Extra pearls in graph theory
Anton Petrunin
TL;DR
This supplement provides additional insights and topics in graph theory, expanding on foundational concepts with discussions on advanced theories, methods, and specific graph classes.
Contribution
It introduces new discussions and perspectives on various advanced topics in graph theory not covered in the original textbook.
Findings
Enhanced understanding of graph polynomials and generating functions
New insights into toroidal graphs and the Rado graph
Extended coverage of probabilistic and algebraic methods in graph theory
Abstract
This is a supplement for "Pearls in graph theory" -- a textbook written by Nora Hartsfield and Gerhard Ringel. We discuss the following topics: Ramsey theory, Probabilistic method, Deletion-contraction formulas, Matrix theorem, Graph-polynomials, Generating functions, Minimum spanning trees, Marriage theorem and its relatives, Toroidal graphs, Rado graph, Rewriting systems.
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Taxonomy
TopicsAdvanced Graph Theory Research · Limits and Structures in Graph Theory · Advanced Topology and Set Theory
Extra pearls in graph theory
Anton Petrunin
I used these topics together with ‘‘Pearls in graph theory’’ by Nora Hartsfield and Gerhard Ringel [21] to teach an undergraduate course in graph theory at the Pennsylvania State University. I tried to keep clarity and simplicity on the same level.
Hope that someone will find it useful for something.
I want to thank Semyon Alesker, Alexander Gil, Rostislav Matveyev, Alexei Novikov, Dmitri Panov, and Lukeria Petrunina for help.
Anton Petrunin
This work is licensed under the Creative Commons Attribution-ShareAlike 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/4.0/
Contents
Chapter 1 Introduction
A Terminology
The diagram on the right may describe regular flights of an airline. It has six flights, which serve four airports labeled by , , , and .
For this and similar types of data, mathematicians use the notion of a pseudograph.
Formally, a pseudograph is a finite nonempty set of vertices (in our example a vertex is an airport) and a finite collection of edges; each edge connects two vertices (in the example above, an edge is a regular flight). A pair of vertices can be connected by multiple edges; such edges are called parallel (in our example, it might indicate that the airline operates multiple flights a day between these airports). Also, an edge can connect a vertex to itself; such an edge is called a loop (we might think of it as a sightseeing flight).
Thus, from a mathematical point of view, the diagram above describes a pseudograph with vertices , , , , and six edges, including one loop at and a pair of parallel edges between and .
The number of edges coming from one vertex is called its degree; loops are counted twice. In the example above, the degrees of and are and , respectively.
A vertex with zero degree is called isolated, and a vertex of degree one is called an end vertex.
A pseudograph without loops is also called a multigraph, and a multigraph without parallel edges is also called a graph. Most of the time we will work with graphs.
If and are vertices of a pseudograph , we say that is adjacent to if there is an edge between and . We say that a vertex is incident with an edge if is an end vertex of .
B Wolf, goat, and cabbage
Тhe vertices and edges of the graph might have very different nature. As an example, let us consider the following classic puzzle.
1.1. Puzzle. * A farmer purchased a wolf, a goat, and a cabbage; he needs to cross a river with them. He has a boat, but he can carry only himself and a single one of his purchases: the wolf, the goat, or the cabbage.*
If left unattended together, the wolf would eat the goat, and the goat would eat the cabbage.
The farmer has to carry himself and his purchases to the far bank of the river, leaving each purchase intact. How can he do it?
Solution. Let us denote the farmer by , the river by the wolf by , the goat by , and the cabbage by . For example, means that the wolf and cabbage are on the left bank of the river and the goat with the farmer are on the right bank.
The starting position is ; that is, everyone is on the left bank. The following graph describes all possible positions that can be achieved; each edge is labeled by the transported purchase.
{wgc{*}}{\parallel}{}$${wc}{\parallel}{{*}g}$${wc{*}}{\parallel}{g}$${w}{\parallel}{{*}gc}$${c}{\parallel}{{*}gw}$${cg{*}}{\parallel}{w}$${wg{*}}{\parallel}{c}$${g}{\parallel}{{*}wc}$${g{*}}{\parallel}{wc}$${}{\parallel}{{*}wgc}$$g$$c$$w$$g$$g$$w$$c$$g
This graph shows that the farmer can achieve by legal moves. It solves the problem and also shows that there are exactly two different solutions, assuming that the farmer does not want to repeat the same position twice. ∎
Often, a graph comes with extra structure; for example, labeling of edges and/or vertices as in the example above.
Formally speaking, the described graph representation can be applied to nearly any puzzle. However, it might be impossible to apply due to a huge number of states and transitions; think of Rubik’s cube or the 15 puzzle.
Here is a small variation of another classic puzzle.
1.2. Puzzle. * Four missionaries and four cannibals must cross a river using a boat that can carry at most two people. The boat cannot cross the river by itself with no people on board. If there are missionaries on one bank, they cannot be outnumbered by cannibals; otherwise the missionaries will be eaten.*
Can the missionaries and cannibals safely cross the river?
Let us introduce a notation to describe the states and transitions in this puzzle. The river will be denoted by ; let denote the boat, we will write the number of cannibals on each side of , and the number of missionaries by subscript. For example, means that on the left bank we have four cannibals, two missionaries, and the boat (these two missionaries will be eaten), and on the right bank there are no cannibals and two missionaries.
1.3. Exercise. * Draw the corresponding graph. Conclude that the puzzle has no solutions.*
Chapter 2 Ramsey numbers
In this chapter, we discuss Ramsey numbers. The will be used as an illustration of the next chapter.
A Ramsey numbers
Recall that the Ramsey number is the least positive integer such that every blue-red coloring of edges in the complete graph contains a blue or a red .
Switching colors in the definition shows that for any and . Therefore, we may assume that .
Note that for any positive integer . Indeed, the one-vertex graph has no edges; therefore we can say that all its edges are blue (as well as red and deep green-cyan turquoise at the same time).
2.1. Exercise. * Show that for any positive integer .*
The following table from [35] includes all known values of for :
For example, the table says that . In order to show this, one has to prove two inequalities and . The inequality means that there is a blue-red coloring of edges of that has no monochromatic ; this will be done in Exercise 2D. The inequality means that in any blue-red coloring of there is a monochromatic . The latter follows from via the inequality ➍ ‣ 2C below, and a proof of is given in [21, 4.3].
B Binomial coefficients
In this section, we review properties of binomial coefficients that will be needed further.
Binomial coefficients will be denoted by ; they can be defined as unique numbers satisfying the identity
[TABLE]
for any real numbers and integer . This identity is called binomial expansion. It can be used to derive some identities on binomial coefficients; for example,
[TABLE]
The number plays an important role in combinatorics, as it gives the number of ways that objects can be chosen from different objects. This value can be found explicitly using the formula
[TABLE]
Note that all different ways to choose objects from different objects fall into two categories: (1) those which include the last object — there are of them, and (2) those which do not include it — there are of them. It follows that
[TABLE]
This identity will be used in the proof of Theorem 2C.
C Upper bound
Recall that according to Theorem 4.3.2 in [21], the inequality
[TABLE]
holds for all integers .
In other words, any value in the table above cannot exceed the sum of values in the cells directly above and on the left from it. The inequality ➍ ‣ 2C might be strict; for example,
[TABLE]
2.2. Theorem. * For any positive integers we have that*
[TABLE]
Proof. Set
[TABLE]
so we need to prove the following inequality:
[TABLE]
Note that from ➌ ‣ 2B, we get the identity
[TABLE]
which is similar to the inequality ➍ ‣ 2C.
Further note that for any positive integer . Indeed, , and there is only one choice of [math] objects from the given . Similarly , and there is only one choice of objects from the given .
The above observations make it possible to calculate the values of recursively. The following table provides some of its
values. The inequality ➎ ‣ 2C means that any value in this table cannot exceed the corresponding value in the table for on page 2A. The latter is nearly evident from ➍ ‣ 2C and ➏ ‣ 2C; let us show it formally.
Since
[TABLE]
the inequality ➎ ‣ 2C holds if or .
Assume the inequality ➎ ‣ 2C does not hold for some and . Choose a minimal criminal pair ; that is, a pair with minimal value such that ➎ ‣ 2C does not hold. From above we have that . Since is minimal, we have that
[TABLE]
summing these two inequalities and applying ➍ ‣ 2C together with ➏ ‣ 2C we get ➎ ‣ 2C — a contradiction. ∎
2.3. Corollary. * The inequality*
[TABLE]
holds for any positive integer .
Proof. By ➋ ‣ 2B, we have that . Applying Theorem 2C, we get that
[TABLE]
∎
D Lower bound
In order to show that
[TABLE]
it is sufficient to color the edges of in red and blue so that it has no red and no blue . Equivalently, it is sufficient to decompose into two subgraphs with no isomorphic copies of in the first one and no isomorphic copies of in the second one.
For example, the subgraphs in the decomposition of in the diagram have no monochromatic triangles; this implies that . We already showed that for any decomposition of into two subgraphs, one of the subgraphs has a triangle; that is, .
Similarly, to show that , we need to construct a decomposition of into two subgraphs and such that contains no triangle and contains no . In fact, in any decomposition of into two subgraphs, either the first subgraph contains a triangle or the second contains a . That is, [[, see]p. 82–83]hartsfield-ringel.
Applying ➍ ‣ 2C, we get
[TABLE]
Therefore, to prove that , we need to show that ; that is, we need to construct a decomposition of into two subgraphs with no .
