3.6 Finding sparse -equilibria in low-rank games
3.6.2 Three or more players
2 we obtain
˜
p0C2qˆT ≥ max
r (˜p0C2)r−2
≥ max
r ((¯pC1−δ1k)C2)r−2 from (3.15)
≥ max
r (ˆpC1C2)r−3
≥ pCqˆ 0T −3 for any q0 (3.22)
Combining (3.21) and (3.22), ˆ
pCqˆT ≥pCqˆ 0T −4
Similarly, ˆpRqˆT ≥p0RˆqT −4 for any p0.
We analyze the run-time for the algorithm above in terms of the support parameter kand number of strategiesn, andB. The first step of the algorithm where we “guess”
˜
p0 requires exhaustively going through all possible choices for each component of
˜
p0 in the interval [−B, B] in steps of δ, and similarly for guessing ˜q0. This takes time (2B/δ)2k = (4B2k/)2k. For each choice of (˜p0,q˜0), we will need to solve the linear program above which takes poly(|G|) time. The applications of Carath´eodory’s Theorem also take poly(|G|) and so the running time is as claimed and this completes the proof of the theorem.
it can be expressed as: T =Pp
i=1t(i)1 ⊗t(i)2 ⊗. . .⊗t(i)l where each t(i)j is an n×1vector (i.e. it is the sum of p pure tensors).
It will be convenient to aggregate the information in a p-decomposition of such a tensor T into a `-tuple (Ci1, Ci2, . . . , Ci`) where each Cij is an n×p matrix whose columns are given by
t(1)ij , . . . , t(p)ij
. We will refer to ther-th row of a matrix Cij as Cij(r). We will also define to be the component-wise product of vectors:
Definition 3.6.4 For two vectors v, w∈Rn, define (vw) to be the vector (v1w1, . . . , vnwn)∈Rn.
The following inequality will be useful in obtaining a result for the extension to three or more players.
Lemma 3.6.5 Letx1, x2, . . . , x`,xe1,xe2, . . . ,xe` ∈Rk be vectors satisfying|xi−xei| ≤δ and furthermore, let |xei| ≤B for all i= 1, . . . , `. Then,
hx1,(xe2. . .xe`)i −δ`k(B +δ)`−1 ≤ hx1,(x2x3. . .x`)i ≤ hx1,(xe2. . .xe`)i+
δ`k(B+δ)`−1
Proof.
hx1,(x2x3. . .x`)i ≥ hx1,(xe2−δ1k)x3. . .x`i
≥ hx1,xe2x3. . .x`i −δ(B +δ)`−2hx1,1ki
≥ hx1,xe2x3. . .x`i −δk(B+δ)`−1 ...
≥ hx1,xe2. . .xe`i −(`−1)δk(B+δ)`−1
≥ hx1,xe2. . .xe`i −δ`k(B +δ)`−1
The other side of the inequality can be similarly shown.
We begin by generalizing Lipton et al. [LMM03] to the multiplayer case. The following theorem shows that low rank games in the multiplayer case have sparse Nash equilibria; in the next section we give nontrivial algorithms to find approximate equilibria with this sparsity.
Corollary 3.6.6 Let G = (T1, . . . , T`, n) be an `-player game, and suppose Ti has rank ki. Then there exists a Nash equilibrium (p∗1, . . . , p∗`) with |supp(p∗i)| ≤ 1 + P`
j=1kj for all i.
Proof. Let (q1, . . . , q`) be a Nash equilibrium for G and let (Ci,1, . . . , Ci,`) be the tensor decomposition for Ti. Givenq1, . . . , q`, we can define vectors in Rn:
w1 = (C1,1( q2C1,2 q3C1,3 · · · q`C1,` )T) w2 = (C2,2(q1C2,1 q3C2,3 · · · q`C2,` )T) w3 = (C3,3(q1C3,1 q2C3,2 · · · q`C3,` )T)
...
w` = (C`,`(q1C`,1 q2C`,2 · · · q`−1C`,`−1 )T)
(3.23)
By the definition of a Nash equilibrium, the following conditions are satisfied:
supp(q1) ⊆ {v : (w1)v = max
u (w1)u} supp(q2) ⊆ {v : (w2)v = max
u (w2)u} ...
supp(q`) ⊆ {v : (w`)v = max
u (w`)u}.
