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nnls.f
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double precision FUNCTION DIFF(X,Y)
c
c Function used in tests that depend on machine precision.
c
c The original version of this code was developed by
c Charles L. Lawson and Richard J. Hanson at Jet Propulsion Laboratory
c 1973 JUN 7, and published in the book
c "SOLVING LEAST SQUARES PROBLEMS", Prentice-HalL, 1974.
c Revised FEB 1995 to accompany reprinting of the book by SIAM.
C
double precision X, Y
DIFF=X-Y
RETURN
END
SUBROUTINE G1 (A,B,CTERM,STERM,SIG)
c
C COMPUTE ORTHOGONAL ROTATION MATRIX..
c
c The original version of this code was developed by
c Charles L. Lawson and Richard J. Hanson at Jet Propulsion Laboratory
c 1973 JUN 12, and published in the book
c "SOLVING LEAST SQUARES PROBLEMS", Prentice-HalL, 1974.
c Revised FEB 1995 to accompany reprinting of the book by SIAM.
C
C COMPUTE.. MATRIX (C, S) SO THAT (C, S)(A) = (SQRT(A**2+B**2))
C (-S,C) (-S,C)(B) ( 0 )
C COMPUTE SIG = SQRT(A**2+B**2)
C SIG IS COMPUTED LAST TO ALLOW FOR THE POSSIBILITY THAT
C SIG MAY BE IN THE SAME LOCATION AS A OR B .
C ------------------------------------------------------------------
double precision A, B, CTERM, ONE, SIG, STERM, XR, YR, ZERO
parameter(ONE = 1.0d0, ZERO = 0.0d0)
C ------------------------------------------------------------------
if (abs(A) .gt. abs(B)) then
XR=B/A
YR=sqrt(ONE+XR**2)
CTERM=sign(ONE/YR,A)
STERM=CTERM*XR
SIG=abs(A)*YR
RETURN
endif
if (B .ne. ZERO) then
XR=A/B
YR=sqrt(ONE+XR**2)
STERM=sign(ONE/YR,B)
CTERM=STERM*XR
SIG=abs(B)*YR
RETURN
endif
SIG=ZERO
CTERM=ZERO
STERM=ONE
RETURN
END
C SUBROUTINE H12 (MODE,LPIVOT,L1,M,U,IUE,UP,C,ICE,ICV,NCV)
C
C CONSTRUCTION AND/OR APPLICATION OF A SINGLE
C HOUSEHOLDER TRANSFORMATION.. Q = I + U*(U**T)/B
C
c The original version of this code was developed by
c Charles L. Lawson and Richard J. Hanson at Jet Propulsion Laboratory
c 1973 JUN 12, and published in the book
c "SOLVING LEAST SQUARES PROBLEMS", Prentice-HalL, 1974.
c Revised FEB 1995 to accompany reprinting of the book by SIAM.
C ------------------------------------------------------------------
c Subroutine Arguments
c
C MODE = 1 OR 2 Selects Algorithm H1 to construct and apply a
c Householder transformation, or Algorithm H2 to apply a
c previously constructed transformation.
C LPIVOT IS THE INDEX OF THE PIVOT ELEMENT.
C L1,M IF L1 .LE. M THE TRANSFORMATION WILL BE CONSTRUCTED TO
C ZERO ELEMENTS INDEXED FROM L1 THROUGH M. IF L1 GT. M
C THE SUBROUTINE DOES AN IDENTITY TRANSFORMATION.
C U(),IUE,UP On entry with MODE = 1, U() contains the pivot
c vector. IUE is the storage increment between elements.
c On exit when MODE = 1, U() and UP contain quantities
c defining the vector U of the Householder transformation.
c on entry with MODE = 2, U() and UP should contain
c quantities previously computed with MODE = 1. These will
c not be modified during the entry with MODE = 2.
C C() ON ENTRY with MODE = 1 or 2, C() CONTAINS A MATRIX WHICH
c WILL BE REGARDED AS A SET OF VECTORS TO WHICH THE
c HOUSEHOLDER TRANSFORMATION IS TO BE APPLIED.
c ON EXIT C() CONTAINS THE SET OF TRANSFORMED VECTORS.
