Solving the $r$-pseudoforest Deletion Problem in Time Independent of $r$

Let $H$ be the graph obtained from $G$ by bypassing $v$. Let $N_{G}(v)=\{u,w\}$. We show that $(H,k)$ is a yes-instance if and only if $(G,k)$ is a yes-instance.

On the one hand, assume $(H,k)$ is a yes-instance, and $X$ is a solution of it. Suppose $X\cap\{u,w\}\neq\emptyset$, then $d_{G-X}(v)\leq 1$. By adding $v$ as an isolated vertex or a leaf to $H-X$, we get $G-X$. It follows that $G-X$ is an $r$-pseudoforest since $H-X$ is an $r$-pseudoforest. Thus, $X$ is also a solution of $(G,k)$. Otherwise, $X\cap\{u,w\}=\emptyset$. In this case, the edge $uw$ is in some component of the $r$-pseudoforest $H-X$. By subdividing $uw$ in $H-X$, we get $G-X$, which is still an $r$-pseudoforest. And so $X$ is a solution of $(G,k)$. In both cases, $(G,k)$ is a yes-instance, and $X$ is a solution of it.

On the other hand, assume $(G,k)$ is a yes-instance, and $X$ is a solution of it. Suppose $v\in X$. Consider $X^{\prime}=X\cup\{u\}-\{v\}$. Note that $|X^{\prime}|\leq|X|$. If $u\in X$, then $H-X^{\prime}=G-X$. If $u\not\in X$, then $H-X^{\prime}=G-X-\{u\}$. In both cases, $H-X^{\prime}$ is an $r$-pseudoforest. Thus $X^{\prime}$ is a solution for $(H,k)$, and $(H,k)$ is a yes-instance.

Otherwise, $v\not\in X$. If $d_{G-X}(v)=2$, then we get $H-X$ from $G-X$ by bypassing $v$. If $d_{G-X}(v)=1$, then we get $H-X$ from $G-X$ by deleting $v$. In both cases, the number of edges and vertices decrease by the same amount, thus $H-X$ is also an $r$-pseudoforest. If $d_{G-X}(v)=0$, then we get $H-X$ from $G$ by deleting the isolated vertex $v$. Above all, we know that $(H,X)$ is a yes-instance, and $X$ is a solution of it. ∎

We can make each connected component of an $r$-pseudoforest into a forest by deleting at most $r$ edges. Since $G$ is a connected $r$-pseudoforest, $|E(G)|\leq|V(G)|+r-1$. ∎

Proof of statement 1: Because $F$ is an $r$-pseudoforest, we know that $|E(C)|\leq|V(C)|+r-1$ for each component $C$ in $F$. Since the ratio $\frac{|V(C)|+r-1}{|V(C)|}$ decreases when $|V(C)|$ increases, we have $\frac{|E(C)|}{|V(C)|}\leq\frac{r+2}{3}$ for each component $C$ in $F$ with at least 3 vertices. It follows that $|E(F)|\leq rt_{1}+rt^{\prime}_{1}+(r+1)t_{2}+(r+1)t^{\prime}_{2}+(r+2)s/3$.

Proof of statement 2: Vertices in each $(r+1)$-edge have average degree $r+1$ and the vertex in an $r$-loop has degree $2r$. After exhaustive applications of Reduction Rule 1, there is no $r$-pseudoforest component in $G$. Note that there are $t_{1}$ $r$-loops, $t^{\prime}_{1}$ components consists of one vertex and at least $(r+2)/3$ loops, $t_{2}$ $(r+1)$-edges, $t^{\prime}_{2}$ components consists of two vertices and at least $2(r+2)/3$ edges in $F$. Thus $G$ contains at least $t_{1}$ vertices of degree at least $2r+1$, $t^{\prime}_{1}$ vertices of degree at least $2(r+2)/3+1$ and $t_{2}$ vertices of degree at least $r+2$, $t^{\prime}_{2}$ vertices of degree at least $2(r+2)/3+1$. Combining with the fact that $\delta(G)\geq 3$, we have

Proof of statement 3: Since every edge in $E(G)\setminus E(F)$ is incident to at least one vertex in $X$, we have $|E(G)|\leq|E(F)|+\sum_{v\in X}d(v).$ It follows that

Since $|V(G)|=|V(F)|+|X|$, denoting $|X|$ by $x$, we have

Combining (1) and (2), for any positive constant $c$, we have $\sum_{v\in X}(d(v)-c)\geq m-cn-[(r-c)t_{1}+(r-c)t^{\prime}_{1}+(r+1-2c)t_{2}+(r+1-2c)t^{\prime}_{2}+(r+2-3c)s/3].$ ∎

