Bonyolultsági osztályok paritási érvelések alapján · Abstract In the ﬁrst part of the thesis, we give an overview of complexity classes of search problems. We ﬁrst study

Eötvös Loránd Tudományegyetem

Természettudományi Kar

Varga László

Bonyolultsági osztályokparitási érvelések

alapján

BSc Szakdolgozat

Témavezetők:Végh László

Georgia Institute of Technology, AtlantaKirály Tamás

Operációkutatási Tanszék, ELTE, Budapest

Budapest, 2012.

Eötvös Loránd University

Faculty of Science

László Varga

Complexity classesbased on

Parity argument

BSc Thesis

Supervisor:László Végh

Georgia Institute of Technology, AtlantaTamás Király

Department of Operations Research, ELTE, Budapest

Budapest, 2012.

Bevezető

A dolgozat első részében röviden áttekintjük a keresési problémák alapvető bonyolult-

sági osztályait, majd megvizsgáljuk az úgynevezett totális keresési problémák osztályát

(TFNP), amelybe olyan feladatok tartoznak, melyekre minden bemenethez létezik tanú,

a megoldás létezése garantált. Egy elemi példa egy adott szám prímfaktorizációja.

Papadimitriou [1] cikkében definiálta a TFNP osztály néhány részosztályát aszerint,

hogy milyen elemi érvelésen múlik a létezés bizonyítása. A dolgozatban elsősorban a PPA

osztályt vizsgáljuk, amelyet a paritási érvelések határoznak meg.

A bevezető után megvizsgáljuk az F2 feletti polinomok gyökkeresésének bonyolult-

ságát. Papadimitriou belátta, hogy a Chévalley tétele alapján adódó természetes keresési

probléma PPA-ban van. A dolgozat egyik fő eredményeként megmutatjuk, hogy a Kom-

binatorikus Nullstellensatzra is hasonló állítás igaz.

Végül, bizonyítjuk Olson tételének egy érdekes általánosítását, illetve megvizsgáljuk

ennek a tételnek egy gráfelméleti alkalmazását, amelyet először Alon és társai tanulmá-

nyoztak [4] cikkükben. Továbbá, belátjuk, hogy az állításoknak megfelelő keresési prob-

lémák PPA-ban vannak és mutatunk egy polinomiális algoritmust egy speciális esetben.

Köszönetnyilvánítás

Köszönettel tartozom témavezetőmnek, Végh Lászlónak az érdekes témajavaslatáért és

rengeteg segítségéért, amelyet nyújtott ezen dolgozat elkészítése során. Továbbá köszönöm

Király Tamásnak a hasznos javaslatait, illetve szüleimnek és barátnőmnek a támogatásukat

az elmúlt hónapokban.

Abstract

In the first part of the thesis, we give an overview of complexity classes of search

problems. We first study the class TFNP, the class of total search problems, where the

existence of a polynomial time checkable witness is guaranteed for every input. A funda-

mental example is finding a prime factorization of a given number.

In [1], Papadimitriou introduced several subclasses of TFNP, that correspond to certain

argument styles for proving existence. We mainly focus on the class PPA (Polynomial

Parity Argument), but we also mention the classes PPAD, PPADS, PPP and PLS.

After this introduction, we study the complexity of finding roots of polynomials over

F2. Papadimitriou showed that Chévalley’s theorem is contained in the class PPA. We

present a proof showing that the Combinatorial Nullstellensatz by Alon is also contained

in PPA.

Then, we present an extension of Olson’s theorem, and its application in graph theory

via Chévalley’s theorem which was first studied in [4]. The corresponding search problems

belong to PPA and we show a polynomial algorithm for a special case.

Acknowledgements

Many thanks to my supervisor, László Végh to introduce these questions and this area

of combinatorics and his help to answer my questions, especially, despite the time-lag. Also,

thanks to Tamás Király for his useful suggestions and to my girlfriend and my parents for

their support.

Contents

1 Introduction 6

1.1 Total Search Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.1 Some problems in TFNP . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.2 General facts about TFNP . . . . . . . . . . . . . . . . . . . . . . . 10

1.2 Parity Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.1 On undirected graphs, the class PPA . . . . . . . . . . . . . . . . . . 12

1.2.2 On directed graphs, the class PPAD and the class PPADS . . . . . . 17

1.3 Other subclasses of TFNP . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.4 The relative complexity of these classes . . . . . . . . . . . . . . . . . . . . 19

2 Complexity of finding roots over F2 21

2.1 Complexity in general . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2 Complexity of Combinatorial Nullstellensatz . . . . . . . . . . . . . . . . . . 23

2.3 A consequence of Combinatorial Nullstellensatz . . . . . . . . . . . . . . . . 26

3 Subgraphs modulo prime power 27

3.1 A generalisation of Olson’s theorem . . . . . . . . . . . . . . . . . . . . . . . 28

3.2 Inequalities for the quantity F (d,Q) . . . . . . . . . . . . . . . . . . . . . . 31

3.3 pd-divisible subgraphs and Q-subgraphs . . . . . . . . . . . . . . . . . . . . 34

3.4 Complexity aspects of subgraphs . . . . . . . . . . . . . . . . . . . . . . . . 36

3.4.1 Complexity of Olson’s problem . . . . . . . . . . . . . . . . . . . . . 36

3.4.2 Complexity of subgraphs modulo 2d . . . . . . . . . . . . . . . . . . 37

3.4.3 A special case: even subgraph of a hypergraph . . . . . . . . . . . . 38

Chapter 1

Introduction

The complexity of decision problems is a well-studied area of complexity theory. Let Σ

be a finite alphabet containing the empty character ∗, and Σ0 = Σ− ∗. These decision

problems ask whether a word x ∈ Σ∗0 is in a language L ⊆ Σ∗0. It often means that a

particular object exists in the input or not. If the question x ∈ L can be decided for every

x ∈ Σ∗0 in polynomial time with a deterministic Turing machine, it is said to be in P. If it

can be decided polynomial time with a non-deterministic Turing machine, it is said to be

in NP. This is equivalent to that there exists a polynomial witness. We assume familiarity

with the fundaments of complexity theory. For an introduction, we refer the reader to [5].

In a slightly different, but equivalent approach, let R ⊆ Σ∗0×Σ∗0 be a polynomial-time

recognizable and polynomially balanced relation. Polynomial-time recognizibilty means

that there is a polynomial-time algorithm which determines for an x, y ∈ Σ∗0 whether

(x, y) ∈ R. Furthermore, polynomially balancedness means that the size of y can be

bounded by a polynomial of |x|.

This relation R defines a decision problem ΥR as follows: given an x ∈ Σ∗0, if there exists

y ∈ Σ∗0 such that (x, y) ∈ R, reply "yes", and reply "no" otherwise. This y corresponds to

the witness for the input x. The class of all such problems coincides with NP. The class

of such problems in NP which can be solved in polynomial time coincides with P.

However, not only to determine the existence of a particular object, but to find it might

become more complex. These problems called search or function problems can be defined

and classified similarly to decision problems.

6

CHAPTER 1. INTRODUCTION

The relation R defines a computational search problem ΠR as follows: given an x ∈ Σ∗0,

find any y ∈ Σ∗0 such that (x, y) ∈ R, if such a y exists, and reply "no" otherwise. The

class of all such problems is denoted FNP. The set of such problems in FNP which can be

solved in polynomial time is called FP.

In addition, a search problem ΠR is polynomially reducible to ΠS if there exist poly-

nomially computable functions, denoted by f and g, such that for any input x ∈ Σ∗0,

(x, g(y)) ∈ R if and only if (f(x), y) ∈ S.

It means that for any input to ΠR, one can compute an input to ΠS such that if there

is a solution for it, it is easy to compute a solution for the original input to ΠR.

1.1 Total Search Problems

In [6], Papadimitriou defined a very interesting subclass of FNP: the class of total

search problems. A relation R is called total, if for every x ∈ Σ∗0 there is always a y ∈ Σ∗0such that (x, y) ∈ R. The set of such problems in FNP which are defined by a total relation

is called TFNP.

That is, the decision problem corresponding to a total relation is meaningless, it is

known that there are solutions, we only have to find one of them.

1.1.1 Some problems in TFNP

There are many interesting problems in TFNP. A simple, famous example for a TFNP

problem is based on the well-known unique prime factorization theorem, also known as

the fundamental theorem of arithmetic.

Theorem 1.1 (Unique Prime Factorization) Any integer greater than 1 can be writ-

ten as a product of prime numbers. This decomposition is unique up to ordering the factors.

It is an important question in applications how one can find this unique factorization.

At all, how one can determine that N is prime or composite? Surprisingly, it was shown

only in 2002 in [7] that this decision problem can be solved in polynomial time in a

deterministic way. But what about factorization?

Factoring. Given an integer N , find prime numbers p1, p2, . . . , pk and α1, α2, . . . , αk

positive integers such that N = pα11 · p

α22 · · · · · p

αkk .

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Proposition 1.2 Factoring is in TFNP.

Proof. If N and its prime factorization is given, one can check in polynomial time if the

numbers in factorization are prime and the product equals to N . So Factoring is in

FNP. Due to the unique prime factorization theorem, for every N there is a factorization,

so the problem is total and this shows the proposition.

In contrast, no polynomial time algorithm is known for the factorization problem, and

due to its difficulty, it is applied in cyptography, e.g. in RSA. However, it is not known to

be FNP-comlpete problem either.

Another well-known example in TFNP is based on Sperner’s theorem.

Theorem 1.3 (Sperner) We are given a triangle RGB and a triangulation T of it. The

vertices of the triangulation are coloured by three colours (red, green and blue) in such a

way that

(i) R, G and B are coloured respestively with red, green and blue.

(ii) Each vertex on a side of triangle is coloured with one of the colours of the endpoints

of the side.

Then the number of triangles in T whose three vertices are coloured with the three

different colours is odd. Specially, there exists such a triangle.

Proof. We call a triangle of T colourful if its three

vertices are coloured with the three different colours,

and an edge is nice if its endpoints are coloured with

different colours.

