Independent dominating sets in graphs of girth five - CiteSeerX
Independent dominating sets in graphs of girth five Ararat Harutyunyan∗
Paul Horn†
Jacques Verstraete‡
March 9, 2009
Abstract One of the first results in probabilistic combinatorics is that every n-vertex graph of . In minimum degree at least d contains a dominating set of size at most n(1+log(d+1)) d+1 this paper, we show that every n-vertex d-regular graph of girth at least five contains an independent dominating set of size at most n(logdd+c) for some constant c. Apart from the value of the constant c, this result is best possible in view of known results on independent dominating sets in random d-regular graphs. In addition, the condition that the graph has girth at least five is necessary since the smallest independent dominating set in an n-vertex graph whose components are all isomorphic to Kd,d is exactly n2 .
1
Introduction
An early result using the probabilistic method is that every n-vertex graph of minimum . This result is due degree at least d contains a dominating set of size at most n(1+log(d+1)) d+1 independently to Lov´ asz [6], Payan [9] and Arnautov [2]. In this paper, we prove the following: Theorem 1. Let n, d be positive integers such that n > 10d2 log2 d. Then there exists an absolute constant c such that any n-vertex d-regular graph of girth at least five contains an independent dominating set of size at most n(logdd+c) . We make no attempt to optimize the constant c here. This theorem is best possible in view of random d-regular graphs: asymptotically almost surely, all independent dominating sets in ∗
Department of Mathematics and Statistics, McGill University Department of Mathematics, University of California at San Diego. Email: [email protected] ‡ Department of Mathematics, University of California at San Diego, 9500 Gilman Drive, La Jolla, California 92093-0112, USA; Research supported by an Alfred P. Sloan Research Fellowship and an NSF Grant DMS 0800704. Email: [email protected] †
1
a random d-regular graph on n vertices have size at least n(logdd+b) for some constant b – more precise results on independent dominating sets in random d-regular graphs for fixed values of d can be found in Duckworth and Wormald [4]. In fact, looking at the proof of Theorem 1, our result generalizes the results of Duckworth and Wormald for large d, since a random d-regular graph has a constant number of cycles of length at most four with high probability. n Also, the girth five requirement is necessary: a graph consisting of 2d disjoint copies of the complete bipartite graph Kd,d , when 2d divides n, has no independent dominating set of size less than n2 . Regularity is also a necessary requirement, as we shall give examples of graphs where the degrees vary by a constant factor and where the smallest independent dominating d for some absolute constant δ > 0. An interesting open question is to set has size (1 + δ) n log d determine whether the theorem holds for d of order n1/2 – for example it is easy to see that the theorem holds in the Erd˝ os-R´enyi polarity graphs [5], which are maximal graph without cycles of length four and in the bipartite incidence graphs of projective planes on n points – these are d-regular n by n bipartite graphs where d(d − 1) = n − 1. We believe the following is true: Conjecture 1. Let G be a d-regular graph on n vertices containing no cycle of length four. Then for some constant a, G contains an independent dominating set of size at most n(logdd+a) .
1.1
Notation and Terminology
If G is a graph then for a set S ⊂ V (G) let ∂S denote the set of vertices not in S which are adjacent to at least one vertex in S. For v ∈ V (G)\S, the S-degree of v is the number of neighbours of v in S. We write u ↔ v to denote that vertices u and v are adjacent.
1.2
Organization
The rest of the paper is organized as follows: In section 2, we introduce the process which we use to complete the proof. In section 3, we prove the lemmas necessary to control the process. The proof of Theorem 1 is given in section 4. Throughout the paper, several standard concentration inequalities and probabilistic tools are used. These are stated precisely in an appendix; Proposition 1 is the Chernoff bounds, Propositions 2 and 3 are variants of Hoeffding’s inequality, the later including an exceptional probability. Proposition 4 is the Lov´ asz Local Lemma.
2
The Process
For an n-vertex d-regular graph G = G0 of girth at least five, we run the following process. Let S0 = C0 = ∅ and X0 = V (G) and d0 = d. Suppose that graph Gt is defined, where Q Q V (Gt ) = St ∪∂St ∪Xt ∪Ct . In addition, define real numbers dt = d ti=1 qi and nt = n ti=1 qi ,
2
where 1/2
t := 210 dt qt := e
−1/e
log3/2 dt t−1 , 1− dt−1
(1) (2)
and 0 = 0. At time t, we randomly and independently select vertices from Xt with probability 1 dt . Let St+1 be the set of selected vertices of Xt which have no selected neighbours. Define Ct+1 := Ct ∪ Qt+1 \(∂St+1 ∩ Ct )
(3)
Xt+1 := Xt \(Qt+1 ∪ St+1 ∪ ∂St+1 )
(4)
where a vertex v ∈ Xt \(St+1 ∪ ∂St+1 ) is placed in Qt+1 with probability qt+1 (v) defined so that P(v 6∈ ∂St+1 )(1 − qt+1 (v)) = qt+1 . (5) Roughly speaking, one can think of the process as follows. The vertices in Xt are those from which we build our independent dominating set. The set St+1 consists of the vertices we add to our independent dominating set at time t. After the vertices are chosen, we remove them and their neighbours from Xt . The role of Ct+1 is largely technical, one can think of it as a trashcan that allows us to normalize the probabilities at each step. The set Xt+1 consists of those vertices that are not in, or adjacent to, our current independent dominating set and not in our trashcan Ct+1 . Let dt (v) denote the number of neighbours of a vertex v ∈ V (Gt ) that are in Xt , and γt (v) denote the number of neighbours of a vertex v ∈ V (Gt ) that are in Ct . We shall use martingale concentration inequalities and the Lov´asz Local Lemma [7] to show that with positive probability, all of the following events occur provided dt is sufficiently large: |dt+1 (v) − dt+1 | ≤ t+1
∀ v ∈ Xt+1 ∪ Ct+1
γt+1 (v) ≤ 100t+1
∀ v ∈ Xt+1 ∪ Ct+1
t+1 nt+1 dt+1 ) ( n t+1 nt+1 nt+1 |St+1 − | ≤ 3 max ,p . ed d2t+1 dt+1 d |Ct+1 | ≤ 200
|Xt+1 − nt+1 | ≤ 20
nt+1 . dt+1
(6) (7) (8)
(9) (10)
assuming these hold for t. Throughout we will assume dt > K where K is a constant. We will 200 see that K = ee suffices, and we shall not optimize the value of K. For technical reasons, we have to stop the process at a particular time T = e log d − a for some constant a, as the concentration inequalities are no longer applicable when the degrees of vertices of Gt become 3
too small. The desired independent dominating set will consist of the union S of all sets St : t ≤ T together with any maximal independent set in the graph obtained by removing S ∪ ∂S.
3
Analysis of degrees and sets
In this section we turn to ensuring that (6)–(10) hold with high probability at time t + 1 assuming that they hold at time t. This combined with the local lemma will allow us to complete the inductive step of our proof.
3.1
Survival probability
Suppose we are given the graph Gt and the sets Xt and Ct and St+1 . We say that a vertex survives if it is not in ∂St+1 . The following lemma shows that the survival probability of each vertex v ∈ (Ct ∪ Xt )\St+1 is very close to e−1/e , and always at least qt . In particular, qt (v) is well-defined by (5). Lemma 1. Let v ∈ (Xt ∪ Ct )\St+1 . Then for dt > K, t t 1− ≤ e1/e P(v survives) ≤ 1 + . dt dt Proof. We write u ↔ v to mean u and v are adjacent in Gt . First note that Y 1 Y 1 P(v survives) = . 1− 1− dt w↔u dt u↔v u∈St
w∈St
By (6), there are at least dt + t terms in the first product and at most dt − t in the second product. Suppose that dt is large enough to satisfy t < dt . Then, using log(1 − x) ≤ −x for 0 < x < 1, (1 − x)y ≥ 1 − xy for y > 1 and (1 − x1 )x−1 ≥ e−1 for x > 1, 1 1 dt +t P(v survives) ≤ exp (dt − t ) · log 1 − 1− dt dt t 1 dt +t ≤ exp − 1 − · 1− dt dt 1 t 1 t +1 ≤ exp − · 1 − · 1− e dt dt 1 t t + 1 ≤ exp − · 1 − · 1− e dt dt 1 2 + 1 t ≤ exp − · exp . e edt It is straightforward to see that since dt > K, we easily have 2 + 1 t t exp K, then
√
qt+1 t ≤ e1/(4e) t+1 .
Proof. Note that: √
p qt+1 t = 2 qt+1 dt log3/2 dt = t+1 10
log dt log dt+1
3/2
= t+1
log dt log dt + log qt+1
3/2 .
and the result follows from the definition of qt and dt , certainly for dt > K.