The required decomposition is given on the diagram. The constructed decomposition is rationally symmetric; the first subgraph contains the chords of angle lengths 1, 2, 4, and 8 and the second contains all the chords of angle lengths 3, 5, 6, and 7.
2.4. Exercise. * Show that*
- (a)
In the decomposition of above, the left graph contains no triangle, and the right graph contains no . 2. (b)
In the decomposition of , neither graph contains any .
For larger values and , the problem of finding the exact lower bound for quickly becomes too hard. Even getting a reasonable estimate is challenging. In the next chapter we will show how to obtain such an estimate by using probability.
Chapter 3 Probabilistic method
We assume that the reader is familiar with the notion of a random variable and expected value, at least on an intuitive level. An introductory part in any textbook on probability should be sufficient; see for example [12, 30, 40].
A Markov’s inequality
A random variable is a real number that depends on a random event. We will consider random variables that take only finitely many values.
Two random variables are called independent if the occurrence of one does not affect the other.
The expected value is the average of a large number of independently selected outcomes of the random variable. The expected value of a random variable will be denoted by . Suppose a random variable takes only values with probabilities , respectively, so . Then
[TABLE]
For example, if is the result of rolling a die, then it takes each value with probability ; therefore
[TABLE]
3.1. Claim. * For any two random variables and and real constant we have*
[TABLE]
Moreover, if and are independent, then
[TABLE]
(We assume that all expected values in the formulas are well defined.)
3.2. Markov’s inequality. * Suppose is a nonnegative random variable and . Denote by the probability of the event . Then*
[TABLE]
Proof. Consider another random variable such that if and otherwise.
Note that and therefore
[TABLE]
The random variable takes the value with probability and [math] with probability . By ➊ ‣ 3A,
[TABLE]
whence ➋ ‣ 3A follows. ∎
B Probabilistic method
The probabilistic method allows us to prove the existence of graphs with certain properties without explicitly constructing them. The idea is to show that if one randomly chooses a graph or its coloring from a specified class, then the probability that the result has the needed property is more than zero. This implies that a graph with required the property exists.
Despite using probability, the final conclusion is determined for certain, without any possible error.
3.3. Theorem. * Assume that the inequality*
[TABLE]
holds for a pair of positive integers and . Then .
Proof. We need to show that the complete graph admits a coloring of edges in red and blue such that it has no monochromatic subgraph isomorphic to .
Let us color the edges randomly; that is, color each edge independently in red or blue with equal chances.
Fix a set of vertices. Define the random variable to be if every edge between the vertices in has the same color; otherwise set .
The random variable may take the values [math] or . The expected value of is the probability that ; that is, all of the edges in have the same color. The probability that all the edges with the ends in are blue is , and with the same probability all the edges are red. Since these two possibilities exclude each other,
[TABLE]
This holds for any -vertex subset of the vertices of .
Note that the number of monochromatic -subgraphs in is the sum of over all possible -vertex subsets . The total number of such subsets is . By 3A,
[TABLE]
Denote by the probability that a random coloring of has at least one monochromatic ’s. By Markov’s inequality,
[TABLE]
Since the number of monochromatic ’s is an integer, it follows that a random coloring of has no monochromatic ’s with probability at least . If , it implies that this probability is positive; in particular, at least one edge-coloring of has no monochromatic . That is, if then there is a coloring with no monochromatic -subgraph. ∎
The following corollary implies that the function grows at least exponentially.
3.4. Corollary. * .*
Proof. Set ; that is, is the largest integer .
Note that
[TABLE]
and
[TABLE]
Therefore,
[TABLE]
By Theorem 3B, we get that . ∎
3.5. Exercise. * Assume the edges of the complete graph are colored randomly, so each edge is colored independently in red or blue with equal chances. Show that the expected number of monochromatic Hamiltonian cycles in is larger than .*
Remark. One might think that the exercise alone is sufficient to conclude that most of the colorings of have a monochromatic Hamiltonian cycle. Let us show that it is not that easy. (It is still true that the probability of the existence of a monochromatic coloring is close to 1, but the proof requires more work.)
The total number of colorings of is . Therefore, in principle, it might happen that of the colorings have no monochromatic Hamiltonian cycles and of the colorings contain all the monochromatic Hamiltonian cycles. To keep the expected value above , this of the colorings should have fewer than of monochromatic cycles on average; the latter does not seem impossible since the total number of Hamiltonian cycles in is .
C Counting proof
In this section, we rewrite the proof of Theorem 3B without the use of probability. We do this to affirm that the probabilistic method provides a real proof, without any possible error.
In principle, any probabilistic proof admits such a translation, but in most cases, the translation is less intuitive.
Counting proof of 3B. The graph has edges. Each edge can be colored in blue or red; therefore the total number of different colorings is
[TABLE]
Fix a subgraph isomorphic to in . Note that this graph is red in different colorings and blue in colorings.
There are different subgraphs isomorphic to in . Therefore, the total number of monochromatic ’s in all the colorings is
[TABLE]
If , then by the pigeonhole principle, there is a coloring with no monochromatic . Hence the result. ∎
D Graph of -cube
In this section, we give another classic application of the probabilistic method.
Let denote the graph of the -dimensional cube; has vertices, each labeled with a sequence of length consisting of zeros and ones. Тwo vertices are adjacent if their labels differ in only one digit.
The graph is shown on the diagram. Note that each vertex of has degree .
Recall that the distance between two vertices in a graph is the length of a shortest path connecting the vertices.
Note that the distance between two vertices in is the number of different digits in their sequences. For example, two sequences and have different digits at positions , , and , and the corresponding vertices at distance from each other.
3.6. Problem. * Suppose denotes the maximal number of vertices in at a distance more than from each other. Then grows exponentially in ; moreover, .*
To solve the problem, one has to construct a set with at least vertices in that are far from each other. It is challenging to construct such a set explicitly. Instead, we will show that if one randomly selects that many vertices, they are far from each other with positive probability. To choose a random vertex in , one can toss a coin times, writing 1 for a head and 0 for a tail, and then select the vertex labeled by the obtained sequence.
The following exercise guides you to a solution of the problem. The same argument shows that for any coefficient , the maximal number of vertices in at a distance larger than from each other grows exponentially in . According to Exercise 3D, the case is very different.
3.7. Exercise. * Let denote the probability that two randomly chosen vertices in at a distance between them.*
- (a)
Use Claim 3D to show that
[TABLE] 2. (b)
Assume vertices in are fixed and is a random vertex. Show that is the expected number of vertices that lies at a distance from . Use Markov’s inequality to show that lies at a distance larger than from each of with probability at least . 3. (c)
Apply a and b to show that there are at least vertices in at a distance larger than from each other.
3.8. Claim. * The probability of obtaining fewer than one-third heads after fair tosses of a coin decays exponentially in ; in fact for any .*
Proof. Let us introduce independent random variables ; each returns the number of heads after the -th toss of the coin; in particular, each takes values [math] or with the probability each. We need to show that the probability of the event is less than .
Consider the random variable
[TABLE]
Note that is the probability of the event that . Also, . By Markov’s inequality, we get that
[TABLE]
The random variable takes the two values and with the probability of each. Therefore,
[TABLE]
Note that
[TABLE]
Since the random variables are independent, 3A implies that
[TABLE]
It follows that
[TABLE]
∎
3.9. Advanced exercise. **
- (a)
Show that contains at most vertices at a distance larger than from each other. 2. (b)
Show that contains at most vertices at a distance at least from each other.
E Remarks
The existence of Ramsey number for any and is the first result in the so-called Ramsey theory. A typical theorem in this theory states that any large object of a certain type contains a very ordered piece of a given size. We recommend the book by Matthew Katz and Jan Reimann [27] on the subject.
Corollaries 2C and 3B imply that
[TABLE]
It is unknown if these inequalities can be essentially improved.111This question might look insignificant at first glance, but it is considered one of the major problems in combinatorics [16]. More precisely, it is unknown whether there are constants and such that the inequality
[TABLE]
holds for any . Similarly, it is unknown whether there are constants and such that the inequality
[TABLE]
holds for any .
The probabilistic method was introduced by Paul Erdős. It finds applications in many areas of mathematics, not only in graph theory.
Note that the probabilistic method is nonconstructive; if it is used to prove the existence of a certain object, then it is still uncontrollably hard to describe a concrete example.
More involved examples of proofs based on the probabilistic method deal with typical properties of random graphs.
To describe the concept, let us consider the following random process that generates a graph with vertices.
Fix a positive integer . Consider a graph with the vertices labeled by , where the existence of an edge between every pair of vertices is decided independently by flipping a coin.
Note that the described process depends only on , and as a result, we can get a graph isomorphic to any given graph with vertices.
3.10. Exercise. * Let be a graph with vertices. Denote by the probability that is isomorphic to . Show that*
[TABLE]
Suppose ; describe graphs and such that these inequalities become equalities.
Fix a property of a graph (for example, connectedness) and denote by the probability that has this property. We say that the property is typical if as .
3.11. Exercise. * Show that random graphs typically have a diameter of 2. That is, the probability that has a diameter of 2 converges to 1 as .*
Note that from the exercise above, it follows that in the described random process, the random graphs are typically connected.
3.12. Exercise. * Show that random graphs typically have a subgraph isomorphic to . That is, the probability that has a subgraph isomorphic to converges to 1 as .*
The following theorem gives a deeper illustration of the probabilistic method with the use of typical properties; a proof can be found in [1, Chapter 44].
