randomized algorithms (2-sat & max-3-sat) (14/02/2008)

Indian Institute of Technology, Delhi

Randomized Algorithms(2-SAT & MAX-3-SAT)

(14/02/2008)

Sandhya S. Pillai(2007MCS3120)Sunita Sharma(2007MCS2927)

Advanced Algorithms Course (CSL 758)Prof. Kavitha Telikepalli

Prof. Naveen Garg


Agenda•Introduction•2-SAT Problem•Monte Carlo vs. Las Vegas methods•Analysis of 2-SAT•Random Walks and Markov inequality•3-SAT Problem•Why 3-SAT is NP-Hard?•Max 3-SAT Problem randomization & de-

randomization•References


Randomized Algorithm

In addition to input, algorithm takes a source of random

numbers and makes random choices during execution.

Behavior can vary even on a fixed input.


SAT

Satisfiability (SAT) is the problem of deciding

whether a boolean formula in propositional logic has

an assignment that evaluates to true. SAT occurs as a

problem and is a tool in applications (e.g. Artificial

Intelligence and circuit design) and it is considered a

fundamental problem in theory, since many problems

can be naturally reduced to it and it is the 'mother' of

NP-complete problems.


2-SAT Problemφ = (x1 V~x2) ^ (x1 V ~x3) ^ (~x1 V x3) ^ (~x1 V ~x2)The k-SAT problem is the variant of SAT, in which each clause consists of exactly k distinct literals. For k >= 3, k-SAT is NP hard, but for k = 1 and 2, there are polynomial time solutions. For k = 1, this solution is trivialBut for k = 2, it is slightly tricky. There is a much

easier random algorithm.For a system with literals Xi, 0 <= i < m, and clauses

Cj, 0 <= j < n, it goes as follows:SAT:Given a boolean formula ф in

CNF,determine if ф is satisfiable or not.


Example

Given a Boolean formula φ in CNF, determine, whether φ

is satisfiable or not.!

Ex: AND of clauses

(A1 V A

2 V A

3 V A

4)Λ (Ā

1 V Ā

3 V Ā

4) Λ (A

1 V Ā

2 V Ā

3 )

Each clause has at least one literal set to true.

If Ā1= true, A

2 = true then

Clause 1 and 2 are satisfiable but the Clause 3 requires

A3 literal set to be false for it to be a satisfiable one.


Input : A Boolean formula φ in conjunctive normal form with exactly two distinct literals in every clause. E.g., φ = (x1 V~x2) ^ (x1 V ~x3) ^ (~x1 V x3) ^ (~x1 V ~x2)1)/* Start with an arbitrary initial assignment to the literals*/for (i = 0; i < m; i++) xi = true; Let us call this assignment A

2)/* here function number_satisfied() returns the number of satisfied clauses for the current assignment */for (t = 0; t < T && number_satisfied(x, n) < n; t++) select an arbitrary non-satisfied clause CJ;

randomly and uniformly pick one of the literals xi in CJ ; xi = (xi + 1) mod 2;

3) If (number_satisfied(x, n) == n) report that the set of clauses is satisfiable ; else report that the set of clauses is not satisfiable

Algorithm


•Function number_satisfied() can be computed in linear time,

some savings might be achieved by only keeping track of the

changes between rounds. However, computation time is not so

much the issue here, the main point is answering the question

how large T must be taken to be reasonable sure that this

Monte-Carlo algorithm gives the correct answer for satisfiable

systems

•There is a chance of error and we need to bound that.


Monte Carlo vs. Las Vegas methods• Definition A Las Vegas algorithm is a randomized

algorithms that always return the correct result. The only variant is that it’s running time might change between executions.– An example for a Las Vegas algorithm is the QuickSort

algorithm.

• Definition A Monte Carlo algorithm is a randomized algorithm that might output an incorrect result. However, the probability of error can be diminished by repeated executions of the algorithm.– The MinCut algorithm is an example of a Monte Carlo

algorithm.


Bounding the Error in Algorithm

•This algorithm will always give correct non-satisfiable instances of ф

•But if ф is satisfiable Then we need to fix value of T such that this algorithm says “not satisfiable “ with a probability <=1/4

• we will assume in the following that there is a satisfiable assignment to the xi Let call it S There may even be several

such assignments, but we concentrate on a single one, which we will refer to as the correct assignment.

