CS 637: Mathematics-II

Lecture 16: Search Reducible to Decision, EXP, NEXP, and coNP
Instructor: Dr. Debasish Pattanayak

Search Reducible to Decision

Decision vs Search

A decision problem asks for one bit: yes or no. A search problem asks us to output a witness, path, assignment, or other object.

\[ \text{Decision: } x \in L? \qquad \text{Search: find } y \text{ such that } R(x,y)=1. \]

Self-Reducibility Idea

For many NP problems, search reduces to repeated calls to the decision version. Example: for SAT, ask whether the formula is satisfiable with \(x_1=0\); if yes, keep that choice, otherwise set \(x_1=1\), and continue variable by variable.

\[ \text{SAT-search} \le_T^p \text{SAT-decision}. \]

Example: Independent Set Search from Decision

Suppose we have an oracle for the decision problem

\[ \operatorname{INDSET} = \{\langle G,k \rangle : G \text{ has an independent set of size } k\}. \]

To recover an actual independent set of size \(k\), first ask whether \((G,k)\) is a yes-instance. Then examine vertices one by one. For a vertex \(v\), ask whether \(G-v\) still has an independent set of size \(k\).

  1. If the answer is yes, discard \(v\); it is not needed.
  2. If the answer is no, then \(v\) must belong to some size-\(k\) independent set. Output \(v\), delete \(v\) and all its neighbors, and reduce the target from \(k\) to \(k-1\).

Repeating this process reconstructs a full independent set using only polynomially many decision queries.

\[ \text{INDSET-search} \le_T^p \text{INDSET-decision}. \]

Template for Recovering a Witness

  1. Ask whether a solution exists.
  2. Fix one bit or one local choice.
  3. Ask whether a solution still exists under that restriction.
  4. Repeat until the entire witness is reconstructed.

This needs only polynomially many decision-oracle queries when the witness length is polynomially bounded.

Why This Matters

If \(P=NP\), then not only can we decide SAT quickly, we can also find a satisfying assignment quickly by self-reduction.

The Classes EXP and NEXP

Definitions

\[ EXP = \bigcup_{c \ge 1} DTIME(2^{n^c}), \qquad NEXP = \bigcup_{c \ge 1} NTIME(2^{n^c}). \]

These are the exponential-time analogues of \(P\) and \(NP\).

Interpretation

Class Machine type Time bound
\(EXP\) deterministic TM \(2^{n^{O(1)}}\)
\(NEXP\) nondeterministic TM \(2^{n^{O(1)}}\)

Basic Inclusion

\[ EXP \subseteq NEXP \]

because every deterministic machine is a special case of a nondeterministic one.

If \(P = NP\), Then \(EXP = NEXP\)

Claim

If \(P=NP\), then every nondeterministic exponential-time computation can be deterministically simulated in exponential time. Hence

\[ P = NP \implies EXP = NEXP. \]

Proof Strategy

Let \(L \in NEXP\). Then some NTM decides \(L\) in time \(2^{n^c}\). Define a padded language

\[ L' = \{\, x \# 1^{2^{|x|^c}} : x \in L \,\}. \]

The padding makes the original exponential time bound polynomial in the new input length.

Why the Padded Language Lies in NP

Let \[ N = |x \# 1^{2^{|x|^c}}|. \] Then \(N = \Theta(2^{|x|^c})\). On input \(x \# 1^t\) with \(t=2^{|x|^c}\), a nondeterministic machine can simulate the original \(NEXP\) machine for at most \(t\) steps. Since \(t = O(N)\), this simulation takes polynomial time in the padded input length. Thus \(L' \in NP\).

If \(P=NP\), then \(L' \in P\).

Recovering \(L\) from \(L'\)

To decide whether \(x \in L\), explicitly construct the padded string \(x \# 1^{2^{|x|^c}}\) and run the polynomial-time decider for \(L'\). Both the construction of the padding and the decider take time polynomial in the padded length, hence total time \[ 2^{|x|^{O(1)}}. \]

\[ L \in EXP. \]

Since \(L\) was arbitrary in \(NEXP\), we get \(NEXP \subseteq EXP\). Together with \(EXP \subseteq NEXP\), this proves \(EXP=NEXP\).

The Class coNP

Definition

\[ coNP = \{ L : \overline{L} \in NP \}. \]

Equivalently, \(L \in coNP\) if its no-instances have polynomial-size certificates verifiable in polynomial time.

Canonical Example

\(\operatorname{TAUT} = \{\varphi : \varphi \text{ is true under every assignment}\}\) lies in \(coNP\), because its complement is

\[ \overline{\operatorname{TAUT}} = \operatorname{SAT}_{\neg}, \]

the set of formulas having a falsifying assignment, and that assignment serves as an NP witness.

Known Relations

  • \(P \subseteq NP \cap coNP\).
  • It is unknown whether \(NP = coNP\).
  • If \(P=NP\), then automatically \(P=NP=coNP\).

Certificate View

\(NP\) talks about short proofs of membership. \(coNP\) talks about short proofs of non-membership. The open question \(NP \stackrel{?}{=} coNP\) asks whether both kinds of proof are equally powerful.

Lecture Summary

Main Messages

  • For many NP problems, search reduces to decision via self-reduction.
  • \(EXP\) and \(NEXP\) are the exponential-time analogues of \(P\) and \(NP\).
  • The padding argument shows \(P=NP \Rightarrow EXP=NEXP\).
  • \(coNP\) is the complement class of \(NP\), and \(NP \stackrel{?}{=} coNP\) remains open.
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