The Predecessor-Existence Problem for $k$-Reversible Processes

The case of a single (necessarily locked) MCS is trivial. If, on the other hand, more than one MCS exists in $G$, then clearly the two conditions suffice for $Y^{\prime}$ to exist and be as stated: in one time step from $Y^{\prime}$, the state of every vertex in a locked MCS remains unchanged and that of every vertex in an unlocked MCS changes, thus yielding $Y$.

It remains for necessity to be shown. We do this by noting that, should the first condition fail and at least two locked MCSs be neighbors, any prospective $Y^{\prime}$ would have to differ from $Y$ in all vertices of each of these MCSs, but the presence of locked vertices in them would make it impossible for $Y$ to be obtained in one time step. Should the second condition be the one to fail and at least one vertex in an unlocked MCS have neighbors outside its MCS only in locked MCSs, any prospective $Y^{\prime}$ would have to differ from $Y$ in all vertices of such an unlocked MCS. Once again it would be impossible to obtain $Y$ in one time step due to the locked vertices. It follows that both conditions are necessary for $Y^{\prime}$ to exist. ∎

The problem is trivially in NP. We proceed with the reduction from another variation of the 3Sat problem, known as 3Sat Exactly-One [8], which is NP-complete and asks whether there exists an assignment of variables satisfying each clause by exactly one positive literal.

The reduction is simple and consists of inverting all literals in all clauses of an instance $S$ of 3Sat Exactly-One, resulting in an instance $S^{\prime}$ of 3Sat Exactly-Two. It is easy to check that a solution for $S$ directly gives a solution for $S^{\prime}$, and conversely, a solution for $S^{\prime}$ directly gives a solution for $S$. ∎

Given two configurations $Y$ and $Y^{\prime}$, verifying whether $Y^{\prime}$ is a predecessor configuration of $Y$ is straightforward and can be done by simulating one step of the $k$-reversible process starting with configuration $Y^{\prime}$. Simulating one step of the process takes $O(n+m)$ time and the final comparison between the resulting configuration and $Y$ takes $O(n)$ time; thus, Pre($k$) is in NP. The remainder of the proof is a reduction from 3Sat Exactly-Two, which by Lemma 2 is NP-complete.

Let $S$ be an arbitrary instance of 3Sat Exactly-Two with the $M$ clauses $c_{1},c_{2},\dots,c_{M}$ and the $N$ variables $x_{1},x_{2},\dots,x_{N}$. We construct an instance $(G,Y)$ of Pre$(k)$ from $S$ as follows.

Vertex set $V(G)$ is the union of:

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  $\{x_{i},\neg{x_{i}}\}$, for each variable $x_{i}$ in $S$;

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  $\{z_{i},z^{\prime}_{i}\}$, for each variable $x_{i}$ in $S$;

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  $\{u_{i,1},\dots,u_{i,2k-3}\}$, for each variable $x_{i}$ in $S$;

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  $\{p_{i,1},\dots,p_{i,2k-3}\}$, for each variable $x_{i}$ in $S$;

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  $\{w_{i,1},\dots,w_{i,k-2}\}$, for each variable $x_{i}$ in $S$, provided $k>2$;

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  $\{w^{\prime}_{i,1},\dots,w^{\prime}_{i,k-2}\}$, for each variable $x_{i}$ in $S$, provided $k>2$;

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  $\{c_{i},c^{\prime}_{i}\}$, for each clause $c_{i}$ in $S$;

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  $\{b_{i,1},\dots,b_{i,k-2}\}$, for each clause $c_{i}$ in $S$, provided $k>2$;

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  $\{b^{\prime}_{i,1},\dots,b^{\prime}_{i,k-1}\}$, for each clause $c_{i}$ in $S$.

Vertices $x_{i}$ and $\neg{x_{i}}$ are called literal vertices and vertices $c_{i}$ and $c^{\prime}_{i}$ are called clause vertices. If $x$ is a neighbor of $u$ and $x$ is a literal vertex, then we say that $x$ is a literal neighbor of $u$. Similarly, if $x$ is a neighbor of $u$ and $x$ is a clause vertex, then $x$ is a clause neighbor of $u$.

