A Note on the Power of Non-Deterministic Circuits with Gate Restrictions

TL;DR

This paper explores how non-deterministic circuits with limited gate sets compare to deterministic ones, revealing a dichotomy where their relative power depends on the gate base, with implications for computational complexity.

Contribution

It provides a clear characterization of when non-deterministic circuits outperform deterministic circuits under gate restrictions, based on the strength of the base gates.

Findings

Non-deterministic circuits are equivalent to deterministic ones over weak gate bases.

Over stronger bases, non-deterministic circuits are super-polynomially more efficient.

A precise boundary between the two regimes is established.

Abstract

We investigate the power of non-deterministic circuits over restricted sets of base gates. We note that the power of non-deterministic circuits exhibit a dichotomy, in the following sense: For weak enough bases, non-deterministic circuits are no more powerful than deterministic circuits, and for the remaining bases, non-deterministic circuits are super polynomial more efficient than deterministic circuits (under the assumption that $P/poly \neq NP/poly$). Moreover, we give a precise characterization of the borderline between the two situations.

Full text

1 Introduction

In this note we are interested in finding evidence in favor of the belief that non-deterministic circuits are more powerful than deterministic circuits, and $P/poly\neq NP/poly$. Except for the well-known result that the polynomial hierarchy collapses if $P/poly=NP/poly$ [2], there seems to be little formal evidence in this direction. For example, we do not know of a function family $\{f_{n}\}_{n\geq 1}$ having non-deterministic circuit complexity $kn$, and deterministic circuit complexity of at least $(k+\epsilon)n$.

A natural path for making progress on this question is to prove that non-deterministic circuits over some restricted set of base gates $G$ (i.e., non-deterministic $G$-circuits) are more powerful than deterministic $G$-circuits. Unfortunately, we are not aware of any results in this direction either. On the contrary, for one of the most well investigated restriction on $G$, namely monotone circuits (i.e., $G=\{\land,\lor,0,1\}$), it is well known that non-deterministic $G$-circuits are no more powerful than deterministic $G$-circuits.

Given a finite set of base gates $G$, let $[G]$ denote the set of all functions/gates that can be implemented by circuits over $G$ (i.e., implemented by a $G$-circuit). Sets of the form $[G]$ are called clones, and Post [3] classified all $[G]$ over the Boolean domain. Sets $[G]$ form a lattice under set inclusion and the cardinality of the lattice is countable infinite. For example, the class of all monotone Boolean functions can be computed by circuits consisting of only $\land$ and $\lor$ gates and constants [math] and $1$, or in other words, the clone of all monotone Boolean functions is $[\land,\lor,0,1]$. For more information on Post’s lattice, see [1].

By making heavy use of Post’s lattice, we are able to make the following observations: If the constants $0,1$ are in the base $G$, and $G$ is not a full base (i.e., $[G]\neq[\land,\lor,\neg]$), then non-deterministic $G$-circuits are no more powerful than deterministic $G$-circuits. Furthermore, if the base $G$ is monotone, linear, or self-dual, then non-deterministic $G$-circuits are no more powerful than deterministic $G$-circuits. For the remaining bases $G$, non-deterministic $G$-circuits are super polynomial more efficient than deterministic $G$-circuits (under the assumption that $P/poly\neq NP/poly$).

1.1 Preliminaries

A Boolean circuit is a directed acyclic graph with three types of labeled vertices: sources (in-degree [math]) labeled $x_{1},\dots,x_{n}$, a sink (the output), and vertices with in-degree $k>0$ are gates labeled by Boolean functions on $k$ inputs. A non-deterministic circuit has, in addition to the inputs $x=(x_{1},\dots,x_{n})$, a set of “non-deterministic” inputs $y=(y_{1},\dots,y_{m})$. A non-deterministic circuit $C$ accepts input $x$ if there exists $y$ such that the circuit output $1$ on $(x,y)$. Let $|C|$ denote the number of gates of a circuit $C$.