Suppose there is a solution $X$ of $r$-pseudoforest deletion on $(G,k)$, which satisfies $X\cap V_{10k}=\emptyset$ and $|X|\leq k$. We show a contradiction by counting the number of edges in $G$. Let $F=G-X$ be the resulting $r$-pseudoforest after deleting $X$ from $G$. Suppose $F$ contains $t_{1}$ $r$-loops, $t^{\prime}_{1}$ components consisting of one vertex and at least $(r+2)/3$ loops, $t_{2}$ $(r+1)$-edges, and $t^{\prime}_{2}$ components consisting of two vertices and at least $2(r+2)/3$ edges. Denote $|V(F)|=t_{1}+t^{\prime}_{1}+2t_{2}+2t^{\prime}_{2}+s$. In the following, we assume that $n=t_{1}+t^{\prime}_{1}+2t_{2}+2t^{\prime}_{2}+s+x\geq 51k$, as otherwise, we may solve the problem via brute force. The assumption also implies that $t_{1}+t^{\prime}_{1}+2t_{2}+2t^{\prime}_{2}+s\geq 50k$.

By the choice of $V_{10k}$, the degree of each vertex in $X$ is at most $d(v_{10k})$. Since $|X|\leq k$, it follows that

By the definition of $V_{10k}$ and the assumption that $X\cap V_{10k}=\emptyset$, we have $X\subseteq\{v_{10k+1},v_{10k+2},...,v_{n}\}$ and so

Combining inequalities (3) and (4), we have

Since $\sum_{v\in V(G)}d(v)=2m$, it follows that

Thus $2m-cn\geq$ $11\{m-cn-[(r-c)t_{1}+(r-c)t^{\prime}_{1}+(r+1-2c)t_{2}+(r+1-2c)t^{\prime}_{2}+(r+2-3c)s/3]\}.$

According to Statement 2 in Lemma 3, we have $10cn\geq 4.5[3n+(2r+1)t_{1}+(2(r+2)/3+1)t^{\prime}_{1}+(r+2)t_{2}+(2(r+2)/3+1)t^{\prime}_{2}]-11[(r-c)t_{1}+(r-c)t^{\prime}_{1}+(r+1-2c)t_{2}+(r+1-2c)t^{\prime}_{2}+(r+2-3c)s/3].$

Therefore, $(10c-13.5)n\geq[4.5(2r+1)-11(r-c)]t_{1}+[(3(r+2)+4.5)-11(r-c)]t^{\prime}_{1}+[4.5(r+2)-11(r+1-2c)]t_{2}+[(3(r+2)+4.5)-11(r+1-2c)]t^{\prime}_{2}-11(r+2-3c)s/3.$

Set $c=10r$ and simplify the above inequality, we get $(100r-13.5)(t_{1}+t^{\prime}_{1}+2t_{2}+2t^{\prime}_{2}+s+x)\geq(108r+4.5)t_{1}+(102r+10.5)t^{\prime}_{1}+(213.5r-2)t_{2}+(212r-0.5)t^{\prime}_{2}+11(29r-2)s/3.$

It follows that $(100r-13.5)x\geq(18+8r)t_{1}+(2r+24)t^{\prime}_{1}+(13.5r+25)t_{2}+(12r+26.5)t^{\prime}_{2}+(19r/3+37/6)s\geq(8t_{1}+2t^{\prime}_{1}+13.5t_{2}+12t^{\prime}_{2}+19/3s)r\geq 2(n-x)r\geq 2(51-1)kr\geq 100kr$, a contradiction to the fact that $x\leq k$.

The above analysis shows that the assumption $X\cap V_{10k}=\emptyset$ is not correct, thus every solution of $(G,k)$ must intersect with $V_{10k}$. ∎

Let $(G,k)$ be a given instance of the $r$-pseudoforest deletion problem, where $G$ is an undirected graph and $k$ is an integer. Each recursive call in Step 4 of Algorithm 1 decreases the parameter by 1, and thus the height of the search tree is at most $k^{\prime}$. At each step, the problem branches into at most $10k^{\prime}$ subproblems. Hence, the number of leaves in the search tree is at most $(10k^{\prime})^{k^{\prime}}\leq(10k)^{k}$. It follows that Algorithm 1 solves the problem of $r$-pseudoforest deletion and runs in time $O^{*}((10k)^{k})$. ∎