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Implementation of parallel randomized algorithm
for skew-symmetric matrix game
Ajay Shankar Bidyarthy
July 13, 2012
Parallel randomized algorithm for symmetric games
Consider skew-symmetric matrix games Anxn, A = −Aᵀ. The value
of the game is v? = 0, the case of optimal strategies of both
players coinside, and (?) simplifies to the problem of computing an
x such that Ax ≤ εe, x ∈ S = {x ∈ Rn|eᵀx = 1, x ≥ 0} .
The algorithm given below finds an ε-optimal strategy x for A in
time O(f (n)) with probability ≥ 1/2, where f (n) is a polynomial.
We assume that ε ∈ (0, 1], and n ≥ 8.
Algorithm
[1]. Initialize: X = 0, U = 0, X ,U ∈ Rn, p = e/n, t = 0.
[2]. Repeat:
[3]. Iteration count: t = t + 1..
[4]. Random direction: pick a random k ∈ {1, ..., n} with
probability pk .
[5]. X − update: Xk = Xk + 1.
[6]. U − update: For i = 1 : n do in parallel Ui = Ui + aik .
[7]. p − update: For i = 1 : n do in parallel
pi = piexp{ ε2aik}/∑n
j=1 pjexp{ε2ajk}.
[8]. Stopping criterion: If U/t ≤ εe, Output: x = X/t and halt.
4th step can be computed as follows:
[1]. For i = 1, 2, ..., n calculate cummulative sum
Si = p1 + p2 + ...+ pi .
[2]. Generate an uniform random number z ∈ [0, 1].
[3]. Pick a random number k ∈ {1, 2, ..., n} such that
Sk−1 < z ≤ Sk .
Theorem:
i) The stopping criterion U/t ≤ εe guarantees the ε− optimality
of the output vector x i.e. x = X/t
Theorem:
i) The stopping criterion U/t ≤ εe guarantees the ε− optimality
of the output vector x i.e. x = X/t
Proof: Let X (t) and U(t) denote the n − vectors X and U
respectively at the end of iteration t ≥ 1. According to 5th and
6th step of algorithm we have
[5]. X − update: Xk = Xk + 1.
[6]. U − update: For i = 1 : n do in parallel Ui = Ui + aik .
It maintain the invariants X (t)/t ∈ S = {x ∈ Rn|eᵀx = 1, x ≥ 0}
and U(t) = AX (t).
Theorem:
ii) Algorithm halts in t? = 4ε−2ln(n) iterations with probability
≥ 1/2.
Proof: According to 5th and 6th step of algorithm we have
[6]. U − update: For i = 1 : n do in parallel Ui = Ui + aik .
[7]. p − update: For i = 1 : n do in parallel
pi = piexp{ ε2aik}/∑n
j=1 pjexp{ε2ajk}.
we can see that
pi (t) = Pi (t)/∑n
j=1 Pj(t), for t = 1, 2, ...,
where
Pi (t) = exp{εUi (t)/2}, i = 1, 2, ..., n.
Proof continue...
Now, define the potential function Φ(t) =∑n
i=1 Pi (t).
If the algorithm selects an index k ∈ {1, 2, ..., n}in step 4 at
iteration t, then
Φ(t + 1) =∑n
i=1 Pi (t + 1) =∑n
i=1 exp{εUi (t + 1)/2}
=∑n
i=1 exp{εUi (t)/2 + εaik/2}
=∑n
i=1 Pi (t)exp{εaik/2}
= Φ(t)∑n
i=1 pi (t)exp{εaik/2}.
And since such a k is selected with probability pk(t), conclude that
the expected value of Φ(t + 1) is given by
E (Φ(t + 1)) = Φ(t)∑n
i ,k=1 pi (t)pk(t)exp{εaik/2}.
Proof continue...
By our assumption that aik ∈ [−1, 1], we have
exp{εaik/2} ≤ 1 + ε2aik + ε2
6 for all εin(0, 1].
Since A is skew-symmetric,
∑ni ,k=1 pi (t)pk(t)aik = 0.
We also have that
∑ni ,k=1 pi (t)pk(t) = (
∑ni=1 pi (t))2 = 1 for any p(t) ∈ S .
Hence
E [Φ(t + 1)] ≤ E [Φ(t)](1 + ε2/6)
Proof continue...
which implies
E [Φ(t)] ≤ Φ(0)(1 + ε2/6)t = n(1 + ε2/6)t ≤ nexp{tε2/6}.
After t? = 4ε−2ln(n) iterations we have E [Φ(t?)] ≤ n5/3. By the
Markov inequality, the latter implies that Φ(t?) ≤ 2n5/3 with
probability ≥ 1/2. Since n ≥ 8 by assumption,
Φ(t?) ≤ n2 with probability ≥ 1/2.
On the other hand, if Φ(t?) ≤ n2, then by the definition
Φ(t) =∑n
i=1 Pi (t) we have Pi (t?) ≤ n2 for all i . By
pi (t) = exp{εUi (t)/2}, i = 1, 2, ..., n. this is equivalent to
ε2Ui (t
?) ≤ 2ln(n), i = 1, 2, ..., n,
Proof continue...
or equivalent to
U(t?)
t? ≤ 4ln(n)t?ε e = εe,
which means that the stopping criterion (step 8) of algorithm is
satisfied at the t?th iteration of algorithm with probability ≥ 1/2.