C ICE STORAGE INCREMENT BETWEEN ELEMENTS OF VECTORS IN C().
C ICV STORAGE INCREMENT BETWEEN VECTORS IN C().
C NCV NUMBER OF VECTORS IN C() TO BE TRANSFORMED. IF NCV .LE. 0
C NO OPERATIONS WILL BE DONE ON C().
C ------------------------------------------------------------------
SUBROUTINE H12 (MODE,LPIVOT,L1,M,U,IUE,UP,C,ICE,ICV,NCV)
C ------------------------------------------------------------------
integer I, I2, I3, I4, ICE, ICV, INCR, IUE, J
integer L1, LPIVOT, M, MODE, NCV
double precision B, C(*), CL, CLINV, ONE, SM
c double precision U(IUE,M)
double precision U(IUE,*)
double precision UP
parameter(ONE = 1.0d0)
C ------------------------------------------------------------------
IF (0.GE.LPIVOT.OR.LPIVOT.GE.L1.OR.L1.GT.M) RETURN
CL=abs(U(1,LPIVOT))
IF (MODE.EQ.2) GO TO 60
C ****** CONSTRUCT THE TRANSFORMATION. ******
DO 10 J=L1,M
10 CL=MAX(abs(U(1,J)),CL)
IF (CL) 130,130,20
20 CLINV=ONE/CL
SM=(U(1,LPIVOT)*CLINV)**2
DO 30 J=L1,M
30 SM=SM+(U(1,J)*CLINV)**2
CL=CL*SQRT(SM)
IF (U(1,LPIVOT)) 50,50,40
40 CL=-CL
50 UP=U(1,LPIVOT)-CL
U(1,LPIVOT)=CL
GO TO 70
C ****** APPLY THE TRANSFORMATION I+U*(U**T)/B TO C. ******
C
60 IF (CL) 130,130,70
70 IF (NCV.LE.0) RETURN
B= UP*U(1,LPIVOT)
C B MUST BE NONPOSITIVE HERE. IF B = 0., RETURN.
C
IF (B) 80,130,130
80 B=ONE/B
I2=1-ICV+ICE*(LPIVOT-1)
INCR=ICE*(L1-LPIVOT)
DO 120 J=1,NCV
I2=I2+ICV
I3=I2+INCR
I4=I3
SM=C(I2)*UP
DO 90 I=L1,M
SM=SM+C(I3)*U(1,I)
90 I3=I3+ICE
IF (SM) 100,120,100
100 SM=SM*B
C(I2)=C(I2)+SM*UP
DO 110 I=L1,M
C(I4)=C(I4)+SM*U(1,I)
110 I4=I4+ICE
120 CONTINUE
130 RETURN
END
C SUBROUTINE NNLS (A,MDA,M,N,B,X,RNORM,W,ZZ,INDEX,MODE)
C
C Algorithm NNLS: NONNEGATIVE LEAST SQUARES
C
c The original version of this code was developed by
c Charles L. Lawson and Richard J. Hanson at Jet Propulsion Laboratory
c 1973 JUN 15, and published in the book
c "SOLVING LEAST SQUARES PROBLEMS", Prentice-HalL, 1974.
c Revised FEB 1995 to accompany reprinting of the book by SIAM.
c
C GIVEN AN M BY N MATRIX, A, AND AN M-VECTOR, B, COMPUTE AN
C N-VECTOR, X, THAT SOLVES THE LEAST SQUARES PROBLEM
C
C A * X = B SUBJECT TO X .GE. 0
C ------------------------------------------------------------------
c Subroutine Arguments
c
C A(),MDA,M,N MDA IS THE FIRST DIMENSIONING PARAMETER FOR THE
C ARRAY, A(). ON ENTRY A() CONTAINS THE M BY N
C MATRIX, A. ON EXIT A() CONTAINS
C THE PRODUCT MATRIX, Q*A , WHERE Q IS AN
C M BY M ORTHOGONAL MATRIX GENERATED IMPLICITLY BY
C THIS SUBROUTINE.
