Chapter 6

Analytic and Probabilistic Methods

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Problem 6.1.1

Find the asymptotics of

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(1)

the sums from Problem 1.1.6 (1);

(2)

the sums from Problem 1.1.6 (2);

(3)

the sums from Problem 1.1.6 (3);

(4)

the number AnA_n of subsets of the set {1,2,…,n}\left\{ 1, 2, \ldots , n\right\} containing no two consecutive numbers;

(5)

the same as in part (4), for three consecutive numbers.

In the answer you may use the function xP(a,b)x_P(a,b) which, given numbers a,ba, b and a polynomial PP having a unique root on the interval [a,b][a,b], returns this root.

Problem 6.1.2

Find the asymptotics of the maximum number of edges in a graph with nn vertices containing no kk-clique. Here k=kn=o(n)k = k_n = o(n).

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Problem 6.1.3

Prove the following relations, assuming in the asymptotics that n→∞n \to \infty and kk is fixed. (The number \exH(n)\ex_H(n) denotes the maximum number of edges in a graph on nn vertices that does not contain HH as a subgraph; it is defined in Problem 2.7.6.)

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(1)

\exPk(n)≳k−22⋅n\ex_{P_k}(n) \gtrsim \dfrac {k-2}{2} \cdot n.

(2)

\exC2k+1(n)≳n24\ex_{C_{2k+1}}(n) \gtrsim \dfrac {n^2}{4}.

(3)

\exK1,k(n)∼k−12⋅n\ex_{K_{1,k}}(n) \sim \dfrac {k-1}{2} \cdot n.

Remark.
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The famous Erdős--Stone--Simonovits theorem states that for any fixed HH such that χ(H)>2\chi (H) > 2, as n→∞n \to \infty we have \exH(n)∼n22χ(H)−2χ(H)−1\ex_H(n) \sim \dfrac {n^2}{2}\dfrac {\chi (H)-2}{\chi (H)-1}. (For bipartite HH it is only known that \exH(n)=o(n2)\ex_H(n) = o(n^2).) That is, if we forbid a graph to contain some fixed subgraph HH, then the fraction of edges that can be drawn, among all possible edges, is determined by the chromatic number of the graph HH. It is surprising that the chromatic number arises in this problem! (This theorem may not be used in solving Problem 6.1.3 (2).)

Problem 6.1.4
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(1)

n2(2+1n)n=(2+o(1))nn^2\left(2+\dfrac {1}{n}\right)^n = (2+o(1))^n. By definition, this means that there exists a function ψ(n)=o(1)\psi (n) = o(1) such that

n2(2+1n)n=(2+ψ(n))n; n^2\left(2+\frac{1}{n}\right)^n = (2+\psi (n))^n;

or, equivalently,

n2(2+1n)nn−2=o(1). \sqrt[n]{n^2\left(2+\frac{1}{n}\right)^n} - 2 = o(1).
(2)

3n(2+1n)n=(2+o(1))n3^{\sqrt{n}}\left(2+\dfrac {1}{n}\right)^n = (2+o(1))^n.

Problem 6.1.5
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(1)

Find the asymptotics of (n[n/2])n\sqrt[n]{\dbinom {n}{[n/2]}} (cf. Problems 1.4.3 (2) and 6.1.10 (2)).

(2)

2nn+1<(n[n/2])<2n\dfrac {2^n}{n+1} < \dbinom {n}{[n/2]} < 2^n.

(3)

Find the asymptotics of (3mm)m\sqrt[m]{\dbinom {3m}{m}}.

(4)

33m3m+1<22m(3mm)<33m\dfrac {3^{3m}}{3m+1} < 2^{2m}\dbinom {3m}{m} < 3^{3m}.

(5)

(n[an])=(a−a(1−a)a−1+o(1))n\dbinom {n}{[an]} = (a^{-a}(1-a)^{a-1}+o(1))^n, where 0<a<10 < a < 1.

(6)

n![a1n]!…[asn]!=(e−a1ln⁡a1−…−asln⁡as+o(1))n\dfrac {n!}{[a_1n]!\ldots [a_sn]!} = (e^{-a_1\ln a_1 - \ldots - a_s\ln a_s}+o(1))^n, where a1+…+as=1a_1+\ldots +a_s=1, 0<ak<10 < a_k < 1.

Problem 6.1.6
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(1)

Find the asymptotics of ln⁡(n!)\ln (n!).

(2)

Find the asymptotics of n!n\sqrt[n]{n!}.

(3)

nne−n+1⩽n!⩽nn+1e−n+1n^ne^{-n+1} \leqslant n! \leqslant n^{n+1}e^{-n+1};

(4)

n!⩽nne−n+1nn! \leqslant n^ne^{-n+1}\sqrt{n}.

(5)

Stirling’s formula. n!∼nne−n2πnn! \sim n^ne^{-n}\sqrt{2\pi n}, i.e., lim⁡n→∞n!nne−n2πn=1\lim_{n \to \infty } \dfrac {n!}{n^ne^{-n}\sqrt{2\pi n}} = 1.

Problem 6.1.7
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(1)

(nk)<nkk!\dbinom {n}{k} < \dfrac {n^k}{k!}.

