Odd Colorings of Sparse Graphs

The first statement is obvious, so now we prove the second. If $G$ is bipartite, then $G$ has a proper 2-coloring $\varphi$. If the degree of each vertex is either 0 or odd, then $\varphi$ is also an odd 2-coloring of $G$. If $G$ is not bipartite, then $G$ has no proper 2-coloring, so clearly $\chi_{o}(G)>2$. Suppose instead $G$ is bipartite, but some vertex $v$ has positive even degree. Now the component of $G$ containing $v$ has only a single 2-coloring (up to permuting colors), but this 2-coloring is not odd since $N(v)$ has only a single color, which appears an even number of times. ∎

If $3|n$, then we can repeat the colors $1,2,3,1,2,3,\ldots$ around $C_{n}$ to get an odd 3-coloring. Suppose instead that $3\nmid n$. By the Pigeonhole principle, in every proper 3-coloring $\varphi$ some color appears twice on the neighborhood of some vertex $v$, so $\varphi$ is not odd. Thus, $\chi_{o}(C_{n})\geq 4$. If $3\nmid n$ and $n\neq 5$, then we can begin with $1,2,3,4$ or $1,2,3,4,1,2,3,4$ and continue with $1,2,3,1,2,3\ldots$. Finally, it is easy to check that each proper 4-coloring $\varphi$ of $C_{5}$ uses the same color on both neighbors of some vertex, so $\varphi$ is not an odd coloring. Thus, $\chi_{o}(C_{5})=5$. ∎

Let $T$ be a tree. We use induction on $|T|$. The case $|T|=1$ is trivial. Now suppose that $|T|\geq 2$. Let $v$ be a leaf of $T$. Let $T^{\prime}:=T-v$ and note that $T^{\prime}$ is a tree. By hypothesis, $T^{\prime}$ has an odd 3-coloring $\varphi$. Let $w$ be the neighbor of $v$ in $T$. To extend $\varphi$ to an odd 3-coloring of $T$, we color $v$ with a color outside $\{\varphi(w),\varphi_{o}(w)\}$. This gives the desired odd 3-coloring of $T$.

If $\mathrm{mad}(G)<2$, then $G$ is a forest. So the second statement follows from the first. Finally, the third statement follows from Proposition 2. ∎

Fix a plane embedding of a planar graph $G$ with girth at least $g$. Since each subgraph of $G$ also has girth at least $g$, it suffices to show that $2|E(G)|/|V(G)|<2g/(g-2)$. Summing the lengths of all facial walks gives $2|E(G)|\geq g|F|$, where $F$ is the set of all faces. To prove the proposition, we substitute this inequality into Euler’s formula and solve for $2|E(G)|/|V(G)|$. ∎

For each $n\geq 1$, let $\epsilon_{n}:=8/(n+1)$. To show that $\mathrm{mad}(K^{*}_{n})=4-\epsilon_{n}$, we give a fractional orientation of $K^{*}_{n}$ where each vertex has indegree exactly $2-\epsilon_{n}/2$. We orient each edge with fraction $1-\epsilon_{n}/4$ towards its endpoint of degree 2 and fraction $\epsilon_{n}/4$ toward its endpoint of degree $n$. Each 2-vertex has indegree $2(1-\epsilon_{n}/4)=2-\epsilon_{n}/2$. Each $(n-1)$-vertex has indegree $(n-1)(\epsilon_{n}/4)=(n-1)(2/(n+1))=(2n-2)/(n+1)=2-4/(n+1)=2-\epsilon_{n}/2$. Thus, $\mathrm{mad}(G)=4-\epsilon_{n}$, as desired.

In an odd coloring of $K^{*}_{n}$, all $(n-1)$-vertices must get distinct colors. So $\chi_{o}(K^{*}_{n})\geq n$. Given any coloring where all $(n-1)$-vertices get distinct colors, it is easy to extend to an odd $n$-coloring of $K^{*}_{n}$. Thus, as claimed, $\chi_{o}(K^{*}_{n})=n$. ∎

Let $\epsilon_{n}:=8/(n+1)$ and $G_{n}:=K^{*}_{n}$. Now $\chi_{o}(G_{n})=n=(n+1)-1=8/\epsilon_{n}-1$. ∎

The second statement follows directly from Corollary 6.

