1. Background
Deoxyribonucleic acid (DNA) is a molecule that carries most of the genetic instructions used in the development, functioning, and reproduction of all known living organisms and many viruses. Just one gram of DNA can hold about 680 peta bytes (6.8* 1017) equal to one billion ( 1 ) Compact Disc (CD), which take 163,000 centuries to be listened.
Considering parallelism aspect, it also is capable of parallelizing 3*10^{14} molecules at a time ( 2 ), and this enormous parallelism persuades the researchers to build a model by the use of DNA strands to solve NPhard problems ( 3 ). Considering the similarities in biological and mathematical operations, and the facts that DNA is stable and also predictable in its reactions, this macromolecule can be used to encode information for mathematical systems.
In 1994, Adelman ( 4 ) designed the first model of molecular computation by solving a sevennode instances of Hamiltonian path problem, and then showed the enormous parallel power of DNA computation. In 1995, Lipton ( 5 ) solved satisfiability problem by Adelman’s method and designed a new method called AdelmanLipton model, which was consisted of three levels as follows:
a) Producing specific sequences and putting them in a test tube.
b) Performing a number of operations on the DNA sequences inside the tube and removing unwanted sequences, which certainly cannot be considered as the final answer of the problem.
c) Reading and reporting the DNA sequences, if any exists, in the final tube as final answer.
Ouyang ( 6 ) solved Maximal clique problem in 1997 by designing Restriction enzyme model. Roweis designed sticker model ( 7 ) in 1998. The Selfassembly model was designed by Winfree in 1998 ( 8 ). The Surface based model ( 9 ) was designed by Smith in 1998. And finally, Sakomoto proposed the Hairpin model in 2000 ( 10 ).
As the other applications of DNA computing we can point to Zhang's image encryption algorithm ( 11 ) or Patel multiple image encryption ( 12 ) which is based on DNA computing or novel data encryption scheme which is proposed by Patro ( 13 ) by using DNA computing.
Many efforts have been done by researchers to solve NPhard problems via these models such as knapsack ( 14 ), nqueen ( 15 ), binpacking ( 16 ), maximal clique problem ( 17 ), graph vertex coloring problem ( 18 ), maximal matching problem ( 19 ), maximal connected subgraph problem ( 20 ), maximum complete bipartite subgraph ( 21 ), maximum independent set problem ( 22 ), maximum matching problem ( 23 ), maximum kcolourable subgraph ( 24 ), hamiltonian path ( 25 , 26 ), minimum ksuppliers ( 27 ).
2. Objectives
Assignment is another famous NPhard problem ( 28 ), which has been solved in different versions by researchers as Wang ( 29 ) who solved the unbalanced version of assignment, or Yang ( 30 ) who solved the quadratic version of assignment.
Based on AdlemanLipton model, we proposed a parallel DNA algorithm with O(n2) time complexity for assignment problem, and O(n) initial DNA strand in comparison with Onn initial DNA strands of similar solutions such as that was proposed by Wang et al ( 31 ).
3. Materials and Methods
In Assignment problem, we have n jobs, which can be performed by n people. Each of the people will do each one of the jobs with a specific wage. The objective of this study was to assign each job to one person in such a way that the total wages become minimized.
As an example, consider four teachers (T_{1}T_{4}) from four different cities (C_{T1}C_{T4}) and also four universities (U1U4) in four different other cities (C_{U1}C_{U4}).
Table 1 shows the ticket price for each flight from cities C_{T1}C_{T4} to cities C_{U1}C_{U4}. The goal is to send each teacher to one of the universities in a way that the total price of purchased ticket becomes minimized.
City  C_{T1}  C_{T1}  C_{T1}  C_{T1} 

