A DNA-based Algorithm for the Solution of One-In-Three 3-SAT Problem
Nung-Yue Shi and Chih-Ping Chu
Department of Computer Science and Information Engineering,
National Cheng Kung University,
Tainan City 701, Taiwan, Republic of China
E-mail: p7890112@ccmail.ncku.edu.tw
Abstract
and
chucp@csie.ncku.edu.tw
find all true assignment (3 SAT problem) and find
One-In-Three
(1IN3)
solutions
on
DNA-based
Satisfiability problem is given a Boolean formula,
Supercomputing. The complexity of the presented
and decide if a satisfying truth assignment exists.
DNA-based solution is discussed in section three. And
( x12 
x5 )  ( x 24  x3  x13  x 9 )  … 
the simulated experiment is applied to verify
x12 )  ( x17  x8  x18 ) is an example of
solving the One-In-Three 3-SAT problem in section
(
correction of the proposed DNA-based algorithm for
four.
Boolean formula. k-SAT means that each clause has
exactly k literals. One-In-Three (1IN3) 3-SAT
Key Words: Satisfiability problem, 3-SAT problem,
problem is defined by Garey and Johnson in 1979 as
NP-complete problem, One-In-Three 3-SAT problem,
follows: There is a set V of variables and a collection
Molecular Solution, DNA-based Supercomputing.
C of clauses over V such that each clause has 3 literals.
And the question is : Is there a truth assignment for V
1. Introduction
such that each clause in C has exactly one true literal?
The words “computer” reminds us a lot of
Note that the only difference with 3-SAT is that each
electronic components like semiconductor memory,
clause in C should be one of the three forms which
CPU, hard disk …etc. But must the computer be this
could be 100, 010, or 001 since it has three literals and
way? Adleman propose a new concept of computation
exaclly one true literal ( the other two literals are of
in the molecular level at his paper in 1998 [1]. The
course false ) .
DNA molecules with their sequences of adenine,
thymine, guanine and cytosine ( represented by the
In this paper, we will use molecular solution to
letters A, T, G, C ) can be used to store information
and perform computation. But actually how is the
6. DNA synthesis. Nowadays, we could ask a
biology related to computer science [5]?
commercial DNA synthesis facility to make the DNA
We could make a molecular computer with the
sequence. Just in a few days, we will receive a test
tube containing about 1018 molecules of DNA which
tools as the following:
is the sequence we ask.
1. Watson-Crick complements. Two strands of DNA
will anneal to form a famous double helix if the
The above six techniques is the basis of
respective base meets its Watson-Crick complements
Adleman-Lipton DNA computing modle.
which are C matches G and A matches T. Of course, if
From which, Adleman developed eight bio-molecular
a molecule of DNA meets another DNA molecule
instructions to perform bio-molecular programs. That
which is not its complement, then they will not
is what we mention next section.
anneal.
2. Ligases. Ligases bond the splitted DNA molecule
2. background
together. For example, DNA ligase will take two
strands of DNA molecule and covalently connect
A test tube contains molecules of DNA which is
them into a single strand. In fact, ligase is used by the
a finite set over the alphabet {A, C, G, T}, we could
cell to repair the broken DNA strands.
perform the following operations [2  10 ]:
1. Append-tail. Given a tube T and a binary digit xj,
3. Nucleases. Nucleases would cut nucleic acid of a
the operation, "Append-tail", will append xj onto the
DNA molecule. For example, nucleases would look
end of every data stored in the tube T. The formal
for a predetermined sequences of bases of a strand of
representation for the operation is written as
DNA molecules, if found, would cut the DNA strands
"Append-tail(T, xj)".
into two pieces.
2. Amplify. Given a tube T, the operation “Amplify(T,
4. Polymerases. Polymerases copy information from
T1, T2)” will produce two new tubes T1 and T2 so that
one DNA molecule into the other. Furthermore, DNA
T1 and T2 are totally a copy of T (T1 and T2 are now
polymerases
identical) and T becomes an empty tube.
will
make
a
Watson-Crick
complementary copy from a DNA strand template. In
fact, if we tell it where to start—that is a primer
3. Merge. Given n tubes T1  Tn, the merge operation
provided by a short piece of DNA strand, DNA
is to merge data stored in any n tubes into one tube,
polymerase will begin adding bases to the primer to
without any change in the individual data. The formal
create a complementary copy of the template.
representation for the merge operation is written as
"(T1, , Tn)", where (T1, , Tn) = T1    Tn.