3.13. Theorem. * Given positive integers and , there is a graph with girth at least and a chromatic number at least .*
Chapter 4 Deletion and contraction
A Definitions
Let be a pseudograph with a marked edge . Denote by the pseudograph obtained from by deleting , and by the pseudograph obtained from by contracting the edge to a point; see diagram.
Assume is a graph; that is, has no loops and no parallel edges. In this case, is also a graph. However, might have parallel edges, but no loops; that is, is a multigraph.
If is a multigraph, then so is . If the edge is parallel to in , then in becomes a loop; that is, is a pseudograph in general.
B Number of spanning trees
Recall that denotes the number of spanning trees in the pseudograph .
An edge in a connected graph is called the bridge, if deletion of makes the graph disconnected; in this case, the remaining graph has two connected components which are called banks.
4.1. Exercise. * Assume that the graph contains a bridge between banks and . Show that*
[TABLE]
4.2. Deletion-plus-contraction formula. *
Let be an edge in the pseudograph . Assume is not a loop, then the following identity holds*
[TABLE]
It is convenient to write the identity ➊ ‣ 4B using a diagram as on the picture; the arrows point from one multigraph to multigraphs with the same total number of spanning trees; the edge is marked in .
Proof. Note that the spanning trees of can be subdivided into two groups: (1) those which contain the edge and (2) those which do not. For the trees in the first group, the contraction of to a point gives a spanning tree in , while the trees in the second group are also spanning trees in .
Moreover, both of the described correspondences are one-to-one. Hence the formula follows. ∎
A spanning tree cannot have loops, so removing all loops does not change the number of spanning trees. Let us state it.
4.3. Claim. * If is a loop in a pseudograph , then*
[TABLE]
The proof of the following claim uses the deletion-plus-contraction formula.
4.4. Claim. * If one removes an end vertex from a pseudograph , then in the obtained graph the number of spanning trees remains unchanged; that is,*
[TABLE]
Proof. Denote by the only edge incident to . Note that the graph is not connected, since the vertex is isolated. In particular, . On the other hand, . Therefore, the deletion-plus-contraction formula ➊ ‣ 4B implies ➋ ‣ 4B. ∎
On the diagrams, we may use a two-sided arrow ‘‘’’ for the graphs with equal numbers of the spanning trees. For example, using the deletion-plus-contraction formula together with the claims, we can draw the following diagram, which in particular implies the following identity:
[TABLE]
Note that the deletion-plus-contraction formula gives an algorithm to calculate the value for a given pseudograph . Indeed, for any edge , both graphs and have a smaller number of edges. That is, the deletion-plus-contraction formula reduces the problem of finding the number of the trees to simpler graphs; applying this formula a few times we can reduce the question to a collection of graphs with an evident answer for each. In the next section we will show how it works.
C Fans and their relatives
Recall that Fibonacci numbers are defined using the recursive identity with . The sequence of Fibonacci numbers starts as
[TABLE]
The graphs of the following type are called fans;
a fan with vertex will be denoted by .
4.5. Theorem. * .*
Proof. Applying the deletion-plus-contraction formula, we can draw the following infinite diagram.
(We ignore loops and end vertices since they do not change the number of spanning trees.) In addition to the fans we use its variations , which differ from by an extra parallel edge.
Set and . From the diagram, we get the following two recursive relations:
[TABLE]
That is, in the sequence
[TABLE]
every number starting from is the sum of the previous two.
Further note that has two vertices connected by a unique edge, and has two vertices connected by a pair of parallel edges. Hence and and therefore
[TABLE]
for any .∎
Comments. We can deduce a recursive relation for , without using :
[TABLE]
This is a special case of the so-called constant-recursive sequences. The general term of constant-recursive sequences can be expressed by a closed formula — read [23] if you wonder how. In our case, it is
[TABLE]
Since is an integer and for any , a shorter formula can be written
[TABLE]
where denotes floor of ; that is, is the maximal integer that does not exceed .
4.6. Exercise. * Consider the sequence of zig-zag graphs of the following type:*
Show that for any .
4.7. Exercise. * Let us denote by the number of spanning trees in the -step ladder ; that is, in the graph of the following type:*
Apply the method we used for the fans to show that the sequence satisfies the following linear recursive relation:
[TABLE]
Note that and ; applying the exercise, we could calculate the first numbers of the sequence :
[TABLE]
The following exercise is analogous, but more complicated.
4.8. Advanced exercise. * Recall that a wheel is the graph of following type:*
Show that the sequence satisfies the following recursive relation:
[TABLE]
Using the exercise above and applying induction, one can show that
[TABLE]
for any . The numbers are called Lucas numbers; they pop up in combinatorics as often as Fibonacci numbers.
D Remarks
The deletion-plus-contraction formula together with Kirchhoff’s rules was used in the solution of the so-called squaring the square problem. The history of this problem and its solution are discussed in a book by Martin Gardiner [14, Chapter 17].
The proof of the recurrence relation above is given by Mohammad Hassan Shirdareh Haghighi and Khodakhast Bibak [39]; this problem is also discussed in the book by Ronald Graham, Donald Knuth, and Oren Patashnik [17].
Chapter 5 Matrix theorem
A Adjacency matrix
Let us describe a way to encode a given multigraph with vertices by an matrix. First, enumerate the vertices of the multigraph by numbers from to ; such a multigraph will be called labeled. Consider the matrix with the component equal to the number of edges from the -th vertex to the -th vertex of .
This matrix is called the adjacency matrix of . Note that is symmetric; that is, for any pair . Also, the diagonal components of vanish; that is, for any .
For example, for the labeled multigraph shown on the diagram, we get the following adjacency matrix:
[TABLE]
5.1. Exercise. * Let be the adjacency matrix of a labeled multigraph. Show that the components of the -th power is the number of walks of length in the graph from vertex to vertex .*
B Kirchhoff minor
In this section, we construct a special matrix called the Kirchhoff minor, associated with a pseudograph, and discuss its basic properties. This matrix will be used in the next section in a formula for the number of spanning trees in a pseudograph . Since loops do not change the number of spanning trees, we can remove all of them. In other words, we can (and will) always assume that is a multigraph.
Fix a multigraph and consider its adjacency matrix ; it is a symmetric matrix with zeros on the diagonal.
Revert the signs of the components of and exchange the zeros on the diagonal for the degrees of the corresponding vertices. The obtained matrix is called the Kirchhoff matrix, Laplacian matrix or admittance matrix of the graph . 2. 2.
Delete from the last column and the last row; the obtained matrix will be called the Kirchhoff minor of the labeled pseudograph .
For example, the labeled multigraph on the diagram has the following Kirchhoff matrix and Kirchhoff minor:
[TABLE]
5.2. Exercise. * Show that in any Kirchhoff matrix the sum of the components in each row or column vanishes. Conclude that*
[TABLE]
5.3. Exercise. * Draw a labeled pseudograph with the following Kirchhoff minor:*
[TABLE]
5.4. Exercise. * Show that the sum of all components in every column of the Kirchhoff minor is nonnegative.*
Moreover, the sum of all components in the -th column vanishes if and only if the -th vertex is not adjacent to the last vertex.
Relabeling. Let us understand what happens with Kirchhoff minor and its determinant as we swap two labels distinct from the last one.
For example, if we swap the labels and in the graph above, we get another labeling shown on the diagram. Then the corresponding Kirchhoff minor will be
[TABLE]
which is obtained from by swapping columns and followed by swapping rows and .
Note that swapping a pair of columns or rows changes the sign of the determinant. Therefore, swapping one pair of rows and one pair of columns does not change the determinant. Summarizing we get the following:
5.5. Observation. * Assume is a labeled graph with vertices, and is its Kirchhoff minor. If we swap two labels , then the corresponding Kirchhoff minor can be obtained from by swapping columns and and then swapping rows and . In particular,*
[TABLE]
Deletion and contraction. Let us understand what happens with the Kirchhoff minor if we delete or contract an edge in the labeled multigraph. (If after the contraction of an edge we get loops, we remove it; this way we obtain a multigraph.)
Assume an edge connects the first and the last vertex of a labeled multigraph as shown in the following example:
1$$2$$3$$4$$e\pgfmathresultpt
1$$2$$3$$4\pgfmathresultpt
2$$3$$4\pgfmathresultpt
Note that deleting only reduces the corner component of by one, while contracting it removes the first row and column. That is, since
[TABLE]
we have
[TABLE]
Summarizing the above discussion, we get the following:
5.6. Observation. * Assume is an edge of a labeled multigraph between the first and last vertices and is the Kirchhoff minor of . Then*
- (a)
the Kirchhoff minor of can be obtained from by subtracting 1 from the corner element with index (1,1); 2. (b)
the Kirchhoff minor of can be obtained from by removing the first row and the first column in .
In particular, applying the cofactor expansion of a determinant, we get that
[TABLE]
Note that the last formula resembles the deletion-plus-contraction formula. This observation will be a key to the proof of the matrix theorem; see next section.
C Matrix theorem
5.7. Matrix theorem. * Let be the Kirchhoff minor of a labeled multigraph with at least two vertices. Then*
[TABLE]
where denotes the number of spanning trees in .
Proof. Denote by the degree of the last vertex in .
Assume . Then is not connected and therefore . On the other hand, the sum in each row of vanishes (compare to Exercise 5B). Hence the sum of all columns in vanish; in particular, the columns in are linearly dependent and hence . Hence the equality ➊ ‣ 5C holds if .
