I have an elliptic curve defined by
y^2 = x^3 + 1062282974404935987005872930817*x + 1204388198013706813607478558721 over Finite Field of size 2017313518945563799802055961909.
And I want to get a point on this curve of order 3569809307570934983774171.
How can I get it?
Easily you can see that the order of E, is 2017313518945565643070719128784. The main stage is that you find the generator(s) of E, and with SageMath they are:
H_1 = (651721743085147348480059087840, 277924022187240437411690075386)
and
H_2 = (364767631279436218861124076682, 0)
But you can see that the order of H_1 is 1008656759472782821535359564392, so
G = 565104 * H_1
G = (1144674520220442511918931779419, 850803345221750997044804585048)
is the desired point on E.
Related
I have this triangle:
I'm trying to get the vertex value highlighted in the green circle in order to draw that red line. Is there any equation that I can use to extract that value?
The centroid vertex G = (x=5.5, y=1.5)
The other vertex B = (x=0, y=1)
and the last vertex C = (x=7, y=0)
Any help would be appreciated. I know it might be a 5th grade math but I can't think of a way to calculate this point.
If you throw away the majority of the triangle and just keep the vector B->G and the vector B->C then this problem shows itself to be a "vector projection" problem.
These are solved analytically using the dot product of the 2 vectors and are well documented elsewhere.
Took me 2 days to figure this out, you basically need to get the slopes for the base vector and the altitude vector (centroid), then solve this equation: y = m * x + b for both vectors (the base + altitude). Then you'll get 2 different equations that you need to use substitution to get the x first then apply that value to the 2nd equation to get the y. For more information watch this youtube tutorial:
https://www.youtube.com/watch?v=VuEbWkF5lcM
Here's the solution in PHP (pseudo) if anyone is interested:
//slope of base
$m1 = getSlope(baseVector);
//slope of altitude (invert and divide it by 1)
$m2 = 1/-$m1;
//points
$x1 = $baseVector->x;
$y1 = $baseVector->y;
//Centroid vertex
$x2 = $center['x'];
$y2 = $center['y'];
//altitude equation: y = m * x + b
//eq1: y1 = (m1 * x1) + b1 then find b1
$b1 = -($m1 * $x1) + $y1;
//equation: y = ($m1 * x) + $b1
//eq2: y2 = (m2 * x2) + b2 then find b2
$b2 = -($m2 * $x2) + $y2;
//equation: y = ($m2 * x) + $b2;
//substitute eq1 into eq2 and find x
//merge the equations (move the Xs to the left side and numbers on the right side)
$Xs = $m1 - $m2; //left side (number of Xs)
$Bs = $b2 - $b1; //right side
$x = $Bs / $Xs; //get x number
$y = ($m2 * $x) + $b2; //get y number
I've got a situation in which I have 2 circles (C1 and C2)
and i need to find the line equation for the line that is tangent to both of these circles.
So as far as i'm aware, given a single point (P1) and C2's point and radius it is possible to quite easily get 2 possible points of tangency for C2 and P1 to make 2 line equations. But as i don't have P1, only the knowledge that the point will be one of a possible 2 points on C1, i'm not sure how to calculate this.
I assume it will be something along the lines of getting the 2 tangent line equations of C1 that are equal to the same of C2.
Both circles can have any radius, they could be the same or they could be hugely different. They will also never overlap (they can still touch though). And I'm looking for the 2 possible internal tangents.
Oh, and also, visuals would be very helpful haha :)
Let O be the intersection point between the line through the centers and the tangent.
Let d be the distance between the centers and h1, h2 be the distances between O and the centers. By similarity, these are proportional to the radii.
Hence,
h1 / h2 = r1 / r2 = m,
h1 + h2 = d,
giving
h1 = m d / (1 + m),
h2 = d / (1 + m).
Then the coordinates of O are found by interpolating between the centers
xo = (h2.x1 + h1.x2) / d
yo = (h2.y1 + h1.y2) / d
and the angle of the tangent is that of the line through the centers plus or minus the angle between this line and the tangent,
a = arctan((y2 - y1)/(x2 - x1)) +/- arcsin(r1 / h1).