Lemma 3. Let v ∈ Xt+1 ∪ Ct+1 and dt > K. Then P(|dt+1 (v) − dt+1 | > t+1 ) ≤ d−100 . t Proof. We begin by estimating Edt+1 (v). Using the definitions of qt and t , and Lemma 3, we have |Edt+1 (v) − dt+1 | = |dt (v)qt+1 − dt+1 | ≤ qt+1 t ≤ e−1/4e t+1 . (11) for dt > K. We wish to obtain concentration on dt+1 (v). If our girth were larger, then dt+1 would actually be the sum of independent indicators, and we could apply the Chernoff bounds, but with girth 5 the indicators that various vertices are in Xt+1 . Instead we must use martingale inequalities to concentrate dt+1 . An additional complication is that the natural exposure martingale is not actually Lipschitz, so we in fact must use a variety of Hoeffding’s inequality that allows an exceptional probability. This concentration inequality is recorded in the appendix as Proposition 3. The following argument is repeated several times in the following lemmas, so the interested reader should pay particularly close attention to the setup here to make future reading easier. In our process we make two choices independently for each vertex, whether or not to initially select the vertex and whether or not to place it in Qt+1 . For convenience, we make both of these choices for each vertex without regard for the other (that is, we flip the coin to see whether a vertex would be in Qt+1 regardless of whether it actually is placed there). Let Iu be the indicator that u is selected at time t, that it is one of the vertices initially chosen with probability d1t , and let Ju be the indicator that u would be placed in Qt+1 . 5
Note that this choice is made independently of Iu . Also note that Iu does not indicate whether u ∈ St+1 , as St+1 consists of selected vertices where no neighbours are selected. We now set up our filtration for an exposure martingale. Fix an ordering of the Xt -neighbours of v, namely u1 , u2 , . . . , udt (v) . Let Ai be the set of all random variables Iw , Jw such that w 6= ui is at distance at most two from ui in Gt \{v}. Define σ-fields Fi and random variables Yj : j = 1, 2, . . . , i as follows Fi = σ(Au1 , Au2 . . . , Aui )
and
Yj = E[dt+1 (v)|Fj ]
Then Y = (Yj )j≥1 is an exposure martingale [1], with Y0 = Edt+1 (v) and Ydt (v) = dt+1 (v). Furthermore, the only way that |Yi − Yi+1 | ≥ log dt is if some vertex at distance at most two from ui+1 in Gt − {v} has at least log dt chosen neighbors. Then X µ = P(|Yi − Yi+1 | ≥ log dt ) dt + t 1 log dt ≤ (dt + t )3 log dt dt 3 log d t (dt + t ) (dt + t ) < log dt dt blog dt c! = (dt + t )3 ·
(1 +
t log dt dt )
blog dt c!
K. Thus Y is (log dt )-Lipschitz with exceptional (2dt )20 probability at most µ ≤ blog dt c! . We will now apply Proposition 3 with λ = t+1 , ν = dt + t and ρ = 6. Note that 0 ≤ dt+1 (v) ≤ dt + t = ν and, as dt > K, we have νρµ < t+1 . Let δ = t+1 − e−1/e t . Then by definition of t , δ 2 > 20d2t log3 dt . By (11) and Proposition 3, (2dt )20 δ2 + blog dt c! 3dt log2 dt 400d log3 d (2dt )20 t t + ≤ d−100 , < 2 exp − t 2 blog d c! 3dt log dt t
P(|dt+1 (v) − dt+1 | > t+1 ) ≤ 2 exp −
for dt > K. This is a point at which the double exponential nature of K arises: so that 220 d121 < blog dt c!. t
3.3
Concentration of C-degrees
To control the size of the set Ct , we need control of the Ct -degrees at each step. In particular, we wish to show that the Ct -degree of a vertex never gets too large. We start with C0 = ∅. Recall, denote the Ct -degree of a vertex v by γt (v). For all vertices, γ0 (v) = γ1 (v) = 0, as C0 = ∅ and no normalization of probabilities is required at the first step. In the next lemma, we deal with the degrees γt (v) for t ≥ 2.
6
Lemma 4. Let v ∈ Xt+1 ∪ Ct+1 and suppose γt (u) < 16t for u ∈ Xt−1 \St . Then for dt > K, P(γt+1 (v) < 100t+1 ) ≤ d−100 . t Proof. Let γt+1 (v, Q) and γt+1 (v, C) respectively denote the Qt+1 -degree and the Ct+1 \Qt+1 t degree of v, so that γt+1 (v) = γt+1 (v, Q) + γt+1 (v, C). Since 0 ≤ qt (u) ≤ 2 dt for u ∈ Xt , 2t dt (v) dt 2t (dt + t ) ≤ dt 22 ≤ 2t + t = 2t + 221 log3 dt . dt
Eγt+1 (v, Q) ≤
By the Chernoff Bound, Proposition 1 in the appendix, since dt > K, 1 P(γt+1 (v, Q) > 3t ) ≤ d−100 2 t Also, by definition of t we have for dt > K that 3t < 4t+1 , so 1 . P(γt+1 (v, Q) > 4t+1 ) ≤ d−100 2 t From Lemma 1, for each u ∈ Ct , 1−
t t ≤ e1/e P(u ∈ ∂St+1 ) ≤ 1 + . dt dt
Therefore, using the assumption γt (v) < 8t , we obtain
|Eγt+1 (v, C) − qt γt (v)| ≤
2t γt (v) = 221 log3 dt . dt
To show concentration of γt+1 (v, C), we use the identical exposure martingale as in the proof of Lemma 3 above, and then apply Proposition 3. Thus we obtain 1 P(|γt+1 (v, C) − qt γt (v)| > t+1 ) ≤ d−100 . 2 t for dt > K. Note that this implies that with probability at most 12 d−100 that t √ γt+1 (v, C) ≤ qt γt (v) + t+1 ≤ 100 qt e1/4e t+1 + t+1 < 96t+1 . where we apply Lemma 2 using our assumption dt > K. Combining with our bound on γt+1 (v, Q), and applying the triangle inequality we have that P(γt+1 (v) > 100t+1 ) < d−100 . t This proves the lemma.
7
3.4
Concentration of sizes of sets
To complete the proof, we need not only control over the degrees of vertices at each step, but the evolving size of our sets Xt , Ct , and St . Here we suppose that at time t, our conditions (6)–(10) are satisfied; then we show that |St+1 | and |Xt+1 | are highly concentrated, while providing an upper bound on the size of Ct+1 . In order to do this, we must break into two cases depending on the relationship between dt and nt . If dt is large compared with nt (say 1/20 dt > nt ) then we need to use a slightly different method to achieve tight concentration. Lemmas 5 and 6 address the case where dt is small, and large respectively. Lemma 5. Suppose (6) − (10) hold at time t. Then 2n nt+1 t P |Xt+1 | − nt+1 > 20 ≤ 2 exp 4 dt+1 dt n 3t+1 nt+1 nt log3 dt P |St+1 | − > ≤ 2 exp − ed d2t+1 d5t
(12) (13)
so long as dt is sufficiently large. Proof. We begin by noting that P(v ∈ St+1 ) = ≤ ≤ ≤
1 1 dt (v) 1− dt dt 1 dt −t 1 1− dt dt t 1 exp −1 + dt dt t 1 + , edt d2t
with a lower bound following similarly, giving 1 ≤ t . P(v ∈ St+1 ) − edt d2t
(14)
First we show concentration of |Xt+1 | = |Xt | − |Xt ∩ (St ∪ Qt+1 ∪ ∂St+1 )|. Note that for a
8
vertex v ∈ Xt , |P(v 6∈ Xt+1 ) − (1 − qt+1 )| = |P(v ∈ St+1 ∪ ∂St+1 ∪ ∂Qt+1 ) − (1 − qt+1 )| = |P(v ∈ St+1 ) + P(v ∈ (∂St+1 ∪ ∂Qt+1 ∩ S¯t+1 )) − (1 + qt+1 )| P(v ∈ (∂S ∪ ∂Q |v ∈ 6 S ) t+1 t+1 t+1 − (1 + qt+1 ) = P(v ∈ St+1 ) + P(v 6∈ St+1 ) 1 − qt = P(v ∈ St+1 ) + − (1 + qt+1 ) P(v 6∈ St+1 ) ≤ (1 + qt )P(v ∈ St+1 ) + ≤
qt P(v ∈ St+1 ) 1 − P(v 6∈ St+1 )
6 . edt
where the last inequality follows from (14) assuming dt > K and noting that qt ≤ 1. This gives us that E|Xt+1 | − nt+1 ≤ E|Xt+1 | − qt+1 |Xt | + qt+1 |Xt | − qt+1 nt nt 6|Xt | + 20qt+1 ≤ edt dt nt+1 5nt+1 + 20e−1/e ≤ dt+1 dt+1 nt+1 ≤ 19 . dt+1
We set up a martingale in the following way: fix an ordering of the vertices in Xt , namely v1 , v2 , . . . , vnt . Let Ij denote the indicator that vj is selected, and let Jj denote the indicator that vj is chosen to be in Q (as in the proof of lemma 3 these are independent) and let t Fj = σ(I1 , J1 , I2 , J2 , . . . , Ij , Jj ) and Zj = E[|Xt+1 ||Fj ]. Then Z = (Zj )nj=0 is a martingale, with Z0 = E|St+1 | and Znt = |St+1 |. Noting that knowing Ij and Jj only tells us information within the neighborhood of vertex vj , this martingale is (dt + )-Lipschitz, which, for convenience, we √ note implies that it is 2dt -Lipschitz assuming dt > K. By Hoeffding’s Inequality, Proposition t+1 2, with λ = ndt+1 = ndtt nt+1 P |Xt+1 | − nt+1 > 20 ≤ P |Xt+1 | − E|Xt+1 | > λ dt+1 2λ2 ≤ exp − 2nt d2t nt ≤ exp − 4 . dt 9
We now show that |St+1 | is concentrated, the proof being quite similar. By (14) n |Xt | |Xt | nt + − E|St+1 | − ≤ E|St+1 | − ed edt edt edt t nt ≤ 2 2 dt t+1 nt+1 ≤ 2 d2t+1 since dt > K and noting that Lemma 2.
nt+1 dt+1
=
nt dt
=
n d;
note that the last inequality follows using
We set up a martingale as before: fix an ordering of the vertices in Xt , namely v1 , v2 , . . . , vnt . Let Ij denote the indicator that vj is selected and let Fj = σ(I1 , I2 , . . . , Ij ) and Wj = t E[|St+1 ||Fj ]. Then W = (Zj )nj=0 is a martingale, with W0 = E|St+1 | and Znt = |St+1 |. √ nt+1 Again W is 2dt -Lipschitz, so by Proposition 2, with λ = dt n2 t ≤ t+1 , d2 t
t+1
t+1 nt+1 nt+1 P |St+1 | − ≥3 ≤ P(||St+1 | − E|St+1 || ≥ λ) edt+1 d2t+1 2λ2 ≤ exp − 2nt d2t 2 n2 ≤ exp − t t6 nt dt 220 n log3 d t t ≤ exp − . d5t This completes the proof of concentration of |St+1 |.