As usual, we denote by and the number of vertices and edges in ; by the assumption we have that .
Assume ; that is, has two vertices and parallel edges connecting them. Clearly, . Further note that ; that is, the Kirchhoff minor is a matrix with a single component . In particular, and therefore the equality ➊ ‣ 5C holds.
Assume the equality ➊ ‣ 5C does not hold in general; choose a minimal criminal graph ; that is, a graph that minimize the value among the graphs violating ➊ ‣ 5C.
From above we have that and . Note that we may assume that the first and last vertices of are adjacent; otherwise permute pair of labels and some and apply Observation 5B. Denote by the edge between the first and last vertex.
Note that the total number of vertices and edges in the pseudographs and are smaller than . Therefore, we have that
[TABLE]
Applying these two identities together with the deletion-plus-contraction formula and Observation 5B, we get that
[TABLE]
that is, the identity ➊ ‣ 5C holds for — a contradiction. ∎
5.8. Exercise. * Fix a labeling for each of the following graphs, find its Kirchhoff minor and use the matrix theorem to find the number of spanning trees.*
- (a)
; 2. (b)
; 3. (c)
.
(Use https://matrix.reshish.com/determinant.php, or any other matrix calculator.)
D Calculation of determinants
In this section, we recall key properties of the determinant which will be used in the next section.
Let be an -matrix; that is, a table , filled with numbers called components of the matrix. The determinant is a polynomial of the components of , satisfying the following conditions:
The unit matrix has determinant 1; that is,
[TABLE] 2. 2.
If we multiply each component of one of the rows of by a number , then for the obtained matrix , we have
[TABLE] 3. 3.
If one adds (or subtracts) term-by-term components in one row to another row, then the obtained matrix has the same determinant; that is
[TABLE]
These three conditions define the determinant in a uniquely. We will not give a proof of the statement; it is not evident and not complicated (sooner or later you will have to learn it if it is not done already).
5.9. Exercise. * Show that the following property follows from the properties above.*
Interchanging any pair of rows of a matrix multiplies its determinant by ; that is, if a matrix is obtained from a matrix by permuting two of its rows, then
[TABLE]
The determinant of -matrix can be written explicitly as a sum of terms. For example,
[TABLE]
is the determinant of the matrix
[TABLE]
However, the properties described above give a more convenient and faster way to calculate the determinant, especially for larger .
Let us show it in one example which will be needed in the next section.
[TABLE]
Let us describe what we used on each line above:
property 3 three times — we add to the first row each of the remaining rows; 2. 2.
property 3 three times — we add the first row to each of the remaining three rows; 3. 3.
property 2 three times; 4. 4.
property 3 three times — we subtract from the first row the remaining three rows; 5. 5.
property 1.
E Cayley formula
Recall that a complete graph is a graph where each pair of vertices is connected by an edge; a complete graph with vertices is denoted by .
Note that every vertex of has degree . Therefore, the Kirchhoff minor in the matrix formula ➊ ‣ 5C for is the following -matrix:
[TABLE]
Following the steps at the end of the previous section, we get
[TABLE]
That is,
[TABLE]
Therefore, applying the matrix theorem, we get the following:
5.10. Cayley formula. **
[TABLE]
that is, the number of spanning trees in the complete graph is .
5.11. Exercise. * Show for any edge in .*
5.12. Exercise. * Use the matrix theorem to show that*
[TABLE]
F Remarks
There is a strong connection between counting spanning trees of a given graph, calculations of currents in an electric chain and random walks; a good survey is given in the book by Peter Doyle and Laurie Snell [7]. Let us give some examples.
Assume that the graph describes an electric chain; each edge has resistance one Ohm, and a battery is connected to the vertices and . Assume that the total current between these vertices is one Ampere. The following procedure calculates the current along a given edge .
Fix a direction of . Note that any spanning tree of has exactly one the following three properties: (1) the edge appears on the (necessarily unique) path from to in with a positive orientation, (2) the edge appears on the path from to in with a negative orientation, (3) the edge does not appear on the path from to in . Denote by , , and the number of the trees in each group. Clearly,
[TABLE]
Then the current can be calculated using the following formula:
[TABLE]
This statement can be proved by checking Kirchhoff’s rules for the currents calculated by this formula.
There are many other applications of Kirchhoff’s rules to graph theory. For example, they can be used to prove the Euler’s formula
[TABLE]
where , , and denote the number of vertices, edges, and regions in a plane drawing of graphs [31].
Chapter 6 Graph-polynomials
Counting problems often lead to an organized collection of numbers. Sometimes it is convenient to consider a polynomial with these numbers as coefficients. If it is done in a smart way, then the algebraic structure of the obtained polynomial reflects the original combinatorial structure of the graph.
A Chromatic polynomial
Denote by the number of different colorings of the graph in colors such that the ends of each edge get different colors.
6.1. Exercise. * Assume that a graph has exactly two connected components and . Show that*
[TABLE]
for any .
6.2. Exercise. * Show that for any integer ,*
[TABLE]
where denotes a wheel with spokes and is a cycle of length .
6.3. Deletion-minus-contraction formula. * Let be an edge in a pseudograph . Then*
[TABLE]
Proof. The valid colorings of can be divided into two groups: (1) those where the ends of the edge get different colors — these remain valid colorings of and (2) those where the ends of get the same color — each of such colorings corresponds to a unique coloring of . Hence
[TABLE]
which is equivalent to the deletion-minus-contraction formula ➊ ‣ 6A. ∎
Note that if the pseudograph has loops, then for any . Indeed, in a valid coloring, the ends of a loop should get different colors, which is impossible.
The latter can also be proved using the deletion-minus-contraction formula. Indeed, if is a loop in , then . Therefore, and
[TABLE]
Similarly, removing a parallel edge from a pseudograph does not change the value for any . Indeed, if has a parallel edge , then in the edge becomes a loop. Therefore, for any and by the deletion-minus-contraction formula we get that
[TABLE]
The same identity can be seen directly — any admissible coloring of is also admissible in — since the ends of get different colors, so does .
Summarizing the discussion above: the problem of finding for a pseudograph can be reduced to the case when is a graph; that is, has no loops and no parallel edges. Indeed, if has a loop, then for all . Further, removing one of the parallel edges from does not change .
Recall that polynomial of is an expression of the following type
[TABLE]
with constants , which are called coefficients of the polynomial. The coefficient is called the free term of the polynomial. If , it is called the leading coefficient of ; in this case is the degree of . If the leading coefficient is 1, then the polynomial is called monic.
6.4. Theorem. * Let be a pseudograph with vertices. Then is a polynomial with integer coefficients and a vanishing free term.*
Moreover, if has a loop, then ; otherwise, is monic and has degree .
Based on this result we can call the chromatic polynomial of the graph . The deletion-minus-contraction formula will play the central role in the proof.
Proof. As usual, denote by and the number of vertices and edges in . To prove the first part, we will use induction on .
As the base case, consider the null graph ; that is, the graph with vertices and no edges. Since has no edges, any coloring of is admissible. We have choices for each of vertices therefore
[TABLE]
In particular, the function is given by a monic polynomial of degree with integer coefficients and a vanishing free term.
Assume that the first statement holds for all pseudographs with at most edges. Fix a pseudograph with edges. Applying the deletion-minus-contraction formula for some edge in , we get that
[TABLE]
Note that the pseudographs and have edges. By the induction hypothesis, and are polynomials with integer coefficients and vanishing free terms. Hence ➋ ‣ 6A implies the same for .
If has a loop, then , as has no valid colorings. It remains to show that if has no loops, then is a monic polynomial of degree .
Assume that the statement holds for any multigraph with at most edges and at most vertices.
Fix a multigraph with vertices and edges. Note that is a multigraph with vertices and edges. By assumption, its chromatic polynomial is monic of degree .
Further the pseudograph has vertices, and its chromatic polynomial either vanishes or has degree . In both cases the difference is a monic polynomial of degree . It remains to apply ➋ ‣ 6A. ∎
6.5. Advanced exercise. * Let be a graph with vertices, edges, and connected components. Show that*
[TABLE]
where are positive integers and .
6.6. Exercise. * Use induction and the deletion-minus-contraction formula to show that*
- (a)
* for any tree with edges;* 2. (b)
* for the cycle of length .* 3. (c)
if is a graph with vertices, then for any . 4. (d)
, where denotes the -spine fan, defined on page 4C. 5. (e)
, where denotes the -step ladder, defined on page 4C.
6.7. Exercise. * Show that a graph is a tree if and only if*
[TABLE]
for some positive integer .
6.8. Exercise. * Show that*
[TABLE]
Remark. Note that for any graph with vertices, we have
[TABLE]
for any . Since both polynomials
[TABLE]
are monic of degree , it follows that so is .
Hence Exercise 6A leads to an alternative way to prove the second statement in Theorem 6A.
6.9. Exercise. * Construct a pair of nonisomorphic graphs with equal chromatic polynomials.*
B Matching polynomial
Recall that a matching in a graph is a set of edges without common vertices.
Given an integer , denote by the number of matchings with edges in the graph .
Note that for a graph with vertices and edges, we have , , and if , then . The maximal integer such that is called the matching number of . The expression
[TABLE]
is called the matching polynomial of .