It is easy to see that a colourful triangle has 3

nice edges, and a triangle which is not colourful has

0 or 2 nice edges. That is, colourful triangles have

odd, and the others have even nice edges. Obviously,

∑triangle t∈T

number of nice edges of t =∑

nice edge enumber of triangles containing edge e

On the one hand, on each side there is on odd number of nice edges, because the

endpoints of the side have different colours. Consequently, the number of nice edges on

the three sides is odd, so the above expression is odd.

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On the other hand, the number of colourful triangles has the same parity as the above

expressions, because a triangle is colourful if and only if it has an odd number of nice

edges.

Thus, the number of colourful triangles is odd.

This theorem implies a natural computational problem. Given a triangle RGB with

a triangulation and a proper colouring, find a colourful triangle. However, in this form,

there is a polynomial algorithm, because the number of triangles is polynomially bounded

in the number of vertices of the triangulation, hence, in the size of input.

The computational problem with the standard triangulation is studied, for instance,

in [1]. In the standard triangulation each vertex can be described by a triple (r, g, b) where

r, g, b ≥ 0 and r + g + b = n. (Each side of RGB triangle is divided into n segments.)

Sperner 2D. We are given n and a polynomial

algorithm (in the size of n: logn) M which com-

putes the colour of the point described by p =

(r, g, b). (M computes a right coluring if M(p) ∈

red, green, blue, r = 0 implies M(p) 6= red,

g = 0 implies M(p) 6= green and b = 0 im-

plies M(p) 6= blue.) Find such a, b, c points that

they are neighbours and M(a),M(b),M(c) =

red, green, blue.

Due to Sperner’s theorem and polynomiality of

M , the below proposition immediately follows.

Proposition 1.4 Sperner 2D is in TFNP.

We close this section with an algebraic problem in TFNP which is based on Chévalley’s

theorem, usually also called Chévalley-Warning theorem.

Theorem 1.5 (Chévalley) Let F be a finite field with characteristic p. Let p1, p2, . . . , pm

be polynomials in n variables over F. Suppose that∑mi=1 deg(pi) < n. Then, the number

of common solutions of the polynomial equation system pi(x1, . . . , xn) = 0 (i = 1 . . .m) is

divisible by p. In particular, if there is a solution, there exists another.

Proof. Denote (x1, . . . , xn) by x and (x1, . . . , xj−1, xj+1, . . . , xn) by x−j . Let

9


Q =∑

x∈Fn

(1− p1(x)p−1) · (1− p2(x)p−1) · · · · · (1− pm(x)p−1)

It is easy to see that Q is the number of solutions modulo p, because due to Fermat’s

theorem, pi(x)p−1 = 1 if and only if pi(x)p−1 6= 0. Therefore, each term of the sum is

precisely 1 if x is a solution, and 0 otherwise.

However, after decomposition, Q is the sum of terms q, where each q is a product of

some term from the pp−1i ’s, so q(x) =

∏nj=1 x

tjj . Because of the condition

∑mi=1 deg(pi) < n,

there exists such an xj variable that occurs in this term q with tj < p− 1. Hence,

∑x∈Fn

q(x) =∑xj∈F

∑x−j=∈Fn−1

q(x) =∑xj∈F

xtjj ·

∑x−j∈Fn−1

xt11 · . . . xtj−1j−1 · x

tj+1j+1 · · · · · x

tnn

Due to the fundamental theorem of symmetric polynomials, it is easy to see that∑xj∈F x

tjj is zero modulo p for tj < p− 1, so

∑x∈Fn q(x) = 0 and it shows that Q is zero

modulo p and then, it shows the theorem.

This theorem implies the following computational problem and that it belongs to

TFNP.

Chévalley over F. Given polynomials p1, p2, . . . , pm in n variables over the finite field

F with∑mi=1 deg(pi) < n. Also, we are given a root (c1, c2, . . . , cn) ∈ F of the equation

system pi(x) = 0 (i = 1, . . . ,m). Find another root.

Proposition 1.6 Chévalley over F is in TFNP.

1.1.2 General facts about TFNP

Similarly to FP and FNP, one can define F(NP ∩ coNP) as the set of functional variants

of NP ∩ coNP problems. The connection between these problems and problems in TFNP

is analysed in [6], and the following two propositions were proved.

Proposition 1.7 TFNP coincides with F(NP ∩ coNP).

Proof. F(NP ∩ coNP) ⊆ TFNP: if a search problem is in F(NP ∩ coNP), there exists

two polynomially balanced relation Ryes and Rno such that for each x either there is an

y with (x, y) ∈ Ryes (in the case when the answer for the corresponding decision problem

with input x is yes) or there is a z with (x, z) ∈ Rno (in the case when the answer is no).

10


So R = Ryes ∪Rno is a total, polynomially balanced relation, so the problem with R is in

TFNP.

TFNP ⊆ F(NP ∩ coNP): Let ΠR be a problem in TFNP which is defined by a total

relation R. R is also polynomially balanced, so there exists two polynomially balanced

relation Ryes := R and Rno := ∅ such that for each x either there is an y with (x, y) ∈ Ryesor there is a z with (x, z) ∈ Rno. The latter case never happens beacuse of totality of R.

So ΠR is also in F(NP ∩ coNP).

Proposition 1.8 There is an FNP-complete problem in TFNP if and only if NP=coNP.

Proof. ⇐: Due to above proposition, if NP=coNP, then TFNP=F(NP∩coNP)=FNP, so

every problem in FNP is also in TFNP, specially, a FNP-complete problem also belongs

to TFNP.

⇒: Suppose that there is a problem ΠR in TFNP which is FNP-complete. So any

problem in FNP is polynomially reducible to ΠR which is in TFNP. Hence, for any problem

in FNP and for such an input that there does not exist a solution, there is a certificate of

non-existence. And this implies that NP = coNP .

1.2 Parity Arguments

In mathematics, the truth of a theorem often depends on the truth of an elementary

fact, it is based on a trivial statement.

A well-known Nash theorem is an outstanding example. This theorem from game theory

depends on Brouwer’s fixpoint theorem and it is based on Sperner’s theorem. The truth

of the Sperner’s theorem now depends on an elementary parity observation. A detailed

proof of the Nash theorem can be found in [8].

There are many such elementary facts in combinatorics and in graph theory. For in-

stance, the following facts are mentioned in [1]:

Local Search Every finite directed acyclic graph has a sink.

Parity Argument Every finite graph has an even number of odd-degree nodes.

Even Leaves Argument Every graph of maximum degree two has an even number of

leaves.

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Pigeonhole Principle There are some objects and some classes and each object belongs

to exactly one class. If the number of objects is more than the number of classes,

there exists a class which has at least two objects.

In [1], Papadimitriou presents his idea about these observations: classify the problems

by that elementary facts which they depend on. In terms of complexity theory, which are

the problems that can be solved in polynomial time, if the natural problem corresponding

to the elementary fact can be solved in polynomial time. We focus on complexity classes

based on the parity argument, however, we mention other similar classes.

1.2.1 On undirected graphs, the class PPA

In [1], Papadimitriou defines the class PPA (Polynomial Parity Argument). We define

it in a slightly different way, but with the same thoughts. First, we present the basic

problems of PPA.

A problem is in PPA if and only if it is polynomially reducible to the following problem.

End of the Line with Degree at most Two. Given an integer n and an undirected

finite graph G = (V,E) in which each node has at most two incident edges. The graph is

described in the following way: it can be assumed that each node has an unique code from

Σn, that is V ⊆ Σn. Furthermore, there is a polynomial algorithm (in n) which computes

for a node its neighbours: for a v ∈ V , it lists the neighbours of v. At last, there is a node

ε which has exactly one incident edge. Find another node which has exactly one incident

edge.

Note that the input is not a complete list of the nodes and edges, so the number of

nodes in the graph may be exponentially large in n. The name of the problem refers to

its trivial solution: from the given node ε, we can start and move to the next node, the

neighbour of the actual node. At last, at the end of the line, we find another node which

has exactly one incident edge. Unfortunately, this trivial algorithm terminates only after

exponential time in the worst case.

However, the same problem without the degree condition is equivalent to End of the

Line with Degree at most Two, and usually it can be used more easily. First, we

study the problem with polynomially bounded degrees.

End of the Line with Polynomially Bounded Degrees. Given an integer n and

an undirected finite graph G = (V,E) in which the degree of each node is polynomially

12


bounded (in n). The graph is described in the following way: it can be assumed that each

node has an unique code from Σn, that is V ⊆ Σn. Furthermore, there is a polynomial

algorithm (in n) which computes for a node its neighbours. At last, there is a node ε which

has an odd number of incident edges. Find another such node.

Proposition 1.9 (Chessplayer algorithm) End of the Line with Polynomially

Bounded Degrees is polynomially reducible to End of the Line with Degree at

most Two. Consequently, it is in PPA.

Proof. Suppose that a graph G is given in which the degree of each node is polynomially

bounded. We construct a new graph Γ in which each node has at most two edges, and the

odd-degree nodes precisely correspond to the odd degree-nodes of G.

For a node v with k edges, dk2e nodes are constructed in the following way. These nodes

are denoted by (v, i) for i = 1 . . . , dk2e and (v, i) is connected to (w, j) if and only if v and

w are connected in the original graph G and this edge is the 2ith or (2i + 1)th incident

edge to v and the 2jth or (2j+ 1)th incident edge to w. Obviously, each degree is at most

two and a node v is odd-degree with degree k if and only if (v, k) has only one edge. And

this construction shows the reduction.

Definition 1.10 φ is called a pairing function, if φ : V ×V → V such that for each node

v ∈ V it pairs up neighbours of v. That is, if v is an even-degree node, then for every node

w with v, w ∈ E there is another node w′ 6= v such that v, w′ ∈ E, φ(v, w) = w′ and

φ(v, w′) = w. If v has odd degree, there exists exactly one node w such that v, w ∈ E

and φ(v, w) = w.

End of the Line. Given an integer n and an undirected finite graph G = (V,E).