5. Gel electrophoresis. A solution of DNA molecules
is placed in one end of gel, and we applied electric
4. Extract. Given a tube T and a binary digit xk, the
current to the gel. This process separates DNA strands
extract operation will produce two tubes +(T, xk) and
by length.
(T, xk), where +(T, xk) is all of the data in T which
contain xk and (T, xk) is all of the data in T which do
not contain xk.
called Boolean satisfiability problem whose instance
is a Boolean expression written using only AND, OR,
5. Detect. Given a tube T, the detect operation is used
NOT, variables and parantheses. A more formal
to check whether any a data is included in T or not. If
definition of satisfiability problem is: There is a set U
at least one data is included in T we have “yes”, and if
of variables and a collection C of clauses over U, is
no data is included in T we have “no“. The formal
there a satisfying truth assignment for C?
representation for the operation is written as
The problem remains NP-complete even if all
“Detect(T)“.
expressions are written in conjunctive normal form
with 3 variables per clause (3-CNF), yielding the
6. Discard. Given a tube T, the discard operation will
3-SAT problem. 3-satisfiability is a special case of
discard T. The formal representation for the operation
k-satisfiability (k-SAT) when each clause contains
is written as “Discard(T)“.
7. Read. Given a tube T, the read operation is used to
describe any a data, which is contained in T. Even if T
exactly k=3 literals. For example, E = ( x1 
x2
 x3 )  ( x1  x3  x 4 ). Note that each
contains many different data, the operation can give
clause has exactly 3 literals, that is why we call it
an explicit description of exactly one of them. The
3-SAT. One-In-Three (1in3) 3-SAT problem is defined
formal representation for the operation is written as
as follows.
“read(T)“.
Definition 3-1:
8. Append-head. Given a tube T and a binary digit xj,
Instance: A set V of logical variables and a collection
the operation, "Append-head", will append xj onto the
C of clauses over V such that each clause
head of every data stored in the tube T. The formal
has 3 literals.
representation for the operation is written as
Question: Is there a truth assignment for V such that
“Append-head(T, xj) “.
3. Molecular solution of One-In-Three
(1IN3) 3-SAT problem
each clause in C has exactly one true literal?
For example, V = ( x1, x2, x3, x4 ) and C = ( x1 
x 2  x3 )  ( x1  x3  x 4 ). Suppose x1 is
3.1 Definition of One-In-Three (1IN3)
the leftmost bit and x4 is the rightmost bit, we could
3-SAT problem
find all truth assignments are { 1000, 1001, 1010,
1011, 1100, 1101, 1110, 1111, 0000, 0001, 0100, 0101,
Satisfiability is the first NP-complete problem which
0011}. There are only 3 assignments ( { 1110, 0100,
determine if the variables of a given Boolean formula
0011 } ) such that each clause in C has exactly one
can be assigned in such a way that it makes the
true literal.
formula evaluate to be true. If there is no such
0011 }.
We get final answer T = { 1110, 0100,
assignment found, we say that the function is
unsatisfiable, otherwise it is satisfiable. Satisfiability
problem is also a decision problem which is also
3.2 Generate DNA-based algorithm to
solve One-In-Three (1IN3) 3-SAT
problem
(11)
Given that x1, x2, x3, x4 are 4 logical variables
For a = 1 to |C| do begin
(12)
For b = 1 to |Ca| do begin
(13)
If vba = xj then begin
x 2  x3 )
(14)
Tb = +(T, vba=1)
(15)
T = (T, vba=1)
 ( x1  x3  x 4 ). We also define vba is a
(16)
(17)
else begin
logical variable which is the xbth bit in the ath
clause.
(18)
Tb = +(T, vba = 0)
Suppose that x1 is the leftmost bit and x4 is the
(19)
T = (T, vba = 0)
rightmost bit. Basically our algorithm contains 2
(20)
blocks of codes, the first block will produce solution
(21)
spaces and generate truth assignments of the 3-SAT
(22)
and
f ( x1, x2, x3, x4 ) = C = ( x1 
end
End for
Discard (T)
|Ca |
problem [2]. The second block is moreover separated
(23)
into three parts.
end
First part would collect truth
T   (Tb )
b 1
assignments which make the clause be “100”. Second
(24)
part would collect truth assignments which make the
(25) * second block
clause be “010”. Third
(26) *first part- truth assignment which make *
part would collect truth
Endfor
*
assignments which make the clause be “001”. Then go
(27) * clause “100” is in T1 *
to the second round (for a = 2, that is, the second
(28)
clause ), and do the same thing again.
