I want to implement Pollard_Lambda for finding discrete log of an elliptic curve point in sage.
for dividing elliptic points in three sections I need to compare y coordinates of elliptic point.So is there any function in sage witch can separate our x and y coordinates of elliptic point pleaze tell me
---Santosh Javheri
It is actually fairly simple to divide a point on an elliptic curve into its x and y coordinates. Here's how it goes for example, on a 'random' Elliptic Curve over a finite field Fq:
q = (2 ** 255) - 19
E = EllipticCurve(GF(q),[0,486662,0,1,0])
point = E([yourXCoordinate,yourYCoordinate]) #any point you'd like on E
x,y = point.xy() #the function you asked for
Hope that helps
Related
I have coordinates corresponding to a set of 2D contours, each corresponding to different heights. These contours do not draw out a perfect ellipsoid in 3D, and instead what I would like to do is to find the best fitting ellipsoid. I do not have any knowledge on the origin of this ellipsoid.
My first thought was to incorporate some type of least squares algorithm, where I find the ellipsoid parameters that minimize the distance between points. I imagine this would be quite expensive and not too far from a brute force approach. I am convinced there is a more elegant and efficient way of doing this. If there is an existing library that handles this (preferably in Python) that would be even better.
I have already seen a related question (Fitting an ellipsoid to 3D data points), but figured I would ask again as it has been over a decade since that post.
So you have a set of (x,y) values for each contour, which describe a portion of an ellipse (blue dots below).
The best fit ellipse is described by the general equation
A x^2 + B y^2 + 2C x y + 2D x + 2E y = 1
and once the coefficients (A,B,C,D,E) are found, the ellipse of fully described. See below in how to find the the curve coordinates (x,y) from the coefficients and a parameter t=0 .. 1.
To find the coefficients of the ellipse, form 5 vectors, each a column of a n×5 matrix Q
for i = 1 to n
Q(i,1) = x(i)^2
Q(i,2) = y(i)^2
Q(i,3) = 2*x(i)*y(i)
Q(i,4) = 2*x(i)
Q(i,5) = 2*y(i)
next i
and a vector K filled with 1 for the right-hand side
for i = 1 to n
K(i) = 1.0
next i
Find the coefficients using a least-squares fit with some linear algebra
[A,B,C,D,E] = inv(tr(Q)*Q)*tr(Q)*K
where tr(Q) is the transpose of Q and * is matrix/vector product
Now we need to extract the geometric properties of the ellipse from the coefficient. I want to have a the semi-major axis, b the semi-minor axis, φ the rotation angle, xc the x-axis center, yc the y-axis center.
xc = -(B*D-C*E)/(A*B-(C^2))
yc = -(A*E-C*D)/(A*B-(C^2))
φ = atan( 2*C/(A-B) )/2
a = SQRT(2*(A*(B+E^2)+B*D^2-C*(C+2*D*E))/((A*B-C^2)*(A+B-SQRT((A-B)^2+4*C^2))))
b = SQRT(2*(A*(B+E^2)+B*D^2-C*(C+2*D*E))/((A*B-C^2)*(A+B+SQRT((A-B)^2+4*C^2))))
Finally to plot the ellipse you need to generate a set of points (x,y) from the curve parameter t=0..1 using the above 5 coefficients.
Generate the centered aligned coordinates (u,v) with
u = a*cos(2*π*t)
v = b*sin(2*π*t)
Generate the centered rotated coordinates (x',y') with
x' = u*cos(φ) - v*sin(φ)
y' = u*sin(φ) + v*cos(φ)
Generate the ellipse coordinates (x,y) with
x = x' + xc
y = y' + yc
The result is observed above in the first picture.
Now for the total solution, each 2D slice would have its own ellipse. But all the slices would not generate an ellipsoid this way.
Extending the above into 3D coordinates (x,y,z) is doable, but the math is quite involved and I feel [SO] is not a good place to develop such an algorithm. You can hack it together, by finding the average center for each slice (weighted by the ellipse area π*a*b). Additionally, the rotation angle should be the same for all contours, and so another averaging is needed. Finally, the major and minor axis values would fall on an elliptical curve along the z-axis and it would require another least-fit solution. This one is driven by the equation
(x/a)^2 + (y/b)^2 + (z/c)^2 = 1
but rather in the aligned coordinates (u,v,w)
(u/a)^2 + (v/b)^2 + (w/c)^2 = 1
Suppose E:y^2=x^3+Ax+B mod p, I have two questions?
how can I draw the graph of E with pari-gp.
how can I get the list of all points over the E.
thank you for all.
To define an Elliptic Curve with SageMath use
E = EllipticCurve(GF(131),[0,1,0,1,0])
print(E)
and outputs
Elliptic Curve defined by y^2 = x^3 + x^2 + x over Finite Field of size 131
In your case ( simplified Weierstrass from)
E = EllipticCurve(GF(p),[A,B])
will be enough.
To plot a curve
E.plot()
is enough
To iterate the points
for T in E.points():
print(T)
is enough.
Try online on SageMathCell.
And notice the symmetry!