The graph is described in the following way: it can be assumed that each node has an

unique code from Σn, that is V ⊆ Σn. Furthermore, there is a polynomial edge recognition

algorithm and a polynomial algorithm which computes φ(v, w) for a node v and one of its

neighbours w. At last, there is a node ε which has odd edges and there is a node δ which

shows it: φ(ε, δ) = δ. Find another node which has odd edges.

Proposition 1.11 End of the Line is polynomially reducible End of the Line with

Degree at most Two. Consequently, it is in PPA.

Proof. Suppose that given a graph G and there is a polynomial edge recognition algorithm

and a pairing function which can be computed in polynomial time. We construct a new

13


graph Γ in which each degree is at most two, and the odd-degree nodes precisely correspond

to the odd degree-nodes of G.

The nodes of the new graph Γ correspond to the pairs (v, w) where v, w ∈ V and two

nodes e and f of Γ are connected if and only if e = (v, w), f = (v, w′) and φ(v, w) =

w′, φ(v, w′) = w or v, w ∈ E and e = (v, w), f = (w, v).

The nodes of Γ have at most two neighbours and the nodes with degree one correspond

to edges which do not have neighbours and it shows that the node of the original graph

has odd-degree. Then, the neighbours of a node of Γ can be computed easily due to the

pairing function, and (ε, δ) would be the standard leaf. This shows the reduction.

Usually, the last version of the problem is used to prove belonging to PPA.

Examples for PPA

In [1], Papadimitriou proved that some problems are in PPA.

First, we present his proof for Chévalley MOD 2, although, another proof will be

shown in 2.5 in a different way, reducing to Combinatorial Nullstellensatz MOD

2.

Proposition 1.12 Chévalley MOD 2 is polynomially reducible to End of the Line.

Consequently, Chévalley MOD 2 is in PPA.

The constructed graph Γ for the equation system p1(x) = x3 + x1x2 + x1x3 + x3x4, p2(x) = 1 + x2 + x4

and the path whose endpoint corresponds to another solution for the given root (0, 1, 0, 0).

Proof. Let Q(x) =∏mi=1(1 + pi(x)). An s is a solution of the equation system if and

only if Q(s) = 1. We construct a biparite graph Γ in which the odd-degree node precisely

corresponds to the common solution of the equation system.

14


On the left side of Γ there are the terms of Q. Each term can be represented by an

m-tuple (a1, a2, . . . , am) such that the term is the product of the aith monomials from

(1 + pi(x)).

On the other side there are all the vectors in 0, 1n. A vector x is connected to the

term t if and only if t(x) = 1.

Obviously, Q(x) = 1 if and only if the node corresponding to x is odd-degree. On the

other hand, each node has even-degree, because the degree of Q is less than n, so there is

a variable that does not appear in the term t of Q. So the odd-degree nodes of Γ precisely

correspond to the common solution vectors of the equation system.

However, the degree of the nodes may be exponentially large, so we must present a

pairing function. For a term t, let ϕ(t, (x1, x2, . . . , xn)) = (x1, x2, . . . , 1−xj , . . . , xn) where

j is the smallest index such that xj does not appear in t. On the other hand, if Q is zero

at x, that is, x has even-degree, there is a smallest i such that 1 + pi(x) = 0. Then we can

pair up the monomials of 1+pi(x) = 0 which are one at x. Denote this pairing by ϕ. Then

let the mate of t represented by (a1, a2, . . . , am) at x be the term which is represented by

(a1, a2, . . . , ϕ(ai), . . . , am).

If Q(x) = 1, we can also pair up the monomials of 1 + p1(x) which are one at x, one

does not have a mate. Denote this pairing by ϕ. Then let the mate of t represented by

(a1, a2, . . . , am) at x be the term which is represented by (a1, a2, . . . , ϕ(ai), . . . , am).Obviously,

this is a pairing function and it completes the proof.

Another interesting problem which was studied by Papadimitriou is the Another

Hamiltonian Path. Lovász proved that number of Hamiltionian paths in a graph and

its complement is always even. It is true for directed and undirected graphs. We present

Papadimitiou’s proof for PPA, that also shows Lovász’s theorem.

Another Hamiltonian Path. Given a (directed or undirected) graph G and a Hamil-

tonian path in it. Find another Hamiltonian path in G or in its complement G.

Proposition 1.13 Another Hamiltonian Path is polynomially reducible to End of

the Line. Consequently, Another Hamiltonian Path is in PPA.

Proof. Denote the number of vertices of G by n.

We can construct a biparite graph Γ whose odd-degree nodes precisely correspond to

the Hamiltonian path of G or G. On the one side of Γ there are the Hamiltonian path of

15


the complete graph on vertices of G. These are actually the permutations of the vertices.

On the other side of Γ there are the subgraphs of G (a subset of edges of G).

A subgraph s is connected to the Hamiltonian path p if and only if s is a proper

subgraph of p. Note that most subgraphs do not have any neighbours.

The path p is adjacent to such nodes on the other side that correspond to the subgraphs

of p ∩G. It is even (e. g. a power of 2), except p ∩G = ∅ or |p ∩G| = n− 1 and there are

only 2n−1 − 1 proper subgraph of p. In the first case, the path p is a Hamiltonian path of

G and in the second case it is a Hamiltonian path of G. So the odd-degree nodes of this

side precisely correspond to the Hamiltonian path of G or G.

On the other hand, a subgraph s is connected to its all possible proper completions

to a Hamiltonian path. If s does not consist of vertex-disjoint paths, it does not have

neighbours in Γ. If subgraph s has ν edges and s consists of µ point-disjoint paths, but s

is not a Hamilton path, s has (n−ν)!2 · 2µ completions. (s has n− ν components and we can

order these components in (n− ν)! different ways and each component can be chosen by

forward or backward but in this way we count each Hamiltonian path twice.) So the node

of corresponding to s has (n−ν)!2 · 2µ neighbours. It is easy to check that this expression

is even, except the case µ = 1, ν = n− 1 (it corresponds to Hamilton path) and the case

µ = 0, ν = 0, n ≤ 3. So if n ≥ 4, each node correspondig to subgraph s has even-degree.

However, the degree of the nodes may have exponentially large degrees, so we must

exhibit a pairing function. For a Hamiltonian path h which does not have odd-degree,

there are 2ν neighbours, all the subset of h ∩ G. These nodes are easy to pair up. For a

subgraph s which has nonzero degree, the completions easily can be paired up by reversing

such a path of h that contains the edge with the smallest index. And this completes the

proof.

It is an interesting question whether these problems are PPA-complete. At present,

any natural PPA-complete problem is not known, only a variant of Sperner lemma on

non-orientable manifolds is known to be PPA-complete ([12]).

In the next chapters, we will study new problems and prove that they are in PPA.

Combinatorial Nullstellensatz MOD 2 is suspected to be PPA-complete, because

it seems fairly general algrebraic theorem and for example, Chévalley MOD 2 is reducibe

to it. However, we cannot prove it yet.

16


1.2.2 On directed graphs, the class PPAD and the class PPADS

There is natural version of parity argument on directed graph: if each node has at most

one edge out and one edge in., and there is a source, there is another node which is sink

or source.

Directed End of the Line. Given an integer n and an directed finite graph G = (V,E)

in which each node has both in- and out-degree at most 1. The graph is described in the

following way: it can be supposed that each node has an unique code from Σn, that is

V ⊆ Σn. Furthermore, there is a polynomial algorithm (in n) which computes for a node

v its in- and out-neighbours. At last, there is a node ε which has exactly one edge out.

Find another node which has exactly one incident edge, that is, its outdegree+indegree is

pecisely one.

In [1], Papadimitriou defines the class PPAD (Polynomial Parity Argument on Directed

Graphs). A problem is in PPAD if and only if it is polynomially reducible to Directed

End of the Line.

One of the most natural problem in PPAD is based on the Sperner’s theorem.

Proposition 1.14 Sperner 2D is polynomially reducible to Directed End of the

Line. Consequently, Sperner 2D is in PPAD.

Proof. Add some edges to the side RG as in the figure. Then,

we construct a directed graph whose nontrivial odd-degree

nodes correspond to the colourful triangles. Let the nodes be

such triangles of the triangulation that have red-green edges

and let two triangles be connected if they have a common

red-green edge, and this arc is directed corresponding to the

clockwise direction of red-green edge. The outermost triangle

is the node ε. Obviously, based on the M which computes the colour of the vertices in

the triangulation, one can make such a polynomial algorithm that computes for a node its

neighbours in the constructed graph.

Sperner 2D can be generalized in 3 or more dimensions. At first, the standard tetra-

hedronization has to be defined. Then, these problems are also in PPAD. More details in

[1].

One can define the proper computational search problem for Brouwer’s fixpont theorem

and Nash theorem, it is a well-studied area nowadays. It turns out that all these problems

17


are PPAD-complete problems. For instance, the two-dimensional Nash problem is studied

in [13].

Also note that in Directed End of the Line only the existence of a sink is guaran-

teed. Similarly, one can define the problem Directed End of the Line with Sink in

which a sink is asked to find. Problems which are polynomially reducible to this problem

are in the class PPADS (Polynomial Parity Argument on Directed Graphs with Sink).

Directed End of the Line with Sink. Given an integer n and an directed finite

graph G = (V,E) in which each node has at most one incoming edge and one outgoing

edge. The graph is described in the following way: it can be supposed that each node has

an unique code from Σn, that is V ⊆ Σn. Furthermore, there is a polynomial algorithm

(in n) which computes for a node v its neighbours, the node from which there is an edge

to v and the node to which there is an edge to v. At last, there is a node ε which has

exactly one edge out. Find another node which has exactly one edge in.

1.3 Other subclasses of TFNP

There are many other interesting subclasses of TFNP based on elementary proof styles.

We mention two other of them: the class PLS and PPP.

Polynomial Local Search, the class PLS

Sink in a DAG. Given an integer n and an directed finite graph G = (V,E) in which

the degree of each node is polynomially bounded. The graph is described in the following

way: it can be supposed that each node has an unique code from Σn, that is V ⊆ Σn.