C_{U1}  40  140  90  180 
C_{U2}  120  20  70  110 
C_{U3}  80  25  25  115 
C_{U4}  80  180  130  140 
We can represent Table 1 as a cost matrix:
[ 40 140 90 180 120 20 70 110 80 25 25 115 80 180 130 40 ]
One possible assignment is as follow:
[ 40 140 90 180 120 20 70 110 80 25 25 115 80 180 130 40 ]
Where selected flights highlighted as yellow. The total cost of this assignment is: 90$+110$+25$+80$=305$. But the best possible assignment is:
[ 40 140 90 180 120 20 70 110 80 25 25 115 80 180 130 40 ]
Which will cost 40$+20$+25$+140$=225$ and means to send T_{1} to U_{1}, T_{2} to U_{2}, T_{3} to U_{3}, and T_{4} to U_{4}.
3.1. AdelmanLipton Model:
Different operations of AdelmanLipton model:
 Merge (T_{1}, T_{2}): The content of tube T_{1} will be merged with T_{2} and will also be moved into T_{1}.
 Copy (T_{1}, T_{2}): The content of T_{1} will be copied into the empty tube T_{2}.
 Detect (T): Determining whether any sequence exists in tube T or not.
 Extract (T_{1}, X, T_{2}): The subset of T_{1}, which contains the X sequence will be transferred into T_{2}.
 Select (T_{1}, X, T_{2}): The subset of T_{1} containing sequences longer than X will be transferred into T_{2}.
 Selecting (T_{1}, L, T_{2}): It will select sequences of size L from tube T_{1} and put them in tube T_{2}.
 SelectMax (T_{1}, T_{2}): The longest sequence in tube T_{1} will be chosen and moved into tube T_{2}.
 SelectMin (T_{1}, T_{2}): The shortest sequence in tube T_{1} will be chosen and moved into tube T_{2}.
 Annealing (T): ssDNA will be converted to dsDNA by freezing the tube.
 Denaturing (T): dsDNA will be converted to ssDNA by freezing the tube.
 Discard (T): The content of tube T will be eliminated.
 PCR (T, T_{1}): A lot of copies of DNA strands will be made and placed in tube T_{1}.
 Append Tail (T, Z): Sequence Z will be appended to the end of all sequences in tube T.
 Read (T): A sequence of tube T will be read.
3.2. DNA Algorithm for Assignment Problem
First of all, all n^{n} combinations of people and jobs will be generated as different DNA strands. Then, the strands having all people will be separated and the others will be discarded. After that, among remained stands, those that have all jobs will be separated and the others will be discarded. The shortest sequences will be regarded as the final solutions.
The proposed algorithm consists of six steps:
 Building 2n unique10mer sequences (sequence with length 10) for each of n people and n jobs.
 Building all n^{2} possible combinations of jobs with people and then adding C_{i}mer sequences of G nucleotides (C_{i} is the cost or profit of performing J_{i} by P_{i}).
 Construct all n^{n} possible combinations of choosing n jobs by n people.
 Eliminate all sequences which do not consist of all people.
 Among remaining sequences of previous step, keep only that were consisted of all jobs.
 Consider the shortest sequence as the optimal solution.
3.3. Strand Design:
First of all, for each person (P_{i}) and each job (J_{i}) of assignment problem produce a distinct 10mer sequence. We can produce 59049 different 10mer sequences using 3 nucleotides a, t, c. Among them, we should select the sequences, which have the most difference with each other and then put these 2n sequences in 2n tube and perform the Polymerase Chain Reaction (PCR) on them. PCR, developed in 1983 by Kary Mullis ( 32 ), is a molecular biology technology, which generates a lot of copies from a single piece of DNA sequence. We have to generate n2 sequences by dot product of n jobs to n person’s sequences.
Finally, equal to the cost of assigning a person to a job, we should add C_{i}mer sequences of G nucleotides to the related sequences. For making the length of sequences shorter according to the costs, it is easily possible to make a rule. Such as subtracting the minimum cost can form the cost belonged to all people.
As an example, in our problem, the minimum ticket price was 20$ and the costs were differed at least 5$ with each other. So, we subtracted 20 from all ticket prices and then added a G for each5$. As you can see in Table 2 for the four teachers, we used four different 10mer, and for the four universities, we used four others different 10mer.
DNA Sequence  DNA Sequence  