After all
(29)
clauses are done, the answer is left on T such that each
(30)
T1 = +(T, v1a = 1)
clause has exactly one true literal.
(31)
T2 = (T, v1a = 1)
We
propose
the
following
For a = 1 to |C| do begin
If
v1a = xj then begin
DNA-based
(32)
end
algorithm to solve One-In-Three (1IN3) 3-SAT
(33)
else begin
problem.
(34)
T1 = +(T, v1a = 0)
Algorithm 1: Solving One-In-Three (1IN3) 3-SAT
(35)
T2 = (T, v1a = 0)
problem for n logical variables and a collection C of
(36)
end
clauses over n
(37)
If
v2a = xj then begin
(38)
T3 = (T1, v2a = 0)
T1 = +(T1, v2a = 0)
(1)
* first block (generate truth assignments) *
(39)
(2)
Append-tail(T1, z11).
(40)
(3)
Append-tail(T2, z10).
(41)
(4)
T = (T1, T2).
(42)
T3 = (T1, v2a = 1)
(5)
For k = 2 to n
(43)
T1 = +(T1, v2a = 1)
(6)
Amplify(T, T1, T2).
(44)
end
(7)
Append-tail(T1, zk1).
(45)
(8)
Append-tail(T2, zk0).
(46)
T4 = (T1, v3a = 0)
(9)
T = (T1, T2).
(47)
T1 = +(T1, v3a = 0)
(10)
EndFor
(48)
end
else begin
If
v3a = xj then begin
end
(49)
else begin
(88)
T9 = +(T, v1a = 1)
T10 = (T, v1a = 1)
(50)
T4 = (T1, v3a = 1)
(89)
(51)
T1 = +(T1, v3a = 1)
(90)
(52)
end
(91)
(53)
T =  (T2, T3, T4)
(92)
T11 = (T9, v2a = 0)
(54)*second part-truth assignment which make*
(93)
T9 = +(T9, v2a = 0)
(55) * clause “010” is in T5 *
(94)
(56)
If
v1a = xj then begin
end
If
v2a = xj then begin
end
(95)
else begin
(57)
T5 = +(T,
0)
(96)
T11 = (T9, v2a = 1)
(58)
T6 = (T, v1a = 0)
(97)
T9 = +(T9, v2a = 1)
(98)
end
v1a =
(59)
end
If
v3a = xj then begin
(60)
else begin
(99)
(61)
T5 = +(T, v1a = 1)
(100)
T12 = (T9, v3a = 1)
(62)
T6 = (T, v1a = 1)
(101)
T9 = +(T9, v3a = 1)
(63)
end
(102)
end
v2a = xj then begin
(103)
(65)
T7 = (T5, v2a = 1)
(104)
T12 = (T9, v3a = 0)
(66)
T5 = +(T5, v2a = 1)
(105)
T9 = +(T9, v3a = 0)
(106)
end
(64)
If
(67)
end
(68)
else begin
(107)
else begin
Discard (T)
(70)
T7 = (T5, v2a = 0)
(108)
(69)
T5 = +(T5, v2a = 0)
(109)
Endfor
(71)
end
(110)
EndAlgorithm
(72)
If
v3a = xj then begin
(74)
T8 = (T5, v3a = 0)
(73)
T5 = +(T5, v3a = 0)
(75)
end
(76)
else begin
(77)
T8 = (T5, v3a = 1)
(78)
T5 = +(T5, v3a = 1)
(79)
end
(80)
T =  (T6, T7, T8)
T =  ( T1, T5, T9 )
The answer is in test tube T.
