An Attempt to Position the German Political Parties on a Tree for 2013 and 2017

Lemma

Assume we have a party tree with labels of 0, 12, 1 on all party vertices, and on all dummy vertices of degree 4 or more, and on all dummy vertices adjacent to other dummy vertices. If this labeling is monotonous, then it is monotonously extendable.

Proof

All we need to show is that the labeling is monotonously extendable for any unlabeled dummy vertex x of degree 3 with all three neighbors y₁, y₂, y₃ party vertices. Let ℓ(y₁), ℓ(y₂), and ℓ(y₃) be the labeling of these vertices.

Let us first note that if we have a path over y_i and y_j, i≠j∈{1, 2, 3}, that was monotonous before labeling x and if we label x with the label of y_i or y_j, then this path remains monotonous.

Therefore if at least two of the labelings of y₁, y₂, y₃ are the same, say ℓ(y₁)=ℓ(y₂), then we use this repeatedly occurring label as the label of x and are done by the previous remark.

If ℓ(y₁), ℓ(y₂), and ℓ(y₃) are all different, say ℓ(y₁)=0, ℓ(y2)=12, ℓ(y₃)=1, then we can choose ℓ(x)=12. By the remark above all that has to be checked is whether paths over y₁ and y₃, that have been monotonous before labeling x are still monotonous after labeling x. But such a path was obviously increasing …≤ℓ(y₁)<ℓ(y₃)≤… and remains increasing …≤ℓ(y₁) <ℓ(x) <ℓ(y₃)≤….

Proposition

Let T≺_dH for two n-party trees T and H. We assume we have corresponding partial labelings of all the party vertices in both trees, and all dummy vertices are unlabeled. Note that if x is a party vertex, x and the contraction vertex c are considered corresponding vertices, and have therefore the same labels.

If this partial labeling is monotonously extendable in T, then it is also monotonously extendable in H.

Proof

Let xy be an edge of H, with y a dummy vertex, and let T be obtained from H by contracting this edge into a vertex c. We take a monotonous extension (ℓ) of the labeling of T, labeling all vertices of T, and label all corresponding dummy vertices of H in the same way. If x and c are dummy vertices, we label x the same as c, ℓ(x)=ℓ(c). By our construction this is also the case if x and c are both party vertices. The only unlabeled vertex in H so far is then y. Every path in H not containing y corresponds to a path in T with exactly the same labels, and is therefore already monotonuously labeled.

Now we label y in the same way c was labeled in T, ℓ(y)=ℓ(c). We take any path in H containing y and have two cases. Either this path does not contain x, then it corresponds 1-1 to a path in T with c instead of x. Therefore the labeling extension in H is monotonuous then. The other case is that this path does contain x, then it contains the edge xy, with both labels equal, ℓ(x)=ℓ(y)=ℓ(c). Contracting this edge xy, we get a corresponding path in T, which is by our assumption monotonuos. Since replacing a label by two identical consecutive copies of that label does not destroy monotony, this path is monotonuosly labeled then too.