Furthermore, there is a polynomial algorithm (in n) which computes for a node v its

neighbours, and a real number c(v). Find a node which does not have such neighbour that

has a greater value.

A problem is in PLS if and only if it is polynomially reducible to Sink in a DAG.

This subclass of TFNP is based on the elemantary fact called Local Search: Every finite

directed acyclic graph has a sink. It has already been defined in [6].

Proposition 1.15 Any combinatorial optimatization problem on a finite state space is

polynomially reducible to Sink in a DAG. Consequently, these problems are in PLS.

18


Polynomial Pigeonhole Principle, the class PPP

Pigeonhole Circuit. Given an integer n and a polynomially computable function

f : 0, 1n → 0, 1n. Find an input x ∈ 0, 1n such that f(x) = 0n or find two inputs

x, y ∈ 0, 1n such that x 6= y and f(x) = f(y).

A problem is in PPP if and only if it is polynomially reducible to Pigeonhole Cir-

cuit. This subclass of TFNP was also defined in [1] and it is based on the elemantary fact

called Pigeonhole Principle.

The following problem is a natural example for PPP, but it is not known to be PPA-

complete or whether it can be solved in polynomial time.

Equal Sums. Given n positive integers such that their sum is less than 2n. Find two

subset with the same sum.

Proposition 1.16 Equal Sums is polynomially reducible to Pigeonhole Circuit.

Consequently, Equal Sums is in PPP.

1.4 The relative complexity of these classes

There are many connections between these com-

plexity classes which are based on inefficient proofs

of existence. In [9], these connections are studied,

they focus on separation theorems. We present only

the reductions on the basic problems and the inclu-

sions of these classes in the next few statements.

The figure shows these known inclusions of these

subclasses of TFNP.

By ignoring the information about the direction,

the next proposition follows. If we find a leaf in the

undirected graph it correspond to a source or sink

in the original directed graph.

Proposition 1.17 Directed End of the Line is polynomially reducible End of the

Line with Degree at most Two. Consequently, PPAD ⊆ PPA.

19


Similarly, it is easy to see the following reduction. If we find a sink, it is a node with

total degree one.

Proposition 1.18 Directed End of the Line is polynomially reducible Directed

End of the Line with Sink. Consequently, PPAD ⊆ PPADS.

Furthermore, te following is a bit surprising, but simple reduction which reveals con-

nection between the parity argument and the pigeonhole principle.

Proposition 1.19 Directed End of the Line with Sink is polynomially reducible

Pigeonhole Circuit Consequently, PPADS ⊆ PPP.

Proof. It can be supposed that nodes of the given graph are encoded by vectors from

0, 1n and let ε be (0, 0, . . . , 0). Then, let f(v) = (0, 0, . . . , 0) if v is a sink, f(v) = u if

there is an arc from v to u, and let f(v) = v if v is an isolated node in the original graph.

It is easy to see that if we find a vector x such that f(x) = (0, 0, . . . , 0), it corresponds to

a sink. In addition there is no v 6= w such that f(v) = f(w) 6= (0, 0, . . . , 0) because in this

case f(v) would have at least two incoming edges.

20

Chapter 2

Complexity of finding roots over

F2

How difficult is finding roots of a polynomial? Over R it usually seems an impossible

task to find the exact roots, moreover, even determining the number of roots seems a

difficult task. What about polynomials over the simplest algebraic field, F2? Obviously, it

is not too complex: we should substitute only the element 0 and the element 1. We are

done beacuse the number of possible roots is tiny.

However, what about a multivariable polynomial? Does it have any roots? How difficult

is finding them? We study these questions over F2.

We will show that determining whether a general multivariable polynomial has any

roots or not is an NP-complete problem. This is proved via reducing the well-known NP-

complete problem, 3-sat.

Furthermore, there is an interesting theorem that gives a sufficient condition to exis-

tence of a root of a multivariable polynomial over an arbitrary field F . The Combinatorial

Nullstellensatz, presented by Alon in [2], is a general algebraic theorem, but it has numer-

ous applications in Combinatorial Number Theory, in Graph Theory and in Combinatorics,

too.

However, Combinatorial Nullstellensatz is not known to be solvable in polynomial time.

An open question in [10] is about the complexity of Combinatorial Nullstellensatz and its

relationship to the class PPA and Chévalley’s theroem.

In this chapter, we present our answer to these questions: Combinatorial Nullstellensatz

is in PPA, and Chévalley’s theorem is polynomially reducible to it.

21

CHAPTER 2. COMPLEXITY OF FINDING ROOTS OVER F2

2.1 Complexity in general

Root of a Polynomial MOD 2. Let pij(i = 1, . . . , k, j = 1, . . . ,m(i)) be polynomials

in F2[x1, . . . xn] and assume k,m(i) and the size of every pij (the number of monomials) is

polynomially bounded in n. Let f =∑ki=1

(∏m(i)j=1 pij

). Determine if this polynomial does

have any root.

Note that the size of f might be exponentially large, even though, we can describe it

in polynomial space.

Proposition 2.1 3-sat is polynomially reducible to Root of a Polynomial MOD 2.

Consequently, Root of a Polynomial MOD 2 is NP-complete.

Proof. Suppose that a 3-sat problem is given in the form S(y11, y12, . . . , yk3) =∧ki=1(yi1∨

yi2 ∨ yi3) where yij(i = 1, . . . , k, j = 1, 2, 3) are variables in T, F. Let xij be a variable

in F2 corresponding to yij . Let f be the following polynomial in F2[x11, x12, . . . , xk3]:

f(x11, x12, . . . , xk3) = 1 +k∏i=1

(xi1 + xi2 + xi3 + xi1xi2 + xi2xi3 + xi3xi1 + xi1xi2xi3)

It is easy to check that if element 1 corresponds to T and element 0 corresponds to F ,

yi1 ∨ yi2 ∨ yi3 = T if and only if xi1 + xi2 + xi3 + xi1xi2 + xi2xi3 + xi3xi1 + xi1xi2xi3 = 1,

so S(y11, y12, . . . , yk3) = T if and only if f(x11, x12, . . . , xk3) = 0.

So if Root of a Polynomial MOD 2 can be solved in polynomial time, 3-sat also

can be solved in polynomial time by constructing the above f polynomial. This shows the

reduction.

22


2.2 Complexity of Combinatorial Nullstellensatz

Theorem 2.2 (Combinatorial Nullstellensatz, Alon) Let F be an arbitrary field, and

let f = f(x1, x2, . . . xn) be a polynomial in F [x1, . . . xn]. Suppose the degree of f is∑ni=1 ti,

where each ti is a nonnegative integer, and suppose the coefficient of∏ni=1 x

tii is nonzero.

Then, if S1, S2, . . . , Sn are subsets of F with |Si| > ti, there exists an (s1, s2, . . . , sn) ∈

S1 × S2 × · · · × Sn such that f(s1, s2, . . . , sn) 6= 0.

As we have already mentioned, we study the complexity of Combinatorial Nullstellen-

satz over F2. This answers an open question in [10].

A vector (s1, s2, . . . , sn) ∈ S1 × S2 × · · · × Sn is called appropriate if it satisfises the

condition of Combinatorial Nullstellensatz: f(s1, s2, . . . , sn) 6= 0.

Note that if ti = 0 for some indices, then we could choose |Si| = 1, so in an appropriate

(s1, s2, . . . , sn) vector, we have to choose the only element of Si to si and therefore, we could

replace xi by the only element of Si in f . Hence, we may assume that Si = F2 for every

index i and the problem is finding an (s1, s2, . . . , sn) ∈ Fn2 such that f(s1, s2, . . . sn) 6= 0.

Also, note that if the size of f (the number of terms) is polynomially bounded, there

is a polynomial algorithm for finding an appropriate vector.

Combinatorial Nullstellensatz MOD 2 with Polynomially Bounded Terms.

Given a polynomial f in n variables with polynomially bounded number of terms. Suppose

that the degree of f is n and the coefficient of x1x2 . . . xn is nonzero. Find an appropriate

vector (s1, s2, . . . , sn) ∈ Fn2 such that f(s1, s2, . . . , sn) 6= 0.

Proposition 2.3 Combinatorial Nullstellensatz MOD 2 with Polynomially

Bounded Terms can be solved in polynomial time.

Proof. First, we can replace the term xti1i1xti2i2. . . x

tikik

to xi1xi2 . . . xik , because these are

equal due to the fact 0t = 0, 1t = 1 in F2.

Substitute 0 and 1 to x1: let g(x2, . . . , xn) = f(0, x2, . . . , xn) and h(x2, . . . , xn) =

f(1, x2, . . . , xn). If in f the coefficient of x2x3 . . . xn is nonzero, then in g the coefficient

of x2x3 . . . xn will be also nonzero. If it is zero, in h the coefficient of x2x3 . . . xn will be

nonzero. Then, substitute 0 and 1 to x2 and in one of them the coefficient of x3x4 . . . xn

will be nonzero, and so on.

At last, we obtain a constant nonzero polynomial, and this means that the substitution

(s1, s2, . . . , sn) ∈ Fn2 vector is appropriate.

23


Hence, we will deal with the case where the size of f might be exponentially large,

however, the description of f is polynomially bounded.

Combinatorial Nullstellensatz MOD 2. Given a polynomial f in n variables in

a general form f =∑ki=1

(∏m(i)j=1 pij

), where pij is a polynomial in F2[x1, . . . xn], k,m(i)

and the size of pij (the number of monomials) is polynomially bounded. Suppose that the

degree of f is n and the coefficient of x1x2 . . . xn is nonzero and suppose that there is

a polynomial pairing function which can pair up terms x1x2 . . . xn. Find an appropriate

vector (s1, s2, . . . , sn) ∈ Fn2 such that f(s1, s2, . . . , sn) 6= 0.

The pairing function is given in the following way. Each term is the product of mono-

mials from the given polynomials, so each term in ith block can be represented by an

(m(i) + 1)-tuple of integers: (i, ai,1, . . . , ai,m(i)). The first coordinate shows that the term

is from which block, and the other coordinates show that the term is the product of which

monomials. (So the same term might have more than one occurrence and these occurrences

are represented by different (m(i) + 1)-tuples.)

