Understanding Polynomial Equivalence in Class P

1. CSCI 2670Introduction to Theory of Computing November 16, 2005

2. Announcement • Homework due next Tuesday (11/22) • 7.3 a, 7.4, 7.6 (union only), 7.9, 7.12 • 1st edition • 7.3a, 7.4, 7.6 (union only), 7.10, 7.12 November 16, 2005

3. Last class • Relationship of time complexity for different TM models • If a problem can be solved in O(t(n)) time on a multi-tape TM, it can be solved in O(t2(n)) time on a single-tape TM • If a problem can be solved in O(t(n)) time on a nondeterministic TM, it can be solved in 2O(t(n)) time on a deterministic TM November 16, 2005

4. Polynomial vs. exponential time • We distinguish between algorithms that have polynomial running time and those that have exponential running time • Assume a single tape deterministic TM • Polynomial functions – even ones with large exponents – grow less quickly than exponential functions • We can only process large data sets with polynomial running time algorithms November 16, 2005

5. Polynomial equivalence • Two algorithms A1 and A2 are polynomially equivalent if we can simulate A2 using A1 with only a polynomial increase in running time November 16, 2005

6. The class P • P is the class of languages that are decidable in polynomial time on a single-tape Turing machine P = k TIME(nk) • P “roughly corresponds” to the problems that are realistically solvable on a computer November 16, 2005

7. Size of input: Important consideration • The running time is measured in terms of the size of the input • If we increase the input size can that make the problem seem more efficient • E.g., if we represent integers in unary instead of binary • We consider only reasonable encodings November 16, 2005

8. A problem in class P • Binary tree query • Given a binary search tree T and a key k, find the node in T with key(node) = k • How do we show this problem is in class P? • Write an algorithm and show that the algorithm has running time O(nk) for some k November 16, 2005

9. Binary search M = “On input <G,k> • Let node = root(G) • Do while key(node) <> k • If key(node) < k • If right(node) == NIL • return NIL • Else • node = right(node) • Else • If left(node) == NIL • return NIL • Else • node = left(node) • Return node November 16, 2005

10. Execution time • Worst case running time? • O(|nodes|) • Occurs if tree is unbalanced • Is this O(n)? • Yes … any reasonable encoding will have an entry for each node November 16, 2005

11. Path problem • PATH = {<G,s,t> | G is a directed graph that has a path from s to t>} • PATH  P • How would we show this? • Mark all nodes that can be reached from s • If t gets marked, accept • If t doesn’t get marked, reject November 16, 2005

12. Solution to path problem M = “On input <G,s,t>, where G is a digraph with nodes s and t: • Place a mark on node s • Repeat the following until no additional nodes are marked • Scan all edges of G • If an edge (a,b) is found where a is marked and b is unmarked, mark b • If t is marked accept • Else reject” November 16, 2005

13. Verify solution is in P • Is M a decider? • Yes • How long will it take M to run? • Loop takes O(|edges|) to execute each time • Loop is executed O(|nodes|) times • Total time is O(|edges|x|nodes|) • Is this polynomial in the length of the input? • Yes • This is o(n2) where n = lentgh of input November 16, 2005