3.3 The Power of the DNA Algorithm to
Solve
One-In-Three
(1IN3)
3-SAT
problem
The example for V = ( x1, x2, x3, x4 ) and f ( x1, x2,
x3, x4 ) = C = ( x1  x 2
 x3 )  ( x1  x3
(81) *third part-truth assignment which make*
 x 4 ) in subsection 3.1 is applied to show the power
(82) * clause “001” is in T9 *
of Algorithm 1. The first execution of Step (13)
(83)
If
v1a = xj then begin
through Step (20) when a = 1 and b = 1, we put the
(84)
T9 = +(T, v1a = 0)
subset whose leftmost encoding bit is 1 on T1, and put
(85)
T10 = (T, v1a = 0)
the subset whose leftmost encoding bit is 0 on T, so
(86)
(87)
end
else begin
we get T1= {1000, 1001, 1010, 1011, 1100, 1101, 1110,
1111 } and T = { 0000, 0001, 0010, 0011, 0100, 0101,
0110, 0111 }. Next, the second execution of Step (13)
1000, 1001, 1101 } and T2 = {0000, 0001,0100, 0101,
through Step (20) when a = 1 and b = 2, we get T2 =
0011}. Next, after the execution of step (37) through
{ 0000, 0001, 0010, 0011 } and T = { 0100, 0101,
Step (44), we get T3 = {1010, 1011, 1000, 1001 } and
0110, 0111}. Then, the third execution of Step (13)
T1 = {1110 , 1111, 1100 , 1101}. Then, after the
through Step (20) when a = 1 and b = 3, we obtain T3
execution of step (45) through Step (52), we obtain T4
= {0100, 0101} and T = { 0100, 0111}. Because the
= {1100, 1101} and T1 = {1110, 1111}. The first
first outer loop is ended, the first execution of Step (22)
execution of Step (53) is applied to merge test tube T2,
is applied to discard test tube T and the first execution
T3, T4 into T, we get T = {0000, 0001, 0100, 0101,
of Step (23) is applied to merge test tube T1, T2, T3
0011, 1010, 1011, 1100, 1000, 1001, 1101 }.
into T, we get T = {1000, 1001, 1010, 1011, 1100,
1101, 1110, 1111, 0000, 0001, 0010, 0011, 0100,
0101}.
Then, after the execution of step (56) through step
(63), we get T5 = {0000, 0001, 0100, 0101, 0011 }
For the second outer loop of The forth execution
and T6 = { 1010, 1011, 1100, 1000, 1001, 1101 }.
of Step (13) through Step (20) when a = 2 and b = 1,
After the execution of step (64) through step (71), we
we put the subset whose leftmost encoding bit is 1 on
get T7 = { 0100, 0101 } and T5 = { 0000, 0001, 0011 }.
T1, and put the subset whose leftmost encoding bit is 0
After the execution of step (72) through step (79), we
on T, so we get T1= {1000, 1001, 1010, 1011, 1100,
get T8 = {0000, 0001} and T5 = {0011}. At step (80),
1101, 1110, 1111 } and T = { 0000, 0001, 0010, 0011,
we merge T6, T7, T8 into T and get T = {1010, 1011,
0100, 0101, }. Next, the fifth execution of Step (13)
1100, 0100, 0101, 0000, 0001, 1000, 1001, 1101 }.
through Step (20) when a = 2 and b = 2, we get T2 =
Then, after the execution of step (83) through step
{ 0000, 0001, 0100, 0101 } and T = { 0010, 0011 }.
(90), we get T9 = { 0100, 0101, 0000, 0001 } and T 10
Then, the sixth execution of Step (13) through Step
= {1010, 1011, 1100, 1000, 1001, 1101 }. After the
(20) when a = 2 and b = 3, we obtain T3 = {0011} and
execution of step (91) through step (98), we get T 11 =
T = { 0010 }. Because the second outer loop is ended,
{ 0000, 0001 } and T9 = { 0100, 0101 }. After the
the second execution of Step (22) is applied to discard
execution of step (99) through step (106), we get T 12 =
test tube T and the second execution of Step (23) is
 and T9 = {0100, 0101 }. At step (107), we discard T.
applied to merge test tube T1, T2, T3 into T, we get T =
At step (108), we merge T1, T5, T9 into T and get T =
{1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111,
{ 1110, 1111, 0011, 0100, 0101 } which is the answer
0000, 0001, 0100, 0101, 0011 } are the truth
when a = 1.
assignments.
