Anyone can help me to improve this code?

Dear friends; I have a code which is not suitable for big matrices or unlimited matrice. I would like to make it better because some times i have a very big matrices such as A = (100,100)

A = zeros(n,n) matrix.

R = [ 0 0 0 1 1] vector

c=2

if R(5,1) == 1

    A(5,5) = c
    A(5,4) = - c  
    A(4,5) = - c 
else 
    A(5,5) = 0    
    A(5,4) = 0  
    A(4,5) = 0
end
if R(4,1) == 1
    A(4,4) = c + A(5,5)
    A(3,4) = - c  
    A(4,3) = - c
else 
    A(4,4) = 0    
    A(3,4) = 0  
    A(4,3) = 0
end
.
.
.
ans
A =

     0     0     0     0     0
     0     0     0     0     0
     0     0     0    -2     0
     0     0    -2     4    -2
     0     0     0    -2     2

Thanks in advance, your help always appreciated

Answer 1

[Andrei Bobrov]

[EDIT]

n = 5;           % 
R = [0 0 0 1 1]';% eg
c = 2;           %

A1 = spdiags(-c*R,1,n,n);
A = A1 + A1' + spdiags(c*(R + [R(2:end);0]),0,n,n);
full(A)

OR

Rin(:,[1 3 2]) = c*[-[circshift(R,-1) R],R + [R(2:end);0]];
A = spdiags(Rin,-1:1,n,n);
full(A)

OR

Rin(:,[1 3 2]) = c*[-[circshift(R,-1) R],conv(R,[1;1],'same')];
A = spdiags(Rin,-1:1,n,n);
full(A)

[Brwa]

Thank you very much you you done very well. I really appreciate your help and effort. and thanks to David Sanchez too.

wish you all the best guys

[Andrei Bobrov]

EDITed

[Brwa]

still have the same mistake, the diagonal line should not be sum of

A(1,1) must not equal A(5,5)+A(4,4)+A(3,3)+A(2,2)+c

if R =[ 1 1 1 1 1]

A(5,5) must equa = c

A(4,4) must equal A(5,5) + c

A(3,3) must equal A(4,4) + c

A(2,2) must equal A(3,3) + c

A(1,1) must equal A(2,2) + c

And if

if R =[ 1 1 1 0 1]

A(5,5) must equa = c

A(4,4) must equal A(5,5) + 0

A(3,3) must equal 0 + c this means we only consider about original number of A(4,4) when R =[ 1 1 1 0 1] the fourth value is 0 so A(4,4) = 0 therefore, A(3,3) = 0 + c = c

A(2,2) must equal A(3,3) + c

A(1,1) must equal A(2,2) + c

Answer 2

[David Sanchez]

Ii this what you need?

for k=1:n-1
  if R(k,1) == 1
    A(k,k)= c + A(k+1,k+1);
  else 
    A(k,k) = A(k+1,k+1);
end

[Brwa]

Dear Aavid Sanchez thanks for your help but your code is also not working

[David Sanchez]

for k=1:(n-1)
  if R(k) == 1
    A(k,k)= c + A(k+1,k+1);
  else 
    A(k,k) = A(k+1,k+1);
end

Do you get an error message or aren't you getting what you want?

[David Sanchez]

I see, you have to start the for loop by the end.

c=2;
if R(5,1) == 1

      A(5,5) = c
      A(5,4) = - c  
      A(4,5) = - c 
  else 
      A(5,5) = 0    
      A(5,4) = 0  
      A(4,5) = 0
  end

for k=(n-1):-1:1
  if R(k) == 1
    A(k,k)= c + A(k+1,k+1);
  else 
    A(k,k) = A(k+1,k+1);
end

[Brwa]

oh, my god still not correct, but I still appreciate your help very very much.

the problem is for R= [ 1 1 1 0 1]

A(3,3) must = c not 2c because A(4,4) =0 so A(3,3) = c + 0 = c

Answer 3

[Brwa]

Thanks for your helps, Andrei Bobrov and David Sanchez. you guys are so nice. This is my code. it works good, to be honest i learnt alot from you two. but if someone can ake it shorter that will be great.

n=5;

A = zeros(n,n);

R = [1;1;0;1;1];

c = 2;

for i = n-1 : -1 : 2;

if R(i,1) == 1 && R(i+1,1) ==1

    A(i,i) = 2*c      ;
    A(i,i-1) = - c  ;
    A(i-1,i) = - c  ; 
elseif  R(i,1) == 1 && R(i+1,1) ~=1  
        A(i,i) = c      ;
        A(i,i-1) = - c  ;
        A(i-1,i) = - c  ;
elseif  R(i,1) ~= 1 && R(i+1,1) ==1
        A(i,i) = c      ;
        A(i,i-1) = 0  ;
        A(i-1,i) = 0  ;
elseif  R(i,1) ~= 1
        A(i,i) = 0      ;
        A(i,i-1) = 0    ;
        A(i-1,i) = 0    ; 
end
    if R(1,1)== 1 && R(2,1) == 1
       A(1,1) = 2*c ;
    elseif R(1,1) == 1 || R(2,1) == 1
       A(1,1) = c;
    else 
       A(1,1) = 0;
    end
    if R(n,1) ==1 && R(n-1,1) ==1
       A(n,n) = c ; 
       A(n-1,n) = -c;
       A(n,n-1) = -c;
       A(n-1,n-1) = 2*c;
    elseif  R(n,1) ==1
       A(n,n) = c ; 
       A(n-1,n) = -c;
       A(n,n-1) = -c;
       A(n-1,n-1) = c;
    end

end

Thank you again