T_{1}  5' TCTATAACTA 3'  U_{1}  5' AACTACATTT 3' 
T_{2}  5' TAACACTATT 3'  U_{2}  5' CATCATTTAC 3' 
T_{3}  5' ACTAATCTCT 3'  U_{3}  U35' ATTACTCCTA 3' 
T_{4}  5' CACACATCTA 3'  U_{4}  5' TACCTCAACT 3' 
Then, as it is shown in Table 3, we should combine all possible ways and then add one G for each 5$ to all of the sequences and also remove 4G from all of them because of subtracting minimum.
Length  DNA Sequence  

T_{1}U_{1}  40  5' TCTATAACTAAACTACATTTGGGGGGGGGGGGGGGGGGGG 3' 
T_{1}U_{2}  32  5' TCTATAACTACATCATTTACGGGGGGGGGGGG 3' 
T_{1}U_{3}  24  5' TCTATAACTAATTACTCCTAGGGG 3' 
T_{1}U_{4}  32  5' TCTATAACTATACCTCAACTGGGGGGGGGGGG 3' 
T_{2}U_{1}  20  5' TAACACTATTAACTACATTT 3' 
T_{2}U_{2}  52  5' TAACACTATTCATCATTTACGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG 3' 
T_{2}U_{3}  44  5' TAACACTATTATTACTCCTAGGGGGGGGGGGGGGGGGGGGGGGG 3' 
T_{2}U_{4}  21  5' TAACACTATTTACCTCAACTG 3' 
T_{3}U_{1}  30  5' ACTAATCTCTAACTACATTTGGGGGGGGGG 3' 
T_{3}U_{2}  42  5' ACTAATCTCTCATCATTTACGGGGGGGGGGGGGGGGGGGGGG 3' 
T_{3}U_{3}  34  5' ACTAATCTCTATTACTCCTAGGGGGGGGGGGGGG 3' 
T_{3}U_{3}  21  5' ACTAATCTCTTACCTCAACTG 3' 
T_{4}U_{1}  38  5' CACACATCTAAACTACATTTGGGGGGGGGGGGGGGGGG 3' 
T_{4}U_{2}  44  5' CACACATCTACATCATTTACGGGGGGGGGGGGGGGGGGGGGGGG 3' 
T_{4}U_{3}  52  5' CACACATCTAATTACTCCTAGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG 3' 
T_{4}U_{4}  39  5' CACACATCTATACCTCAACTGGGGGGGGGGGGGGGGGGG 3' 
3.4. Sample Space:
Put each of the related strands to the first person in a separated tube.

For (i in 1 to n)
 ° Merge (T, T_{i})
 ° Discard (T_{i})

For (i in 2 to n)

° For (j in 1 to n)
 ▪ PCR (T, T_{j})
 ▪ AppendTail (T_{j}, P_{i} J_{j})
 ▪ Merge (Temp, T_{j})
 ▪ Discard (T_{j})
 ° Discard (T)
 ° Copy (Temp, T)
 ° Discard (Temp)

° For (j in 1 to n)
The trace of the top algorithm for three people and three job is shown as an example in Table 4.
Eliminate all sequences, which do not consist of all people.
Round  Tube T_{1}  Tube T_{2}  Tube T_{3}  Tube T 

