Novel analysis of transition probabilities in randomized k sat algorithm

International Journal in Foundations of Computer Science & Technology (IJFCST), Vol.4, No.6, November 2014
NOVEL ANALYSIS OF TRANSITION PROBABILITIES
IN RANDOMIZED K-SAT ALGORITHM
Abolfazl Javan1, Mohsen Jamalabdollahi2, and Ali Moeini3
1, 3 Department of Algorithms and Computation, Faculty of Engineering Science, School
of Engineering, University of Tehran, Tehran, Iran
2 Department of Electrical and Computer Engineering, Michigan Technological
University, Houghton, USA
ABSTRACT
In this paper, we propose a new analysis for randomized 2-SAT and 3-SAT algorithms, and show that we
could determine more precise boundaries for transition probability of Markov chain using Karnaugh map.
In our analysis we will show the probability that the selected literal has been flipped correctly is so close
to 2 3 and 4 7 , respectively for 2-SAT and 3-SAT with large number of variables. Then we will extend our
result to k-SAT and show that both transition probability of Markov chain in randomized algorithm for k-
SAT approaches to 0.5. Finally we use this result to determine the probability and complexity of finding the
satisfying assignment for randomized k-SAT algorithm. It will be shown that the probability of finding
satisfying assignment and its complexity respectively are within a polynomial factor of (0.9272n ) and
(1.0785n ) for satisfiable 3-SAT with n variables (for n  ).
KEYWORDS
SAT, random walk, satisfiability, Markov chain, Karnaugh map
1. INTRODUCTION
In k-SAT problem there are n Boolean variables x1 , x2 ,..., xn and an assignment to those
variables is a vector n
a  (a1, a2,...,an){0,1} . A clause C of length k is a disjunction
( ... ) 1 2 k c  l  l   l of literals, where a literal is either a variable i x or its negation xi , for
1 i  n . A k-CNF formula over variables x1 , x2 ,..., xn is the conjunction of m clauses
m c ,c ,...,c 1 2 where each clause is the disjunction of k literals, and a k-CNF formula is called
satisfiable if and only if there is an assignment of variables to truth values so that every clause
contains at least one true literal.
In random walk method, the goal is to find a satisfying assignment for given formula. The
algorithm is started with selecting an arbitrary assignment. If the assignment satisfies the formula
then we reach our goal. But if it does not, there will be some false clause with given assignment.
We randomly choose a clause between those false clauses and flip the value of one literal on that
clause. By flipping one of the literal we will have a new assignment which should consider as
candidate of our answer.
The k-SAT problem has been studied in many works, and various algorithms have been
proposed [1]. In the worst case, k-SAT problem needs to search through all 2n possible
DOI:10.5121/ijfcst.2014.4601 1

assignments where n represents the number of variables in the input formula. However some
efficient algorithm have been proposed for solving k-SAT in time less than O(2n ) [2-10].
Papadimitriou in [2] proposed the random-walk algorithm (RWA) for 2-SAT problem and showed
it could be solved in O(n2 ) . Moreover, Schöning in [3] and [4] proposed the randomized method
for 3-SAT which effectively has an expected running time of O(3 4n)  O(1.334n ) .
The main novelty of Schöning’s algorithm is to restart the algorithm after every 3n flips.
Since then, several bounds O(1.3302n ) [5], O(1.3290n ) [6], O(1.324n ) [7] and O(1.32216n )
[8] have been achieved with modifications of Schöning’s proposed method. Recently some other
approaches have been proposed for improving the complexity of the algorithm like Iwama et al.
[9] with O(1.32113n) and Hertli et al. [10] with O(1.32065n ) . However, in all of them, the
minimum probability of successful transition in Markov chain has been considered. The novelty
and improvement which has been proposed by Schöning for 3-SAT, made it potential for k-SAT
solution in general.
In Schöning’s method we should repeat this procedure 3n times until we reach the satisfying
assignment. If t X denotes the number of corrected literals at step t , then the stochastic process
X0 , X1 , X2 ,..., Xt ,... is a Markov chain with states 0,1,2,…,n which indicate the number of
correct variables in given assignment. If the value of all variables in selected assignment were
false or we reach to this state at step t , we will definitely correct one literal by flipping its value
at step t 1, therefore,
( 1 1| 0) 1    P Xt Xt (1)
Schöning in [3, 4] has analyzed that the transition probability of satisfiable k-SAT is at least 1 k ,
from state j to state j 1 and at most k 1 k for transmission to state j 1. Then, the
transition probabilities of Markov chain for 3-SAT has been defined as [11]:
2
( 1| ) 1 1      P X j X j t t (2)
3
( 1| ) 2 1      P X j X j t t (3)
3
In this paper we will show that we can find more precise boundaries for these transition
probabilities in Markov chain. At first we propose our idea for 2-SAT and 3-SAT, and then we will
extend it to popular k-SAT. To do this in section 2, the randomized algorithm for 2-SAT and 3-
SAT is reviewed. Moreover, we present the Karnaugh map for the selected clause. The Karnaugh
map helps us to find all possible combination of variables of the selected clause.
2. ANALYSIS OF 2-SAT
A Boolean formula which is given as the conjunction of a set of clauses with exactly two literals
per clause, can be an input to 2-SAT problem with n variables. A conventional method to find a
solution for 2-SAT problem is to start with an assignment, identify a clause that is not satisfied
and change the assignment, so that clause became satisfied. Suppose that the selected clause is
(A B) . Presume that the selected literal is A and its value is flipped to true. We are looking to

