Lecture 9: Complexity Theory -- Time Complexity¶

Let M be a standard DTM that halts on all inputs.

The running time of M is the function \(f: \mathbb{N} \to \mathbb{N}\) [input length] \(\to\) [number of steps].
On any input of length \(n\), M halts within \(f(n)\) steps.

Theorm¶

\(DTIME(f(n)) = \{L | L\) is decided by some standard DTM with running time \(O(t(n))\}\)¶

\(\{0^k1^k | k \geq 0\}\) \(\in\) \(DTIME(nlogn)\)
\(\{0^k1^k | k \geq 0\}\) \(\in\) \(DTIME(n)\)

The Cobham-Edmonds Thesis¶

Any "reasonable" and "general" deterministic model of computation is polynomially related.

\(P = \{L | L\) is decided by some standard DTM with running time \(O(n^k)\}\)
\(P = \cup_{k \geq 0} DTIME(n^k)\)

1. Every context-free language is in P.¶

Step 1: Chonsky Normal Form

For any CFG G, there is an equivalent CFG G' in Chonsky Normal Form.

Chonsky Normal Form: Every production is of the form \(A \to BC\) or \(A \to a\).

Given a string \(x = x_1x_2...x_n\), we want to know whether \(x \in L(G)\).

Enumerate all derivations of length \(\leq 2|w|-1\).
We can determine in \(|R|^{2|w|-1}\) steps.

这可以用二叉树的性质证明
But we cannot determine whether \(x \in L(G)\) in polynomial time.

Step 2: Use Dynamic Programming

Subproblem: For \(1 \leq i \leq j \leq n\) Define, \(T[i,j] = \{A \in V-\Sigma | A \to^* a_{i}...a_j\}\)

这是可以生成子串的非终结符的集合

Base Case: \(T[i,i] = \{A \in V-\Sigma | A \to a_i\}\)
Recurrence: \(T[i,j] = \cup_{i \leq k < j} \{A \in V-\Sigma | A \to BC, B \in T[i,k], C \in T[k+1,j]\}\)

Subproblem: \(O(n^2)\) subproblems, * each subproblem can be solved : * Enumerate k from i to j-1: \(O(n)\) * \(A \to BC\) can be checked in \(O(|R|)\) * \(B \in T[i,k]\) and \(C \in T[k+1,j]\) can be checked in \(O(|V|-|\Sigma|)\)

Total time: \(O(n^3|R|(||V|-|\Sigma|)^2)\)
Therefore, we can determine in \(O(n^3)\) time.

2. NP¶

Let U be a non-deterministic TM, running time of U is a function \(f: \mathbb{N} \to \mathbb{N}\).

For any input of length \(n\), Every branch of U halts within \(f(n)\) steps.

NP = \(\{L | L\) is decided by some non-deterministic TM U with polynomial running time \(O(n^k)\}\)

3. Polynomial Verifiable¶

A language L is polynomial verifiable if there exists a polynomial time verifier V such that:

It Satisfies: Efficient, Soundness, Completeness
\(V\) must have polynomial running time.
For any \(x \in A\), there exists a \(y\) such that \(V(x,y) = \text{accept}\) WITH \(|y| \leq poly(|x|)\)

4. A language L is in NP if and only if L is polynomial verifiable.¶

4.1 A language L is in NP \(\Rightarrow\) L is polynomial verifiable

If L is in NP, there exists a non-deterministic TM U that decides L in polynomial time.

Construct a verifier V:

certificate y = the branch of U that accepts x

V = on input x, y:

Run U on x deterministically on the guidance of y
If U accepts, accept
Else, reject

4.2 A language L is polynomial verifiable \(\Rightarrow\) L is in NP

If L is polynomial verifiable, Construct a non-deterministic TM U:

U = on input x:

Non-deterministically generate a certificate y of length \(\leq poly(|x|)\)
Run V on x, y
If V accepts, accept
Else, reject

Since V is polynomial time, U can decide L in polynomial time.

SAT Problem¶

Let X be a set of boolean variables

literal: \(x_i\) or \(\overline{x_i}\)
clause: disjunction of literals: \(x_1 \vee x_2 \vee \overline{x_3}\)
boolean formula: conjunction of clauses: \((x_1 \vee x_2) \wedge (\overline{x_1} \vee x_3)\)
Truth assignment of X is a function \(f: X \to \{0,1\}\)
T satisfies F if F is true under T

SAT Problem: Given a boolean formula F, does there exist a truth assignment T that satisfies F?

2.1 Non-deterministic Polynomial Time

M = on input F:

Non-deterministically guess a truth assignment T
If T satisfies F, accept
Otherwise, reject

2.2 Polynomial Verifiable

1. SAT is in P if and only if P = NP¶

Polynomial Time Reduction¶

Let A, B be two languages, A can be polynomially reduced to B [\(A \leq_p B\)] if there is a computable function f: \(\Sigma^* \to \Sigma^*\) such that:

\(x \in A \Leftrightarrow f(x) \in B\)
\(f\) is computable in polynomial time

Lemma 1¶

If A \(\leq_p B\) and B is in P, then A is in P.

\(x \rightarrow f(x) \rightarrow\) B \(\rightarrow\) P

NP-Completeness¶

A language A is NP-complete if: 1. A is in NP 2. For any language B in NP, \(B \leq_p A\)

Cook Levin Theorem¶

SAT is NP-complete.

Proof:

See link

Lemma 2¶

If A is NP-complete and \(A\) \(\leq_p B\), then B is NP-complete.

Proof:

B is in NP
For any language C in NP, \(C \leq_p A \leq_p B\)
Therefore, B is NP-complete.

3-SAT¶

Cliques¶

Given a graph G = (V, E), a clique is a subset of V such that every pair of vertices in the subset is connected by an edge.

CLIQUES = \(\{<G, k> | G\) has a clique of size \(k\}\)

Theorem: CLIQUES is NP-complete.

Proof: 3-SAT \(\leq_p\) CLIQUES See link

Vertex Cover¶

Given a graph G = (V, E), a vertex cover is a subset of V such that every edge in E is incident to at least one vertex in the subset.

VERTEX-COVER = \(\{<G, k> | G\) has a vertex cover of size \(k\}\)

最后更新: 2024年12月16日 00:12:48
创建日期: 2024年12月16日 00:12:48