We're now in position to complete the proof of 3-coloring for non-stars in the case of oddly many leaves. Of the two consecutive-leaf buds for each Smaller tree on each side (including the bud and two leaves of the Picture that bud as being at the center of the To handle that case, it's helpful to introduce a tiny bit of terminology: Let's call the vertices that leaves are connected to "buds."Ĭonsecutively through those buds' leaves. It's only if there is an odd number leaves that we have to do something fancier. Here is a corrected version:Īs nvcleemp noted (at MathOveflow), one should start with a 2-coloring (Black and White) of the tree and, if the number of leaves is even, simply change every other leaf around the cycle to a third color (Red). The proof I posted here (and at MathOverflow) yesterday is flawed. However, this method does not tell you ahead of time which trees require only three colors. Incidentally, if $T$ has no degree two vertices, then $G(T)$ is called a Halin graph with or without the degree two restriction, it has bounded treewidth, so an alternative method for finding an optimal coloring in linear time is to use dynamic programming. The graph is the graph of a triangular cylinder, which can be three-colored. In the remaining base case, $T$ has exactly two internal nodes, each adjacent to two leaves. On the other hand, if $p$, $q$, and $r$ use three different colors, then $v$ can be given $q$'s color, the left child of $v$ can be given $r$'s color, and the right child of $v$ can be given $p$'s color. If $p$, $q$, and $r$ use only two colors among them, then the two children of $v$ can be colored in alternation with $q$ and $r$ using the same two colors, and $v$ can be given the remaining third color. Let $p$, $q$, and $r$ be the parent of $v$, the node to the left of $v$'s left child in the leaf cycle, and the node to the right of $v$'s right child in the leaf cycle $q$ and $r$ are adjacent in the smaller graph, so must have different colors. If $v$ has exactly two children, and there are three or more internal nodes, then remove $v$ and its two children from $T$, form the graph by connecting the remaining leaves into a cycle, and then color it. The remaining children of $v$ must alternate in color and reinserting the removed children maintains the same alternation. If $v$ has three or more children, then the tree can be reduced in size by removing two consecutive children, forming the graph by connecting the remaining leaves into a cycle, and then coloring the smaller graph. If $v$ has one child, then the graph can be reduced down to nothing by repeatedly removing vertices of degree at most two, and three-colored by reinserting the vertices in the reverse order and giving them a color distinct from their two neighbors. If you orient the tree from an arbitrarily chosen root node, there always exists an internal node $v$ whose children are all leaves. To prove that every tree with two or more internal nodes gives you a 3-colorable graph, use induction on the number of nodes of the tree.
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