the probability that our decision (flipping A from false to true) would increase the number of
correct variables by one or not.
3
Figure 1. The Karnaugh map of selected clause (A B) in 2-SAT
Figure 1, shows the Karnaugh map [12] of this selected clause. Although the problem may have
more than one satisfying assignment, the goal is to find only one assignment that satisfies the
function. That combination of variables of selected clause which is the same as literals of true
assignment, could be anywhere in the blocks of Karnaugh map expect for block number ‘0’
where all literals are false for selected clause (A B) . For 2-SAT there are only 3 possible
combination of variables which are (AB, AB, AB) . Each of these combinations represents a block
of Karnaugh map which is probable for literals of true assignment. Therefore, for all possible
selected clauses we have at least one block which the true assignment could not have the same
literals (improbable block) and 2k 1 probable blocks in Karnaugh map of variables of selected
clause. We consider this case as first case which all blocks of Karnaugh map (expect improbable
block) are probable for literals of true assignment.
Moreover, we consider a second case which there are some other clauses in formula (with same
literals to selected clause) that may omit probable blocks in Karnaugh map (i.e. if the formula
also contains clause (A B) , then the block number 1 in Figure 1 is turn to an improbable block
for satisfying assignment). As an introduction of the second case, in the following, we first
analyze the first case and then will come back to second one.
2.1 First Case
Returning to our example of 2-SAT, given a selected clause (A B) for random-walk algorithm,
if we flip the value of literal A to true, the value of correct literals will be increased if the
satisfying assignment is placed at blocks 2 or 3 where literal A has the value of true. Same as
literal A, if we choose literal B and flip its value to true, the value of correct literals will be
increased if the satisfying assignment is placed at blocks 1 or 3 where literal B has the value of
true. Therefore for 2-SAT, when we choose a literal from an arbitrary selected clause and flip its
value we have 2 chance for success and one for fail. Then, the transition probabilities of Markov
chain for 2-SAT (first case) has been defined as:
( 1| ) 2 1      P X j X j t t (4)
3
( 1| ) 1 1      P X j X j t t (5)
3

2.2 Second Case
Suppose  and  respectively represent the position of satisfying assignment and the total
number of improbable blocks (1   2k 1 because we have at least one improbable blocks
and satisfying assignment) in given standard Karnaugh map of selected clause. When we flip the
value of literal A to true, the value of correct literals will be increased if the satisfying
assignment is placed in one of blocks 2 and 3 where literal A has the value of true. Then we
have:
4

P X j X j P P i P
(   1|  )  (   2,3)  (   2,3 | 

)
3
t t i
i i i

i z
i
P ( 2,3 | i , z ) P ( z | i )
P
1
3
1
1
     



    
(6)
Where i 
denotes the set of position(s) of i improbable block(s) in given standard Karnaugh