You can write the implicit equation of the tangent as
cos(a).y - sin(a).x = cos(a).yo - sin(a).xo.
(source: imag.fr)
So we are going to use a homothetic transformation. If the circles C and C' have respectively centres O and O', and radius r and r', then we know there exists a unique homothetic transformation with centre J and ratio a, such that :
a = |JO|/|JO'| = r/r'
Noting AB is the vector from A to B, and |z| the norm of a vector z.
Hence you get J, knowing that it is between O and O' which we both already know.
Then with u the projection of JR on JO', and v the decomposition on its orthogonal, and considering the sine s and cosine c of the angle formed by O'JR, we have
|u| = |JR| * c
|v| = |JR| * s
c^2 + s^2 = 1
And finally because the triangle JRO' is right-angled in R :
s = r' / |JO|'
Putting all of this together, we get :
J = O + OO' / |OO'| * a / (a+1)
if |OJ| == r and |O'J| == r' then
return the orthogonal line to (OO') passing through J
|JR| = √( |JO'|^ - r'^2 )
s = r' / |JO'|
c = √( 1 - s^2 )
u = c * |JR| * OO' / |OO'|
w = (-u.y, u.x) % any orthogonal vector to u
v = s * |JR| * w / |w|
return lines corresponding to parametric equations J+t*(u+v) and J+t*(u-v)
I'm currently working on a game in LBP2 that has modify the way a controller gives input. This question:
How can I convert coordinates on a square to coordinates on a circle?
Has helped me quite a lot with what I am doing, but I do have one problem. I need the inverse function of the one they give. They go from square -> circle, and I've tried searching all over for how to map a circle to a square.
The function given in the previous question is:
xCircle = xSquare * sqrt(1 - 0.5*ySquare^2)
yCircle = ySquare * sqrt(1 - 0.5*xSquare^2)
From Mapping a Square to a Circle
My question is given xCircle and yCircle... how do I find xSquare and ySquare?
I've tried all of the algebra I know, filled up two pages of notes, tried to get wolfram alpha to get the inverse functions, but this problem is beyond my abilities.
Thank you for taking a look.
x = ½ √( 2 + u² - v² + 2u√2 ) - ½ √( 2 + u² - v² - 2u√2 )
y = ½ √( 2 - u² + v² + 2v√2 ) - ½ √( 2 - u² + v² - 2v√2 )
Note on notation: I'm using x = xSquare , y = ySquare, u = xCircle and v = yCircle;
i.e. (u,v) are circular disc coordinates and (x,y) are square coordinates.
For a C++ implementation of the equations, go to
http://squircular.blogspot.com/2015/09/mapping-circle-to-square.html
See http://squircular.blogspot.com
for more example images.
Also, see http://arxiv.org/abs/1509.06344 for the proof/derivation
This mapping is the inverse of
u = x √( 1 - ½ y² )
v = y √( 1 - ½ x² )
P.S. The mapping is not unique. There are other mappings out there. The picture below illustrates the non-uniqueness of the mapping.
if you have xCircle and yCircle that means that you're on a circle with radius R = sqrt(xCircle^2 + yCircle^2). Now you need to extend that circle to a square with half-side = R,
if (xCircle < yCircle)
ySquare = R, xSquare = xCircle * R/yCircle
else
xSquare = R, ySquare = yCircle * R/xCircle
this is for the first quadrant, for others you need some trivial tweaking with the signs
There are many ways you could do this; here's one simple way.
Imagine a circle of radius R centred on the origin, and a square of side 2R centred on the origin, we want to map all of the points within and on the boundary of the circle (with coordinates (x,y)) to points within and on the boundary of the square. Note that we can also describe points within the circle using polar coordinates (r, ø) (that's supposed to be a phi), where
x = r cos ø,
y = r sin ø
(ie r^2 = x^2 + y^2 and r <= 1). Then imagine other coordinates x' = a(ø) x = a(ø) r cos ø, and y' = a(ø) y (ie, we decide that a won't depend on r).