The matching polynomial conveniently organizes the numbers so we can work with all of them simultaneously. For example, the degree of is the matching number of and the total number of matchings in is its value at :
[TABLE]
6.10. Exercise. * Assume that a graph has exactly two connected components and . Show that*
[TABLE]
for any .
6.11. Exercise. * Show that the values*
[TABLE]
equal the number of matchings with even and odd numbers of edges, respectively.
Assume is an edge in a graph . Recall that the graph is obtained by deleting from . Let us denote by the graph obtained by deleting the vertices of with all their edges from ; that is, if connects two vertices and , then
[TABLE]
The following exercise is analogous to the deletion-contraction formulas 6A and 4B.
6.12. Exercise. * Let be a graph.*
- (a)
Show that
[TABLE]
for any edge in . 2. (b)
Use part a to show that the matching polynomials of complete graphs satisfy the following recursive relation:
[TABLE] 3. (c)
Use b to calculate for .
C Spanning-tree polynomial
Consider a connected graph with vertices; assume .
Let us prepare independent variables , one for each vertex of . For each spanning tree in consider the monomial
[TABLE]
where denotes the degree of the -th vertex in .
The tree has edges and therefore . It follows that the total degree of the monomial is .
The sum of these monomials is a polynomial of degree of variables . This polynomial will be called the spanning-tree polynomial of and it will be denoted by .
For example, the graph shown on the diagram has three spanning trees; each is obtained by deleting one of the edges in the cycle . Abusing notation slightly, let us use the same label for a vertex in and for the corresponding variable. The monomial for the tree obtained by deleting the edge is . Indeed, in this tree the vertex has degree 3, vertices and have degree and the remaining vertices , , and have degree 1. The monomials for the other two trees are and . Therefore,
[TABLE]
Note that does not depend on , , and since the corresponding vertices have degree 1 in any spanning tree of .
Note that for any graph ; that is, is the total number of spanning trees in . Indeed, each spanning tree in contributes one monomial to , and each monomial contributes to the value . The following exercise shows that the polynomial keeps a lot more information about spanning trees in .
6.13. Exercise. * Let be the spanning-tree polynomial of a graph . Show the following:*
- (a)
* is the number of spanning trees with a leaf at the first vertex.* 2. (b)
The coefficient of in front of equals the number of paths of length connecting -th and -th vertices. 3. (c)
The partial derivative
[TABLE]
is the number of spanning trees in with degree 2 at the first vertex. 4. (d)
The two values
[TABLE]
are the numbers of spanning trees in with odd or even degrees at the first vertex, respectively.
6.14. Theorem. **
[TABLE]
where is the complete graph with vertices.
Note that the theorem above generalizes the Cayley formula (5E). Indeed,
[TABLE]
In the proof we will use the following algebraic lemma; its proof is left to the reader.
6.15. Lemma. * Let be a polynomial. Assume*
[TABLE]
Then is divisible by ; that is, there is a polynomial such that
[TABLE]
Proof of 6C. Let us apply induction on ; the base case is evident.
Assume that the statement holds for ; that is,
[TABLE]
We need to show that
[TABLE]
First, let us show that the equality holds if ; that is,
[TABLE]
Indeed, is the sum of all monomials in without . Each of these monomials corresponds to a spanning tree in with ; in other words, has a leaf at .111We use as a label of a vertex in and as the corresponding variable. Note that the tree is obtained from another tree with the vertices by adding an edge from to for some .
Note that the monomial in that corresponds to equals to the product of times the monomial in that corresponds to . To get the sum of all monomials in without , we need to sum up these products for all and all the monomials in ; this way we get
[TABLE]
By ➌ ‣ 6C, the latter equals
[TABLE]
which implies ➎ ‣ 6C.
Now assume ➍ ‣ 6C does not hold. Denote by the left-hand side in ➍ ‣ 6C. Observe that is a symmetric homogeneous polynomial of degree . That is, any permutation of values does not change and each monomial in has total degree .
By Lemma 6C and ➎ ‣ 6C, we get that is divisible by . Since is symmetric it is divisible by each ; that is, there is a polynomial such that
[TABLE]
Since , the total degree of is at least . But has degree , a contradiction. ∎
6.16. Exercise. * Show that the number of spanning trees in with degree at the first vertex equals .*
6.17. Exercise. * Assume that the vertices of the left part of have corresponding variables and the vertices in the right part have corresponding variables . Show that*
[TABLE]
Conclude that .
D Remarks
Very good expository papers on chromatic polynomials are written by Ronald Read [36] and Alexandr Evnin [11]. Matching polynomials are discussed in a paper by Christopher Godsil and Ivan Gutman [15].
Our discussion of spanning-tree polynomials is based on a modification by Fedor Petrov [34] of the original proof by Arthur Cayley [5].
Chapter 7 Generating functions
In this chapter, we discuss one connection between graph theory and power series; we prove the exponential formula (Theorem 7C) and discuss its applications. This material is similar to Chapter 6, but more challenging.
A Exponential generating functions
The power series
[TABLE]
is called the exponential generating function of the sequence
If the series converges in some neighborhood of zero, then it defines a function which remembers all information of the sequence . The latter follows since
[TABLE]
that is, the -th derivative of at [math] equals .
However, without assuming the convergence, we can treat as a formal power series. We are about to describe how to add, multiply, take the derivative, and do other operations with formal power series.
Sum and product. Consider two exponential generating functions
[TABLE]
We will write
[TABLE]
if the power series and are obtained from and by opening the parentheses of these formulas and combining like terms.
It is straightforward to check that is the exponential generating function for the sequence
[TABLE]
The product is also exponential generating function for the sequence
[TABLE]
7.1. Exercise. * Assume is the exponential generating function of the sequence Show that corresponds to the sequence .*
Composition. Once we define addition and multiplication of power series we can also substitute one power series into another. For example, if , then the expression
[TABLE]
is another power series obtained by substituting for in
[TABLE]
It is harder to express the sequence corresponding to in terms of , but it is easy to find the first few terms. Since we assume , we have
[TABLE]
Derivative. The derivative of
[TABLE]
is defined as the following formal power series
[TABLE]
Note that coincides with the ordinary derivative of if the latter converges.
Also, is the exponential generating function of the sequence obtained from the original sequence by deleting the zero-term and shifting the indexes by 1.
7.2. Exercise. * Let be the exponential generating function of the sequence Describe the sequence with the exponential generating function*
[TABLE]
Calculus. If converges and
[TABLE]
then we have
[TABLE]
Also by taking the derivative of we get that
[TABLE]
These identities have a perfect meaning in terms of formal power series and they still hold without convergence. We will not prove it, but the proof is not difficult.
B Fibonacci numbers
Recall that Fibonacci numbers, denoted as , are defined using the recursive identity with initial conditions and .
7.3. Exercise. * Let be the exponential generating function of Fibonacci numbers .*
- (a)
Show that it satisfies the following differential equation
[TABLE] 2. (b)
Conclude that
[TABLE] 3. (c)
Use the identity ➊ ‣ 7A to derive the Binet’s formula:
[TABLE]
C Exponential formula
Fix a set of graphs . Denote by the number of spanning subgraphs of isomorphic to one of the graphs in . Let
[TABLE]
be the exponential generating function of the sequence .
7.4. Theorem. * Let be a set of connected graphs.*
- (a)
Fix a positive integer . Denote by the number of spanning subgraphs of with exactly connected components such that each component is isomorphic to one of the graphs in . Then
[TABLE]
where is the exponential generating function of the sequence . 2. (b)
Let be the number of all spanning subgraphs of such that each connected component is isomorphic to one of the graphs in . Let be the exponential generating function of the sequence . Then
[TABLE]
Taking the logarithm and derivative of the formula in b, we get the following:
7.5. Corollary. * Assume and as in Theorem 7Cb. Then*
[TABLE]
The second formula in this corollary provides a recursive formula for the corresponding sequences which will be important later.
Proof of 7C; a. Denote by the number of spanning subgraphs of which have ordered connected components and each connected component is isomorphic to one of the graphs in . Let be the corresponding generating function.
Note that for each graph described above there are ways to order its components. Therefore, for any and
[TABLE]
Hence it is sufficient to show that
[TABLE]
To prove the latter identity, we apply induction on and the multiplication formula ➋ ‣ 7A for exponential generating functions. The base case is evident.
Assuming that the identity ➍ ‣ 7C holds for ; we need to show that
[TABLE]
Assume that a spanning graph with ordered connected components of is given. Denote by the number of vertices in the first components. There are ways to choose these vertices among vertices of . For each choice, we have ways to choose a spanning subgraph with components in it. The last component has vertices and we have ways to choose a subgraph from . All together, we get
[TABLE]
spanning graphs with ordered connected components and vertices in the first components. Summing it up for all , we get the multiplication formula ➋ ‣ 7A for exponential generating functions; hence ➎ ‣ 7C follows.
b*.* To find we need to add the numbers of spanning graphs with components. That is,
[TABLE]
The last equality follows from the Taylor expansion for ; see ➌ ‣ 7A. ∎
D Perfect matchings
Recall that a perfect matching is a 1-factor of the graph. In other words, it is a set of isolated edges that covers all the vertices. Note that if a graph admits a perfect matching, then the number of its vertices is even.
The double factorial is defined as the product of all the integers from up to some non-negative integer that have the same parity (odd or even) as ; the double factorial of is denoted by . For example,
[TABLE]
7.6. Exercise. * Let denote the number of perfect matchings in . Show that*
- (a)
; 2. (b)
* for any odd ;* 3. (c)
* for any integer .* 4. (d)
Conclude that and for any odd .