2 
1
i
1
map of selected clause with total number of elements, say z , equals to 


k
. For instance, in
Figure 1 we have 1 {0}, 2 {1,2,3}, 3 {(1,2),(1,3),(2,3)}       . Then for all elements in i 
we
have
1

( ) 2 1
   P z i
1





k
i 
. The main issue in Equation (6) is the calculation of P(  i) for
all n and m . we have:
P1  P( 1)  P({(A  B),(A  B),(A  B)} Fm | (A  B)Fm ) (7)
P P P A B F A B A B F A B F
       
 (   2)  ((  )  ,{(  ),(  )}  | (  ) 
) 2
m m m
P A B F A B A B F A B F
(( ) ,{( ),( )} | ( ) )
       
m m m
(( ) ,{( ),( )} | ( ) )
m m m
(8)
P P P A B F A B A B F A B F
       
 (   3)  ((  )  ,{(  ),(  )}  | (  ) 
) 3
m m m
(( ) ,{( ),( )} | ( ) )
       
m m m
(( ) ,{( ),( )} | ( ) )
m m m
(9)
Note that

4
1
1
i
i P
.
Then we can define:
P Number of F that include A B but do not include A B A B A B

4
M
   
 

1

m
M
  

(  ) {(  ),( , ),( 
)}


1
1
( )
1
m
Number of F that include A B
m
m
(10)

M 2k n (11)
M  4 2n . As we have considered a satisfiable formula, then it does not
5
Where M is the total number of clauses for k-SAT with n variables and we have [11-13]:
 
 
 k
Then for 2-SAT we have  
 
include all clauses {(A  B),(A  B),(A  B), (A  B)} or all combinations of other clauses,
simultaneously. Then we should adjust Equation (10) as:

 
4


M k
n
M
1
  
 
 
 
 
 
 
 




2
m k
k
M
m
m
P
2
1
1
1 (12)
Like what we defined for P( 1) in (10) we can same results for   i as:
1,2,...2 1
M k
M 2
k
2
1


 


n
  
1
2
 





 
 
 

 
 

for i k
m k
k
M
m
m i
Pi (13)
then we have:
1
1
2 1
1
1
2
1
( 1| ) 2
P X j X j P P P
t t
P
1 1 2 3





      


        

3
1
2
3
0
3
1
3
2
3
2
3
3
i
i
(14)
In Figure 2, the probability that the selected literal has been flipped correctly (or successful
transition) is depicted using Equations (13) and (14). It can be seen that by increasing of n or
decreasing of m , the value of successful transition increases because the probability that each
clause has been placed in formula on condition that other clauses with same combination of
variables has not been placed increases.
Although this probability decreases when m increases, but the increase of n is more effective.
Therefore we can see from Figure 2 which the probability of successful transition is so close to
2 3 for large n . Nevertheless with fixed n as we increase the number of m this probability
decrease and reach to its lower band, say 0.5. But there is an important point we should regard.
As we have considered that the formula is satisfiable therefore m should be limited to
satisfiability upper threshold. On the other hand we cannot consider that m increases as much as
it reaches M . Geordet in [13] has proposed an upper bound (m  (1 (ln n)0.5 )n for number of
clauses that we can have for a satisfiable 2-SAT formula with high probability. By considering
this upper bound and large n we can see that the probability of successful transition is so close to
2 3 . Therefore, the probability that our decision (flipping the value of A from false to true)
decreases the value of correct literals by one is close to 1 2 3 1 3 and we can define new
boundaries for probability of transition from state j to state j 1 and j 1 in Markov chain for
a satisfiable 2-SAT as:

6
( 1| ) 2 0.5
P X j X j for large n and m n n t t (1 (ln ) )
3
1

       (15)
( 1| ) 1 0.5
P X j X j for large n and m n n t t (1 (ln ) )
3
1

       (16)
0 10 20 30 40 50
0.67
0.66
0.65
0.64
0.63
0.62
0.61
0.6
0.59
0.58
Number of Clauses in Formula
P(Xt+1=j+1|Xt=j)
n= 20
n=50
n=100
n=200
Figure 2. Probability that the selected literal has been flipped correctly for 2-SAT with different number of
variables
3. ANALYSIS OF 3-SAT
Like what we have stated for 2-SAT, we choose and arbitrary assignment for the satisfiable 3-
SAT. After analyzing the satisfiability of an arbitrary clause, we randomly choose a clause
between those unsatisfied clauses, say (A B  C) , and pick one literal and flip its value. Again
presume that the selected literal is A and its value is flipped to true. Note that the selected clause
could be any combination of literals A,B,C and for each of these combinations one of the
blocks in Karnaugh map is improbable block and 2k 1 probable.
Figure 3. The Karnaugh map of selected clause (A B C) in 3-SAT

