Theoretical and Experimental DNA Computation (Natural ...

106 4 Complexity Issues
ACC[] := bin(2n);
ACC[] := ADD(ACC[],FETCH(M[···],bin(n + k)));
M[···]:=STORE(ACC[],cond,bin(3n + k));
ACC[] := FETCH(M[···],FETCH(M[···],bin(3n + k)));
ACC[] := ADD(ACC[],FETCH(M[···],bin(2n + k)));
M[···]:=STORE(ACC[],cond,bin(2n + k));
ACC[] := bin(n);
ACC[] := ADD(ACC[],FETCH(M[···],bin(n + k)));
M[···]:=STORE(ACC[],cond,bin(3n + k));
ACC[] := FETCH(M[···],FETCH(M[···],bin(3n + k)));
M[···]:=STORE(ACC[],cond,bin(n + k));
Total size = n × (12Size(FETCH)+6Size(STORE) +4Size(ADD) +
Size(SUB)+Size(cond − eval)) = O(n 2 log n).
Depth is at most (12Depth(FETCH)+6Depth(STORE)+4Depth(ADD)+
Depth(SUB)+Depth(cond)) = O(log n).
This instruction, however, is repeated log n times, giving a total size of
O((n log n) 2 ) and depth O(log 2 n).
Final instruction
Input: M[1],...,M[5n], each consisting of r bits
Output: M[1],...,M[5n] representing the contents of memory after the n
processors have executed the first instruction.
ACC[] := FETCH(M[···],bin(2n + k));
M[···]:=STORE(ACC[], 1,bin(4n + k));
Total size = n×(Size(FETCH)+Size(STORE)) = O(n 2 log n). Total depth
= Depth(FETCH)+Depth(STORE)=O(log n).
In total we have a combinational circuit for list ranking which has size
O((n log n) 2 ) and depth O(log 2 n).
4.13 Summary
In this chapter we have emphasized the rôle that complexity considerations
are likely to play in the identification of “killer applications” for DNA computation.
We have examined how time complexities have been estimated within
the literature. We have shown that these are often likely to be inadequate
from a realistic point of view. In particular, many authors implicitly assume
that arbitrarily large numbers of laboratory assistants or robotic arms are
available for the mechanical handling of tubes of DNA. This has often led to
serious underestimates of the resources required to complete a computation.