C B() ON ENTRY B() CONTAINS THE M-VECTOR, B. ON EXIT B() CON-
C TAINS Q*B.
C X() ON ENTRY X() NEED NOT BE INITIALIZED. ON EXIT X() WILL
C CONTAIN THE SOLUTION VECTOR.
C RNORM ON EXIT RNORM CONTAINS THE EUCLIDEAN NORM OF THE
C RESIDUAL VECTOR.
C W() AN N-ARRAY OF WORKING SPACE. ON EXIT W() WILL CONTAIN
C THE DUAL SOLUTION VECTOR. W WILL SATISFY W(I) = 0.
C FOR ALL I IN SET P AND W(I) .LE. 0. FOR ALL I IN SET Z
C ZZ() AN M-ARRAY OF WORKING SPACE.
C INDEX() AN INTEGER WORKING ARRAY OF LENGTH AT LEAST N.
C ON EXIT THE CONTENTS OF THIS ARRAY DEFINE THE SETS
C P AND Z AS FOLLOWS..
C
C INDEX(1) THRU INDEX(NSETP) = SET P.
C INDEX(IZ1) THRU INDEX(IZ2) = SET Z.
C IZ1 = NSETP + 1 = NPP1
C IZ2 = N
C MODE THIS IS A SUCCESS-FAILURE FLAG WITH THE FOLLOWING
C MEANINGS.
C 1 THE SOLUTION HAS BEEN COMPUTED SUCCESSFULLY.
C 2 THE DIMENSIONS OF THE PROBLEM ARE BAD.
C EITHER M .LE. 0 OR N .LE. 0.
C 3 ITERATION COUNT EXCEEDED. MORE THAN 3*N ITERATIONS.
C
C ------------------------------------------------------------------
SUBROUTINE NNLS (A,MDA,M,N,B,X,RNORM,W,ZZ,INDEX,MODE)
C ------------------------------------------------------------------
integer I, II, IP, ITER, ITMAX, IZ, IZ1, IZ2, IZMAX, J, JJ, JZ, L
integer M, MDA, MODE,N, NPP1, NSETP, RTNKEY
c integer INDEX(N)
c double precision A(MDA,N), B(M), W(N), X(N), ZZ(M)
integer INDEX(*)
double precision A(MDA,*), B(*), W(*), X(*), ZZ(*)
double precision ALPHA, ASAVE, CC, DIFF, DUMMY, FACTOR, RNORM
double precision SM, SS, T, TEMP, TWO, UNORM, UP, WMAX
double precision ZERO, ZTEST
parameter(FACTOR = 0.01d0)
parameter(TWO = 2.0d0, ZERO = 0.0d0)
C ------------------------------------------------------------------
MODE=1
IF (M .le. 0 .or. N .le. 0) then
MODE=2
RETURN
endif
ITER=0
ITMAX=3*N
C
C INITIALIZE THE ARRAYS INDEX() AND X().
C
DO 20 I=1,N
X(I)=ZERO
20 INDEX(I)=I
C
IZ2=N
IZ1=1
NSETP=0
NPP1=1
C ****** MAIN LOOP BEGINS HERE ******
30 CONTINUE
C QUIT IF ALL COEFFICIENTS ARE ALREADY IN THE SOLUTION.
C OR IF M COLS OF A HAVE BEEN TRIANGULARIZED.
C
IF (IZ1 .GT.IZ2.OR.NSETP.GE.M) GO TO 350
C
C COMPUTE COMPONENTS OF THE DUAL (NEGATIVE GRADIENT) VECTOR W().
C
DO 50 IZ=IZ1,IZ2
J=INDEX(IZ)
SM=ZERO
DO 40 L=NPP1,M
40 SM=SM+A(L,J)*B(L)
W(J)=SM
50 continue
C FIND LARGEST POSITIVE W(J).
60 continue
WMAX=ZERO
DO 70 IZ=IZ1,IZ2
J=INDEX(IZ)
IF (W(J) .gt. WMAX) then
WMAX=W(J)
IZMAX=IZ
endif
70 CONTINUE
C
C IF WMAX .LE. 0. GO TO TERMINATION.