Edge set $E(G)$ is the union of:

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  $\{(x_{i},z_{i}),(x_{i},z^{\prime}_{i}),(\neg{x_{i}},z_{i}),(\neg{x_{i}},z^{\prime}_{i})\}$, for each variable $x_{i}$ in $S$;

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  $\{(x_{i},u_{i,1}),\dots,(x_{i},u_{i,2k-3})\}$, for each variable $x_{i}$ in $S$;

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  $\{(\neg{x_{i}},p_{i,1}),\dots,(\neg{x_{i}},p_{i,2k-3})\}$, for each variable $x_{i}$ in $S$;

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  $\{(z_{i},w_{i,1}),\dots,(z_{i},w_{i,k-2})\}$, for each variable $x_{i}$ in $S$, provided $k>2$;

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  $\{(z^{\prime}_{i},w^{\prime}_{i,1}),\dots,(z^{\prime}_{i},w^{\prime}_{i,k-2})\}$, for each variable $x_{i}$ in $S$, provided $k>2$;

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  $\{(c^{\prime}_{j},b^{\prime}_{j,1}),\dots,(c^{\prime}_{j},b^{\prime}_{j,k-1})\}$, for each clause $c_{j}$ in $S$;

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  $\{(c_{j},b_{j,1}),\dots,(c_{j},b_{j,k-2})\}$, for each clause $c_{j}$ in $S$, provided $k>2$;

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  $\{(c_{j},x_{i}),(c^{\prime}_{j},x_{i})\}$, for each literal $x_{i}$ occurring in clause $c_{j}$;

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  $\{(c_{j},\neg{x_{i}}),(c^{\prime}_{j},\neg{x_{i}})\}$, for each literal $\neg{x_{i}}$ occurring in clause $c_{j}$.

We finish the construction by defining the target configuration $Y$:

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  $Y(x_{i})=Y(\neg{x_{i}})=+1$, for $1\leq i\leq N$;

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  $Y(z_{i})=+1$, $Y(z^{\prime}_{i})=-1$, for $1\leq i\leq N$;

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  $Y(u_{i,j})=Y(p_{i,j})=+1$, for $1\leq i\leq N$ and $1\leq j\leq k-1$;

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  $Y(u_{i,j})=Y(p_{i,j})=-1$, for $1\leq i\leq N$ and $k\leq j\leq 2k-3$, provided $k>2$;

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  $Y(w_{i,j})=-1$, $Y(w^{\prime}_{i,j})=+1$, for $1\leq i\leq N$ and $1\leq j\leq k-2$, provided $k>2$;

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  $Y(c_{i})=+1$, $Y(c^{\prime}_{i})=-1$, for $1\leq i\leq M$;

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  $Y(b_{i,j})=-1$, for $1\leq i\leq M$ and $1\leq j\leq k-2$, provided $k>2$;

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  $Y(b^{\prime}_{i,j})=-1$, for $1\leq i\leq M$ and $1\leq j\leq k-1$, provided $k>2$.

Figure 1 illustrates the case of $k=3$, $M=1$, and $N=3$, the single clause being $c_{1}=x_{1}\vee\neg{x_{2}}\vee\neg{x_{3}}$.

For each variable in $S$, at most $6k-6$ vertices are created, and for each clause at most $2k-1$ vertices, resulting in at most $n=N(6k-6)+M(2k-1)$ vertices. Likewise, the total number of edges is at most $m=N(6k-6)+M(2k+3)$. Since $k$ is a constant, we have a polynomial-time reduction.

We proceed by showing that if $S$ is satisfiable then $Y$ has at least one predecessor configuration. In fact, given any satisfying truth assignment for $S$, we can construct a predecessor configuration $Y^{\prime}$ of $Y$ in the following manner:

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  $Y^{\prime}(x_{i})=+1$, if variable $x_{i}$ is true in the given assignment;

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  $Y^{\prime}(x_{i})=-1$, if variable $x_{i}$ is false in the given assignment;

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  $Y^{\prime}(\neg{x_{i}})=-Y^{\prime}(x_{i})$;

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  $Y^{\prime}({z_{i}})=+1$, $Y^{\prime}(z^{\prime}_{i})=-1$, for $1\leq i\leq N$;

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  $Y^{\prime}(u_{i,j})=Y^{\prime}(p_{i,j})=+1$, for $1\leq i\leq N$ and $1\leq j\leq k-1$;

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  $Y^{\prime}(u_{i,j})=Y^{\prime}(p_{i,j})=-1$, for $1\leq i\leq N$ and $k\leq j\leq 2k-3$;

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  $Y^{\prime}(w_{i,j})=-1$, $Y^{\prime}(w^{\prime}_{i,j})=+1$, for $1\leq i\leq N$ and $1\leq j\leq k-2$, provided $k>2$;

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  $Y^{\prime}(c_{i})=Y^{\prime}(c^{\prime}_{i})=+1$, for $1\leq i\leq M$;

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  $Y^{\prime}(b_{i,j})=-1$, for $1\leq i\leq M$ and $1\leq j\leq k-2$, provided $k>2$;

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  $Y^{\prime}(b^{\prime}_{i,j})=-1$, for $1\leq i\leq M$ and $1\leq j\leq k-1$.

Figure 2 shows predecessor configuration $Y^{\prime}$ for the $Y$ given in Figure 1. As an example, we have let all of $x_{1}$, $\neg{x_{2}}$, and $x_{3}$ be true in the satisfying assignment.

