Upload
ajay-bidyarthy
View
87
Download
1
Embed Size (px)
Citation preview
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.
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.
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide
third Slide
content of the third slide