x1)
(both are practically the same with exchange of indices 1 <--> 2).
For a line equation Y = aX + b, you can calculate a = tan(alpha).
So if line1 is defined as x, y and alpha, the equation is Y = tan(alpha) * X + b.
Now to find b, you need a point on your line. This point has coordinates (x, y).
y = ax + b
b = y - ax
So you line equation is:
Y = tan(alpha) * X + (y - tan(alpha) * x)
Now you only have to solve the lines equation:
Y = a1 * X + b1
Y = a2 * X + b2
Which is:
a1 * X + b1 = a2 * X + b2
(a1 - a2) * X = b2 - b1
X = (b2 - b1) / (a1 - a2)
Right now you can calculate Y too.
So if we replace, we obtain:
X = ((y2 - tan(alpha2) * x2) - (y1 - tan(alpha1) * x1)) / (tan(alpha1) - tan(alpha2)
Simplified:
X = (y2 - y1 - tan(alpha2) * x2 + tan(alpha1) * x1)) / (tan(alpha1) - tan(alpha2)
And then:
Y = tan(alpha1) * X + (y - tan(alpha1) * x
Suppose we are given a training dataset {yᵢ, xᵢ}, for i = 1, ..., n, where yᵢ can either be -1 or 1 and xᵢ can be e.g. a 2D or 3D point.
In general, when the input points are linearly separable, the SVM model can be defined as follows
min 1/2*||w||²
w,b
subject to the constraints (for i = 1, ..., n)
yᵢ*(w*xᵢ - b) >= 1
This is often called the hard-margin SVM model, which is thus a constrained minimization problem, where the unknowns are w and b. We can also omit 1/2 in the function to be minimized, given it's just a constant.
Now, the documentation about Matlab's quadprog states
x = quadprog(H, f, A, b) minimizes 1/2*x'*H*x + f'*x subject to the restrictions A*x ≤ b. A is a matrix of doubles, and b is a vector of doubles.
We can implement the hard-margin SVM model using quadprog function, to get the weight vector w, as follows
H becomes an identity matrix.
f' becomes a zeros matrix.
A is the left-hand side of the constraints
b is equal to -1 because the original constraint had >= 1, it becomes <= -1 when we multiply with -1 on both sides.
Now, I am trying to implement a soft-margin SVM model. The minimization equation here is
min (1/2)*||w||² + C*(∑ ζᵢ)
w,b
subject to the constraints (for i = 1, ..., n)
yᵢ*(w*xᵢ - b) >= 1 - ζᵢ
such that ζᵢ >= 0, where ∑ is the summation symbol, ζᵢ = max(0, 1 - yᵢ*(w*xᵢ - b)) and C is a hyper-parameter.
How can this optimization problem be solved using the Matlab's quadprog function? It's not clear to me how the equation should be mapped to the parameters of the quadprog function.
The "primal" form of the soft-margin SVM model (i.e. the definition above) can be converted to a "dual" form. I did that, and I am able to get the Lagrange variable values (in the dual form). However, I would like to know if I can use quadprog to solve directly the primal form without needing to convert it to the dual form.
I don't see how it can be a problem. Let z be our vector of (2n + 1) variables:
z = (w, eps, b)
Then, H becomes diagonal matrix with first n values on the diagonal equal to 1 and the last n + 1 set to zero:
H = diag([ones(1, n), zeros(1, n + 1)])
Vector f can be expressed as:
f = [zeros(1, n), C * ones(1, n), 0]'
First set of constrains becomes:
Aineq = [A1, eye(n), zeros(n, 1)]
bineq = ones(n, 1)
where A1 is a the same matrix as in primal form.
Second set of constraints becomes lower bounds:
lb = (inf(n, 1), zeros(n, 1), inf(n, 1))
Then you can call MATLAB:
z = quadprog(H, f, Aineq, bineq, [], [], lb);
P.S. I can be mistaken in some small details, but the general idea is right.
I wanted to clarify #vharavy answer because you could get lost while trying to deduce what 'n' means in his code. Here is my version according to his answer and SVM wikipedia article. I assume we have a file named "test.dat" which holds coordinates of test points and their class membership in the last column.
Example content of "test.dat" with 3D points:
-3,-3,-2,-1
-1,3,2,1
5,4,1,1
1,1,1,1
-2,5,4,1
6,0,1,1
-5,-5,-3,-1
0,-6,1,-1
-7,-2,-2,-1
Here is the code:
data = readtable("test.dat");
tableSize = size(data);
numOfPoints = tableSize(1);
dimension = tableSize(2) - 1;
PointsCoords = data(:, 1:dimension);
PointsSide = data.(dimension+1);
C = 0.5; %can be changed
n = dimension;
m = numOfPoints; %can be also interpretet as number of constraints
%z = [w, eps, b]; number of variables in 'z' is equal to n + m + 1
H = diag([ones(1, n), zeros(1, m + 1)]);
f = [zeros(1, n), C * ones(1, m), 0];
Aineq = [-diag(PointsSide)*table2array(PointsCoords), -eye(m), PointsSide];
bineq = -ones(m, 1);
lb = [-inf(1, n), zeros(1, m), -inf];
z = quadprog(H, f, Aineq, bineq, [], [], lb);
If let z = (w; w0; eps)T be a the long vector with n+1+m elements.(m the number of points)
Then,
H= diag([ones(1,n),zeros(1,m+1)]).
