3-bit Counter - Using solely (nested) for loops

Hi there,

I am trying to write a way to create an 3 column 8-bit RGB array to capture all 15 million+ possible colors - without storing the values, dynamically on an arduino (end code will be in C). I want to approach this problem by scaling down and using a language I am comfortable with - i.e. MATLAB.

So I want to understand a very rudimentary 3 bit counter, 2^3 = 8 rows, 3 columns - without using MATLAB's inbulit flipud, or dec2bin functions.

Only nested for loops.

What is the most logical way of approaching this problem? There will be a pattern of numbers always recurring in a sequence set by the column value as shown below - is it smart to utilize this sequence? I would really appreciate any pointers!

**I believe this problem (with regards to the RGB matrix) is documented, however, I want to exhaust all possible methods to do it by myself - before looking up resources

4 2 1 repitition (2^col)
______
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

[DGM]

It's been forever since I used Matlab to write code for AVR, but I was never comfortable with the ambiguity and overhead that came with it. Not knowing how it's going to try to compile, couldn't you just make a simple incrementing loop operating on a uint8 scalar? You're going to be occupying an entire 8 bits anyway, even if you only use the first 3 on the output. I don't see the reason to try to nest anything or treat the bits as separate numbers within the counter.

[Hans123]

Thanks for that perspective, could you elaborate a little on how I can have 3 seperate columns of varying 8bits without using loops? If I name the columns A,B and C - C will have values from 0 to 256 for the first 257 rows while both A and B have 0 - shown below.

Won't it be intuitive to have a nested for loop iterating each time spitting out a row?

A B C
0 0 0 {1st row}
0 0 1 {2nd row}
......
0 0 (256) {257th row}
0 1 0
0 1 1
....
0 1 (256)
0 2 0
0 2 1
.....
...
...
(256) (256) (256) {16,777,216th row}

[Rik]

I don't understand what you want to do exactly. Do you want code that produces the entire 3x16777216 array?

mod might be helpful in some way, depending on what you want to do exactly.

[DGM]

Okay, I misunderstood what you're trying to do. I thought you were trying to make a 3-bit counter by incrementing three unique 1-bit numbers. You're actually dealing with columns of 8-bit numbers. Sorry. I guess that makes more sense.

I guess it's be better to call it a 3-digit base-256 counter ... but they aren't really "digits"

Answer 1

[DGM]

I guess the simple way would be something like this:

% i'm going to assume your counter goes [R G B]
% i.e. B is the least significant term of the counter
for R=uint8(0:255)
    for G=uint8(0:255)
        for B=uint8(0:255)
            % do whatever you want with R,G,and B
            [R G B]
        end
    end
end

Although I have no idea how that will compile. It might be naive of me to try to coax it into only using 8 bits.

That would get you a sequence, but it's not really a counter that can be incremented externally. For that, you could make a function:

function [R G B]=testcounter(R,G,B,n)
    if B<(n-1)
        B=B+1;
    else
        B=0;
        if G<(n-1)
            G=G+1;
        else
            G=0;
            if R<(n-1)
                R=R+1;
            else
                R=0;
            end
        end
    end
end

Testing it like so:

B=0;
G=0;
R=0;

% counter base
% using a small number is good for demo
% set this to 256 for uint8, obviously
n=3; 

for k=1:(n^3+1) % show counter roll over
    [R G B]
    [R G B]=testcounter(R,G,B,n);
end

[Hans123]

this is an absolute genius method; I can't thank you enough.

I wish I can develop this thought process - took me 2 days of thinking to figure out I need help!

Cheers!