Analysis of Markov chain algorithms on spanning trees, rooted ...

J. Fehrenbach and L. Rüschendorf 11
Similary, for p ′ a ∈ (A ′ i) 0≤i≤l ∈ P Y X a and p ′ b = (B′ i) 0≤i≤l ∈ P
Y X b with
A ′ i =
{
B
′
i , 0 ≤ i ≤ j
B ′ i ⊕ {a, b}, j < i ≤ l
where again j ∈ {0, . . . , l − 1} with b ∈ A ′ j ⊕ A ′ j+1 holds F G (p ′ a) = F G (p ′ b ).
Pro<strong>of</strong>: By definition <strong>of</strong> X a and X b , one can easily see that the graphs M a :=
(V, X a ∪ Y )/(X a ∩ Y ) and M b := (V, X b ∪ Y )/(X b ∩ Y ) are isomorph. If a
connects the nodes w and t and b the nodes w and s in M, then I : M a → M b
given by
⎧
⎨ v, v /∈ {v a , s}
I(v) := t, v = v a
⎩
v b , v = s
und I(e) :=
{ e, e ≠ b
a, e = b
for all nodes v and edges e in M a defines an isomorphism. Figure ?? shows an
example.
t
✉
❚
✔✔
✔ ✔ ❚
a ❚
w
✑ ✉ ❚
✑✔ ✉
✔ ✔ ◗ ◗◗
❚
✑✑ c b ❚
◗❚ ✉
u
s
M
v a
✉
❅ ❅
✉
b ❅
❅ ✉
u
s
M a = (M/a) \ {c}
t
✉
❅ ❅
a ❅
✉
❅ ✉
u
v b
M b = (M/b) \ {c}
Figure 3: An example for a graph M from Lemma 3.1. The resulting graphs
M a and M b are isomorph.
The number <strong>of</strong> nodes in M a is the same as that in M b . The pro<strong>of</strong> <strong>of</strong> Lemma
3.1 is given by induction over |M a |:
For |M a | = 2 the path p a = (A i ) 0≤i≤1 given by A 0 = X a and A 1 = Y is the
only path in P X a Y
with a positive weight in F G
, i.e. F G
(p a ) = 1. The same holds
for p b = (B i ) 0≤i≤1 given by B 0 = X b = (A 0 \ {a}) ∪ {b} and B 1 = Y = A 1 . So
in this case the lemma is proved.
If |M a | = l + 1 let D a be the set <strong>of</strong> nodes in M a <strong>of</strong> minimal degree d min
and D b the analogous set <strong>of</strong> nodes in M b . Each node in D a corresponds to a
node in D b via the isomorphism I.
The value <strong>of</strong> F G
(p a ) is based by construction on the paths in V(p a ). For
v ∈ D a each path p ′ a ∈ P ∩ V(p X a v Yv
a) corresponds to a path p ′ b ∈ P ∩ V(p
Xu b b ),
Yu
u := I(v) ∈ D b . Is a an edge at v in M a , so b is an edge at u in M b . This leads to
Xv a = Xv b and Y v = Y u and, therefore, p ′ a = p ′ b . Otherwise, if a is not an edge at
v we have v = u. In this case either X a arises from Xv a and Y v = Y or Xv a = X a
and Y arises from Y v by exchanging two edges. The same holds for Xv b and Y v .
This modification can also be done at X and Y and for the resulting spanning