• The above process can be modeled as a random walk. The graph is the line graph with n nodes: node y is connected to node y - 1 and node y + 1, as far as these indices are at least 1 and at most n.


Random WalkLet G = (V,E) be a connected, undirected graph. A

random walk on G, starting from

vertex s ∈ V , is the random process defined as follows:

u := s

repeat for T steps

choose a neighbor v of u uniformly at random.u := vGiven G, this is obviously trivial to implement on a

computer.


Consider a particle moving in an one-dimensional line. At each point in time, the particle will move either1 step to the right with probability p or 1 step to theleft with probability 1 − p.Analysis: let A 2 0,1n be any satisfying assignment

With probability at least ½ distance to A is reducedWith probability at most ½ distance to A is increased

0 n


Randomized 2-SAT Analysis•Distance can never be larger than n

– if it starts at some 0 < i < n

– Dominated by a walk where

– With probability exactly ½ distance to A is reduced

– With probability exactly ½ distance to A is increased


Analysis contd..

•Let us define a random variable Xi

Xi=# of steps to reach state n starting from state i

Xi=1 + # of steps to reach state n starting from state

i+1 with probably ½ (Xi+1)

Xi=1 + # of steps to reach state n starting from state i-1

with probably ½ ( Xi-1)

this is a memory less property There for start from the

current state is a fresh start.


Analysis contd..

E[Xi]=1/2 E[1+Xi+1]+1/2E[1+ Xi-1]

= 1+(E[ Xi-1 ]+E[Xi+1])/2

Let Si=E[ Xi ] Then

Si=1+( Si-1+Si+1)/2

Sn-1=1+( Sn-2 + Sn)/2


S0=S1+1 ............(1)

2S1=S0+S2+2 .............................(2)

.

.

2Sn-1=Sn-2+2..............................(n)

adding equations 1 to n we get

Sn-1=2(n-1)+1=2n-1

Sn-2=(2n-1)+(2n-3)

.

S0=(2n-1)+(2n-3)+..............+3+1=n2

Analysis contd..


Analysis contd..Let us define another random variable Y

Y= # of variables whose truth value is same in

current assignment(A) & satisfiable assignment(S)

Movement of Y is similar to

i-1 ← i → i+1

Now we have to fix T


Markov Inequality

Markov's inequality gives an upper bound for the probability

that a non-negative function of a random variable is greater

than or equal to some positive constant

Proposition 1:For any non-negative random variable Y and

any real

number k >0 we have

Pr [Y>=k] <= E [Y] / k

As an example let k = 2*E[ Y ]. Then the above says

Pr [Y>= 2 *E[ Y ] ] <= 1/2. Namely, if you move out to

twice the expectation, you can have only half the area under the

curve to your right. This is quite intuitive.


Proof for Markov Inequality

E[Y]=∑ Pr[Y=y]*y

= ∑ Pr[Y=y]*y+∑ Pr[Y=y]*y y<k y>=k

>= 0 + k* Pr [ Y>=k ]

=Pr [ Y>=k ]<=E [ Y ] / k

Therefore T=4n2


Probability

Pr[φ is satisfiable but algorithm returns unsatisfiable]

=Pr[our algorithm does not reach state n in 4n2 steps]

= Pr[ Z>=4E[Z] <= ¼ ]

Z=# of steps to reach state n from our start state

E[Z]<= n2


Practical Example of Satisfiability

Circuit satisfiability is a good example of this problem that we

don't know how to solve in polynomial time. In this problem, the

input is a boolean circuit: a collection of and, or, and not gates

connected by wires. We will assume that there are no loops in the

circuit (so no delay lines or ip-ops). The input to the circuit is a

set of m boolean (true/false) values x1; : : : ; xm. The output is a

single boolean value. Given specific input values, we can

calculate the output in polynomial (actually, linear) time using

depth-first-search and evaluating the output of each gate in

constant time.


Why NP hard ?

