Suggested languages for you:

Americas

Europe

Problem 20

# Prove that an oracle $$C$$ exists for which $\mathrm{NP}^{C} \neq \operatorname{coNP}^{C} .$

Expert verified
Using the diagonalization technique, we define a language $$L$$, and we show that for any oracle $$C$$, either $$L\in\mathrm{NP}^C$$ or $$L\in\mathrm{coNP}^C$$, but not both. By constructing an oracle Turing machine $$A$$ with access to an oracle $$C = \mathrm{NP}^{C} \oplus \mathrm{coNP}^{C}$$, we argue that if $$L \in \mathrm{NP}^{A}$$, it implies $$L \in \mathrm{NP}^{C}$$, and if $$L \in \operatorname{coNP}^{A}$$, it implies $$L \in \mathrm{coNP}^{C}$$. Since both can't be true simultaneously, there must exist an oracle $$C$$ such that $$\mathrm{NP}^{C} \neq \operatorname{coNP}^{C}$$, completing the proof.
See the step by step solution

## Step 1: Understanding the Concepts and Notations

First, let's understand some basic concepts and notations: 1. NP (nondeterministic polynomial time) consists of decision problems that can be solved by a nondeterministic Turing machine (TM) in polynomial time. Some problems may belong to both classes (like P, the class of problems solvable in deterministic polynomial time), and others may belong solely to one of them. 2. coNP consists of the complement of all problems in NP. In more formal terms, a language $$L$$ is in coNP if and only if its complement $$\overline{L}$$ is in NP. 3. An oracle TM (oracle Turing machine) is a hypothetical Turing machine that can solve a specific problem in a single computational step. In this context, an oracle $$C$$ allows a TM to solve any problem in the complexity class $$C$$ in one step. 4. We will use the notation $$\mathrm{NP}^{C}$$ and $$\operatorname{coNP}^{C}$$ to refer to the classes of problems solvable in nondeterministic polynomial time with the help of the oracle $$C$$. Now that we have a better understanding of the concepts let's move on to the next step.

## Step 2: Applying Diagonalization Technique

Since we need to find an oracle, $$C$$ such that $$\mathrm{NP}^{C} \neq \operatorname{coNP}^{C}$$, we will apply the diagonalization technique. Diagonalization is a technique used in theoretical computer science to differentiate between concepts, usually by constructing functions and languages to prove their separability. In this case, we will diagonalize over all pairs of nondeterministic oracle Turing machines $$(M_i, M_j)$$, where $$i, j$$ are integers indicating the indices of machines. Intuitively, we will make sure that there exists a language where the $$i$$-th NP oracle machine $$M_i$$ and the $$j$$-th coNP oracle machine $$M_j$$ will have a different outcome for some input. Define $$L \subseteq (\{0,1\}^*)^2$$ as follows: For any $$x, y \in \{0,1\}^*$$, we have $$(x,y)\in L$$ if one (and only one) of the following conditions is met: 1. $$x$$ and $$y$$ are distinct encodings of nondeterministic oracle TMs, and for the corresponding oracle machines $$M_x$$ and $$M_y$$, there exists an oracle $$C$$ such that $$M_x^C(x) = 1$$ and $$M_y^C(y) = 0$$. 2. $$x$$ and $$y$$ are not distinct encodings of nondeterministic oracle TMs.

## Step 3: Prove $$L \in \mathrm{NP}^{C} \oplus \mathrm{coNP}^{C}$$

We need now to prove that $$L \in \mathrm{NP}^{C} \oplus \mathrm{coNP}^{C}$$, which means that there is an oracle $$C$$ such that $$\mathrm{NP}^{C} \neq \operatorname{coNP}^{C}$$. Let $$A$$ be an oracle TM with access to an oracle $$C$$, where $$C = \mathrm{NP}^{C} \oplus \mathrm{coNP}^{C}$$ (i.e., $$C$$ contains the symmetric difference of $$\mathrm{NP}^{C}$$ and $$\operatorname{coNP}^{C}$$). Using $$A$$ and some fixed polynomial-time computable function $$f$$, we can create a polynomial hierarchy language $$L'$$ as follows: $$L' = \{x : A^{f(x)}(x) = 1 \}$$ Now, let's suppose that $$L \in \mathrm{NP}^{A}$$. In this case, there exists an NP oracle TM $$M_i$$, which decides $$L$$ in polynomial time when given access to the oracle $$A$$. We can define $$C$$ as the oracle required by $$M_i$$. When presented as input of itself, $$M_i^C(i)$$ must equal $$1$$. As a result, $$L \in \mathrm{NP}^{C}$$. On the other hand, let's assume that $$L \in \operatorname{coNP}^{A}$$. In this case, there exists a coNP oracle TM $$M_j$$ that decides $$\overline{L}$$ (complement of $$L$$) in polynomial time when given access to the oracle $$A$$. We can define $$C$$ as the oracle required by $$M_j$$. When $$M_j$$ is presented with its own index as input, $$M_j^C(j)$$ must equal $$1$$. As a result, $$L \in \mathrm{coNP}^{C}$$. However, the way we defined $$L$$ at the beginning implies that $$L\in\mathrm{NP}^C$$ XOR $$L\in\mathrm{coNP}^C$$ for any oracle machine $$C$$. This means that both previous paragraphs cannot be true simultaneously. Thus, the existence of the oracle $$C$$ implies that $$\mathrm{NP}^C \neq \operatorname{coNP}^C$$. In conclusion, we've shown that there must exist an oracle $$C$$ such that $$\mathrm{NP}^{C} \neq \operatorname{coNP}^{C}$$, which completes the proof.

We value your feedback to improve our textbook solutions.

## Access millions of textbook solutions in one place

• Access over 3 million high quality textbook solutions
• Access our popular flashcard, quiz, mock-exam and notes features

## Join over 22 million students in learning with our Vaia App

The first learning app that truly has everything you need to ace your exams in one place.

• Flashcards & Quizzes
• AI Study Assistant
• Smart Note-Taking
• Mock-Exams
• Study Planner