Table of Contents
Fetching ...

5-Coloring Planar Graphs with a Color Class of Order at Most $|V|/6$

Yuta Inoue, Ken-ichi Kawarabayashi, Atsuyuki Miyashita

TL;DR

The paper proves that every planar graph admits a 5-coloring with a color class of size at most $|V|/6$, achieved via a human-checkable framework of $1/6$-reducibility and a discharging argument. It introduces a compact set of reducible configurations (including 10 families) and five discharging rules, enabling a constructive $O(|V|^2)$ algorithm to find the required partition into five independent sets. This work bridges the gap between the Four Color Theorem and the Five Color Theorem by delivering a nontrivial, verifiable bound and a practical coloring method, while suggesting avenues for tightening the bound or exploring dual/related problems. The results highlight how a carefully designed reducibility-discharge scheme can yield strong, constructive colorings without computer-assisted proofs.

Abstract

We show that any planar graph $G=(V,E)$ has a 5-coloring such that one color class contains at most $|V|/6$ vertices. In other words, there exists a partition of $V$ into five independent sets $\{V_1, \cdots, V_5\}$ such that $|V_5| \leq |V| / 6$. Our proof yields an $O(|V|^2)$-time algorithm to find such a partition, and unlike the Four Color Theorem, our proof is fully verifiable without computer assistance.

5-Coloring Planar Graphs with a Color Class of Order at Most $|V|/6$

TL;DR

The paper proves that every planar graph admits a 5-coloring with a color class of size at most , achieved via a human-checkable framework of -reducibility and a discharging argument. It introduces a compact set of reducible configurations (including 10 families) and five discharging rules, enabling a constructive algorithm to find the required partition into five independent sets. This work bridges the gap between the Four Color Theorem and the Five Color Theorem by delivering a nontrivial, verifiable bound and a practical coloring method, while suggesting avenues for tightening the bound or exploring dual/related problems. The results highlight how a carefully designed reducibility-discharge scheme can yield strong, constructive colorings without computer-assisted proofs.

Abstract

We show that any planar graph has a 5-coloring such that one color class contains at most vertices. In other words, there exists a partition of into five independent sets such that . Our proof yields an -time algorithm to find such a partition, and unlike the Four Color Theorem, our proof is fully verifiable without computer assistance.
Paper Structure (16 sections, 16 theorems, 1 equation, 12 figures)

This paper contains 16 sections, 16 theorems, 1 equation, 12 figures.

Key Result

Theorem 1.1

Any planar graph $G=(V,E)$ has a 5-coloring such that one color class contains at most $|V|/6$ vertices. In other words, there exists a partition of $V$ into five independent sets $\{V_1, \cdots, V_5\}$ such that $|V_5| \leq |V| / 6$.

Figures (12)

  • Figure 1: The chart for the Heesch notation. We depict vertices of degree 5 by a filled circle, 6 by a dot, 7 by an unfilled circle, 8 by a square, and 9 by a triangle.
  • Figure 2: A drawing of an example configuration $\mathcal{H} = (H, d_G|_{V_H}, \Pi|_{V_H})$. The vertices are depicted by Heesch notation (corresponding to the value of $d_G$, which is $d_G(v_1) = 8, d_G(v_3) = 7, d_G(v_6) = 5, d_G(v_2) = d_G(v_4) = d_G(v_5) = 6$). The gray lines represent halfedges in $\Pi_H$. For example, $\pi_{v_1}^H$ is the cyclic permutation $(v_1v_5, v_1v_6, v_1v_2, f_{11}, f_{12}, f_{13}, f_{14}, f_{15})$, $\pi_{v_2}^H$ is the cyclic permutation $(v_2v_1, v_2v_6, v_2v_3, f_{21}, f_{22}, f_{23})$, and so on. (Here, $f_{ij}$ is some halfedge connected to $v_i$).
  • Figure 3: An example of two configurations having isomorphic graphs and matching degrees but different local rotation on the central vertex.
  • Figure 4: A visualization of how the example configuration should be 5-colored, shown from left to right. The red vertex (labeled (0)) of each graph is the vertex that we color with the fifth color, and the gray vertices are the vertices that we know have a neighboring vertex of the fifth color (hence, having $1$ less degree compared to the previous graphs). We can confirm that all the non-red-vertices form a 0-reducible configuration, by deleting the 5 vertices in labeled order from $\mathbf{1}$ to $\mathbf{5}$ and by seeing that they are degree 4 or less at the time of deletion.
  • Figure 5: The two cases to consider in \ref{['lem:five-wheel']}.
  • ...and 7 more figures

Theorems & Definitions (29)

  • Theorem 1.1
  • Definition 2.1
  • Definition 3.1
  • Definition 3.2
  • Lemma 3.3
  • Lemma 3.4
  • Lemma 3.5
  • proof
  • Lemma 3.6
  • proof
  • ...and 19 more