The pairing function as defined in Definition 1.10 pairs up from an extra node w these

tuples (i, ai,1, . . . , ai,m(i)) which corresponds to terms x1x2 . . . xn.

Proposition 2.4 Combinatorial Nullstellensatz MOD 2 is polynomially reducible

to End of the Line. Consequently, Combinatorial Nullstellensatz MOD 2 is in

PPA.

Proof. We shall construct a graph Γ whose odd-degree nodes precisely correspond to

appropriate vectors s such that f(s) 6= 0 and furthermore, an extra node w: the standard

leaf.

The graph is bipartite. The nodes of the left side are all the vectors in Fn2 and the extra

node w. The nodes of the other side are the terms of the polynomial f =∑ki=1

(∏m(i)j=1 pij

).

Each term is represented in the above way.

There is an edge between vector x and term t if and only if t(x) = 1, and there is an

edge between node w and term t if and only if t(x) = x1x2 . . . xn.

It is easy to see that any vector x is appropriate if and only if its degree is odd. The

extra node w has also odd degree because the coefficient of x1x2 . . . xn is nonzero due to

the conditions.

All nodes in the other side have even degree. In Γ the degree of t(x) = x1x2 . . . xn

terms are precisely 2, because it is connected only to the (1, 1, . . . , 1) vector and to the

24


extra w node. Let t be any other term, and let xl be a variable not appearing in t. Then,

if t is connected to (s1, s2, . . . , 0, . . . , sn), it is also connected to (s1, s2, . . . , 1, . . . , sn), so

the degree of these nodes are even.

Therefore, odd-degree nodes are precisely s vectors such that f(s) 6= 0 and the extra

w node.

However, the nodes of this graph have exponentially large degrees, so to apply Pa-

padimitriou’s chessplayer algorithm, we must exhibit a pairing function between the edges

out of a node.

For a node corresponding to the term t(x) 6= x1x2 . . . xn, we pair up the vector x for

which t(x) = 1 to (x1, x2, . . . , 1 − xl, . . . xn) where xl is such a variable which does not

appear in t. (We choose the smallest such index l.) The degree of other nodes in this side

is only 2, its edges can be simply paired up.

For a vector x which is not an appropriate vector, we should pair up the terms such

that t(x) = 1. Suppose that t is a term of the block g =∏m(i)j=1 pij for an index i. If g(x) = 0,

there is a j such that pij(x) = 0. Pick the smallest such j. There is an even number of

monomials of pij such that pij(x) = 1. We pair these monomials by a pairing function φi.

Then the mate of term (i, ai1, . . . , aij , . . . , ai,m(i)) is (i, ai1, . . . , φi(aij), . . . , ai,m(i)).

It is a more complicated case when g(x) = 1. Since x is not a right vector, there is

an even number of indices l, such that(∏m(l)

j=1 plj)is 1 at x. We pair these blocks by

a pairing function φ. So, for i and every 1 ≤ j ≤ m(i), pij is 1 at x, and we can pair

the monomials of pij with pij(x) = 1 by a pairing function φij . One of them does not

have a mate, denote its index by ωij . If t is represented by (i, ωi1, . . . , ωi,m(i)), its mate is

(φ(i), ωφ(i),1, . . . , ωφ(i),m(φ(i))). Otherwise there is a j such that aij 6= ωij . Pick the smallest

such j. Then the mate of (i, ai1, . . . , ai,m(i)) is (i, ai1, . . . , φij(aij), . . . , ai,m(i)).

Observe that this gives a bijection and a correct pairing function.

At last, we pair up the terms, which are connected to the extra node w. These are the

terms x1x2 . . . xn. Due to the conditions, there is a polynomial algorithm which can pair

up these terms x1x2 . . . xn, so it can pair up the nodes which are connected to the extra

node w.

We presented a polynomial algorithm that computes the mate of an edge out of a node,

so the proof is complete.

25


2.3 A consequence of Combinatorial Nullstellensatz

We recall the following search problem. It is known to be in PPA, we have already

presented its proof in Proposition 1.12, however, it can be proved by using Combinatorial

Nullstellensatz, too.

Chévalley MOD 2. Given polynomials p1, p2, . . . , pm in F2[x1, . . . xn] such that∑mi=1 deg(pi) <

n. Also, given a root (c1, c2, . . . , cn) ∈ Fn2 of the equation system pi(x) = 0 (i = 1, . . . ,m).

Chévalley’s theorem states that the number of roots is even. Find another root.

In [2], applying the Combinatorial Nullstellensatz, Alon proved Chévalley’s theorem for

arbitrary prime p. This proof shows that Chévalley MOD 2 is reducible to Combina-

torial Nullstellensatz MOD 2. This reduction also shows that Chévalley MOD

2 is in PPA.

Proposition 2.5 Chévalley MOD 2 is reducible to Combinatorial Nullstellen-

satz MOD 2.

Proof. Given polynomials p1, p2, . . . , pm and the root c = (c1, c2, . . . , cn) ∈ Fn2 , we shall

construct the polynomial f as follows:

f =m∏i=1

(1 + pi) +n∏i=1

(xi − ci + 1)

Obviously, f is in the correct form and the coefficient of x1x2 . . . xn is 1, and there

is only one term x1x2 . . . xn, and thus the pairing function is obvious. So the condi-

tions of Combinatorial Nullstellensatz MOD 2 hold. However, the vectors s =

(s1, s2, . . . , sn) ∈ Fn2 such that f(s) 6= 0 are precisely the roots of the equation system

different from c.

Indeed, if s = c, the first product is 1, beacuse c is a root of equation system, then

pi(s) = 0 for all i, and also the second product is 1, so f(s) = 1 + 1 = 0. If s 6= c, clearly,

the second product is 0. If s is a root of equation system, the first product is 1, otherwise

it is 0, and then if it is a root, f(s) = 1, and f(s) = 0 otherwise.

This shows that Chévalley MOD 2 is reducible to Combinatorial Nullstellen-

satz MOD 2.

26

Chapter 3

Subgraphs modulo prime power

Olson’s theorem is an interesting result in Combinatorial Number Theory and it has

an exciting applicaton in Graph Theory. Olson proved the following theorem in [3] for

arbitrary prime p.

Theorem 3.1 (Olson) Let n,m be positive integers and let p be a prime. Let d1 ≥ d2 ≥

· · · ≥ dn nonnegative integers and for 1 ≤ i ≤ n, 1 ≤ j ≤ m let aij be an integer. If

m >∑ni=1(pdi − 1), then there is a subset ∅ 6= J ⊆ 1, 2, . . . ,m such that

∑j∈J aij is

divisible by pdi for every 1 ≤ i ≤ n.

Olson’s theorem is tight in the following sense: for arbitrary numbers di, if m =∑ni=1(pdi − 1) there exists integers 1 ≤ i ≤ n, 1 ≤ j ≤ m such that the proper non-

trival subset does not exist.

Alon, Friedland and Kalai presented the application in Graph Theory and proved a

generalisation of Olson’s theorem in [4].

Theorem 3.2 Let cardp(Q) denote the number of distinct elements in Q modulo p. Let

n,m be positve integers and let p be a prime. Let d1 ≥ d2 ≥ · · · ≥ dn nonnegative integers

and for 1 ≤ i ≤ n, 1 ≤ j ≤ m let aij be an integer. Then for arbitrary subset Qi of

0, 1, . . . , pdi − 1 (1 ≤ i ≤ n), if m >∑ni=1

(pdi − cardp(Qi)

), then there is a subset

∅ 6= J ⊆ 1, 2, . . . ,m such that∑j∈J

aij ≡ qi (mod pdi) for some qi ∈ Qi for every 1 ≤ i ≤ n

However, this theorem is usually not tight. We present another generalisation of Olson’s

theorem applying Chévalley’s theorem. Our result gives stronger inequalities than Alon’s

result, moreover, in some cases, we present tight values.

27

CHAPTER 3. SUBGRAPHS MODULO PRIME POWER

We investigate the following general problem.

Let us be given a prime p and d1 ≥ d2 ≥ · · · ≥ dn nonnegative integers andQ1, Q2, . . . , Qn

sets such that each of them contains zero and Qi ⊆ 0, 1, . . . , pdi − 1 for every 1 ≤ i ≤ n.

Denote (d1, d2, . . . , dn) by d and (Q1, Q2, . . . , Qn) by Q.

Determine the minimum value F (d,Q) such that for every m > F (d,Q) and for arbi-

trary integers aij (1 ≤ i ≤ n, 1 ≤ j ≤ m) there exists a nontrivial subset J ⊆ 1, 2, . . . ,m

that fulfills the following conditon:

∑j∈J

aij ≡ qi (mod pdi) for some qi ∈ Qi for every 1 ≤ i ≤ n. (♣)

Conversely, we are looking for the maximum value m = F (d,Q) such that there exists

such integers aij (1 ≤ i ≤ n, 1 ≤ j ≤ m) that any nontrivial subset J ⊆ 1, 2, . . . ,m does

not fulfill (♣).

Let 0 = (0, 0, . . . , 0). Then, Olson’s theorem states that F (d,0) =∑ni=1(pdi−1).

Similarly, we can formalize the generalisation by Alon. Their theorem states that

F (d,Q) ≤∑ni=1

(pdi − cardp(Qi)

).

After these questions, we mention an application of Olson’s theorem (and its exten-

sions) in Graph Theory. Similar problems can be studied, but determining the exact value

seems more complicated.

Then, we study the complexity of these problems. We prove that the search problem

Olson MOD 2d is reducible to Chévalley MOD 2, hence Olson MOD 2d is in PPA.

In addition, we discuss the search problem 2d-divisible subgraph, which is also in PPA.

Then, we present a polynomial time algorithm for a special case, the Even Subgraph

of a Hypergraph.

3.1 A generalisation of Olson’s theorem

At first, we investigate the residue of binomial coefficients modulo p. Lucas’ theorem

gives a nice relationship between the residue of(cb

)and the form of c and b in base p.

