Working in Swift, ARTKit / SceneKit
I have a line AB in 3d and I have xyz coordinates of both points A and B.
I also have a point C and I know its xyz coordinates too.
Now, I want to find out the xyz coordinates of point D on line AB; given that CD is perpendicular to AB.
What would be a simple way to do it in Swift.
Parameterize the line AB with a scalar t:
P(t) = A + (B - A) * t`
The point D = P(t) is such that CD is perpendicular to AB, i.e. their dot product is zero:
dot(C - D, B - A) = 0
dot(C - A - (B - A) * t, B - A) = 0
dot(C - A, B - A) = t * dot(B - A, B - A)
// Substitute value of t
--> D = A + (B - A) * dot(C - A, B - A) / dot(B - A, B - A)
Swift code:
var BmA = B - A
var CmA = C - A
var t = dot(CmA, BmA) / dot(BmA, BmA)
var D = A + BmA * t;
Related
I'm trying to make a function that takes 3 points as arguments. The first two of which represent two points on a line. The third one represents another point, outside of that line. Suppose a perpendicular through the third point on the line defined by the first two points. Now what I want to do, is calculate that intersection. I've come up with this procedure so far, but somehow it works only like 50% of the time. Could somebody figure out what I'm doing wrong here?
def calculateIntersection(p1: (Double, Double), p2: (Double, Double), c: (Double, Double)): (Double, Double) = {
var intersection: (Double, Double) = null
// CASE 1: line is vertical
if(p1._1 == p2._1) {
intersection = (p1._1, c._2)
}
// CASE 2: line is horizontal
else if(p1._2 == p2._2) {
intersection = (c._1, p1._2)
}
// CASE 3: line is neither vertical, nor horizontal
else {
val slope1: Double = (p2._2 - p1._2) / (p2._1 - p1._1) // slope of the line
val slope2: Double = pow(slope1, -1) * -1 // slope of the perpendicular
val intercept1: Double = p1._2 - (slope1 * p1._1) // y-intercept of the line
val intercept2: Double = c._2 - (slope2 * c._1) // y-intercept of the perpendicular
intersection = ((intercept2 - intercept1) / (slope1 - slope2),
slope1 * ((intercept2 - intercept1) / (slope1 - slope2)) + intercept1)
}
intersection
}
Given the following definitions:
type Point = (Double, Double)
implicit class PointOps(p: Point) {
def +(other: Point) = (p._1 + other._1, p._2 + other._2)
def -(other: Point) = (p._1 - other._1, p._2 - other._2)
def dot(other: Point) = p._1 * other._1 + p._2 * other._2
def *(scalar: Double) = (p._1 * scalar, p._2 * scalar)
def normSquare: Double = p._1 * p._1 + p._2 * p._2
}
meaning that
a + b // is vector addition
a - b // is vector subtraction
a dot b // is the dot product (scalar product)
a * f // is multiplication of a vector `a` with a scalar factor `f`
a.normSquare // is the squared length of a vector
you obtain the projection of a point p on line going through points line1 and line2 as follows:
/** Projects point `p` on line going through two points `line1` and `line2`. */
def projectPointOnLine(line1: Point, line2: Point, p: Point): Point = {
val v = p - line1
val d = line2 - line1
line1 + d * ((v dot d) / d.normSquare)
}
Example:
println(projectPointOnLine((-1.0, 10.0), (7.0, 4.0), (6.0, 11.0)))
gives
(3.0, 7.0)
This works in 3D (or n-D) in exactly the same way.
Some math behind that (How to derive it from scratch)
(notation as above)
We have three points: l1 and l2 for the line and p for the target point.
We want to project the point p orthogonally onto the line that goes through l1 and l2 (assuming l1 != l2).
Let d = l2 - l1 be the direction from l1 to l2. Then every point on the line can be represented as
l1 + d * t
with some scalar factor t. Now we want to find a t such that the vector connecting p and l1 + d * t is orthogonal to d, that is:
(p - (l1 + d * t)) dot d == 0
Recall that
(v1 + v2) dot v3 = (v1 dot v3) + (v2 dot v3)
for all vectors v1, v2, v3, and that
(v1 * s) dot v2 = (v1 dot v2) * s
for scalar factors s. Using this and the definition v.normSquared = v dot v, we obtain:
(p - l1 - d * t) dot d
= (p - l1) dot d - (d dot d) * t
= (p - l1) dot d - d.normSquare * t
and this should become 0. Resolving for t gives:
t = ((p - l1) dot d) / d.normSquare
and this is exactly the formula that is used in the code.
(Thanks at SergGr for adding an initial sketch of the derivation)
In literature, papers and a book (e.g., [1] and [2]), I found different equations for the Izhikevich Neuron model.