A family of non-deterministic circuits $\{C_{n}\}_{n\geq 1}$, with $C_{n}$ having $n$ (ordinary) inputs, decide a language $L$ if each $C_{n}$ decide $L_{n}$ (i.e., $C_{n}$ accepts $x$ if and only if $|x|=n$ and $x\in L$). The class $NP/poly$ is defined as the class of languages decidable by non-deterministic circuit families $\{C_{n}\}$, with $|C_{n}|\leq poly(n)$. Recall that $P/poly$ is the class of languages decidable by (deterministic) circuit families $\{C_{n}\}$, with $|C_{n}|\leq poly(n)$.

By a circuit we always mean a deterministic circuit, unless it is explicitly said to be a non-deterministic circuit, or it is clear from the context. We assume all sets of base gates $G$ to be finite. By abusing notation slightly, if a language $L$ is decidable by deterministic $G$-circuits, we denote this by $L\in[G]$.

Definition 1

We say that a set of base gates $G$ lack non-deterministic power, if any language $L\in[G]$ that has non-deterministic $G$-circuit complexity $s(n)$, has deterministic $G$-circuit complexity $O(s(n))$.

Definition 2

We say that a set of base gates $G$ has full non-deterministic power if, under the assumption that $P/poly\neq NP/poly$, there is a language $L\in[G]$ that has polynomial non-deterministic $G$-circuit complexity, but does not have polynomial deterministic $G$-circuit complexity.

2 Classification

Proposition 3

Given finite sets of base gates $G_{1}$ and $G_{2}$ with $[G_{1}]=[G_{2}]$, then $G_{1}$ lack non-deterministic power if and only if $G_{2}$ lack non-deterministic power.

Proof.

If $G_{1}\subseteq[G_{2}]$, then every gate $g(x_{1},\dots,x_{k})\in G_{1}$ has an implementation of size $c^{k}$ using gates from $G_{2}$. Hence, we can convert any $G_{1}$-circuit into an equivalent $G_{2}$-circuit without blowing up the size more than a constant factor. Similarly, if $G_{2}\subseteq[G_{1}]$, then any $G_{2}$-circuit can be converted to an equivalent $G_{1}$-circuit without increasing the size more than a constant factor. The same holds for non-deterministic circuits, and the result follows. ∎

Corollary 4

Given finite sets of base gates $G_{1}$ and $G_{2}$ with $[G_{1}]=[G_{2}]$, then $G_{1}$ has full non-deterministic power if and only if $G_{2}$ has full non-deterministic power.

Proof.

By the proof of Proposition 3. ∎

Definition 5

A Boolean function $f(x_{1},\dots,x_{n})$ is said to be self-dual if $f(x_{1},\dots,x_{n})=\overline{f(\overline{x_{1}},\dots,\overline{x_{n}})}$ for all $x_{1},\dots,x_{n}\in\{0,1\}$, where $\overline{0}=1$ and $\overline{1}=0$.

From Post’s classification of Boolean clones [3] we know that the function $d(x_{1},x_{2},x_{3})=(x_{1}\land\neg x_{2})\lor(\neg x_{2}\land\neg x_{3})\lor(\neg x_{3}\land x_{1})$ generates the clone of all self-dual Boolean functions. In other words, any self-dual Boolean function can be computed by a circuit consisting of $d(x_{1},x_{2},x_{3})$-gates (from now on referred to as $d$-gates).

Definition 6

A set of base gates $G$ is said to be

• •

monotone if $G\subseteq[\land,\lor,0,1]$,

• •

linear if $G\subseteq[\oplus,1]$, and

• •

self-dual if $G\subseteq[d(x_{1},x_{2},x_{3})]$.

Proposition 7

If $G$ is self-dual, then $G$ lacks non-deterministic power.

Proof.