And indeed any (qb1, . . . ,qb`) satisfying these conditions (when the wi are defined rel- ative the theqbi) is a Nash equilibrium.
Set s = P`
j=1kj. Now, by Carath´eodory’s Theorem there exists qb1 ∈ Rn with supp(qb1)⊆supp(q1) and|supp(qb1)| ≤s+ 1 for which:
qb1(C2,1|C3,1|C4,1| · · · |C`,1) =q1(C2,1|C3,1| · · · |C`,1),
since the right-hand-side is a convex combination of vectors inRs
0 fors0 ≤s. Similarly, there exists qb2 ∈Rn with supp(qb2)⊆supp(q2) and |supp(qb2)| ≤s+ 1 for which:
qb2(C1,2|C3,2|C4,2| · · · |C`,2) = q2(C1,2|C3,2|C4,2| · · · |C`,2), since the right-hand-side is a convex combination of vectors in Rs
0 for s0 ≤s.
A symmetric argument gives us a sparse qbi from each qi. Moreover, these qbi produce precisely the same vectors w1, w2, . . . , w` via (3.23). And, since supp(qbi) ⊆ supp(qi) for all i, the strategies qbi satisfy the above conditions for being a Nash equilibrium, assuming the original qi did.
We now turn to algorithms for finding sparse approximate equilibria, with three or more players. We first consider the case when the tensor decomposition is known.
This is not an unnatural assumption: in§3.6.3, we describe a class of games for which the tensor decomposition is naturally given by the description of the payoff functions.
Theorem 3.6.7 Let G = (T1, . . . , T`, n) be an `-player game, and suppose we are given a k-decomposition of Ti = (Ci1, . . . , Ci`) where each of the Cij is an n ×k matrix with integer values in [−B, B] for i, j = 1, . . . , `. Then for every >0, there is a deterministic procedure P running in time
(2B)`k`/k(`−1)`
poly(|G|)
that returns a 4-Nash equilibrium (p1, p2, . . . , p`) with |supp(pi)| ≤1 +`k for all i.
Proof. As in the two player case, our first step is to “guess” a collection of vectors to within δ = k`(2B) `−1 precision. We describe the “correct” guess relative to an (arbitrary) Nash equilibrium (p1, . . . , p`) forG.
1. Let (p011, . . . , p01`) = (p1C11, p2C12, . . . , p`C1`) be an`-tuple of 1×kvectors. Note that since all entries in (C11, C12, . . . , C1`) lie in [−B, B], entries in (p011, . . . , p01`) also lie in [−B, B].
2. Guess an (`−1)-tuple (pf012, . . . ,pf01`) such that
|(p01i)j −(p]01i)j| ≤δ i= 2, . . . , `;j = 1. . . , k
3. For C11(r), ther-th row inC11 let αr =
C11
pf012pf013. . .pf01`
T
r
; r= 1, . . . , n S1 = n
r|max
t αt−2≤αr ≤max
t αto
4. By repeating steps 1 through 3 on T2, . . . , T` we can similarly obtainS2, . . . ,S`. 5. Find (¯p1, . . . ,p¯`) satisfying the following linear program
|(¯pjCmj)r−(gpmj)r| ≤δ r= 1, . . . , k;j = 1. . . , `;
m= 1, . . . , j−1, j+ 1, . . . , ` (3.24) (¯pj)i≥0; i= 1, . . . , n;j = 1, . . . , ` (3.25)
n
X
i=1
(¯pj)i = 1; j = 1, . . . , ` (3.26) (¯pj)i = 0; i /∈ Sj;j = 1, . . . , ` (3.27)
6. Let Di = (C1,i|C2,i |. . .|Ci−1,i |Ci+1,i |. . .|C`,i) for i = 1, . . . , `. v = ¯piDi is a convex combination of column vectors in Di that are in an`k-dimensional subspace. Apply Carath´eodory’s Theorem (Theorem 3.5.1) to obtain pbi such that v =pbiDi for i= 1, . . . , ` and supp(pbi)≤`k+ 1.
7. Return (pb1, . . . ,pb`).
As before, in order to prove the correctness of the above algorithm we postulate the following claims:
Claim 3.6.8 A feasible solution to the linear program formulated in step 5 of the algorithm exists.
