Transcript Randomized Algorithms - 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
Indian Institute of Technology, Delhi
Agenda
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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
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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.
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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.
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2-SAT Problem
φ = (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 trivial
But 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.
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Example
Given a Boolean formula φ in CNF, determine,
whether φ is satisfiable or not.!
Ex: AND of clauses
(A1 V A2 V A3 V A4)Λ (Ā1 V Ā3 V Ā4) Λ (A1 V Ā2 V Ā3 )
Each clause has at least one literal set to true.
If Ā1= true, A2 = 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.
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Algorithm
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}
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•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 MonteCarlo algorithm gives the correct answer for satisfiable systems
•There is a chance of error and we need to bound that.
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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.
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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.
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Random Walk
Let 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 := v
Given G, this is obviously trivial to implement on
a computer.
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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 the
left with probability 1 − p.
Analysis: let A 2 {0,1}n be any satisfying assignment
With probability at least ½ distance to A is reduced
With probability at most ½ distance to A is
increased
0
n
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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
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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.
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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
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Analysis contd..
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
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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
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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.
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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
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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
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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.
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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.
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3 CNF SAT
A special case of SAT that is incredibly useful in proving NPhardness 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.
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For example:
(A1 V A2 V A3 )Λ (Ā1 V Ā3 V A4) Λ (A1 V Ā2 V A3 )
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.
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Proof
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.
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3. Change every gate clause into a CNF formula. There are
only three types of clauses, one for each type of gate:
A1 = A2 ΛA3 -> (A1 V Ā2 V Ā3 ) Λ(Ā1 V A2 ) Λ (Ā1 V A1)
A1 = A2 V A3 -> ( Ā1 V A2 V A3) Λ (A1 V Ā2 ) Λ ( A1 V
Ā3 )
A1 = Ā2 -> (A1 VA2 ) Λ ( Ā1 V Ā2 )
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-> ( A1 V e V u) Λ (A1V ē V u) Λ (A1 V e V ū) Λ (A1 V
ēVū)
A1 V A2 -> (A1V A2 V e) Λ (A1 V A2V ē )
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Example
If we start with the above example formula , we obtain the following 3CNF
formula.
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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
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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 NPcomplete.
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Max 3-CNF
This problem is to find an assignment which
maximizes the number of satisfiable clauses.
Let us take for example
(A1 V A2 V Ā3) ^ (A1 V Ā2 VA4) ^ ( A2 V Ā3 V A4)
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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
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There may be dependent clauses too.
Eg. (A1 V A2 V A3) Λ ( Ā1 V A2 V A3)
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]
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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
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Approximation Algorithm
Approximation algorithms are algorithms used to find
approximate solutions to optimization problems.
Approximation algorithms are often associated with NPhard problems; since it is unlikely that there can ever be
efficient polynomial time exact algorithms solving NPhard problems, one settles for polynomial time suboptimal 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.
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Derandomization
De randomization : First devise a randomized algorithm then
argue that it can be derandomized to yield a deterministic
algorithm
Consider the following formula.
(A1 V Ā2 V A3) Λ ( Ā2 V Ā3 V A1) Λ( Ā3 V A2 V Ā1 )
The assignment for the literals in the above clauses can be
drawn in the form of a binary tree.
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Assignment of the Variables
A1=0
A2=0
A3=0
A1=1
A2=1
A2=0
A2=1
A3=0 A3=1
A3=0 A3=1
A3=1
A3=0 A3=1
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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
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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 Ā2 V A3) ... satisfied.
( Ā2 V Ā3 V A1) ... satisfied
( Ā3 V A 2 V
Ā1 ) ... not satisfied. ->
By discarding Ā1 from the clause we obtain ( A2 V Ā3 ).
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.
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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.
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• 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.
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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
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Thank You
-Sandhya S. Pillai (2007MCS3120)
-Sunita Sharma (2007MCS2927)
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