Keean Schupke k.schupke at imperial.ac.uk
Thu Sep 1 10:17:39 EDT 2005

```Thanks for that, altough I have completely rewritten it! Here's the new
implementation:

unify :: Subst -> (Term,Term) -> [Subst]
unify sigma (s,t) = let
s' = if isVar s then subst s sigma else s
t' = if isVar t then subst t sigma else t
in if isVar s' && s'==t' then [sigma] else if isFunc s' &&
isFunc t'
then if fname s' == fname t' && arity s' == arity t'
then unify' sigma (terms s') (terms t')
else []
else if not (isVar s)
then unify sigma (t',s')
else [s' ~> t' : sigma]

unify' :: Subst -> [Term] -> [Term] -> [Subst]
unify' s (t0:ts) (u0:us) = case unify s (t0,u0) of
s@(_:_) -> unify' (concat s) ts us
_ -> []
unify' s [] [] = [s]
unify' _ _ _ = []

Once again, thoughts or improvements greatly appreciated...

Regards,
Keean.

Fergus Henderson wrote:

>You should delete the line above.  It's not needed and could cause serious
>efficiency problems.  With that line present, unifying two lists
>of length N which differ only in the last element would take time
>proportional to N squared, but without it, the time should be linear in N.
>
>
>
>>   unify s (Var x,t) = [(x,t):s] -- no occurs check
>>   unify s (t,Var x) = [(x,t):s] -- no occurs check
>>
>>
>
>These are not right; you need to look up the variable x
>in the substitution s, and if it is already bound, then
>you need to unify what it is bound to with the term t.
>
>
>

```