* Last revised 4/29/2010*

In this lecture we define , we state the versus problem, we prove that its formulation for search problems is equivalent to its formulation for decision problems, and we prove the time hierarchy theorem, which remains essentially the only known theorem which allows to show that certain problems are not in .

**1. Computational Problems **

In a *computational problem*, we are given an *input* that, without loss of generality, we assume to be encoded over the alphabet , and we want to return as *output* a solution satisfying some property: a computational problem is then described by the property that the output has to satisfy given the input.

In this course we will deal with four types of computational problems: *decision* problems, *search* problems, *optimization* problems, and *counting* problems.\footnote{ This distinction is useful and natural, but it is also arbitrary: in fact every problem can be seen as a search problem} For the moment, we will discuss decision and search problem.

In a *decision* problem, given an input , we are required to give a YES/NO answer. That is, in a decision problem we are only asked to verify whether the input satisfies a certain property. An example of decision problem is the 3-coloring problem: given an undirected graph, determine whether there is a way to assign a “color” chosen from to each vertex in such a way that no two adjacent vertices have the same color.

A convenient way to *specify* a decision problem is to give the set of inputs for which the answer is YES. A subset of is also called a *language*, so, with the previous convention, every decision problem can be specified using a language (and every language specifies a decision problem). For example, if we call 3COL the subset of containing (descriptions of) 3-colorable graphs, then 3COL is the language that specifies the 3-coloring problem. From now on, we will talk about decision problems and languages interchangeably.

In a *search* problem, given an input we want to compute some answer that is in some relation to , if such a exists. Thus, a search problem is specified by a relation , where if and only if is an admissible answer given .

Consider for example the search version of the 3-coloring problem: here given an undirected graph we want to find, if it exists, a coloring of the vertices, such that for every we have . This is different (and more demanding) than the decision version, because beyond being asked to determine whether such a exists, we are also asked to construct it, if it exists. Formally, the 3-coloring problem is specified by the relation that contains all the pairs where is a 3-colorable graph and is a valid 3-coloring of .

**2. P and NP **

In most of this course, we will study the *asymptotic* complexity of problems. Instead of considering, say, the time required to solve 3-coloring on graphs with nodes on some particular model of computation, we will ask what is the best asymptotic running time of an algorithm that solves 3-coloring on all instances. In fact, we will be much less ambitious, and we will just ask whether there is a “feasible” asymptotic algorithm for 3-coloring. Here feasible refers more to the rate of growth than to the running time of specific instances of reasonable size.

A standard convention is to call an algorithm “feasible” if it runs in polynomial time, i.e. if there is some polynomial such that the algorithm runs in time at most on inputs of length .

We denote by the class of decision problems that are solvable in polynomial time.

We say that a search problem defined by a relation is a search problem if the relation is efficiently computable and such that solutions, if they exist, are short. Formally, is an search problem if there is a polynomial time algorithm that, given and , decides whether , and if there is a polynomial such that if then .

We say that a decision problem is an decision problem if there is some relation such that if and only if there is a such that . Equivalently, a decision problem is an decision problem if there is a polynomial time algorithm and a polynomial such that if and only if there is a , such that accepts.

We denote by the class of decision problems. The class has the following alternative characterization, from which it takes its name. ( stands for *N*ondeterministic *P*olynomial time.)

Theorem 1is the set of decision problems that are solvable in polynomial time by a non-deterministic Turing machine.

*Proof:* Suppose that is solvable in polynomial time by a non-deterministic Turing machine : then we can define the relation such that if and only if is a transcript of an accepting computation of on input (that is, the sequence of configurations of an accepting branch of on input ) and it’s easy to prove that is an relation and that is in according to our first definition. Suppose that is in according to our first definition and that is the corresponding relation. Then, on input , a non-deterministic Turing machine can guess a string of length less than and then accept if and only if . Such a machine can be implemented to run in non-deterministic polynomial time and it decides .

The reason why , as a complexity class, is defined as a class of decision problems rather than search problems is that one can develop a cleaner theory for decision problems (the definition of reduction, for example, is simpler), and the complexity of decision problems in completely characterizes the complexity of search problems.

Theorem 2For every search problem there is an decision problem such that if the decision problem is solvable in time then the search problem is solvable in time . In particular, if and only if every search problem is solvable in polynomial time.

*Proof:* Let be an relation defining an search problem. Then consider the following decision problem:

- Input: a pair of strings
- Goal: determine if there is a string such that , where denotes string concatenation.

