I need to draw an elliptic curve over the finite field F17(in other words, I want to draw some specific dots on the curve), but somehow I don't get it right.
The curve is defined by the equation:
y^2 = x^3 +x + 1 (mod 17)
I tried the way below, but it can't work.
for x = 0:16, plot(x, mod(sqrt(x^3+x+1), 16),'r')', end
Can someone help ?
[Update]
According to Nathan and Bill's suggestions, here is a slightly modified version.
x = 0:18
plot(mod(x,16), mod(sqrt(x.^3+x+1), 16),'ro')
However, I feel the figure is WRONG , e.g.,y is not an integer when x=4 .
You have to test all points that fulfill the equation y^2 = x^3 +x + 1 (mod 17). Since it is a finite field, you cannot simply take the square root on the right side.
This is how I would go about it:
a=0:16 %all points of your finite field
left_side = mod(a.^2,17) %left side of the equation
right_side = mod(a.^3+a+1,17) %right side of the equation
points = [];
%testing if left and right side are the same
%(you could probably do something nicer here)
for i = 1:length(right_side)
I = find(left_side == right_side(i));
for j=1:length(I)
points = [points;a(i),a(I(j))];
end
end
plot(points(:,1),points(:,2),'ro')
set(gca,'XTick',0:1:16)
set(gca,'YTick',0:1:16)
grid on;
Matlab works with vectors natively.
your syntax was close, but needs to be vectorized:
x = 0:16
plot(x, mod(sqrt(x.^3+x+1), 16),'r')
Note the . in x.^3. This tells Matlab to cube each element of x individually, as opposed to raising the vector x to the 3rd power, which doesn't mean anything.
You can use this code if you want to plot on Real numbers:
syms x y;
v=y^2-x^3-x-1;
ezplot(v, [-1,3,-5,5]);
But, for plot in modulo, at first you can write below code;
X=[]; for x=[0:16], z=[x; mod(x^3+x+1,17)]; X=[X, z]; end, X,
Then, you can plot X with a coordinate matrix.
Related
I want to solve my differential equation and plot velocity vectors but I am having some trouble with that. I tried this:
syms y(x);
ode = (1+exp(x))*y*diff(y,x)-2*exp(x) == 0;
ySol = dsolve(ode)
[X,Y] = meshgrid(-2:.2:2);
Z = 2*exp(X)/((1+exp(X)).*Y);
[DX,DY] = gradient(Z,.2,.2);
figure
contour(X,Y,Z)
hold on
quiver(X,Y,DX,DY)
hold off
and I get this error:
Warning: Matrix is singular to working precision.
Warning: Contour not rendered for non-finite ZData
It is probably something simple that I do not see but I am just starting using Matlab and I cold not find a right way to do my task. Please help me...
EDIT
As bconrad suggested, I changed my Z function like this:
Z = 2*exp(X)/((1+exp(X)).*Y);
and the previous errors are fixed. However, my prime goal, to plot velocity vectors is not accomplished yet because I get a graph like this:
Don’t have the ability to check at the moment, but I reckon you want an element by element division in that line. You’re missing a dot on the division, try
Z = 2*exp(X)./((1+exp(X)).*Y);
I took a closer look once at my station. The zero-division mentioned by Pablo forces inf's in Z, so quiver get's confused when scaling the vectors (understandably) and just doesn't show them. Try this (with the ode part removed):
[X,Y] = meshgrid(-2 : .2 : 2);
Z = 2 * exp(X) ./ ((1 + exp(X)) .* Y);
Z(isinf(Z)) = nan; % To avoid 0-division problems
[DX, DY] = gradient(Z, .2, .2);
figure
contour(X, Y, Z, 30, 'k')
hold on
quiver(X, Y, DX, DY, 6)
hold off
I've done 3 things here:
Added the line Z(isinf(Z)) = nan; forcing infinite values to be essentially ignored by quiver
Added the arguments 30, 'k' to the contour function to show 30 lines, and make them black (a bit more visible)
Added the argument 6 to the quiver function. This overrides the automatic length-scaling of the vectors.