If dt is large, for example dt ≥ n1/20 , then we can improve the above lemma as follows. 1/20
, n > 10d2 log2 d and (6)−(10) hold at time t, Then for dt > K, nt+1 P |Xt+1 | − nt+1 > 20 ≤ 2e−5/2 (15) dt+1 ( ) nt t+1 nt+1 nt+1 p P |St+1 | − > 3 max , ≤ 2e−5/2 . (16) edt dt+1 dt+1 d
Lemma 6. Suppose dt ≥ nt
Proof. We begin by showing the concentration for |Xt+1 |, using the same martingales as in the proof of 5. We note that exposing a vertex in our martingale changes the value by more than log dt only if at least log dt neighbors of v are chosen. Thus the set of bad outcomes where such an event occurs has probability at most (following calculations in Lemma 3), noting nt < d20 t , dt + t 1 (2dt )40 nt < ≤ d−100 , t log dt dt blog dt c! 10
since dt > K. Here again we require K to be double exponential. The martingale Z in the proof of Lemma 5 is thus log dt -Lipschitz with exceptional probability µ = d−100 . We now choose λ = ndtt and it is easy to verify since |Xt+1 | ≤ |Xt | that the conditions t t+1 . of Lemma 3 are satisfied with νρµ < λ when ν = nt + 20 ndtt , ρ = 1000 and λ = 20 ndt+1
nt+1 P |Xt+1 | − nt+1 ≥ 20 dt+1
≤ P( |Xt+1 | − E|Xt+1 | ≥ λ) n2t+1 ≤ 2 exp − + d100 ≤ 2e−5/2 . t nt d2t+1 log(dt )2
for dt > K. Concentration for |St+1 | in this setting is similar. The martingale W in the proof of Lemma 5 is log dt -Lipschitz with exceptional probability at most d−100 , with the same bound in the t conditions on Proposition 3 as above. By Proposition 3, taking λ = max
t+1 nt+1 , d2t+1
√nt+1 ddt+1
and , ν = nt + 20 ndtt and µ = d−100 t
ρ = 1000 −n2t+1
+ d−100 . t 2dt+1 dnt log2 dt n ≤ 2 exp − 2 + d−100 ≤ 2e−5/2 t 2 3d log dt
P (||St+1 | − E|St+1 || ≥ λ) ≤ exp
completing the second inequality, using the bound on E|St+1 | obtained in Lemma 5.
Lemma 7. Suppose that (6) − (10) hold at time t. Then for dt > K, t+1 nt+1 nt P |Ct+1 | > 200 ≤ 2 exp − 6 2 . dt+1 10 dt
(17)
Proof. Here we show concentration of |Ct+1 | = |Ct | − |Ct ∩ ∂St+1 | + |Qt+1 |. Using the upper and lower bounds for survival probability in Lemma 1, we have that qt+1 (v) ≤ 1 −
1− 1+
t dt t dt
≤3
t dt
so by (10) we have that 4t nt . dt By the Chernoff bounds, Proposition 1, since vertices are independently chosen to be in Qt+1 , n 16t+1 nt+1 8t nt t t P |Qt+1 | > ≤ P |Qt+1 | > ≤ exp − . dt+1 dt 2dt E|Qt+1 | ≤
11
nt+1 ≥ Here the first inequality comes from the fact that 2 t+1 dt+1
t nt dt
for dt > K.
By the upper bound for survival probability in Lemma 1, we have that t t+1 nt+1 −1/e 1+ |Ct | − E|Ct ∩ ∂St+1 | ≤ e |Ct | ≤ 200e−1/(2e) , dt dt+1 with the last inequalities holding for dt > K and using the condition (8). To get concentration for |Ct ∩ ∂St+1 | we apply a martingale argument, similar to those above. We use Hoeffding’s inequality, Proposition 2, for concentration. Fix an ordering on the vertices in ∂Ct ∩ Xt , namely v1 , v2 , . . . , vr . We denote by Ij the indicator that vj ∈ St+1 and let Fj = σ(I1 , I2 , . . . , Ij ) denote the σ-field generated by I1 , I2 , . . . , Ij . Let W = |Ct ∩ ∂St+1 |, and let (Wi )ri=0 = E[W |Fi ]. Then W is an exposure martingale. Let ci = |∂{vi } ∩ Ct |. Then P ci ≤ (dt + t )|Ct | since dt (u) ≤ (dt + t ) for u ∈ Ct . Furthermore, each ci ≤ 100t as (7) nt+1 holds at time t. By Proposition 2, setting λ = t+1 dt+1 , and noting that |Ct | > 200λ, we have λ2 t λ2 −1/e ≤ exp − P W ≥e 1+ |Ct | + λ ≤ exp − Pr dt 200t (dt + t )|Ct | 2 i=1 c2i λ2 ≤ exp − 400t dt |Ct | nt ≤ exp − 6 2 10 dt nt+1 Finally, we note that if |Ct+1 | ≥ 200 t+1 dt+1 , then either
|Qt+1 | > 16
t+1 nt+1 dt+1
or
|Ct | − |Ct ∩ ∂S| ≥ (1 + 200e−1/2e )
t+1 nt+1 . dt+1
Therefore P(|Ct+1 | > 200
n nt t+1 nt+1 nt t t ) < exp − + exp − 6 2 ≤ 2 exp − 6 2 . dt+1 2dt 10 dt 10 dt
This completes the proof.
4
Proof of Main Theorem
Proof. The spirit of our proof is inductive: we run the process described above. At each step, we insure that (with high probability) individual vertices are good – a vertex v is good if it satisfies (6), and bad otherwise. We note that the change of degrees of vertices are only dependent if the vertices are within distance five of each other. This gives us a limited dependence, allowing us to use the Lov´asz Local Lemma (Proposition 4) to say that, with positive probability, no vertex is bad. That is, there exists a choice of St+1 and Qt+1 such
12
that the resulting graph at time t + 1 satisfies all our conditions. We shall stop our process before dt < K. Recall the conditions which we wish to hold: |dt+1 (v) − dt+1 | ≤ t+1
∀ v ∈ Xt+1 ∪ Ct+1
γt+1 (v) ≤ 100t+1
∀ v ∈ Xt+1 ∪ Ct+1
t+1 nt+1 dt+1 ( ) n t+1 nt+1 nt+1 |St+1 − | ≤ 3 max ,p . ed d2t+1 dt+1 d |Ct+1 | ≤ 200
|Xt+1 − nt+1 | ≤ 20
nt+1 . dt+1
(18) (19) (20)
(21) (22)
Clearly at time t = 0, the conditions (18)–(22) are satisfied. Let us assume that they are satisfied at time t. Let Av be the event that either (18) or (19) fails at vertex v at time t + 1. Note that as the conditions are satisfied at time t, the combined C and X degrees of all vertices are less than 4dt , and hence Av is mutually dependent with at most (4dt )5 other events Au . By the Lov´ asz Local Lemma, Proposition 4, since P(Av ) ≤ 2d−100 , t 5
P(Ai ) ≤ d−50 (1 − d−50 )(4dt ) t t
and there are, by induction and the sizes guaranteed by (20) and (22), at most 2nt vertices combined in X and C over which we want control, all degrees are good with probability at least 2nt 2nt ≥ exp − (1 − d−50 ) t d50 t +1 all degrees are good at time t + 1. Case 1. If dt ≤ n1/20 , Lemmas 5 and 7 implies |St+1 |, |Xt+1 | and |Ct+1 | are all within the required bounds with probability at least 1 − 3 exp(−nt /d5t ) provided dt > K. Since
2nt nt > 3 exp(− 5 ) exp − 50 dt + 1 dt
for dt > K, conditions (18) – (22) hold with positive probability at time t + 1. Case 2. In the event that dt > n1/20 note that Lemmas 6 and 7 and our condition that n > 5d2t log2 dt imply that the sizes of C and X and number of vertices picked are all correct with probability 1 − 5e−5/2 (once again using the weakest of our concentration bound). In 13
1 this case, note that exp(−2nt /(d50 t + 1)) > e2 and again we have that conditions (18) – (22) hold with positive probability at time t + 1. Therefore, there is some choice of St+1 and Ct+1 such that all of conditions (18) – (22) are satisfied at time t + 1. We condition on such a good occurrence and continue.