In order to map the boundary of the circle (r = 1) to the boundary of the square (x' = R), we must have, for ø < 45deg, x' = a(ø) R cos ø = R, so we must have a(ø) = 1/cos ø. Similarly, for 45 < ø < 90 we must have the boundary of the circle map to y' = R, giving a(ø) = 1/sin ø in that region. Continuing round the circle, we see that a(ø) must always be positive, so the final mapping from the circle to the square is
x' = a(ø) x,
y' = a(ø) y
where
ø = |arctan y/x| = arctan |y/x|
and
a(ø) = 1/cos ø, when ø <= 45 deg (ie, when x < y), and
a(ø) = 1/sin ø, when ø > 45 deg.
That immediately gives you the mapping in the other direction. If you have coordinates (x', y') on the square (where x' <= R and y' <= R), then
x = x'/a(ø)
y = y'/a(ø)
with a(ø) as above.
A much simpler mapping, though, is to calculate the (r, ø) for the desired position on the circle, and map that to x' = r and y' = ø. That also maps every point in the circle into a rectangle, and vice versa, and might have better properties, depending on what you want to do.
So that's the real question: what is it you're actually aiming to do here?
I was implementing the solution above but the results are not satisfiying.
The square coordinates are not exact.
Here is a simple counter-example:
Consider the point (x,y)=(0.75, 1) on the square.
We map it to the circle with (u,v)=(0.53, 0.85) on the circle.
Applying the expression above we get the new square coordinates
(x',y')=(u/v,r)=(0.625543242, 1) with r=(u^2+v^2)^(1/2).
This point is close but not the expected precise solution.
I solved a root finding problem in order to get the inverse expression of the mapping from square to circle like above.
you need to solve the system equations like above:
I) u = x*(1-y^2/2)^(1/2)
II) v = y*(1-x^2/2)^(1/2)
One ends up with 8 root points as solution. One of the roots I implemented into Excel-VBA which I present here below and it works very fine.
' given the circle coordinates (u,v) caluclates the x coordinate on the square
Function circ2sqrX(u As Double, v As Double) As Double
Dim r As Double, signX As Double, u2 As Double, v2 As Double, uuvv As Double, temp1 As Double
u2 = u * u
v2 = v * v
r = Sqr(u2 + v2)
signX = 1
If v = 0 Or u = 0 Then
circ2sqrX = u
Exit Function
End If
If u < 0 Then
signX = -1
End If
If Abs(u) = Abs(v) And r = 1 Then
circ2sqrX = signX
Exit Function
End If
uuvv = (u2 - v2) * (u2 - v2) / 4
temp1 = 2 * Sqr(uuvv - u2 - v2 + 1)
circ2sqrX = -((temp1 - u2 + v2 - 2) * Sqr(temp1 + u2 - v2 + 2)) / (4 * u)
End Function
' given the circle coordinates (u,v) caluclates the y coordinate on the square
' make use of symetrie property
Function circ2sqrY(u As Double, v As Double) As Double
circ2sqrY=circ2sqrX(v,u)
End Function
how can i calculate the polynomial that has the tangent lines (1) y = x where x = 1, and (2) y = 1 where x = 365
I realize this may not be the proper forum but I figured somebody here could answer this in jiffy.
Also, I am not looking for an algorithm to answer this. I'd just like like to see the process.
Thanks.
I guess I should have mentioned that i'm writing an algorithm for scaling the y-axis of flotr graph
The specification of the curve can be expressed as four constraints:
y(1) = 1, y'(1) = 1 => tangent is (y=x) when x=1
y(365) = 1, y'(365) = 0 => tangent is (y=1) when x=365
We therefore need a family of curves with at least four degrees of freedom to match these constraints; the simplest type of polynomial is a cubic,
y = a*x^3 + b*x^2 + c*x + d
y' = 3*a*x^2 + 2*b*x + c
and the constraints give the following equations for the parameters:
a + b + c + d = 1
3*a + 2*b + c = 1
48627125*a + 133225*b + 365*c + d = 1
399675*a + 730*b + c = 0
I'm too old and too lazy to solve these myself, so I googled a linear equation solver to give the answer:
a = 1/132496, b = -731/132496, c = 133955/132496, d = -729/132496
I will post this type of question in mathoverflow.net next time. thanks
my solution in javascript was to adapt the equation of a circle:
var radius = Math.pow((2*Math.pow(365, 2)), 1/2);
var t = 365; //offset
this.tMax = (Math.pow(Math.pow(r, 2) - Math.pow(x, 2), 1/2) - t) * (t / (r - t)) + 1;
the above equation has the above specified asymptotes. it is part of a step polynomial for scaling an axis for a flotr graph.