You'll want to play with the arguments in the contour and quiver functions to make your figure appear as you'd like.
PS: There is a handy arrow function on the file exchange that I find gives better control when creating vector field plots. See https://www.mathworks.com/matlabcentral/fileexchange/278-arrow - the ratings do it justice.
I am working to create a helix in Matlab.
Going by the below code:
t = 0:pi/50:20*pi;
(Can you please explain me this syntax or we have to follow this everytime when creating a helix?)
st = sin(t);
ct = cos(t);
plot3(st,ct,t)
As maximum efficiency in a helix angle is between 40 and 45 degrees, if I want to enter the angle as 42, how is it possible in the code?
It would be very helpful if anyone can share their opinion on this
TIA
What happens in the code is merely an execution of the parametric mathematical description of a helix, which you can read up on wikipedia as
x(t) = cos(t)
y(t) = sin(t)
z(t) = t
The first line of your code generates a vector for values of t from 0 to 20pi in steps of pi/50 (i.e., 1000 steps). Since every 2pi means one full rotation (cos and sin are 2pi-periodic), it coincides to 10 turns of the helix (if you want to change this, let t run up to 2*pi*NumberOfRotations). The other two lines generate corresponding vectors for x and y. plot3 plots a line in 3-D where x and y are passed and as argument for z we pass t since z=t.
To change the slope of the helix, use the more general description given by
x(t) = a*cos(t)
y(t) = a*sin(t)
z(t) = b*t
where a is the radius and b/a is the slope. To get 42° use b = a*atand(42). To make sure the aspect ratio is correct in display, use axis equal; after the plot and maybe axis vis3d; if you want to turn it around.
I want to plot relations like y^2=x^2(x+3) in MATLAB without using ezplot or doing algebra to find each branch of the function.
Does anyone know how I can do this? I usually create a linspace and then create a function over the linspace. For example
x=linspace(-pi,pi,1001);
f=sin(x);
plot(x,f)
Can I do something similar for the relation I have provided?
What you could do is use solve and allow MATLAB's symbolic solver to symbolically solve for an expression of y in terms of x. Once you do this, you can use subs to substitute values of x into the expression found from solve and plot all of these together. Bear in mind that you will need to cast the result of subs with double because you want the numerical result of the substitution. Not doing this will still leave the answer in MATLAB's symbolic format, and it is incompatible for use when you want to plot the final points on your graph.
Also, what you'll need to do is that given equations like what you have posted above, you may have to loop over each solution, substitute your values of x into each, then add them to the plot.
Something like the following. Here, you also have control over the domain as you have desired:
syms x y;
eqn = solve('y^2 == x^2*(x+3)', 'y'); %// Solve for y, as an expression of x
xval = linspace(-1, 1, 1000);
%// Spawn a blank figure and remember stuff as we throw it in
figure;
hold on;
%// For as many solutions as we have...
for idx = 1 : numel(eqn)
%// Substitute our values of x into each solution
yval = double(subs(eqn(idx), xval));
%// Plot the points
plot(xval, yval);
end
%// Add a grid
grid;
Take special care of how I used solve. I specified y because I want to solve for y, which will give me an expression in terms of x. x is our independent variable, and so this is important. I then specify a grid of x points from -1 to 1 - exactly 1000 points actually. I spawn a blank figure, then for as many solutions to the equation that we have, we determine the output y values for each solution we have given the x values that I made earlier. I then plot these on a graph of these points. Note that I used hold on to add more points with each invocation to plot. If I didn't do this, the figure would refresh itself and only remember the most recent call to plot. You want to put all of the points on here generated from all of the solution. For some neatness, I threw a grid in.