Let T = min{t : dt < K}. We stop the process at time T ; we’ve already verified that we can run the process to this point. Note that T = e log d − a for some constant a. We now take S −1 St , plus a maximal independent set in the graph our independent dominating set to be Tt=0 on XT ∪ CT . We verify that this set has the claimed size. Note that since |St | = (1 + o(1)) nd , S d it is clear that |St | ≤ (1 + o(1)) n log but we wish to verify the claimed error term. First d observe that: T −1 −1 TX X t nt nt n 3 max ,√ |St | − T ≤ ed d2t dt d t=0 t=0 ! T −1 √ T −1 n X t X dt √ ≤ 3 + d d d t=0 t=0 t n = O , d P noting that both sums are bounded by geometric sums of the form t e−t/4e by the definition of dt and t and hence converge. Next, note that since dT > K, the fact that (20) holds imply that |CT | is O( nd ), and the fact that ndt nt = d n and (22) implies that |XT | is O( d ) so the maximal independent set in both of these has size O( nd ). Therefore we have found the desired independent dominating set.
5
Concluding remarks
• In this paper, we showed that if G is a d-regular graph without 4-cycles, then so long as n > 5d2 log2 d, then G contains an independent dominating set of size n(logdd+c) for some constant c. While the constant 5 is not optimal (a more careful reading of the proof will likely allow one to improve this), the order of the lower bound on n as a function of d seems to be harder to improve. The real problem comes from concentration of |St+1 | in Lemma 6. When the process starts here, concentration around nd is as good as we can get; if n = o(d2 log2 d), we would have to concentrate around something larger order than nd , already ruining the term log d of order nd . If we are satisfied with an independent dominating set of size n(1+o(1)) , then d n it is enough to assume that d2 log d → ∞. • In the introduction it was mentioned that an n-vertex graph of minimum degree at least d . However, it may not be possible to find an has a dominating set of size at most n(1+log(d+1)) d+1 14
log d independent dominating set of size n(1+o(1)) , so the condition of regularity in Theorem 1 d is necessary. To see this, we take a random graph Gn,2d/n where d/ log n → ∞ on a set of n2 vertices, and then take an independent set of size n2 and join each of its vertices to d randomly selected vertices in Gn,2d/n . If the resulting graph is G, then with high probability G has minimum degree d and maximum degree roughly 3d. The expected number of cycles of length 4 at most four on each vertex is O( dn ), so with high probability, if d = o(n1/3 ), we can delete some edges without affecting the degrees of the vertices by much. Then, any independent set on the random graph side must be fairly small, and leaves most of the independent set side uncovered. An interesting question is whether an analogue of this theorem does occur in random graphs with some degree sequence, or alternately graphs with a good spectral gap (for example, spectral gap of the normalized Laplacian).
• At its heart our method is a variant of methods like the R¨odl nibble, and the differential equation method of Wormald. In a follow-up paper, we will explore a similar process on hypergraphs, looking at matching and covering problems. • If G is an n by n bipartite graph which is d-regular and has no 4-cycles, then a simpler proof than that of the main theorem shows that we can find an independent dominating set of size at most 2n(logd d+c) for some constant c. To see this, we can avoid using the set C and directly pick vertices from each side: we are able to pick vertices in the top part with probability log d−3 log log d – much higher than in our process in the proof of the main theorem – and show d that the degrees of vertices at the top are then highly concentrated at log3 d. Notice that the surviving vertices at the bottom all have degree d. Then we apply the local lemma as in the proof of our main theorem to conclude that all the vertices have degree close to log3 d at the top. Repeating this at the bottom, we choose vertices at the bottom with probability log d−3 log log d . After applying concentration inequalities and the local lemma, we see that we log3 d d have picked close to 2n log vertices and all the vertices at the top and at the bottom have d 3 degree close to log d. Essentially, in one step we have reduced the problem to a bipartite graph which is close to (log3 d)-regular and where each part has close to n logdlog d vertices. Now we repeat this whole procedure roughly log∗ d times to get the required independent dominating set.
6
Appendix : Concentration Inequalities
In this section, we present the inequalities which we will need from probability theory involving concentration of measure. All of these inequalities deal with upper bounds for expressions of the form P(|X − EX| > λ) where X is a random variable and λ is a real number. The most basic inequality of this type for binomial distributions is the Chernoff Bound [1]:
15
Proposition 1. (Chernoff Bound) Let Xi , i = 1, 2, . . . , n be independent random variables P with Bernouilli distributions with means pi , i ∈ [n], and let p = n−1 pi and h ≥ 0. Then h2 n P(Sn > (p + h)n) ≤ exp − 2a h2 n P(Sn < (p − h)n) ≤ exp − , 2b where Sn = X1 + X2 + · · · + Xn and a is the maximum of α(1 − α) for p ≤ α ≤ p + h and b is the maximum of β(1 − β) for p − h ≤ β ≤ p. In particular, if X has binomial distribution with probability p and mean pn, then for any ε < 1, ε2 pn . P(|X − pn| > εpn) < 2 exp − 2 Most of the inequalities we will need concern martingales. One of the most fundamental martingale inequalities is Hoeffding’s Inequality. Further refinements and generalizations of this inequality may be found in McDiarmid [8]. In the following two propositions, c = (c1 , c2 , . . . , cn ) P is a vector of positive real numbers and kck2 denotes ni=1 c2i . Proposition 2. (Hoeffding’s Inequality) Let (Xi )ni=1 be a martingale with difference sequence (Yi )ni=1 , where −ai ≤ Yi ≤ −ai + ci for some function ai on (Ω, Fi−1 ). Then for λ ≥ 0 2λ2 2λ2 P(Xn > E(Xn ) + λ) ≤ exp − and P(X < E(X ) − λ) ≤ exp − . n n kck2 kck2 The following martingale concentration inequality, similar to the version derived by Chung and Lu [3], is used frequently and is a minor modification of Hoeffding’s Inequality. For completeness, we give a complete proof. Proposition 3. Let (Xi )ni=1 be a martingale with difference sequence (Yi )ni=1 , and suppose n X
P(|Yi | > ci ) ≤ µ
and
a < Xn ≤ a + ν
i=1
for some a, ν, µ ∈ R. Then for any positive real numbers λ and ρ > 1 satisfying λ > νρµ, (ρ − 1)2 λ2 + µ. P(|Xn − EXn | > λ) ≤ 2 exp − 2ρ2 kck2 Proof. Let T = min{i : |Yi | > ci } – by convention, the minimum over the empty set is infinite, in which case we write T = ∞. Let T ∧i = min{i, T }. Then it is well-known (see Williams [10]) that (XT ∧i )ni=1 is a martingale. Let (Zi )ni=1 be the difference sequence of that martingale. By definition of T , |Zi | ≤ ci . By Proposition 2, with X = XT ∧n , (ρ − 1)λ (ρ − 1)2 λ2 P(|X − E(X)| > ) ≤ 2 exp − . ρ 2ρ2 kck2
16
Finally we note that P(X 6= Xn ) ≤ µ and, since a ≤ Xn ≤ a + ν, |E(X) − E(Xn )| ≤ νµ < λ/ρ. This completes the proof as P(|Xn − E(Xn )| > λ) ≤ P(|X − E(X)| > (ρ − 1)λ/ρ) + P(X 6= Xn ) (ρ − 1)2 λ2 + µ. ≤ 2 exp − 2ρ2 kck2 We call a martingale (Xi )ni=1 satisfying the conditions of Proposition 3 with ci = c ∈ R a c-Lipschitz martingale with exceptional probability µ. The final probabilistic result we need, the Lov´asz Local Lemma [7], states that if there is limited local dependence between a set of rare events, then there is a positive probability that none of the events occurs. Proposition 4. (Lov´ asz Local Lemma) Let A1 , A2 , . . . , An be events in some probability space and suppose that for some set Ji ⊂ [n], Ai is mutually independent of {Aj : j 6∈ Ji ∪ {i}}. If Q there exist real numbers γi ∈ [0, 1) such that P(Ai ) ≤ γi j∈Ji (1 − γj ), then P
n \ i=1
n Y Ai ≥ (1 − γi ) > 0. i=1
References [1] N. Alon, and J. Spencer. The probabilistic method, John Wiley and Sons Inc., New York, 2008. [2] V.I. Arnautov, Estimation of the exterior stability number of a graph by means of the minimal degree of the vertices (in Russian), Prikl. Mat. i Programmirovanie 11 (1974), 3–8. [3] F. Chung and L. Lu, Coupling on-line and off-line analyses for random power law graphs. Internet Math. 1 (2004), no. 4, 409–461. [4] W. Duckworth and N. Wormald, On the independent domination number of random regular graphs, combinatorics, Probability and Computing, 15 (2006) 513–522. [5] P. Erd˝ os, and A. R´enyi, On a problem in te theory of graphs, Publ. Math. Inst. Hungar. Acad Sci. 7A (1962), 623–641. [6] L. Lov´ asz, On the ratio of optimal and integral fractional covers, Disc. Math 13 (1975), 383–390.
17
[7] P. Erd˝ os and L. Lov´ asz. Problems and results and results on 3-chromatic hypergraphs and some related questions, in: ”Infinite and Finite Sets”, (A. Hajnal et. al. eds), Colloq. Math. Soc. J. Bolyai 11, (1975) 609–627. [8] C. McDiarmid, Concentration, Probabilistic methods for algorithmic discrete mathematics, 195–248, Algorithms Combin., 16 Springer, Berlin. [9] C. Payan, Sur le nombre d’absorption d’un graphe simple (in French) Proc. Colloq. Th. Graphes (Paris 1974), C.C.E.R.O 17 (1975), 307–317. [10] D. Williams, Probability with Martingales. Cambridge Mathematical Textbooks. Cambridge University Press, Cambridge, 1991. xvi+251 pp.