In [1], I found this:
v' = 0.04v^2 + 5v + 140 - u + I
u' = a(bv - u)
if v >= 30 mV => v = c; u = u + d
In [2], I found this:
C*v' = k(v - v_r)(v - v_t) - u + I_in
u' = a{b(v - v_r) - u}
if v >= v_peak => v = c; u = u + d
I tried to go from the second set of equations to the first set of equations, ignoring the C. I found this:
k = 0.04; v_r = -82.6556; v_t = -42.3444
However, this gave me complete garbage. Thus, my question is as follows:
Why are there 2 different sets of equations, and which one should I take if I want to simulate Izhikevich neurons?
Bonus question #1: How do I go from set 2 to set 1?
Bonus question #2: Am I correct to state that 'I' is in [pA]?
References:
[1] Simple Model of Spiking Neurons - E.M. Izhikevich (2003)
[2] Dynamical Systems in neuroscience - E.M. Izhikevich (2007)
1) question #1: go from set 2 to set 1:
1.1) create a new variable V:
V = C(v - v_r)
1.2) you obtain:
(v - v_r) = V / C
v = (V / C) + v_r
v' = (V' / C) + A
1.3) replace (1.2) in set 2 and you'll obtain set 1 with new coefficients:
K = k/(C^2) = 0.04
D = (k/C)(v_r - v_t) = 5
E = -CA = 140
B = b/C
V' = KV^2 + DV + E -u +I
u' = a(BV' - u)
2) question #2: is 'I' expressed as [pA]?
yes, it is. v is [mV] and each time step is 1ms
I want to calculate Fourier series for some function func.
I build this method:
function y = CalcFourier(accurate, func, a, b, val_x)
f = #(x) eval(func);
% calculate coefficients
a0 = (2 / (b - a)) * calcArea(func, a , b);
an = (2 / (b - a)) * calcArea(strcat(func, '*cos(2*n*pi*x / (b - a))'), a , b);
an = (2 / (b - a)) * calcArea(strcat(func, '*sin(2*n*pi*x / (b - a))'), a , b);
partial = 0;
an_f = #(n) an;
bn_f = #(n) bn;
for n = 1:accurate
partial = partial + an_f(n)* cos(2*n*pi*val_x / (b - a)) + bn_f(n) * sin(2*n*pi*val_x / (b - a));
end
y = (a0 / 2) + partial;
end
And this - to approximate the coefficient's:
function area = calcArea(func, a, b)
f = #(x) eval(func);
area = (a - b) * (f(a) - f(b)) / 2;
end
On line an = (2 / (b - a)) * calcArea(strcat(func, '*cos(2*n*pi*x / (b - a))'), a , b); I'm getting error:
??? Error using ==> eval
Undefined function or variable 'n'.
Error in ==> calcArea>#(x)eval(func) at 2
f = #(x) eval(func);
Error in ==> calcArea at 3
area = (a - b) * (f(a) - f(b)) / 2;
Error in ==> CalcFourier at 5
an = (2 / (b - a)) * calcArea(strcat(func,
'*cos(2*n*pi*x / (b - a))'), a , b);
>>
Is there any option to declate n as "some constant"? Thanks!
You try to use a variable called n in line 4 of your code. However at that time n is not defined yet, that only happens in the for loop. (Tip: Use dbstop if error at all times, that way you can spot the problem more easily).
Though I don't fully grasp what you are doing I believe you need something like this:
n=1 at the start of your CalcFourier function. Of course you can also choose to input n as a variable, or to move the corresponding line to a place where n is actually defined.
Furthermore you seem to use n in calcArea, but you don't try to pass it to the function at all.
All of this would be easier to find if you avoided the use of eval, perhaps you can try creating the function without it, and then matlab will more easily guide you to the problems in your code.
if the symbolic toolbox is available it can be used to declare symbolic variables , which can be treated as 'some variable' and substituted with a value later.
however a few changes should be made for it to be implemented, generally converting anonymous functions to symbolic functions and any function of n to symbolic functions. And finally the answer produced will need to be converted from a symbolic value to some more easy to handle value e.g. double
quickly implementing this to your code as follows;
function y = test(accurate, func, a, b, val_x)
syms n x % declare symbolic variables
%f = symfun(eval(func),x); commented out as not used
The two lines above show the declaration of symbolic variables and the syntax for creating a symbolic function
% calculate coefficients
a0 = symfun((2 / (b - a)) * calcArea(func, a , b),x);
an = symfun((2 / (b - a)) * calcArea(strcat(func, '*cos(2*n*pi*x / (b - a))'),...
... a , b),[x n]);
bn = symfun((2 / (b - a)) * calcArea(strcat(func, '*sin(2*n*pi*x / (b - a))'),...
... a , b),[x n]);
partial = 0;
the function definitions in in your code are combined into the lines above, note they functions are of x and n, the substitution of x_val is done later here...
for n = 1:accurate
partial = partial + an(val_x,n)* cos(2*n*pi*val_x / (b - a)) +...
