# Is the halting problem solvable for NPDAs?

After the total silence in response to my last question, I am rethinking my assumptions. DPDAs are, of course, solvable, and I believe that their loops can be found in the manner I described in my prior question:

1. arrive at the same state as you were in previously
2. with the same top symbol as you had last time
3. without consuming anything on the stack, and
4. without consuming any input.

But is my last question actually unanswerable because we cannot, in fact, determine whether a NPDA will halt? Is the halting problem even solvable for NPDAs?