18
Paul Horn†
Jacques Verstraete‡
March 9, 2009
Abstract One of the first results in probabilistic combinatorics is that every n-vertex graph of . In minimum degree at least d contains a dominating set of size at most n(1+log(d+1)) d+1 this paper, we show that every n-vertex d-regular graph of girth at least five contains an independent dominating set of size at most n(logdd+c) for some constant c. Apart from the value of the constant c, this result is best possible in view of known results on independent dominating sets in random d-regular graphs. In addition, the condition that the graph has girth at least five is necessary since the smallest independent dominating set in an n-vertex graph whose components are all isomorphic to Kd,d is exactly n2 .
1
Introduction
An early result using the probabilistic method is that every n-vertex graph of minimum . This result is due degree at least d contains a dominating set of size at most n(1+log(d+1)) d+1 independently to Lov´ asz [6], Payan [9] and Arnautov [2]. In this paper, we prove the following: Theorem 1. Let n, d be positive integers such that n > 10d2 log2 d. Then there exists an absolute constant c such that any n-vertex d-regular graph of girth at least five contains an independent dominating set of size at most n(logdd+c) . We make no attempt to optimize the constant c here. This theorem is best possible in view of random d-regular graphs: asymptotically almost surely, all independent dominating sets in ∗
Department of Mathematics and Statistics, McGill University Department of Mathematics, University of California at San Diego. Email: [email protected] ‡ Department of Mathematics, University of California at San Diego, 9500 Gilman Drive, La Jolla, California 92093-0112, USA; Research supported by an Alfred P. Sloan Research Fellowship and an NSF Grant DMS 0800704. Email: [email protected] †
1
a random d-regular graph on n vertices have size at least n(logdd+b) for some constant b – more precise results on independent dominating sets in random d-regular graphs for fixed values of d can be found in Duckworth and Wormald [4]. In fact, looking at the proof of Theorem 1, our result generalizes the results of Duckworth and Wormald for large d, since a random d-regular graph has a constant number of cycles of length at most four with high probability. n Also, the girth five requirement is necessary: a graph consisting of 2d disjoint copies of the complete bipartite graph Kd,d , when 2d divides n, has no independent dominating set of size less than n2 . Regularity is also a necessary requirement, as we shall give examples of graphs where the degrees vary by a constant factor and where the smallest independent dominating d for some absolute constant δ > 0. An interesting open question is to set has size (1 + δ) n log d determine whether the theorem holds for d of order n1/2 – for example it is easy to see that the theorem holds in the Erd˝ os-R´enyi polarity graphs [5], which are maximal graph without cycles of length four and in the bipartite incidence graphs of projective planes on n points – these are d-regular n by n bipartite graphs where d(d − 1) = n − 1. We believe the following is true: Conjecture 1. Let G be a d-regular graph on n vertices containing no cycle of length four. Then for some constant a, G contains an independent dominating set of size at most n(logdd+a) .
1.1
Notation and Terminology
If G is a graph then for a set S ⊂ V (G) let ∂S denote the set of vertices not in S which are adjacent to at least one vertex in S. For v ∈ V (G)\S, the S-degree of v is the number of neighbours of v in S. We write u ↔ v to denote that vertices u and v are adjacent.
1.2
Organization
The rest of the paper is organized as follows: In section 2, we introduce the process which we use to complete the proof. In section 3, we prove the lemmas necessary to control the process. The proof of Theorem 1 is given in section 4. Throughout the paper, several standard concentration inequalities and probabilistic tools are used. These are stated precisely in an appendix; Proposition 1 is the Chernoff bounds, Propositions 2 and 3 are variants of Hoeffding’s inequality, the later including an exceptional probability. Proposition 4 is the Lov´ asz Local Lemma.
2
The Process
For an n-vertex d-regular graph G = G0 of girth at least five, we run the following process. Let S0 = C0 = ∅ and X0 = V (G) and d0 = d. Suppose that graph Gt is defined, where Q Q V (Gt ) = St ∪∂St ∪Xt ∪Ct . In addition, define real numbers dt = d ti=1 qi and nt = n ti=1 qi ,
2
where 1/2
t := 210 dt qt := e
−1/e
log3/2 dt t−1 , 1− dt−1
(1) (2)
and 0 = 0. At time t, we randomly and independently select vertices from Xt with probability 1 dt . Let St+1 be the set of selected vertices of Xt which have no selected neighbours. Define Ct+1 := Ct ∪ Qt+1 \(∂St+1 ∩ Ct )
(3)
Xt+1 := Xt \(Qt+1 ∪ St+1 ∪ ∂St+1 )
(4)
where a vertex v ∈ Xt \(St+1 ∪ ∂St+1 ) is placed in Qt+1 with probability qt+1 (v) defined so that P(v 6∈ ∂St+1 )(1 − qt+1 (v)) = qt+1 . (5) Roughly speaking, one can think of the process as follows. The vertices in Xt are those from which we build our independent dominating set. The set St+1 consists of the vertices we add to our independent dominating set at time t. After the vertices are chosen, we remove them and their neighbours from Xt . The role of Ct+1 is largely technical, one can think of it as a trashcan that allows us to normalize the probabilities at each step. The set Xt+1 consists of those vertices that are not in, or adjacent to, our current independent dominating set and not in our trashcan Ct+1 . Let dt (v) denote the number of neighbours of a vertex v ∈ V (Gt ) that are in Xt , and γt (v) denote the number of neighbours of a vertex v ∈ V (Gt ) that are in Ct . We shall use martingale concentration inequalities and the Lov´asz Local Lemma [7] to show that with positive probability, all of the following events occur provided dt is sufficiently large: |dt+1 (v) − dt+1 | ≤ t+1
∀ v ∈ Xt+1 ∪ Ct+1
γt+1 (v) ≤ 100t+1
∀ v ∈ Xt+1 ∪ Ct+1
t+1 nt+1 dt+1 ) ( n t+1 nt+1 nt+1 |St+1 − | ≤ 3 max ,p . ed d2t+1 dt+1 d |Ct+1 | ≤ 200
|Xt+1 − nt+1 | ≤ 20
nt+1 . dt+1
(6) (7) (8)
(9) (10)
assuming these hold for t. Throughout we will assume dt > K where K is a constant. We will 200 see that K = ee suffices, and we shall not optimize the value of K. For technical reasons, we have to stop the process at a particular time T = e log d − a for some constant a, as the concentration inequalities are no longer applicable when the degrees of vertices of Gt become 3
too small. The desired independent dominating set will consist of the union S of all sets St : t ≤ T together with any maximal independent set in the graph obtained by removing S ∪ ∂S.
3
Analysis of degrees and sets
In this section we turn to ensuring that (6)–(10) hold with high probability at time t + 1 assuming that they hold at time t. This combined with the local lemma will allow us to complete the inductive step of our proof.
3.1
Survival probability
Suppose we are given the graph Gt and the sets Xt and Ct and St+1 . We say that a vertex survives if it is not in ∂St+1 . The following lemma shows that the survival probability of each vertex v ∈ (Ct ∪ Xt )\St+1 is very close to e−1/e , and always at least qt . In particular, qt (v) is well-defined by (5). Lemma 1. Let v ∈ (Xt ∪ Ct )\St+1 . Then for dt > K, t t 1− ≤ e1/e P(v survives) ≤ 1 + . dt dt Proof. We write u ↔ v to mean u and v are adjacent in Gt . First note that Y 1 Y 1 P(v survives) = . 1− 1− dt w↔u dt u↔v u∈St
w∈St
By (6), there are at least dt + t terms in the first product and at most dt − t in the second product. Suppose that dt is large enough to satisfy t < dt . Then, using log(1 − x) ≤ −x for 0 < x < 1, (1 − x)y ≥ 1 − xy for y > 1 and (1 − x1 )x−1 ≥ e−1 for x > 1, 1 1 dt +t P(v survives) ≤ exp (dt − t ) · log 1 − 1− dt dt t 1 dt +t ≤ exp − 1 − · 1− dt dt 1 t 1 t +1 ≤ exp − · 1 − · 1− e dt dt 1 t t + 1 ≤ exp − · 1 − · 1− e dt dt 1 2 + 1 t ≤ exp − · exp . e edt It is straightforward to see that since dt > K, we easily have 2 + 1 t t exp K, then
√
qt+1 t ≤ e1/(4e) t+1 .
Proof. Note that: √
p qt+1 t = 2 qt+1 dt log3/2 dt = t+1 10
log dt log dt+1
3/2
= t+1
log dt log dt + log qt+1
3/2 .
and the result follows from the definition of qt and dt , certainly for dt > K.
