Converting FORTRAN code for Finite Element Analysis - Dealing with a Go To Command

Hello everyone,

I'm working on a problem and trying to come to grips with finite element analysis. I'm working from "A Simple Introduction to Finite Element Electromagnetic Problems", MATTHEW N. 0. SADIKU, IEEE TRANSACTIONS ON EDUCATION, VOL. 32, NO. 2, MAY 198. In it some sample FORTRAN code is provided for one of the example problems. My attempt is then to translate this program into MATLAB however I'm struggling to deal with the Go To statements.

The FORTRAN program is:

EDIT: Updated Attempt At Code:

%Nodes:
N(1,:) = [0,0];N(2,:) = [0.2,0];N(3,:) = [0.4,0];N(4,:) = [0.6,0];N(5,:) = [0.8,0];N(6,:) = [1.0,0];
N(7,:) = [0,0.2];N(8,:) = [0.2,0.2];N(9,:) = [0.4,0.2];N(10,:) = [0.6,0.2];N(11,:) = [0.8,0.2];
N(12,:) = [0,0.4];N(13,:) = [0.2,0.4];N(14,:) = [0.4,0.4];N(15,:) = [0.6,0.4];N(16,:) = [0,0.6];
N(17,:) = [0.2,0.6];N(18,:) = [0.4,0.6];N(19,:) = [0,0.8];N(20,:) = [0.2,0.8];N(21,:) = [0,1];

X = N(:,1); Y = N(:,2);

%Triangles:
TRI = [1,2,7;2,8,7;2,3,8;3,9,8;3,4,9;4,10,9;4,5,10;5,11,10;5,6,11;7,8,12;...
    8,13,12;8,9,13;9,14,13;9,10,14;10,15,14;10,11,15;12,13,16;13,17,16;...
    13,14,17;14,18,17;14,15,18;16,17,19;17,20,19;17,18,20;19,20,21];
%Fixed (perscribed) Potentials [Vf,Node Number]
Vp = [0,1;0,2;0,3;0,4;0,5;50,6;100,11;100,15;100,18;100,20;50,21;0,19;0,16;0,12;0,7];
ND = size(N,1);   %Number of Nodes
NE = size(TRI,1); %Number of Elements
NP = size(Vp,1);  %Number of Fixed Nodes
NF = ND-NP;       %Number of Free Nodes

%Preallocations:
Cg = zeros(ND,ND);
B = zeros(ND,1);    %-[Cfp][Vf] rearranged into a matrix with 1's at p nodes

for I = 1:NE    %Over Each Element {I}
    nodes = TRI(I,:);
    Xe = X(nodes); Ye = Y(nodes);
    
    %Local Coefficient Matrix:
    
    P(1) = Ye(2) - Ye(3); Q(1) = Xe(3) - Xe(2);
    P(2) = Ye(3) - Ye(1); Q(2) = Xe(1) - Xe(3);
    P(3) = Ye(1) - Ye(2); Q(3) = Xe(2) - Xe(1);
    
    
    A{I} = 0.5.*(P(2)*Q(3) - P(3)*Q(2));   %Saves Area of Specific Element
    
    
    for i = 1:3
        for j = 1:3
            C(i,j) = (1./(4*A{I})).*(P(i).*P(j) +Q(i).*Q(j));
            %Calculates Individual Element Coefficient Matrix
        end
    end
    clearvars i j
    %Tested & Working To Here
    
    %Global Coefficient Matrix:
    
    for j = 1:3 %Runs over local node numbers for rows:
        for L = 1:3
            IR = TRI(I,j);  %Row Index
            IC = TRI(I,L);  %Column Index
            
            for k = 1:NP
                
                if IR == Vp(k,2)    %If Row is at a fixed node:
                    Cg(IR,IR) = 1;
                    B(IR,1)=Vp(k,1);
                    
                end
            end
            for k = 1:NP
                if IC == Vp(k,2)    %If Column is at a fixed node:
                    B(IR,1) = B(IR,1) - C(j,L)*Vp(k,1);
                    %if IC & IR statements are not true do this
                    
                end
            end
            if IR ~= Vp(k,2)  && IC ~= Vp(k,2)
                Cg(IR,IC) = Cg(IR,IC) + C(j,L);
            end
        end
        
        
    end
end
Vsol = Cg  \ B;

The code above does not reproduce the expected potentials when the inverse is performed (V = Cg \ B). The resulting B matrix is however tested and correct. The issue comes therefore in the calculation of the global coefficient matrix, Cg.

For rows where IR == Vp(k,2) is true, the row

Cg(IR,:) = [0,...,Irth Position, 0,,,] e.g IR = 1, Cg(1,:) = [1,0,0,0] for a system with 4 nodes

And the program appears to set these 1's correctly but I'm also getting non-zero values where I don't expect them in the resulting matrix.

Thanks for the time

[Witold Waldman]

Is the completed MATLAB code for the sample FORTRAN program available?

[Ben Barrowes]

If you did have the compilable fortran code, I could use my f2matlab (fileexchange) then remgoto (homegrown tool) on your code (refactors all goto's).

Helping you is complicated by the fact that you don't use the same variable names and structure as the original code. But in general, "while", "break", and "continue" in matlab can reproduce the behavior of 100% of legal fortran goto's. PM me for more help. Otherwise, good luck!

Answer 1

[Walter Roberson]

The general pattern for needing to exit two layers of looping is:

need_to_exit_loop = false;
for first_variable = start_value : end_value
    for second_variable = start_second : end_second
        if condition_to_leave
            need_to_exit_loop = true;
            break;
        end
    end
    if need_to_exit_loop
        break;
    end
end

[ADSW121365]

I'm not sure if this helps me or not. My understanding maybe the issue. Think the break command maybe the way forward I'm just unsure of the specific application.

for I = 1:NE    %Over Each Element {I}
%Calculate Local Coefficient Matrix First C = size([ND x ND])

    %Global Coefficient Matrix:
    
    for j = 1:3 %Runs over local node numbers for rows:
        IR = TRI(I,j);
        
        for k = 1:NP
            if IR == Vp(k,2)
                %Do Calculations: 
                Cg(IR,IR) = 1;
                B(IR,1) = Vp(k,1);
                %At this point should break such j loop is closed I THINK
                %k loop should run to completion such Cg, B are assigned for all IR
            end
        end
        
        %if above isn't met for rows begin next loop for columns 
        % (i.e this only runs when IR =~ Vk(2,k))
        for L = 1:3
            IC = TRI(I,L);
            for k = 1:NP
                if IC == Vp(k,2)
                    B(IR,1) = B(IR,1) - C(j,L).*Vp(k,1);
                    %Again should break to close j loop 
                    %k loop should run to completion such B is assigned for all IC
                end
            end
        
        
        %if IC & IR statements are not true do this
        Cg(IR,IC) = Cg(IR,IC) + C(j,L);
        end
    end
end

[ADSW121365]

The more I think about it, the more I think it wants me to skip to the second from final end ( the one closing the j loop) if either condition is met, after it does the calculation but I'm honestly not certain as FORTRAN is a language I have zero experience in.