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function est = mynormest1 (L, U, P, Q)
%MYNORMEST1 estimate norm(A,1), using LU factorization (L*U = P*A*Q).
%
% Example:
% est = mynormest1 (L, U, P, Q)
% See also: testall
% Copyright 2006-2007, Timothy A. Davis.
% http://www.cise.ufl.edu/research/sparse
n = size (L,1) ;
est = 0 ;
S = zeros (n,1) ;
for k = 1:5
if k == 1
x = ones (n,1) / n ;
else
j = find (abs (x) == max (abs (x))) ;
j = j (1) ;
x = zeros (n,1) ;
x (j) = 1 ;
% fprintf ('eka: k %d j %d est %g\n', k, j, est) ;
end
% x=A\x, but use the existing P*A*Q=L*U factorization
x = Q * (U \ (L \ (P*x))) ;
est_old = est ;
est = sum (abs (x)) ;
unchanged = 1 ;
for i = 1:n
if (x (i) >= 0)
s = 1 ;
else
s = -1 ;
end
if (s ~= S (i))
S (i) = s ;
unchanged = 0 ;
end
end
if (any (S ~= signum (x)))
S' %#ok
signum(x)' %#ok
error ('Hey!') ;
end
if k > 1 & (est <= est_old | unchanged) %#ok
break ;
end
x = S ;
% x=A'\x, but use the existing P*A*Q=L*U factorization
x = P' * (L' \ (U' \ (Q'*x))) ;
if k > 1
jnew = find (abs (x) == max (abs (x))) ;
if (jnew == j)
break ;
end
end
end
for k = 1:n
x (k) = power (-1, k+1) * (1 + ((k-1)/(n-1))) ;
end
% x=A\x, but use the existing P*A*Q=L*U factorization
x = Q * (U \ (L \ (P*x))) ;
est_new = 2 * sum (abs (x)) / (3 * n) ;
if (est_new > est)
est = est_new ;
end
function s = signum (x)
%SIGNUM compute sign of x
s = ones (length (x),1) ;
s (find (x < 0)) = -1 ; %#ok
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