How do I do Pyramid plot

So I did this code, it shows a plot that looks like a roof with the colors black and white.

Now I need help with a similar code but that shows a pyramid, not a 3D pyramid. I put an attachment for you guys to see what I'm looking for. Thanks :))

M=uint8(zeros(512,512));
[r,c]=size(M);

for j=1:c
    if j < 257
        M(:,j)=j-1;
    else
        M(:,j)=c-j;
    end
end
figure, imshow(M)

Answer 1

[Image Analyst]

What about using row? Your code only checks the column, so of course every row is the same. You need to consider the row. Have a loop over the rows and use the row to compute the true value. Here is a snippet to get you started:

M = zeros(512,512, 'uint8');
[rows, columns] = size(M);

for row = 1 : rows
	for col = 1 : columns
		% You need to fix the lines below to set the value properly.
		% Just think hard enough about it and you'll figure it out.
		if col < 257
			value = col-1;  % What about row.  It needs to be involved!
		else
			value = columns-col;
		end
		M(row, col) = value;
	end
end
hFig = figure;
imshow(M, []);
hFig.WindowState = 'maximized'

[Diogo Costa]

I can´t figure it out, I tried but i can´t achieve what it supposed to achieved

[Bjorn Gustavsson]

Take it step-by-step. How did you produce the figure in the image above? Did you manage to make one in the y-direction?

[Diogo Costa]

I produced this code to create the image above

M=uint8(zeros(512,512));

[r,c]=size(M);

for j=1:c

if j < 257

M(:,j)=j-1;

else

M(:,j)=c-j;

end

end

figure, imshow(M)

[Image Analyst]

Well, I see you totally ignored my advice. Why??? Where is your loop over rows, and how does the M value depend on the row? Please take my advice and you will have your homework problem solved. I assume you're not allowed to turn in someone else's solution as your own.

[Diogo Costa]

No, i didn't try that way, i'm working on that now. I understand i need a certain rows to be white but i can figure it out the numbers of the matrix to make a diagonal line :/

[Image Analyst]

What if you do that "if" block you did for col just after the row for loop and before the col for loop starts? And if you let M be a double to start with, instead of a uint8? Get a row value and a col value. The bottom part of the code would then look like:

		finalValue = colValue * rowValue;
		M(row, col) = finalValue;
	end
end
hFig = figure;
% Now convert the double M to uint8
M = uint8(255 * mat2gray(M));
imshow(M, []);
impixelinfo
hFig.WindowState = 'maximized'

[Diogo Costa]

I add this

if row < 257

valor = row-1;

else

valor = rows - row;

end

and did this to the image

[Diogo Costa]

I'm closer i know

clear all; close all; clc;

M = zeros(512,512);
[rows, colunas] = size(M);
for row = 1 : rows
	for col = 1 : colunas
		 
		if col < 257
			colvalor = col-1;  
		else
			colvalor = colunas-col;
        end
        if row < 257
            rowvalor = row-1;
        else
            rowvalor = rows - row;
        end
            finalvalor = colvalor * rowvalor;
            M(row, col) = finalvalor;
	end
end
hFig = figure;
M = uint8(255 * mat2gray(M));
imshow(M, []);
impixelinfo
hFig.WindowState = 'maximized'

[Diogo Costa]

i need that to be a cross X not a plus sign + :(

[Diogo Costa]

I'm portuguese btw, sry for the language on the code

[Image Analyst]

Right, but I'm sure you'll figure it out - you're a smart engineer.

[Diogo Costa]

I managed to do this, any final tip ? I'm not that good programming :((

clear all; close all; clc;

M = zeros(512,512);
[rows, colunas] = size(M);

for row = 1 : rows
	for col = 1 : colunas
		 
		if col < 257
			valorcol = col - 1;  
		else
			valorcol = colunas - col;
        end
        if row < 257
            rowvalor = row - 1;
        else
            rowvalor = rows - row;
        end
            valorfinal = valorcol + rowvalor;
            M(row, col) = valorfinal;
	end
end

hFig = figure;
M = uint8(255 * mat2gray(M));
imshow(M, []);
impixelinfo
hFig.WindowState = 'maximized';

[Image Analyst]

You almost had it but you need to use min() instead of plus when you compute valorfinal:

M = zeros(512,512);
[rows, colunas] = size(M);

for row = 1 : rows
	% Get the row value.
	if row < 257
		rowvalor = row - 1;
	else
		rowvalor = rows - row;
	end
	for col = 1 : colunas
		% Get the column value.
		if col < 257
			valorcol = col - 1;
		else
			valorcol = colunas - col;
		end
		% Make M the minimum.
		valorfinal = min(valorcol, rowvalor);
		M(row, col) = valorfinal;
	end
end

hFig = figure;
M = uint8(255 * mat2gray(M));
imshow(M, []);
impixelinfo
hFig.WindowState = 'maximized';

[Diogo Costa]

Thank you so much, you were so helpfull :))

[Bjorn Gustavsson]

Now that you have an iterative solution you might also take a look at this matrix-based version:

x = -21:21;
[x,y] = meshgrid(x,x);
imagesc(min(21-abs(x),21-abs(y)))
% or for a +-signed roof:
imagesc(min(21-abs(x+y),21-abs(x-y)))

At some stage you should start thinking about solving the problems with matrix-based operations too - that is one of the strengths of matlab.

Answer 2

[Bjorn Gustavsson]

Surely this is a homework task?

Regardless, after you've figured out loops and conditional statements, you should start to think about more matrix based operations.

For example if you want to assign one value to every element in a column of a matrix you can do that in one step:

idx_col2fill = 23;
M = ones(512);
M(:,idx_col2fill) = 0;

Then you should have a look at special matrices that can be used for a range of purposes. Check for example diag:

imagesc(diag(32))

and simple matrix-manipulation operations such as flipud, fliplr, and the transpose operators: ' and .'

HTH

[Diogo Costa]

Ye it is, i'm searching for some help :((

It need to seem like this:

but instead of that it needs to be a cross that looks like a pyramid seen from above

[Bjorn Gustavsson]

But now you're well on your way! Take pen and paper, figure out how you can combine this variation in the x-direction with a similar variation in the y-direction, and when you've at least figured out how to do the same shape in the y-direction you should take a good look at the min and max functions.

HTH