Randomized Algorithms (2-SAT

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Randomized Algorithms(2-SAT MAX-3-SAT)(14/02/2
008)

• Sandhya S. Pillai(2007MCS3120)
• Sunita Sharma(2007MCS2927)
• Advanced Algorithms Course (CSL 758)
• Prof. Kavitha Telikepalli Prof. Naveen Garg

2
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

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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

• f (x1 Vx2) (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 gt 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 lt i lt m, and
  clauses Cj, 0 lt j lt n, it goes as follows
• SATGiven a boolean formula ? in
  CNF,determine if ? is satisfiable or not.

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Example

• Given a Boolean formula f in CNF, determine,
  whether f is satisfiable or not.!
• Ex AND of clauses
• (A1 V A2 V A3 V A4)? (A1 V A3 V A4) ? (A1 V
  A2 V A3 )
• Each clause has at least one literal set to true.
• If A1 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 f in conjunctive normal
  form with exactly two distinct literals in every
  clause. E.g., f (x1 Vx2) (x1 V x3) (x1
  V x3) (x1 V x2)
• 1)/ Start with an arbitrary initial assignment
  to the literals/
• for (i 0 i lt 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 lt T number_satisfied(x, n) lt 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 Monte-Carlo 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 its 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 ?
• But if ? is satisfiable Then we need to fix value
  of T such that this algorithm says not
  satisfiable with a probability lt1/4
• 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 0,1n be any satisfying
assignment With probability at least ½ distance
to A is reduced With probability at most ½
distance to A is increased
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Randomized 2-SAT Analysis

• Distance can never be larger than n
• if it starts at some 0 lt i lt 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
• Xi1 of steps to reach state n starting from
  state i1 with probably ½ (Xi1)
• Xi1 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..

• EXi1/2 E1Xi11/2E1 Xi-1
• 1(E Xi-1 EXi1)/2
• Let SiE Xi Then
• Si1( Si-1Si1)/2
• Sn-11( Sn-2 Sn)/2

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Analysis contd..

• S0S11 ............(1)
• 2S1S0S22 .............................(2)
• .
• .
• 2Sn-1Sn-22..............................(n)
• adding equations 1 to n we get
• Sn-12(n-1)12n-1
• Sn-2(2n-1)(2n-3)
• .
• S0(2n-1)(2n-3)..............31n2

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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 ? i1
• 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 1For any
non-negative random variable Y and any
real number k gt0 we have Pr Ygtk lt E Y /
k As an example let k 2E Y . Then the above
says Pr Ygt 2 E Y lt 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
EY? PrYyy ? PrYyy? PrYyy
yltk ygtk gt 0 k Pr Ygtk Pr Ygtk
ltE Y / k Therefore T4n2
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Probability

• Prf is satisfiable but algorithm returns
  unsatisfiable Prour algorithm does not reach
  state n in 4n2 steps
• Pr Zgt4EZ lt ¼
• Z of steps to reach state n from our start
  state
• EZlt 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 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.

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For example

• (A1 V A2 V A3 )? (A1 V A3 V A4) ? (A1 V A2
  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 gt 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 ?A3 -gt (A1 V A2 V A3 ) ?(A1 V A2 )
  ? (A1 V A1)
• A1 A2 V A3 -gt ( A1 V A2 V A3) ? (A1 V
  A2 ) ? ( A1 V A3 )
• A1 A2 -gt (A1 VA2 ) ? ( A1 V A2 )
• 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-gt ( A1 V e V u) ? (A1V e V u) ? (A1 V e
  V u) ? (A1 V e V u )
• A1 V A2 -gt (A1V A2 V e) ? (A1 V A2V e )

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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) lt O(n) T3 SAT(O(n)) gt T3 SAT(n) gt
  TCSAT(( ? (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.

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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 A3) (A1 V A2 VA4) ( A2 V A3 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) ? ( A1 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
• EX EX1 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 ? 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,
• EX EX1 EX2......... EXm
• 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 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.

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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 A2 V A3) ? ( A2 V A3 V A1) ?( A3 V A2
  V A1 )
• 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
A10
A11
A20
A21
A21
A20
A30
A31
A30
A30
A31
A31
A31
A30
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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 21/82.65

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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 A10, A1 1
• if A1 1 , compute the satisfiable clauses
  probability.
• i.e. (A1 V A2 V A3) ... satisfied.
• ( A2 V A3 V A1) ... satisfied
• ( A3 V A2 V A1 ) ... not
  satisfied. -gt
• By discarding A1 from the clause we obtain (
  A2 V A3 ).
• 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) 11/4. 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
  A20 and A21 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)