Index exceeds the number of array elements (1)

%When I try to perform a row interchange for 'b' vector It tells me it exceeds the no of array elements. This is an attempt to perform partial pivoting for matrix A and vector b separately.

Thanks

A = [1 2 3;4 2 -1; 3 2 3];
b = [1; 3; 2];
k = 1;
A = pivoting(A,k)
function [A, b, flag] = pivoting(A, b, k)
n = size(A,1);
flag = 0; 
for k=1:n-1
    [big, i]=max(abs(A(k:n,k)))
    ipr=i+k-1
    if ipr~=k
        A([k,ipr],:)=A([ipr,k],:)
        b([k,ipr])=b([ipr,k])
    end
end  
end

Answer 1

[Kojiro Saito]

It's because you're trying to manipulate b in line,

b([k,ipr])=b([ipr,k])

but, you're inputting as follows.

k = 1;
A = pivoting(A,k)

so, in function pivoting, b (second input variable) is scalar (1).

You just need to input three variables.

% A = pivoting(A,k)
A = pivoting(A,b,k)

[Jonas Freiheit]

That worked for me but, Is there another way to solve this program successfully without changing the pivoting function and only the code in the for loop?

Thanks for the help Kojiro!

[Kojiro Saito]

It didn't mean changing pivoting function. You're calling pivoting with two inputs (A and k),

A = [1 2 3;4 2 -1; 3 2 3];
b = [1; 3; 2];
k = 1;
A = pivoting(A,k)

but I suggested that you would call pivoting with three inputs (A, b and k)

A = [1 2 3;4 2 -1; 3 2 3];
b = [1; 3; 2];
k = 1;
A = pivoting(A,b,k)

[Jonas Freiheit]

Sorry what I meant was is it possible to do it with only calling pivoting with two inputs (A and k)?

[Kojiro Saito]

I see. Actually, k is initialized in pivoting function, so you don't need to input it.

The following would work.

A = [1 2 3;4 2 -1; 3 2 3];
b = [1; 3; 2];
%k = 1;
A = pivoting(A)

function [A, flag] = pivoting(A)
n = size(A,1);
flag = 0; 
for k=1:n-1
    [big, i]=max(abs(A(k:n,k)))
    ipr=i+k-1
    if ipr~=k
        A([k,ipr],:)=A([ipr,k],:)
        %b([k,ipr])=b([ipr,k])
    end
end  
end

[Jonas Freiheit]

Sorry sorry, I mean only by changing this code:

for k=1:n-1

[big, i]=max(abs(A(k:n,k)))

ipr=i+k-1

if ipr~=k

A([k,ipr],:)=A([ipr,k],:)

b([k,ipr])=b([ipr,k])

end

end

end

Thanks, don't mean to annoy you, I'm trying to learn how to change the source code without changing the pivoting function.

[Kojiro Saito]

The following is what you want to do?

A = [1 2 3;4 2 -1; 3 2 3];
b = [1; 3; 2];
%k = 1;
n = size(A,1);
for k=1:n-1
    [big, i]=max(abs(A(k:n,k )))
    ipr=i+k-1
    if ipr~=k
        A([k,ipr],:)=A([ipr,k ],:)
        b([k,ipr])=b([ipr,k ])
    end
end

[Jonas Freiheit]

Yeah, including this code with only changing from the bold text code.

A = [1 2 3;4 2 -1; 3 2 3];

b = [1; 3; 2];

k = 1;

A = pivoting(A,k)

function [A, b, flag] = pivoting(A, b, k)

n = size(A,1);

flag = 0;

for k=1:n-1 %Here

[big, i]=max(abs(A(k:n,k)))

ipr=i+k-1

if ipr~=k

A([k,ipr],:)=A([ipr,k],:)

b([k,ipr])=b([ipr,k])

end

end

end %End

Thanks, Kojiro

[Kojiro Saito]

It's not possible only changing the bold text code because the funtion pivoting requires three inputs, so you need to modify input variables so that it allows two inputs.

[Jonas Freiheit]

Alright that makes sense,

Thanks for the help Kojiro!