dedicated to the memory of dear friend Dick Schelp (1936-2010) and - - PowerPoint PPT Presentation

dedicated to the memory of dear friend Dick Schelp (1936-2010)

and remembering my years in Louisville (1986-1987, 1994-2001)

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 1 / 29

from the irregularity strength of graphs to the degree irregularity of random hypergraphs

Jen˝

• Lehel

The University of Memphis

May 12, 2011

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 1 / 29

graph: G = (V , E) simple: no multiple edges hypergraph: H = (V , E) r–uniform hg. |e| = r degree: d(v) = |{e ∈ E | v ∈ e}| (degree) irregular hg. - no degrees repeat d(x) = d(y) for any distinct x, y ∈ V there is no irregular simple graph (2-uniform hg.), since a degree must repeat [Behzad, Chartrand, 1967]

irregularity strengh of graphs

assign pos. integer weights (multiplicities) to the edges such that the weighted degrees are distinct; what is the largest weight we must use? [Chartrand, Jacobson, Lehel, Oellermann, Ruiz, Saba, 1986/1988]

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 2 / 29

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 3 / 29

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 r = 3, n = 6 there exists an irregular simple r-uniform hg. with n vertices, for every r ≥ 3 and n ≥ r + 3 [Gy´ arf´ as, Jacobson, Kinch, Lehel, Schelp, 1989]

irregularity of a random hg.

what is the probability that a random r-uniform hg has no repeated degrees?

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 4 / 29

content

1 graphs - irregularity strength

lower and upper bounds trees regular graphs

2 hypergraphs - degree repetition in random r-uniform hg

probability of repeating degrees for r ≥ 6 formula for all r

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 5 / 29

irregularity strength

edge k-weighting: w : E − → {1, . . . , k} weighted degree: w(x) = {w(e) | x ∈ e} irregular weighting: w(x) = w(y) for any distinct x, y ∈ V irregularity strength: s(G) = min{k | w is an irreg. k-weighting}

examples

s(G) < ∞ provided G has at most 1 isolated vertex and no isolated edge, s(K3) = 3, s(Kn) = 3, s(2P3) = 5, s(P4k) = 2k, s(C4k) = 2k + 1

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 6 / 29

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 7 / 29

lower bound

given G and an irregular k-weighting with k = s(G), the largest weighted degree is k ·∆(G) (here ∆(G) is the max degree in G) – since all the n weighted degrees are distinct, n ≤ k · ∆(G), thus we have s(G) ≥ n/∆(G) – a similarly count gives ni ≤ s(G) · i − i + 1, where ni is the number of vertices of degree i in G, and hence s(G) ≥ (ni + i − 1)/i leading to:

Proposition.

If ni is the number of vertices of degree i in G, then s(G) ≥ max

d0≤d

(nd0 + nd0+1 + · · · + nd) + d0 − 1 d [Jacobson, Lehel, 1986]

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 8 / 29

upper bounds

given G, with n vertices, e edges ”naive” bound s(G) ≤ 2e−1 ”greedy” bound s(G) ≤ 2n − 3 [Chartrand et al., 1986/1988] ”congruence method” s(G) ≤ n + 1 [Aigner, Triesch, 1990] ”spanning tree” bound s(G) ≤ n − 1 [Jacobson, Lehel, 1986] for connected G and n ≥ 4

Theorem.

If G = K3, and has no isolated edges and vertices, then s(G) ≤ n − 1 [Nierhoff, 1998/2000]

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 9 / 29

the congruence method

Zm is the additive group (of integers mod m) reservation on V : a labeling ρ : V → Zm given a forest (V , F), a reservation ρ on V is feasible: if there is an edge weighting w : F → {1, . . . , m}, call it a realization, such that w(x) ≡ ρ(x) (mod m), for every x ∈ V

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 10 / 29

the congruence method

given G = (V , E) with n vertices, let m = n − 1, and let (V , F) be a spanning forest of G idea: if there was a feasible reservation ρ : V → Zm, then a realization w extended with w(e) = m, for every e ∈ E \ F, is an irregular m-weighting, thus s(G) ≤ m = n − 1 the proof consists of an algorithm that builds feasible reservations for the subtree components of (V , F) this stepwise construction of feasible reservations uses decompositions

• f Zm

because some vertices might have the same reservation, further steps must be done to resolve those repetitions by altering the realization s(G) = n − 1 for G = K4, 2P3, K1,n−1, and ... ?