Figure 3, shows the Karnaugh map of these three variables. The probability that satisfying
assignment place in each other blocks (blocks number 1 through 7) are same and equals to 1 7 .
When we flip the value of literal A to true, the number of true literals will be increased if the
satisfying assignment is placed in one of blocks 4 to 7 where literal A has the value of true.
Therefore for 3-SAT, when we choose a literal from an arbitrary selected clause and flip its value
we have 4 chance for success and 3 for fail. Then, the transition probabilities of Markov chain for
3-SAT (first case) has been defined as:
( 1| ) 3 1      P X j X j t t (18)
7
( 1| ) 4 1      P X j X j t t
7
(17)
7
What we have analyzed represents an especial case of satisfiability problem where we assume
that there is no clause with the same literals as selected clause in formula. But for most of them
there is no guarantee for such presumption. In the following we will analyze the second case
which there is at least (not exactly) one improbable block in Karnaugh map of selected clause.
If  denotes the position of satisfying assignment in Karnaugh map which is showed in
Figure 3, then by using the upper bound which is proposed by Janson (m  4.596n) [14] for the
number of clauses that we can have for a satisfiable 3-SAT, then with large n and m  4.6n (we
can also consider the lower bound m  3.52n for 3-SAT proposed in [15] for more reliable
formula), we have (see index for proof):
7

P ( X  j  1| X  j )  P {4    7}  P (4    7 | 

i )
P
t t i
1

1

7
7
i
(4 7 | , ) ( | ) 4
 
P i z P z i P P
        
1
    
i i i i
1
 
4
7
7
i z i
(19)
0 10 20 30 40 50
0.6
0.58
0.56
0.54
0.52
0.5
0.48
0.46
0.44
0.42
0.4
P(Xt+1=j+1|Xt=j)
n=10
n=20
n=30
n=50
Figure 4. Probability that the selected literal has been flipped correctly for 3-SAT with different number of
variables

8
where i P
is defined in Equation (13). Therefore, the probability that our decision (flipping the
value of A from false to true) decreases the value of correct literals by one is 1 4 7  3 7 and
we can define new boundaries for probability of transition from state j to state j 1 and j 1
in Markov chain for 3-SAT problem as:
i P
Figure 4 shows the probability of successful transition for a satisfiable 3-SAT with different
number of variables using Equations (13) and (19). We should note that although the value of
successful transition probability decreases from 4 7 till m reaches its upper bound, but there are
some unsatisfiable formulas with those m which should be considered. Therefore, if the number
of all unsatisfiable formula subtracted from the denominator of Equation (13) then we have actual
value of and the probability of successful transition will be close to 4 7 .
Note that the Karnaugh map is analyzing the possible combination of three literals in our
selected unsatisfied clause and does not care about other possible value for other n  3 literals in
problem. On the other hand, each block in Karnaugh map represents all 2n3 possible assignment
which the value of literals A, B and C at those assignments are fixed to the number of that block.
Therefore, all possible satisfying assignments randomly distributed in probable blocks with same
probability.
4. ANALYSIS OF K-SAT
In k-SAT, the Karnaugh map of unsatisfied clause has 2k blocks which one of them represents
the unsatisfied combination of literals. Then we have 2k 1 blocks that the satisfying assignment
could place independently in each of them with the same probability. When we pick one literal
from unsatisfied clause and flip its value, then our decision may increase the number of correct
literals by one if the satisfying assignment is placed in one of blocks where the value of selected
literal is true. The total number of blocks in Karnaugh map that the selected literal could be true
equals 2k 1 . Therefore the transition probability of Markov chain for k-SAT are defined as:
k
2 1