(2)

ln⁡(n−1)(n−2)…(n−k)nk=−k(k+1)2n(1+O(kn))\ln \dfrac {(n-1)(n-2)\ldots (n-k)}{n^k} = -\dfrac {k(k+1)}{2n}\left(1+O\left(\dfrac {k}{n}\right)\right) for k=kn<n2k=k_n < \dfrac {n}{2}.

This means that there exists a function ψ(n)=O(kn)\psi (n) = O\left(\dfrac {k}{n}\right) such that

ln⁡(n−1)(n−2)…(n−k)nk=−k(k+1)2n(1+ψ(n)); \ln \frac{(n-1)(n-2)\ldots (n-k)}{n^k} = -\frac{k(k+1)}{2n}(1+\psi (n));

or, equivalently,

−1−2nk(k+1)ln⁡(n−1)(n−2)…(n−k)nk=O(kn). -1 - \frac{2n}{k(k+1)}\ln \frac{(n-1)(n-2)\ldots (n-k)}{n^k} = O\left(\frac{k}{n}\right).
(3)

(nk)=nkk!e−k(k−1)2n+O(k3/n2)\dbinom {n}{k} = \dfrac {n^k}{k!}e^{-\frac{k(k-1)}{2n}+O(k^3/n^2)} for k=kn<n2k=k_n < \dfrac {n}{2}.

(Formulate yourself what e−k(k−1)2n+O(k3/n2)e^{-\frac{k(k-1)}{2n}+O(k^3/n^2)} means here.)

(4)

(nk)∼nkk!\dbinom {n}{k} \sim \dfrac {n^k}{k!} for k=kn=o(n)k=k_n=o(\sqrt{n}).

(5)

(2nn−k)/(2nn)=e−k2n(1+o(1))\dbinom {2n}{n-k}\Big/\dbinom {2n}{n} = e^{-\frac{k^2}{n}(1+o(1))} for k=kn=o(n)k=k_n=o(n).

(Formulate yourself what e−k2n(1+o(1))e^{-\frac{k^2}{n}(1+o(1))} means here.)

Informally, part (4) means that for k≪nk \ll \sqrt{n} the probability of getting exactly kk heads in nn coin tosses is approximately equal to nkk!2−n\dfrac {n^k}{k!}2^{-n}. For the informal remarks on this and the next part, an intuitive understanding of what probability is suffices.

Informally, part (5) means that for k≪nk \ll n the probability PkP_k of getting exactly n−kn-k heads in 2n2n coin tosses is approximately equal to P0e−k2/nP_0e^{-k^2/n} (the normal distribution).

Problem 6.1.8
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(1)

Is it true that the notations eo(n)e^{o(n)} and o(en)o(e^n) are ``equivalent''?

That is, is it true that for any function f:Z→(0,+∞)f : \mathbb {Z} \to (0,+\infty ) the conditions lim⁡n→∞ln⁡f(n)n=0\lim_{n\to \infty }\dfrac {\ln f(n)}{n}=0 and lim⁡n→∞f(n)e−n=0\lim_{n\to \infty }f(n)e^{-n}=0 are equivalent?

(2)

Find functions f,g:Z→(0,+∞)f,g : \mathbb {Z} \to (0,+\infty ) such that f(n)∼g(n)f(n)\sim g(n), but ef(n)≠O(eg(n))e^{f(n)} \neq O(e^{g(n)}).

(3)

Can functions f,g:Z→(0,+∞)f,g : \mathbb {Z} \to (0,+\infty ) simultaneously satisfy the relations f(n)=o(g(n))f(n)=o(g(n)) and g(n)=o(f(n))g(n)=o(f(n))?

(4)

Can functions f,g:Z→(0,+∞)f,g : \mathbb {Z} \to (0,+\infty ) simultaneously satisfy the relations f(n)=O(g(n))f(n)=O(g(n)) and g(n)=O(f(n))g(n)=O(f(n))?

(5)

Do the two relations of part (4) imply that f(n)∼g(n)f(n)\sim g(n)?

Problem 6.1.9
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(1)

Which function grows faster: x(xx)x^{(x^x)} or (x!)(2x)(x!)^{(2^x)}?

That is, find lim⁡x→∞x(xx)(x!)(−2x)\lim_{x\to \infty } x^{(x^x)}(x!)^{(-2^x)}.

(2)

Does there exist a function ψ(n)=o(1)\psi (n) = o(1) such that (2+ψ(n))n×2−ne−n→∞(2+\psi (n))^n \times 2^{-n}e^{-\sqrt{n}} \to \infty?

(As in any mathematical problem, the answer must be justified: give an example of such a function, or prove its existence, or prove that no such function exists.)

Problem 6.1.10

In Problems 6.1.10 (2, 3, 4, 5, 6, 7), unlike the other problems, one may use Stirling’s formula 6.1.6 (5) without proof.

Find the asymptotics of

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(1)

ln⁡(n2n)\ln \dbinom {n^2}{n};

(2)

(n[n/2])\dbinom {n}{[n/2]};

(3)

(n2n)\dbinom {n^2}{n};

(4)

(n[nα])\dbinom {n}{[n^\alpha ]}, α∈(0,1)\alpha \in (0,1);

(5)

(2n−1)!!:=(2n−1)⋅(2n−3)⋅…⋅3⋅1(2n-1)!! := (2n-1)\cdot (2n-3)\cdot \ldots \cdot 3\cdot 1;

(6)

∑k=0n(nk)2\displaystyle \sum_{k=0}^{n}\dbinom {n}{k}^2;

(7)

∑k=0n(nk)4\displaystyle \sum_{k=0}^{n}\dbinom {n}{k}^4.