Let us present a proof of the second part of Exercise 7Dd based on Theorem 7C.
Solution. Denote by the number of perfect matchings in and let be the corresponding exponential generating function.
Note that a perfect matching can be defined as a spanning subgraph such that each connected component is isomorphic to . So we can apply the formula in Theorem 7C for the set consisting of only one graph .
Note that if contains a spanning subgraph isomorphic to , then . It follows that and for . Therefore,
[TABLE]
[TABLE]
That is,
[TABLE]
for any positive integer . In particular,
[TABLE]
That is, for any odd and for even . ∎
Remark. Note that by Corollary 7C, we also have
[TABLE]
which is equivalent to the recursive identity
[TABLE]
E All matchings
Now let be the set of two graphs and . Evidently, . Further, we have that for all since contains no spanning subgraphs isomorphic to or .
Therefore, the exponential generating function of the sequence is a polynomial of degree 2
[TABLE]
Note that a matching in a graph can be identified with a spanning subgraph with all connected components isomorphic to or . If we denote by the number of all matchings and by the corresponding exponential generating function, then by Theorem 7Cb, we get that
[TABLE]
Applying Corollary 7C, we also have
[TABLE]
The latter is equivalent to the following recursive formula for :
[TABLE]
Since and , we can easily find first the few terms of this sequence:
[TABLE]
7.7. Exercise. * Prove formula ➏ ‣ 7E directly — without using generating functions. Compare to Exercise 6Bb.*
F Two-factors
Let be the set of all cycles.
Note that a -factor of a graph can be defined as a spanning subgraph with components isomorphic to cycles. Denote by and the number of -factors and spanning cycles111Also known as Hamiltonian cycles. in respectively. Let and be the corresponding exponential generating functions.
7.8. Exercise. **
- (a)
Show that and
[TABLE]
for . In particular,
[TABLE] 2. (b)
Use part a and the following identity from 7C
[TABLE]
to find . 3. (c)
Count the number of 2-factors in by hand, and compare it to the result in part b. 4. (d)
Use part a to conclude that
[TABLE] 5. (e)
Use part d and Theorem 7Cb* to show that*
[TABLE]
G Counting spanning forests
Recall that a forest is a graph without cycles. Assume we want to count the number of spanning forests in ; denote its number by and by the number of connected spanning forests. That is, is the number of spanning trees in . For example, has the following 7 spanning forests; therefore .
By Corollary 7C, the following identity
[TABLE]
holds for the corresponding exponential generating functions.
According to the Cayley theorem, ; in particular,
[TABLE]
Applying the product formula ➋ ‣ 7A, we can use to calculate recurrently:
[TABLE]
It is instructive to check by hand that there are exactly spanning forests in .
For the general term of , no simple formula is known; however, the recursive formula above provides a sufficiently fast way to calculate its terms.
H Counting connected subgraphs
Let be the number of all spanning subgraphs of and be the number of connected spanning subgraphs of . All 4 connected spanning subgraphs of are shown on the diagram; therefore .
Assume and are the corresponding exponential generating functions. These two series diverge for all ; nevertheless, the formula for formal power series in Theorem 7Cb still holds, and by Corollary 7C we can write
[TABLE]
Note that ; indeed, to describe a subgraph of we can choose any subset of edges of , and is the total number of these independent choices. In particular, the first few terms of are
[TABLE]
Applying the product formula ➋ ‣ 7A, we can calculate the first few terms of :
[TABLE]
Note that in the previous section we found from , and now we go in the opposite direction. No closed formula is known for , but the recursive formula gives a sufficiently good way to calculate it.
I Remarks
The method of generating functions was introduced and widely used by Leonard Euler; the term generating function was coined later by Pierre Laplace. For more on the subject, we recommend the classic book by Frank Harary and Edgar Palmer [20].
Chapter 8 Minimum spanning trees
A Optimization problems
We say that a graph has weighted edges if each edge in is labeled by a real number, called its weight. In this case the weight of a subgraph (briefly ) is defined as the sum of the weights of its edges.
An optimization problem typically asks to minimize the weight of a subgraph of a certain type. A classic example is the so-called traveling salesman problem. It asks to find a Hamiltonian cycle with minimal weight (if it exists in the graph).
No fast algorithm is known to solve this problem and it is expected that no fast algorithm exists. We say fast if the required time depends polynomially on the number of vertices in the graph. The best-known algorithms require exponential time, which is just a bit better than brute force checking of all possible Hamiltonian cycles.
In fact, the traveling salesman problem is a classic example of the so-called NP-hard problems; it means that if you find a fast algorithm to solve it, then you solve the P=NP problem — the most important question in modern mathematics.
The so-called nearest neighbor algorithm gives a heuristic way to find a Hamiltonian cycle with small (but not necessarily the smallest) weight. In the following description, we assume that the graph is complete.
- (i)
Start at an arbitrary vertex. 2. (ii)
Find the edge with minimal weight connecting the current vertex and an unvisited vertex and move to . 3. (iii)
Repeat step ii until all vertices are visited and then return to the original vertex. This way we walked along a Hamiltonian cycle.
The nearest neighbor algorithm is an example of greedy algorithms; in other words, it chooses the cheapest step at each stage. The cheap choices at the beginning might force it to make expansive choices at the end, so it may not find an optimal solution. In fact a greedy algorithm might give the worst solution. For example, if we start from the vertex in the shown weighted graph, then the nearest neighbor algorithm produces the cycle , which is the worst — its total weight is , and the other two Hamiltonian cycles have weights and .
Let us list a few other classic examples of optimization problems. Unlike the traveling salesman problem, there are fast algorithms which solve them.
The shortest path problem asks for a path with minimal weight, connecting two given vertices. It can be thaut of as optimal driving directions between two locations.
The assignment problem asks for a matching of a given size with minimal weight, in a bipartite graph. It can be used to minimize the total cost of work, assuming that available workers charge different prices for each task. This problem is closely related to the subject of the next chapter.
The minimum spanning tree problem asks for a spanning tree with minimal weight. For example, we may think of minimizing the cost to build a computer network between given locations. This problem will be the main subject of the remaining sections of this chapter.
B Neighbors of a spanning tree
Let be a connected graph. Suppose and are spanning trees in . If
[TABLE]
for some edges and , then we say that is a neighbor of .
8.1. Exercise. * Let be a connected graph that is not a tree. Suppose that any two distinct spanning trees in are neighbors. Show that has exactly one cycle.*
8.2. Theorem. * Let be a graph with weighted edges. A spanning tree in has minimal weight if and only if*
[TABLE]
for any neighbor of .
8.3. Exchange lemma. * Let and be spanning trees in a graph . Then for any edge in that is not in , there is an edge in , but not in such that the subgraphs*
[TABLE]
are spanning trees in .
Proof. Note that has two components, denote them by and .
Let be a path in that connects the vertices of . Note that the path starts in and ends in . Therefore, one of the edges of , say , connects to .
The subgraph has a cycle created by and . Removing the edge from this cycle leaves a connected subgraph . Evidently, and have the same number of vertices and edges; since is a tree, so is .
Further . Since connects two connected subgraphs and , the resulting subgraph is connected. Again and have the same number of vertices and edges; since is a tree, so is . ∎
Proof of 8B. The ‘‘only-if’’ part is trivial — if has minimal weight among all spanning trees, then, in particular, it has to be minimal among its neighbors. It remains to show the ‘‘if’’ part.
Suppose that a spanning tree has the minimal weight among its neighbors. Consider a minimum weight spanning tree in ; if there is more than one, then we assume in addition that shares with the maximal number of edges.
Arguing by contradiction, assume is not a minimum weight spanning tree; in this case . Then there is an edge in , but not in . Let be an edge in provided by the exchange lemma; in particular, and are spanning trees.
Since is a neighbor of , we have
[TABLE]
Since has minimal weight, we have an equality in the last inequality. That is, is another minimum weight spanning tree in . By construction, has one more common edge with ; the latter contradicts the choice of . ∎
8.4. Exercise. * Let be a graph with weighted edges and a minimal weight spanning tree in .*
Suppose that for each edge , we change its weight from to . Show that remains a minimal weight spanning tree in with the new weights.
C Kruskal’s algorithm
Kruskal’s algorithm finds a minimum weight spanning tree in a given connected graph. If the graph is not connected, the algorithm finds a spanning tree in each component of the graph. Let us describe it informally.
- (i)
Suppose is a graph with weighted edges. Start with the spanning forest in formed by all vertices and no edges. 2. (ii)
Remove from an edge with minimal weight. If this edge connects different trees in , then add it to . 3. (iii)
Repeat the procedure while has any remaining edges.
Note that Kruskal’s algorithm is greedy; it chooses the cheapest step at each stage.
8.5. Theorem. * Kruskal’s algorithm finds a minimum weight spanning tree in any connected pseudograph with weighted edges.*
Proof. Evidently, the subgraph remains a forest at all stages. If at the end, has more than one component, then the original graph could not have edges connecting these components; otherwise, the algorithm would choose it at some stage. Therefore, the resulting graph is a tree.
Suppose the spanning tree produced by Kruskal’s algorithm does not have minimal total weight. Then does not satisfy Theorem 8B; that is, there is a neighbor of such that
[TABLE]
Therefore, .