For the second loop when a = 2, after the
The truth assignments T = {1000, 1001, 1010,
execution of step (29) through step (36), we get T 1 =
1011, 1100, 1101, 1110, 1111, 0000, 0001, 0100, 0101,
{1110, 1111 } and T2 = { 0011, 0100, 0101 }. After
0011 }. We keep on tracing second block of codes
the execution of step (37) through step (44), we get T 3
which would find the truth assignment such that each
=  and T1 = { 1110, 1111 }. After the execution of
clause has exactly one true literal. The first execution
step (45) through step (52), we get T 4 = {1111 } and
of Step (29) through Step (36) when a = 1, we put the
T1 = { 1110 }. The second execution of Step (53) is
assignment whose leftmost encoding bit is 1 on T1,
applied to merge test tube T2, T3, T4 into T, we get T =
and put the assignment whose leftmost encoding bit is
{0011, 0100, 0101, 1111 }.
0 on T2, so we get T1= {1110, 1111, 1010, 1011, 1100,
Then, after the execution of step (56) through step
(63), we get T5 = {0011, 0100, 0101 } and T 6 =
use 2*9*p = (18p) “extraction” operations, (p )
{ 1111 }. After the execution of step (64) through step
“discard” operations, and (3p ) merge operations.
(71), we get T7 = {0011 } and T5 = { 0100, 0101 }.
Therefore, from the analysis above, it is inferred that
After the execution of step (72) through step (79), we
the time complexity of Algorithm 1 is O(24p) with
get T8 = {0101 } and T5 = {0100 }. The second
“extract” operation, O(2p) with “discard” operation,
execution of Step (80) is applied to merge test tube T6,
O(2n) with “append” operation,
T7, T8 into T, we get T = { 1111, 0011, 0101 }.
“merge” operation, and O(n-1) with “amplify”
Then, after the execution of step (83) through step
O(n+4p) with
operation in the Adleman-Lipton model.
(90), we get T9 = { 0011, 0101 } and T 10 = { 1111 }.
Theorem 4-2: A set V of n logical variables and a
After the execution of step (91) through step (98), we
collection C of clauses which are {C1, C2, …, Cp}
get T11 = {0101 } and T9 = { 0011 }. After the
over n. The One-In-Three (1IN3) 3-SAT problem for
execution of step (99) through step (106), we get T 12 =
C and V can be solved with O( 2n ) library strands in
 and T9 = { 0011}. And then we discard T and merge
the Adleman-Lipton model.
T1, T5, T9 into T which is the final answer = T = {1110,
Theorem 4-3: A set V of n logical variables and a
0100, 0011 }.
collection C of clauses which are {C1, C2, …, Cp}
over n. The One-In-Three (1IN3) 3-SAT problem for
C and V can be solved with O( n ) tubes in the
Adleman-Lipton model.
4. The Complexity of Algorithm 1
Theorem 4-4: A set V of n logical variables and a
The following theorems describe time complexity and
collection C of clauses which are {C1, C2, …, Cp}
volume complexity of Algorithm 1, numbers of test
over n. The One-In-Three (1IN3) 3-SAT problem for
tube used and the longest library strand in solution
C and V can be solved with the longest library strand,
space in Algorithm 1.
O( 15*n +15*p ), in the Adleman-Lipton model.
Theorem 4-1: A set V of n logical variables and a
collection C of clauses which are {C1, C2, …, Cp}
5.Simulated Experimental Results
over n. The One-In-Three (1IN3) 3-SAT problem for
C and V can be solved in O(24p) with “extract”
operation, O(2p) with “discard” operation, O(2n) with
“append” operation, O(n+4p) with “merge” operation,
and
O(n-1)
with
“amplify”
operation
in
the
Adleman-Lipton model.
Consider the example for V = ( x1, x2, x3, x4 ) and
f ( x1, x2, x3, x4 ) = C = ( x1  x 2
 x3 )  ( x1
 x3  x 4 ) in subsection 3.1, DNA sequences
Proof:
generated by Adleman’s program are shown in table
Algorithm 1 can be applied to solve the One-In-Three
5-1. Adleman program is also used to calculate the
(1IN3) 3-SAT problem for C and V.
From the first
enthalpy, entropy, and free energy for binding of each
block of codes in algorithm 1, it is obvious that we use
probe to its corresponding region on a library strand.
2*3*p = (6p) “extraction” operations, (p ) “discard”
The energy used are shown in table 5-2.
operations, ( 2*n ) “append” operations and ( n+p )
“merge” operations, and ( n-1) “amplify” operation.