This is what I get:
Ok I was about to write my answer and I just saw that #rayryeng proposed a similar idea (Good job Ray!) but here it goes. The idea is also to use solve to get an expression for y, then convert the symbolic function to an anonymous function and then plot it. The code is general for any number of solutions you get from solve:
clear
clc
close all
syms x y
FunXY = y^2 == x^2*(x+3);
%//Use solve to solve for y.
Y = solve(FunXY,y);
%// Create anonymous functions, stored in a cell array.
NumSol = numel(Y); %// Number of solutions.
G = cell(1,NumSol);
for k = 1:NumSol
G{k} = matlabFunction(Y(k))
end
%// Plot the functions...
figure
hold on
for PlotCounter = 1:NumSol
fplot(G{PlotCounter},[-pi,pi])
end
hold off
The result is the following:
n = 1000;
[x y] = meshgrid(linspace(-3,3,n),linspace(-3,3,n));
z = nan(n,n);
z = (y .^ 2 <= x .^2 .* (x + 3) + .1);
z = z & (y .^ 2 >= x .^2 .* (x + 3) - .1);
contour(x,y,z)
It's probably not what you want, but I it's pretty cool!
Can someone share a technique using MATLAB to plot the surface f(x,y)=(21/4)x^2y over the region x^2 <= y <= 1?
Also, if anyone is aware of some tutorials or links that would help with this type of problem, could you please share them?
Thanks.
Here is another approach:
%%
close all
x=linspace(-1,1,40);
g1=x.^2;
g2=ones(1,40);
y=[];
n=20;
for k=0:n
y=[y;g1+(g2-g1)*k/n];
end
x=x(ones(1,n+1),:);
z=21/4*x.^2.*y;
meshz(x,y,z)
axis tight
xlabel('x-axis')
ylabel('y-axis')
view(136,42)
And the result:
And finally, you can map the region (-1,1)x(0,1) in the uv-plane into the region bounded by $y=x^2 and y=1 in the xy-plane with the parametrization:
f(u,v) = (u\sqrt{v},v)
Capture from: https://math.stackexchange.com/questions/823168/transform-rectangular-region-to-region-bounded-by-y-1-and-y-x2
This code produces the same image shown above:
close all
[u,v]=meshgrid(linspace(-1,1,40),linspace(0,1,20));
x=u.*sqrt(v);
y=v;
z=21/4*x.^2.*y;
meshz(x,y,z)
axis tight
xlabel('x-axis')
ylabel('y-axis')
view(136,42)
First off, let's look at your valid region of values. This is telling us that y >= x^2 and also y <= 1. This means that your y values need to be on the positive plane bounded by the parabola x^2 and they also must be less than or equal to 1. In other words, your y values must be bound within the area dictated from y = x^2 to y = 1. Pictorially, your y values are bounded within this shape:
As such, your x values must also be bound between -1 and 1. Therefore, your actual boundaries are: -1 <= x <= 1 and 0 <= y <= 1. However, this only locates our boundaries for x and y but it doesn't handle where the plot has valid values. We'll tackle that later.
Now that we have that established, you can use ezsurf to plot surface plots in MATLAB that are dictated by a 2D equation.
You call ezsurf like so:
ezsurf(FUN, [XMIN,XMAX,YMIN,YMAX]);
FUN is a function or a string that contains the equation you want, and XMIN,XMAX,YMIN,YMAX contain the lowest and highest x and y values you want to plot. Plotting without these values assumes a span from -2*pi to 2*pi in both dimensions. As such, let's create a new function that will handle when we have valid values, and when we don't. Use this code, and save it to a new file called myfun.m. Make sure you save this to your current Working Directory.
function z = myfun(x,y)
z = (21/4)*x.^2.*y;
z(~(x.^2 <= y & y <= 1)) = nan;
end
This will allow you to take a series of x and y values and output values that are dictated by the 2D equation that you have given us. Any values that don't satisfy the condition of x^2 <= y <= 1, you set them to NaN. ezsurf will not plot NaN values.
