rom-based computations: quantum versus classical b.c. travaglione, m.a.nielsen, h.m. wiseman, and a....

ROM-based computations: quantum versus classical

B.C. Travaglione, M.A.Nielsen, H.M. Wiseman, and A. Ambainis

Quantum research Focused mainly on time requirements

e.g. Shor’s algorithm Deutsch-Jozsa algorithm Grover’s algorithm

Space complexity important also!

Space Complexity Number of (qu)bits required to solve a

problem ROM-Based

Read only memory Writable memory

Space complexity - function of writable memory

This paper Compares space complexity of “error-

free, reversible quantum and classical computation”

Summary ROM-Based computing Universality

1 qubit Quantum ROM computer 2 bit classical ROM computers

Time efficiency

ROM-Based Computation

u1 to uj doesn’t change at any time during computation(Read only bits)

fi is a boolean mapping from a combination of ‘u’s

Classical ROM Computations Sequence of arbitrary classical

reversible gates Gates can be controlled by one of the j

ROM bits FANOUT increases complexity, and bit

reducing activities (e.g. AND) can be simulated.

Quantum ROM computations Similarly, quantum gates controlled by

up to one ROM bit applied to n qubits Note that fi is boolean, so qubits must

end up in computational basis state.

Why only one control? Arbitrary number of controls on

quantum and classical gates can be broken down to gates with 2 and 3 controls respectively

This adds unnecessary complexity as conditional bits must be writable

This constraint does not affect these results

Notation



Time efficiency

Universality There are 2^(2j) possible distinct

boolean propositions. A universal computer can achieve any

of these. We will show that one writable qubit is

sufficient for quantum case, 2 bits necessary for classical.

Method Well known that AND and NOT are

sufficient to express any boolean proposition.

AND and XOR are also sufficient since “NOT a” can be replaced with “a XOR 1”

Show that any writable (qu)bit can be transformed from |0> to any of the 2^(2j) different boolean propositions

Method Sufficient to show that we can transform

|f> to |f u1u2u3…um > where f is an arbitrary boolean function and

m {1,2…j}



Time efficiency

One Writable Qubit Universal We will use only Pauli matrices:

As well as X-1/2 , X1/2 , Z-1/2 and Z1/2.

One Writable Qubit Universal denotes that the operator W is

controlled by the ROM bit ui.

performs a bit flip iff ui = uj = 1

|f> to |f uiuj >

One Writable Qubit Universal Easy to see with circuit diagram

One Writable Qubit Universal Also Bloch sphere helps visualize

One Writable Qubit Universal We can add more bits to the conjunction

by recursively substituting gates For example, substituting with

Which makes Z essentially controlled by uj and uk, causes our qubit to be flipped iff ui = uj = uk = 1.

One Writable Qubit Universal After substitution we have:

|f> to |f uiujuk > Continuing like this, we can create a

sequence of gates that transforms

|f> to |f u1u2u3…um >



Time efficiency

One Writable Bit Universal? The only operations on one classical bit

are NOT and CNOT. Cannot achieve |f> to |f u1u2u3…um >

with any combination of NOT or CNOT gates with one input. NOT UNIVERSAL.

Are 2 bits universal?

Two Writable Bits Universal We use these 4 gates in our proof

Two Writable Bits Universal These correspond to these equations

Two Writable Bits Universal We will now prove that using these

functions we can transform the inputs |a>|b> to |a>|b u1u2 … uj >

Two Writable Bits Universal

Let S0 denote the N(1) gate above We can show that the sequence

performs the transform

Two Writable Bits Universal If we iterate this to m-1, we will get

Using this , we can come up with a sequence of gates

Two Writable Bits Universal Which results in the transform

This shows that two writable bits is universal by our definition.



Time efficiency

Quantum Time Efficiency Recall that

Quantum Time Efficiency Substituting

we are able to transform

Quantum Time Efficiency This generalizes to replacing each

with

And each with

Quantum Time Efficiency Thus we can take the AND of up to 2k

ROM bits using exactly 4k ROM calls.

Classical Time Efficiency 3 writable bit classical computers can

do this efficiently, but not 2 writable bits. Conjecture: It requires O(2j) ROM calls

for a 2 writable bit computer to perform

Classical Time Efficiency In a non-reversible setting however, the

classical two bit computer requires only j ROM calls as is shown in the following circuit. (O indicates re-initialization)

Conclusion In error-free non-reversible ROM computing: Quantum computers more space efficient

than their classical counterparts only requiring 1 writable qubit to be universal.

Conjecture: Minimal QC can calculate certain boolean functions exponentially faster than the minimal classical ROM computer.

rom-based computations: quantum versus classical b.c. travaglione, m.a.nielsen, h.m. wiseman, and a....

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