Problem 6.1.11

Find the asymptotics of the function s=s(n)s = s(n) defined by

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(1)

ss=ns^s = n;

(2)

ss3=ns^{s^3} = n;

(3)

s(n):=max⁡{k:k!⩽n}s(n) := \max \left\{ k : k! \leqslant n\right\};

(4)

s(n):=min⁡{m∈N:(nm)<2(m2)}s(n) := \min \left\{ m \in \mathbb {N} : \dbinom {n}{m} < 2^{\binom {m}{2}}\right\} (cf. Problem 4.1.5);

(5)

s(n):=min⁡{m∈N:2mm>n}s(n) := \min \left\{ m \in \mathbb {N} : \dfrac {2^m}{m} > n\right\} (the function 2m/m2^m/m arises as the complexity of realizing Boolean functions);

(6)

s(n):=min⁡{m∈N:(m[m/2])>n}s(n) := \min \left\{ m \in \mathbb {N} : \dbinom {m}{[m/2]} > n\right\} (cf. Problem 1.4.3 (2)).

Problem 6.1.12

In the answers you may use constants given as sums of series. Find the asymptotics of

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(1)

the number of linear subspaces of Z2n\mathbb {Z}_2^n (see Problem 1.4.7 and the definition preceding it);

(2)

the number of unicyclic graphs with nn vertices (see Problem 2.2.5 (2) and the definition preceding it).

Problem 6.2.1
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(1)

For each of 100 types of work, a company has exactly 8 specialists. An employee may be a specialist in several types of work; the assignment of specialists to types of work is known to the person who schedules days off. Each employee must be given a day off on either Saturday or Sunday. Prove that this can be done so that on both Saturday and Sunday, for each type of work, a specialist in it is at work. (This is Problem 1.6.7.)

(2)

For each of several types of work, a company has exactly 8 specialists. (Now the number of types of work need not be 100.) Each type of work shares specialists with at most 30 other types. Each employee must be given a day off on either Saturday or Sunday. Prove that this can be done so that on both Saturday and Sunday, for each type of work, a specialist in it is at work.

Problem 6.2.2
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(1)

200 students from 10 groups, each of which has 20 students, stand in a circle. Prove that one can choose a leader in each group so that no two leaders stand next to each other.

(2)

The same for 1600 students from 100 groups, each of which has 16 students.

Problem 6.2.3
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(1)

Prove that the first 8 positive integers can be colored in 2 colors so that there is no monochromatic arithmetic progression of length 3.

(2)

Prove that the first 15 million positive integers can be colored in 2 colors so that there is no monochromatic arithmetic progression of length 32.

Problem 6.2.4
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(1)

Prove that for any M∈RM \in \mathbb {R} all real numbers can be colored in 2 colors so that for any x∈Rx \in \mathbb {R} the numbers xx and x+Mx+M have different colors.

(2)

Prove that for any 25 distinct numbers M1,…,M25∈RM_1,\ldots ,M_{25} \in \mathbb {R} all real numbers can be colored in 3 colors so that for any x∈Rx \in \mathbb {R} the numbers x+M1,…,x+M25x+M_1,\ldots ,x+M_{25} include numbers of each of the three colors.

Problem 6.2.5

Every inhabitant of a town is either healthy or sick, and also either rich or poor. Wealth and health are independent, i.e. the proportion of healthy rich inhabitants among the rich equals the proportion of healthy inhabitants among all inhabitants. It is known that there is a rich townsperson and there is a healthy townsperson. Must there be a rich healthy townsperson?

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Problem 6.2.6

Are the following subsets dependent? (We call subsets dependent if they are not independent.)

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(1)

In the set of all squares of a chessboard, the subset of squares in its first three rows and the subset of squares in its last four columns.

(2)

The subsets {1,2}⊂{1,2,3,4}\left\{ 1,2\right\} \subset \left\{ 1,2,3,4\right\} and {1,3}⊂{1,2,3,4}\left\{ 1,3\right\} \subset \left\{ 1,2,3,4\right\}.

(3)

The subsets {1,2}⊂{1,2,3,4,5,6}\left\{ 1,2\right\} \subset \left\{ 1,2,3,4,5,6\right\} and {1,3}⊂{1,2,3,4,5,6}\left\{ 1,3\right\} \subset \left\{ 1,2,3,4,5,6\right\}.

Problem 6.2.7

Are the following subsets of the set of integers from 1 to 105 dependent?

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(1)

The subset of numbers divisible by 5 and the subset of numbers divisible by 7.

(2)

The subset of numbers divisible by 15 and the subset of numbers divisible by 21.

(3)

The subset of numbers divisible by 15 and the subset of numbers divisible by 5.

(4)

The subset of numbers divisible by 10 and the subset of numbers divisible by 7.

Problem 6.2.8

(Cf. the remark after Problem 6.2.1 (2).) Are the following subsets of the set of all colorings of the numbers 1,2,…,4001,2,\ldots ,400 in two colors dependent?

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(1)

The subset of colorings for which {1,2,…,8}\left\{ 1,2,\ldots ,8\right\} is monochromatic and the subset of colorings for which {11,12,…,18}\left\{ 11,12,\ldots ,18\right\} is monochromatic.

