3.2 Location Problem Models in Tree Space
3.2.1 Split Sets of Optimal Solutions
The first thing we establish for the tree space location problems are criteria for splits of optimal solutions. In order to prove the correctness of the criteria we need the following lemma that shows that the distance between two trees reduces when a non-common split is removed from one of the trees.
Lemma 3.2. Suppose that for
X = ({e1, . . . , en1},(w1, . . . , wn1)), T = ({f1, . . . , fn2},(w1, . . . , wn2))∈ Tn there exists a split fl such that fl∈Split(T)\Split(X).
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Let T0 = ({f1, . . . , fl−1, fl+1, . . . , fn2},(w1, . . . , wl−1, wl+1, . . . , wn2)) ∈ Tn, i.e., the tree which is derived by taking X and removing fl. Then d(X, T)< d(X, T0).
Proof. Suppose that the geodesic Γ(X, T) is given by the path P = (X =T0, T1, . . . , Tk=T), i.e., d(X, T) = L(Γ(X, T)) =Pk
j=1ktj −tj−1k2. Sincefl∈Split(T) = Split(Tk) but fl 6∈Split(X) = Split(T0), there exists some i∈1, . . . , k such that fl ∈Split(Tj) for j ≥i and fl 6∈Split(Tj) forj < i.
With this we define a pathP0 fromX toT0 by modifying the sequence of trees that yield the geodesic path. Let P0 = (X =T0, T1, . . . , Ti−1, Ti0, Ti+10 , Tk0 =T0) where we
Having this auxiliary lemma we can show that there are certain splits that have to be contained in Split(T) for any optimal tree T.
Theorem 3.3. Let e ∈
M
T
i=1
Split(Ti). Then e ∈ Split(X∗) for all optimal solutions X∗ ∈ Tn to the median problem (PMed) and e ∈ Split(X∗) for all optimal solutions X∗ ∈ Tn to the center problem (Pmax).
Proof. In order to prove that any optimal solution X∗ to (PMed) or (Pmax) has to satisfy e ∈ Split(X∗), we show that any tree X ∈ Tn with e 6∈ Split(X) is not optimal.
First, assume that for X ∈ Tn there exists some split s ∈ Split(X) such that s is incompatible with e. Apply Lemma 3.2 to X and X0 where X0 is derived from X by removings. This yields
d(X0, Ti)< d(X, Ti) ∀i= 1, . . . , M.
Thus,
M
X
i=1
d(X0, Ti)<
M
X
i=1
d(X, Ti) and maxM
i=1 d(X0, Ti)<maxM
i=1 d(X, Ti) soX cannot be an optimal solution for (PMed) or (Pmax).
Now, let X = ((s1, . . . , sn1),(w1, . . . , wn1)) be a tree for which no s ∈ Split(X) is incompatible with e. If e ∈ Split(X) then the statement holds, so we assume that e 6∈ Split(X). As no split in Split(X) is incompatible with e this implies that Split(X)∪{e}is a compatible set of splits and recalling Definition 2.18, eis a double compatible split ofX and Ti for all i= 1, . . . , M.
Now define c := mini=1,...,MwTei > 0, i.e., c is the minimal length of the split e among all Ti. We now construct a tree X0 for which d(X0, Ti) < d(X, Ti) for all i = 1, . . . , M. Let X0 = ((s1, . . . , sn1, e),(w1, . . . , wn1, c)) and fix a tree Ti. As e is a double compatible split of Ti and X as well as for Ti and X0 it holds that Split(X)\C= Split(X0)\C. So if (A,B) = ((A1, . . . , Ak),(B1, . . . , Bk)) is a support of the geodesic fromXtoT, i.e., satisfies (P1)−(P3), then it also is a support of the geodesic fromX0 to T, as the weights of all splits s∈Split(X)\C = Split(X0)\C coincide. Now, apply the length formula for the geodesic from Theorem 2.21 for both trees:
d(X, Ti) = v u u t
k
X
l=1
(kAlk2+kBlk2)2+X
s∈C
(wsTi −wsX)2
d(X0, Ti) = v u u t
k
X
l=1
(kAlk2+kBlk2)2+X
s∈C
(wsTi −wsX0)2
As the support and all split lengths ofX and X0 are the same except for the split e∈C, we get
d(X, Ti)2−d(X0, Ti)2 = (weTi −weX)2−(weTi−wXe 0)2 >0
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due to
(wTei−wXe )2 = (weTi)2 >(weTi −c)2 = (weTi −weX0)2,
where (wTei)2 > (wTei −c)2 follows from the choice of c. Hence, we get d(X, Ti) >
d(X0, Ti) for all i= 1, . . . , M which implies
M
X
i=1
d(X0, Ti)<
M
X
i=1
d(X, Ti) and maxM
i=1 d(X0, Ti)<maxM
i=1 d(X, Ti), soX cannot be an optimal solution for (PMed) or (Pmax).
We have seen that there are certain splits that are always in the split sets of optimal trees. The counterpart of this idea is to ask which splits are never contained in the split set of optimal trees.
Theorem 3.4. Let e 6∈ ∪Mi=1Split(Ti). Then e6∈Split(X∗) for any optimal solution X∗ to (PMed) and e 6∈Split(X∗) for any optimal solution X∗ ∈ Tn to (Pmax).
Proof. Let X = ((s1, . . . , snX),(w1, . . . , wnX))∈ Tn be such that e ∈ Split(X), i.e., e=si for somei∈ {1, . . . , nX}. Then define X0 by removing si,
X0 = ((s1, . . . , si−1, si+1, . . . , snX),(w1, . . . , wi−1, wi+1, wnX)),
as in 3.2. By the prerequisite it holds thate6∈Split(Ti) fori= 1, . . . , M . Applying Lemma 3.2 then yields
d(X0, Ti)< d(X, Ti) ∀i= 1, . . . , M
soX cannot be optimal for (PMed) or (Pmax) as X0 yields a lower objective.
