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 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