(2)

The subset of colorings for which {1,2,…,8}\left\{ 1,2,\ldots ,8\right\} is not monochromatic and the subset of colorings for which {11,12,…,18}\left\{ 11,12,\ldots ,18\right\} is not monochromatic (cf. Problem 6.2.1 (2)).

(3)

The subset of colorings for which {1,2,…,8}\left\{ 1,2,\ldots ,8\right\} is monochromatic and the subset of colorings for which {6,7,…,13}\left\{ 6,7,\ldots ,13\right\} is monochromatic.

Problem 6.2.9

Subsets AA and BB of a finite set are independent if and only if AA and B‾\overline{B} are independent.

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Problem 6.2.10
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(1)

Must there be a rich, healthy, smart citizen if in a city the proportion of rich citizens is greater than 23\dfrac {2}{3}, the proportion of healthy ones is greater than 23\dfrac {2}{3}, and the proportion of smart ones is greater than 23\dfrac {2}{3}?

(2)

The same question if the city has a rich citizen, a healthy citizen and a smart citizen, wealth, health and intelligence are pairwise independent, and the proportion of rich healthy smart citizens among the rich healthy ones is the same as the proportion of smart citizens among all inhabitants. (Together with the pairwise independence condition, the latter is called mutual independence.)

(3)

The same question if more than half of the citizens are rich, more than half are healthy, more than half are smart, wealth and intelligence are independent, and health and intelligence are independent.

Problem 6.2.10 shows that the stronger the condition characterizing the independence of several sets, the smaller the proportion of each set that suffices to guarantee that their intersection is nonempty. Moreover, the most interesting result (6.2.10 (3)) is obtained in the middle'' between the extreme conditions — the complete absence of independence (6.2.10 (1)) and mutual independence (6.2.10 (2)). This often happens: the most useful considerations lie between the extreme'' points of view.

Problem 6.2.11
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(1)

Let A1,A2,A3,A4A_1,A_2,A_3,A_4 be subsets of a 720-element set, each of which has more than 480 elements. If AkA_k and Ak+1A_{k+1} are independent for every k=1,2,3k=1,2,3, then A1∩A2∩A3∩A4≠∅A_1\cap A_2\cap A_3\cap A_4 \neq \emptyset.

(2)

Let n⩾2n \geqslant 2 and let A1,A2,…,AnA_1,A_2,\ldots ,A_n be subsets of a finite set, the proportion of each of which is greater than 1−1n−11-\dfrac {1}{n-1}. If AkA_k and Ak+1A_{k+1} are independent for every k=1,2,…,n−1k=1,2,\ldots ,n-1, then A1∩A2∩…∩An≠∅A_1\cap A_2\cap \ldots \cap A_n \neq \emptyset.

For more on independence, see [KZP].

Problem 6.2.12

Give an example of subsets A,B1,B2A, B_1, B_2 of a finite set that are

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(1)

pairwise independent, but such that AA is not independent of the collection B1,B2B_1, B_2;

(2)

not pairwise independent, but such that AA is independent of the collection B1,B2B_1, B_2.

Problem 6.2.13

Denote by MM the family of all colorings of the set {1,2,…,400}\left\{ 1,2,\ldots ,400\right\} in two colors. For a subset α⊂{1,2,…,400}\alpha \subset \left\{ 1,2,\ldots ,400\right\}, denote by Aα⊂MA_\alpha \subset M the subset of those colorings for which α\alpha is monochromatic. Then A{1,2,…,8}A_{\left\{ 1,2,\ldots ,8\right\} } is independent of the collection {Aα:α⊂{9,10,…,400}}\left\{ A_\alpha : \alpha \subset \left\{ 9,10,\ldots ,400\right\} \right\}. (Cf. the remark after Problem 6.2.1 (2).)

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Problem 6.2.14

The following conditions on subsets A,B1,…,BkA, B_1,\ldots ,B_k are equivalent:

  • AA is independent of the collection B1,…,BkB_1,\ldots ,B_k;

  • A‾\overline{A} is independent of the collection B1,…,BkB_1,\ldots ,B_k;

  • AA is independent of the collection B1‾,…,Bk‾\overline{B_1},\ldots ,\overline{B_k}.

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Problem 6.2.15
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(1)

The Lovász local lemma in symmetric form. Let A1,…,AnA_1,\ldots ,A_n be subsets of a finite set. If for some dd and every kk the proportion of the subset AkA_k is at least 1−14d1-\dfrac {1}{4d} and there exists a collection of at least n−dn-d subsets AjA_j of which AkA_k is independent, then A1∩…∩An≠∅A_1\cap \ldots \cap A_n \neq \emptyset.

(2)

For d>2d > 2, the statement of part (1) remains true if 1−14d1-\dfrac {1}{4d} is replaced by 1−1e(d+1)1-\dfrac {1}{e(d+1)}.

(3)

If ak=1a_k=1 for every k⩽0k\leqslant 0 and for some d⩾2d\geqslant 2 the inequality ak+1⩾ak−ak−d4da_{k+1}\geqslant a_k-\dfrac {a_{k-d}}{4d} holds for every k⩾0k\geqslant 0, then ak>0a_k > 0 for every kk.

