3. Informational Digraphs
Let
be an OR game. If
and
for positive integer
, then we say that
is a
proper initial segment of
h. We define binary relation ≺ on the set
as follows. For
,
, we say that
if there exist
and
such that
is a proper initial segment of
. While von Stengel and Forges (2008) [
3] used this binary relation in the context of a two-player game satisfying perfect recall, our ordering can be applied to any general OR game. From the pair
, we define an
informational digraph of
by the directed graph
, where
and
≺. A sequence of vertices
, where
and
for each
is called a
walk from
to
if
and
. When there is a
walk from
to
, we say that vertex
is
reachable from
and we denote this
reachability relation of digraph
G by
so that
if
is reachable from
. We call walk
from
to
the
flow of information from information set
to information set
in game
. We say that
G is
asymmetric if for all
,
,
implies
. Note that asymmetry implies irreflexivity. We say that digraphs
and
have the same reachability relation if
and
. We say that
H is a
transitive reduction of
G if
H is the digraph with the least number of edges that has the same reachability relation as
G, that is, for all
such that
we have
.
is a
maximal walk from
to
if it is not a proper subsequence of any walk from
to
. Walk
is a proper subsequence of walk
if
is a subsequence of
while
is not a subsequence of
.
We say that has an informational cycle if its informational digraph has some vertex and walk from to itself satisfying the following: For all , imply , except at and where . Although the graph may have longer cycles with repeated vertices, the condition used in this definition is necessary for any kind of cycle to exist in the graph. We say that G is acyclic if it does not contain any informational cycle. We say that informational digraph does not contain redundant information if its transitive reduction is unique.
Lemma 1. Let be an acyclic informational digraph. If G has a walk from to for some , then there exists some maximal walk from to .
Proof. Fix any and in V and let denote the set of all walks from to . We define binary relation ⊲ on the set as follows. For , , we say that if is a subsequence of . Clearly, ⊲ is a reflexive and transitive. Moreover, it is anti-symmetric since G is acyclical Note that if G were cyclical, there could exist infinite-length walks and such that is a subsequence of and is a subsequence of , where and is a cyclic permutation of .
Then, it follows that ⊲ is a partial order. Now, since G is acyclical and by the completeness of OR games (axiom OR3), every chain of has an upper bound in . Therefore, by Zorn’s Lemma, has a maximal element. Note that the maximal walk need not be unique. □
The following claims are direct consequences of Lemma 1 and the definition of transitive reduction.
Claim 1. Let be an acyclic informational digraph and let be a transitive reduction of G. If α is a walk in from to for some , then it is also a maximal walk from to .
Claim 2. Let be an acyclic informational digraph. Suppose further that and are two transitive reductions of G. If α is a maximal walk from I to in H for some , then it is also a maximal walk from I to in .
We say that OR game
can be
time structured if and only if there exists a real valued function
f on the vertices of its informational digraph satisfying:
implies
. The following result is a direct consequence of the fact that acyclic digraphs admit a topological sorting (see Kahn (1962) [
7]).
Theorem 1. G is acyclic if there exists a real valued function f acting on its vertices satisfying: implies .
Theorem 2. Let Ω be an OR game with informational digraph . Then, Ω can be time structured if G is (i) asymmetric and (ii) does not contain redundant information.
Proof. By Theorem 1, it suffices to show that G is acyclic if (i) and (ii) hold.
(Only if). Suppose that G is acyclic, then (i) is trivially satisfied. To show (ii), we consider any two transitive reductions of G, say H and . We claim that . Indeed, suppose that there exist vertices x and y such that . Then, by Claim 1 is a maximal walk from x to y in H, and by Claim 2 it is also a maximal walk from x to y in . Hence, .
(If) Suppose that (i) and (ii) are satisfied. (i) ensures that G cannot have cycles of length . Suppose by contradiction that G had some cycle , where . Then, there exists some cyclic permutation of such that is mapped to while is not mapped to through the permutation. One can then construct graph such that (a) and (b) contains walk , but not walk . As a result, there exists some transitive reduction of G denoted by A, which contains and some other transitive reduction of G, denoted by B, which contains . Since , the digraph contains redundant information, a contradiction. □
Example 1. We consider Figure 6 on page 1012 of [3]. Using the convention that denotes the nth information set of player m, we can construct informational digraph , where and The above is a negative example of Theorem 2 as is an informational cycle. We can obtain a positive example of the theorem by removing edge from E. The digraph will then become acyclic and its transitive reduction is given by , where .