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 11 / 29

bound improvements

better bounds if G is regular or G is a tree different from a star

Theorem.

If G is regular and n ≥ 3, then s(G) < n/2 + 9 [Faudree, Lehel, 1987]

Theorem.

If a tree T with n ≥ 3 has a matching that contains ℓ leaves, then s(T) ≤ n − ℓ, except possibly when ℓ = n/2 − 1 and T is equipartite. [Aigner, Triesch, 1990] Corollaries. s(T) ≤ n − 2 if T is not a star s(T) ≤ n − 3 if T is not a star and not a path Pn, n ≤ 6,

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 12 / 29

trees

if there are nj vertices of degree j, then s(G) ≥ Λ(G) = max

d0≤d

(nd0 + nd0+1 + · · · + nd) + d0 − 1 d for a tree T, either Λ(T) = n1 or Λ(T) = ⌈(n1 + n2)/2⌉ . does s(T) stay ”close” to Λ(T)? If T is a full d-ary tree, for d = 2, or 3 then s(T) = Λ(T) = n1 [Cammack, Schelp, Schrag,1991] If n ≥ 3, and n2 = 0, then s(T) = Λ(T) = n1 [Amar, Togni, 1998] If n1 ≥ 3, and every pair of vertices of degree not equal to 2 are at a distance at least 8, then s(T) = Λ(T) = ⌈(n1 + n2)/2⌉ [Ferrara, Gould, Karo` nski, Pfender, 2010]

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 13 / 29

for several graph families s(G) ≤ Λ(G) + c (paths, cycles, wheels, k-cubes, grids, Tur´ an graphs, union of cliques,...) not true for disconnected graphs:

Theorem.

Λ(tP3) = 2t, meanwhile ⌈(15t − 1)/7⌉ ≤ s(tP3) ≤ ⌈(15t − 1)/7⌉ + 1 [Kinch,Lehel, 1990] Is s(T) ≤ Λ(T) + c true for connected graphs with some constant c? In particular, is it true for trees, perhaps with c = 1? [Ebert, Hemmeter, Lazebnik, Woldar, 1990]

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 14 / 29

Theorem.

Let Tt be obtaned from a path P2t+1 by attaching pendant edges to alternating interior vertices. Then Λ(T) = t + 2 = n1, meanwhile lim

t→∞ s(Tt) = 11− √ 5 8

· n1 > 1.095 · Λ(T) [Bohman,Kravitz, 2004]

problems

find further families of trees with s(T) = Λ(T) find the smallest factor f such that s(T) ≤ f · Λ(T), for all trees

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 15 / 29

regular graphs

let G be a d-regular graph with n vertices

Proposition.

if d = n − 1 then G = Kn and s(G) = 3 if d ≤ n − 2 then Λ(G) = ⌈(n + d − 1)/d⌉ ≥ 3 if d = n − 2 then G = Kn − n

2 · K2 and s(G) = 3

[Gy´ arf´ as, 1989]

Theorem.

If d = n − 3 or n − 4, and G = K3,3, then s(G) = 3 (s(K3,3) = 4) [Amar, 1988/1993]

Theorem.

If d = 2 then s(G) ≤ ⌈n/2⌉ + 2 [Faudree, Jacobson,Lehel, Schelp, 1989]

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 16 / 29

problem

is there a constant c = c(d) such that s(G) ≤ n

d + c, for all d-regular G?

recall: Λ(G) = ⌈ n+d−1

d

⌉ [ Faudree, Jacobson, Kinch, Lehel, 1987/1991] s(G) ≤ 48 n

d + 1, for d ≤ √n

s(G) ≤ 240 n

d · log n + 1, for d > √n

[Frieze, Gould, Karo` nski, Pfender, 2002] a two-stage strategy was used for the proof