Remark.
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Here is a formulation of part (1) in probabilistic language, which is not used in what follows. Let a probability space be given and let A1,…,AnA_1,\ldots ,A_n be events. Suppose that for some dd and every kk the probability of the event AkA_k is at least 1−14d1-\dfrac {1}{4d} and there exists a collection of at least n−dn-d events of which AkA_k is independent. Then the probability of the event A1∩…∩AnA_1\cap \ldots \cap A_n is positive.

Problem 6.2.16

Suppose we are given a number

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(1)

k⩾10k \geqslant 10;

(2)

k=9k=9

and a family of kk-element subsets of a finite set MM. If each element of MM is contained in exactly kk subsets of the family, then there exists a coloring of MM in two colors for which every subset of the family contains elements of both colors. (That is, the chromatic number of any kk-uniform kk-regular hypergraph equals two for k⩾9k\geqslant 9. Cf. Problem 6.2.1 (2).)

Problem 6.2.17

Several subsets of a finite set are chosen. Each of them has at least 3 elements. Each of them intersects at most aia_i of the chosen ii-element subsets. If ∑iai2−i⩽1/8\displaystyle \sum_i a_i2^{-i} \leqslant 1/8, then the elements of the given set can be colored in two colors so that every chosen subset contains elements of both colors.

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Problem 6.2.18
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(1)

For any partition of the vertex set of a cycle of length 16n16n into nn sets of 16 vertices each, one can choose one vertex from each set so that there are no edges between the chosen nn vertices.

(2)

The same for 11n11n vertices.

(3)

In a graph, the degree of every vertex does not exceed Δ\Delta. All the vertices are colored in rr colors. There are at least 2eΔ+12e\Delta +1 vertices of each color. Then one can choose rr vertices of different colors, no two of which are joined by an edge.

Problem 6.2.19
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(1)

Every kk-element arithmetic progression in {1,2,…,n}\left\{ 1,2,\ldots ,n\right\} is intersected by at most k2[nk−1]k^2\left[\dfrac {n}{k-1}\right] other such progressions.

(2)

For every positive integer kk there exists a coloring of the first [2k−3(k−1)k2]\left[\dfrac {2^{k-3}(k-1)}{k^2}\right] positive integers in 2 colors for which there is no monochromatic kk-element arithmetic progression.

(3)

Every kk-element arithmetic progression in {1,2,…,n}\left\{ 1,2,\ldots ,n\right\} is intersected by at most nknk other such progressions.

(4)

For every positive integer kk there exists a coloring of the first [2k−3k]\left[\dfrac {2^{k-3}}{k}\right] positive integers in 2 colors for which there is no monochromatic kk-element arithmetic progression.

Problem 6.2.20
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(1)

If X⊂RX\subset \mathbb {R} is a finite set and m,rm,r are positive integers such that 4rm(m−1)(1−1r)m<14rm(m-1)\left(1-\dfrac {1}{r}\right)^m < 1, then for any mm-element subset M⊂RM\subset \mathbb {R} there exists a coloring of R\mathbb {R} in rr colors such that for any x∈Xx\in X the set x+M:={x+a:a∈M}x+M := \left\{ x+a : a\in M\right\} contains points of each of the rr colors.

(2)

The same for X=ZX=\mathbb {Z}.

(3)

The same for X=RX=\mathbb {R}.

Problem 6.2.21
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(1)

If (n2)(kn−2)+1<2(n2)−1/e\dbinom {n}{2}\dbinom {k}{n-2}+1 < 2^{\binom {n}{2}-1}/e, then R(n,n)>kR(n,n) > k, where R(n,n)R(n,n) is the Ramsey number (Sec. 4.1).

(2)

R(n,n)≳2e−1n2n/2R(n,n) \gtrsim \sqrt{2}e^{-1}n2^{n/2}. (Cf. Problem 4.1.5.)

Problem 6.2.22

There are several colors. To each vertex of some graph a list of at least 10d10d of these colors is assigned, where d>1d>1. For any vertex vv and any color from its list, there are at most dd neighbors of the vertex vv whose lists contain this color. Then one can color each vertex of the graph with some color from its list so that the endpoints of every edge have different colors.

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Problem 6.2.23

In a directed graph, at most Δ\Delta edges enter each vertex and at least δ\delta edges leave each vertex. Then for any positive integer k⩽11−(4δΔ)−1/δk \leqslant \dfrac {1}{1-(4\delta \Delta )^{-1/\delta }} there exists a directed cycle of length divisible by kk.

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Problem 6.2.24

The cells of an n×nn\times n board are colored in several colors. There are at most n−116\dfrac {n-1}{16} cells of each color. Then one can place nn pairwise non-attacking rooks on the board so that they stand on cells of different colors.

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Problem 6.2.25

A CNF formula is a conjunction of a collection of disjunctions of some of the variables x1,…,xnx_1,\ldots ,x_n or their negations. If each factor'' of a CNF formula has exactly $k$ summands'' and each ``factor'' has common variables with at most 2k−22^{k-2} others, then the Boolean function defined by the formula is not identically zero.

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Remark.
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One of the central problems in computer science is the kk-satisfiability problem (kk-SAT problem): for a given CNF formula each disjunction of which contains exactly kk variables, determine whether the Boolean function it defines is identically zero. For k=2k=2 there is a polynomial algorithm solving it. For larger kk, no fast algorithms answering this question are known. Constructing such an algorithm, or proving that none exists, is equivalent to solving the famous open problem ``P≠NP\mathrm{P} \neq \mathrm{NP}''.

