Past exam of the laser science and quantum information processing course of the University of Oxford / 2019 / 5 / a / Solution

Allow clean workspace bits, as in standard reversible computation. Boolean algebra expresses any Boolean function as XORs of ANDs and constants. A Toffoli gate maps $(a,b,c)$ to $(a,b,c\oplus ab)$, so an initially zero target computes an AND without erasing the inputs. A CNOT gate accumulates XORs, and a NOT gate supplies complemented bits or constants. Intermediate products can be computed into workspace, XORed into an output, and then uncomputed by reversing the workspace computation. This realizes the reversible embedding $(x,y)\mapsto (x,y\oplus f(x))$ for any Boolean function and, with reversible clean-up, universal reversible classical computation.

The Fredkin transformation has control $a$ and targets $b,c$. A network of three gates, shown in i, iswhere the sequence is executed left to right. After the first gate $c_1=c\oplus b$; after the second $b_1=b\oplus a(c\oplus b)$; and the last gives $c_2=c\oplus b\oplus b_1$. If $a=0$, $(b_1,c_2)=(b,c)$; if $a=1$, $(b_1,c_2)=(c,b)$. Thus the circuit is exactly a Fredkin gate.

The workspace qualification matters for the unrestricted universality claim. On four or more wires, every embedded NOT, CNOT and Toffoli gate is an even permutation of all computational states. Without ancillary wires they cannot implement, for example, a permutation swapping just $0000$ and $0001$ while fixing the other four-bit strings. Clean ancillas remove this obstruction on the designated input subspace.