Evaluation of Tree Functions

05‐09‐2015

PARALLEL AND DISTRIBUTED ALGORITHMS BY DEBDEEP MUKHOPADHYAY AND ABHISHEK SOMANI http://cse.iitkgp.ac.in/~debdeep/courses_iitkgp/PAlgo/index.htm

THE EULER TOUR TECHNIQUE: EVALUATION OF TREE FUNCTIONS

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AN OVERVIEW

When is a problem parallelizable? The complexity classes P and NC and the notion of P-completeness. The Circuit Value Problem is P-complete.

WHEN IS A PROBLEM PARALLELIZABLE?

A problem is parallelizable if it can be solved very fast with a reasonable number of processors. By very fast, we mean polylogarithmic time, i.e., O(logk n) time for some fixed constant k and input size n. Another requirement is that the number of processors used should be a polynomial in the input size, i.e., O(nc) for a fixed constant c.

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THE CLASS NC

We now discuss the complexity class NC. This is Nick’s class in honour of Nick Pippenger. The class NC consists of all the problems that can be solved by polylogarithmic-time algorithms on a PRAM, using a polynomial number of processors.

THE CLASS NC

The class NC is defined to be the set of all language L such that: on all inputs of length n, L can be recognized by a PRAM algorithm running in O(logk n) time.

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P AND NC

Let P be the class of all languages that can be recognized by a deterministic Turing machine within a polynomial number of steps. P can be defined as the class of all problems that can be solved on a RAM in polynomial time.

P AND NC

Obviously, NC  P since we can take any NC algorithm and convert it into a polynomial time sequential algorithm. However, it is not known whether P  NC. Most people believe that P  NC.

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P AND NC

If P  NC, then there must be problems in P that cannot be solved efficiently in parallel. The class of P-complete problems are most likely candidates of problems that are not in NC.

P AND NC

The situation is similar to the relationship between the class NP and P. If someone can prove that there is a polynomial-time algorithm for at least one NP-complete problem, then there are polynomial time algorithms for all NP- complete problems.

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REDUCIBILITY

Similarly, if there is a polylogarithmic-time algorithm for at least one P-complete problem, then there are polylogarithmic- time algorithms for all P-complete problems. How do we prove a problem to be P- complete?

REDUCIBILITY

Let L1 and L2 be two languages. The language L1 is NC-reducible to the language L2 if: there exists an NC-algorithm that transforms an arbitrary input u1 for L1 into an input u2 for L2 such that:

u1  L1 if and only if u2  L2.

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THE NOTION OF P- COMPLETENESS

A language L is P-complete if: 1. L  P, and 2. every language in P is NC-reducible to P.

The next question is: Is there a P-complete problem?

THE CIRCUIT VALUE PROBLEM (CVP)

We are given a Boolean circuit consisting of NOT gates and two-valued AND and OR gates. The problem is to determine the output of the Boolean circuit for a given set of inputs. More precisely, our Boolean circuit C is specified by a sequence of gates

C = < g1, g2, …, gn >.

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THE CIRCUIT VALUE PROBLEM (CVP)

Each gi is: •either an input, or

•gi = gj  gk, or gi = gj  gk , or gi =  gj for j, k < i.

CVP IS P-COMPLETE

We need to prove that: CVP is in P, and An arbitrary language L  P is NC-reducible to CVP. CVP is clearly in P. Given an n-gate CVP, we can evaluate the gates sequentially in O(n) time by evaluating g1, g2, …, gn.

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CVP IS P-COMPLETE

We next show that an arbitrary language L  P is NC-reducible to CVP. We provide an NC algorithm that takes an arbitrary instance IL of L, and maps IL into an instance ICVP of CVP such that,

IL has a yes answer if and only if ICVP has a value 1.

TURING MACHINE

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TURING MACHINE

Since L  P , there exists a one-tape deterministic Turing machine M that accepts L in polynomial time T(n), for an input size n. We can assume that the input bits appear in n consecutive tape cells numbered 1 to n and all other cells are initially blank. The tape head is initially scanning cell 1 and after T(n) steps, the output is written into cell 1.

TURING MACHINE

Let Q = { q1,…, qs } be the set of states of M and q1 is the initial state.

Let  = { a1,…, am } be the tape alphabet. The state transition function is given by:  : Q Q {L , R }

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TURING MACHINE

We will show how to construct a Boolean circuit C such that the value of C is 1 if and only if M accepts L, i.e., cell 1 contains 1 at time T(n). Note that, M can only access T(n) cells in T(n) time. Hence, all cell indices i will be in the range 1  i  T(n). All time steps are within the range 0  t  T(n).

There are only s states in M hence, all states qj are within the range 1  j  s.

SOME TOOLS

The circuit C will be defined by the following Boolean functions: H(i , t), such that H(i , t) = 1 if and only if the head scans cell i at time t. C(i , j, t), such that C(i , j, t) = 1 if and only if cell

i contains character aj at time t. S(k , t), such that S(k , t) = 1 if and only if the

state of M is qk at time t.

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SOME TOOLS

Our circuit C will consist of T(n) levels. For each time step of the Turing machine M, one level of our circuit will emulate the behavior of M. Level l will contain the gates computing the Boolean function H(i , l), C(i , j, l), S(k , l) for 1  i  T(n). Note that, we don’t know exactly how M recognizes L, but in our circuit C, we capture all possible steps of M at every level.

SOME TOOLS

The number of gates at every level is T(n). We will first specify H(i , 0), C(i , j, 0), S(k , 0). These are the inputs to our circuit. We will then show how to express H(i , t + 1), C(i , j, t + 1), S(k , t + 1) in terms of H(*, t), C(*, *, t), S(*, t). This will give us a correct description of the circuit C.