Lemma 3. Let v ∈ Xt+1 ∪ Ct+1 and dt > K. Then P(|dt+1 (v) − dt+1 | > t+1 ) ≤ d−100 . t Proof. We begin by estimating Edt+1 (v). Using the definitions of qt and t , and Lemma 3, we have |Edt+1 (v) − dt+1 | = |dt (v)qt+1 − dt+1 | ≤ qt+1 t ≤ e−1/4e t+1 . (11) for dt > K. We wish to obtain concentration on dt+1 (v). If our girth were larger, then dt+1 would actually be the sum of independent indicators, and we could apply the Chernoff bounds, but with girth 5 the indicators that various vertices are in Xt+1 . Instead we must use martingale inequalities to concentrate dt+1 . An additional complication is that the natural exposure martingale is not actually Lipschitz, so we in fact must use a variety of Hoeffding’s inequality that allows an exceptional probability. This concentration inequality is recorded in the appendix as Proposition 3. The following argument is repeated several times in the following lemmas, so the interested reader should pay particularly close attention to the setup here to make future reading easier. In our process we make two choices independently for each vertex, whether or not to initially select the vertex and whether or not to place it in Qt+1 . For convenience, we make both of these choices for each vertex without regard for the other (that is, we flip the coin to see whether a vertex would be in Qt+1 regardless of whether it actually is placed there). Let Iu be the indicator that u is selected at time t, that it is one of the vertices initially chosen with probability d1t , and let Ju be the indicator that u would be placed in Qt+1 . 5
Note that this choice is made independently of Iu . Also note that Iu does not indicate whether u ∈ St+1 , as St+1 consists of selected vertices where no neighbours are selected. We now set up our filtration for an exposure martingale. Fix an ordering of the Xt -neighbours of v, namely u1 , u2 , . . . , udt (v) . Let Ai be the set of all random variables Iw , Jw such that w 6= ui is at distance at most two from ui in Gt \{v}. Define σ-fields Fi and random variables Yj : j = 1, 2, . . . , i as follows Fi = σ(Au1 , Au2 . . . , Aui )
and
Yj = E[dt+1 (v)|Fj ]
Then Y = (Yj )j≥1 is an exposure martingale [1], with Y0 = Edt+1 (v) and Ydt (v) = dt+1 (v). Furthermore, the only way that |Yi − Yi+1 | ≥ log dt is if some vertex at distance at most two from ui+1 in Gt − {v} has at least log dt chosen neighbors. Then X µ = P(|Yi − Yi+1 | ≥ log dt ) dt + t 1 log dt ≤ (dt + t )3 log dt dt 3 log d t (dt + t ) (dt + t ) < log dt dt blog dt c! = (dt + t )3 ·
(1 +
t log dt dt )
blog dt c!
K. Thus Y is (log dt )-Lipschitz with exceptional (2dt )20 probability at most µ ≤ blog dt c! . We will now apply Proposition 3 with λ = t+1 , ν = dt + t and ρ = 6. Note that 0 ≤ dt+1 (v) ≤ dt + t = ν and, as dt > K, we have νρµ < t+1 . Let δ = t+1 − e−1/e t . Then by definition of t , δ 2 > 20d2t log3 dt . By (11) and Proposition 3, (2dt )20 δ2 + blog dt c! 3dt log2 dt 400d log3 d (2dt )20 t t + ≤ d−100 , < 2 exp − t 2 blog d c! 3dt log dt t
P(|dt+1 (v) − dt+1 | > t+1 ) ≤ 2 exp −
for dt > K. This is a point at which the double exponential nature of K arises: so that 220 d121 < blog dt c!. t
3.3
Concentration of C-degrees
To control the size of the set Ct , we need control of the Ct -degrees at each step. In particular, we wish to show that the Ct -degree of a vertex never gets too large. We start with C0 = ∅. Recall, denote the Ct -degree of a vertex v by γt (v). For all vertices, γ0 (v) = γ1 (v) = 0, as C0 = ∅ and no normalization of probabilities is required at the first step. In the next lemma, we deal with the degrees γt (v) for t ≥ 2.
6
Lemma 4. Let v ∈ Xt+1 ∪ Ct+1 and suppose γt (u) < 16t for u ∈ Xt−1 \St . Then for dt > K, P(γt+1 (v) < 100t+1 ) ≤ d−100 . t Proof. Let γt+1 (v, Q) and γt+1 (v, C) respectively denote the Qt+1 -degree and the Ct+1 \Qt+1 t degree of v, so that γt+1 (v) = γt+1 (v, Q) + γt+1 (v, C). Since 0 ≤ qt (u) ≤ 2 dt for u ∈ Xt , 2t dt (v) dt 2t (dt + t ) ≤ dt 22 ≤ 2t + t = 2t + 221 log3 dt . dt
Eγt+1 (v, Q) ≤
By the Chernoff Bound, Proposition 1 in the appendix, since dt > K, 1 P(γt+1 (v, Q) > 3t ) ≤ d−100 2 t Also, by definition of t we have for dt > K that 3t < 4t+1 , so 1 . P(γt+1 (v, Q) > 4t+1 ) ≤ d−100 2 t From Lemma 1, for each u ∈ Ct , 1−
t t ≤ e1/e P(u ∈ ∂St+1 ) ≤ 1 + . dt dt
Therefore, using the assumption γt (v) < 8t , we obtain
|Eγt+1 (v, C) − qt γt (v)| ≤
2t γt (v) = 221 log3 dt . dt
To show concentration of γt+1 (v, C), we use the identical exposure martingale as in the proof of Lemma 3 above, and then apply Proposition 3. Thus we obtain 1 P(|γt+1 (v, C) − qt γt (v)| > t+1 ) ≤ d−100 . 2 t for dt > K. Note that this implies that with probability at most 12 d−100 that t √ γt+1 (v, C) ≤ qt γt (v) + t+1 ≤ 100 qt e1/4e t+1 + t+1 < 96t+1 . where we apply Lemma 2 using our assumption dt > K. Combining with our bound on γt+1 (v, Q), and applying the triangle inequality we have that P(γt+1 (v) > 100t+1 ) < d−100 . t This proves the lemma.
7
3.4
Concentration of sizes of sets
To complete the proof, we need not only control over the degrees of vertices at each step, but the evolving size of our sets Xt , Ct , and St . Here we suppose that at time t, our conditions (6)–(10) are satisfied; then we show that |St+1 | and |Xt+1 | are highly concentrated, while providing an upper bound on the size of Ct+1 . In order to do this, we must break into two cases depending on the relationship between dt and nt . If dt is large compared with nt (say 1/20 dt > nt ) then we need to use a slightly different method to achieve tight concentration. Lemmas 5 and 6 address the case where dt is small, and large respectively. Lemma 5. Suppose (6) − (10) hold at time t. Then 2n nt+1 t P |Xt+1 | − nt+1 > 20 ≤ 2 exp 4 dt+1 dt n 3t+1 nt+1 nt log3 dt P |St+1 | − > ≤ 2 exp − ed d2t+1 d5t
(12) (13)
so long as dt is sufficiently large. Proof. We begin by noting that P(v ∈ St+1 ) = ≤ ≤ ≤
1 1 dt (v) 1− dt dt 1 dt −t 1 1− dt dt t 1 exp −1 + dt dt t 1 + , edt d2t
with a lower bound following similarly, giving 1 ≤ t . P(v ∈ St+1 ) − edt d2t
(14)
First we show concentration of |Xt+1 | = |Xt | − |Xt ∩ (St ∪ Qt+1 ∪ ∂St+1 )|. Note that for a
8
vertex v ∈ Xt , |P(v 6∈ Xt+1 ) − (1 − qt+1 )| = |P(v ∈ St+1 ∪ ∂St+1 ∪ ∂Qt+1 ) − (1 − qt+1 )| = |P(v ∈ St+1 ) + P(v ∈ (∂St+1 ∪ ∂Qt+1 ∩ S¯t+1 )) − (1 + qt+1 )| P(v ∈ (∂S ∪ ∂Q |v ∈ 6 S ) t+1 t+1 t+1 − (1 + qt+1 ) = P(v ∈ St+1 ) + P(v 6∈ St+1 ) 1 − qt = P(v ∈ St+1 ) + − (1 + qt+1 ) P(v 6∈ St+1 ) ≤ (1 + qt )P(v ∈ St+1 ) + ≤
qt P(v ∈ St+1 ) 1 − P(v 6∈ St+1 )
6 . edt
where the last inequality follows from (14) assuming dt > K and noting that qt ≤ 1. This gives us that E|Xt+1 | − nt+1 ≤ E|Xt+1 | − qt+1 |Xt | + qt+1 |Xt | − qt+1 nt nt 6|Xt | + 20qt+1 ≤ edt dt nt+1 5nt+1 + 20e−1/e ≤ dt+1 dt+1 nt+1 ≤ 19 . dt+1
We set up a martingale in the following way: fix an ordering of the vertices in Xt , namely v1 , v2 , . . . , vnt . Let Ij denote the indicator that vj is selected, and let Jj denote the indicator that vj is chosen to be in Q (as in the proof of lemma 3 these are independent) and let t Fj = σ(I1 , J1 , I2 , J2 , . . . , Ij , Jj ) and Zj = E[|Xt+1 ||Fj ]. Then Z = (Zj )nj=0 is a martingale, with Z0 = E|St+1 | and Znt = |St+1 |. Noting that knowing Ij and Jj only tells us information within the neighborhood of vertex vj , this martingale is (dt + )-Lipschitz, which, for convenience, we √ note implies that it is 2dt -Lipschitz assuming dt > K. By Hoeffding’s Inequality, Proposition t+1 2, with λ = ndt+1 = ndtt nt+1 P |Xt+1 | − nt+1 > 20 ≤ P |Xt+1 | − E|Xt+1 | > λ dt+1 2λ2 ≤ exp − 2nt d2t nt ≤ exp − 4 . dt 9
We now show that |St+1 | is concentrated, the proof being quite similar. By (14) n |Xt | |Xt | nt + − E|St+1 | − ≤ E|St+1 | − ed edt edt edt t nt ≤ 2 2 dt t+1 nt+1 ≤ 2 d2t+1 since dt > K and noting that Lemma 2.
nt+1 dt+1
=
nt dt
=
n d;
note that the last inequality follows using
We set up a martingale as before: fix an ordering of the vertices in Xt , namely v1 , v2 , . . . , vnt . Let Ij denote the indicator that vj is selected and let Fj = σ(I1 , I2 , . . . , Ij ) and Wj = t E[|St+1 ||Fj ]. Then W = (Zj )nj=0 is a martingale, with W0 = E|St+1 | and Znt = |St+1 |. √ nt+1 Again W is 2dt -Lipschitz, so by Proposition 2, with λ = dt n2 t ≤ t+1 , d2 t
t+1
t+1 nt+1 nt+1 P |St+1 | − ≥3 ≤ P(||St+1 | − E|St+1 || ≥ λ) edt+1 d2t+1 2λ2 ≤ exp − 2nt d2t 2 n2 ≤ exp − t t6 nt dt 220 n log3 d t t ≤ exp − . d5t This completes the proof of concentration of |St+1 |.
