Theoretical and Experimental DNA Computation (Natural ...

100 4 Complexity Issues
Size(STI)=O(S(n)logS(n)); Depth(STI)=O(log S(n))
We then cascade the R(n) combinational circuits produced in the preceding
construction stage (Fig. 4.10) and transform this resulting circuit into one
over the basis NAND. This completes the description of the translation of
the program of instructions to a combinational circuit.
In total, we have,
Theorem 1. Let A be a CREW P-RAM algorithm using P (n) processors and
S(n) memory locations (where S(n) ≥ P (n)) and taking T (n) parallel time.
Then there is a combinational logic circuit, C, computing exactly the same
function as A and satisfying
Size(C) =O(T (n)P (n)S(n)logS(n)); Depth(C) =O(T (n)logS(n)).
Proof. The modification of the control program allows A to be simulated by
at most R(n) blocks which simulate each parallel instantiation (c.f. Fig. 4.10).
InoneblockthereareatmostP (n) different circuits that update the S(n)
locations in the global memory. Each such circuit is a translation of a program
comprising O(Tk(n)) instructions, where Tk(n) is the runtime of the program
runoneachprocessorduringthekth cycle of the control program. Letting
Ck(n) denote the circuit simulating a program in the kth cycle, we have
R(n) �
R(n) �
Depth(C) ≤k=1
Depth(Ck(n)) ≤ O(log S(n)) k=1 O(Tk(n))
= O(T (n)logS(n))
For the circuit size analysis,
R(n) �
Size(C) ≤ P (n) k=1 Size(Ck(n)) ≤ O(P (n)T (n)S(n)logS(n)). ⊓⊔
Corollary 1: IfA is an NC algorithm, i.e., requires O(n k ) processors and
O(log r n) time, then the circuit generated by the translation above has polynomially
many gates and polylogarithmic depth.
Proof: Immediate from Theorem 1, given that S(n) mustbepolynomialinn.
⊓⊔
4.11 Assessment
So far we have presented a detailed, full-scale translation from CREW P-
RAM algorithms to operations on DNA. The volume of DNA required and
the running time of the DNA realization are within a factor log S of the