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INPUT GATES FOR OUR CIRCUIT

For t = 0, the tape head is scanning cell 1, the input bits are stored in cell 1 to n and M is in state q1. H(i , 0)= 1if i = 1, 0 otherwise. Since, the head is scanning cell 1.

C(i , j, 0) = 1 if cell i contains initially aj. S(k , 0) = 1, if k = 1, 0 otherwise.

GATES AT A GENERAL LEVEL

Let us consider constructing H(i , t +1) when we know H(*, t), C(*, *, t), S(*, t). H(i , t +1) = 1 if and only if the head scans cell i at time t +1. Since the Turing machine M moves its head only one cell at each time step, It must be either on cell i - 1 or on cell i + 1 at time t. Let us consider the case when the head is at cell i -1at time t

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GATES AT A GENERAL LEVEL

H(i , t +1) = 1 if and only if the head moves right and goes to cell i at time t +1.

Suppose the content of cell i -1is aj and the Turing machine M is in state qk. Hence, to move the head to cell i, our transition function  should be such that:  (qk, aj) = ( , , R).

That is, if the current state is qk and cell i -1 contains character aj, the head moves right.

GATES AT A GENERAL LEVEL

Suppose, IR is the set of all pairs

{(k, j):  (qk, aj) = ( , , R)}. We do not know the exact movement of the head of M when M recognizes L. So, we have to take into account all possible state transitions that may result in the head moving to cell i at time t + 1.

Consider a pair (k, j) IR.

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GATES AT A GENERAL LEVEL

For the pair (k, j) IR, the action of the Turing machine M can be written as: C(i -1, j, t)  S(k , t) Hence, if  the head of M is at cell i –1

 cell i – 1 contains the character aj, and  M is in state qk, The head of M will move to the cell i if and only if: H(i -1, t)  (C(i -1, j, t)  S(k , t)).

GATES AT A GENERAL LEVEL

But we do not know the state of M or the content of cell i -1at time t. Hence, we have to consider all possible states and all possible characters in cell i -1. We can write this as:

This takes into account all possible characters aj in cell i - 1 and all possible states qk of M so that the head moves to cell i.

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GATES AT A GENERAL LEVEL

Similarly, we have to take into account the cases when the head of M is in the cell i +1 at time t and moves to cell i at time t +1. The complete specification is:

A VIEW OF THE CIRCUIT

A view of the circuit for H(i,t+1)

H(i,t+1) H(i+1,t+1)

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COMPLEXITY OF CREATING H(I,T +1)

Hence, H(i , t +1) can be expressed as a sequence of AND and OR gates of length O(|Q|||) = sm. If we consider the number of states |Q| = s of M and the size of the alphabet || = m as constants. We can compute all the H(i , t +1)s (remember, 1  i  T(n)) in O(1) time using (T(n))2 processors. This is clearly an NC algorithm.

C(I,J,T +1)

Recall that C(i , j, t +1) represents the fact that cell i contains aj at time t +1. Hence, C(i , j, t +1) = 1 if either,

 cell i contains aj and the tape head does not scan cell i at time t, or,

 the tape head writes aj in cell i at time t +1.

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C(I,J,T +1)

Hence, C(i , j, t +1) = where,

Hence, C(i , j, t +1) can be expressed as a constant- length sequence of AND and OR gates. All the C(i , j, t +1) s can be computed in O(1) time using O((T(n))2) processors.

A VIEW OF THE CIRCUIT

A view of the circuit for C(i,j,t+1)

C(i,j,t+1) C(i+1,j,t+1)

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S(K,T +1)

Recall that, S(k, t +1) = 1 if and only if the state of M is qk at time t +1. Let (k’, j) is a pair such that, .

In other words, if M is in state qk’, and the current character on the tape is aj, then M goes to state qk. Then,

S(K,T +1)

Since, 1  i  T(n), the function S(k, t +1) can be expressed by a sequence of AND and OR gates of length T(n). Such sequences can be generated in O(log n) time using O((T(n))2) processors. This concludes the construction of the CVP that is equivalent to a Turing machine M accepting an arbitrary language L  P.

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THE VALUE OF THE CIRCUIT

The value of the circuit will be given by C(1, *, T(n)). It is clear from the description that the circuit can be constructed by an NC algorithm, since T(n)is a polynomial in n. The output of the circuit is 1 if and only if M recognizes L. Since L is an arbitrary language in P, it follows that CVP is P-complete.

IMPLICATIONS

If L1 is NC-reducible to L2 and L2 is NC-reducible to L3, then

L1 is NC-reducible to L3. That is NC-reducibility is transitive. Let L be a P-complete problem. If L  NC, then NC = P.

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PROVING OTHER PROBLEMS P-COMPLETE

Let a language L be P-complete. If L is NC- reducible to another language L’  P, then L’ is also P-complete. This is used for proving P-completeness of many other problems. For example if we can reduce the CVP problem to another problem in P, that problem will also be P-complete. We conclude with a list of P-complete problems.

PROVING OTHER PROBLEMS P-COMPLETE

Ordered depth-first search : Given a directed graph G = (V, E) specified by adjacency lists and three vertices s, u, v, determine whether vertex u is visited before vertex v in the depth-first traversal of G starting at s.

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PROVING OTHER PROBLEMS P-COMPLETE

Maximum flow : Given a network with integer-valued capacities and two distinguished vertices, the source s and the sink t, determine whether the value of the maximum flow is odd.

PROVING OTHER PROBLEMS P-COMPLETE

Linear inequalities : Given an n x m matrix, and an n-dimensional vector b, all entries being integers, determine whether there exists a rational vector x such that Ax  b.