Draw a surface using Gaston with a nonrectangular domain

New to Julia
,
1

Dear all,
I have the code:

using Gaston

X = LinRange(-1,1, 30) # 50
Y = LinRange(-1,1, 30)


x = [u for u in X, v in Y]
y = [v for u in X, v in Y]
z = [u^2+v^2  for u in X, v in Y]

#
surf(x, y, z, w = :l, lc = :turquoise,
       Axes(xlabel = :x, ylabel = :y, zlabel = :z,zrange= "[*<0:1<*]",xrange= "[*<-1:1<*]",yrange= "[*<-1:1<*]",
           #view= "60, 45, 1, 1",
            xtics = """("-0.5" -0.5, "0.0" 0, "+0.5" 0.5) offset .25,-0.5""",
            ytics = """("-0.5" -0.5, "0.0" 0, "+0.5" 0.5) offset -1.25,-0.5""",
            view = "75, 30, 0.7, 2.1",
            size = "1.1,1.",
            xyplane = "at 0",
            #pm3d = "depthorder",
            hidden3d = :on,
            key = :off))

Supose that I want to draw this surface in a nonrectangular domain, eg, for all u \in [-1,1] and 0 \leq v \leq 1-u. How can I adapt the code above?

2

I am not sure I understand the second inequality, since u is a range, not a constant. However, I think this example may help you:

Gnuplot does not need a rectangular domain to plot; all it needs is a list of coordinates. Let’s say we want to plot a circle; we can do it as follows:

Θ = LinRange(0, 2π, 100);
x1 = [cos(θ) for θ in Θ];
y1 = [sin(θ) for θ in Θ];
z1 = [1 for θ in Θ];
surf(x1, y1, z1)

This plot results:
image
Let’s add an oval:

x2 = [2cos(θ)+1 for θ in Θ];
y2 = [sin(θ)+1 for θ in Θ];
z2 = [2 for θ in Θ];
surf!(x2, y2, z2)

with the result:
image

Both plots have clearly non-rectangular domains. It’s often more convenient to think about sets of arbitrary (x,y,z) coordinates rather than a 2-D function evaluated on a matrix.

If you want a surface, then you can put the sets of coordinates in the rows of a matrix. Gnuplot will then draw lines between the corresponding sets of points:

surf([x1';x2'], [y1';y2'], [z1';z2'])

results in:
image
which is, again, a plot on a non-rectangular domain.

1 Like
3

Thank you so much for your response.
However, you don’t underestood my question.

Lets to think in a domain in \mathbb{R}^2 where 0 \leq x \leq 1 and 0 \leq y \leq x. Do you agree with me that we don’t have an rectangle? The region is a triangle, whose vertex are the points (0,0), (1,0) and (1,1).

My intention is plot the graph of a given function z=f(x,y) where (x,y) are in this triangle (the extremes of this region are not constants). Given a x in the rangle [0,1], the y-value varies in the range [0,x].

We usually plot a surface when the extremes are constants. But, in this case, how can we proceed?

What I need to adapt in my code to plot this suface?
Thank you.

4

Got it. I think the easiest way is to use the fact that gnuplot ignores NaNs when plotting.

f(x,y) = y < x ? cos(8x) + sin(8y) : NaN
x = LinRange(0, 1, 100);
y = LinRange(0, 1, 100);
surf(x, y, f, w = :pm3d)

produces this plot:
image
with a triangular domain.

The downside of this approach is that you have to redefine (or wrap) the function to return NaN for values outside the range you’re interested in. Pinging @lazarusA; he might know a better way.

7

The initial code is the following:

using Gaston

f(x,y) = y < x && y > x^2 ? x^2+y^2 : NaN
x = LinRange(-1, 1, 100);
y = LinRange(-1, 1, 100);

surf!(x, y, f,w = :pm3d)

But I want to plot without use the structure if-else. Is there another way to do this? Thank you!

8

I can’t think of any. Any particular reason why you want to avoid it?

9

If you have a little points, the is a problem; see below:

image
image696×576 48 KB

a little strange…

10

The region is a triangle,

Then I would do a Mesh, unfortunately today is already late. Maybe tomorrow I can take a look at it again.

11

Thank you so much!
Can be tomorrow.

12

Is the problem that the calculation is inefficient because most points are ignored? Or is it something else?

13

A more number of points, the calculation is ineficient.
To produce the figure below, I have used 1000 points on x and 1000 points in y.
I have a solution. But, it isn’t the better way…

image