1 first find a ”pre-weighting” w : E → {1, 2, 3} such that no degree

repeats more than M ∼ α · (n/d) times (or M ∼ α · (n/d) log n-times) /probabilistic stage/

2 after multiplying each w(e) by M there will be enough ”room” to be

able to ”shake off” identical degrees by modifying edge weights to

• btain an irregular (8M + 1)-weighting

/deterministic Lemma/

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 17 / 29

similar two-stage strategy with a ”pre-weighting” w : E → {1, 2} leads to s(G) ≤ 48 n

d + 6, for d ≥ 104/3 · n2/3 · (log n)1/3

[Cuckler, Lazebnik, 2008] the best bound so far is obtained by replacing the probabilistic stage with a deterministic construction:

Theorem.

If G is d-regular with n vertices, then s(G) ≤ 16 n

d + 6

[Przybylo, 2009]

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 18 / 29

random uniform hg.

almost all hypergraphs have no repeated degrees, i.e. a.a. hg are irregular [Gy´ arf´ as, Jacobson, Kinch, Lehel, Schelp, 1992] random r - uniform hg Hr(n, p):

• n an n element (labeled) vertex set V

each of the n

r

• r-sets is taken as an edge

independently and uniformly with probability p.

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 19 / 29

estimating P{d(x) = d(y) for some x, y ∈ V }

1 first compute P{d(x) = d(y)} for FIXED x, y ∈ V

”split degrees” d∗(x) = {e ∈ E | x ∈ e, y / ∈ e} d∗(y) = {e ∈ E | y ∈ e, x / ∈ e} d(x) = d(y) iff d∗(x) = d∗(y) = s, for some 0 ≤ s ≤ K = n−2

r−1

• ∼ nr−1

P{d∗(x) = s} = K

s

• ps(1 − p)K−s

P{d∗(x) = s = d∗(y)} = P{d∗(x) = s} · P{d∗(y) = s} P{d∗(x) = s = d∗(y)} = K

s

2p2s(1 − p)2K−2s

2 we have the expected number of ”bad pairs”

E{#x, y ∈ V , d(x) = d(y)} = n

2

K

• s=0

K

s

2p2s(1 − p)2K−2s

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 20 / 29

the case p = 1/2 and r ≥ 6

Proposition.

for p = 1/2 and r ≥ 6, P{Hr(n, p) has repeated degrees} − → 0 as n − → ∞

Proof.

claim: the probability of a bad pair approaches zero. we verify that the expected number of bad pairs tends to 0, then we are done by Markov’s

• inequality. with p = 1/2 and using the Stirling-formula, and K ∼ nr−1 :

E{#x, y ∈ V , d(x) = d(y)} = n

2

K

• s=0

K

s

2p2s(1 − p)2K−2s = n

2

2K

K

1

22K ∼

n

2 22K √ Kπ

• 1

22K

∼ c · n(5−r)/2 → 0, for r ≥ 6.

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 21 / 29

Corollary.

for p = 1/2 and r ≥ 6, P{Hr(n, p) is irregular} − → 1 as n − → ∞, for p = 1/2 and r ≥ 6 – what if p = pn is a function of n? – how about the small ranks r = 3, 4, 5? answered recently with Balister, Bollob´ as, and Morayne [2010, 2011]

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 22 / 29

Theorem.

for fixed r ≥ 3, as n → ∞, P{Hr(n, pn) is irregular} →            if pn(1 − pn) · nr−5 → 0 fr(c) if pn(1 − pn) · nr−5 → c ∈ (0, ∞) 1 if pn(1 − pn) · nr−5 → ∞ where c ∈ (0, ∞), and fr(c) = exp

• −

1 2 √ 2

• (r−1)!

