Reversible Causal Graph Dynamics

Abstract

Causal Graph Dynamics extend Cellular Automata to arbitrary, bounded-degree, time-varying graphs. The whole graph evolves in discrete time steps, and this global evolution is required to have a number of physics-like symmetries: shift-invariance (it acts everywhere the same) and causality (information has a bounded speed of propagation). We study a further physics-like symmetry, namely reversibility. We extend a fundamental result on reversible cellular automata by proving that the inverse of a causal graph dynamics is a causal graph dynamics. We also address the question of the evolution of the structure of the graphs under reversible causal graph dynamics, showing that any reversible causal graph dynamics preserves the size of all but a finite number of graphs.

A Proofs of Sections 4 and 5

1.1 A.1 Proofs of Section 4

Lemma 1

(Shift-equivalence classes isometry). Let $X\in \mathcal{X}_{\varSigma ,\varDelta ,\pi }$ be a graph. If $C_1\subseteq {V(X)}$ and $C_2\subseteq {V(X)}$ are two shift-equivalence classes of X, then $|C_1|=|C_2|$.

Proof

Consider two shift-equivalent and distinct vertices u and v in X. Consider a path w. The vertices u.w and v.w are shift-equivalent and distinct. More generally, if we have n shift-equivalent and distinct vertices $v_1,...,v_n$, any vertex $u=v_1.w$ will be shift-equivalent to $v_2.w,...,v_n.w$ and distinct from all of them, hence the equivalence classes are all of the same size.$\Box$

Lemma 2

(Invertible CGD preserves shift-equivalence classes). Let $(F,R_\bullet )$ be a shift-invariant dynamics over $\mathcal {X}_{\varSigma ,\varDelta ,\pi }$, such that F is a bijection. Then for any X and any $u,v\in X$, $u \approx v$ if and only if $R_X(u) \approx R_X(v)$.

Proof

$u\approx v$ expresses $X_u=X_v$, which by bijectivity of F is equivalent to $F(X_u)=F(X_v)$ and hence $F(X)_{R_X(u)}=F(X)_{R_X(v)}$. This in turn is expressed by $R_X(u) \approx R_X(v)$.$\Box$

Lemma 3

(Properties of primal extensions). Any primal extension ${}^\Box X$ is asymmetric.

Proof

As ${}^\Box X$ has a prime number of vertices, by Lemma 1, its has either one single equivalence class of maximal size or only trivial equivalence classes. As the primal extension adds at least two vertices and that these vertices have different degree (1 for the last vertex on the line, and 2 for its only neighbor), ${}^\Box X$ contains at least two non equivalent vertices, hence the first result.$\Box$

Theorem 1

(Invertible implies almost-vertex-preserving). Let $(F,R_\bullet )$ be a CGD over $\mathcal {X}_{\varSigma ,\varDelta ,\pi }$, such that F is a bijection. Then there exists a bound p, such that for any graph X, if $|X| > p$ then $R_X$ is bijective.

Proof

When $|\pi |\le 1$, $\mathcal {X}_{\varSigma ,\varDelta ,\pi }$ is finite so the theorem is trivial. So we assume in the rest of the proof that $|\pi |>1$.

[Finite graphs] First we prove the result for any finite graph. By contradiction, assume that there exists a sequence of finite graphs $(X(n))_{n\in \mathbb {N}}$ such that |X(n)| diverges and such that for all n, $R_{X(n)}$ is not bijective. As this sequence is infinite, we have that one of the two following cases is verified an infinite number of n:

• $R_{X(n)}$ is not surjective,

• $R_{X(n)}$ is not injective.