Given a non-deterministic $d$-circuit $C(x,y)$ with $x=(x_{1},\dots,x_{n})$ and $y=(y_{1}\dots,y_{m})$ computing $f$, recall that $f(x)=1$ if and only if there is a $y$ such that $C(x,y)=1$. Assume there is $y$ and $y^{\prime}$ such that $C(x,y)=1$ and $C(x,y^{\prime})=0$, then $C(\overline{x},\overline{y^{\prime}})=1$, because $C$ consists of $d$-gates which are self-dual. Hence, $f(x)=1=f(\overline{x})$ which is impossible since $f$ is self-dual. Thus, if $f(x)=1$, then $C(x,y)=1$ for all $y$. To construct the equivalent deterministic $d$-circuit $C^{\prime}(x)$ we replace all $y_{i}$ inputs in $C(x,y)$ with $x_{1}$ (i.e., we replace each non-deterministic variables by the ordinary variable $x_{1}$). Note that the more natural transformation of replacing the $y_{i}$’s by constants, does not work, since the resulting circuit is then not necessarily a $d$-circuit. ∎

Proposition 8

If $G$ is monotone, then $G$ lacks non-deterministic power.

Proof.

The result follows from the fact that a non-deterministic monotone $G$-circuit $C(x,y)$ outputs $1$ on input $x,y$ if and only if $C(x,1)=1$. That is, given a non-deterministic monotone $G$-circuit $C(x,y)$ we can construct an equivalent deterministic monotone $G$-circuit, without increasing the size, by replacing all $y$ variables with the constant $1$. ∎

Proposition 9

If $G$ is linear, then $G$ lacks non-deterministic power.

Proof.

Any function $L\in[G]$ for a linear base $G$ can be computed by a deterministic $G$-circuit of size $O(n)$. ∎

Proposition 10

Any $G$ such that $\{x\land(y\lor\neg z)\}\subseteq[G]$ has full non-deterministic power.

Proof.

By Post’s lattice, $[G,0,1]=[\land,\lor,\neg]$. Hence, any circuit (over any finite basis) can be converted into an equivalent $\{G,0,1\}$ circuit $C$, without blowing up the size more than a constant factor. Note that $x\land(y\lor z)\in[G]$, and consider the $G$-circuit $C^{\prime}=x^{\prime}\land(C\lor x^{\prime\prime})$, where all $1$’s and [math]’s in $C$ have been replaced by $x^{\prime}$ and $x^{\prime\prime}$ respectively. The transformation can be carried out both for deterministic and non-deterministic circuits.

Given $L$, having polynomial non-deterministic complexity and super polynomial deterministic complexity over the full basis (such an $L$ exist under the assumption that $P/poly\neq NP/poly$). Consider, $L^{\prime}=\{(x,1,0)\mid x\in L\}\cup\{(x,1,1)\mid x\in\{0,1\}^{*}\}$. If $C_{n}(x,y)$ is a family of non-deterministic circuits deciding $L$, then $C_{n}^{\prime}(x,x^{\prime},x^{\prime\prime},y)$ (as defined above) is a family of non-deterministic circuits over the basis $[G]$ deciding $L^{\prime}$. Hence, $L^{\prime}$ has polynomial non-deterministic complexity over the basis $[G]$. Assume towards contradiction that $L^{\prime}$ has polynomial deterministic complexity over the basis $[G]$, i.e., there is a polynomial size circuit family $C^{\prime}_{n}(x,x^{\prime},x^{\prime\prime})$ deciding $L^{\prime}$. Then, $C^{\prime}_{n}(x,1,0)$ is a polynomial size circuit family over the full basis $[\land,\lor,\neg]$ deciding $L$, contradicting that $L$ has super polynomial deterministic circuit complexity. ∎

Proposition 11

Any $G$ such that $\{x\lor(y\land\neg z)\}\subseteq[G]$ has full non-deterministic power.

Proof.

By Post’s lattice, $[G,0,1]=[\land,\lor,\neg]$. Hence, any circuit (over any finite basis) can be converted into an equivalent $\{G,0,1\}$ circuit $C$, without blowing up the size more than a constant factor. Note that $x\lor(y\land z)\in[G]$, and consider the $G$-circuit $C^{\prime}=x^{\prime}\lor(C\land x^{\prime\prime})$, where all [math]’s and $1$’s in $C$ have been replaced by $x^{\prime}$ and $x^{\prime\prime}$ respectively. The transformation can be carried out both for deterministic and non-deterministic circuits.

