COMS W3261 CS Theory
Lecture 19: The Classes P and NP
1. Time Complexity
- We now turn our attention to recursive languages and ask the question "How quickly
can a Turing machine determine whether a given input string w is in a given
recursive language L?"
- We say that a function T(n) is O(f(n))
[read "big-O of f(n)"]
if there exists an integer
n0 and a constant c > 0 such that for all
integers n ≥ n0, we have
T(n) ≤ cf(n).
- Big-O notation allows us to ignore constant factors and just focus
on asymptotic behavior.
E.g., T(n) = 2n2 is O(n2).
- Big-O notation allows us to ignore low-order terms.
E.g., T(n) = 2n2 + 3n + 4 is
- A Turing machine M is of time complexity T(n) if on any input of
length n, M halts after making at most T(n) moves.
- Note that time complexity is a worst-case measure. If a
Turing machine M is of time complexity T(n), then it
makes at most T(n) moves on any input of length n.
2. The Classes P and NP
- We define P to be the set of languages L such that
L = L(M) for some
deterministic Turing machine M of time complexity T(n) where
T(n) is a polynomial function of n.
- A problem that cannot be solved in polynomial time is said to be intractable.
- NP is the set of languages L such that L = L(M) for some
nondeterministic Turing machine M where on any input of length n,
there are no sequences of more than T(n) moves of M
where T(n) is a polynomial function of n.
- At present, it is not known whether all problems in NP can be solved in
polynomial time. The question of whether P = NP is one of the most important
open problems in computer science and mathematics.
- The Clay Mathematics Institute has identified the P vs NP question
as one of its seven
Millennium Prize Problems and offers a one million dollar prize for its solution.
3. Polynomial-Time Reductions
- In our study of problems that can be solved in polynomial time,
we restrict ourselves to polynomial-time reductions.
- A polynomial-time reduction is an algorithm that maps any
instance I of problem A into an instance J of problem B
in a number of steps that is a polynomial function of the
length of I such that I is in A iff J is in B.
- Note that in a polynomial-time reduction the length of J must be a polynomial
function of the length of I.
4. NP-Complete Problems
- A language L is NP-complete if
A language satisfying condition (2) is said to be NP-hard.
If A is NP-complete, B is in NP, and there is a polynomial-time
reduction of A to B, then B is NP-complete.
If any one NP-complete problem is in P, then P = NP.
Examples of problems in P
- L is in NP and
- For every language L' in NP there is a polynomial-time
reduction of L' to L.
Examples of NP-complete problems
- Determining whether a set of integers has a subset whose sum is negative.
- Finding the cheapest path between a pair of nodes in a graph.
- Lots of others.
Examples of problems in NP not known to be in P or to be NP-complete
- Determining whether a set of integers has a nonempty subset whose sum is zero
Note that we can verify a zero-sum subset in deterministic polynomial time,
but finding if one exists is difficult.
- Finding a cycle in a graph that contains each node exactly once
("the Hamiltonian cycle problem").
Again, note that verifying that a cycle is Hamiltonian is easy, but finding a
Hamiltonian cycle is hard.
- Lots of others.
5. Boolean Expressions
- Boolean expressions are generated by the following CFG:
E → E ∨ T | T
T → T ∧ F | F
F → ( E ) | ¬ F | var
var represents a variable whose value can be
either 1 (for true) or 0 (for false).
The operator ∧ stands for logical AND, ∨ for logical OR,
and ¬ for logical NOT.
Note that the grammar gives the operator ¬ the highest precedence,
then ∧, then ∨.
6. The Satisfiability Problem
- A truth assignment for a boolean expression assigns either
the value true (1) or the value false (0)
to each of the variables in the expression.
- The value E(T) of an expression E given a truth assignment
the result of evaluating E with each variable x in E replaced
- A truth assignment T satisfies E if E(T) = 1.
- An expression E is satisfiable if there exists a truth assignment T
that satisfies E.
- The satisfiability problem (SAT) is to determine whether a given
boolean expression is satisfiable.
- The value of E = x ∧ ¬(y ∨ z)
given the truth assignment
T(x) = 1, T(y) = 0, T(x) = 0 is 1.
[1 ∧ ¬(0 ∨ 0) = 1].
Thus, E is satisfiable.
- The expression E = x ∧ (¬x ∨ y) ∧
¬y is not satisfiable
because none of the four truth assignments to the variables x and y causes
E to have the value 1.
- We shall shortly prove that SAT is NP-complete.
7. Practice Problems
- Is the function nlog n polynomial
or exponential, where log is to the base 2?
- Show that the function 2n2 is O(n2).
- Show that the function
2n2 + 3n + 4 is O(n2).
- Show that if f(n) and g(n) are polynomial functions,
then f(g(n)) is a polynomial function.
- If M is a nondeterministic Turing machine of time complexity T(n),
then show that M can be simulated by a deterministic Turing machine of time complexity
- Show that P and NP closed under the following operations:
- Kleene closure
- Show that P is closed under complementation.
- List all satisfying truth assignments for
x ∧ (y ∧ ¬x) ∧ (z ∨ ¬y).
8. Reading Assignment