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Chapter 8: Problem 4

Show that PSPACE is closed under the operations union, complementation, and star.

Short Answer

Expert verified
PSPACE, the set of decision problems solvable with polynomial memory, is closed under union, complementation, and star operations. Demonstrating this requires constructing polynomial space-bounded Turing machines for resultant problems given such Turing machines for input problems. For the union of two decision problems A and B in PSPACE, construct Turing machine M3 to simulate M1 on input x, accept if M1 does, simulate M2 if M1 doesn't accept, and accept if M2 does. Finally, if neither accepts, M3 rejects. This process maintains polynomial space complexity, showing the union of A and B is also in PSPACE. For complementation, a Turing machine M2 for the complement of problem A simulates M1 on input x and inverts the decision - if M1 accepts, M2 rejects, and vice versa. As M1's simulation maintains polynomial space-bound, the complement is also in PSPACE. For the star operation, Turing machine M2 accepts input x if M1 accepts all substrings in every possible partition of x. If M1 rejects any substring, the partition counter increases. If all partitions are checked without M1 accepting all substrings in any, M2 rejects x. The polynomial space complexity remains, so the star problem for A is in PSPACE. Hence, PSPACE is closed under the operations of union, complementation, and star-operation.

Step by step solution

01

1. Union

Given two decision problems A and B in PSPACE, we need to show that their union C (i.e., C = A 鈭� B) is also in PSPACE. Let M1 and M2 be the polynomial space-bounded Turing machines for A and B respectively. We can construct a new Turing machine M3 for C as follows: 1. On input x, simulate M1 on x, and if M1 accepts, then M3 accepts. 2. If M1 does not accept, simulate M2 on x, and if M2 accepts, then M3 accepts. 3. If neither M1 nor M2 accept, then M3 rejects. Since M1 and M2 are polynomial space-bounded, simulating them one after the other does not increase the space complexity beyond a polynomial, and M3 also uses polynomial space. Therefore, their union C is in PSPACE.
02

2. Complementation

Given a decision problem A in PSPACE, we need to show that its complement A虆 (i.e., A虆 = {x | x 鈭� A}) is also in PSPACE. Let M1 be the polynomial space-bounded Turing machine for A. We can construct a new Turing machine M2 for A虆 as follows: 1. On input x, simulate M1 on x. 2. If M1 accepts x, then M2 rejects. 3. If M1 rejects x, then M2 accepts. Since M1 is polynomial space-bounded, simulating it in M2 does not increase the space complexity beyond a polynomial. Therefore, the complement A虆 is also in PSPACE.
03

3. Star

Given a decision problem A in PSPACE, we need to show that its star problem A* (i.e., A* = {x1x2鈥n | xi 鈭� A for all i, n 鈮� 0}) is also in PSPACE. Let M1 be the polynomial space-bounded Turing machine for A. We can construct a new Turing machine M2 for A* as follows: 1. On input x, initialize a counter to 0. 2. For each possible partition of x into substrings x1, x2, 鈥�, xn: a. Simulate M1 with each substring as input. If M1 accepts all substrings, then M2 accepts x and halts. b. If M1 rejects any substring, increment the counter and continue with the next possible partition. 3. If all partitions have been checked and M1 did not accept all substrings for any of them, then M2 rejects x. Since M1 is polynomial space-bounded, simulating it for multiple substrings of x does not increase the space complexity beyond a polynomial. M2's space complexity is dominated by the space needed to store the partition information and for M1's simulation. Therefore, the star problem A* is also in PSPACE. In conclusion, we have shown that PSPACE is closed under union, complementation, and the star operation, meeting the requirements of the exercise.

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Key Concepts

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

Computational Complexity Theory
Computational complexity theory is a framework to systematically study the efficiency and resource requirements of algorithms, which are the basic processes for solving problems. In simpler terms, it's all about examining how much computational resources, like time and memory, are required to perform a task. The main focus lies in classifying problems based on the resources they require for a computer to solve them.

Within this sphere, problems are often categorized by complexity classes such as PSPACE, which you might have encountered in your studies. PSPACE, in particular, refers to the set of decision problems that can be solved by a Turing machine using a polynomial amount of space. Understanding how different operations affect this class 鈥� for example, whether they preserve the class' limitations on space 鈥� is a crucial concept in this field. Consequently, when a textbook exercise asks if PSPACE is closed under certain operations like union, complementation, and star, it's inquiring whether these operations maintain the polynomial space constraint.
Turing Machines
A Turing machine is a mathematical model of computation that defines an abstract machine, which manipulates symbols on a strip of tape according to a set of rules. In essence, Turing machines are a cornerstone concept and a standard way to formalize the notion of computation in complexity theory. A Turing machine has a 'head' which reads and writes symbols on the tape, and a 'state register' that stores the state of the Turing machine.

One way to visualize a Turing machine is as a person in a very long library aisle, filled with bookshelves, where each book has a single symbol on its spine. The person can move forwards or backwards, change the symbol on a book, and their actions are determined by a set of rules they hold. Turing machines are used to describe algorithms and assess their efficiency 鈥� in terms of space and time 鈥� for solving decision problems. The machines built for PSPACE problems will always use an amount of tape (space) that's polynomial with respect to the size of the input.
Polynomial Space-Bounded Computation
Polynomial space-bounded computation describes algorithms that solve problems without exceeding a memory limit that grows polynomially with the size of the input. For instance, if an algorithm takes an input of size 'n' and the space it requires doesn't exceed some constant times 'n^k' (where 'k' is a fixed exponent), it's considered space-bounded by a polynomial.

The significance of this constraint lies in its practical implications: problems solvable within polynomial space are considered tractable, at least in terms of memory usage. Space-bounded computation is a useful measure because unlike time, space (memory) can be reused throughout the computation. This is precisely why poly-time Turing machines are so central to complexity theory; they provide a standard by which we can categorize problems, like those in PSPACE, that exhibit this polynomial space-bound characteristic.
Decision Problems
Decision problems are questions with binary answers 鈥� typically 'yes' or 'no'. In the context of computational problems, they're generally problems that ask whether a particular input satisfies certain conditions. For example, determining if a given number is prime or checking if there's a path between two points in a network are decision problems.

These problems are fundamental to theoretical computer science because they provide a clear way to measure the efficiency of algorithms 鈥� through the lens of 'can this problem be solved within certain resource constraints?' Decision problems are often the subject of complexity classes, such as PSPACE, as they serve as the basic units that these classes aim to group and constrain. Therefore, when looking at whether PSPACE is closed under union, complementation, or star operations, we're asking if we can still decide the 鈥榶es鈥� or 鈥榥o鈥� answer of combined or modified problems without exceeding polynomial space.

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Most popular questions from this chapter

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.

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.)

Let \(A\) be the language of properly nested parentheses. For example, \((())\) and \((()(()))()\) are in \(A\), but ) ( is not. Show that \(A\) is in \(\mathrm{L}\).

Let \(B\) be the language of properly nested parentheses and brackets. For example, ( \([()()]()[])\) is in \(B\) but \(([)]\) is not. Show that \(B\) is in \(\mathrm{L}\).
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