Problem 6.2.26
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(1)

Lovász local lemma. *Let A1,…,AnA_1,\ldots ,A_n be subsets of a finite set, J1,…,Jn⊂{1,…,n}J_1,\ldots ,J_n\subset \left\{ 1,\ldots ,n\right\}, and γ1,…,γn∈(0,1)\gamma_1,\ldots ,\gamma_n\in (0,1). Suppose also that for any kk

  • the proportion of the subset AkA_k is at least 1−(1−γk)∏j∉Jkγj1-(1-\gamma_k)\displaystyle \prod_{j\notin J_k}\gamma_j;

  • the set AkA_k is independent of the collection {Aj:j∈Jk}\left\{ A_j : j\in J_k\right\}.

Then the proportion of the intersection ⋂k=1nAk\displaystyle \bigcap_{k=1}^{n}A_k is at least ∏k=1nγk>0\displaystyle \prod_{k=1}^{n}\gamma_k > 0.*

(2)

There exists c>0c>0 such that R(3,n)>cnnR(3,n) > cn\sqrt{n} for every nn.

Remark.
?

Here is a formulation of part (1) in probabilistic language, which is not used in what follows. Let a probability space be given, let A1,…,AnA_1,\ldots ,A_n be events, J1,…,Jn⊂{1,…,n}J_1,\ldots ,J_n\subset \left\{ 1,\ldots ,n\right\}, and γ1,…,γn∈(0,1)\gamma_1,\ldots ,\gamma_n\in (0,1). Suppose that for any kk the probability of the event AkA_k is at least 1−(1−γk)∏j∉Jkγj1-(1-\gamma_k)\displaystyle \prod_{j\notin J_k}\gamma_j and the event AkA_k is independent of the collection {Aj:j∈Jk}\left\{ A_j : j\in J_k\right\}. Then the probability of the event A1∩…∩AnA_1\cap \ldots \cap A_n is at least ∏j=1mγj\displaystyle \prod_{j=1}^{m}\gamma_j.

Remark. Using more involved computations, one derives from the Lovász local lemma that R(3,n)>c1n2/ln⁡2nR(3,n) > c_1n^2/\ln^2 n. Moreover, this is the ``best'' that can be squeezed out of the Lovász local lemma. The inequality R(3,n)>c2n2/ln⁡nR(3,n) > c_2n^2/\ln n (Kim's theorem) is also known. Its proof, instead of the Lovász local lemma, uses quasirandom graphs, sharp concentration inequalities, etc.

Problem 6.3.1

If in a graph G=(V,E)G=(V,E) with nn vertices the minimum vertex degree equals δ\delta, then

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(1)

for any p∈(0,1)p\in (0,1) there exists a set of vertices A⊂VA\subset V such that the union of AA and the set of all vertices not joined to any vertex of AA contains at most np+n(1−p)δ+1np+n(1-p)^{\delta +1} vertices;

(2)

there exists a set of vertices D⊂VD\subset V such that every vertex of V∖DV\setminus D is joined by an edge to some vertex of DD and ∣D∣⩽n1+ln⁡(δ+1)δ+1\left|D\right|\leqslant n\dfrac {1+\ln (\delta +1)}{\delta +1}.

Problem 6.3.2
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(1)

If (km)p(m2)+(kn)(1−p)(n2)<1\dbinom {k}{m}p^{\binom {m}{2}}+\dbinom {k}{n}(1-p)^{\binom {n}{2}} < 1 for some p∈(0,1)p\in (0,1), then R(m,n)>kR(m,n) > k (here R(m,n)R(m,n) are the Ramsey numbers, see Sec. 4.1).

(2)

R(4,n)⩾Ω(n2ln⁡2n)R(4,n) \geqslant \Omega \left(\dfrac {n^2}{\ln^2 n}\right) (we write g⩾Ω(f)g \geqslant \Omega (f) if f=O(g)f=O(g)).

Problem 6.3.3
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(1)

Cherchez la femme. At a Russian-French meeting there were no representatives of other countries. The total amount of money held by the French turned out to be greater than the total amount of money held by the Russians, and the total amount of money held by the women turned out to be greater than the total amount of money held by the men. Must there have been a Frenchwoman at the meeting?

(2)

Banknotes of different denominations and from different countries are packed into two suitcases. The average value of a banknote equals 100 rubles. The total number of banknotes in the left suitcase is greater than in the right one. Must there be a banknote worth at most 200 rubles in the left suitcase? (Cf. Markov's inequality 6.3.9 (1).)

(3)

For any integers l,q>0l,q>0 there exists a graph that contains no cycles of length less than ll and cannot be properly colored in qq colors. (See the definition of a proper coloring in Sec. 3.1.)

Problem 6.3.4

For given nn and pp, the probability of having kk vertices with no edges between them is less than ekln⁡n−pk(k−1)/2e^{k\ln n - pk(k-1)/2}.

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Remark.
?

Fix p∈(0,1)p\in (0,1) and call the probability of a graph (in the Erdős--Rényi model, or probability space) with nn vertices {1,2,…,n}\left\{ 1,2,\ldots ,n\right\} and rr edges the number P(G)=Pp(G):=pr(1−p)n(n−1)2−r\mathbb {P}\left(G\right)=\mathbb {P}_{p}\left(G\right) := p^r(1-p)^{\frac{n(n-1)}{2}-r}. The probability of a family (or, which is the same, a property) of graphs with vertices 1,2,…,n1,2,\ldots ,n is the sum of the probabilities of the graphs belonging to it.