$\bullet$ $[R_{X(n)}$ not surjective]. There exists a vertex $v' \notin {\text {im}}R_{X(n)}$. Without loss of generality, we can assume that $|v'|<b$ where b is the bound from the boundedness property of F. We will now consider a particular primal extension of F(X(n)), ${}^\Box F(X(n))$, where the chosen vertex in F(X(n)) is the furthest away from the pointed vertex $\varepsilon$. Indeed, if F(X(n)) is large enough, a vertex lying at maximal distance of $\varepsilon$ in F(X(n)) either has a free port or is part of a cycle, and thus is a valid vertex to perform the primal extension. Indeed, if this vertex has no free port, then any of its edge can be removed without splitting the graph, as it would contradict its maximality – therefore it is in a cycle. Now, consider the graph $Y(n)=F^{-1}({}^\Box F(X(n)))$. Using uniform continuity of $F^{-1}$ and $R_\bullet$, and the fact that |X(n)| is as big as we want, we have that there exists an index n and a radius r such that $Y(n)^r=X(n)^r$ and $R^b_{Y(n)^r}=R^b_{X(n)^r}$. As F(Y(n)) is asymmetric by construction, $v'\in {\text {im}}R_{Y(n)^r}^b$ which contradicts $v' \notin {\text {im}}R_{X(n)}$.

$\bullet$ $[R_{X(n)}$ not injective]. There exist two vertices $u,v \in X(n)$ such that $R_{X(n)}(u)=R_{X(n)}(v)$ and $u\ne v$. Without loss of generality, we can assume that $u =\varepsilon$ as F is shift-invariant. According to Lemma 2, we have that $\varepsilon \approx v$. Moreover, using the uniform continuity of $R_\bullet$, we have that, as $R_{X(n)}(v)=R_{X(n)}(\varepsilon )=\varepsilon$, there exists a radius l, which does not depend on n, such that $|v| <l$ . Let us consider a primal extension of X(n), ${}^\Box X(n)$, where the primal extension has been performed at maximal distance from $\varepsilon$, by the same argument as in the previous $\bullet$. In this graph, $\varepsilon$ and v are not shift-equivalent and thus, $R_{{}^\Box X(n)}(\varepsilon )\ne R_{{}^\Box X(n)}(v)$ . By continuity of $R_\bullet$, we have that there exists a radius $r>l$ such that $R_{{}^\Box X(n)^r}^0=R_{ X(n)^r}^0$ for a large enough n, hence $R_{{}^\Box X(n)^r}^0(v)=R_{ X(n)^r}^0(v)=\varepsilon$, which contradicts $R_{{}^\Box X(n)}(\varepsilon )\ne R_{{}^\Box X(n)}(v)$.

[Infinite graphs] Now we show that the result on finite graphs can be extended to infinite graphs, proving that for any infinite graph $R_X$ is bijective:

$\bullet$ $[R_X$ injective ]. By contradiction. Take X infinite such that there is $u\ne v$ and $R_X(u)=R_X(v)$. Without loss of generality we can take $u=\varepsilon$, i.e. $v\ne \varepsilon$ and $R_X(v)=\varepsilon$. By continuity of $R_\bullet$, there exists a radius r, which we can take larger than |v| and p, such that $R_{X}=R_{X^r}$. Then $R_{X^r}(v)=R_X(v)=\varepsilon$, thus $R_{X^r}$ is not injective in spite of $X^r$ being finite and larger than p, leading to a contradiction.

$\bullet$ $[R_X$ surjective ]. By contradiction. Take X infinite such that there is $v'$ in F(X) and $v'\notin {\text {im}}R_X$. By boundedness, there exists $u'\in F(X)$ such that $u'$ lies at distance less than b of $v'$. Using shift-invariance, we can assume without loss of generality that $u'=\varepsilon$, hence, $|v'|<b$. By continuity of $R_\bullet$, there exists a radius r, which we can take larger than p, such that the images of $R_{X}$ and $R_{X^r}$ coincide over the disk of radius b. Then, $v'\notin {\text {im}}R_X$ implies $v'\notin {\text {im}}R_{X^r}$, thus $R_{X^r}$ is not surjective in spite of $X^r$ being finite and larger than p, leading to a contradiction.

