Introduction to Complexity

• Let P be the set of problems for which a polynomial time solution exists.
  • These are the problems we consider to be "solvable"
  • This set is closed under addition and multiplication
    • let $a_1, a_2 \in P$
    • Then $a_1 + a_2 \in P$ (What does this mean?)
    • Then $a_1 * a_2 \in P$ (What does this mean?)
• A decision problem is a problem that has a yes/no answer.
• In CLRS
  • They refer to a problem by a name
  • PATH is an example of a decision problem.
  • Given a Graph G=(V, E), and two vertices $u,v \in V$ does a path exist from u to v?
  • Any solution to PATH will return a yes if such a path exists and no if it does not.
• They state that optimization problems can be recast as a decision problem with a bound.
• We will focus on decision problems for this topic.
• Polynomial time verification
  • Given
    • A graph G
    • A path p in G
    • A cost k, this is the bound
  • Can we verify that this is a solution to PATH in polynomial time?
    • Clearly we can do this in the length of p time.
      • Let $p = \{u, n_1, n_2, ... n_{m-1}, v\}$
      • let $u = n_0$ and $v = n_m$
      • For each pair $(n_i, n_{i+1}), 0 \le i \lt k$ check to see if $(n_i, n_{i+1}) \in E$
      • Check that $k = \sum\limits_0^{m-1}w(n_i, n_{i+1})$.
• The Hamiltonian Cycle problem (HAM_CYCLE)
  • Given an undirected graph G=(V,E)
  • Does there exist a path p that starts at a node, visits every other node once, and returns to the starting node without traversing any edge twice?
  • Some images stolen from wikipedia
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  • What would it mean to verify a solution to HAM_CYCLE?
  • Can this be done in polynomial time?
  • By the way, we don't know a good solution to HAM_CYCLE
  • HAM_PATH is just to find a path that visits all nodes exactly once where no edge is traversed twice.
  • HAM_PATH is also hard.
• The class NP is the set of all decision problems which can be verified in polynomial time.
  • IE Given a solution, can you check to see if the solution is right in polynomial time?
• We have seen that HAM-CYCLE $\in$ NP
• What about problems in P?
  • Any problem in P has an algorithm that can solve it in Polynomial time.
  • Therefore given an input to such a problem and a solution, just solve the problem in polynomial time and check the solution (remember this is a yes/no thing)
• Therefore $P \subset NP$
• It is a long standing problem, does $P = NP$?
  • The standard belief is that this is not the case.