... bn(val_x,n) * sin(2*n*pi*val_x / (b - a));
end
The for loop which now replaces the symbolic n with values and calls the symbolic functions with x_val and each n value
y = (a0 / 2) + partial;
y = double(y);
end
Finally the solution is calculated and then converted to double;
Disclaimer: I have not checked if this code generates the correct solution, however I hope it gives you enough information to understand what has been changed and why to carry out the process given in your code above, using the symbolic toolbox to address the issue...
I'm working on a function that takes a 1xn vector x as input and returns a nxn matrix L.
I'd like to speed things up by vectorizing the loops, but there's a catch that puzzles me: loop index b depends on loop index a. Any help would be appreciated.
x = x(:);
n = length(x);
L = zeros(n, n);
for a = 1 : n,
for b = 1 : a-1,
c = b+1 : a-1;
if all(x(c)' < x(b) + (x(a) - x(b)) * ((b - c)/(b-a))),
L(a,b) = 1;
end
end
end
From a quick test, it looks like you are doing something with the lower triangle only. You might be able to vectorize using ugly tricks like ind2sub and arrayfun similar to this
tril_lin_idx = find(tril(ones(n), -1));
[A, B] = ind2sub([n,n], tril_lin_idx);
C = arrayfun(#(a,b) b+1 : a-1, A, B, 'uniformoutput', false); %cell array
f = #(a,b,c) all(x(c{:})' < x(b) + (x(a) - x(b)) * ((b - c{:})/(b-a)));
L = zeros(n, n);
L(tril_lin_idx) = arrayfun(f, A, B, C);
I cannot test it, since I do not have x and I don't know the expected result. I normally like vectorized solutions, but this is maybe pushing it a bit too much :). I would stick to your explicit for-loop, which might be much clearer and which Matlab's JIT should be able to speed up easily. You could replace the if with L(a,b) = all(...).
Edit1
Updated version, to prevent wasting ~ n^3 space on C:
tril_lin_idx = find(tril(ones(n), -1));
[A, B] = ind2sub([n,n], tril_lin_idx);
c = #(a,b) b+1 : a-1;
f = #(a,b) all(x(c(a, b))' < x(b) + (x(a) - x(b)) * ((b - c(a, b))/(b-a)));
L = zeros(n, n);
L(tril_lin_idx) = arrayfun(f, A, B);
Edit2
Slight variant, which does not use ind2sub and which should be more easy to modify in case b would depend in a more complex way on a. I inlined c for speed, it seems that especially calling the function handles is expensive.
[A,B] = ndgrid(1:n);
v = B<A; % which elements to evaluate
f = #(a,b) all(x(b+1:a-1)' < x(b) + (x(a) - x(b)) * ((b - (b+1:a-1))/(b-a)));
L = false(n);
L(v) = arrayfun(f, A(v), B(v));
If I understand your problem correctly, L(a, b) == 1 if for any c with a < c < b, (c, x(c)) is “below” the line connecting (a, x(a)) and (b, x(b)), right?
It is not a vectorization, but I found the other approach. Rather than comparing all c with a < c < b for each new b, I saved the maximum slope from a to c in (a, b), and used it for (a, b + 1). (I tried with only one direction, but I think that using both directions is also possible.)
x = x(:);
n = length(x);
L = zeros(n);
for a = 1:(n - 1)
L(a, a + 1) = 1;
maxSlope = x(a + 1) - x(a);
for b = (a + 2):n
currSlope = (x(b) - x(a)) / (b - a);
if currSlope > maxSlope
maxSlope = currSlope;
L(a, b) = 1;
end
end
end
I don't know your data, but with some random data, the result is the same with original code (with transpose).
An esoteric answer: You could do the calculations for every a,b,c from 1:n, exclude the don't cares, and then do the all along the c dimension.
[a, b, c] = ndgrid(1:n, 1:n, 1:n);
La = x(c)' < x(b) + (x(a) - x(b)) .* ((b - c)./(b-a));
La(b >= a | c <= b | c >= a) = true;
L = all(La, 3);
Though the jit would probably do just fine with the for loops since they do very little.
Edit: still uses all of the memory, but with less maths
[A, B, C] = ndgrid(1:n, 1:n, 1:n);
valid = B < A & C > B & C < A;
a = A(valid); b = B(valid); c = C(valid);
La = true(size(A));
La(valid) = x(c)' < x(b) + (x(a) - x(b)) .* ((b - c)./(b-a));
L = all(La, 3);
Edit2: alternate last line to add the clause that c of no elements is true
L = all(La,3) | ~any(valid,3);
Say I have the following matrix equation
X - B*X*C = D
Where,
X: 3 by 5, to be solved;
B: 3 by 3;
C: 5 by 5;
D: 3 by 5;
Is there any convenient method that I can use for solving the system? fsolve?
In case B or C are invertible, you can check the matrix cookbook section 5.1.10 deals with similar settings:
X * inv(C) - B * X = D * inv(C)
Can be translated to
x = inv( kron( eye, -B ) + kron( inv(C)', eye ) ) * d
where x and d are vector-stack of X and D respectively.
You can use MATLAB's dlyap function:
X = dlyap(B,C,D)