Suppose the above decision problem is solvable in time by an algorithm . Then consider the following algorithm for the search problem ( denotes the empty string):

- Input
- if , then return “
*no solution*” - while
- if then
- else

- return

If there is no solution for , then the algorithm discovers that at the beginning. Otherwise, every iteration of the *while* loop increases the length of while maintaining the invariant that is a prefix of a valid solution to . Since is an relation, such a solution must be of length polynomial in the length of , and so the loop terminates after a polynomial number of iterations, with a valid solution.

For a function , we define by the set of decision problems that are solvable by a deterministic Turing machine within time on inputs of length , and by the set of decision problems that are solvable by a non-deterministic Turing machine within time on inputs of length . Hence, and .

**3. Diagonalization **

Diagonalization is essentially the only way we know of proving separations between complexity classes. The basic principle is the same as in Cantor’s proof that the set of real numbers is not countable. First note that if the set of real numbers in the range is countable then the set of infinite binary sequences is countable: we can identify a real number in with its binary expansion . If we had an enumeration of real numbers, then we would also have an enumeration of infinite binary string. (The only thing to watch for is that some real numbers may have two possible binary representations, like and .)

So, suppose towards a contradiction that the set of infinite binary sequences were countable, and let be the -th bit of the -th infinite binary sequence. Then define the sequence whose -th bits is . This is a well-defined sequence but there can be no such that , because differs from in the -th bit.

Similarly, we can prove that the Halting problem is undecidable by considering the following decision problem : on input , the description of a Turing machine, answer NO if halts and accepts and YES otherwise. The above problem is decidable if the Halting problem is decidable. However, suppose where decidable and let be a Turing machine that solves , then halts and accepts if and only if does not halt and accept, which is a contradiction.

It is easy to do something similar with time-bounded computations.

Definition 3 (Time constructible functions)A function is time constructible if there is an algorithm that, on input , computes the value in time

All polynomials are time constructible, as well as all combinations of exponential, polynomial, and root functions such as , , , and so on. Indeed, one needs to work quite hard to come with an example of a function that is not time constructible. One example of a function that is not time constructible is the following: define to equal if the number written in binary encodes a turing machine that halts on all inputs, and define to equal otherwise.

Theorem 4 (Time hierarchy theorem — simplified version)For every time-constructible function , there is a language such that every algorithm that decides must run in time on infinitely many inputs, but .

*Proof:* The theorem holds for any model of computation which is simulable within the model. For concreteness, we give the proof for the case of Turing machine. We define the language

That is, an input to the problem is a pair , where is a Turing machine and is a possible input for , and the goal of the problem is to determine whether , on input , halts and rejects in at most steps, where is the length of .

First, we note that is decidable in time polynomial in on inputs of length . Given an input , we compute , where is the length of in time (here we are using the time-constructibility of ) and then we run a simulation of the computation of on input . If the simulation halts and reject within steps, then we accept. If the simulation accepts within steps, or is still running after steps, then we reject.

It remains to argue that every machine that decides must run in time for infinitely many inputs.

Let be a machine that decides and let be any string. Suppose that , on input , halts after steps. Then if , on input , rejects, then , by definition of , and so is incorrectly deciding on input . If , on input accepts, then , and so is again incorrect. In either case we reach a contradiction to the assumption that decides .

It follows that halts after steps on all the inputs of the form , of which there are infinitely many.

We define two more complexity classes in order to derive certain simple consequences of the time hierarchy theorem.

Corollary 5and

*Proof:* Apply the time hierarchy theorem to , and let be the resulting language. Then is decidable in time , and so it is in . Suppose that , and so that there is a machine that decides in time at most on inputs of length , where is a polynomial. Since for all but finitely many values of , contradicting the property of guaranteed by the time hierarchy theorem.

Apply the time hierarchy theorem to , and let be the resulting language. Then is decidable in time , and so , but we cannot have or else the machine deciding in time would run in time for all but a finite number of inputs.

## 3 comments

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April 25, 2010 at 12:37 am

borisThank you for posting this.

I think there is a typo in Theorem 2 – “if the decision problem is solvable in time {t(n)} then the decision problem is solvable in time {O(n^{O(1)} \cdot t(n))}”.

April 25, 2010 at 1:14 am

Sune JakobsenThanks for a great post.

There is an unfinished sentence just before theorem 4:

“Indeed, one needs to work quite hard to”

April 25, 2010 at 8:57 am

Marcus RittGreat post.

To be compatible with the algorithm for the search problem, shouldn’t the decision problem in the proof of theorem 2 decide if there’s a suffix z such that ?