Need help with HTBasic code

[Anonymous]

I have this code:

X=C+1! STARTS MATRIX ANALYSIS
Y=C+2
ALLOCATE Ya(Y,Y),Za(Y,Y)
MAT Za=(0)
REDIM Qa(Y)
MAT Qa=(0)
REDIM Xa(Y)
FOR I=1 TO No
L=D(I*2-1)
R=D(I*2)
Qa(L)=Qa(L)+Obs(1,I)
Qa(R)=Qa(R)-Obs(1,I)
Za(L,L)=Za(L,L)+1
Za(R,R)=Za(R,R)+1
Za(L,R)=Za(L,R)-1
Za(R,L)=Za(R,L)-1
Za(X,L)=Za(X,L)+1
Za(L,X)=Za(X,L)
Za(X,R)=Za(X,R)-1
Za(R,X)=Za(X,R)
Za(X,X)=Za(X,X)+1
Qa(X)=Qa(X)+Obs(1,I)
NEXT I
FOR I=1 TO C
Za(Y,I)=Cel(4,I)
Za(I,Y)=Cel(4,I)
NEXT I
MAT Ya=INV(Za)
MAT Xa=Ya*Qa
DEALLOCATE Ya(*),Za(*)
W=0
FOR I=1 TO No
L=D(I*2-1)
R=D(I*2)
Wa(I)=Obs(1,I)-(Xa(L)-Xa(R)+Xa(X))
W=W+Wa(I)^2
NEXT I
Stddev=SQR(ABS(W/(No-C)))
Ab_comp=Xa(C+1)

where C = 4, and No = 6

can anyone help, to disassemble or understand. As I understand it, the matrix is compiled here

[ZeRider]

Hi,
Advice : Add some PRINT in your code to see the different data values at each step
You will see if data are extracted from array well defined in your code : For example, L=D(I*2-1) and R=D(I*2) are always 0 (?)
or if array dimensions are compatible with your loop …

[Anonymous]

yes, i have 0 for L and R

[Anonymous]

and add this code to the begining

C=4
RESTORE C0
IF C=3 THEN RESTORE C3
IF C=4 THEN RESTORE C4
C0: DATA 0
C3: DATA 6,1,2,1,3,2,3,2,1,3,1,3,2
C4: DATA 12,1,2,1,3,2,3,2,4,3,4,3,1,3,2,4,2,4,1,4,3,2,1,1,4
READ No
FOR I=1 TO No
READ D(2*I-1),D(2*I)
NEXT I

[Author]

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