Denote by the forest at the stage right before adding to it. Note that is a subgraph of . Since connects different trees in , it connects different trees in . But , therefore the algorithm should reject at this stage — a contradiction. ∎
D Other algorithms
Let us describe two more algorithms that find a minimal weight spanning tree in a given connected graph . For simplicity, we assume that has distinct weights.
Prim’s algorithm.
- (i)
Start with a tree formed by a single vertex in . 2. (ii)
Choose an edge with the minimal weight that connects to a vertex not yet in the tree. Add to . 3. (iii)
Repeat step ii until all vertices are in the tree .
Borůvka’s algorithm.
- (i)
Start with the forest formed by all vertices of and no edges. 2. (ii)
For each component of , find an edge with minimal weight that connects to another component of ; add to . 3. (iii)
Repeat step ii until has one component.
8.6. Exercise. * Show that both (a) Prim’s algorithm and (b) Borůvka’s algorithm produce a minimum weight spanning tree for a given connected graph with all different weights.*
8.7. Exercise. * Compare Kruskal’s algorithm, Prim’s algorithm, and Borůvka’s algorithm. Which one should work faster? When and why?*
8.8. Exercise. * Assume is a connected graph with weighted edges and all weights are different. Let be a minimum weight spanning tree in .*
- (a)
Suppose is an end vertex in and is its adjacent edge. Show that belongs to . 2. (b)
Suppose is a cycle in and is an edge in with maximal weight. Show that does not belong to . 3. (c)
Come up with a minimum-spanning-tree algorithm based on observations a and b.
E Remarks
The NP-hardness of the traveling salesman problem follows from a result by Richard Karp [26]. The discussed algorithms are named after Joseph Kruskal, Robert Prim and Otakar Borůvka. Joseph Kruskal also described the so-called reverse-delete algorithm [29], which is closely relevant to Exercise 8D. Prim’s algorithm was first discovered by Vojtěch Jarník, rediscovered independently two times: by Robert Prim and by Edsger Dijkstra.
For more on optimization problems, read the book by Dieter Jungnickel [24]. A reader-friendly introduction to the P=NP problem is given in the book by Thomas Cormen, Charles Leiserson, Ronald Rivest, and Clifford Stein [6, Chapter 34].
Chapter 9 Marriage theorem
and its relatives
A Alternating and augmenting paths
Recall that a matching in a graph is a set of edges without common vertices.
Let be a matching in a graph . Suppose a path in alternates between edges from and edges not from , then is called an -alternating path.
If an alternating path connects two unmatched vertices of , then it is called -augmenting. An -augmenting path can be used to improve the matching ; namely by deleting all the edges of in and adding the remaining edges of , we obtain a new matching with more edges. This construction implies the following:
9.1. Observation. * Assume is a graph and is a maximal matching in . Then has no -augmenting paths.*
On the diagrams we denote the edges in by solid lines and the remaining edges by dashed lines.
9.2. Exercise. * Connect two unmatched vertices on the diagram by an augmenting path. Use it to construct a larger matching.*
Recall that a bigraph stands for a bipartite graph.
9.3. Exercise. * Let be a matching in a bigraph . Suppose is an -augmenting path. Show that has its ends in the opposite parts of .*
The following theorem states the converse to the observation.
9.4. Theorem. * Assume is a matching in a graph . If is not maximal, then contains an -augmenting path.*
This theorem implies that the Hungarian algorithm [21, Section 7.2] produces a maximal matching. Indeed, the Hungarian algorithm checks all -alternating paths in the graph and it will find an -augmenting path if it exists. Once it is found it could be used to improve the matching . Therefore, by the theorem, theHungarian algorithm improves any nonmaximal matching; in other words, it can stop only at a maximal matching.
Proof. Suppose is a nonmaximal matching; that is, there is a matching with a larger number of edges. Consider the subgraph of formed by all edges in and .
Note that each component of is either an -alternating path or a cycle. Each cycle in has the same number of edges from and . Since , the one component of has to be a path that starts and ends with an edge in . Evidently, such a path is -augmenting. ∎
B Marriage theorem
Assume that is a graph and is a subset of its vertex set. We say that a matching of covers if any vertex in is incident to an edge in .
Given a set of vertices in a graph , the set of all vertices adjacent to at least one of the vertices in will be called the set of neighbors of . Note that if is a bigraph and lies in the left part, then lies in the right part.
9.5. Marriage theorem. * Let and be the left and right parts of a bigraph . There is a matching that covers if and only if for any , we have*
[TABLE]
where is the set of all neighbors of .
Proof. Assume that a matching is covering . Note that for any set , the set of its neighbors includes the vertices matched with . In particular,
[TABLE]
it proves the ‘‘only if’’ part.
Consider a maximal matching of . To prove the ‘‘if’’ part, it is sufficient to show that covers . Arguing by contradiction, assume that there is a vertex in that is not incident to any edge in .
Consider the maximal set of vertices in which are reachable from by -alternating paths. Denote by and the set of left and right vertices in respectively.
Since is maximal, is the set of neighbors of . According to Observation 9A, the matching provides a bijection between and . In particular,
[TABLE]
the latter contradicts the assumption. ∎
9.6. Exercise. * Assume is an -regular bigraph; . Show that*
- (a)
* admits a 1-factor;* 2. (b)
the edge chromatic number of is ; in other words, can be decomposed into -factors.
In Exercise 9B one has to assume that is bipartite. Indeed, the following exercise states that the graph shown on the diagram has no 1-factor, while it is 3-regular.
9.7. Exercise. * Prove that the graph shown on the diagram has no 1-factor.*
9.8. Exercise. * Children from 25 countries, 10 kids from each, decided to stand in a rectangular formation with 25 rows of 10 children in each row. Show that you can always choose one child from each row so that all 25 of them will be from different countries.*
9.9. Exercise. * The sons of the king divided the kingdom between each other into 23 parts of equal area — one for each son. Later a new son was born. The king proposed a new subdivision into 24 equal parts and gave one of the parts to the newborn son.*
Show that each of the 23 older sons can choose a part of land in the new subdivision which overlaps with his old part.
9.10. Exercise. * A table is filled with nonnegative numbers. Assume that the sum in each column and each row is 1. Show that one can choose cells with positive numbers that do not share columns and rows.*
9.11. Advanced exercise. * In a group of people, for some fixed and any , any girls like at least boys in total. Show that then all but girls may get married to the boys they like.*
C Vertex covers
A set of vertices in a graph is called a vertex cover if any edge is incident to at least one of the vertices in .
9.12. Theorem. * In any bigraph, the number of edges in a maximal matching equals the number of vertices in a minimal vertex cover.*
It is instructive to do the following exercise before reading the proof.
9.13. Exercise. * Let be a maximal matching in a bigraph . Assume two unmatched vertices and lie on the opposite parts of . Show that no pair of -alternating paths starting from and can have a common vertex.*
On the following diagram, a maximal matching is marked by solid lines; the remaining edges of the graph are marked by dashed lines. The vertices of the cover are marked in black and the remaining vertices in white; the unmatched vertices are marked by a cross.
Proof. Fix a bigraph ; denote by and its left and right parts. Let be a matching, and be a vertex cover in .
By the definition of vertex cover, any edge in is incident to at least one vertex in . Therefore,
[TABLE]
That is, the number of vertices in any vertex cover is at least as large as the number of edges in any matching .
Now assume that is a maximal matching. Let us construct a vertex cover such that .
Denote by and the set of left and right unmatched vertices (these are marked by a cross on the diagram). Denote by and the set of vertices in which can be reached by -alternating paths starting from and from , respectively.
Note that and do not overlap. Otherwise, there would be an -augmenting path from to (compare to Exercise 9C). Therefore, is not maximal — a contradiction.
Further note that if is an edge in , then both of its end vertices lie either in or , or neither.
Let us construct the set by taking one incident vertex (left or right) of each edge in by the following rule: if connects vertices in , then include its right vertex in ; otherwise include its left vertex. Since has exactly one vertex incident to each edge of , we have
[TABLE]
It remains to prove that is a vertex cover; that is, at least one vertex of any edge in is in .
Note that if the left vertex of lies in , then is an edge on an -alternating path starting from . Therefore, the right vertex of also lies in .
Therefore, it is sufficient to consider only the following three cases:
The edge has its right vertex in and its left vertex outside of . In this case, both vertices of lie in .
The edge connects vertices in . In this case, the right vertex of is in .
The edge connects vertices outside of . In this case, the left vertex of is in . ∎
9.14. Exercise. * A few squares on a chessboard are marked. Show that the minimal number of rows and columns that cover all marked squares is the same as the maximal number of rooks on the marked squares that do not threaten each other.*
D Edge cover
A collection of edges in a graph is called an edge cover if every vertex is incident with at least one of the edges in .
On the diagram, two edge covers of the same graph are marked in solid lines.
9.15. Theorem. * Let be a connected graph with vertices and . Assume that a minimal edge cover of contains edges, and a maximal matching of contains edges. Then*
[TABLE]
The following exercise will guide you thru the proof.
9.16. Exercise. * Suppose , , , , , and are as in the theorem. Denote by the number of components in .*
- (a)
Show that contains no paths of length 3 and no triangle. Conclude that each component of is a star; that is, it is isomorphic to for some . Use it to show that
[TABLE] 2. (b)
Choose edges, one in each component of . Observe that it is a matching. Conclude that . Use a to show that
[TABLE] 3. (c)
Consider a subgraph of formed by all edges in the matching and one edge incident to each unmatched vertex in . Observe that is a vertex cover with edges. Conclude that
[TABLE] 4. (d)
Observe that b and c imply the theorem.