By construction, and considering configuration $Y^{\prime}$ as described above, each vertex $x_{i}$ or $\neg{x_{i}}$ has at least $k$ neighbors in state $+1$ in $Y^{\prime}$ and exactly $k-1$ neighbors in state $-1$. This condition is sufficient to guarantee that every literal vertex reaches state $+1$ in one time step, regardless of its state in configuration $Y^{\prime}$.

Each of vertices $u_{i,j}$, $p_{i,j}$, $w_{i,j}$, $w^{\prime}_{i,j}$ , $b_{i,j}$, and $b^{\prime}_{i,j}$ has only one neighbor, and will obviously keep its state in the next configuration. Each of vertices $z_{i}$ and $z^{\prime}_{i}$ has at most $k-1$ neighbors in the opposite state, since vertices $x_{i}$ and $\neg{x_{i}}$ have mutually opposite states. So $z_{i}$ and $z^{\prime}_{i}$ remain unchanged as well.

In order for configuration $Y$ to be obtained after one time step, in configuration $Y^{\prime}$ no vertex $c_{i}$ can have more than one literal neighbor in state $-1$, which would lead to state $-1$ for $c_{i}$ in the next configuration. Since $S$ is satisfiable, there are exactly two positive literals for each clause in $S$, and by construction of $Y^{\prime}$, each vertex $c_{i}$ has exactly one literal neighbor in state $-1$. Similarly, every vertex $c^{\prime}_{i}$ needs at least one literal neighbor in state $-1$, and as the construction guarantees, $c^{\prime}_{i}$ has exactly one literal neighbor in state $-1$.

Hence, given the configuration $Y^{\prime}$ as constructed, we see that the next configuration is precisely configuration $Y$. We then conclude that whenever $S$ is satisfiable, $Y$ has at least one predecessor configuration.

Conversely, we now show that if $Y$ has at least one predecessor configuration then $S$ is satisfiable.

In any predecessor configuration of $Y$, vertices $u_{i,j}$, $p_{i,j}$, $w_{i,j}$, $w^{\prime}_{i,j}$, $b_{i,j}$, and $b^{\prime}_{i,j}$ should all be in the same states as in configuration $Y$, since they all have only one neighbor each.

Vertices $z_{i}$ and $z^{\prime}_{i}$ should also have the same states as in configuration $Y$ in any predecessor configuration. For suppose that, in a predecessor configuration, $z_{i}$ is in state $-1$; then necessarily $x_{i}$ and $\neg{x_{i}}$ should be in state $+1$, and consequently vertex $z^{\prime}_{i}$ would reach state $+1$ in the next configuration, which is different from its state in configuration $Y$. An analogous argument holds for vertex $z^{\prime}_{i}$. This condition also forces $x_{i}$ to have a state different than $\neg{x_{i}}$ in any predecessor configuration of $Y$, since if both have the same state then in the next configuration $z_{i}$ and $z^{\prime}_{i}$ will also have the same state.

Each vertex $c_{i}$ must have state $+1$ in a predecessor configuration of $Y$, otherwise every literal neighbor of $c_{i}$ would need to have state $-1$ in the predecessor configuration and consequently vertex $c_{i}$ would not change to state $+1$, which is its state in $Y$. Also, it must be the case that at least two of the literal neighbors of $c_{i}$ have state $+1$, otherwise $c_{i}$ would have state $-1$ in the next configuration, thus not matching configuration $Y$.

Each vertex $c^{\prime}_{i}$ has the same literal neighbors as vertex $c_{i}$. As we know that some of these neighbors have state $+1$ in a predecessor configuration, $c^{\prime}_{i}$ must have state $+1$, otherwise all of these neighbors would need to have state $-1$ in the predecessor configuration, contradicting the fact that some of them have state $+1$. Along with the restriction imposed by $c_{i}$, we have that exactly two of the three literal vertices associated with clause $c_{i}$ must have state $+1$ in a predecessor configuration.

Hence, given any predecessor configuration of $Y$, we can associate with any literal the value true if its vertex has state $+1$, and value false otherwise. The construction guarantees that this will satisfy every clause with exactly two positive literals and that no opposite literals will have the same assignment.

We conclude that if $Y$ has a predecessor configuration then $S$ is satisfiable. ∎

The graph constructed in the proof of Theorem 3 is bipartite for the node sets $\bigcup_{i,j}\{x_{i},\neg{x_{i}},b_{i,j},b^{\prime}_{i,j},w_{i,j},w^{\prime}_{i,j}\}$ and $\bigcup_{i,j}\{c_{i},c^{\prime}_{i},u_{i,j},p_{i,j},z_{i},z^{\prime}_{i}\}$. ∎