Given $L$, having polynomial non-deterministic complexity and super polynomial deterministic complexity over the full basis. Consider, $L^{\prime}=\{(x,0,1)\mid x\in L\}\cup\{(x,1,x^{\prime\prime})\mid x\in\{0,1\}^{*}\;and\;x^{\prime\prime}\in\{0,1\}\}$. If $C_{n}(x,y)$ is a family of non-deterministic circuits deciding $L$, then $C_{n}^{\prime}(x,x^{\prime},x^{\prime\prime},y)$ is a family of non-deterministic circuits over the basis $[G]$ deciding $L^{\prime}$. Hence, $L^{\prime}$ has polynomial non-deterministic complexity over the basis $[G]$. Assume towards contradiction that $L^{\prime}$ has polynomial deterministic complexity over the basis $[G]$, i.e., there is a polynomial size circuit family $C^{\prime}_{n}(x,x^{\prime},x^{\prime\prime})$ deciding $L^{\prime}$. Then, $C^{\prime}_{n}(x,0,1)$ is a polynomial size circuit family over the full basis $[\land,\lor,\neg]$ deciding $L$, contradicting that $L$ has super polynomial deterministic circuit complexity. ∎

Proposition 12

If $G$ is self-dual, monotone, or linear, then $G$ lack non-deterministic power. All other $G$ have full non-deterministic power.

Proof.

By the results above and inspection of Post’s lattice. In particular, if $G$ is neither self-dual, monotone, nor linear, then by Post’s lattice $\{x\land(y\lor\neg z)\}\subseteq[G]$ or $\{x\lor(y\land\neg z)\}\subseteq[G]$. ∎

Corollary 13

Any $G$ such that $\{0,1\}\subseteq[G]$ and $[G]\neq[\land,\lor,\neg]$, lacks non-deterministic power.

Proof.

By the results above and inspection of Post’s lattice. ∎

3 Final Remarks

Coming back to our original motivation for studying non-deterministic versus deterministic $G$-circuit complexity. We note that $G=\{x\land(y\lor\neg z)\}$ and $G=\{x\lor(y\land\neg z)\}$ are the two weakest bases for which it is possible that non-deterministic $G$-circuits are more powerful than deterministic $G$-circuits (indeed, this is the case assuming $P/poly\neq NP/poly$). Unfortunately, as the following two propositions show, it is probably somewhat challenging to prove super polynomial lower bounds for deterministic $G$-circuits (with $G=\{x\land(y\lor\neg z)\}$ or $G=\{x\lor(y\land\neg z)\}$), since this implies that $P/poly\neq NP/poly$.

Proposition 14

Let $L\in G=[x\land(y\lor\neg z)]$ with deterministic circuit complexity $s(n)$ over the full basis $[\land,\lor,\neg]$. Then, $L$ has $O(s(n))$ deterministic $G$-circuit complexity.

Proof.

First note that $G$ is $1$-reproducing, which means that all constant $1$ vectors are in $L$ (i.e., $(1,\dots,1)\in L$). We first show that such an $L$ can be computed by $\{\land,\lor,x\oplus y\oplus 1\}$ circuits of size $O(s(n))$ if it can be computed by $[\land,\lor,\neg]$ circuits of size $s(n)$. Hence, it suffices to show how to simulate $\neg$ gates with $x\oplus y\oplus 1$ gates. Given a $\{\land,\lor,\neg\}$ circuit for $L_{n}$ with inputs $x_{1},\dots,x_{n}$ we construct an equivalent circuit $C_{n}$ by replacing all $\neg$ gates with $x\oplus y\oplus 1$ gates, where the first input is the original input to the $\neg$ gate and the second input is the output of $x_{1}\land\dots\land x_{n}$.

