Pre-alocating an array, storing data in them and dynamically using them

Hello, I have this code shown below. matrices A, x and C_N are mainly used for data storage which I can acess later on for further calaculations. My problem is this, when t>R, then the code picks the value A(i-1) which is basically the previous value which is then used for further calculations down the code. This only works when the size of the vectors A, x and C_N is equal to 4,1; in this specfic case. That is, when A=zeros(4,1). I realized that this wokrs because when i=4, the condition t>R is satistfied. However, this brings a problem because I dont know in subsequent values of R, at what value of i will the condition t>R be satisfied. For example when the sizes of the mentioned vectors are (10,1) as shown with R=3.303, the code returns matrix Km=[] an empty matrix which implies that k0=0 which is not true, ko should be exactly one. Since A(4-1)=2.331 which whe operated on returns value of 1. How can I solve this? Note that the sizes of A, x and C-N can only be approximated I can know the exact size that will be required. Thank you.

A=zeros(10,1);

x=zeros(10,1);

C_N=zeros(10,1);

R=3.303;

r=1;

x(1)=0;%2*r*sind(acosd((R-r)/(2*r*(fix(R)-1))));

A(1)=R;

C_N(1)=fix(R/r);

for i=2:numel(A)

x(i)=2*r*sind(acosd((A(i-1)-r)/(2*r*(fix(A(i-1)/r)-1))));

A(i)=R*sind(acosd(sum(x)/R));

C_N(i)=2*fix(A(i)/r);

t=sum(x);

if t>R

A(i)=0;

x(i)=0;

C_N(i)=0;

%tr=A(3)

k0=0.5*A(i-1);

k1=k0-fix(k0);

Km=[1:1:fix(0.5*A(i-1))]';

if k1==0

for j=1:numel(Km)

corr(j)=2*fix(2*(sqrt(R^2-(Km(j)*r+0.18)^2)-sum(x))/r);

if j==numel(Km) && corr(j)<1 && numel(Km)==2

corr(j)=fix(2*(sqrt(R^2-(0)^2)-sum(x))/r);

C_Ncorr=sum(corr);

C_N(i-1)=C_Ncorr;

elseif j==numel(Km)

C_Ncorr=sum(corr);

disp( C_Ncorr);

C_N(i-1)=C_Ncorr;

end

end

else

for j=1:numel(Km)

corr(j)=2*fix(2*(sqrt(R^2-(2*Km(j)*r+0.36)^2)-sum(x))/r);

if j==numel(Km)

corrlast=fix(2*(sqrt(R^2-(0)^2)-sum(x))/r);

C_Ncorr=sum(corr)+corrlast;

C_N(i-1)=C_Ncorr;

end

end

end

%C_N(i)=C_Ncorr;

end

% display(C_N(1:i));

end

Answer 1

[Stephen23]

I think the problem is that you're checking the condition t > R at a fixed position in your loop, but you don't know beforehand which iteration i will satisfy this condition when you change the vector sizes or the value of R.

The solution is to break out of the loop as soon as the condition t > R is met, rather than continuing to iterate through all elements. Here's a modified version of your code:

A=zeros(10,1);
x=zeros(10,1);
C_N=zeros(10,1);
R=3.303;
r=1;
x(1)=0;%2*r*sind(acosd((R-r)/(2*r*(fix(R)-1))));
A(1)=R;
C_N(1)=fix(R/r);

for i=2:numel(A)
    x(i)=2*r*sind(acosd((A(i-1)-r)/(2*r*(fix(A(i-1)/r)-1))));
    A(i)=R*sind(acosd(sum(x)/R));
    C_N(i)=2*fix(A(i)/r);
    t=sum(x);
    
    if t>R
        A(i)=0;
        x(i)=0;
        C_N(i)=0;
        
        % Use A(i-1) which contains the previous valid value
        k0=0.5*A(i-1);
        k1=k0-fix(k0);
        Km=[1:1:fix(0.5*A(i-1))]';
        
        if k1==0
            for j=1:numel(Km)
                corr(j)=2*fix(2*(sqrt(R^2-(Km(j)*r+0.18)^2)-sum(x))/r);
                if j==numel(Km) && corr(j)<1 && numel(Km)==2
                    corr(j)=fix(2*(sqrt(R^2-(0)^2)-sum(x))/r);
                    C_Ncorr=sum(corr);
                    C_N(i-1)=C_Ncorr;
                elseif j==numel(Km)
                    C_Ncorr=sum(corr);
                    disp(C_Ncorr);
                    C_N(i-1)=C_Ncorr;
                end
            end
        else
            for j=1:numel(Km)
                corr(j)=2*fix(2*(sqrt(R^2-(2*Km(j)*r+0.36)^2)-sum(x))/r);
                if j==numel(Km)
                    corrlast=fix(2*(sqrt(R^2-(0)^2)-sum(x))/r);
                    C_Ncorr=sum(corr)+corrlast;
                    C_N(i-1)=C_Ncorr;
                end
            end
        end
        
        % Break out of the main loop once condition is met
        break;
    end
    % display(C_N(1:i));
end

% Display results
fprintf('Final k0 value: %.3f\n', k0);

Final k0 value: 1.163

fprintf('Km matrix:\n');

Km matrix:

disp(Km);

     1

fprintf('Valid entries in A: %d\n', sum(A~=0));

Valid entries in A: 3

fprintf('A values:\n');

A values:

disp(A(A~=0));

    3.3030
    2.8698
    2.3260