LNCS 2950 - Aspects of Molecular Computing (Frontmatter Pages)

A DNA Algorithm for the Hamiltonian Path Problem 293
– Step 0: (t =0)(pre-loading). Put into each chamber Fj, enough copies of the
strands associated with vertices vi for which there exists an edge (i, j) ∈ A.
– Steps 1 to (n − 1): (from t =1tot =(n− 1)).
• Computations: In each time unit, for all 1 ≤ i ≤ n, in parallel a filtering
in all Fi and an append operation in all Ci is performed.
• Movement (pumping) of strands from step t to step t+1. Input(Fj �
)=
(t+1)
Output(Ci ) for all (i, j) ∈ A. The inlet of each chamber Fj in time
(t)
t + 1 is the union of the outlets of chambers Ci in time t such that
(i, j) ∈ A.
– Readout: After step (n−1) collects the strands in the output of chambers Ci.
If there are no output strands, the graph has no Hamiltonian paths. If there
are output strands in chamber Ci, there are Hamiltonian paths ending with
vertex vi; we need yet to determine the initial vertex of the strands in the
output of each chamber Ci. The initial vertex of each Hamiltonian path and
the exact sequence of vertices could be determined with a DNA sequencing
microchip.
2.3 Example
We show an example of the execution of the algorithm for a graph with four vertices.
This graph has 3 Hamiltonian paths: 2143, 2134 and 1234. Figure 1 shows
the graph and the associated microfluidic device. Table 1 shows the contents of
each chamber in each step. Remember that Input(Fj )= (t+1) � Output(Ci )for
(t)
all (i, j) ∈ A. In step 3 (time t =3=n−1), we must check if there is any output
in any chamber Ci. In this case, for this graph, there are outputs in chambers
C3 and C4 so we know there are Hamiltonian paths ending at vertices v3 and v4
respectively. To find out the first vertex and the exact order of the other vertices
in each path, we need to sequence these strands: paths 2143, 2134, and 1234 are
found.
t =1 t =2 t =3 t =4
F1 = {2} F1 = {12} F1 = {−} F1 = {−}
C1 = {21} C1 = {−} C1 = {−} C1 = {−}
F2 = {1} F2 = {21} F2 = {−} F2 = {−}
C2 = {12} C2 = {−} C2 = {−} C2 = {−}
F3 = {1, 2, 4} F3 = {21, 12} F3 = {214, 134, 234} F3 = {2134, 1234}
C3 = {13, 23, 43} C3 = {213, 123} C3 = {2143} C3 = {−}
F4 = {1, 3} F4 = {21, 13, 23, 43} F4 = {213, 123, 143} F4 = {2143}
C4 = {14, 34} C4 = {214, 134, 234} C4 = {2134, 1234} C4 = {−}
Table 1. Running of the algorithm. In time t = 3 we must readout the output
of the chambers Ci. In bold, strands codifying Hamiltonian paths.