f = [zeros(1; n + 1); ones(1;m)]
The inequality constraints can be specified as :
A = -diag(y)[X; ones(m; 1); zeroes(m;m)] -[zeros(m,n+1),eye(m)],
where X is the n x m input matrix in the primal form.Out of the 2 parts for A, the first part is for w0 and the second part is for eps.
b = ones(m,1)
The equality constraints :
Aeq = zeros(1,n+1 +m)
beq = 0
Bounds:
lb = [-inf*ones(n+1,1); zeros(m,1)]
ub = [inf*ones(n+1+m,1)]
Now, z=quadprog(H,f,A,b,Aeq,beq,lb,ub)
Complete code. The idea is the same as above.
n = size(X,1);
m = size(X,2);
H = diag([ones(1, m), zeros(1, n + 1)]);
f = [zeros(1,m+1) c*ones(1,n)]';
p = diag(Y) * X;
A = -[p Y eye(n)];
B = -ones(n,1);
lb = [-inf * ones(m+1,1) ;zeros(n,1)];
z = quadprog(H,f,A,B,[],[],lb);
w = z(1:m,:);
b = z(m+1:m+1,:);
eps = z(m+2:m+n+1,:);
I have a question connected to this code:
t = -20:0.1:20;
plot3(zeros(size(t)),t,-t.^2);
grid on
hold on
i = 1;
h = plot3([0 0],[0 t(i)],[0 -t(i)^2],'r');
h1 = plot3([-1 0],[0 0],[-400 -200],'g');
for(i=2:length(t))
set(h,'xdata',[-1 0],'ydata',[0 t(i)],'zdata',[-400 -t(i)^2]);
pause(0.01);
end
In this code, I plot two intersecting lines. H1, and H2. H1 is fixed, H2 moves as a function of time. H2 happens to trace a parabola in this example, but its movement could be arbitrary.
How can I calculate and draw the bisector of the angle between these two intersecting lines for every position of the line H2? I would like to see in the plot the bisector and the line H2 moving at the same time.
Solving this problem for one position of H2 is sufficient, since it will be the same procedure for all orientations of H2 relative to H1.
I am not a geometry genius, there is likely an easier way to do this. As of now, no one has responded though, so this will be something.
You have three points in three space:
Let A be the common vertice of the two line segments.
Let B and C be two known points on the two line segments.
Choose an arbitrary distance r where
r <= distance from A to B
r <= distance from A to C
Measure from A along line segment AB a distance of r. This is point RB
Measure from A along line segment AC a distance or r. This is point RC
Find the mid point of line segment connecting RB and RC. This is point M
Line segment AM is the angular bisector of angle CAB.
Each of these steps should be relatively easy to accomplish.
Here is basically MatlabDoug's method with some improvement on the determination of the point he calls M.
t = -20:0.1:20;
plot3(zeros(size(t)),t,-t.^2);
grid on
hold on
v1 = [1 0 200];
v1 = v1/norm(v1);
i = 1;
h = plot3([-1 0],[0 t(i)],[-400 -t(i)^2],'r');
h1 = plot3([-1 0],[0 0],[-400 -200],'g');
l = norm([1 t(i) -t(i)^2+400]);
p = l*v1 + [-1 0 -400];
v2 = (p + [0 t(i) -t(i)^2])/2 - [-1 0 -400];
p2 = [-1 0 -400] + v2/v2(1);
h2 = plot3([-1 p2(1)],[0 p2(2)],[-400 p2(3)],'m');
pause(0.1)
for(i=2:length(t))
l = norm([1 t(i) -t(i)^2+400]);
p = l*v1 + [-1 0 -400];
v2 = (p + [0 t(i) -t(i)^2])/2 - [-1 0 -400];
p2 = [-1 0 -400] + v2/v2(1);
set(h,'xdata',[-1 0],'ydata',[0 t(i)],'zdata',[-400 -t(i)^2]);
set(h2,'xdata',[-1 p2(1)],'ydata',[0 p2(2)],'zdata',[-400 p2(3)]);
pause;
end
I just use the following:
Find the normalized vectors AB, and AC, where A is the common point of the segments.
V = (AB + AC) * 0.5 // produces the direction vector that bisects AB and AC.
Normalize V, then do A + V * length to get the line segment of the desired length that starts at the common point.
(Note that this method does not work on 3 points along a line to produce a perpendicular bisector, it will yield a vector with no length in that case)
I have added a C# implementation (in the XZ plane using Unity 3D Vector3 struct) that handles Perpendicular and Reflex bisectors in case someone that knows MATLAB would translate it.
public Vector3 GetBisector(Vector3 center, Vector3 first, Vector3 second)
{
Vector3 firstDir = (first - center).normalized;
Vector3 secondDir = (second - center).normalized;
Vector3 result = ((firstDir + secondDir) * 0.5f).normalized;
if (IsGreaterThan180(-firstDir, secondDir))
{
// make into a reflex vector
(result.x, result.z) = (-result.x, -result.z);
}
if (result.sqrMagnitude < 0.99f)
{
// we have a colinear set of lines.
// return the perpendicular bisector.
result = Vector3.Cross(Vector3.up, -firstDir).normalized;
}
return result;
}
bool IsGreaterThan180(Vector3 dir, Vector3 dir2)
{
// < 0.0 for clockwise ordering
return (dir2.x * dir.z - dir2.z * dir.x) < 0.0f;
}
Also note that the returned bisector is a vector of unit length. Using "center + bisector * length" could be used to place it into worldspace.