Game Development Reference
In-Depth Information
Further, note that if u
V is a good vertex, then there is at least one
u where u is good. Similarly, if u
optimal u
V is good then for every
u we have that u is good, and if u
optimal transition u
V is good
G , where the
u then u is good. Hence, we can define a game
and u
V consists of all good vertices of G , and for all u, u
V we
set of vertices
G iff u
have that ( u, u ) is a transition of
u is an optimal transition of G .
G are the same as in G . Now we prove the
The transition probabilities in
following three claims:
V we have that val ( u, G )= val ( u, G ).
(a) For every u
σ,π
v
( Reach ( T, G ))
val ( v, G )= .
(b)
σ
Σ G
π
Π G :
P
σ,π
v
(c)
σ
Σ G
π
Π G :
P
( Reach ( T,G ))
.
Note that Claim (c) is exactly (5.9). We start by proving Claim (a). Let
u ∈
V . Due to Proposition 5.6, there is a MD strategy π ∈ Π G which is
optimal minimising in every vertex of G (particularly in u ) and selects
only the optimal transitions. Hence, the strategy π can also be used in the
restricted game G and thus we obtain val ( u, G ) ≤ val ( u, G ). Now suppose
that val ( u, G ) < val ( u, G ). By applying Proposition 5.6 to
G , there is an
optimal minimising MD strategy π
Π G . Further, for every vertex t of
G which is not good there is a strategy π t
Π G such that for every σ
σ,π t ( Reach ( T,G )) < val ( u, G ) (this follows immediately
from (5.10)). Now consider a strategy π
Σ G we have that
P
Π G which for every play of G
initiated in u behaves in the following way:
G ,
As long as player
uses only the transitions of G that are preserved in
the strategy π behaves exactly like the strategy π .
r which is not a transition in G
for the first time, then the strategy π starts to behave either like the
optimal minimising strategy π or the strategy π r , depending on whether
r is good or not (observe that if r is good, then val ( r ,G ) < val ( r, G )
because r
When player
uses a transition r
G ).
r is not a transition of
Now it is easy to check that for every σ
Σ G
we have that
σ,π u ( Reach ( T,G )) < val ( u, G ), which contradicts the assumption that u is
good.
Now we prove Claim (b). Due to Proposition 5.5, for every u ∈
P
V we
can fix a strategy σ u
Σ G and n u
1 such that for every π
Π G we
( Reach n u ( T, G )) > val ( u, G ) / 2. For every k
σ u
u
have that
0, let B ( k )
be the set of all vertices u reachable from v in G via a path of length
exactly k which does not visit T . Observe that B ( k ) is finite because G is
finitely branching. Further, for every i
P
0 we define a bound m i inductively
Search Nedrilad ::




Custom Search