1 1
 
(   1|  )  (2   2  1)  (2   2  1| 
)
 1 1

i
i
k k k k
t t P X j X j P  P   i P (22)
2 1
 

P k  k  i z  P z   i P (23)
        
1
(2 1 2 1| , ) ( | )
k
i z
i i i
Like what we have shown for k  2,3 , for a satisfiable k-SAT it can be shown that:
1
 
k
( 1| ) 2
    
k i



 P X j X j P for k
i
k
t t :
2
2 1
2 1
1
1
1 (24)

M . Using these new boundaries for
9
1
 
( 1| ) 2 
1
k
    
k i



 P X j X j P for k
i
k
t t :
2
2 1
2 1
1
1
1 (25)
Where i P
is defined in Equation (13). Figure 5 shows the probability of successful transition for a
satisfiable 3-SAT with different number of variables using Equations (13) and (24). Figure 5
shows that the probability of successful transition reaches to 0.5 as k increases. For Figure 5 we
consider n  5k and the number of k and m is limited (for instance at k  10, we depict P for
m<27) due to the limitation of calculating the value of  
 
m
transition probability the Markov chain for k-SAT is depicted in Figure 6.
5. ANALYSIS OF SUCCESS PROBABILITY AND COMPLEXITY OF
K-SAT
Here, we aim to analysis the probability that the algorithm finds the satisfying assignment (or
some other satisfying assignment is found). Using this probability we can demonstrate the
expected number of independent repetitions of the procedure until we find a satisfying assignment
[3].
If Xt denotes the number of corrected literals at step t , then the transition probability from start
(t  0) to step j in Markov chain (or the probability that our selected assignments has j correct
literals), say X  j 0 , equals to:
n j for j n
j n X P j X P n    
 
(  )  (   )  2 : 0 0 0 (26)
Suppose that we aim to run the algorithm for at most 3n times like [3]. Then the probability that
the algorithm finds the satisfying assignment at final step or before it (t  3n) could be defined
as [3-4, 16]:
t q j n

  
 
(   3 ) 
2 , (27)
n
j
i j
n
P X n for t n
0
Where qi j , denotes the probability that we reach to final step or X n t  , if we started from state
n  j at Markov chain, and we limit the algorithm to move i (i  j) steps in wrong direction
(with this, we limit the number of steps to 3n). Therefore we need i  j step in the right direction
to reach state n from n  j and the probability can be estimated as follows:
        i
(28)
j
P(X n for all t 2i j | X n j) q 2i j (1 )
0 , 
i
i j



t i j i j P P
0
  

Further we can lower bound the above sum by its largest term by considering i  j , then:
j q   
P ( 1 P ) r ( 1 ) 2 , j
3 3
  (29)
 
j j r r
2 2 3 3

i j r P P
  


1 (30)
( 3 ) 2 3 (32)
( 3 ) 1 (33)
10
For 0   1 2 , we have [5]:
r
r
r r

 

 


1 1








  
 
 
    
1
1
8 (1 )
Then we can write:
r
3
 
1
 
 2 3 1 3

q P (1 P
)
i , j 2 3 r



2
8 (1 )
(31)
By inserting Equation (20) into (16) we have:
t P P j n


   



 
  
n
j
j
n
P X n for t n
0
3
2 3 1 3
2 3 (1 )
2
With j  n 3 we have:
n

2
3
   2 3  1 3

P X n for t n P P
t p n


7 3
(1 )
2
( )
Where p(n) is a polynomial factor of n and therefore, the probability that we reach the final
state (find the satisfying statement) for k-SAT with large n is:
n
(1 2
k
k
2
k
k
1
p n
P

 


 



  

 


 
1 3
1 2 3 1
2
3
7 3
)
2 1
2 1
2
( )
(34)
0 10 20 30 40 50
0.52
0.518
0.516
0.514
0.512
0.51
0.508
0.506
0.504
0.502
0.5
P(Xt+1=j+1|Xt=j)
k=5
k=6
k=8
k=10
Figure 5 Probability that the selected literal has been flipped correctly for k-SAT with n  5k

11
Figure 6.Markov chain for k-SAT (k,n) and its transition probabilities
For 3-SAT the probability of finding satisfying statement and its complexity respectively are:

1
1 3
7 3
) 1
7
4
3
2
  for n
p n p n p n
    

 
P 
n

 