1.2 A.2 Proofs of Section 5

Lemma 4

If $(F,R_\bullet )$ is an invertible, shift-invariant dynamics such that for all X, $R_X$ is a bijection, then $(F^{-1},S_\bullet )$ is a shift-invariant dynamics, with $S_{Y}=(R_{F^{-1}(Y)})^{-1}$.

Proof

Consider Y and $u'.v'\in Y$. Take X and $u.v\in X$ such that $F(X)=Y$, $R_X(u)=u'$ and $R_X(u.v)=u'.v'$. We have: ${F^{-1}(Y_{u'})} ={F^{-1}(F(X)_{R_X(u)})}={F^{-1}(F(X_u))}={X_{(R_X)^{-1}(u')}}={F^{-1}(Y)_{S_Y(u')}}$. Moreover, take $v\in X_u$ such that $R_X(u.v)=R_X(u).R_{X_u}(v)=u'.v'$. We have: $S_Y(u'.v') =(R_X)^{-1}(R_X(u.v)) =u.v =(R_{X})^{-1}(u').(R_{X_u})^{-1}(v')= S_Y(u').S_{Y_{u'}}(v')$.    $\square$

Theorem 2

(Invertible implies reversible). If $(F,R_\bullet )$ is an invertible CGD, then $(F,R_\bullet )$ is reversible.

Proof

Continuity of $F^{-1}$ is directly given by the continuity of F together with the compactness of $\mathcal {X}_{\varSigma ,\varDelta ,\pi }$. Its boundedness derives either from the bijectivity of $R_X$ for $|X|>p$ or from the finiteness of X when $|X|>p$.

We must construct $S_\bullet$. For $|F(X)|=|X|>p$, we know that $R_X$ is bijective and we let $S_{F(X)}=R_X^{-1}$. For $|X|\le p$, we will proceed in two steps. First, we will construct an appropriate $S_{F(X)}$ for X. Second, we will make consistent choices for $S_{F(X)_{u'}}$ so that $S_\bullet$ is shift invariant.

We write $\tilde{u}$ for the shift-equivalence class of u in X. For all $v'\in F(X)$, we make the arbitrary choice $S_{F(X)}(\tilde{v'})=v$, where v is such that its image $R_X(v)$ is shift equivalent to $v'$ in F(X), i.e. $R_X(v)\approx v'$. For this X, we have enforced $\approx$-compatibility. Then we make consistent choices for $S_{F(X)_{u'}}$. This is obtained by demanding that $S_{F(X)_{u'}}(\widetilde{\overline{u'}.v'})=\overline{u}.v$. Indeed, this accomplishes shift-invariance because $S_{F(X)_{u'}}(v')=S_{F(X)_{u'}}(\overline{u'}.u'.v')=\varepsilon .v'=v'$ implying the equality: $S_{F(X)}(u'.v')=u.v=S_{F(X)}(u').S_{F(X)_{u'}}(v')$. Moreover, $S_{F(X)_{u'}}$ is itself shift-invariant because: $S_{F(X)_{u'.v'}}(w')=S_{F(X)_{u'.v'}}(\overline{u'.v'}.u'.v'.w')=\overline{u.v}.u.v.w=w$, and $S_{F(X)_{u'}}(v')=v$ implying that $S_{F(X)_{u'}}(v'.w')=v.w=S_{F(X)_{u'}}(v').S_{F(X)_{u'.v'}}(w')$ , and $\approx$-compatible because $v'\approx w'$ implies $S_{F(X)_{u'}}(v')=S_{F(X)_{u'}}(w')$, and thus $S_{F(X)_{u'}}(v')\approx S_{F(X)_{u'}}(w')$.

Continuity of the constructed $S_\bullet$ is due to the continuity of $R_\bullet$ and the finiteness of p.

Shift-invariance of $(F^{-1},S_\bullet )$ follows from $\approx$-compatibility of $S_\bullet$ and shift-invariance of $(F,R_\bullet )$, because $F^{-1}(F(X)_u')=X_v$ where v is such that $R_X(v)\approx u'$, hence $F^{-1}(F(X)_u')=X_{S_{F(X)}(u')}$.    $\square$