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imtql1.f
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1996-09-28
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subroutine imtql1(n,d,e,ierr)
c
integer i,j,l,m,n,ii,mml,ierr
double precision d(n),e(n)
double precision b,c,f,g,p,r,s,tst1,tst2,pythag
c
c this subroutine is a translation of the algol procedure imtql1,
c num. math. 12, 377-383(1968) by martin and wilkinson,
c as modified in num. math. 15, 450(1970) by dubrulle.
c handbook for auto. comp., vol.ii-linear algebra, 241-248(1971).
c
c this subroutine finds the eigenvalues of a symmetric
c tridiagonal matrix by the implicit ql method.
c
c on input
c
c n is the order of the matrix.
c
c d contains the diagonal elements of the input matrix.
c
c e contains the subdiagonal elements of the input matrix
c in its last n-1 positions. e(1) is arbitrary.
c
c on output
c
c d contains the eigenvalues in ascending order. if an
c error exit is made, the eigenvalues are correct and
c ordered for indices 1,2,...ierr-1, but may not be
c the smallest eigenvalues.
c
c e has been destroyed.
c
c ierr is set to
c zero for normal return,
c j if the j-th eigenvalue has not been
c determined after 30 iterations.
c
c calls pythag for dsqrt(a*a + b*b) .
c
c questions and comments should be directed to burton s. garbow,
c mathematics and computer science div, argonne national laboratory
c
c this version dated august 1983.
c
c ------------------------------------------------------------------
c
ierr = 0
if (n .eq. 1) go to 1001
c
do 100 i = 2, n
100 e(i-1) = e(i)
c
e(n) = 0.0d0
c
do 290 l = 1, n
j = 0
c .......... look for small sub-diagonal element ..........
105 do 110 m = l, n
if (m .eq. n) go to 120
tst1 = dabs(d(m)) + dabs(d(m+1))
tst2 = tst1 + dabs(e(m))
if (tst2 .eq. tst1) go to 120
110 continue
c
120 p = d(l)
if (m .eq. l) go to 215
if (j .eq. 30) go to 1000
j = j + 1
c .......... form shift ..........
g = (d(l+1) - p) / (2.0d0 * e(l))
r = pythag(g,1.0d0)
g = d(m) - p + e(l) / (g + dsign(r,g))
s = 1.0d0
c = 1.0d0
p = 0.0d0
mml = m - l
c .......... for i=m-1 step -1 until l do -- ..........
do 200 ii = 1, mml
i = m - ii
f = s * e(i)
b = c * e(i)
r = pythag(f,g)
e(i+1) = r
if (r .eq. 0.0d0) go to 210
s = f / r
c = g / r
g = d(i+1) - p
r = (d(i) - g) * s + 2.0d0 * c * b
p = s * r
d(i+1) = g + p
g = c * r - b
200 continue
c
d(l) = d(l) - p
e(l) = g
e(m) = 0.0d0
go to 105
c .......... recover from underflow ..........
210 d(i+1) = d(i+1) - p
e(m) = 0.0d0
go to 105
c .......... order eigenvalues ..........
215 if (l .eq. 1) go to 250
c .......... for i=l step -1 until 2 do -- ..........
do 230 ii = 2, l
i = l + 2 - ii
if (p .ge. d(i-1)) go to 270
d(i) = d(i-1)
230 continue
c
250 i = 1
270 d(i) = p
290 continue
c
go to 1001
c .......... set error -- no convergence to an
c eigenvalue after 30 iterations ..........
1000 ierr = l
1001 return
end