A random variable is a function defined on the set of graphs with vertices 1,2,…,n1,2,\ldots ,n.

For example, the number of edges of a graph is a random variable.

Let a random variable YY take kk distinct values y1,…,yky_1,\ldots ,y_k. Then the mathematical expectation (expectation) of the random variable YY is its ``weighted average''

E[Y]:=∑s=1kysP(Y−1(ys)), \mathbb {E}\left[Y\right] := \sum _{s=1}^{k}y_s\mathbb {P}\left(Y^{-1}(y_s)\right),

where Y−1(ys)Y^{-1}(y_s) is the set of all graphs GG for which Y(G)=ysY(G)=y_s. The latter probability is denoted by P(Y=ys)\mathbb {P}\left(Y=y_s\right).

Problem 6.3.5

For given nn and pp, find the expectation of the number of

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(1)

isolated vertices;

(2)

triangles;

(3)

kk-cliques;

(4)

kk-cliques that are connected components;

(5)

Hamiltonian cycles;

(6)

non-self-intersecting cycles of length kk;

(7)

non-self-intersecting cycles of length kk that are connected components with exactly kk edges;

(8)

trees with kk vertices;

(9)

tree components of a given size kk, i.e., trees with kk vertices that are connected components.

Problem 6.3.6

For a given pp, find the asymptotics (for constant kk and n→∞n\to \infty) of the function E(k)[Y]:=E[Y(Y−1)…(Y−k+1)]\mathbb {E}_{(k)}\left[Y\right] := \mathbb {E}\left[Y(Y-1)\ldots (Y-k+1)\right] (i.e., of the kk-th factorial moment), if YY is the number of isolated vertices.

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Problem 6.3.7

For given nn and pp, find the variance of the number of

?
(1)

isolated vertices;

(2)

triangles.

Problem 6.3.8

Prove that for any random variables XX and YY the following properties hold:

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(1)

E[X+Y]=E[X]+E[Y]\mathbb {E}\left[X+Y\right] = \mathbb {E}\left[X\right]+\mathbb {E}\left[Y\right];

(2)

Var⁡[X+Y]=Var⁡[X]+Var⁡[Y]\operatorname {Var}\left[X+Y\right] = \operatorname {Var}\left[X\right]+\operatorname {Var}\left[Y\right] if XX and YY are independent (i.e., for any x,yx,y we have P(X=x,Y=y)=P(X=x)⋅P(Y=y)\mathbb {P}\left(X=x,Y=y\right)=\mathbb {P}\left(X=x\right)\cdot \mathbb {P}\left(Y=y\right)).

Problem 6.3.9

Let XX be a random variable (as defined before Problem 6.3.4) and a>0a>0.

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(1)

Markov’s inequality. P(∣X∣>a)⩽E[∣X∣]/a\mathbb {P}\left(\left|X\right|>a\right) \leqslant \mathbb {E}\left[\left|X\right|\right]/a. (Cf. Problem 6.3.3 (1).)

(2)

Chebyshev’s inequality. P(∣X−E[X]∣>a)⩽Var⁡[X]/a2\mathbb {P}\left(\left|X-\mathbb {E}\left[X\right]\right|>a\right) \leqslant \operatorname {Var}\left[X\right]/a^2.

An event AnA_n occurs asymptotically almost surely (or with asymptotic probability 1) with respect to a sequence f(n)f(n) if Pf(n)(An)→1\mathbb {P}_{f(n)}\left(A_n\right) \to 1. A commonly used shorthand: for p(n)=f(n)p(n)=f(n) the event AnA_n occurs a.a.s. (formally, this phrase makes no sense, since it means if $p(n)=f(n)$, then the event $A_n$ occurs a.a.s.'', while without specifying the sequence $f(n)$ the phrase the event AnA_n occurs a.a.s.'' cannot be properly defined).

Recall that here nn is the number of vertices of the graph.

Problem 6.3.10

For p(n)=1/(2n)p(n)=1/(2n)

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(1)

a.a.s.there are more than n/2n/2 isolated vertices;

(2)

for some C>0C>0, a.a.s.every connected component has fewer than Cln⁡nC\ln n vertices (specialists say: fewer than O(ln⁡n)O(\ln n) vertices);

(3)

a.a.s.every connected component is a tree or a unicyclic graph;

(4)

for some C>0C>0, a.a.s.there are fewer than CC unicyclic components.

Problem 6.3.11
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(1)

For p(n)=o(n−3/2)p(n)=o(n^{-3/2}), a.a.s.the edges are pairwise disjoint.

(2)

For p=p(n)=o(n−3/2)p=p(n)=o(n^{-3/2}) and pn2→∞pn^2\to \infty, there exists a function r=r(n)=o(pn2)r=r(n)=o(pn^2) such that a.a.s.the number of vertices of degree 1 is greater than pn2−rpn^2-r and less than pn2+rpn^2+r, and the degrees of all the other vertices are zero.

Problem 6.3.12

If c>1c>1 (0<c<10<c<1), then for p(n)=cln⁡n/np(n)=c\ln n/n a.a.s.the random graph is connected (disconnected).

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(1)
(2)
(3)
(4)
(5)
Remark.
?