Now, call ezsurf like so:
ezsurf(#myfun, [-1,1,0,1]);
You thus get:
This will spawn a new figure for you, and there are some tools at the top that will allow you interact with your 3D plot. For instance, you can use the rotation tool that's at the top bar beside the hand to rotate your figure around and see what this looks like. Click on this tool, then left click your mouse and hold the left mouse button anywhere within the surface plot. You can drag around, changing the azimuth and the latitude to get the perspective that you want.
Edit: June 4th, 2014
Noting your comments, we can decrease the jagged edges by increasing the number of points in the plot. As such, you can append a final parameter to ezsurf which is N, the number of points to add in each dimension. Increasing the number of points will decrease the width in between each point and so the plot will look smoother. The default value of N is 60 in both dimensions. Let's try increasing the amount of points in each dimension to 100.
ezsurf(#myfun, [-1,1,0,1], 100);
Your plot will look like:
Hope this helps!
Try the following to make the required function, compute the values, and plot only the region that is desired:
% Make the function. You could put this in a file by itself, if you wanted.
f = #(x,y) (21/4)*x.^2.*y;
[X Y] = meshgrid(linspace(0,1));
Z = f(X,Y);
% compute the values we want to plot:
valsToPlot = (X.^2 <= Y) & (Y <= 1);
% remove the values that we don't want to plot:
X(~valsToPlot) = nan;
Y(~valsToPlot) = nan;
Z(~valsToPlot) = nan;
% And... plot.
figure(59382);
clf;
surf(X,Y,Z);
This should be an easy one. I am trying to find the coordinates of a point on a straight line. I am implementing in MATLAB. I know, the coordinates of the endpoints and the distance from one of the point.
I am using the following formula for calculating the coordinates (please note, I cannot use mid-point formula, as the distance can vary).
I am getting the wrong results when the slope is negative. Can you please suggest, what are the conditions, that needs to be considered for using this formula? Am not aware of any other formula as well.
That's too complicated solution for such a simple task. Use direct vector computations:
function P = point_on_line(A, B, AP)
D = B - A;
P = A + D / norm(D) * AP;
end
Call like this:
P = point_on_line([x1 y1], [x2 y2], len);
x = P(1);
y = P(2);
Ask if you need any clarifications.
Nothing wrong with your solution, but you need to take care of quadrant ambiguities when you take the arctangent to compute the angle θ.
There is a nice solution for that in most programming languages: atan2. Thus:
%// Your points (fill in any values)
A = [-10 0];
B = [-1 -1];
%// Use atan2!
th = atan2( B(2)-A(2) , B(1)-A(1) );
%// Distance from A to the point of interest
AP = sqrt( (B(2)-A(2))^2 + (B(1)-A(1))^2 ) / 2;
%// The point of interest
C = [
A(1) + AP*cos( th )
A(2) + AP*sin( th )];
%// Verify correctness with plots
figure(1), clf, hold on
line([A(1); B(1)], [A(2); B(2)])
plot(...
A(1), A(2), 'r.',...
B(1), B(2), 'b.',...
C(1), C(2), 'k.', 'markersize', 20)
In general, whenever and wherever you need to take an arctangent, use atan2 and not atan. The normal atan is only for cases where you don't know the individual components of the division y/x.
Note that your solution is not extensible to 3D, whereas the vector solutions proposed by the others here are. So in general I would indeed advise you to start working with vectors. Not only is it a lot simpler in many circumstances, it is also more versatile.
This is a wrong approach to the solution, as the solution is not unique. There are two points on you line with the same distance AP from the point A: one going left and another one going right.
The are infinite approaches to solve this, I prefer vector notation.
vector ab is a 2x1 matlab matrix:
ab = B-A
abN is the normalized vector
abN = ab/norm(ab)
stepping from A in the abN direction a distance of d (in your case AP) is:
A + abN*d
hope it helped.
Ohad