Observe that since $\{\land,\lor,x\oplus y\oplus 1\}$ is $1$ reproducing, $C_{n}(1,\dots,1)=1$ (as it should, since $(1,\dots,1)\in L_{n}$). For every $(x_{1},\dots,x_{n})\neq(1,\dots,1)$, the $x\oplus y\oplus 1$ gates works like a $\neg$ gates on its first input (since $x_{1}\land\dots\land x_{n}=0$).

By Post’s lattice we know that $[\land,\lor,x\oplus y\oplus 1]=[G,1]$, and hence $L$ has circuit complexity $O(s(n))$ over $\{G,1\}$, via a circuit family $C_{n}$. The base $G=[x\land(y\lor\neg z)]$ is $1$-separating, which means that for any $G$-circuit $C(x_{1},\dots,x_{n})$ there is an $1\leq i\leq n$, such that $x_{i}=1$ for all $\{(x_{1},\dots,x_{n})\mid C(x_{1},\dots,x_{n})=1\}$. Since $L\in G$ and $G$ is $1$-separating, there is for each $L_{n}$, an $1\leq i\leq n$ such that $x_{i}=1$ for all $(x_{1},\dots,x_{n})\in L_{n}$. Hence, $C^{\prime}_{n}=x_{i}\land C_{n}$ (replacing all occurrences of $1$ in $C_{n}$ by $x_{i}$) is a $G$-circuit family of size $O(s(n))$ deciding the language $L$. ∎

Proposition 15

Let $L\in G=[x\lor(y\land\neg z)]$ with deterministic circuit complexity $s(n)$ over the full basis $[\land,\lor,\neg]$. Then, $L$ has $O(s(n))$ deterministic $G$-circuit complexity.

Proof.

First note that $G$ is [math]-reproducing, which means that all constant [math] vectors are in $L$ (i.e., $(0,\dots,0)\in L$). We first show that such an $L$ can be computed by $\{\land,\lor,\oplus\}$ circuits of size $O(s(n))$ if it can be computed by $[\land,\lor,\neg]$ circuits of size $s(n)$. Hence, it suffices to show how to simulate $\neg$ gates with $\oplus$ gates. Given a $\{\land,\lor,\neg\}$ circuit for $L_{n}$ with inputs $x_{1},\dots,x_{n}$ we construct an equivalent circuit $C_{n}$ by replacing all $\neg$ gates with $\oplus$ gates, where the first input is the original input to the $\neg$ gate and the second input is the output of $x_{1}\lor\dots\lor x_{n}$.

Observe that since $\{\land,\lor,\oplus\}$ is [math] reproducing, $C_{n}(0,\dots,0)=0$ (as it should, since $(0,\dots,0)\in L_{n}$). For every $(x_{1},\dots,x_{n})\neq(0,\dots,0)$, the $\oplus$ gates works like a $\neg$ gates on its first input (since $x_{1}\lor\dots\lor x_{n}=1$).

By Post’s lattice we know that $[\land,\lor,\oplus]=[G,0]$, and hence $L$ has circuit complexity $O(s(n))$ over $\{G,0\}$, via a circuit family $C_{n}$. The base $G=[x\lor(y\land\neg z)]$ is [math]-separating, which means that for any $G$-circuit $C(x_{1},\dots,x_{n})$ there is an $1\leq i\leq n$, such that $x_{i}=0$ for all $\{(x_{1},\dots,x_{n})\mid C(x_{1},\dots,x_{n})=0\}$. Since $L\in G$ and $G$ is [math]-separating, there is for each $L_{n}$, an $1\leq i\leq n$ such that $x_{i}=0$ for all $(x_{1},\dots,x_{n})\notin L_{n}$. Hence, $C^{\prime}_{n}=x_{i}\lor C_{n}$ (replacing all occurrences of [math] in $C_{n}$ by $x_{i}$) is a $G$-circuit family of size $O(s(n))$ deciding the language $L$. ∎