well, you are missing data (you need another point to determine the polynomial)
a*(x-1)^2+b*(x-1)+c=y-1
a*(x-365)^2+b*(x-365)+c=y-1
you can solve the exact answer for b
but A depends on C (or vv)
and your question is off topic anyways, and you need to revise your algebra
I am working on a geometry problem that requires finding the intersection of two parabolic arcs in any rotation. I was able to intesect a line and a parabolic arc by rotating the plane to align the arc with an axis, but two parabolas cannot both align with an axis. I am working on deriving the formulas, but I would like to know if there is a resource already available for this.
I'd first define the equation for the parabolic arc in 2D without rotations:
x(t) = ax² + bx + c
y(t) = t;
You can now apply the rotation by building a rotation matrix:
s = sin(angle)
c = cos(angle)
matrix = | c -s |
| s c |
Apply that matrix and you'll get the rotated parametric equation:
x' (t) = x(t) * c - s*t;
y' (t) = x(t) * s + c*t;
This will give you two equations (for x and y) of your parabolic arcs.
Do that for both of your rotated arcs and subtract them. This gives you an equation like this:
xa'(t) = rotated equation of arc1 in x
ya'(t) = rotated equation of arc1 in y.
xb'(t) = rotated equation of arc2 in x
yb'(t) = rotated equation of arc2 in y.
t1 = parametric value of arc1
t2 = parametric value of arc2
0 = xa'(t1) - xb'(t2)
0 = ya'(t1) - yb'(t2)
Each of these equation is just a order 2 polynomial. These are easy to solve.
To find the intersection points you solve the above equation (e.g. find the roots).
You'll get up to two roots for each axis. Any root that is equal on x and y is an intersection point between the curves.
Getting the position is easy now: Just plug the root into your parametric equation and you can directly get x and y.
Unfortunately, the general answer requires solution of a fourth-order polynomial. If we transform coordinates so one of the two parabolas is in the standard form y=x^2, then the second parabola satisfies (ax+by)^2+cx+dy+e==0. To find the intersection, solve both simultaneously. Substituting in y=x^2 we see that the result is a fourth-order polynomial: (ax+bx^2)^2+cx+dx^2+e==0. Nils solution therefore won't work (his mistake: each one is a 2nd order polynomial in each variable separately, but together they're not).
It's easy if you have a CAS at hand.
See the solution in Mathematica.
Choose one parabola and change coordinates so its equation becomes y(x)=a x^2 (Normal form).
The other parabola will have the general form:
A x^2 + B x y + CC y^2 + DD x + EE y + F == 0
where B^2-4 A C ==0 (so it's a parabola)
Let's solve a numeric case:
p = {a -> 1, A -> 1, B -> 2, CC -> 1, DD -> 1, EE -> -1, F -> 1};
p1 = {ToRules#N#Reduce[
(A x^2 + B x y + CC y^2 + DD x + EE y +F /. {y -> a x^2 } /. p) == 0, x]}
{{x -> -2.11769}, {x -> -0.641445},
{x -> 0.379567- 0.76948 I},
{x -> 0.379567+ 0.76948 I}}
Let's plot it:
Show[{
Plot[a x^2 /. p, {x, -10, 10}, PlotRange -> {{-10, 10}, {-5, 5}}],
ContourPlot[(A x^2 + B x y + CC y^2 + DD x + EE y + F /. p) ==
0, {x, -10, 10}, {y, -10, 10}],
Graphics[{
PointSize[Large], Pink, Point[{x, x^2} /. p /. p1[[1]]],
PointSize[Large], Pink, Point[{x, x^2} /. p /. p1[[2]]]
}]}]
The general solution involves calculating the roots of:
4 A F + 4 A DD x + (4 A^2 + 4 a A EE) x^2 + 4 a A B x^3 + a^2 B^2 x^4 == 0
Which is done easily in any CAS.