Structural Optimal Jacobian Accumulation and Minimum Edge Count are NP-Complete Under Vertex Elimination

We first show containment in NP. Note that a total elimination sequence is a permutation of $I$ and therefore can be encoded in polynomial space. Moreover, given such a sequence, we can compute its cost in polynomial time by computing the vertex eliminations one after another.

To show hardness, we reduce from Vertex Cover. It is well known that Vertex Cover is NP-hard and cannot be solved in $2^{o(n+m)}$ time unless the ETH fails [13, 14]. We will provide a linear reduction from Vertex Cover to Structural Optimal Jacobian Accumulation thereby proving the theorem. To this end, let $(G=(V,E),k)$ be an input instance of Vertex Cover. Let $n=|V|$ and $m=|E|$. We create an equivalent instance $(D=(S\uplus I,A),k^{\prime})$ of Structural Optimal Jacobian Accumulation as follows. For each vertex $v\in V$, we create five vertices $v_{1},v_{2},v_{3},v_{4},v_{5}$. The vertices $v_{1},v_{4}$ and $v_{5}$ are contained in $S$ for all $v\in V$. Vertices $v_{2}$ and $v_{3}$ are contained in $I$. Next, we add the set $\left\{(v_{1},v_{2}),(v_{2},v_{3}),(v_{2},v_{4}),(v_{2},v_{5}),(v_{3},v_{4}),(v_{3},v_{5})\right\}$ of arcs to $A$ for each $v\in V$. Finally, for each edge $\{u,v\}\in E$, we add the arcs $(u_{1},v_{3}),(u_{2},v_{3}),(v_{1},u_{3})$ and $(v_{2},u_{3})$ to $A$. Notice that every arc goes from a lower-indexed vertex to a higher-indexed vertex. Hence, the constructed digraph is a DAG. To finish the construction, we set $k^{\prime}=6m+4n+k.$ An illustration of the construction is depicted in Figure 1.

Since the reduction can clearly be computed in polynomial time, it only remains to show correctness. We proceed to show that $(G,k)$ is a yes-instance of Vertex Cover if and only if the constructed instance $(D,k^{\prime})$ is a yes-instance of Structural Optimal Jacobian Accumulation.

First, assume that $G$ contains a vertex cover $C$ of size at most $k$. We show that eliminating all vertices $v_{2}$ with $v\in C$ first, then $u_{3}$ for all $u\in V\setminus C$, followed by $u_{2}$ for all $u\in V\setminus C$, and finally all vertices $v_{3}$ for each $v\in C$ results in a total cost of $k^{\prime}$. To see this, note that the cost of eliminating $v_{2}$ for any vertex $v\in C$ is $\deg(v)+3$ as $v_{2}$ has a single in-neighbor $v_{1}$, three out-neighbors $v_{3},v_{4}$, and $v_{5}$, and $\deg(v)$ out-neighbors $u_{3}$ (one for each $u\in N(v)$). The cost of eliminating $u_{3}$ for any $u\notin C$ afterwards is $2\deg(u)+2$ as by construction $u_{3}$ has in-neighbors $\{w_{1}\mid w\in N(u)\}\cup\{u_{2}\}$, two out-neighbors $u_{4},u_{5}$ and no in-neighbors in $\{w_{2}\mid w\in N(u)\}$, as each $w\in N(u)$ is by definition in $C$ and hence the corresponding $w_{2}$ has been eliminated before. The cost of eliminating $u_{2}$ for $u\in V\setminus C$ is afterwards $\deg(u)+2$ as $u_{2}$ has the single in-neighbor $u_{1}$, two out-neighbors $u_{4}$ and $u_{5}$ and $\deg(u)$ out-neighbors $\{v_{3}\mid v\in C\}$ (note that $u$ cannot have neighbors in $V\setminus C$ since $C$ is a vertex cover and $u\notin C$). Finally, the cost of eliminating $v_{3}$ for any $v\in C$ is $2\deg(v)+2$ since $v_{3}$ has $\deg(v)+1$ in-neighbors $\{w_{1}\mid w\in N[v]\}$ and two out-neighbors $v_{4}$ and $v_{5}$. Summing these costs over all vertices and applying the handshake lemma²²2The handshake lemma states that the sum of vertex degrees is twice the number of edges [5]. gives a total cost of

This shows that the constructed instance $(D,k^{\prime})$ is a yes-instance of Structural Optimal Jacobian Accumulation.