Pari-GP
From a tutorial
a=ffgen(P,’a)
Es = ellinit([a^4,a^6],a);
I just want to sample cartesian points from an elliptic curve in $(x,y)$ form. Given one of x or y, I want to be able to calculate the other's possibilities.
Examples: I have EllipticCurve('11a') and I want to know where it intersects the x-axis in (x,y) coordinates. What about where the curve intersects x=5? Or where it intersects y=3?
Can somebody assist me in doing this?
You just need to solve a polynomial equation in one variable. The factor function, or the .roots method of polynomials can do that for you.
I have line like in 2D defined by ax+by+c = 0 so (a,b,c).
I need to compute a polar representation of this line like Hough approach with rho an theta.
How to do this?
A line in cartesian coordinates is not as easily represented in polar coordinates.
You can simply substitute x,y with their respective polar equivalents, r*cos(theta), r*sin(theta), giving you
a*r*cos(theta) + b*r*sin(theta) + c = 0
This implicit equation is not as easy to figure out, however. But, if you first convert your implicit line equation to a parametric vector equation of the form (x,y) = R(t) = R0 + t*V, where R0,V are cartesian vectors which you can derive from a,b,c, you can then write
(r*cos(theta), r*sin(theta)) = R0 + t*V
and solve this system of equations for r and theta in terms of t.
However, polar coordinates are not the same as the Hough transform.
In the Hough system, the line is defined by the length rho of a perpendicular line that crosses (0,0) , which is theta = atan(b/a). Figuring out rho seems more difficult at first, but this tutorial explains it.
I'd like to generate random points being located on the surface of an n-dimensional torus. I have found formulas for how to generate the points on the surface of a 3-dimensional torus:
x = (c + a * cos(v)) * cos(u)
y = (c + a * cos(v)) * sin(u)
z = a * sin(v)
u, v ∈ [0, 2 * pi); c, a > 0.
My question is now: how to extend this formulas to n dimensions. Any help on the matter would be much appreciated.
I guess that you can do this recursively. Start with a full orthonormal basis of your vector space, and let the current location be the origin. At each step, choose a point in the plane spanned by the first two coordinate vectors, i.e. take w1 = cos(t)*v1 + sin(t)*v2. Shift the other basis vectors, i.e. w2 = v3, w3 = v4, …. Also take a step from your current position in the direction w1, with the radius r1 chosen up front. When you only have a single basis vector remaining, then the current point is a point on the n-dimensional torus of the outermost recursive call.
Note that while the above may be used to choose points randomly, it won't choose them uniformly. That would likely be a much harder question, and you definitely should ask about the math of that on Math SE or perhaps on Cross Validated (Statistics SE) to get the math right before you worry about implementation.
An n-torus (n being the dimensionality of the surface of the torus; a bagel or doughnut is therefore a 2-torus, not a 3-torus) is a smooth mapping of an n-rectangle. One way to approach this is to generate points on the rectangle and then map them onto the torus. Aside from the problem of figuring out how to map a rectangle onto a torus (I don't know it off-hand), there is the problem that the resulting distribution of points on the torus is not uniform even if the distribution of points is uniform on the rectangle. But there must be a way to adjust the distribution on the rectangle to make it uniform on the torus.
Merely generating u and v uniformly will not necessarily sample uniformly from a torus surface. An additional step is needed.
J.F. Williamson, "Random selection of points distributed on curved surfaces", Physics in Medicine & Biology 32(10), 1987, describes a general method of choosing a uniformly random point on a parametric surface. It is an acceptance/rejection method that accepts or rejects each candidate point depending on its stretch factor (norm-of-gradient). To use this method for a parametric surface, several things have to be known about the surface, namely—
x(u, v), y(u, v) and z(u, v), which are functions that generate 3-dimensional coordinates from two dimensional coordinates u and v,
The ranges of u and v,
g(point), the norm of the gradient ("stretch factor") at each point on the surface, and
gmax, the maximum value of g for the entire surface.
For the 3-dimensional torus with the parameterization you give in your question, g and gmax are the following:
g(u, v) = a * (c + cos(v) * a).
gmax = a * (a + c).
The algorithm to generate a uniform random point on the surface of a 3-dimensional torus with torus radius c and tube radius a is then as follows (where RNDEXCRANGE(x,y) returns a number in [x,y) uniformly at random, and RNDRANGE(x,y) returns a number in [x,y] uniformly at random):
// Maximum stretch factor for torus
gmax = a * (a + c)
while true
u = RNDEXCRANGE(0, pi * 2)
v = RNDEXCRANGE(0, pi * 2)
x = cos(u)*(c+cos(v)*a)
y = sin(u)*(c+cos(v)*a)
z = sin(v)*a
// Norm of gradient (stretch factor)
g = a*abs(c+cos(v)*a)
if g >= RNDRANGE(0, gmax)
// Accept the point
return [x, y, z]
end
end
If you have n-dimensional torus generating formulas, a similar approach can be used to generate uniform random points on that torus (accept a candidate point if norm-of-gradient equals or exceeds a random number in [0, gmax), where gmax is the maximum norm-of-gradient).