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Chapter 8: Problem 6

Show that any PSPACE-hard language is also NP-hard.

Short Answer

Expert verified
To show that any PSPACE-hard language is also NP-hard, we need to understand the connection between PSPACE and NP. Since NP is a subset of PSPACE, any language A' in NP can also be reduced in polynomial time to a PSPACE-hard language L because the same holds for languages in PSPACE. This implies that L is at least as hard as any problem in NP, making it NP-hard. Therefore, any PSPACE-hard language is also NP-hard.

Step by step solution

01

Definition of PSPACE-hard

A language L is PSPACE-hard if every language A in PSPACE, the space of problems solvable by a Turing machine using a polynomial amount of space, can be reduced to L in polynomial time. In other words, L is at least as hard as the hardest problems in PSPACE.
02

Definition of NP-hard

A language L is NP-hard if every language A in NP, the set of problems solvable by a non-deterministic Turing machine in polynomial time, can be reduced to L in polynomial time. In simpler terms, L is at least as hard as the hardest problems in NP.
03

Reduction

Polynomial-time reduction is a transformation that converts instances of one problem to instances of another problem. It is commonly used to show that problems are of equivalent difficulty. If a language A is reducible to language B, it means that any instance of A can be transformed to an instance of B in polynomial time.
04

Understanding the Connection Between PSPACE and NP

PSPACE and NP are two complexity classes representing problems solvable in polynomial space and non-deterministic polynomial time, respectively. It is known that NP is a subset of PSPACE (i.e., every problem in NP is also in PSPACE). This is because a non-deterministic Turing machine can be simulated by a deterministic Turing machine using an exponential amount of time but only a polynomial amount of space since the number of possible branches is polynomial.
05

Proving that PSPACE-hard Languages are NP-hard

Now we want to show that any PSPACE-hard language is also NP-hard. Let L be a PSPACE-hard language. By definition, every language A in PSPACE can be reduced to L in polynomial time. Since NP is a subset of PSPACE, every language A' in NP can also be reduced to L in polynomial time as it also holds for languages in PSPACE. Thus, L is at least as hard as any problem in NP, making it NP-hard. So, any PSPACE-hard language is also NP-hard.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

PSPACE-hard
To grasp the concept of PSPACE-hardness, we first need to understand PSPACE itself. PSPACE is the class of problems that can be solved by a Turing machine using a polynomial amount of space. The phrase "polynomial space" means the amount of memory or space that the machine uses relative to the size of the input grows polynomially.

A problem is considered PSPACE-hard if every other problem in PSPACE can be transformed into this problem using a conversion method called polynomial-time reduction. Essentially, PSPACE-hard problems are at least as difficult as the most challenging problems in PSPACE. This means solving a PSPACE-hard problem could theoretically help us solve any problem within the PSPACE class, by transforming it into the PSPACE-hard problem.
NP-hard
NP-hardness highlights the difficulty and complexity of a problem even further. NP, or non-deterministic polynomial time, is a class of decision problems for which a given solution can be verified in polynomial time. An NP-hard problem, however, does not necessarily belong to the NP class.

A problem is NP-hard if every problem in the NP class can be transformed into it using polynomial-time reduction. In simpler terms, if you manage to solve an NP-hard problem efficiently (i.e., in polynomial time), you can solve every problem in the NP class efficiently. This makes NP-hard problems a central focus in the realm of computational complexity as they represent some of the most challenging problems to tackle computationally.
Polynomial-time reduction
Polynomial-time reduction is a key concept in computational complexity that allows us to compare the difficulties of different problems. It involves transforming one problem into another in such a way that the transformation process itself uses only polynomial time.

This type of reduction is significant because it helps show that two problems are similar in terms of their complexity. If problem A can be reduced to problem B in polynomial time, solving B effectively means that A can also be solved just as efficiently. This is why polynomial-time reductions are often used to determine whether a problem belongs to a particular complexity class.
Complexity classes
Complexity classes are categorizations of problems based on the resources required to solve them, such as time and space. Two commonly discussed complexity classes are PSPACE and NP mentioned earlier.