For the first statement, let $k:=\lfloor 8/\epsilon\rfloor+2$^††margin: $k$ and note that $k\geq 7$. Our proof is by induction on $|V(G)|$. The base case is when $|V(G)|=1$, so $\chi_{o}(G)=1$. Instead assume $|V(G)|\geq 2$. If $G$ contains a 1-vertex $v$ with neighbor $w$, then $G-v$ has an odd $k$-coloring $\varphi$ and we can extend $\varphi$ to $v$ by coloring $v$ to avoid $\varphi(w)$ and $\varphi_{o}(w)$. If $G$ contains a 3-vertex $v$, then denote $N(v)$ by $\{w_{1},w_{2},w_{3}\}$. Now $G-v$ has an odd $k$-coloring $\varphi$ and we can extend $\varphi$ to $v$ by coloring $v$ to avoid $\{\varphi(w_{1}),\varphi(w_{2}),\varphi(w_{3}),\varphi_{o}(w_{1}),\varphi_{o}(w_{2}),\varphi_{o}(w_{3})\}$. So we assume that $G$ has neither 1-vertices nor 3-vertices. Suppose that $G$ has adjacent 2-vertices $v_{1}$ and $v_{2}$, and denote the remaining neighbors of $v_{1}$ and $v_{2}$, respectively, by $v_{0}$ and $v_{3}$. Now $G-\{v_{1},v_{2}\}$ has an odd $k$-coloring $\varphi$. To extend this to $v_{1}$ and $v_{2}$, we first color $v_{1}$ to avoid $\{\varphi(v_{0}),\varphi_{o}(v_{0}),\varphi(v_{3})\}$ and then color $v_{2}$ to avoid the new color on $v_{1}$ as well as $\{\varphi(v_{0}),\varphi(v_{3}),\varphi_{o}(v_{3})\}$.

Since $\mathrm{mad}(G)<4$, we know that $\delta(G)=2$. We use discharging to show that some vertex $v$ with $d(v)\leq 8/\epsilon-2$ has “many” 2-neighbors. Similar to the case of adjacent 2-vertices above, we will delete $v$ and its 2-neighbors, find an odd $k$-coloring $\varphi$ for this smaller graph, and extend $\varphi$ to all of $G$. Let $x:=1-\epsilon/2$^††margin: $x$ . We give each vertex $v$ initial charge $d(v)$ and use the following single discharging rule: Each 2-vertex takes charge $x$ from each neighbor. Since $G$ does not have adjacent 2-vertices, each 2-vertex finishes with charge $2+2x=4-\epsilon$. For each $4^{+}$-vertex $v$, let $d_{2}(v)$^††margin: $d_{2}(v)$ denote the number of 2-neighbors of $v$. Now $v$ finishes with charge $d(v)-xd_{2}(v)$. If $d(v)-xd_{2}(v)>2+2x=4-\epsilon$ for each $v$, then we contradict the assumption $\mathrm{mad}(G)\leq 4-\epsilon$. So there exists $v$ such that $d(v)-xd_{2}(v)\leq 2+2x$.

Form $G^{\prime}$ from $G$ by deleting $v$ and all of its 2-neighbors. By induction, $G^{\prime}$ has an odd $k$-coloring $\varphi$. To extend $\varphi$ to $G$, we first color $v$ to avoid the colors on the colored neighbors (in $G$) of its deleted 2-neighbors (in $G^{\prime}$) as well as, for each $4^{+}$-neighbor $w$, to avoid $\varphi(w)$ and $\varphi_{o}(w)$. For convenience, we denote this color used on $v$ by $\varphi(v)$. Then we color each 2-neighbor $x$ of $v$, with other neighbor $y$, to avoid $\{\varphi(v),\varphi_{o}(v),\varphi(y),\varphi_{o}(y)\}$. This gives an odd coloring $\varphi^{\prime}$ of $G$. We must only show that $\varphi^{\prime}$ uses at most $k$ colors. In total, $v$ must avoid at most $2d(v)-d_{2}(v)$ colors. So it will suffice to show that $k\geq 1+(2d(v)-d_{2}(v))$. Note that $1+2d(v)-d_{2}(v)<1+d(v)+(d(v)-xd_{2}(v))\leq 1+d(v)+2+2x<1+d(v)+4$. Since $1+2d(v)-d_{2}(v)$ is an integer, $1+2d(v)-d_{2}(v)\leq d(v)+4$. So now we must bound $d(v)$. Clearly, $d(v)-d(v)x\leq d(v)-d_{2}(v)x\leq 2+2x$. So $d(v)\leq(2+2x)/(1-x)=(4-\epsilon)/(\epsilon/2)=8/\epsilon-2$. Since $d(v)$ is an integer, $d(v)\leq\lfloor 8/\epsilon\rfloor-2$. Thus, $d(v)+4\leq(\lfloor 8/\epsilon\rfloor-2)+4=\lfloor 8/\epsilon\rfloor+2=k$. ∎