The circuit satisfiability problem asks, given a circuit, whether

there is an input that makes the circuit output True, or

conversely, whether the circuit always outputs False. Nobody

knows how to solve this problem faster than just trying all 2m

possible inputs to the circuit, but this requires exponential

time. On the other hand, nobody has ever proved that this is

the best we can do; maybe there's a clever algorithm that

nobody has discovered yet!

Hence this comes under NP hard problem.


3 CNF SAT

A special case of SAT that is incredibly useful in proving NP-

hardness results is 3SAT (or 3-CNF-SAT).

A boolean formula is in conjunctive normal form (CNF) if it

is a conjunction (and) of several clauses, each of which is the

disjunction (or) of several literals, each of which is either a

variable or its negation.


For example:

(A1 V A

2 V A

3 )Λ (Ā

1 V Ā

3 V A

4) Λ (A

1 V Ā

2 V A

3 )

Given such a boolean formula, can we come up with

an algorithm, that is polynomial in time ? The answer

to this question is NO!!! Hence this is NP hard

problem.


We could prove that 3SAT is NP-hard by a reduction from the

more general SAT problem,but it's easier just to start over

from scratch, with a boolean circuit. We perform the reduction

in several stages.

1. Make sure every and and or gate has only two inputs. If

any gate has k > 2 inputs, replace it with a binary tree of k-1

two-input gates.

2. Write down the circuit as a formula, with one clause per

gate. This is just the previous reduction.

Proof


3. Change every gate clause into a CNF formula. There are

only three types of clauses, one for each type of gate:

A1 = A

2 ΛA

3 -> (A

1 V Ā

2 V Ā

3 ) Λ(Ā

1 V A

2 ) Λ (Ā

1 V A

1)

A1 = A

2 V A

3 -> ( Ā

1 V A

2 V A

3) Λ (A

1 V Ā

2 ) Λ ( A

1 V

Ā3 )

A1 = Ā

2 -> (A

1 VA

2 ) Λ ( Ā

1 V Ā

2 )

4. Make sure every clause has exactly three literals. Introduce

new variables into each one- and two-literal clause, and

expand it into two clauses as follows:

A1-> ( A

1 V e V u) Λ (A

1V ē V u) Λ (A

1 V e V ū) Λ (A

1 V

ē V ū )

A1 V A

2 -> (A

1V A

2 V e) Λ (A

1 V A

2V ē )


Example

If we start with the above example formula , we obtain the following 3CNF

formula.


Although the 3CNF formula is complicated than the

original one at first glance, it's actually only a constant

factor larger. Even if the formula were larger than the

circuit by a polynomial, like n373, we would have a valid

reduction.

The formula is satisfiable if and only if the original circuit

is satisfiable. As with the more general SAT problem, the

formula is only a constant factor larger than any

reasonable description of the original circuit, and the

reduction can be carried out in polynomial time. Thus, we

have a polynomial-time reduction from circuit satisfiability

to 3SAT


TCSAT

(n) <= O(n) + T3 SAT

(O(n)) => T3 SAT

(n) >=

TCSAT

(( ᾨ (n)) - O(n)

So 3SAT is NP-hard. Finally, since 3SAT is a

special case of SAT, it is also in NP, so 3SAT is NP-

complete.


Max 3-CNF

This problem is to find an assignment which

maximizes the number of satisfiable clauses.

Let us take for example

(A1 V A

2 V Ā

3) ^ (A

1 V Ā

2 VA

4) ^ ( A

2 V Ā

3 V A

4)


Randomized Algorithm for Max 3-CNF

Set each variable to true with probability 1/2

independently. (i.e. for instance, toss the coin and if head,

set the variable to true and if tail, set the variable to false).

If a clause is not satisfied, this means all the 3 variables in

the clause are false.

Prob. that a clause is not satisfied = 1/2 * 1/2 * 1/2

(since each variable is independently set to false)

Prob. that a clause is satisfied = 1 - Prob. that a clause is

not satisfied

= 1 - 1/8 = 7/8


There may be dependent clauses too.

Eg. (A1 V A

2 V A

3) Λ ( Ā

1 V A

2 V A

3)

In the above two clauses the first clause is true and hence the second

clause is bound to be true if x2 = true or x3 = true or both are true.

Each clause is satisfied with prob. 7/8

Let m be the no. of clauses.

Expected number of satisfiable clauses can be found as follows.