PSPACE includes problems solvable by a Turing machine using polynomial space, while NP includes those problems where solutions can be verified in polynomial time. There are other complexity classes like P, which encompasses problems solvable in polynomial time, and EXP, which includes problems solvable in exponential time.

Understanding how these classes relate to each other is crucial. For instance, it's known that NP is a subset of PSPACE. This means every problem in NP can also be solved in polynomial space. By studying these classes and their relationships, we can better understand the inherent difficulty of computational problems.

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Most popular questions from this chapter

For any positive integer \(x\), let \(x^{\mathcal{R}}\) be the integer whose binary representation is the reverse of the binary representation of \(x\). (Assume no leading os in the binary representation of \(x .)\) Define the function \(\mathcal{R}^{+}: \mathcal{N} \longrightarrow \mathcal{N}\) where \(\mathcal{R}^{+}(x)=x+x^{\mathcal{R}} .\) a. Let \(A_{2}=\left\\{\langle x, y\rangle \mid \mathcal{R}^{+}(x)=y\right\\}\). Show \(A_{2} \in \mathrm{L}\). b. Let \(A_{3}=\left\\{\langle x, y\rangle \mid \mathcal{R}^{+}\left(\mathcal{R}^{+}(x)\right)=y\right\\} .\) Show \(A_{3} \in \mathrm{L}\).

Show that for any function \(f: \mathcal{N} \longrightarrow \mathcal{R}^{+}\), where \(f(n) \geq n\), the space complexity class \(\operatorname{SPACE}(f(n))\) is the same whether you define the class by using the singletape TM model or the two-tape read-only input TM model.

The game of Nim is played with a collection of piles of sticks. In one move, a player may remove any nonzero number of sticks from a single pile. The players alternately take turns making moves. The player who removes the very last stick loses. Say that we have a game position in Nim with \(k\) piles containing \(s_{1}, \ldots, s_{k}\) sticks. Call the position balanced if each column of bits contains an even number of \(1 \mathrm{~s}\) when each of the numbers \(s_{i}\) is written in binary, and the binary numbers are written as rows of a matrix aligned at the low order bits. Prove the following two facts. a. Starting in an unbalanced position, a single move exists that changes the position into a balanced one. b. Starting in a balanced position, every single move changes the position into an unbalanced one. Let \(N I M=\left\\{\left\langle s_{1}, \ldots, s_{k}\right\rangle \mid\right.\) each \(s_{i}\) is a binary number and Player I has a winning strategy in the Nim game starting at this position \(\\} .\) Use the preceding facts about balanced positions to show that \(N I M \in \mathrm{L}\).

Define \(U P A T H\) to be the counterpart of \(P A T H\) for undirected graphs. Show that \(\overline{B I P A R T I T E} \leq_{\mathrm{L}} U P A T H .\) (Note: In fact, we can prove \(U P A T H \in \mathrm{L}\), and therefore \(B I P A R T I T E \in \mathrm{L}\), but the algorithm [62] is too difficult to present here.)

The cat-and-mouse game is played by two players, "Cat" and "Mouse," on an arbitrary undirected graph. At a given point, each player occupies a node of the graph. The players take turns moving to a node adjacent to the one that they currently occupy. A special node of the graph is called "Hole." Cat wins if the two players ever occupy the same node. Mouse wins if it reaches the Hole before the preceding happens. The game is a draw if a situation repeats (i.e., the two players simultaneously occupy positions that they simultaneously occupied previously, and it is the same player's turn to move). \(H A P P Y-C A T=\\{\langle G, c, m, h\rangle \mid G, c, m, h\) are respectively a graph, and positions of the Cat, Mouse, and Hole, such that Cat has a winning strategy if Cat moves first \(\\}\). Show that HAPPY-CAT is in P. (Hint: The solution is not complicated and doesn't depend on subtle details in the way the game is defined. Consider the entire game tree. It is exponentially big, but you can search it in polynomial time.)
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