In the other direction, assume that there is an ordering $\sigma$ of the vertices in $I$ resulting in a total cost of at most $k^{\prime}$. Let $J\subseteq V$ be the set of vertices such that for each $v\in J$ it holds that $v_{2}$ is eliminated before $v_{3}$ or $v_{3}$ is eliminated before $u_{2}$ for any $u\in N(v)$ by $\sigma$. We will show that $J$ is a vertex cover of size at most $k$ in $G$. To this end, we first provide a lower bound for the cost of eliminating any vertex, regardless of which vertices have been eliminated previously. Note that $v_{3}$ for any vertex $v\in V$ has two out-neighbors $v_{4}$ and $v_{5}$ in $S$, $\deg(v)$ in-neighbors $\{w_{1}\mid w\in N(v)\}$ in $S$, and one additional in-neighbor which is either $v_{2}$ if $v_{2}$ was not eliminated before or $v_{1}$ if $v_{2}$ was eliminated before. Hence, the cost for eliminating $v_{3}$ is at least $2\deg(v)+2$. Moreover, the cost of eliminating $v_{2}$ for any vertex $v\in V$ is at least $\deg(v)+2$ as $v_{2}$ has the in-neighbor $v_{1}\in S$, two out-neighbors $v_{4},v_{5}\in S$, and for each $w\in N(v)$ at least one additional out-neighbor ($w_{3}$ if $w_{3}$ was not eliminated before or $w_{4}$ and $w_{5}$ if $w_{3}$ was eliminated before). Summing these costs over all vertices (and again applying the handshake lemma) gives a lower bound of $6m+4n=k^{\prime}-k$.

The next step is to prove that $J$ contains at most $k$ vertices. To this end, note that for each vertex $v\in J$, the cost increases by at least one over the analyzed lower bound. If $v_{3}$ is eliminated after $v_{2}$ for some $v\in J$, then the cost of eliminating $v_{2}$ increases by one as $v_{2}$ has the additional out-neighbor $v_{3}$. If $v_{3}$ is eliminated before $u_{2}$ for some $v\in J$ and $u\in N(v)$, then the cost of eliminating $u_{2}$ increases by one as the out-neighbor $v_{3}$ is replaced by the two out-neighbors $v_{4}$ and $v_{5}$. This immediately implies that $|J|\leq k$.

Finally, we show that $J$ is a vertex cover. Assume towards a contradiction that this is not the case. Then, there is some edge $\{u,v\}\in E$ with $u\notin J$ and $v\notin J$. By definition of $J$, it holds that $u_{3}$ is eliminated before $u_{2}$, $u_{3}$ is eliminated after $v_{2}$, $v_{3}$ is eliminated before $v_{2}$, and $v_{3}$ is eliminated after $u_{2}$ by $\sigma$. Note that this implies that $\sigma$ eliminates $u_{3}$ before $u_{2}$, before $v_{3}$, before $v_{2}$, before $u_{3}$, a contradiction. Thus, $J$ is a vertex cover of size at most $k$ and the initial instance of Vertex Cover is therefore a yes-instance. This concludes the proof. ∎

We again start by showing containment in NP. We can encode a (not necessarily total) elimination sequence in polynomial space. Moreover, given such a sequence, we can compute the resulting DAG in polynomial time and verify that it contains at most $k$ edges.

To show hardness, we reduce from Independent Set in 2-degenerate subcubic graphs of minimum girth five, that is, graphs in which each vertex has between two and three neighbors and each cycle has length at least five. This problem is known to be NP-hard and cannot be solved in $2^{o(n+m)}$ time assuming the ETH [15]. We will provide a linear reduction from that problem to Minimum Edge Count to show the theorem. To this end, let $(G=(V=\{v_{1},v_{2},\ldots,v_{n}\},E),k)$ be an instance of Independent Set where each vertex has between two and three neighbors and no cycles of length three or four are present in $G$. We will construct an instance ${(D=(S\uplus I,A),k^{\prime})}$ of Minimum Edge Count. We begin by imposing an arbitrary total order $\pi$ on the vertex set $V$. We partition the vertices into four types based on their degree and the order $\pi$ as follows. Vertices of type 1 have degree 3 and either all neighbors come earlier with respect to $\pi$ or all neighbors come later. Vertices of type 2 have degree 3 and at least one earlier and at least one later neighbor with respect to $\pi$. Vertices of type 3 have degree 2 and either both neighbors come earlier or both neighbors come later with respect to $\pi$. Finally, vertices of type 4 have degree 2 and one of the neighbors comes earlier with respect to $\pi$ while the other neighbor comes later.