We state a result from [B, p.100, Theorem 5.4]. Let pn=p(n)p_n=p(n). For k⩾2k\geqslant 2 denote by Tk=Tk,pnT_k=T_{k,p_n} the number of connected components of the random graph that are trees with kk vertices.

(1)

If pn=o(n−k/(k−1))p_n=o(n^{-k/(k-1)}), then a.a.s.Tk=0T_k=0.

(2)

If lim⁡n→∞pnnk/(k−1)=c>0\lim_{n\to \infty }p_nn^{k/(k-1)}=c>0, then the sequence of random variables Tk=Tk,pnT_k=T_{k,p_n} converges as n→∞n\to \infty to a random variable having the Poisson distribution with parameter λ:=ck−1kk−2/k!\lambda := c^{k-1}k^{k-2}/k!, i.e., for any s∈Zs\in \mathbb {Z}, s⩾0s\geqslant 0, we have lim⁡n→∞P(Tk=s)=λss!e−λ\lim_{n\to \infty }\mathbb {P}\left(T_k=s\right)=\dfrac {\lambda^s}{s!}e^{-\lambda }.

(3)

If nk/(k−1)=o(pn)n^{k/(k-1)}=o(p_n) and lim⁡n→∞(pnkn−ln⁡n−(k−1)ln⁡ln⁡n)=−∞\lim_{n\to \infty }(p_nkn-\ln n-(k-1)\ln \ln n)=-\infty, then lim⁡n→∞P(Tk⩾L)=1\lim_{n\to \infty }\mathbb {P}\left(T_k\geqslant L\right)=1 for any L>0L>0.

(4)

If lim⁡n→∞(pnkn−ln⁡n−(k−1)ln⁡ln⁡n)=x∈R\lim_{n\to \infty }(p_nkn-\ln n-(k-1)\ln \ln n)=x\in \mathbb {R}, then the sequence of random variables Tk=Tk,pnT_k=T_{k,p_n} converges as n→∞n\to \infty to a random variable having the Poisson distribution with parameter λ:=e−x/(k⋅k!)\lambda := e^{-x}/(k\cdot k!).

(5)

If lim⁡n→∞(pnkn−ln⁡n−(k−1)ln⁡ln⁡n)=+∞\lim_{n\to \infty }(p_nkn-\ln n-(k-1)\ln \ln n)=+\infty, then a.a.s.Tk=0T_k=0.

Problem 6.3.13
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(1)

Find at least one function p∗(n)p^*(n) such that

  • for p(n)/p∗(n)→0p(n)/p^*(n)\to 0, a.a.s.the graph contains no triangle,

  • for p(n)/p∗(n)→+∞p(n)/p^*(n)\to +\infty, a.a.s.the graph contains a triangle.

(2)

The same with the triangle replaced by a subgraph isomorphic to K4K_4.

Remark.
?

Such a function p∗p^* is called a threshold probability. A threshold probability exists for every monotone family of graphs. A monotonically increasing (decreasing) family of graphs is a family of graphs that, together with each graph, contains every supergraph (subgraph) of it.

Problem 6.3.14

The chromatic number of the graph is a.a.s.at most

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(1)

one for p(n)=o(1/n2)p(n)=o(1/n^2);

(2)

two for p(n)=o(1/n)p(n)=o(1/n);

(3)

three for p(n)=c/np(n)=c/n, where c<1c<1.

Problem 6.3.15
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(1)

The greedy coloring algorithm (see Problem 3.2.3: the vertices of the graph are processed in some order, and each of them is assigned the smallest color that does not occur among its already colored neighbors), for any positive ε\varepsilon, a.a.s.(for p(n)=1/2p(n)=1/2) errs by a factor of at most 2+ε2+\varepsilon.

(2)

For any ε,δ>0\varepsilon ,\delta >0 there exists a sequence GnG_n of graphs with nn vertices such that, for a random numbering of the vertices of the graph GnG_n (i.e., with the probability of each numbering equal to 1/21/2), the probability that the ratio of the number of colors in the greedy coloring to χ(Gn)\chi (G_n) is greater than n1−εn^{1-\varepsilon } is greater than δ\delta. (In other words, on the one hand, for almost every graph the greedy coloring is good in any numbering, but, on the other hand, there are graphs for which, almost no matter how you number them, the result is still rubbish!)

Remark.
?

See [R3, R4, R5] for more details. In particular, the following results are proved in [R4].

Bollobás’s first theorem. There exists a sequence fn=o(n2log⁡2n)f_n = o\left(\dfrac {n}{2\log_2 n}\right) for which, for p(n)=1/2p(n) = 1/2, a.a.s.∣χ(G)−n2log⁡2n∣<fn\left|\chi (G) - \dfrac {n}{2\log_2 n}\right| < f_n. (This theorem generalizes to practically all values of pp [JLR].)

Bollobás’s second theorem. For any α>2/3\alpha > 2/3 there exist sequences ana_n and bnb_n for which, for p(n)=n−αp(n) = n^{-\alpha }, a.a.s.χ(G)∈{an,bn}\chi (G) \in \left\{ a_n, b_n\right\}. (In this theorem, for some α\alpha the sequences ana_n and bnb_n can be chosen so that lim⁡n→∞an=lim⁡n→∞bn=∞\lim_{n \to \infty } a_n = \lim_{n \to \infty } b_n = \infty.)