Solution Vector for Matrix from Upper Bandwith, transformed vector, and upper triangular result

Rebecca Berkawitz

2016 10 29

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更新时间：2021 8 20

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back_subst_banded.m

Modify the backward substitution algorithm to take advantage of the upper bandwidth of an upper triangular system. No arithmetic should be performed on any element known to be zero.

Inputs:

U The upper triangular result

d The transformed right-hand-side vector

ub The upper bandwidth, i.e., the number of stripes above the matrix diagonal that have non-zero elements.

Outputs:

x The solution vector

. . . .. So far this is what I have:

Function x= back_subst_banded(U, d, ub)

% U= upper triangular result from gaussbandfunc, d= transformed

%right hand vector for gaussbandfunc, ub= upper bandwith

format short

 p = (1:ub)';          % initialize the pivoting vector
s = max(abs(A'));     % compute the scale of each row
for k = 1:(ub-1)
    r = abs(A(p(k),k)/s(p(k)));
    kp = k;
    for i = (k+1):bb
      t = abs(A(p(i),k)/s(p(i))); 
      if t > r,  r = t;  
kp = i; 
      end
    end
    l = p(kp); 
 p(kp) = p(k); 
 p(k) = l;   % interchange p(kp) and p(k) 
    for i = (k+1):ub
      A(p(i),k) = A(p(i),k)/A(p(k),k);
      for j = (k+1):ub
        A(p(i),j) = A(p(i),j)-A(p(i),k)*A(p(k),j);
      end
    end
end
d = zeros(bb,1);    
d(1) = b(p(1));
for i = 2:ub
    d(i) = b(p(i));
    for j = 1:(i-1)
      d(i) = d(i)-A(p(i),j)*d(j);
    end
end
d