Table 5-1: Sequences chosen to represent xk1 and xk0 in
From the second block of codes in Algorithms 1, we
the example for V = ( x1, x2, x3, x4 ) and f ( x1, x2, x3, x4 )
= C = ( x1 
x 2  x3 )  ( x1  x3  x 4 )
Standard
3.86073
10.5438
1.3245
deviation
in subsection 3.1.
Bit
5  3 DNA Sequence
The library strands are shown in table 5-4 and
x 11
CATTCACAAACAATT
represent every possible truth assignments such that
0
TCATTCTCAACAAAA
each clause has exactly one.
x 21
CTCTATTCCTCTCAA
x 20
ACACCCTCTAATCTA
Table 5-4 : DNA sequences chosen to represent
x 31
CATTCACAAACAATT
answers in test tube T.
x 30
TCCTATTTAACTCCC
{0011}
x 41
CTCTATTCCTCTCAA
TCATTCTCAACAAAA ACACCCTCTAATCTA
x 40
TATAACTTTCTCTCT
CATTCACAAACAATT CTCTATTCCTCTCAA- 3
x1
5 -
3 Table 5-2: The energy for binding each probe to its
AGTAAGAGTTGTTTT TGTGGGAGATTAGAT
corresponding region on a library strand.
GTAAGTGTTTGTTAA GAGATAAGGAGAGTT- 5
Entropy
Enthalpy
Bit
Free energy
energy
energy (H)
{0100}
(G)
TCATTCTCAACAAAA CTCTATTCCTCTCAA
(S)
x 11
109.6
285.7
24.4
x 10
104.5
267.5
24.3
x 21
114.2
295.4
26
x 20
102.6
261.2
24.4
x 31
103.9
273.3
22.1
x 30
103.2
265.5
24
x 41
102.4
270.7
21.5
x 40
105.1
271.8
23.9
The program also figured out the average and standard
deviation for the enthalpy, entropy and free energy
5 TCCTATTTAACTCCC TATAACTTTCTCTCT- 3
3 AGTAAGAGTTGTTTT GAGATAAGGAGAGTT
AGGATAAATTGAGGG ATATTGAAAGAGAGA- 5
{1110}
5 CATTCACAAACAATT CTCTATTCCTCTCAA
CATTCACAAACAATT TATAACTTTCTCTCT- 3
3 GTAAGTGTTTGTTAA GAGATAAGGAGAGTT
GTAAGTGTTTGTTAA ATATTGAAAGAGAGA- 5
over all probe/library strand interaction. The energy
levels are shown in table 5-3.
6. Discussion and Conclusion
Table 5-3: The energy over all probe/library strand
In this paper, we propose the DNA-based
interactions.
Average
Enthalpy
Entropy
energy
energy
(H)
(S)
105.688
273.887
algorithm to solve the One-In-Three (1IN3) 3-SAT
Free energy
problem. Nowadays, many NP-complete problems
(G)
which could not be solved by a traditional digital
23.825
computer is now tried to be solved by DNA-based
algorithm. Even so, it is still very difficult to support
Journal of Computational Science and
biological operations using mathematical instructions.
Engineering, Volume 2, Number 1-2, 2006, pp.
In the future, there are still many difficulties to be
72 – 80.
overcome
and
we
hope
that
DNA-based
supercomputing could become a reality someday.
[9] Weng-Long Chang, “Fast Parallel DNA-based
Algorithms for Molecular Computation: the
Set-Partition Problem”, IEEE Transactions on
References
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Sun-Yuan
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[4]
[10]Weng-Long Chang and Minyi Guo, ”Solving the
Mathematics
and
Computation, Volume 203, Issue 2, 15 September
2008, Pages 502-512
[6] Weng-Long Chang, Michael Ho, and Minyi Guo,
"Molecular Solutions for the Subset-sum Problem
on DNA-based Supercomputing", BioSystems
(Elsevier Science), Vol. 73, No. 2, 2004, pp.
117-130.
[7] Weng-Long Chang, Minyi Guo, and Michael Ho,
"Towards solution of the set-splitting problem on
gel-based DNA computing", Future Generation
Computer Systems, Volume: 20, Issue: 5, June 15,
2004, pp. 875-885.
[8] Weng-Long Chang, Michael Ho, Minyi Guo,
Chengfei Liu, “Fast Parallel Bio-molecular
Solutions: the Set-basis Problem”, International
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Applications, pp. 431-436, 2002