🚀 go-pugleaf

If you have a program with a set of variables that enter a loop at time
t, and continue to loop in a cycle of length l, then the method outlined
will run the program in twice the memory at one-third speed minus
housekeeping, and detect the loop between t and t + l, i.e., during its
first cycle.

That's all the practical solution to the halting problem does; interpret
it as you wish. It's practical because the slowdown and memory waste are
practical during debugging, and because t and l are relatively small in
most practical cases. It's a solution to the halting problem because in
the worst case it'll still detect the loop eventually, though I don't
think these theoretical statements matter for the real world.

In article <GESSEL.90Nov14165831@carthage.cs.swarthmore.edu> gessel@carthage.cs.swarthmore.edu (Daniel Mark Gessel) writes:
>    > ]Here are two articles about the practical solution to the halting
>    > ]problem...
> The Halting Problem is _not_ the cycle problem. The Computational Model is
> a Turing Machine, not a computer with finite memory. 

See, this is what I mean about computer scientists. The computational
model I work with is realistic---it's just my mental picture of what can
be coded on lots of real machines. As I said before, it's finite. It has
very, very little to do with a Turing machine.

---Dan