Contour Plot of Rosie Function

Hello,

I am trying to get the same plot as my professor showed in class. He said to use a for loop instead of meshgrid. I keep getting an error that X must not be a scalar. I'm not sure where I'm going wrong. Below is the plot I'm supposed to get followed by my code that is not working.

i=1; j=1;
x1=-1:.01:3;
x2=-1:.01:3;
for a=x1
    for b=x2
        f(i,j)=100*(a^2-b)^2+(1-a)^2;
        j=j+1;
    end
    j=1;
    i=i+1;
end 

clf;figure(1)
contour(a,b,f)

[Adam Danz]

The error means that 'a' and 'b' cannot be scalars (single values) in contour(a,b,f).

Look at the values of the inputs and it should be clear that a contour could not possibly be created based on those values.

You probably mean

contour(x1,x2,f)

but that doesn't reproduce the contour plot in the image.

[Brittany Burns]

I did mean to use x1 and x2 instead of a and b. After that, I'm stuck on getting the correct contour. I've been able to fiddle with it and hand the contours directly to contour and get what's below, but that still is not correct.

i=1; j=1;
conts=[-1 -.5 0 .5 2 3 3.5 3.6 3.7 3.8 3.9 3.98 4 4.02 4.1 4.2 4.5 5 10 ];
x1=-1:.01:3;
x2=-1:.01:3;
for a=x1
    for b=x2
        f(i,j)=100*(a^2-b)^2+(1-a)^2;
        j=j+1;
    end
    j=1;
    i=i+1;
end 

clf;figure(1)
contour(x1,x2,f,conts)

[Adam Danz]

I'm sure it's based on some dataset, function, or algorithm that our prof must have provided at some point.

Answer 1

[Star Strider]

Transpose ‘f’, since ‘x1’ and ‘x2’ appear to be the same (otherwise it would also be necessary to reverse their orders in the argument list for the contour call):

i=1; j=1;
% conts=[-1 -.5 0 .5 2 3 3.5 3.6 3.7 3.8 3.9 3.98 4 4.02 4.1 4.2 4.5 5 10 ];
conts = 10.^(-1:1:3);
x1=-1:.01:3;
x2=-1:.01:3;
for a=x1
    for b=x2
        f(i,j)=100*(a^2-b)^2+(1-a)^2;
        j=j+1;
    end
    j=1;
    i=i+1;
end 
% clf;
figure(1)
contour(x1,x2,f.',conts, 'ShowText','on', 'LineWidth',1.5)
axis('equal')

.