Let Xi be a random variable.

Xi = 1 if ith clause is satisfied.

= 0 otherwise

X = ∑i Xi

Expected value of X

E[X] = E[X1 + X2 + ... + Xm]


Linearity of expectation: Let r be any real number and let X and

X1 , . . . , Xn be random variables on a discrete probability space

Ω such that their expectations all exist. Then the expectation of

rX and of X1 +. . .+Xm exists and we have

E [rX] = rE [X] and E [X1 + . . . + Xn ] = E [X1 ] + . . . +

E [Xn ] .

Since linearity of expectations always holds,

E[X] = E[X1] + E[X2].........+ E[Xm]

= 7/8 + 7/8 + ............+ 7/8

= 7/8* m


Approximation AlgorithmApproximation algorithms are algorithms used to find

approximate solutions to optimization problems.

Approximation algorithms are often associated with NP-

hard problems; since it is unlikely that there can ever be

efficient polynomial time exact algorithms solving NP-

hard problems, one settles for polynomial time sub-

optimal solutions.

This is termed as 7/8th approximation algorithm because,

if the MAX-3SAT is satisfiable, then the expected weight

of the assignment found is at least 7/8 of optimal clauses.


DerandomizationDe randomization : First devise a randomized algorithm then

argue that it can be derandomized to yield a deterministic

algorithm

Consider the following formula.

(A1 V Ā

2 V A

3) Λ ( Ā

2 V Ā

3 V A

1) Λ( Ā

3 V A

2 V

Ā

1 )

The assignment for the literals in the above clauses can be

drawn in the form of a binary tree.


Assignment of the Variables

A1=1A1=0

A2=1A2=0A2=0 A2=1

A3=0A3=1

A3=0 A3=1A3=0 A3=1

A3=0 A3=1


At each leaf node, the number of clauses satisfied by the assignment

of the variables is shown on selection of the particular path towards

that leaf. If n variables, then the tree will consist of n levels.

If n levels are there, 2n leaves are there. Thus 2n assignments are

possible. The average of the numbers on the leaves = 7/8 * m

In 7/8 of the leaves, the clause will be contributing for appearing in

the leaf.There are 2n leaves. Each clause contributes 1 to 7/8 * 2n

leaves

Sum of the numbers on leaves = 7/8 * 2n * m = 21 (Since m = 3 in

the example here)

Avg. of the numbers on leaves = (7/8 * 2n * m)/ 2n = 21/8=2.65


Here, starting from the root, first compute the average at each node. Then

pick the path having greater average.

The process for computing the average is

Consider the root assignment A1=0, A1 = 1

if A1 = 1 , compute the satisfiable clauses probability.

i.e. (A1 V Ā

2 V A

3) ... satisfied.

( Ā2 V Ā

3 V A

1) ... satisfied

( Ā3 V A

2 V

Ā

1 ) ... not satisfied. ->

By discarding Ā1 from the clause we obtain ( A

2 V Ā

3 ).

The probability that this clause is not satisfied is 1/4. So the probabilitythat

this clause is satisfied is ¾ . Thus the total probability turns out to be (1 +

1 + ¾) =1+1/4. This is the average when A1 = 1

Similarly compute the average for A1 = 0. which comes out to be 10/4.

Choose the greater average and move towards that path. So we move

towards A1 = 1 path.


Now compute the average by taking the assignment A2=0 and

A2=1 and move towards the path with greater average.

Thus at each level i, we have to compute the average, with

assignment true, and with false.

Thus, on reaching the leaf, we would have the average

number of clauses which will be satisfiable with the

assignment on the chosen path.


• In the above algorithm greedy approach is been

followed, at each level we check which sub tree will

give the best average & we take decision according to

the current maximum. Thus algorithm may result in an

sub-optimal result.


References

•Randomized Algorithmes: R.Motwani

• Chapters 1,5,6,10

•Introducton to Algorithms: CLRS

• Chapter 5

•A compendium of NP webpage http://www.nada.kth.se/~viggo/problemlist/compendium.html


Thank You

-Sandhya S. Pillai (2007MCS3120)

-Sunita Sharma (2007MCS2927)

randomized algorithms (2-sat & max-3-sat) (14/02/2008)

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