Proof. (Of Claim 3.6.8) We will show that (¯p1 = p∗1,p¯2, . . . ,p¯` = p∗`), the Nash equilibrium strategies for T1, . . . , T` satisfy the linear program. Since p∗i0 = p∗iC11 satisfies |p∗i0[j]−pf0mi[j]| ≤δ for j = 1, . . . , k and for m = [`]− {i}, p∗i satisfies (3.24).
(3.25) and (3.26) follow from ¯pi being a probability distribution. So, it suffices to show that p∗1, . . . , p∗` satisfy (3.27), i.e., for any r ∈ supp(p∗i), r ∈ supp(Si). We will show this to be true for i = 1; it symmetrically follows for i = 2, . . . , `. Note that
|fp01i| ≤B and |p∗iC1i−pf01iC1i| ≤δ. Then, applying Lemma 3.6.5 gives us
C11(p∗2C12. . .p∗`C1`)T
r− ≤
C11(pf012. . .pf01`)T
r
≤ C11(p∗2C12. . .p∗`C1`)T
r+ and likewise,
maxt
C11(pf012. . .pf01`)T
t
− ≤ C11(p∗2C12. . .p∗`C1`)T
r
≤maxt
C11(pf012. . .pf01`)T
t+ Combining these two equations,
maxt
C11(pf012. . .pf01`)T
t
−2 ≤
C11(pf012. . .pf01`)T
r
≤ max
t
C11(pf012. . .pf01`)T
t
Hence r ∈ S1.
Claim 3.6.9 (pb1, . . . ,pb`) as returned by the algorithm is a 4-equilibrium.
Proof. (Of Claim 3.6.9) For the strategy tuple (p1, . . . , p`) and given a decomposition of Ti = (Ci1, . . . , Ci`), the payoff to player i given by Ti(p1, . . . , p`) may be expressed as
Ti(p1, . . . , p`) =piCii p1Ci1p2Ci2. . .pi−1Ci(i−1)pi+1Ci(i+1). . . p`Ci`
T
Put this way, we need to show that for any (p01, . . . , p0`), pb1, . . . ,pbl returned by the algorithm satisfy
pbiCii(pb1Ci1 . . .pdi−1Ci,i−1pdi+1Ci,i+1. . .pb`Ci`)T ≥ p0iCii(pb1Ci1. . .pb`Ci`)T − for i = 1, . . . , `. We will prove this for i = 1, the proof follows symmetrically for i = 2, . . . , `. First, we note that from the application of Carath´eodory’s theorem in step 6,
pb1C11(pb2C12. . .pb`C1`)T = pb1C11(¯p2C12. . .p¯`C1`)T (3.28) Then, applying Lemma 3.6.5
pb1C11(¯p2C12. . .p¯`C1`)T ≥ pb1C11
pf012. . .pf01`T
− (3.29)
By the same argument as before, since we picked only those r in supp(pb1)⊆ S1 such that
C11
pf012. . .pf01`T
r
≥max
r
C11
pf012. . .pf01`T
r
−2 we get
pb1C11
pf012. . .pf01`
T
≥ max
r C11(r)
pf012. . .pf01`
T
−2 (3.30) Applying Lemma 3.6.5 and from Carath´eodory’s Theorem,
maxr
C11
pf012. . .pf01`T
r
≥ max
r
C11(¯p2C12. . .p¯`C1`)T
r
−
= max
r
C11(pb2C12. . .pb`C1`)T
r−
≥ p01C11(bp2C12. . .pb`C1`)T −(3.31) for any p01
Combining (3.28)–(3.31),
pb1C11(pb2C12. . .pb`C1`)T ≥p01C11(pb2C12. . .pb`C1`)T −4 for any p01
The total number of guesses made in step 2. of the procedure is k(` −1)` (k guesses for each component of the (`−1) different vectors in step 2, repeated` times in step 4.). Each component is within [−B, B] and is guessed to δ accuracy, so the guessing takes time
(2B/δ)k(`−1)` = (2B(k`(2B)`−1)/)k(`−1)` = (2B)`k`/k(`−1)`
For each guess, we solve a linear program, taking time poly(|G|), for a total running time as claimed. This completes the proof of the theorem.