If dt is large, for example dt ≥ n1/20 , then we can improve the above lemma as follows. 1/20
, n > 10d2 log2 d and (6)−(10) hold at time t, Then for dt > K, nt+1 P |Xt+1 | − nt+1 > 20 ≤ 2e−5/2 (15) dt+1 ( ) nt t+1 nt+1 nt+1 p P |St+1 | − > 3 max , ≤ 2e−5/2 . (16) edt dt+1 dt+1 d
Lemma 6. Suppose dt ≥ nt
Proof. We begin by showing the concentration for |Xt+1 |, using the same martingales as in the proof of 5. We note that exposing a vertex in our martingale changes the value by more than log dt only if at least log dt neighbors of v are chosen. Thus the set of bad outcomes where such an event occurs has probability at most (following calculations in Lemma 3), noting nt < d20 t , dt + t 1 (2dt )40 nt < ≤ d−100 , t log dt dt blog dt c! 10
since dt > K. Here again we require K to be double exponential. The martingale Z in the proof of Lemma 5 is thus log dt -Lipschitz with exceptional probability µ = d−100 . We now choose λ = ndtt and it is easy to verify since |Xt+1 | ≤ |Xt | that the conditions t t+1 . of Lemma 3 are satisfied with νρµ < λ when ν = nt + 20 ndtt , ρ = 1000 and λ = 20 ndt+1
nt+1 P |Xt+1 | − nt+1 ≥ 20 dt+1
≤ P( |Xt+1 | − E|Xt+1 | ≥ λ) n2t+1 ≤ 2 exp − + d100 ≤ 2e−5/2 . t nt d2t+1 log(dt )2
for dt > K. Concentration for |St+1 | in this setting is similar. The martingale W in the proof of Lemma 5 is log dt -Lipschitz with exceptional probability at most d−100 , with the same bound in the t conditions on Proposition 3 as above. By Proposition 3, taking λ = max
t+1 nt+1 , d2t+1
√nt+1 ddt+1
and , ν = nt + 20 ndtt and µ = d−100 t
ρ = 1000 −n2t+1
+ d−100 . t 2dt+1 dnt log2 dt n ≤ 2 exp − 2 + d−100 ≤ 2e−5/2 t 2 3d log dt
P (||St+1 | − E|St+1 || ≥ λ) ≤ exp
completing the second inequality, using the bound on E|St+1 | obtained in Lemma 5.
Lemma 7. Suppose that (6) − (10) hold at time t. Then for dt > K, t+1 nt+1 nt P |Ct+1 | > 200 ≤ 2 exp − 6 2 . dt+1 10 dt
(17)
Proof. Here we show concentration of |Ct+1 | = |Ct | − |Ct ∩ ∂St+1 | + |Qt+1 |. Using the upper and lower bounds for survival probability in Lemma 1, we have that qt+1 (v) ≤ 1 −
1− 1+
t dt t dt
≤3
t dt
so by (10) we have that 4t nt . dt By the Chernoff bounds, Proposition 1, since vertices are independently chosen to be in Qt+1 , n 16t+1 nt+1 8t nt t t P |Qt+1 | > ≤ P |Qt+1 | > ≤ exp − . dt+1 dt 2dt E|Qt+1 | ≤
11
nt+1 ≥ Here the first inequality comes from the fact that 2 t+1 dt+1
t nt dt
for dt > K.
By the upper bound for survival probability in Lemma 1, we have that t t+1 nt+1 −1/e 1+ |Ct | − E|Ct ∩ ∂St+1 | ≤ e |Ct | ≤ 200e−1/(2e) , dt dt+1 with the last inequalities holding for dt > K and using the condition (8). To get concentration for |Ct ∩ ∂St+1 | we apply a martingale argument, similar to those above. We use Hoeffding’s inequality, Proposition 2, for concentration. Fix an ordering on the vertices in ∂Ct ∩ Xt , namely v1 , v2 , . . . , vr . We denote by Ij the indicator that vj ∈ St+1 and let Fj = σ(I1 , I2 , . . . , Ij ) denote the σ-field generated by I1 , I2 , . . . , Ij . Let W = |Ct ∩ ∂St+1 |, and let (Wi )ri=0 = E[W |Fi ]. Then W is an exposure martingale. Let ci = |∂{vi } ∩ Ct |. Then P ci ≤ (dt + t )|Ct | since dt (u) ≤ (dt + t ) for u ∈ Ct . Furthermore, each ci ≤ 100t as (7) nt+1 holds at time t. By Proposition 2, setting λ = t+1 dt+1 , and noting that |Ct | > 200λ, we have λ2 t λ2 −1/e ≤ exp − P W ≥e 1+ |Ct | + λ ≤ exp − Pr dt 200t (dt + t )|Ct | 2 i=1 c2i λ2 ≤ exp − 400t dt |Ct | nt ≤ exp − 6 2 10 dt nt+1 Finally, we note that if |Ct+1 | ≥ 200 t+1 dt+1 , then either
|Qt+1 | > 16
t+1 nt+1 dt+1
or
|Ct | − |Ct ∩ ∂S| ≥ (1 + 200e−1/2e )
t+1 nt+1 . dt+1
Therefore P(|Ct+1 | > 200
n nt t+1 nt+1 nt t t ) < exp − + exp − 6 2 ≤ 2 exp − 6 2 . dt+1 2dt 10 dt 10 dt
This completes the proof.
4
Proof of Main Theorem
Proof. The spirit of our proof is inductive: we run the process described above. At each step, we insure that (with high probability) individual vertices are good – a vertex v is good if it satisfies (6), and bad otherwise. We note that the change of degrees of vertices are only dependent if the vertices are within distance five of each other. This gives us a limited dependence, allowing us to use the Lov´asz Local Lemma (Proposition 4) to say that, with positive probability, no vertex is bad. That is, there exists a choice of St+1 and Qt+1 such
12
that the resulting graph at time t + 1 satisfies all our conditions. We shall stop our process before dt < K. Recall the conditions which we wish to hold: |dt+1 (v) − dt+1 | ≤ t+1
∀ v ∈ Xt+1 ∪ Ct+1
γt+1 (v) ≤ 100t+1
∀ v ∈ Xt+1 ∪ Ct+1
t+1 nt+1 dt+1 ( ) n t+1 nt+1 nt+1 |St+1 − | ≤ 3 max ,p . ed d2t+1 dt+1 d |Ct+1 | ≤ 200
|Xt+1 − nt+1 | ≤ 20
nt+1 . dt+1
(18) (19) (20)
(21) (22)
Clearly at time t = 0, the conditions (18)–(22) are satisfied. Let us assume that they are satisfied at time t. Let Av be the event that either (18) or (19) fails at vertex v at time t + 1. Note that as the conditions are satisfied at time t, the combined C and X degrees of all vertices are less than 4dt , and hence Av is mutually dependent with at most (4dt )5 other events Au . By the Lov´ asz Local Lemma, Proposition 4, since P(Av ) ≤ 2d−100 , t 5
P(Ai ) ≤ d−50 (1 − d−50 )(4dt ) t t
and there are, by induction and the sizes guaranteed by (20) and (22), at most 2nt vertices combined in X and C over which we want control, all degrees are good with probability at least 2nt 2nt ≥ exp − (1 − d−50 ) t d50 t +1 all degrees are good at time t + 1. Case 1. If dt ≤ n1/20 , Lemmas 5 and 7 implies |St+1 |, |Xt+1 | and |Ct+1 | are all within the required bounds with probability at least 1 − 3 exp(−nt /d5t ) provided dt > K. Since
2nt nt > 3 exp(− 5 ) exp − 50 dt + 1 dt
for dt > K, conditions (18) – (22) hold with positive probability at time t + 1. Case 2. In the event that dt > n1/20 note that Lemmas 6 and 7 and our condition that n > 5d2t log2 dt imply that the sizes of C and X and number of vertices picked are all correct with probability 1 − 5e−5/2 (once again using the weakest of our concentration bound). In 13
1 this case, note that exp(−2nt /(d50 t + 1)) > e2 and again we have that conditions (18) – (22) hold with positive probability at time t + 1. Therefore, there is some choice of St+1 and Ct+1 such that all of conditions (18) – (22) are satisfied at time t + 1. We condition on such a good occurrence and continue.