i=1  P_{1}J_{1}  P_{1}J_{2}  P_{1}J_{3}  P_{1}J_{1},P_{1}J_{2},P_{1}J_{3} 
i=2, j=1  P_{1}J_{1}P_{2}J_{1},P_{1}J_{2} P_{2}J_{1},P_{1}J_{3} P_{2}J_{1}  P_{1}J_{1},P_{1}J_{2},P_{1}J_{3}  
i=2, j=2  P_{1}J_{1}P_{2}J_{2},P_{1}J_{2} P_{2}J_{2}, P_{1}J_{3} P_{2}J_{2}  P_{1}J_{1},P_{1}J_{2},P_{1}J_{3}  
i=2, j=3  P_{1}J_{1}P_{2}J_{3},P_{1}J_{2}P_{2}J_{3},P_{1}J_{3}P_{2}J_{3}  P_{1}J_{1},P_{1}J_{2},P_{1}J_{3}  
i=3, j=1  P_{1}J_{1}P_{2}J_{1}P_{3}J_{1},P_{1} J_{2}P_{2}J_{1}P_{3}J_{1}, P_{1}J_{3}P_{2}J_{1}P_{3}J_{1},P_{1}J_{1}P_{2}J_{2}P_{3}J_{1}, P_{1}J_{2}P_{2}J_{2}P_{3}J_{1},P_{1}J_{3}P_{2}J_{2}P_{3}J_{1}, P_{1}J_{1}P_{2}J_{3}P_{3}J_{1},P_{1}J_{2}P_{2}J_{3}P_{3}J_{1}, P_{1}J_{3}P_{2}J_{3}P_{3}J_{1}  P_{1}J_{1}P_{2}J_{1},P_{1} J_{2}P_{2}J_{1}, P_{1}J_{3}P_{2}J_{1},P_{1}J_{1}P_{2}J_{2}, P_{1}J_{2}P_{2}J_{2},P_{1}J_{3}P_{2}J_{2}, P_{1}J_{1}P_{2}J_{3},P_{1}J_{2}P_{2}J_{3}, P_{1}J_{3}P_{2}J_{3}  
i=3, j=3  P_{1}J_{1}P_{2}J_{1}P_{3}J_{3},P_{1} J_{2}P_{2}J_{1}P_{3}J_{3}, P_{1}J_{3}P_{2}J_{1}P_{3}J_{3},P_{1}J_{1}P_{2}J_{2}P_{3}J_{3}, P_{1}J_{2}P_{2}J_{2}P_{3}J_{3},P_{1}J_{3}P_{2}J_{2}P_{3}J_{3}, P_{1}J_{1}P_{2}J_{3}P_{3}J_{3},P_{1}J_{2}P_{2}J_{3}P_{3}J_{3}, P_{1}J_{3}P_{2}J_{3}P_{3}J_{3}  P_{1}J_{1}P_{2}J_{1},P_{1} J_{2}P_{2}J_{1}, P_{1}J_{3}P_{2}J_{1},P_{1}J_{1}P_{2}J_{2}, P_{1}J_{2}P_{2}J_{2},P_{1}J_{3}P_{2}J_{2}, P_{1}J_{1}P_{2}J_{3},P_{1}J_{2}P_{2}J_{3}, P_{1}J_{3}P_{2}J_{3}  
Final  P_{1}J_{1}P_{2}J_{1}P_{3}J_{1},P_{1}J_{2}P_{2}J_{1}P_{3}J_{1},P_{1}J_{3}P_{2}J_{1}P_{3}J_{1},P_{1}J_{1}P_{2}J_{2}P_{3}J_{1},P_{1}J_{2}P_{2}J_{2}P_{3}J_{1},P_{1}J_{3}P_{2}J_{2}P_{3}J_{1},P_{1}J_{1}P_{2}J_{3}P_{3}J_{1}, P_{1}J_{2}P_{2}J_{3}P_{3}J_{1},P_{1}J_{3}P_{2}J_{3}P_{3}J_{1}P_{1}J_{1}P_{2}J_{1}P_{3}J_{2},P_{1}J_{2}P_{2}J_{1}P_{3}J_{2},P_{1}J_{3}P_{2}J_{1}P_{3}J_{2},P_{1}J_{1}P_{2}J_{2}P_{3}J_{2},P_{1}J_{2}P_{2}J_{2}P_{3}J_{2}, P_{1}J_{3}P_{2}J_{2}P_{3}J_{2},P_{1}J_{1}P_{2}J_{3}P_{3}J_{2},P_{1}J_{2}P_{2}J_{3}P_{3}J_{2},P_{1}J_{3}P_{2}J_{3}P_{3}J_{2},P_{1}J_{1}P_{2}J_{1}P_{3}J_{3},P_{1}J_{2}P_{2}J_{1}P_{3}J_{3},P_{1}J_{3}P_{2}J_{1}P_{3}J_{3}, P_{1}J_{1}P_{2}J_{2}P_{3}J_{3},P_{1}J_{2}P_{2}J_{2}P_{3}J_{3},P_{1}J_{3}P_{2}J_{2}P_{3}J_{3},P_{1}J_{1}P_{2}J_{3}P_{3}J_{3},P_{1}J_{2}P_{2}J_{3}P_{3}J_{3},P_{1}J_{3}P_{2}J_{3}P_{3}J_{3} 

