CS 172 , Spring, 1999
Professor M. Blum
Problem #1
a) Define the number of steps taken by a NDTM on input x.
b) Define the number of steps taken by a NDTM on inputs of length of n.
a) # NDTMi [x] =
case1: min(y belongs to (summation *)) { # DTMi[y,x]} ( this is the deterministic half of the NDTMi)
if there exists y such that DTMi[y,x] accepts (i.e. enters an accepting state)
case2: 1 otherwise
b) # NDTMi(n) = Max{# NDTMi[x]}, |x| = n
Problem #2
Define two (computational) problems p1, p2 to be poly-time equivalent iff it is possible to solve p1 in polynomial time
given an algorithm to solve p2 in polynomial time (p1 <= p2), and vice-versa (p2 <= p1).
Are the following two problems poly-time equivalent?
If so, prove it.
If not, explain why not.
Decision:
Instance: NDTMi, x in {0,1}*, m in unary
........m
(ie 1.....1 = 1 ).
Question: Does NDTMi accept x in m steps? ie does there exist a y in {0,1}* s.t. DTMi accepts (y,x) in m steps, if any
(ie if such y exists);
"NONE" if there is no such y.
1) YES! Decision <= Optimization:
If optimization program returns y, then the Decision program returns YES• If optimization program returns "NONE", then the DEcision program returns NO.•
The running time of the Decision algorithm = running time of the Optimizaton algorithm + O(1).
2) Optimization <= Decision:
If Decision algor returns No, then the Opt. algor returns NONE.• If Dec. algor returns YES, then the Opt. algor must find y. It can do this by finding the bits of y = Yk Y(k-1)
... Yo one at a time starting say with Yo.
•
course, exam #, semester/year (e.g., CS 150, Midterm #2, Fall 1994) CS 172, Solution of Midterm #1, Spring 1999
course, semester/year CS 172 , Spring, 1999exam # Professor M. Blumprofessor (e.g., Professor J. Wawrzynek)1
