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I am trying to define a mathematical framework that starts from a generic function $f$, which I can synthesis as a circuit of generalised Toffoli gates -- i.e. $m$ controls, 1 target.

I would like to read of a framework similar to the one used for the group of linear reversible boolean functions. Which relates to circuits of CNOTs only.

For the case of linear reversible boolean functions, we have $$f: \mathbb{F}_2^n \mapsto \mathbb{F}_2^n,$$ which is also defined by an invertible boolean $n$-by-$n$ matrix.

Assuming a boolean function group defined by a matrix $M_{f}$. I am looking for a decomposition of such matrix, where each component can be translated into a Toffoli.

Hence the steps are:

  1. Defining a boolean function $f$;
  2. defining the matrix operator $M_f$;
  3. defining a method to decompose $M_f$.
Daniele Cuomo
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    Would it not be $S_{2^n}$, the symmetric group on $2^n$ elements, by virtue of the universality of Toffoli gates? I think I'm missing something. – Mark Spinelli Jan 29 '24 at 20:16
  • I don't know to be fair. May you explain what you think in an answer? – Daniele Cuomo Jan 29 '24 at 21:00
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    You should define what a generalized Toffoli gate is. Is it a Toffoli gate that acts on arbitrary triplets of commuting Pauli products instead of specifically on the single qubit Z operators of its controls and the X operator of its target? – Craig Gidney Jan 29 '24 at 22:56
  • It would be also interesting to see how linear reversible boolean function can be synthesised into generalised Toffoli gates, rather than CNOTs. – Daniele Cuomo Jan 30 '24 at 13:49

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Further to your definition, an early result in quantum computing shows that for $m\gt 1$ the generalized Toffoli gate (or C$^m$NOT gate) is computationally universal, at least because judicious choice of the control inputs instantiates AND, OR, or NOT gates.

More importantly any circuit acting on $n$ qubits (or, bits) comprising only CCNOT gates can be represented as a permutation matrix on $2^n$ inputs (and not, as indicated in the question, an $n\times n$ boolean matrix), and further all such $2^n\times 2^n$ matrices can be realized with generalized Toffoli gates.

Each such matrix is a representation of an element of the symmetric group $S_{2^n}$, with composition being the group operation. This is a big group! It has $(2^n)!$ elements.

With three bits (qubits) there are $\Vert S_8\Vert=8!=40320$ such permutation matrices.

Analogously for $m\ge 1$ the generalized Fredkin gate (or C$^m$SWAP gate) is computationally universal, and each circuit can similarly be represented as a permutation matrix; however, for the generalized Fredkin gate, not all such $2^n\times 2^n$ matrices can be realized, because SWAP gates must leave the Hamming weight invariant (the number of $1$'s in the input to the circuit must be the same as the number of $1$'s in the output).

With only three bits (qubits) and not allowing any ancilla, I find that there are only $1\times 3\times 2\times 3\times 1\times 2\times 1\times 1=36$ such matrices; I do not know what, if any, group this corresponds to.

You may also enjoy Gajewski's PhD thesis (PDF), or Aaronson, Grier, and Schaeffer's classification of reversible circuits (arXiv abstract).

Mark Spinelli
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  • Thank you! Please notice that the n-to-n boolean matrix is correct and can encode linear reversible boolean functions. – Daniele Cuomo Jan 30 '24 at 16:24
  • @DanieleCuomo How? Even a truth-table on $n$ bits has $2^n$ rows...? – Mark Spinelli Jan 30 '24 at 16:53
  • See, for example, "Reducing the depth of linear reversible quantum circuits" – Daniele Cuomo Jan 30 '24 at 17:49
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    @DanieleCuomo Oh I think I see your point. An $n$-bit input $x$ gets mapped into an $n$-bit output $f(x)$ through a matrix $A$ such that $f(x)=A x$. But for each of the $2^n$ different inputs $x$ there's a different matrix $A$. – Mark Spinelli Jan 30 '24 at 18:43
  • Exactly. I am looking for something similar. Some compact definition (as $f(x) = Ax$ is), which relates to Toffoli circuits. – Daniele Cuomo Jan 30 '24 at 22:06
  • The Aaronson et al. paper might be your best bet for more info, at least for now. A deep enough Toffoli circuit can achieve an almost any matrix $A$ I bet, but a Fredkin circuit is invariant to the Hamming weight. Aaronson et al. classify a lot of the invariants. – Mark Spinelli Jan 31 '24 at 00:06