How can I fix error matrix dimensions must agree?

This is suppose to produce any shape on a graph and shift it 4 units to the right and 2 units down with one input. I am getting told that there is error in matrix dimensions.

function [Row1,Row2] = problem3(nSides)

r = 1;

theta = pi./nSides .* (1:2:2*nSides-1);

Row1 = r .* cos(theta);

Row2 = r .* sin(theta);

fill(Row1+4,Row2-2,'k');

axis square; grid on;

end

[Geoff Hayes]

Hi Gimil,

At what line are you getting the error message, and what exactly is it? What is your input nSides - a scalar or vector or matrix?

Geoff

[Azzi Abdelmalek]

What is the size of nSides?

[Sara]

You should report the input value you are using and paste here the exact error you get from matlab to get help

[Gimil]

nSides is a scalar, when you can put any number and it will make a shape

[Gimil]

its for a quiz question on cody and states Error: Matrix dimensions must agree. If i just run it in matlab it works fine with no errors

[Gimil]

Here's the question: This question will cover the basics behind translating shapes in MATLAB. Write a function called "problem3" that accept ONE input. This single input will be formatted as follows:

Row 1: X coordinates Row 2: Y coordinates Your function should then shift the shape (any shape) that is created by these coordinates to the right by 4 units and down by 2 units. The function should then return this new shape back to the user. The output of the function should follow the same format as the input.

[Geoff Hayes]

There's the problem - the input is formatted with an x coordinate and a y coordinate i.e. [1;2] for coordinate (1,2) and so is not a scalar. This will raise the error when theta is calculated as pi./nSides is a 2x1 vector and (1:2:2*nSides-1) is a 1x4 vector. The calculation for theta will have to be clarified given that the input is a coordinate...

[Gimil]

So, how could that be done?

[Image Analyst]

My answer below works just fine. Try this:

[Row1,Row2] = problem3(5);

It produces two plots. The original one, and the shifted one. Isn't that what you want?

[Geoff Hayes]

Re-reading the problem definition a second time indicates that the input is not a scalar but a 2xN matrix of N coordinates. And that it is the shape defined by these coordinates that is shifted 4 units down and 2 to the right. The output of this function is another 2xN matrix with the shifted coordinates.

The body of the provided code seems more complex than what the solution to the problem requires.

Answer 1

[Image Analyst]

It's trivial. Just repeat the code after adding 4 and subtracting 2:

function [Row1,Row2] = problem3(nSides)
% First the original, unshifted plot.
subplot(1, 2, 1);
r = 1;
theta = pi./nSides .* (1:2:2*nSides-1);
Row1 = r .* cos(theta);
Row2 = r .* sin(theta);
fill(Row1+4, Row2-2, 'k');
grid on;
xlim([0 9]);
ylim([-5, 0]);
axis equal;

% Now shift 4 units to the right and 2 units down.
subplot(1, 2, 2);
Row1 = Row1 + 4;
Row2 = Row2 - 2;
fill(Row1+4, Row2-2, 'k');
grid on;
xlim([0 9]);
ylim([-5, 0]);
axis equal;

Get rid of the subplot() calls if you want it all on one axes. Put a "hold on" in there if you don't want the second plot to blow away the first one.

[Image Analyst]

Then nSides is a bad name for that input argument. When it's called nSides, you'd think it was the number of sides, and in fact your code does draw an n-sided polygon. So what should the inside of the function be? Should it just plot the coordinates:

x = nSides(:,1);
y = nSides(:, 2);
plot(x, y, 'bo-', 'LineWidth', 2, 'MarkerSize', 15);
hold on;
plot(x+4, y-2, 'bo-', 'LineWidth', 2, 'MarkerSize', 15);