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Theorem 2: If the number of the element in A is n and
the sum of the elements in A is S in the partition problem
can be solved with O(n+(0.25S 2 +2.5S)) biological
operations, O(2 n ) strands, the logest strands is O(n+S)
and O(S) test tubes.
Proof:. Algorithm 1 includes four main steps. From
the algorithm, Line(1), it is very obvious that it takes n
amplify operations, 2n append operations, n merge
operations and three test tubes to construct sticker-based
solution space. From the algorithm, Line(2), it takes n
extract operations, n detect operations, S append
operations, n merge operations, and three test tubes to
construct sticker-based solution space for elements of 2 n
possible subsets to a n-element set A. From the algorithm,
Parallel_Searcher (T 01 , S/2), it takes S(S+2)/8 exact
operations, S(S+2)/8 merge operations, one detect
operation, one discard operation and S/2+2 test tubes.
The length of a DNA strand has never been changed, is
15(n+S).In Line(4), it takes one detect operation and it
takes one read operation in Line(5).
Hence, from the statements mentioned above, it is at
once inferred that the time complexity of Algorithm 1 is
O(n+(0.25S 2 +2.5S).After Line(1) of Algorithm1, the
solution is constructed and there is 2 n strands in it.
Referring to the Michael’s algorithm [8, 9], the number
of the tube used in Algorithm 1 is O(n). The longest
strand in Algorithm 1 is O(n+S).
Ⅳ . EXPERIMENTAL RESULTS BY
SIMULATED DNAS COMPUTING
Consider that a finite set A is {1, 2, 3 } as an example.
S is denoted to represent the sum of the elements in A So
S is 6. The partition problem is to find a subset of A in
which the sum is one half of S. Table 1 shows the strands
pool and every step in T 0 .
After Line(4) in Table 1, we will get the full
solution space of the partition problem above. The
DNA strands are {x 1
0
x 2
0
x 3
0
y 1,1
0
y 2,1
0
y 2,2
0
y 3,1
0
y 3,2
0
y 3,3 0 ;
x 1
0
x 2
0
x 3
1
y 1,1
0
y 2,1
0
y 2,2
0
y 3,1
1
y 3,2
1
y 3,3 1 ; x 1
0
x 2
1
x 3
0
y 1,1
0
y 2,1
1
y 2,2
1
y 3,1
0
y 3,2
0
y 3,3 0 ; x 1
0
x 2
1
x 3
1
y 1,1
0
y 2,1
1
y 2,2
1
y 3,1
1
y 3,2
1
y 3,3 1 ; x 1
1
x 2
0
x 3
0
y 1,1
1
y 2,1
0
y 2,2
0
y 3,1
0
y 3,2 0 , y 3,3 0 ; x 1
1
x 2
0
x 3
1
y 1,1
1
y 2,1
0
y 2,2
0
y 3,1
1
y 3,2
1
y 3,3 1 ; x 1
1
x 2
1
x 3
0
y 1,1
1
y 2,1
1
y 2,2
1
y 3,1
0
y 3,2
0
y 3,3 0 ; x 1
1
x 2
1
x 3
1
y 1,1
1
y 2,1
1
y 2,2
1
y 3,1
1
y 3,2
1
y 3,3 1 }.
Searching the union of the DNA strands by the algorithm
Parallel_Searcher(T 0, S/2) showing in Table 2, we will
get the legal strands of the partition problem. Reading
the DNA sequence in tube T 3, we could get the answer
is {1, 2} and {3}.
Table 1
Every steps of the improved algorithm based on DNA computing
tube T 0
1 { x 0 1 x 0 2 x 0 3 ; x 0 1 x 0 2 x 1 3 ; x 0 1 x 1 2 x 0 3 ; x 0 1 x 1 2 x 1 3 ; x 1 1 x 0 2 x 0 3 ; x 1 1 x 0 2 x 1 3 ; x 1 1 x 1 2 x 0 3 ; x 1 1 x 1 1
2 x 3 . }
2 { x 1 0 x 2 0 x 3 0 y 1,1 0 ; x 1 0 x 2 0 x 3 1 y 1,1 0 ; x 1 0 x 2 1 x 3 0 y 1,1 0 ; x 1 0 x 2 1 x 3 1 y 1,1 0 ; x 1 1 x 2 0 x 3 0 y 1,1 0 ; x 1 1 x 2 0 x 3 1 y 1,1 0 ; x 1 1 x 2 1 x 3 0 y 1,1 0 ; x 1
1
x 2 1 x 3 1 y 1,1 0 . }
3
4
{ x 1 0 x 2 0 x 3 0 y 1,1 0 y 2,1 0 y 2,2 0 ; x 1 0 x 2 0 x 3 1 y 1,1 0 y 2,1 0 y 2,2 0 ; x 1 0 x 2 1 x 3 0 y 1,1 0 y 2,1 1 y 2,2 1 ; x 1 0 x 2 1 x 3 1 y 1,1 0 y 2,1 1 y 2,2 1 ; x 1 1 x 2 0 x 3
0
y 1,1
0
y 2,1 0 y 2,2 0 ; x 1 1 x 2 0 x 3 1 y 1,1 0 y 2,1 0 y 2,2 0 ;
x 1 1 x 2 1 x 3 0 y 1,1 0 y 2,1 1 y 2,2 1 ; x 1 1 x 2 1 x 3 1 y 1,1 0 y 2,1 1 y 2,2 1 . }
0 0 0 0 0 0
{x 1 x 2 x 3 y 1,1 y 2,1 y 2,2 y 0 3,1 y 0 3,2 y 0 3,3 ; x 0 1 x 0 2 x 1 3 y 0 1,1 y 0 2,1 y 0 2,2 y 1 3,1 y 1 3,2 y 1 3,3 ; x 0 1 x 1 2 x 0 3 y 0 1,1 y 1 1 0 0
2,1 y 2,2 y 3,1 y 3,2 y 0 3,3 ;
x 0 1 x 1 2 x 1 3 y 0 1,1 y 1 2,1 y 1 2,2 y 1 3,1 y 1 3,2 y 1 3,3 ; x 1 1 x 0 2 x 0 3 y 1 1,1 y 0 2,1 y 0 2,2 y 0 3,1 y 0 3,2 , y 0 3,3 ; x 1 1 x 0 2 x 1 1
3 y 1,1 y 0 0 1 1
2,1 y 2,2 y 3,1 y 3,2 y 1 3,3 ; x 1 1 x 1 2 x 0 3 y 1 1,1 y 1 2,1 y 1 2,2 y 0 0
3,1 y 3,2
y 0 3,3 ; x 1 1 x 1 2 x 1 3 y 1 1,1 y 1 2,1 y 1 2,2 y 1 3,1 y 1 3,2 y 1 3,3 . }
Table 2
Search the solution space of the element
Tube T 0 T 1 T 2 T 3 T 4
Inital
pool
1
000000;000111
011000;011111
100000;100111
111000;111111
000000;000111
011000;011111
100000;100111 ;
111000; 111111
2 000000;000111
100000;100111 ;
011000;011111
111000; 111111
3 000000;000111 100000;100111 011000; 011111 111000; 111111
4 000000 100000;000111
011000;100111
111000
111000; 011111 111111
5 100000 011000; 000111 111000; 100111
011111;100111 ;
111111
6 011000 000111; 111000
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