C THIS INDICATES SATISFACTION OF THE KUHN-TUCKER CONDITIONS.
C
IF (WMAX .le. ZERO) go to 350
IZ=IZMAX
J=INDEX(IZ)
C
C THE SIGN OF W(J) IS OK FOR J TO BE MOVED TO SET P.
C BEGIN THE TRANSFORMATION AND CHECK NEW DIAGONAL ELEMENT TO AVOID
C NEAR LINEAR DEPENDENCE.
C
ASAVE=A(NPP1,J)
CALL H12 (1,NPP1,NPP1+1,M,A(1,J),1,UP,DUMMY,1,1,0)
UNORM=ZERO
IF (NSETP .ne. 0) then
DO 90 L=1,NSETP
90 UNORM=UNORM+A(L,J)**2
endif
UNORM=sqrt(UNORM)
IF (DIFF(UNORM+ABS(A(NPP1,J))*FACTOR,UNORM) .gt. ZERO) then
C
C COL J IS SUFFICIENTLY INDEPENDENT. COPY B INTO ZZ, UPDATE ZZ
C AND SOLVE FOR ZTEST ( = PROPOSED NEW VALUE FOR X(J) ).
C
DO 120 L=1,M
120 ZZ(L)=B(L)
CALL H12 (2,NPP1,NPP1+1,M,A(1,J),1,UP,ZZ,1,1,1)
ZTEST=ZZ(NPP1)/A(NPP1,J)
C
C SEE IF ZTEST IS POSITIVE
C
IF (ZTEST .gt. ZERO) go to 140
endif
C
C REJECT J AS A CANDIDATE TO BE MOVED FROM SET Z TO SET P.
C RESTORE A(NPP1,J), SET W(J)=0., AND LOOP BACK TO TEST DUAL
C COEFFS AGAIN.
C
A(NPP1,J)=ASAVE
W(J)=ZERO
GO TO 60
C
C THE INDEX J=INDEX(IZ) HAS BEEN SELECTED TO BE MOVED FROM
C SET Z TO SET P. UPDATE B, UPDATE INDICES, APPLY HOUSEHOLDER
C TRANSFORMATIONS TO COLS IN NEW SET Z, ZERO SUBDIAGONAL ELTS IN
C COL J, SET W(J)=0.
C
140 continue
DO 150 L=1,M
150 B(L)=ZZ(L)
C
INDEX(IZ)=INDEX(IZ1)
INDEX(IZ1)=J
IZ1=IZ1+1
NSETP=NPP1
NPP1=NPP1+1
C
IF (IZ1 .le. IZ2) then
DO 160 JZ=IZ1,IZ2
JJ=INDEX(JZ)
CALL H12 (2,NSETP,NPP1,M,A(1,J),1,UP,A(1,JJ),1,MDA,1)
160 continue
endif
C
IF (NSETP .ne. M) then
DO 180 L=NPP1,M
180 A(L,J)=ZERO
endif
C
W(J)=ZERO
C SOLVE THE TRIANGULAR SYSTEM.
C STORE THE SOLUTION TEMPORARILY IN ZZ().
RTNKEY = 1
GO TO 400
200 CONTINUE
C
C ****** SECONDARY LOOP BEGINS HERE ******
C
C ITERATION COUNTER.
C
210 continue
ITER=ITER+1
IF (ITER .gt. ITMAX) then
MODE=3
c write (*,'(/a)') ' NNLS quitting on iteration count.'
GO TO 350
endif
C
C SEE IF ALL NEW CONSTRAINED COEFFS ARE FEASIBLE.
C IF NOT COMPUTE ALPHA.
C
ALPHA=TWO
DO 240 IP=1,NSETP
L=INDEX(IP)
IF (ZZ(IP) .le. ZERO) then
T=-X(L)/(ZZ(IP)-X(L))
IF (ALPHA .gt. T) then
ALPHA=T
JJ=IP
endif
endif
240 CONTINUE
C
C IF ALL NEW CONSTRAINED COEFFS ARE FEASIBLE THEN ALPHA WILL
C STILL = 2. IF SO EXIT FROM SECONDARY LOOP TO MAIN LOOP.