The analogous statements are known for the Fr´echet variance and have been shown in [MOP12], Lemma 5.1. Additionally, the Fr´echet problem has a unique optimal solution, as shown in [Stu03]. We summarize this:
Theorem 3.5. [MOP12],[Stu03] (PF)has a unique optimal solution X∗. Moreover, if e 6∈ ∪Mi=1Split(Ti) then e 6∈ Split(X∗) and if e ∈
M
T
i=1
Split(Ti) it follows that e ∈ Split(X∗).
4 Solving Tree Space Location
Problems by Transformations to Euclidean Location Problems
In the previous chapter we have introduced Location Theory and modeled and de-fined the location problems in tree space that we want to solve. We already men-tioned that it is in general a very difficult task to solve these problems.
Therefore our approach is to first try to understand the structure of simple versions of the problems (Pmax),(PMed),(PF), e.g., problems in low dimensions or problems where we have strong assumptions on the given set of trees T{T1, . . . , TM}. This helps to understand the key difficulties in solving these problems and, if possible, we try to generalize the methods for these easy cases to a more general setting. We start with the lowest-dimensional, non-trivial tree space, that is,T3.
4.1 Location Problems in T
3In this section we completely describe the solutions of the location problems (Pmax), (PMed), (PF) in T3. The solutions to the problems are easy to obtain once we see that the T3 is just a star graph with three infinitely long edges emanating from the center vertex. Hence, the location problems in T3 boil down to well-known location problems on networks.
First we investigatethe structure ofT3: There are only three splits on the set of leaves {0,1,2,3}, these beings1 = ({2,3}|{0,1}),s2 = ({1,3}|{0,2}),s3 = ({1,2}|{0,3}).
Moreover, maximal orthants ofTn consist ofn−2 splits; this means that a maximal set of compatible splits has cardinality 1 in T3. These simple observations have already been stated in Example 2.2.1, where the embedding ofT3 intoR3 was given.
Furthermore, geodesics in T3 are easy to obtain: If two trees X1, X2 are contained in the same orthant, the geodesic is the straight line connecting the two. If X1 and X2 are in different orthants then the geodesic is the cone path, since all three splits inT3 are pairwise incompatible.
Denote Oi := O({si}) for i = 1,2,3 and Ti := {T ∈ T : T ∈ Oi} for i = 1,2,3, which are the facilities in the respective orthants. Moreover, letli = maxT∈T∩Oiktk2
for i = 1,2,3, where t ∈ R3+ is the embedded vector of edge lengths, compare Definition 2.9.
Theorem 4.1. Let T = {T1, . . . , Tm} ⊂ T3 and deploy the notation introduced above. Then the solutions to the problem are characterized as follows:
(Pmax): Let i∈ {1,2,3} be the index with li ≥lj forj 6=i, then the optimal solution is contained inOi with the length of si being li−l2 j wherelj is the second biggest value among l1, l2, l3.
(PMed): If there exists noi∈ {1,2,3} such that |Ti| ≥ |T\Ti| then 0is the unique optimal solution. Otherwise let i be that index and suppose that Ti ={Tj1, . . . , Tjk} is ordered such that ktjik2 ≥ ktji+1k2 and let
d= (|T
i|−|T\Ti|
2 forT even
|Ti|−|T\Ti|+1
2 forT odd
Then for |T| odd, the unique optimal solution is Td and for |T| even the optimal solution is every tree on the edge between Td+1 and Td, where T0 := 0 is the star tree.
(PF): If there exists no i∈ {1,2,3} such that P
T∈Tiktk2 ≥P
T∈T\Tiktk2 then 0 is the unique optimal solution. Otherwise let i be that index, then the unique optimal solution X is contained in Oi and the length of si ∈Split(X) is
1 M
X
T∈Ti
ktk2− X
T∈T\Ti
ktk2
.
These results may be obtained by calculating the objective of the above solutions and showing that the objective value for all non-optimal trees is higher by estimations.
Instead of doing this we use this possibility to point out the equivalence of the above location problems in T3 to specific location problems on networks. We refer to [LPT95] for a nice overview over facility location on networks.
In [LPT95] one is given a set of facilities V = {v1, . . . , vM} that are contained in the network N. Conveniently, [LPT95] defines networks as subsets of Rd with a finite amount of vertices and the edges are the straight line segments connecting the vertices. Our goal is to model our problems in this setup.
As all objective functions are non-decreasing ind(X, Ti), it is clear that there exists no optimal solution inOi with a weight greater than the maximal weight of trees in this orthant li. This way, we may bound the star graph with infinitely long edges:
LetOi :={T ∈ Oi :wsT
i ≤li} for i= 1,2,3 and let T3 :=∪i=1,2,3Oi.
Now simply settingN =χ(T3) andV =χ(T), we may apply Theorem 3.5 [LPT95]
to receive the solutions to the median problem and Theorem 4.4 [LPT95] to receive the unique solution to the center problem as described in Theorem 4.1.
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In order to get the solution to the Fr´echet problem, we note another connection of T3 to a geometrical object called “open book”, [HHL+13]. The authors of [HHL+13]
mention T3 as one of the first examples for open books. Open books consist of
“pages” of dimension d, here the orthants, that are joined along a common “spine”
of dimension d−1. Hence, the T3 is a very simple example of an open book. The solution to the Fr´echet problem as stated in Theorem 4.1 is then gained by applying Theorem 2.9 in combination with Lemma 3.3 from [HHL+13].