Let T = min{t : dt < K}. We stop the process at time T ; we’ve already verified that we can run the process to this point. Note that T = e log d − a for some constant a. We now take S −1 St , plus a maximal independent set in the graph our independent dominating set to be Tt=0 on XT ∪ CT . We verify that this set has the claimed size. Note that since |St | = (1 + o(1)) nd , S d it is clear that |St | ≤ (1 + o(1)) n log but we wish to verify the claimed error term. First d observe that: T −1 −1 TX X t nt nt n 3 max ,√ |St | − T ≤ ed d2t dt d t=0 t=0 ! T −1 √ T −1 n X t X dt √ ≤ 3 + d d d t=0 t=0 t n = O , d P noting that both sums are bounded by geometric sums of the form t e−t/4e by the definition of dt and t and hence converge. Next, note that since dT > K, the fact that (20) holds imply that |CT | is O( nd ), and the fact that ndt nt = d n and (22) implies that |XT | is O( d ) so the maximal independent set in both of these has size O( nd ). Therefore we have found the desired independent dominating set.
5
Concluding remarks
• In this paper, we showed that if G is a d-regular graph without 4-cycles, then so long as n > 5d2 log2 d, then G contains an independent dominating set of size n(logdd+c) for some constant c. While the constant 5 is not optimal (a more careful reading of the proof will likely allow one to improve this), the order of the lower bound on n as a function of d seems to be harder to improve. The real problem comes from concentration of |St+1 | in Lemma 6. When the process starts here, concentration around nd is as good as we can get; if n = o(d2 log2 d), we would have to concentrate around something larger order than nd , already ruining the term log d of order nd . If we are satisfied with an independent dominating set of size n(1+o(1)) , then d n it is enough to assume that d2 log d → ∞. • In the introduction it was mentioned that an n-vertex graph of minimum degree at least d . However, it may not be possible to find an has a dominating set of size at most n(1+log(d+1)) d+1 14
log d independent dominating set of size n(1+o(1)) , so the condition of regularity in Theorem 1 d is necessary. To see this, we take a random graph Gn,2d/n where d/ log n → ∞ on a set of n2 vertices, and then take an independent set of size n2 and join each of its vertices to d randomly selected vertices in Gn,2d/n . If the resulting graph is G, then with high probability G has minimum degree d and maximum degree roughly 3d. The expected number of cycles of length 4 at most four on each vertex is O( dn ), so with high probability, if d = o(n1/3 ), we can delete some edges without affecting the degrees of the vertices by much. Then, any independent set on the random graph side must be fairly small, and leaves most of the independent set side uncovered. An interesting question is whether an analogue of this theorem does occur in random graphs with some degree sequence, or alternately graphs with a good spectral gap (for example, spectral gap of the normalized Laplacian).
• At its heart our method is a variant of methods like the R¨odl nibble, and the differential equation method of Wormald. In a follow-up paper, we will explore a similar process on hypergraphs, looking at matching and covering problems. • If G is an n by n bipartite graph which is d-regular and has no 4-cycles, then a simpler proof than that of the main theorem shows that we can find an independent dominating set of size at most 2n(logd d+c) for some constant c. To see this, we can avoid using the set C and directly pick vertices from each side: we are able to pick vertices in the top part with probability log d−3 log log d – much higher than in our process in the proof of the main theorem – and show d that the degrees of vertices at the top are then highly concentrated at log3 d. Notice that the surviving vertices at the bottom all have degree d. Then we apply the local lemma as in the proof of our main theorem to conclude that all the vertices have degree close to log3 d at the top. Repeating this at the bottom, we choose vertices at the bottom with probability log d−3 log log d . After applying concentration inequalities and the local lemma, we see that we log3 d d have picked close to 2n log vertices and all the vertices at the top and at the bottom have d 3 degree close to log d. Essentially, in one step we have reduced the problem to a bipartite graph which is close to (log3 d)-regular and where each part has close to n logdlog d vertices. Now we repeat this whole procedure roughly log∗ d times to get the required independent dominating set.
6
Appendix : Concentration Inequalities
In this section, we present the inequalities which we will need from probability theory involving concentration of measure. All of these inequalities deal with upper bounds for expressions of the form P(|X − EX| > λ) where X is a random variable and λ is a real number. The most basic inequality of this type for binomial distributions is the Chernoff Bound [1]:
15
Proposition 1. (Chernoff Bound) Let Xi , i = 1, 2, . . . , n be independent random variables P with Bernouilli distributions with means pi , i ∈ [n], and let p = n−1 pi and h ≥ 0. Then h2 n P(Sn > (p + h)n) ≤ exp − 2a h2 n P(Sn < (p − h)n) ≤ exp − , 2b where Sn = X1 + X2 + · · · + Xn and a is the maximum of α(1 − α) for p ≤ α ≤ p + h and b is the maximum of β(1 − β) for p − h ≤ β ≤ p. In particular, if X has binomial distribution with probability p and mean pn, then for any ε < 1, ε2 pn . P(|X − pn| > εpn) < 2 exp − 2 Most of the inequalities we will need concern martingales. One of the most fundamental martingale inequalities is Hoeffding’s Inequality. Further refinements and generalizations of this inequality may be found in McDiarmid [8]. In the following two propositions, c = (c1 , c2 , . . . , cn ) P is a vector of positive real numbers and kck2 denotes ni=1 c2i . Proposition 2. (Hoeffding’s Inequality) Let (Xi )ni=1 be a martingale with difference sequence (Yi )ni=1 , where −ai ≤ Yi ≤ −ai + ci for some function ai on (Ω, Fi−1 ). Then for λ ≥ 0 2λ2 2λ2 P(Xn > E(Xn ) + λ) ≤ exp − and P(X < E(X ) − λ) ≤ exp − . n n kck2 kck2 The following martingale concentration inequality, similar to the version derived by Chung and Lu [3], is used frequently and is a minor modification of Hoeffding’s Inequality. For completeness, we give a complete proof. Proposition 3. Let (Xi )ni=1 be a martingale with difference sequence (Yi )ni=1 , and suppose n X
P(|Yi | > ci ) ≤ µ
and
a < Xn ≤ a + ν
i=1
for some a, ν, µ ∈ R. Then for any positive real numbers λ and ρ > 1 satisfying λ > νρµ, (ρ − 1)2 λ2 + µ. P(|Xn − EXn | > λ) ≤ 2 exp − 2ρ2 kck2 Proof. Let T = min{i : |Yi | > ci } – by convention, the minimum over the empty set is infinite, in which case we write T = ∞. Let T ∧i = min{i, T }. Then it is well-known (see Williams [10]) that (XT ∧i )ni=1 is a martingale. Let (Zi )ni=1 be the difference sequence of that martingale. By definition of T , |Zi | ≤ ci . By Proposition 2, with X = XT ∧n , (ρ − 1)λ (ρ − 1)2 λ2 P(|X − E(X)| > ) ≤ 2 exp − . ρ 2ρ2 kck2
16
Finally we note that P(X 6= Xn ) ≤ µ and, since a ≤ Xn ≤ a + ν, |E(X) − E(Xn )| ≤ νµ < λ/ρ. This completes the proof as P(|Xn − E(Xn )| > λ) ≤ P(|X − E(X)| > (ρ − 1)λ/ρ) + P(X 6= Xn ) (ρ − 1)2 λ2 + µ. ≤ 2 exp − 2ρ2 kck2 We call a martingale (Xi )ni=1 satisfying the conditions of Proposition 3 with ci = c ∈ R a c-Lipschitz martingale with exceptional probability µ. The final probabilistic result we need, the Lov´asz Local Lemma [7], states that if there is limited local dependence between a set of rare events, then there is a positive probability that none of the events occurs. Proposition 4. (Lov´ asz Local Lemma) Let A1 , A2 , . . . , An be events in some probability space and suppose that for some set Ji ⊂ [n], Ai is mutually independent of {Aj : j 6∈ Ji ∪ {i}}. If Q there exist real numbers γi ∈ [0, 1) such that P(Ai ) ≤ γi j∈Ji (1 − γj ), then P
n \ i=1
n Y Ai ≥ (1 − γi ) > 0. i=1
References [1] N. Alon, and J. Spencer. The probabilistic method, John Wiley and Sons Inc., New York, 2008. [2] V.I. Arnautov, Estimation of the exterior stability number of a graph by means of the minimal degree of the vertices (in Russian), Prikl. Mat. i Programmirovanie 11 (1974), 3–8. [3] F. Chung and L. Lu, Coupling on-line and off-line analyses for random power law graphs. Internet Math. 1 (2004), no. 4, 409–461. [4] W. Duckworth and N. Wormald, On the independent domination number of random regular graphs, combinatorics, Probability and Computing, 15 (2006) 513–522. [5] P. Erd˝ os, and A. R´enyi, On a problem in te theory of graphs, Publ. Math. Inst. Hungar. Acad Sci. 7A (1962), 623–641. [6] L. Lov´ asz, On the ratio of optimal and integral fractional covers, Disc. Math 13 (1975), 383–390.
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[7] P. Erd˝ os and L. Lov´ asz. Problems and results and results on 3-chromatic hypergraphs and some related questions, in: ”Infinite and Finite Sets”, (A. Hajnal et. al. eds), Colloq. Math. Soc. J. Bolyai 11, (1975) 609–627. [8] C. McDiarmid, Concentration, Probabilistic methods for algorithmic discrete mathematics, 195–248, Algorithms Combin., 16 Springer, Berlin. [9] C. Payan, Sur le nombre d’absorption d’un graphe simple (in French) Proc. Colloq. Th. Graphes (Paris 1974), C.C.E.R.O 17 (1975), 307–317. [10] D. Williams, Probability with Martingales. Cambridge Mathematical Textbooks. Cambridge University Press, Cambridge, 1991. xvi+251 pp.
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