University of Virginia Computer Science
CS150: Computer Science, Fall 2005
19 October 2005
Alrogithm: A procedure that always terminates.
Proof: A proof of S in an axiomatic system is a sequence of strings, T0, T1, ..., Tn where:
Input: an axiomatic system (a set of axioms and inference rules), a statement S, and a proof P containing n steps of S.
Output: true if P is a valid proof of S; false otherwise.
How much work is a proof checking procedure?Finite-Length Proof Finding Problem
Input: an axiomatic system (a set of axioms and inference rules), a statement S, n (the maximum number of proof steps)
Output: A valid proof of S with no more then n steps if there is one. If there is no proof of S with <= n steps, unprovable.
How much work is a straightforward proof finding procedure?
n — maximum number of steps
r — number of inference rules
At worst, we can try all possible rules at every step:
Proof Finding Problem
Input: an axiomatic system, a statement S.
Output: If S is true, output is a valid proof. If S is not true, output is false.
Can we write an algorithm to solve the proof finding problem?
Is there an algorithm (a procedure that always terminates) that solves the problem?
A problem is decidable is there exists an algorithm that can solve the problem for all possible inputs. It is not necessary to know what that algorithm is to say a problem is decidable, only to know that some algorithm to solve it must exist. For example, chess is a decidable problem, even if we do not yet know an algorithm that solves it.
A problem is undecidable is there is no algorithm that can solve the problem. There might be a procedure, but it is not guaranteed to terminate.
Is the proof finding problem decidable?How can you prove a problem is undecidable?
Input: a procedure P (described by a Scheme program), and the
input to that procedure
Output: true if applying P to input halts (finishes execution), false otherwise.
What if we had halts?(define (halts? procedure input) ...?...)
We can prove the halting problem is undecidable informally by arguing that if we could define halts?, we could use it to define contradict-halts (the input parameter to contradict-halts is not used, but necessary because of how we defined halts?):(define (find-proof S axioms rules) ;; If S is provable, evaluates to a proof of S. ;; Otherwise, evaluates to #f. (if (halts? find-proof-exhaustive S axioms rules) (find-proof-exhaustive S axioms rules) #f))Where (find-proof-exhaustive S axioms rules) is a procedure that tries all possible proofs starting from the axioms that evaluates to a proof if it finds one, and keeps working if it doesn't.
(define (contradict-halts input) (if (halts? contradict-halts null) (infinite-loop) 150))But contradict-halts is non-sensical: if it halts, it loops infinitely; if it doesn't halt, it evaluates to 150 and halts. This means something in the program must not exist. I'm pretty sure 200 exists, and we know how to define infinite-loop, and if seems likely to exist (and its worked well for us so far). So, it must be halts? that cannot exist. Undecidable Problems
If solving a problem P would allow us to solve the halting problem, then P is undecidable — there is no solution to P, since we have proved there is no solution to the halting problem!
There is a remarkably close parallel between the problems of the physicist and those of the cryptographer. The system on which a message is enciphered corresponds to the laws of the universe, the intercepted messages to the evidence available, the keys for a day or a message to important constants which have yet to be determined. The correspondence is very close, but the subject matter of cryptography is very easily dealt with by discrete machinery, physics not so easily.
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CS 150: Computer Science
University of Virginia
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