optimization question

[Matt Gerrans]

> and call this function everywhere instead of doing direct comparisons.
> Then the optimization becomes trivial:
>
>         def eqsub(s, i, j, t):
>                 return (len(t) == j-i) and s[i:j] == t

I think this might add some additional speed (depending on the likelyhood of
matches), since indexing a single character doesn't do a slice (based on
what dis shows):

    def eqsub(s, i, j, t):
           return (len(t) == j-i) and (s[i]==t[0]) and (s[i:j] == t)

- Matt