Theoretical and Experimental DNA Computation (Natural ...

4.12 A Worked Example: The List Ranking Problem 105
1. LDA 2n + k
2. STA 4n + k /* rank(k) :=distance(k)
3. HALT
Combinational circuit realization
First instruction block
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.
For each processor k the following combinational circuit is used:
ACC[] := FETCH(M[···],bin(k));
M[···]:=STORE(ACC[], 1,bin(n + k));
ACC[] := bin(k);
ACC[] := SUB(ACC[],FETCH(M[···],bin(n + k));
M[···]:=STORE(ACC[], 1,bin(3n + k));
ACC := FETCH(M[···],bin(3n + k));
r−1
∧
cond := i=0 (ACC[i] ≡ 0);
ACC[] := bin(1);
M[···]:=STORE(ACC[],cond,bin(2n + k));
ACC[] := bin(0);
M[···]:=STORE(ACC[], ¬cond, bin(2n + k));
Total size of first instruction = n × (3Size(FETCH)+4Size(STORE)+
Size(SUB)+Size(cond − eval)) = O(n 2 log n).
Depth is at most 3Depth(FETCH)+4Depth(STORE)+Depth(SUB)+
Depth(cond − eval) =O(log n).
Second instruction block
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[] := bin(n)
ACC[] := ADD(ACC[],FETCH(M[···],bin(n + k)));
M[···]:=STORE(ACC[], 1,bin(3n + k));
ACC[] := FETCH(M[···],FETCH(M[···],bin(3n + k));
ACC[] := SUB(ACC[],FETCH(M[···],bin(n + k));
M[···]:=STORE(ACC[], 1,bin(3n + k));
ACC[] := FETCH(M[···],bin(3n + k));
r−1
∧
cond := ¬( i=0 (ACC[i] ≡ 0));