number of delayed goals

Hi there,

Is there a rationale for the fact that, according to delayed_goals/{1,2}, 
the total number of delayed goals can be less than the number
of delayed goals for one single variable?

Running Eclipse 5.4 on a Linux box, and having loaded lib(fd), I observe

 [eclipse 3]: X#=Y+Z,delayed_goals(D),length(D,DL).

  X = X{[-10000000..10000000]}
  Y = Y{[-10000000..10000000]}
  Z = Z{[-10000000..10000000]}
  D = [fd_eq([0, -1 * Z{[-10000000..10000000]}, -1 *
Y{[-10000000..10000000]}, 1 * X{[-10000000..10000000]}])]
  DL = 1
  Delayed goals:
        0 - Z{[-10000000..10000000]} - Y{[-10000000..10000000]} +
X{[-10000000..10000000]}#=0

and I'm fine with it. 

I don't understand the reason for the second element that is
returned by delayed_goals/2, however:
  [eclipse 4]: X#=Y+Z,delayed_goals(X,D),length(D,DL).

  Y = Y{[-10000000..10000000]}
  Z = Z{[-10000000..10000000]}
  X = X{[-10000000..10000000]}
  D = [fd_eq([0, 1 * X{[-10000000..10000000]}, -1 *
Z{[-10000000..10000000]}, -1 * Y{[-10000000..10000000]}]), fd_eq([0, 1 * X, -1 * Z, -1 * Y])]
  DL = 2
  Delayed goals:
        0 + X{[-10000000..10000000]} - Z{[-10000000..10000000]} -
Y{[-10000000..10000000]}#=0


I guess it does not affect the soundness of the constraint solver
(in notinstance:~=/2, for instance, identical constraints are also stored
more than once). If, as in my case, the task is to reconstruct goals 
from constraints in the store, this can be worked around, but 
seems like an inconvenience.

Is it possible to get just the first solution?

Thanks,
Alex


-- 
______________________________
Alexander Pretschner
Software&Systems Engineering
Institut fuer Informatik, TU Muenchen
ph. +49 89 28928325
fax +49 89 28925310
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