Theorem 3.3 (Lucas,[11]) Let p be a prime, let c(d−1) . . . c(1)c(0) be the form of c and

b(d−1) . . . b(1)b(0) be the form of b in base p. Then(c

b

)≡(c0b0

)·(c1b1

)· · ·(cd−1bd−1

)(mod p)

28


Proof. The proof is based on the following equation.

c∑b=0

(c

b

)xb = (1 + x)c =

d−1∏r=0

((1 + x)pr

)cr

≡d−1∏r=0

(1 + xp

r)cr

(mod p) =

=d−1∏r=0

cr∑sr=0

(crsr

)xsr·pr

=c∑b=0

∑(s0,s1,...,sd−1):

∑srpr=b

d−1∏r=0

(crsr

)xbThat is, (

c

b

)≡

∑(s0,s1,...,sd−1):

∑srpr=b

d−1∏r=0

(crsr

)(mod p)

However, 0 ≤ sr ≤ cr < p, so the inner sum contains at most one term: sr = br for

every r if br ≤ cr for every r, and it is an empty sum if br > cr for any r In the first case,

the theorem immediately follows. In the latter case the sum over the empty set is zero,

and for any br > cr,(cr

br

)is also zero.

We use the following corollary of Lucas’ theorem.

Corollary 3.4 Let c(d−1) . . . c(1)c(0) be the form of c in base p. Then,(c

pr

)≡ c(r) (mod p)

Proof. It is easy to see that if b = pr then b(r) = 1 and b(i) = 0, so(c

pr

)≡(c(0)

0

)·(c(1)

0

)· · · · ·

(c(r)

1

)· · · · ·

(c(d−1)

0

)≡ c(r) (mod p)

The following definition and lemma show the consequence of Lemma 3.4 and Lucas’s

theorem: how can we encode integers by using polynomials over Fp.

Definition 3.5 Let f =∑ki=1 pi be a polynomial over Z, where each pi is in the form∏n

j=1 xtij

j , with coefficient 1, where each tij is a nonnegative integer. (For instance, we

write x2ixj + x2

ixj + x2ixj instead of 3x2

ixj.)

Let Ψ(r)p (f) ∈ Fp[x1, . . . , xn] be the following polynomial:

∑1≤i1<i2<···<ipr≤k

pi1pi2 . . . pipr

Lemma 3.6 Let f =∑ki=1 pi be a polynomial over Z, where each pi is in the form∏n

j=1 xtij

j , with coefficient 1, where each tij is a nonnegative integer. Let s = (s1, s2, . . . , sn) ∈

29


Fnp be such that pi(s) ∈ 0, 1 for every 1 ≤ i ≤ k. Further, let c = f(s) =∑ki=1

∏nj=1 s

tij

j

and let c(d−1) . . . c(1)c(0) be the form of c in base p. Then,

c(r) = γ ∈ 0, 1, . . . , p− 1 if and only if Ψ(r)p (f)(s) = γ ∈ Fp ∼= 0, 1, . . . , p− 1

Proof. It is easy to see that the number of terms in Ψ(r)p (f) that are 1 at s is precisely( c

pr

). Due to Lemma 3.4,

( cpr

)≡ c(r) (mod p). Hence, Ψ(r)

p (f)(s) equals c(r), because both

are congruent to( cpr

)modulo p and both are in Fp ∼= 0, 1, . . . , p − 1. This shows the

lemma.

Note that deg(Ψ(r)p (f)) ≤ pr · deg(f).

Definition 3.7 Let R be a subset of 0, 1, . . . , d−1. Let us define the set Ω ⊆ 0, 1, 2, . . . , pd−

1 by the following poperty: c ∈ Ω if and only if c(r) = 0 for every r ∈ R in the form of c

in base p. We call Ω the R-zero set modulo pd.

Now, we can prove a generalisation of Olson’s theorem. Let κ(R) = (p− 1)∏r∈R p

r.

Theorem 3.8 (R-zero Olson) Let n,m be positve integers and let p be a prime. Let

d1 ≥ d2 ≥ · · · ≥ dn nonnegative integers and for 1 ≤ i ≤ n, 1 ≤ j ≤ m let aij be

an integer. Let Ri be an arbitrary subset of 0, 1, . . . , di − 1 and denote the Ri-zero set

modulo pdi by Ωi. If m >∑ni=1 κ(Ri), then there is a subset ∅ 6= J ⊆ 1, 2, . . . ,m such

that ∑j∈J

aij ≡ ωi (mod pdi) for some ωi ∈ Ωi for every 1 ≤ i ≤ n.

That is, F (d,Ω) ≤∑ni=1 κ(Ri).

Proof. Let x1, x2, . . . , xm ∈ Fp variables corresponding to the columns of the matrix

whose elements are the aij . For each i we shall construct an equation system whose roots

precisely correspond to ∅ 6= J ⊆ 1, 2, . . . ,m such that∑j∈J aij ≡ ωi (mod pdi) for

some ωi ∈ Ωi. Due to Fermat’s little theorem, xp−1 = 1 if x 6= 0 and xp−1 = 0 if x = 0.

The first case corresponds to j ∈ J and the second case corresponds to j 6∈ J .

Let gi(x) =∑mj=1 aij · x

p−1j . By writing gi in the form where all coefficients are 1, we

can construct the polynomials fir = Ψ(r)p (gi) for every 1 ≤ i ≤ n and every r ∈ Ri.

It is easy to see that deg(fir) ≤ (p− 1) · pr, so the sum of the degree of polynomials is

less than (p− 1)∏r∈Ri

pr = κ(Ri) for every 1 ≤ i ≤ n.

For an s ∈ Fnp , let ci = gi(s). Due to Lemma 3.6, c(r)i = 0 if and only if fir = Ψ(r)

p (gi) is

zero at s, so ci ∈ Ω if and only if s is the solution of the equation system fir(x) = 0 (r ∈ R).

30


For all 1 ≤ i ≤ n, we can construct these equation systems. A root of the whole equa-

tion system precisely corresponds a set J ⊆ 1, 2, . . . ,m that∑j∈J aij ≡ ωi (mod pdi)

for some ωi ∈ Ωi for every 1 ≤ i ≤ n. The sum of the degree of the equations is less

than∑ni=1 κ(Ri) and due to the conditions of the theorem,

∑ni=1 κ(Ri) < m, hence the

conditions of Chévalley’s theorem are true. Therefore, the number of roots are divisible

by p and (0, 0, . . . , 0) is obviously a root. So there is an another root which corresponds

such an ∅ 6= J ⊆ 1, 2, . . . ,m that∑j∈J aij ≡ ωi (mod pdi) for some ωi ∈ Ωi for every

1 ≤ i ≤ n.

Corollary 3.9 Let Ri = 0, 1, . . . , d− 1 and so, Ωi = 0. Then the statement of R-zero

Olson theorem coincides with the statement of Olson’s theorem. So, Olson’s theorem is a

special case of R-zero Olson theorem.

3.2 Inequalities for the quantity F (d, Q)

It is easy to see the following proposition which gives a lower bound.

Proposition 3.10 If d = (d1, d2, . . . , dn) and Q = (Q1, Q2, . . . , Qn), then

F (d1, Q1) + F (d2, Q2) + · · ·+ F (dn, Qn) ≤ F (d,Q)

Proof. If m1 = F (d1, Q1), then there exists m1 integers, a1j (1 ≤ j ≤ m1), such that∑j∈J1 a1j is not in Q1 modulo pd1 for any ∅ 6= J1 ⊆ 1, 2, . . . ,m1. Similarly, there exists

mi = F (di, Qi) integers. Denote these integers by aij where∑i−1l=1 ml < j ≤

∑il=1ml. Let

other elements aij be zero and denote∑nl=1ml by m.

Then, suppose that there is a ∅ 6= J ⊆ 1, 2, . . . ,m such that fulfills the property (♣).

However, now there must exist ∅ 6= Ji such that∑j∈Ji

aij is in Qi modulo pdi , but this

contradicts to the choice of elements aij .

So in this case, m =∑ni=1 F (di, Qi), there is no such nonempty subset J that fulfills

the property (♣), so∑ni=1 F (di, Qi) ≤ F (d,Q).

Proposition 3.11 The condition∑ni=1 κ(Ri) < m is tight in R-zero Olson theorem. Con-

sider arbitrary positive integer n, arbitrary nonnegative integers d1 ≥ d2 ≥ · · · ≥ dn and

arbitrary Ri ⊆ 0, 1, . . . , di − 1 and∑ni=1 κ(Ri) = m. Then, there exist integers aij such

that the statement of Theorem 3.8 does not hold. That is,

F (d,Ω) =n∑i=1

κ(Ri)

31


Proof. Applying Proposition 3.10, we just have to show that F (d,Ω) = κ(R) for the

R-zero sets Ω modulo pd.

It is easy to see that the largest number in the R-zero sets Ω modulo pd is∑r 6∈R

(p− 1) · pr = pd − 1−∑r∈R

(p− 1) · pr = pd − 1− κ(R) ≡ −κ(R)− 1 (mod pd)

So, if n = 1,m = κ(R), and aij = −1 for i = 1, 1 ≤ j ≤ m, for any subset ∅ 6= J ⊆

1, 2, . . . ,m, κ(R) <∑j∈J aij < pd, and this shows the propositon.

The tight example: gray elements are −1, others are 0

For an arbitrary Q, we get an estimate using this Ri-zero subsets.

Proposition 3.12 For arbitrary Qi ⊆ 0, 1, . . . , pdi − 1, let Qi be the largest Qi = Ωi ⊆

Qi such that Ωi is the Ri-zero subset modulo pdi for some Ri ⊆ 0, 1, . . . , di − 1. Then,

F (d,Q) ≤n∑i=1

κ(Ri)

Proof. If m >∑ni=1 κ(Ri), for arbitrary integers aij (1 ≤ i ≤ n, 1 ≤ j ≤ m), there exists

a subset ∅ 6= J ⊆ 1, 2, . . . ,m such that∑j∈J

aij ≡ ωi (mod pdi) for some ωi ∈ Ωi for every 1 ≤ i ≤ n.

