Improved Erdős-Pósa inequalities for odd cycles in planar graphs

TL;DR

Problem: determine the Erdős–Pósa ratio between the odd cycle transversal $\tau(G)$ and the odd cycle packing $\nu(G)$ in planar graphs. Approach: a constructive, cloud-based decomposition in the vertex-face incidence graph $\mathrm{VF}(G)$, combined with face-merging operations and a reduced conflict graph to bound local transversals by local packings, using induction on the number of faces $|\mathcal{F}(G)|$. Result: proves $\tau(G) \le 4\,\nu(G)$ with a polynomial-time algorithm to obtain the transversal and packing; also notes a $(3+\varepsilon)$-approximation for maximum odd cycle packing and a $2.4$-approximation for the transversal in planar graphs, and discusses a general framework for uncrossable cycle families with finite Erdős–Pósa ratios. Significance: sharpens the best-known planar bound from 6 to 4, provides constructive algorithms, and extends the methodology to related cycle families and approximation guarantees.

Abstract

In an undirected graph, the odd cycle packing number is the maximum number of pairwise vertex-disjoint odd cycles. The odd cycle transversal number is the minimum number of vertices that hit every odd cycle. The maximum ratio between transversal and packing number is called Erdős-Pósa ratio. We show that in planar graphs, this ratio does not exceed 4. This improves on the previously best known bound of 6 by Král', Sereni and Stacho.

Figures (5)

• Figure 1: The figure shows a graph $G$ where $V(G)$ are the filled vertices and $E(G)$ are the solid edges. Its vertex-face incidence graph $\text{VF}(G)$ is the graph on all vertices (filled and empty) with the dotted edges. When choosing $T\subseteq V(G)$ to be the filled red vertices, $\text{VF}(G)[\mathcal{F}(G)\cup T]$ decomposes into three connected components, drawn in red, green, and blue, respectively. Hence $T$ is an $\mathcal{F}$-transversal when choosing $\mathcal{F}$ to be the faces of $G$ that are filled in red. Note that $\mathcal{F}$ are exactly the odd faces of $G$; thus $T$ is an odd cycle transversal for $G$ by \ref{['prop:vf_definition_of_transversal']}.
• Figure 2: The left picture shows the faces $F_1$ and $F_2$, which meet at the vertex $v$, before merging. The edges incident to $v$ are ordered counterclockwise. Note that $j_1=3$, while for $j_2$ there are two options: $j_2=6$ and $j_2=8$. The right picture shows the graph obtained after merging $F_1$ and $F_2$ along $v$ for the choice $j_2 = 8$. The large face $F$ corresponds to the face set $\{F_1, F_2\}$.
• Figure 3: Example for a situation as in \ref{['def:reduced_conflict_graph']}, showing the surroundings of the light blue face of a graph $G$: The colored faces are the faces in $\mathcal{F}\subseteq \mathcal{F}(G)$. The relevant vertices of $G$ are drawn in black. The gray dotted lines are the edges in the vertex-face incidence graph $\text{VF}(G)$ that are incident to $\mathcal{F}$ and the black vertices. Then the reduced conflict graph $R$ for $\mathcal{F}$ is constructed on the colored vertices (each representing a face in $\mathcal{F}$). The edges of $R$ are the thick dashed lines (black and light and dark green). Note that $R$ depends on the embedding of $\text{VF}(G)$ since there are two possibilities for embedding the edge between $v_2$ and the orange face. When we choose the light blue face as $F\in \mathcal{F}$ in \ref{['lem:degree_in_reduced_conflict_graph']}, we get $T_1 = \{v_1, v_2\}$. Hence $\mathcal{F}_1$ consists of the light blue, the violet, the orange, the dark blue, and the magenta face. $\mathcal{F}_2$ then consists of the red and the yellow face; we get $T_2 = \{v_3, v_4\}$. The edges in $\delta_R(F)$ that ultimately carry a charge of $\frac{1}{2}$ are colored in light green; the edges with a charge of $1$ are colored in dark green: $v_1$ adds a charge of $\frac{1}{2}$ to each of $e_1$ and $e_2$, $v_2$ adds a charge of $\frac{1}{2}$ to each of $e_2$ and $e_3$, and $v_3$ and $v_4$ add a charge of 1 to $e_4$ and $e_5$, respectively.
• Figure 4: The left part shows the situation of Case 1 in Lemma \ref{['lem:packing_number1']} where $F_2$ and $F_4$ meet in $v^\prime$. The cycle $C$ in $\text{VF}(G)$ is drawn dashed. The right part shows the more complicated Case 2. The embeddings of the cycles $C_1$ (dashed lines) and $C_2$ (dotted lines) both separate the faces $F^*_1(v_3)$ and $F^*_2(v_4)$ from each other.
• Figure 5: Example for two possible situations in Case 2 of the proof of \ref{['lem:clouds']}. The colored faces are the faces in $\mathcal{N}$, the light blue face in the middle is $F$. The dashed lines represent the reduced conflict graph $R$. On the left hand side, we see the case where $\mathcal{B}\neq \emptyset$: $\mathcal{B}$ consists of the red faces that clearly contain $v^*$. $F$ and the orange faces also contain $v^*$. The only 3 faces that do not contain $v^*$ are the green faces, but they contain $v_1$, $v_2$, and $v_3$, respectively, so $\{v_1, v_2, v_3, v^*\}$ is an $\mathcal{F}'$-transversal for any $\mathcal{F}^\prime$. On the right hand side, we see the case where $\mathcal{B} = \emptyset$ and $|\mathcal{F}^\prime| = |\mathcal{F}| = 6$. $R$ contains a perfect matching on $\mathcal{F}$ as indicated by the thick, colored lines. This induces an $\mathcal{F}^\prime$-transversal of size $3$, indicated by the black vertices.

Theorems & Definitions (33)

• Theorem 1
• Definition 2: Odd cycle transversal
• Proposition 3
• proof
• Definition 4: Vertex-face incidence graph
• Definition 5: $\mathcal{F}$-Transversal
• Proposition 6
• proof
• Definition 7: $T$-join
• Proposition 8