C
IF (ALPHA.EQ.TWO) GO TO 330
C
C OTHERWISE USE ALPHA WHICH WILL BE BETWEEN 0. AND 1. TO
C INTERPOLATE BETWEEN THE OLD X AND THE NEW ZZ.
C
DO 250 IP=1,NSETP
L=INDEX(IP)
X(L)=X(L)+ALPHA*(ZZ(IP)-X(L))
250 continue
C
C MODIFY A AND B AND THE INDEX ARRAYS TO MOVE COEFFICIENT I
C FROM SET P TO SET Z.
C
I=INDEX(JJ)
260 continue
X(I)=ZERO
C
IF (JJ .ne. NSETP) then
JJ=JJ+1
DO 280 J=JJ,NSETP
II=INDEX(J)
INDEX(J-1)=II
CALL G1 (A(J-1,II),A(J,II),CC,SS,A(J-1,II))
A(J,II)=ZERO
DO 270 L=1,N
IF (L.NE.II) then
c
c Apply procedure G2 (CC,SS,A(J-1,L),A(J,L))
c
TEMP = A(J-1,L)
A(J-1,L) = CC*TEMP + SS*A(J,L)
A(J,L) =-SS*TEMP + CC*A(J,L)
endif
270 CONTINUE
c
c Apply procedure G2 (CC,SS,B(J-1),B(J))
c
TEMP = B(J-1)
B(J-1) = CC*TEMP + SS*B(J)
B(J) =-SS*TEMP + CC*B(J)
280 continue
endif
c
NPP1=NSETP
NSETP=NSETP-1
IZ1=IZ1-1
INDEX(IZ1)=I
C
C SEE IF THE REMAINING COEFFS IN SET P ARE FEASIBLE. THEY SHOULD
C BE BECAUSE OF THE WAY ALPHA WAS DETERMINED.
C IF ANY ARE INFEASIBLE IT IS DUE TO ROUND-OFF ERROR. ANY
C THAT ARE NONPOSITIVE WILL BE SET TO ZERO
C AND MOVED FROM SET P TO SET Z.
C
DO 300 JJ=1,NSETP
I=INDEX(JJ)
IF (X(I) .le. ZERO) go to 260
300 CONTINUE
C
C COPY B( ) INTO ZZ( ). THEN SOLVE AGAIN AND LOOP BACK.
C
DO 310 I=1,M
310 ZZ(I)=B(I)
RTNKEY = 2
GO TO 400
320 CONTINUE
GO TO 210
C ****** END OF SECONDARY LOOP ******
C
330 continue
DO 340 IP=1,NSETP
I=INDEX(IP)
340 X(I)=ZZ(IP)
C ALL NEW COEFFS ARE POSITIVE. LOOP BACK TO BEGINNING.
GO TO 30
C
C ****** END OF MAIN LOOP ******
C
C COME TO HERE FOR TERMINATION.
C COMPUTE THE NORM OF THE FINAL RESIDUAL VECTOR.
C
350 continue
SM=ZERO
IF (NPP1 .le. M) then
DO 360 I=NPP1,M
360 SM=SM+B(I)**2
else
DO 380 J=1,N
380 W(J)=ZERO
endif
RNORM=sqrt(SM)
RETURN
C
C THE FOLLOWING BLOCK OF CODE IS USED AS AN INTERNAL SUBROUTINE
C TO SOLVE THE TRIANGULAR SYSTEM, PUTTING THE SOLUTION IN ZZ().
C
400 continue
DO 430 L=1,NSETP
IP=NSETP+1-L
IF (L .ne. 1) then
DO 410 II=1,IP
ZZ(II)=ZZ(II)-A(II,JJ)*ZZ(IP+1)
410 continue
endif
JJ=INDEX(IP)
ZZ(IP)=ZZ(IP)/A(IP,JJ)
430 continue
go to (200, 320), RTNKEY
END