Download Lemke-Howson Algorithm for Non-Zero-Sum Games and more Schemes and Mind Maps Commercial Law in PDF only on Docsity!
Game Theory Prof. K. S. Mallikarjuna Rao Department of Industrial Engineering & Operations Research Indian Institute of Technology - Bombay
Lecture 24 Non-Zero-Sum Games: Lemke-Howson Algorithm - III
In the previous lecture we introduced labels for the vertices of the best response polyhedron and the corresponding normalized polytope. We also stated a lemma without a proof. We will go back to that lemma and continue discussing the Lemke-Howson algorithm.
Lemma. In a non degenerate game for a pair of extreme points (x, y) we have |L(x)| = m and |L(y)| = n.
Proof: Game is non-degenerate. Therefore, at most Supp(x) pure strategies can be best responses to x, where x is the normalized vector of x. And similarly, at most Supp(y) pure strategies can be best responses to y.
|L(x)| ≤ |S 1 \ Supp(x)| + |Supp(x)| = m |L(y)| ≤ |S 2 \ Supp(y)| + |Supp(y)| = n
These polytopes are full dimension and (x,y) is an extreme point. Therefore,
|L(x)| ≥ m |L(y)| ≥ n
Hence, this proves that |L(x)| = m and |L(y) = n|. This proves the lemma.
Now, let us look at the following theorem: Theorem. A pair of extreme points (x, y) ∈ P 1 × P 2 \ {( 0 , 0 )} is fully labelled iff the corresponding normalized vector (x, y) is a Nash equilibrium.
Proof. Let (x, y) ∈ P 1 × P 2 \ {( 0 , 0 )} be fully labelled. Let T 1 = Supp(x) and T 2 = Supp(y). For all k ∈ T 1 , x does not have label k as xk > 0. Therefore, y must have label k.
⇒(Ay)k = 1
⇒(Ay)k′ =
yT^1
while
(Ay)k′ ≤
yT^1
∀k′^ ∈ S 1
This implies that k is the best response to y. Further, for k ∈/ T 1 , x does have label k. A label cannot appear twice in a fully labelled pair. This implies that y does not have label k, which in turn implies that k is not the best response to y. Hence, (x, y) is a Nash equilibrium. Conversely, let (x, y) be a Nash equilibrium. This implies,
S 1 \ Supp(x) ∪ Supp(y) ⊆ L(x) S 2 \ Supp(y) ∪ Supp(x) ⊆ L(y)
Therefore, L(x)∪L(y) = S 1 ∪S 2. This is the same as saying that they are fully labelled. This proves the theorem.
Lemke-Howson Algorithm
The idea is to start from the origin and pivot alternatingly in P 1 and P 2 until a labelled pair is found. Let V 1 be the set of extreme points of P 1 and V 2 denote the set of extreme points of P 2.
Let Ei be the set of edges between adjacent extreme points in Vi.
E 1 = {(x, x′) ∈ V 1 ×V 1 : |L(x) ∩ L(x′)| = m − 1 } E 2 = {(y, y′) ∈ V 2 ×V 2 : |L(y) ∩ L(y′)| = n − 1 }
Let V = V 1 ×V 2 and E be given by the following:
{( (x, y), (x′, y)
∈ V ×V |(x, x′) ∈ E 1
(x, y), (x, y′)
∈ V ×V |(y, y′) ∈ E 2
The whole idea here is that if we restrict our attention to extreme points that are almost fully labelled with the possible exception of a particular label i, then there is always a unique way in which we can proceed. We introduce another notation. For i ∈ S 1 ∪ S 2 , let
V i^ = {(x, y) ∈ V : L(x) ∪ L(y) ⊇ S 1 ∪ S 2 \ {i} Ei^ = E ∩ (V i^ ×V i)
Theorem. Let i ∈ S 1 ∪ S 2. Then, V i^ contains ( 0 , 0 ) as well as every (x, y) ∈ V such that (x, y) is Nash Equilibrium. Assuming non-degeneracy, the point ( 0 , 0 ) and the elements of V i^ correspond- ing to an equilibrium have degree one in the graph (V i, Ei) and all other nodes in V i^ have degree two.
Proof. ( 0 , 0 ) and all pairs corresponding to Nash Equilibria are fully labelled. Therefore, the first part is obvious. For the second part, consider (x, y) ∈ V i^ and let (x, y) be the corresponding normalized strategies. Now, as x, y are extreme points, from an earlier lemma that we have seen, |L(x)| = m and |L(y)| = n. If (x, y) = ( 0 , 0 ) or (x, y) is Nash Equilibrium, then (x, y) is fully labelled and L(x) ∩ L(y) = φ. The neighbours of (x, y) are those elements in V i^ that replace i with some other label, those where