 



n n
(0.9272) :
( )
14
( )
(3
7
2
( )
1 3
2 3
7 3
(35)
T (n)  p(n)O(1.0785n ) for : n  (36)
For k,n  we can consider P  2k1 (2k 1)  0.5 then for k-SAT we have:

   

 
3
1
2
 for k n
p n
P
n
: ,
2
( )
10 3
(37)
T (n)  p(n)O(1.1199n ) for : k,n (38)
6. CONCLUSION
Using Karnaugh map of selected unsatisfied clause, we proposed a new analysis of transition
probability in randomized algorithm for a satisfiable k-SAT and showed that we can achieve more
precise boundaries for them. Moreover, we used these new boundaries to find the probability of
increasing of correct literals when we flip the value of selected literal from unsatisfied clause in
randomized algorithm. Finally we used these probabilities to analysis of expected complexity of
reaching satisfied assignment for k-SAT randomized algorithm.
APPENDIX
Here, we calculate the probability that the of satisfying assignment places in one of blocks 4 to 7
based on Karnaugh map of Figure 3. Based on the definition of part 2 we have:
7
        
P{4  7} P(4  7 |  i, z  )P(z  | 
i)P
i i i
i 
1
z
For i  1 the set of possible blocks that are improbable is {0} 1   , the we have:
1 4
7
.
7
 P (4    7 |   1, z   ) P ( z   | 
 1)  4  
1 1 z

12
For i  2 we have {1,2,3,4,5,6,7} 2   , then:
4
.
7
1
7
3
6
1 4
1
7
4
6
1 3
(4 7 | 2, ) ( | 2) 2 2     
 
    
 
z
P      z   P z   
  For i  3 we have:
{ 7 3 to positions of ns combinatio all  
2 1 7}
 
 
then:
4
.
7
1
21
      z
2
5
2 4
1
21
3
5
4
1
3
1
1
21
4
5
2 3
(4 7 | 3, ) ( | 3) 3 3
    
 
    
 
  
 
    
 

P   z  P z  
For i  4 we have:
then:
0 4 4 4
.
7
1
35
1
3
3 4
      z
1
1 3 35
2
3
2 4
2 3
1 1 1
35
4
3 3
(4 7 | 5, ) ( | 5) 5 5
   
 
    
      
    
 
  
 
    
 
  
 

P   z  P z  
} 7 1 5 7
{ 6 to positions of ns combinatio all  
 
 
then:
      z
(4 7 | 6, ) ( | 6) 6 6
0 4 4 4
1
.
7
3
1
21
1
2
3 4
2 3
1 1 2 4 3 3
21
   
 
  
 
    
 
  
 
      
  
 

P   z  P z  
And finally, for i  7 we have:

13
{ 7 7 to positions of ns combinatio all  
6 1 7}
 
 
then:
      z
0 4 4 4
.
7
2 3
1 1 3 4
7
3 3
(4 7 | 7, ) ( | 7) 7 7
   
 
  
 
    
 
  
 

P   z  P z  
Therefore the probability of increasing the correct literals by flipping the value of each literal in
selected unsatisfied clause for 3-SAT is:
.
{4 7} 4
7
7
1 
  