Ωi ⊆ Qi, so this subset J fulfills the property (♣). Hence, F (d,Q) ≤∑ni=1 κ(Ri).

There are some other tricks which give additional estimates to F (d,Q).

Proposition 3.13 Let ciQi = ciqi(mod pdi) : qi ∈ Qi with (ci, p) = 1, so ci is not

divisible by p. Let c ·Q be (c1Q1, c2Q2, . . . , cnQn) where ciQi = ciqi(mod pdi) : qi ∈ Qi

with (ci, p) = 1, so ci is not divisible by p. Then

F (d,Q) = F (d, c ·Q)

Proof. Let a′ij = ci · aij for every 1 ≤ i ≤ n, 1 ≤ j ≤ m. Then, for a subset J ⊆

1, 2, . . . ,m,∑j∈ a

′ij is in ci ·Qi modulo pdi if and only if

∑j∈ aij is in Qi modulo pdi due

to (ci, p) = 1.

32


Hence, there exists a subset ∅ 6= J ⊆ 1, 2, . . . ,m fulfilling property (♣) if and only of

this subset J fulfills the property (♣) with the a′ij integers, and sets ci ·Qi. And it shows

the proposition.

Proposition 3.14 Let Gi be a set of positive integers whose sum is less than pdi. Let

Qi = ∑gi∈G′i

gi : G′i ⊆ Gi. Then

F (d,Q) ≤n∑i=1

pdi − |Gi|

Proof. We can add new columns to the matrix ((aij)). The first |G1| new columns contain

the negative of the elements of G1 in the first row, and zeros in the others. Then, the next

|G2| new columns contains the negative of the elements of G2 in the second row, and

zeros in the others, and so on. If the original matrix has m columns, the new matrix has

m′ = m+∑ni=1 |Gi| columns.

If m >∑ni=1 p

di − |Gi|, then m′ >∑ni=1 p

di , so due to Olson’s theorem, there exists a

subset ∅ 6= J ′ ⊆ 1, 2, . . . ,m′ such that fulfills the property (♣) with the proper elements.

Let Ji be the subset of J which corresponds to the coloumns of the set Gi, let G′i be the

set of these elements, and let J = J ′ \⋃ni=1 Ji. Because Gi contains positive integers and

their sum is less than pdi , J 6= ∅. Then,

∑j∈J

aij ≡∑gi∈G′i

gi ∈ Qi (mod pdi)

This J fulfills the property (♣) So if m >∑ni=1 p

di − |Gi|, arbitrary integers aij there

exists proper subset J . This implies F (d,Q) ≤∑ni=1 p

di − |Gi|.

Note that, the Proposition 3.13, 3.14, 3.10 easily can be extended to modulo k =

(k1, k2, . . . , kn) instead of (pd1 , pd2 , . . . , pdn). We hope these observations and propositions

can be taken to uniform theory in a future work.

In the next section we discuss an application of Olson’s theorem which was published

in [4]. We present their results and then we apply our R-zero Olson’s theorem for a more

general question about subgraphs modulo prime power.

33


3.3 pd-divisible subgraphs and Q-subgraphs

First, we present a simple corollary of the original Olson’s theorem which was proved

in [4]. It will be useful for the discussion of applications of Olson’s theorem.

Corollary 3.15 Let n,m be positve integers and let p be a prime. Let d1 ≥ d2 ≥ · · · ≥

dn ≥ 1 positive integers and for 1 ≤ i ≤ n, 1 ≤ j ≤ m let aij be an integer such that∑ni=1 aij is divisible by p for every j index. If m > pdn−1−1+

∑n−1i=1 (pdi−1), then there is

a subset ∅ 6= J ⊆ 1, 2, . . . ,m such that∑j∈J aij is divisible by pdi for every 1 ≤ i ≤ n.

Proof. 1 ≤ j ≤ m, let bij = aij , if 1 ≤ i ≤ n− 1 and let bnj = 1p (∑ni=1 aij).

According to Olson’s theorem, there is an ∅ 6= J ⊆ 1, 2, . . . ,m such that∑j∈J bij

is divisible by pdi for every 1 ≤ i ≤ n − 1 and∑j∈J bnj is divisible by pdn−1 because

m > pdn−1 − 1 +∑n−1i=1 (pdi − 1).

However,∑j∈J bnj =

∑j∈J

1p (∑ni=1 aij) is divisible by pdn−1 , so

∑j∈J (

∑ni=1 aij) is

divisible by pdn . Because of d1 ≥ d2 ≥ · · · ≥ dn ≥ 1,∑j∈J bij =

∑j∈J aij is divisible by

pdn for every 1 ≤ i ≤ n− 1, so∑j∈J anj is divisible by pdn and we are done.

A graph is called pd-divisible, if every degree is divisible by pd. In [4], Alon, Friedland

and Kalai discussed the following question. Given a prime p and a prime power pd and an

integer n. Which is the smallest value m such that for every graph on n vertices there is

a nonempty pd-divisible subgraph, a nonempty subset of edges such that for every vertex

of the graph, the number of edges which are incident to it is divisible by pd for a p prime.

Conversely, there how many edges can a graph have such that it contains no nontrivial pd-

divisible subgraph. Denote the maximum number of these edges by f(n, pd). The authors

proved the following theorem.

Theorem 3.16 (Alon, Friedland, Kalai) For the maximum number of edges of a graph

G on n vertices that contains no nontrivial pd-divisible subgraph,

f(n, 2d) ≤

(pd − 1) · n if p is an odd prime.

(2d − 1) · n− 2d−1 if p = 2

Proof. Denote the number of edges of G by m. Let aij = 1 if and only if the jth edge is

incident to the ith vertex. (That is, ((aij)) is the incidence matrix of G.)

For an odd prime p, suppose m > (pd − 1) · n. According to Olson’s theorem with

d1 = d2 = · · · = dn = d, there is a nonempty J subset of edges such that∑j∈J aij is

divisible by pd for every i, so there is nontrivial pd-divisible subgraph.

34


For p = 2, suppose m > (2d − 1) · n − 2d−1. According to the Corollary 3.15 with

d1 = d2 = · · · = dn = d, there is a nonempty J subset of edges such that∑j∈J aij is

divisible by pd for every i, so there is nontrivial pd-divisible subgraph. The conditions of

corollary are true, beacuse∑ni=1 aij = 2 for every index j.

Proposition 3.17 If n ≥ 3, there exists graphs that has (pd − 1) · n edges (if p > 2) or

(2d − 1) · n − 2d−1 edges (if p = 2), but does not contain nontrivial pd-divisible subgraph.

So,

f(n, 2d) ≤

(pd − 1) · n if p is an odd prime.

(2d − 1) · n− 2d−1 if p = 2

Proof. For p > 2 let G0 be the graph which is obtained from a triangle by replacing each

edges by pd − 1 parallel edges. For p = 2, let G0 be the similar graph, but replace one

of edges would be by 2d−1 − 1 edges instead of 2d − 1 edges. Then add to G0 n − 3 new

vertices and each of them would be joined by pd − 1 edges to vertices of G0.

It is easy to see that this graph does not contains pd-divisible subgraph, it has n vertices

and (pd − 1) · n edges (if p > 2) or (2d − 1) · n− 2d−1 edges (if p = 2).

More general, we study the following quantity, for hypergraphs instead of graphs, and

arbitrary Q subset of residue instead of divisibility.

Definition 3.18 Let Q is a subset of 0, 1, . . . , pd − 1 containing the 0. A subgraph is

called Q-subgraph modulo pd if every degree in it is in Q modulo pd. Let f(n, pd, Q) be the

maximum number of hyperedges if a hypergraph has n vertices and it has not a nontrivial

Q-subgraph.

Applying R-zero Olson theorem, the following proposition follows.

Proposition 3.19 Let p be a prime, R be a subset of 0, 1, . . . , d−1 and Ω be the R-zero

set modulo pd. Then, the maximum number of hyperedges of a hypergraph G on n vertices

that contains no nontrivial Ω-subgraph is

f(n, pd,Ω) ≤ n · κ(R) = n · (p− 1)∏r∈R

pr.

Proof. Denote the number of edges of G by m. Let aij = 1 if and only if the jth edge is

incident to the ith vertex. (That is, ((aij)) is the incidence matrix of G.)

35


Supposem > n·κ(R). According to R-zero Olson theorem with d1 = d2 = · · · = dn = d,

there is a nonempty J subset of edges such that∑j∈J aij is in Ω modulo pd for every i,

so there is nontrivial Ω-subgraph.

The following open question might be interesting.

Question 3.20 Is f(n, pd,Ω) = n · κ(R) true for every n, d and R ⊆ 0, 1, . . . , d− 1?

Obviously, this is not the most general definition of these problems. We can simi-

larly define the problem with k1, k2, . . . , kn integers, and Q1, Q2, . . . , Qn sets where Qi ⊆

0, 1, . . . , ki − 1 for every 1 ≤ i ≤ n and such subgraph is expected that the degree of

node i is in Qi modulo ki. This general question is left for future work. It easy to see that

f(d,Q) ≤ F (d,Q).

Question 3.21 When does f(d,Q) = F (d,Q) hold?

3.4 Complexity aspects of subgraphs

3.4.1 Complexity of Olson’s problem

Olson MOD 2d. Let us be given the integers n and m such that m > n · (2d − 1) and

given integers aij (1 ≤ i ≤ n, 1 ≤ j ≤ m). Find a ∅ 6= J ⊆ 1, 2, . . . ,m such that∑j∈J aij

is divisible by 2d for every i.

Note that, such subset exists according to Olson’s theorem. Also note that d is not the

part of the input, it is fixed for the problem.

Proposition 3.22 Olson MOD 2d is reducible to Chévalley MOD 2. Consequently,

it is in PPA.

Proof. The proof of R-zero Olson theorem shows that one can construct an equation

system whose roots correspond precisely the right subsets of J , and this equation system

is a proper input of Chévalley MOD 2. So, the only thing we have to check is the

polynomiality of reduction.