For (i in 1 to n)
 °Extract (T, P_{i}, Temp)
 °Discard(T)
 °Copy (Temp, T)
 °Discard (Temp)
Eliminate all sequences, which do not consist of all jobs.

For (i in 1 to n)
 ° Extract (T, Job_{i}, Temp)
 ° Discard(T)
 ° Copy (Temp, T)
 ° Discard (Temp)
Final Answer
 SelectMin (T, Temp)
 Read (Temp)
4. Results
4.1. The solutions of assignment problem with n jobs and individuals can be obtained by above mentioned DNA molecules operations.
Proof. At first, all combinations of the n job assignments were generated in the sample space. Then, the sequences containing all people were kept and this guaranteed that none of the people are jobless. And finally, among the remaining sequences, the sequences containing all the jobs were kept and this guaranteed that all of the jobs will be done at the end. Therefore, for sure, all possible assignments are in final tube, and as the cost of each assignment was attached to each of them in the strain design phase, so the strand with minimum length among the strands of final tube would be the minimum cost assigning solution.
4.2. The time complexity of the proposed algorithm for solving assignment problem with n jobs and individuals is n2.
Proof. As the complexity of every biological operation is O(1) ( 33 ), the time complexity of algorithm can be calculated by adding the time complexity of all steps as follows:
T (Sample space): O(n) for the first loop and On^{2} for the nested loop.
T (Persons): Four biological operations in each step, which become O(1), and in n step it would be O(n).
T (Jobs): Four biological operations in each step, which become O(1), and in n step it would be O(n).
T (Last step): Two biological operations, which become O(1).
So, the time complexity of the algorithm will be arrived by equation 1 as follows:
T(n) = O(n)+O(n^{2})+O(n)+O(n)+O(1)= O(n^{2})
5. Discussion
In the proposed method, by using adenine, thymine, and cytosine; we produced distinct 10mer sequences equal to the number of people and jobs of assignment problem. Then, by using polymerase chain reaction on them, we produced n^{2} sequences equal to the dot product of n persons to n jobs. Finally, by adding guanine equal to the cost of assigning a job to a person to the related sequence, we got the desired sequence.
Wang et al.( 31 ), Kang et al.( 34 ), Shu et al.( 35 ), and Ebrahim et al.( 36 ), Tsaftaris et al.( 37 ), Rashid et al. ( 38 ) all used the same model to solve the assignment problem using n^{n} distinct sequence with the time complexity of O(n^{2}). The proposed algorithm reduced the initial cost to a great extent by using just 2n initial sequences in comparison with n^{n} initial sequences, which is used by the other methods and solve the problem with the same time complexity.
6. Conclusion
Ultraefficient parallelism of the methods of DNA computing in contrast with the obvious limitations in storage, speed, intelligence, and miniaturization of electronic computers is considerable. Although there are still some problems that need a further study in biologic technology, it is still possible to solve a lot of NPhard problems in linear time via DNA computing. In this article, we highlighted a DNA computing model with biological operations based on AdelmanLipton model to solve Assignment problem with the time complexity of O(n^{2}) with just buying 2n 10mer sequences and making the other needed sequences in our lab in comparison with the previous methods which need a lot of different sequences.
We hope that, in near future, molecular computer become usable instead of electronic computers, which can cause DNA computing solutions become applicable.
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