The second statement follows from the first by Proposition 4. So we prove the first. Assume the statement is false, and $G$ is a counterexample with as few vertices as possible. As in the proof of Theorem 1, we assume $G$ has no 1-vertex and $G$ has no adjacent 2-vertices. Suppose $G$ has a 3-vertex $v$ with neighbors $w_{1},w_{2},w_{3}$, where $d(w_{1})=2$. Denote the other neighbor of $w_{1}$ by $x_{1}$. Let $G^{\prime}:=G-\{v,w_{1}\}$. By minimality, $G^{\prime}$ has an odd 6-coloring $\varphi$. To extend $\varphi$ to $G$, we color $v$ to avoid $\{\varphi(w_{1}),\varphi(w_{2}),\varphi(w_{3}),\varphi_{o}(w_{2}),\varphi_{o}(w_{3})\}$ and then color $w_{1}$ to avoid the new color on $v$ and also to avoid $\{\varphi(x_{1}),\varphi_{o}(x_{1})\}$. Thus, no 3-vertex has a 2-neighbor.

Suppose $G$ has a 4-vertex with neighbors $w_{1},w_{2},w_{3},w_{4}$, where $d(w_{1})=d(w_{2})=d(w_{3})=2$. For each $i\in\{1,2,3\}$, denote the other neighbor of $w_{i}$ by $x_{i}$. Let $G^{\prime}:=G-\{v,w_{1},w_{2},w_{3}\}$. By minimality, $G^{\prime}$ has an odd 6-coloring $\varphi$. To extend $\varphi$ to $G$, color $v$ to avoid $\{\varphi(x_{1}),\varphi(x_{2}),\varphi(x_{3})$, $\varphi(w_{4}),\varphi_{o}(w_{4})\}$ and then color each $w_{i}$ to avoid the new color on $v$ and also avoid $\{\varphi(x_{i})$, $\varphi_{o}(x_{i}),\varphi(w_{4})\}$. Thus, no 4-vertex in $G$ has three (or more) 2-neighbors.

Suppose $G$ has a 5-vertex $v$ with five 2-neighbors $w_{1},w_{2},w_{3},w_{4},w_{5}$. For each $i\in\{1,\ldots,5\}$, denote the other neighbor of $w_{i}$ by $x_{i}$. Let $G^{\prime}:=G-\{v,w_{1},w_{2},w_{3},w_{4},w_{5}\}$. By minimality, $G^{\prime}$ has an odd 6-coloring $\varphi$. To extend $\varphi$ to $G$, we color $v$ to avoid $\{\varphi(x_{1}),\varphi(x_{2}),\varphi(x_{3}),\varphi(x_{4}),\varphi(x_{5})\}$. We then color each $w_{i}$ to avoid the new color on $v$, as well as $\{\varphi(x_{i}),\varphi_{o}(x_{i}),\varphi(v)\}$. Thus, no 5-vertex in $G$ has five 2-neighbors.

Now we use discharging to reach a contradiction, which will finish the proof. We give each vertex $v$ initial charge $d(v)$ and use the following single discharging rule: Each 2-vertex takes $1/2$ from each neighbor. We show that each vertex finishes with charge at least 3, which contradicts the hypothesis $\mathrm{mad}(G)<3$. Recall that $\delta(G)\geq 2$.

If $d(v)=2$, then $v$ finishes with $2+2(1/2)=3$. If $d(v)=3$, then $v$ has no 2-neighbors, so $v$ finishes with 3. If $d(v)=4$, then $v$ has at most two 2-neighbors, so $v$ finishes with at least $4-2(1/2)=3$. If $d(v)=5$, then $v$ has at most four 2-neighbors, so $v$ finishes with at least $5-4(1/2)=3$. If $d(v)\geq 6$, then $v$ finishes with at least $d(v)-d(v)/2=d(v)/2\geq 3$, since $d(v)\geq 6$. This finishes the proof. ∎

In the proof of Theorem Theorem 2(a), to contradict the hypothesis $\mathrm{mad}(G)<3$, we showed that each vertex finished with charge at least 3. To contradict the weaker hypothesis $\mathrm{mad}(G)\leq 3$, it suffices to show that also some vertex finishes with charge more than 3. Suppose the contrary.