2π·c

1/2 ∈ (0, 1) [Balister, 2010] for r = 3, 4, the hg is a.s. non-irregular for r = 6, the hg is a.s. irregular with any constant p and there is a threshold in pn separating a.s. non-irregularity from irregularity and r = 5 is a ”threshold” in the rank with computed probability in [0, 1)

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 23 / 29

r=3,4

random 3- or 4-uniform hg is a.s. NOT irregular:

For r = 3, 4, P{Hr(n, pn) has repeated degrees} − → 1 as n − → ∞ for a vertex x of Hr(n, pn), we have 0 ≤ d(x) ≤ N = n−1

r−1

• the degree can be described by the r.v. SN:

the number of successes in N Bernoulli trials with probability of success pn in one trial mean degree: E{SN} = Npn (mean in the b.d. B(n, pn)) central interval: [Npn − ∆, Npn + ∆) let Z∆ be the # of vertices x with d(x) ∈ [Npn − ∆, Npn + ∆)

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 24 / 29

degree repetition in the central interval

in each case, r = 3 and r = 4, with appropriate choice for ∆ = ∆n → ∞ with n → ∞, some degrees must repeate in the central interval of length 2∆ BUT for different reasons!

r=3

P{Z∆ > 2∆} − → 1 as n → ∞, H3(n, pn) a.s. has repeated degrees by the pigeon hole principle

r=4

P{Z∆ > cn √ 2∆} − → 1 as n → ∞, with a sequence cn → ∞ H4(n, pn) a.s. has repeated degrees by the birthday paradox

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 25 / 29

⌣ random ⌣⌣ pigeons principle ⌣

birthday paradox

– from an s element set t elements are selected at random (with repetition) uniformly and independently – if t = cs √s where cs is any sequence with lim

s→∞ cs = ∞, then

some element is a.s. selected more than once r = 4 : P{Z∆ > cn √ 2∆} − → 1 then the birthday paradox is applied with s = 2∆ and cn → ∞ such that c2

n · n−1/2 → 0, and ∆ = c2 n · n

since the distribution of degrees in [Npn − ∆, Npn + ∆) is neither independent nor uniform, several technical steps are required to verify that degree repetition is even more likely in this case

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 26 / 29

lemmas

r = 3 : P{Z∆ > 2∆} − → 1, with ∆ = αn · n → ∞ and αn → 0 r = 4 : P{Z∆ > cn √ 2∆} − → 1, with ∆ and cn as before in both proofs the Central Limit theorem is used Berry – Ess´ een estimation for the rate of convergence plus Chebysev’s inequality (second moment method)

Lemma 1.

For every r ≥ 3, ∆ → ∞ and lim

n→∞ ∆/

• pn(1 − pn) = 0,

E{Z∆}2 ≈

2∆2(r−1)! πnr−3pn(1−pn)

Lemma 2.

For every r ≥ 3, ∆ → ∞ and lim

n→∞ ∆/

• pn(1 − pn) = 0,

E{Z 2

∆} < (1 + o(1))

• E{Z∆} + (2∆ + 4∆2)
• (r−1)!

2πnr−3pn(1−pn)

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 27 / 29

the missing cases

r = 5 r = 6, for any p = pn – Dick Schelp suggested the notion of k–irregularity, i.e. when no vertex degree repeats k times – Paul Balister computed asymptotically the probability of the k-iregularity

• f a random r- uniform hg, for every fixed r ≥ 3 and k ≥ 2, as a function
• f σ2 =

n

r−1

• pn(1 − pn) and n → ∞

– for k = 2, Balister’s general formula implies the earlier results presented for r = 3, 4

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 28 / 29

Theorem.

for fixed r ≥ 3, as n → ∞, P{Hr(n, pn) is irregular} →            if pn(1 − pn) · nr−5 → 0 fr(c) if pn(1 − pn) · nr−5 → c ∈ (0, ∞) 1 if pn(1 − pn) · nr−5 → ∞ where c ∈ (0, ∞), and fr(c) = exp

• −

1 2 √ 2

• (r−1)!

2π·c

1/2 ∈ (0, 1) [Balister, 2010] for r = 3, 4, the hg is a.s. non-irregular for r = 6, the hg is a.s. irregular with any constant p and there is a threshold in pn separating a.s. non-irregularity from irregularity and r = 5 is a ”threshold” in the rank with computed probability in [0, 1)

Jen˝

• Lehel (U of M)

irregularity 24th cumberland/louisville 29 / 29