i
i P
P 
REFERENCES
[1] Jun Gu, Paul W. Purdom, John Franco, and Benjamin W. Wah, (1997), "Algorithms for the
Satisfiability (SAT) Problem: A Survey In Satisfiability Problem, Theory and Applications", Vol. 35
of DIMACS: Series in Discrete Mathematics and Theoretical Computer Science, American
Mathematical Society, pp 19–151.
[2] Christos H. Papadimitriou, (1991), "On selecting a satisfying truth assignment", In Proceeding of the
32nd Ann Symposium on Foundations of Computer Science (FOCS), pp 63–169.
[3] Uwe Schöning. 1999. "A Probabilistic Algorithm for k-SAT and Constraint Satisfaction Problems",
In Proceedings of the 40th Annual Symposium on Foundations of Computer Science (FOCS), IEEE,
pp 410–414.
[4] Uwe Schöning, (2002), "A Probabilistic Algorithm for k-SAT Based on Limited Local Search and
Restart", Algoritmica, Vol. 32, pp 615–623.
[5] Thomas Hofmeister, Uwe Schöning, Rainer Schuler, and Osamu Watanabe, (2002), "A probabilistic
3-SAT algorithm further improved", In Proceedings of the 19th Symposium on Theoretical Aspects of
Computer Science (STACS), 2285, pp 192–202.
[6] Sven Baumer and Rainer Schuler, (2003), "Improving a probabilistic 3-SAT Algorithm by Dynamic
Search and Independent Clause Pairs", Electronic Colloquium on Computational Complexity, Report
No. 10.
[7] Kazuo Iwama and Suguru Tamaki, (2003), "Improved Upper Bounds for 3-SAT", Electronic
Colloquium on Computational Complexity, Report No. 53.
[8] Daniel Rolf, (2006), "Improved Bound for the PPSZ/ Schöning-Algorithm for 3-SAT", Journal on
Satisfiability, Boolean Modeling and Computation, 1:111-122.
[9] Kazuo Iwama, Kazuhisa Seto, Tadashi Takai, and Suguru Tamaki, (2010), "Improved randomized
algorithms for 3-SAT", In Algorithms and Computation, Lecture Notes in Computer Science, 6506,
pp 73-84.
[10 ]Timon Hertli, Robin A. Moser, and Diminik Scheder, (2011), "Improving PPSZ for 3-SAT using
Critical Variables", http://arxiv.org/abs/1009.4830.
[11] Michael Mitzenmacher, Eli Upfal, (2005), "Probability and Computing Randomized Algorithms and
Probabilistic Analysis", Cambridge University Press, First published.
[12] M. Morris Mano, (2002), "Digital Design", Prentice Hall, Third Edition.
[13] Andreas Goerdt, (1996), "A threshold for unsatisfiability", Journal of Computer and System Science,
Vol. 53, No. 3, pp 469-486.
[14] Svante Janson, Yannis C. Stamatiou, and Malvina Vamvakari, (2000), "Bounding the unsatisfiability
threshold of random 3-SAT", Random Structure and Algorithm, Vol. 17, No. 2, pp 103-116.
[15] MohammadTaghi Hajiaghayi, and Gregory B. Sorkin, (2003), "The Satisfiability Threshold of
Random 3-SAT Is at Least 3.52", http://arxiv.org/abs/math/0310193.

[16] Stasys Jukna, (2011), "Extremal Combinatorics: With Applications in Computer Science", Springer,
14
Second Edition.
Authors
Abolfazl Javan:
He received his B.S. in Software Engineering from Ferdowsi University of
Mashhad, Mashhad, Iran, and M.S. degree in Algorithms and Computation from
University of Tehran, Tehran, Iran. He is currently a research assistant at
University of Tehran.
His research interests include Graphs Theory, Randomized Algorithms,
Approximation Algorithms, and Bioinformatics.
Mohsen Jamalabdollahi:
He received his B.S. from University of Mazandaran and M.S. degree from K. N.
Toosi University of Technology, Tehran, Iran. He is currently a Ph.D. student at
Michigan Technological University in electrical engineering.
His research interests include Localization of Wireless Sensor Network,
Randomized Algorithms, Software Defined Radio for Mobile Communications,
Embedded System Design, MIMO-OFDM(A) and Equalization.
Ali Moeini:
He received his PhD in Nonlinear Systems at the University of Sussex, UK, in
1997. He is an Associate Professor of Electrical & Computer Engineering at
Faculty of En¬gineering Science, School of Engineering, the University of Tehran,
and ad joint Professor at Department of Information Technology Management,
Faculty of Management, the University of Tehran in Iran.
His research interests include Formal Methods in Knowledge and Software
Engineer¬ing and Knowledge Management, Soft computing Methods,
Randomized Algorithms, Approximation Algorithms, and Bioinformatics.
He has published many papers in Journals and Conferences on the above mentioned topics.

Novel analysis of transition probabilities in randomized k sat algorithm

More Related Content

What's hot (18)

Viewers also liked (20)

Similar to Novel analysis of transition probabilities in randomized k sat algorithm (20)

More from ijfcstjournal (20)

Recently uploaded (20)

Novel analysis of transition probabilities in randomized k sat algorithm