Given n,m and aij integers, the size of the input is polynomailly bounded in n and in

m. We shall construct the polynomials gi =∑mj=1 aijxj and the fir = Ψ(r)

2 (gi) for every

0 ≤ r < d and 1 ≤ i ≤ n. The size of gi is polynomially bounded in m, so the size of fir is

36


polynomially bounded in m because it has at most(m

2r

)terms, which is a polynomial of m

for fixed r ≤ d. The number of thats is d · n. So the size of whole equation system is also

polynomially bounded.

Hence, the Olson MOD 2d is reducible to Chévalley MOD 2 and Olson MOD

2d is in PPA.

Even-Sum Olson MOD 2d. Given the integers n and m such that m > n ·(2d−1)−2d−1

and given integers aij (1 ≤ i ≤ n, 1 ≤ j ≤ m) such that∑ni=1 aij is even for every j index.

Find an ∅ 6= J ⊆ 1, 2, . . . ,m such that∑j∈J aij is divisible by 2d for every i.

Note that, such subset exists according to Corollary 3.15.

Proposition 3.23 Even-Sum Olson MOD 2d is polynomially reducible to Chévalley

MOD 2. Consequently, it is in PPA.

Proof. The proof of Corollary 3.15 shows the reduction to Olson’s theorem and the

proof of Olson’s theorem shows the reduction to Chévalley MOD 2.

In Olson’s theorem and in the corollary the right J subset are precisely the same and

the roots of constructed equation system correspond the right subsets, and similarly to

the proof of Proposition 3.22, the reduction is polynomially bounded.

Similarly, the following problem is also in PPA.

R-zero Olson MOD 2d. Given R ⊆ 0, 1, . . . , d − 1, the integers n and m such that

m > n · κ(R) and given integers aij (1 ≤ i ≤ n, 1 ≤ j ≤ m). Find a ∅ 6= J ⊆ 1, 2, . . . ,m

such that∑j∈J aij is in Ω modulo 2d for every i where Ω is the R-zero subset modulo 2d.

Note that, such subset exists according to R-zero Olson theorem.

Proposition 3.24 R-zero Olson MOD 2d is reducible to Chévalley MOD 2. Con-

sequently, it is in PPA.

3.4.2 Complexity of subgraphs modulo 2d

According to the application of Olson’s theorem, the following problem can be naturally

asked.

2d-divisible subgraph. Given a G = (V,E) graph, where |V | = n and |E| = m.

Suppose that m > n · (2d − 1) − 2d−1. Find a 2d-divisble subgraph, an ∅ 6= F ⊆ E such

that for every v ∈ V , the number of incident edges of F is divisible by 2d.

37


Note that, such subgraph exists according to Theorem 3.16.

For d = 1, this is the well-known fact that in a graph, if the number of edges is at

least as the number of vertices, there is an even subgraph, namely, there is a cycle in the

graph. To find a cycle in such a graph, there is polynomial algorithm but for d > 1, we do

not know whether it is in P.

Proposition 3.25 2d-divisible subgraph is reducible to Even-Sum Olson MOD 2d.

Consequently, it is in PPA.

Proof. The proof of Theorem 3.16 shows the reduction. For the graphG, we shall construct

the incidence matrix of G, where m > n · (2d − 1) − 2d−1 and∑ni=1 aij = 2 is even for

every j index. So the conditions of Even-Sum Olson MOD 2d are true and the right

∅ 6= J ⊆ 1, 2, . . . ,m subsets precisely correspond to the 2d-divisble subgraphs.

Similarly, 2d-divisible subgraph in Hypergraphs and Ω-subgraph modulo 2d

can be defined and proved that they are in PPA.

3.4.3 A special case: even subgraph of a hypergraph

There is an interesting special case of finding 2d-divisible subgraphs in hypergraphs.

Let G = (V,E) be a hypergraph on |V | = n with |E| = m > n. Then, there is an even

subgraph: there exists an ∅ 6= F ⊆ E such that for every v ∈ V , the number of incident

hyperedges of F is even. More precisely, |e ∈ F : v ∈ e| is even for every v ∈ V .

As we have already mentioned, it is well-known for graph and it holds with m ≤ n

yet. It is worth checking that for hypergraphs the condition m > n is tight: there exists

hypergraph with m = n such that does not contains nontrivial even subgraph.

For example, let V = a, b, c, d and E = a, b, c, a, b, d, a, c, d, b, c, d. It is

easy to check that this hypergraph does not contains nontrivial even subgraph.

Furthermore, because of the above observations, it is PPA, and in what follows, we

show that it can be solved in polynomial time.

Even Subgraph of a Hypergraph. Given a G = (V,E) hypergraph, where |V | = n

and |E| = m. Suppose that m > n. Find an even subgraph: an ∅ 6= F ⊆ E such that for

every v ∈ V , the number of incident hyperedges of F is even.

38


Proposition 3.26 Even Subgraph of a Hypergraph is in P.

Proof. We present a polynomial time algorithm for this search problem.

It is easy to see that if x ∈ Fm2 , 0 = (0, 0, . . . , 0) ∈ Fn2 and A is the incidence matrix of

G, F = e ∈ E : xe = 1 corresponds to an even subgraph if and only if, Ax = 0 over F2.

Due to the condition m > n, according to linear algebra, there exists a nontrivial

solution x and it can be found in polynomial time using e.g. Gaussian elimination.

Note that Gaussian elimination runs in O(n3), though, in graphs an even subgraph

can be found in O(n). It is an interesting question that even subgraph of a hypergraph

can be found more efficiently.

Gaussian elimination can be easily translated to a combinatorial algorithm in the

hypergraph. The below figure shows an example.

If we make a Gaussian elimination over F2 on M , there is only one kind of steps: a row

can be changed to the sum of that and another row over F2 in order to be only one 1 in

each column.

Gauss elimination terminates when such step cannot be made: there are some singular

columns with exactly one 1 and at least one other. This other column can be written

the sum of some singular columns. Then, a solution x can be established: let 1 be in x

corresponding to this singular columns and the other column, and let 0 be in x otherwise.

Changing a row to the sum of that and another row over F2 corresponds in the hy-

pergraph to changing a node to the symmetric difference of that and another node. This

operation v ← w ⊕ v is the following: w will stay in the same edges, and if v, w ∈ e, then

let w 6∈ e, if v ∈ e, w 6∈ e, then let w ∈ e, if v 6∈ e, w ∈ e, then let w ∈ e, if v, w 6∈ e, then let

w 6∈ e. At the end, there is a large hyperedge and some one-element hyperedges that the

large one contains them. The hyperedges corresponding to these one constitute an even

subgraph.

It is an interesting question if there are more special case where a 2d-divisible subgraph

can be found in polynomial time. The simplest non-trivial question is the following.

Question 3.27 Can 4-divisible subgraph be solvable in polynomial time?

39


We close this section with a colourful example for a run of the polynomial algorithm.

v1

v2 v3

v4v5

v6

e1 e2

e3e4e5

e6

e7

v1

v2 v3

v4v5

v6

e1e2

e3e4e5

e6

e7

v1

v2 v3

v4v5

v6

e1e2

e3e4e5

e6

e7

1 0 0 0 0 0 0

1 1 0 0 0 1 0

1 1 1 0 0 0 0

0 0 0 1 1 0 0

0 0 1 0 1 1 1

0 0 1 0 0 0 1

1 0 0 0 0 0 0

0 1 0 0 0 1 0

0 1 1 0 0 0 0

0 0 0 1 1 0 0

0 0 1 0 1 1 1

0 0 1 0 0 0 1

1 0 0 0 0 0 0

0 1 0 0 0 1 0

0 0 1 0 0 1 0

0 0 0 1 1 0 0

0 0 1 0 1 1 1

0 0 1 0 0 0 1

The original hypergraph v2 ← v1 ⊕ v2, v3 ← v1 ⊕ v3 v3 ← v2 ⊕ v3

v1

v2 v3

v4v5

v6

e1e2 e3

e4e5

e6

e7

v1

v2 v3

v4v5

v6

e1e2 e3

e4e5 e6

e7

v1

v2 v3

v4v5 v6

e1e2 e3

e4e5

e6

e7

1 0 0 0 0 0 0

0 1 0 0 0 1 0

0 0 1 0 0 1 0

0 0 0 1 1 0 0

0 0 0 0 1 0 1

0 0 0 0 0 1 1

1 0 0 0 0 0 0

0 1 0 0 0 1 0

0 0 1 0 0 1 0

0 0 0 1 0 0 1

0 0 0 0 1 0 1

0 0 0 0 0 1 1

1 0 0 0 0 0 0

0 1 0 0 0 0 1

0 0 1 0 0 0 1

0 0 0 1 0 0 1

0 0 0 0 1 0 1

0 0 0 0 0 1 1

v5 ← v3 ⊕ v5, v6 ← v3 ⊕ v6 v4 ← v5 ⊕ v4 v2 ← v6 ⊕ v2, v3 ← v6 ⊕ v2

So (0, 1, 1, 1, 1, 1, 1) is a solution of Ax = 0, so e2, e3, e4, e5, e6, e7 is an even-subgraph.

40

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http://www.cs.elte.hu/~lovasz/complexity.pdf

http://www.cs.elte.hu/~lovasz/complexity.pdf

http://www.cs.elte.hu/~veghal/orak/jatekelm-2010dec10.pdf

http://www.cs.elte.hu/~veghal/orak/jatekelm-2010dec10.pdf

http://www.math.uiuc.edu/~west/regs/combnull.html

REFERENCES

[11] N. J. Fine: Binomial coefficients modulo a prime. The American Mathematical

Monthly, Vol. 54, No. 10, Part 1, Dec., 1947, pp. 589-592

[12] M. Grigni: A Sperner Lemma Complete for PPA. Information Processing Letters,

77(5-6) pp. 255-259, March 2001

[13] Chen, Deng: On the Complexity of 2D Discrete Fixed Point Problem. Theoretical

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Documents

Bonyolultsági osztályok paritási érvelések alapján · Abstract In the ﬁrst part of the thesis, we give an overview of complexity classes of search problems. We ﬁrst study