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CS 590 Midterm Practice Name: This is a collection of practice problems for the midterm. The midterm will b e very likely to contain questions very similar to these, or to those in the homework. On the test you will b e allowed to refer to a single 8.5x11 inch (or A4) sheet of notes|with anything on it which you think might b e useful. You will also b e allowed to use as much blank scratch pap er as you need. Note: answers with pro ofs, or at least arguments for your answer, will receive more credit than those without.
CS 590 Theory of Computation. Test Practice
12 March 1993
Problem Practice-1: Let C = fe : e converges on all of the form hx; y i when x and y are relatively primeg. Is C recursive?
Answer: No. It is an index set, so it is non-recursiveness by Rice's theorem.
Problem Practice-2: Let A = fx : 9 y ( y (x) # in x steps g. Is A recursively enumerable?
Answer: Yes. The predicate \ y (x) # in x steps" is decidable. So, A is recursively enumerable by the normal form theorem.
Problem Practice-3: Supp ose the function f is computible. What can you say ab out the set Gf = fhx; y i : f (x) = y g?
Answer: It is r.e., since hx; y i 2 Gf i 9 y (f (x) = y ) and the predicate in parentheses is partially decidable.
Problem Practice-4: Let R(x; y ) b e true i x = y. Is R(x; y ) decidable? decidable?
(Hint: how do es R(x; y ) compare to the problem of recognizing total, constant functions?)
Answer: No. Let c b e such that c is equal to zero everywhere. Supp ose f (x; y ) is the characteristic function for R, and consider the function g (x) = f (x; c). This would b e the caracteristic function for the predicate \ x = 0 everywhere", which is undecidable. So, f must b e undecidable as well. Note, this argument shows that the problem of deciding if a function is equal to zero everywhere reduces to the problem of deciding the predicate R.
Problem Practice-5: Is there a total, computible function k such that if e (x) is the charac- teristic function for the predicate R(x) then k (e) (x) is the characteristic for the predicate \x is prime and R(x)"?
Answer: Yes. Let g (e; x) = pr (x) U (e; x). This is a computible function with parameters e and x, by the universal function theorem. So, by the smn theorem, there is a total, computible function k such that k (e) (x) is the characteristic function in question.
Problem Practice-6: Let K 1 = fe : We 6 = ;g. Prove that K 1 is undecidable.
Answer: K 1 is a non-trivial index set, so the result follows by Rice's theorem. Alternatively, one can show that K m K 1 : Let m b e total and computible such that
m(x) =
1 if x 2 Wx " otherwise
Then
x 2 K i m(x) (y ) = 1 everywhere i Wm(x) 6 = ; i m(x) 2 K 1
Problem Practice-12: Supp ose f is a partial, computible function. Under what conditions is f ^1 computible?
Answer: If f is one to one, then f ^1 = y (f (x) = y ). If it is not, then its inverse isn't even well-de ned.
Problem Practice-13: Prove that dm (K ) 6 = dm (T O T ).
Answer: If T O T m K then T O T would b e r.e., which we know it is not.
Problem Practice-14: Is dm (K ) = dm ( K )?
Answer: No, since K 6 m K (otherwise is would b e r.e., and K would b e recursive).
Problem Practice-15: Prove that m is re exive and transitive.
Answer: See the b o ok.