Now $G$ has no $7^{+}$-vertex and each 6-vertex has six 2-neighbors. Each 5-vertex has exactly four 2-neighbors and each 4-vertex has exactly two 2-neighbors. Now it is straightforward to check that $G$ cannot have any two adjacent $3^{+}$-vertices $v$ and $w$; in each case, we delete $v$, $w$, and their 2-neighbors, color the smaller graph and extend. Thus, $G$ is bipartite, where one part consists of 2-vertices and the other consists of 6-vertices. Form $G^{\prime}$ from $G$ by contracting one edge incident to each 2-vertex. Now $G^{\prime}$ is 6-regular, but no component is $K_{7}$, so $\chi(G^{\prime})\leq 6$ by Brooks’ Theorem. A 6-coloring of $G^{\prime}$ induces a 6-coloring of the 6-vertices in $G$, which we can easily extend to an odd 6-coloring of $G$. ∎

The second statement follows from the first by Proposition 4. So we prove the first. Assume the statement is false, and $G$ is a counterexample with as few vertices as possible. As in the proof of Theorem 1, we assume $G$ has no 1-vertex and $G$ has no adjacent 2-vertices. Suppose $G$ has a 3-vertex $v$ with neighbors $w_{1},w_{2},w_{3}$, where $d(w_{1})=d(w_{2})=2$. Denote the other neighbors of $w_{1}$ and $w_{2}$ by, respectively, $x_{1}$ and $x_{2}$. Let $G^{\prime}:=G-\{v,w_{1},w_{2}\}$. By minimality, $G^{\prime}$ has an odd 5-coloring $\varphi$. To extend $\varphi$ to $G$, we color $v$ to avoid $\{\varphi(w_{1}),\varphi(w_{2}),\varphi(w_{3}),\varphi_{o}(w_{3})\}$ and color each $w_{i}$ to avoid the new color on $v$ and avoid $\{\varphi(x_{i}),\varphi_{o}(x_{i}),\varphi(w_{3})\}$. Thus, each 3-vertex has at most one 2-neighbor. More generally, each 3-vertex has at most one $3^{-}$-neighbor. If not, then we delete $v$ and any 2-neighbors. We color $v$ to avoid the color on each 3-neighbor, the color on the colored neighbor of each (deleted) 2-neighbor, as well as $\{\varphi(w),\varphi_{o}(w)\}$, where $w$ is the third neighbor of $v$. Thus, each 3-vertex has at most one $3^{-}$-neighbor.

Suppose $G$ has a 4-vertex with four 2-neighbors $w_{1},w_{2},w_{3},w_{4}$. For each $i\in\{1,2,3,4\}$, denote the other neighbor of $w_{i}$ by $x_{i}$. Let $G^{\prime}:=G-\{v,w_{1},w_{2},w_{3},w_{4}\}$. By minimality, $G^{\prime}$ has an odd 5-coloring $\varphi$. To extend $\varphi$ to $G$, color $v$ to avoid $\{\varphi(x_{1}),\varphi(x_{2}),\varphi(x_{3})$, $\varphi(x_{4})\}$ and then color each $w_{i}$, in succession, to avoid the new color on $v$ and also avoid $\{\varphi(x_{i})$, $\varphi_{o}(x_{i}),\varphi_{o}(v)\}$. Thus, no 4-vertex in $G$ has four 2-neighbors. Similarly, if each $w_{i}$ has degree at most 3 and at least one $w_{i}$ is a 2-vertex, then we form $G^{\prime}$ from $G$ by deleting $v$ and all its 2-neighbors. Again $G^{\prime}$ has an odd 5-coloring $\varphi$. We extend $\varphi$ to $v$ by avoiding the color on each 3-neighbor and the color on the other neighbor of each 2-neighbor. Finally, we can extend the coloring to all 2-neighbors to get an odd 5-coloring of $G$. Thus, $G$ has no 4-vertex with at least one 2-neighbor and all $3^{-}$-neighbors. Similarly, $G$ does not contain adjacent 4-vertices, $v$ and $w$, each with three 2-neighbors. If so, then $G-(N[v]\cup N[w])$ has an odd 5-coloring. We color $v$ and $w$ with distinct colors that each differ from the colors on the three colored vertices at distance two in $G$. Again, we can extend this coloring to an odd 5-coloring of $G$.

Now we use discharging to reach a contradiction, which will finish the proof. We give each vertex $v$ initial charge $d(v)$ and use the following two discharging rules.

1. (R1)

   Each 2-vertex takes $3/7$ from each neighbor.

2. (R2)

   Each 3-vertex with a 2-neighbor and each 4-vertex with three 2-neighbors takes $1/7$ from each $3^{+}$-neighbor.