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Extra resources for Discrete Mathematics [Lecture notes]
E. in time proportional to the size of G ∩ Int(C). Proof. 6. Let us first explain the notation uses in the pseudocode. Line 2 initializes the path P , which will be used together with a segment of C in order to produce an outer cycle of the recursively smaller problem. P is a path along neighbors of vr along their embedding order. Note that vr is adjacent to vr−1 , v1 , and also to the vertices vr , vr+1 , . . , vr+1 . Hence deg(vr ) is really equal to i + 3. As C is a cycle (with its orientation implicit by notation — in the increasing order of indices) then Cx,y denotes the subpath of C starting at vertex x an ending in vertex y, following the same orientation as C.
Vr along C we have |L(vi )| = 3, and (3) for every vertex x in the interior of C we have |L(x)| = 5. and inductively claim that every smaller near triangulation satisfying above conditions admits a list coloring which extends the precoloring of v1 , v2 . The induction basis is easy. If |V (GC )| = 3, then v1 , v2 , v3 are the only vertices of GC . As L(v3 ) = 3 there exists a color different from both c(v1 ) and c(v2 ) for v3 . The induction step comes in two flavors. Assume first that C has a chord, a pair of two nonconsecutive vertices vi and vj along C which are adjacent.
Vr+ lie on C. Let GC be the near triangulation GC − vr , whose outer face fC is determined by the facial cycle C = v1 , v2 , . . , vr−1 , vr , vr+1 , . . , vr+ . Let us construct an appropriate list assignment L for vertices vr , . . , vr+ to keep the induction going. Let a, b be the colors in L(vr ) different form c(v1 ). For every vertex vr+i let L (vr+i ) be obtained by removing colors a, b from L (vr+1 ) (and possible additional colors to keep L (vr+1 ) of length exactly 3 — this is not entirely necessary as additional options for colors do not make a coloring problem more difficult, but it simplifies the arguments).