We now describe the construction of $D$, depicted in Figure 2. We begin by creating a set $T\subseteq S$ of $4$ sink vertices. Next, for each $v_{i}\in V$, we create a vertex $u_{i}\in I$ as well as two vertex sets $I_{i}\subseteq S$ and $O_{i}\subseteq S$ both of size $4$. We add arcs from each vertex in $I_{i}$ to every vertex in $\{u_{i}\}\cup T$ to $A$. We also add arcs from $u_{i}$ to each vertex in $O_{i}\cup T$ to $A$. Finally, we add arcs from $I_{i}$ to $O_{i}$ based on the type of $v_{i}$ as follows. Note that since $|I_{i}|=|O_{i}|=4$, there are up to $16$ possible arcs from vertices in $I_{i}$ to vertices in $O_{i}$. For type-1 vertices, we add 14 of the possible arcs (it does not matter which arcs we add). For vertices of type 2,3, and 4, we add $16,11,$ and $12$ arcs to $A$, respectively. After completing this procedure for every vertex in $V$, we add an arc $(u_{i},u_{j})$ for any edge $\{v_{i},v_{j}\}\in E$ where $v_{i}$ comes before $v_{j}$ in the order $\pi$. This concludes the construction of the graph $D$. Let $n^{\prime}$ and $m^{\prime}$ be the number of vertices and arcs in $D$. To conclude the construction, we set $k^{\prime}=m^{\prime}-k$. Observe that $D$ is a DAG and the number of vertices and arcs is linear in $n+m$.

Since the reduction can be computed in polynomial time and the constructed instance has linear size in the input instance, it only remains to show correctness. To this end, first assume that there exists an independent set $X\subseteq V$ of size $k$ in $G$. We eliminate the vertices $X^{\prime}=\{u_{i}\mid v_{i}\in X\}$. Note that Proposition 1 ensures that the order in which we eliminate the vertices does not matter. We will show that the resulting graph contains at most $k^{\prime}$ arcs.

First, note that new arcs between vertices $u_{i}$ and $u_{j}$ might be created while eliminating vertices in $X^{\prime}$. However, since $X$ is an independent set, such arcs are only created between vertices where $N(v_{i})\cap N(v_{j})\neq\emptyset$. Since $G$ has girth at least 5, if the elimination of a vertex in $X^{\prime}$ could create the arc $(u_{i},u_{j})$, then this arc was not there initially and was not added by the elimination of a different vertex in $X^{\prime}$. We next show that the elimination of each vertex in $X^{\prime}$ reduces the number of arcs in the graph by exactly one. Let $u_{i}\in X^{\prime}$ be an arbitrary vertex. If $v_{i}$ has degree three, then $u_{i}$ has exactly $15$ incident arcs, where 12 are to or from $I_{i},O_{o}$ and $T$ and three are to vertices $u_{j_{1}},u_{j_{2}},u_{j_{3}}$. The number of new arcs created in this case is $14$ as shown next. For each $\ell\in[3]$, four new arcs are created from vertices in $I_{i}$ to $u_{j_{\ell}}$ if $v_{i}$ comes before $v_{j_{\ell}}$ with respect to $\pi$ and four new arcs from $u_{j_{\ell}}$ to vertices in $O_{i}$ are created otherwise. Hence, in any case, 12 new arcs are created. If $v_{i}$ is of type 1, then 2 arcs are created from vertices in $I_{i}$ to vertices in $O_{i}$. If $v_{i}$ is of type 2, then two additional arcs are created between the vertices $u_{j_{1}},u_{j_{2}}$, and $u_{j_{3}}$. Hence, if $v_{i}$ has degree 3, then the elimination of vertex $u_{i}$ removes $15$ arcs and creates 14 new ones, that is, the number of arcs decreases by one. If $v_{i}$ has degree 2, then $u_{i}$ is incident to exactly $14$ arcs (12 to and from vertices in $I_{i},O_{i}$, and $T$ and two additional arcs to or from vertices $u_{j_{1}}$ and $u_{j_{2}}$). The number of new arcs created in this case is $13$ as shown next. For each $\ell\in[2]$, four new arcs are created from vertices in $I_{i}$ to $u_{j_{\ell}}$ if $v_{i}$ comes before $v_{j_{\ell}}$ with respect to $\pi$ and four new arcs from $u_{j_{\ell}}$ to vertices in $O_{i}$ are created otherwise. In any case, 8 new arcs are created this way. If $v_{i}$ is of type 3, then five arcs are created from vertices in $I_{i}$ to vertices in $O_{i}$. If $v_{i}$ is of type 4, then four additional arcs are created from vertices in $I_{i}$ to vertices in $O_{i}$ and one additional arc is added between $u_{j_{1}}$ and $u_{j_{2}}$. Hence, if $v_{i}$ has degree 2, then the elimination of vertex $u_{i}$ removes $14$ arcs and creates 13 new ones, that is, the number of arcs also decreases by one in this case. Since $k^{\prime}=m^{\prime}-k$ and each of the $k$ removals decreases the number of arcs by one, the resulting graph contains at most $k^{\prime}$ arcs, showing that the constructed instance is a yes-instance.