E Minimal cut
Recall that a directed graph (or briefly a digraph) is a graph where the edges have a direction (also known as orientation) associated with them; that is, an edge in a digraph is defined as an ordered pair of vertices. A choice of directions on every edge of a graph is called its orientation.
9.17. Min-cut theorem. * Let be a digraph. Fix vertices and in . Then the maximal number of directed paths from to which do not have common edges equals the minimal number of edges one can remove from so that there will be no directed path from to .*
Proof. Denote by the maximal number of directed paths from to which do not have common edges. Denote by the minimal number of edges one can remove from to disconnect from ; more precisely, after removing edges from , there will be no directed path from to .
Let be a maximal collection of directed paths from to which have no common edges. Note that in order to disconnect from , we have to cut at least one edge in each path . In particular, .
Consider the new orientation on where each path is oriented backward — from to .
Consider the set of the vertices which are reachable from by directed paths for this new orientation.
Assume contains ; that is, there is a path from to , which can move along only backward.
(Further, the path will be used the same way as the augmenting path in the proof of the marriage theorem. In a sequence of moves, we will improve the collection so that there will be no overlaps. On the diagram, a case with that requires two moves is shown; is marked by a solid line and is marked by a dashed line.)
Since is a maximal collection, overlaps with some of the paths . Without loss of generality, we can assume that first overlaps with — assume it meets at the vertex and leaves it at the vertex . Let us modify the paths and the following way: Instead of the path consider the path that goes along from to and after that goes along to . Instead of the path , consider the trail which goes along from to and after that goes along to .
If the constructed trail is not a path (that is, if visits some vertices several times), then we can discard some circuits from to obtain a genuine path, which we will still denote by .
Note that the obtained collection of paths satisfies the same conditions as the original collection. Further, since we discard the part of from to , the total number of edges in is smaller than in the original collection . Therefore, by repeating the described procedure several times, we get paths without overlaps — a contradiction.
It follows that .
Note that all edges which connect to the remaining vertices of are directed toward . That is, every such edge which comes out of in the original orientation belongs to one of the paths .
Moreover, for each path there is only one such edge. In other words, if a path leaves , then it cannot come back. Otherwise, could be made larger by moving backward along . Therefore, cutting one such edge in each path makes it impossible to leave . In particular, we can disconnect from by cutting edges from ; that is, . ∎
Remark. The described process has the following physical interpretation. Think of each path , and like water pipelines from to .
At each overlap of with another path , the water in and runs in the opposite directions. So we can cut the overlapping edges and reconnect the open ends of the pipes to each other while keeping the water flow from to unchanged. As a result, we get pipes from to with no common edges and possibly some cycles which we can discard. An example of this procedure for two paths and is shown on the diagram; as above, is marked by a solid line and is marked by a dashed line.
9.18. Advanced exercise. * Assume is a bigraph. Let us add two vertices, and , to so that is connected to each vertex in the left part of , and is connected to each vertex in the right part of . Orient the graph from left to right. Denote the obtained digraph by .*
Give another proof of the marriage theorem for a bigraph , applying the min-cut theorem to the digraph .
F Remarks
The marriage theorem was proved by Philip Hall in [19]; it has many applications in all branches of mathematics. The theorem on vertex cover was discovered by Dénes Kőnig [28], and rediscovered by Jenő Egerváry [8]. The theorem on min-cut was proved by Peter Elias, Amiel Feinstein, and Claude Shannon [9]; it was rediscovered by Lester Ford and Delbert Fulkerson [13].
These theorems form a base for the Hungarian algorithm that solves the assignment problem; see page ‣ 8A. This algorithm was found by Harold Kuhn, but, as it was discovered by François Ollivier, essentially the same algorithm was found much earlier by Carl Gustav Jacobi [33].
An extensive overview of the marriage theorem and its relatives is given by Alexandr Evnin in [10].
Chapter 10 Toroidal graphs
Recall that a graph is called planar if it can be drawn on a plane with no crossings; the latter is equivalent to the existence of its drawing on a sphere. Here and further, we say drawing for drawing with no crossings.
In this chapter, we will discuss drawing of graphs on a torus — another surface that can be obtained by revolving a circle about an axis (this is the surface of a donut).
A graph is called toroidal, if it can be drawn on a torus with no crossing.
If one cuts a torus along a parallel and a meridian as shown on the diagram, then the obtained surface can be developed into a square.111Formally speaking, this means that there is a continuous map from a square to torus such that if and only if or and are corresponding points on the opposite sides of the square. It gives a convenient way to describe drawings of graphs on the torus which will be called square diagram. One only has to remember that the corresponding points on opposite sides of the square are identified in the torus; in particular, all four vertices of the square correspond to one point in the torus.
For example, on the given square diagram you see a drawing of the complete graph ; in particular it shows that is toroidal. Note that the edge , after coming to the right side of the square, reappears at the corresponding point of the left side and goes further to . Similarly, the edge comes to the top and reappears at the bottom. At these points, the edges cross the parallel and the meridian.
The following square diagrams show that and are toroidal graphs as well.
10.1. Exercise. * Show that any graph with crossing number 1 is toroidal. That is, if a graph admits a drawing on a sphere with one crossing, then it admits a drawing on a torus with no crossings.*
10.2. Exercise. * Show that each of the following graphs is toroidal; construct a corresponding square diagram in each case.*
*(a) (b) (c) *
A Simple regions
Choose a drawing of a pseudograph on a torus or sphere; it subdivides the surface into regions. A region is called simple if its interior can be parameterized by an open plane disc.222Formally speaking, this means that there is a continuous bijection from the interior of to an open disc in the plane such that its inverse is also continuous.
The annulus shown on the diagram is an example of a nonsimple region. On a sphere it may appear only in drawings of nonconnected graphs. That is, if a drawing of a pseudograph on a sphere has a nonsimple region, then is not connected. This statement should be intuitively obvious, but the proof is not trivial.
Drawings of connected graphs on a torus may have nonsimple regions. For example, in the first drawing of below,
the regions and are simple, and is not (it contains the meridian of the torus). The second drawing of has only two regions, and , and both of them are simple.
Further, we will use the following claim, which should be intuitively obvious. We do not present its proof, but it is not hard; a reader familiar with topology may consider it as an exercise.
10.3. Claim. * Let be a drawing of a pseudograph on a torus or sphere and its simple region. Suppose that a drawing obtained from by adding a new edge in the region .*
- (a)
If both ends of are in , then divides into two simple regions. 2. (b)
If only one the ends of is in , then does not divide and the corresponding region of is simple.
10.4. Advanced exercise. * Suppose that is a drawing of a connected nonlanar graph on the torus. Show that each region of is simple.*
B Euler’s formula
10.5. Theorem. * For any drawing of a connected pseudograph on the torus we have*
[TABLE]
where , , and denote the number of vertices, edges and regions in , respectively.
Moreover, equality holds if all regions of are simple.
Note that the inequality might be strict. For example, for the first drawing of , above we have
[TABLE]
It happens since the region is not simple. For the second drawing of we have the equality
[TABLE]
as it is supposed to be by the second part of the theorem.
In the proof we will use the following lemma which is a simple corollary of Claim 10A. Given a drawing of pseudograph with vertices, edges and regions, set
[TABLE]
10.6. Lemma. * Let be a drawing of a pseudograph on a torus or sphere. Suppose that another drawing of a connected pseudograph that is obtained from by adding a new edge . Then*
[TABLE]
Moreover, if all regions of are simple, then
- (a)
we have equality in ➊ ‣ 10B and 2. (b)
all regions of are simple as well.
Proof. Since is connected, one of the ends of belongs to . Denote by the region of that contains .
If the other end of is not in , then does not divide its region. In this case has an extra vertex and an extra edge, and the number of regions did not change; that is, , , and . Therefore,
[TABLE]
If the other end of is in , then may divide into two regions or may not divide it. According to Claim 10A, the latter may happen only if is not simple. Note that after adding , the number of vertices did not change. That is, , , and the equality holds if is simple. Therefore,
[TABLE]
and the equality holds if is simple.
According to Claim 10A, if is simple, then the region(s) of that correspond to are simple as well. The rest of the regions did not change. Hence b follows. ∎
Proof of 10B. We need to show that
[TABLE]
for any drawing of connected pseudograph on the torus .
Let be a drawing of the pseudograph formed by one meridian and one parallel as shown. Note that it has 1 vertex, 2 edges, and 1 region; therefore
[TABLE]
Without loss of generality, we may assume that the edges of and intersect and have only a finite number of points of intersection. The latter can be achieved by perturbing the drawing of .
Let us subdivide the graphs and by adding a new vertex at every crossing point of the graphs. The obtained graphs, say and , are subgraphs of a bigger graph, say , formed by all edges and vertices of and . Note that adding a vertex on an edge increases the number of vertices and edges by and the number of regions stays the same. Since the subdivision is obtained by adding a finite number of extra vertices, we get that
[TABLE]
By construction, is connected. If , then there is an edge of that is not in but has one of its ends in . By adding to and applying the procedure recursively, we will obtain in a finite number of steps. Applying Lemma 10B at each step, we get
[TABLE]
Analogously, can be obtained from in a finite number of steps by adding one edge at a time. Since the only region of is simple, the obtained drawings will have only simple regions. Therefore, applying Lemma 10B at each step, we get
[TABLE]
Finally, ➌ ‣ 10B, ➍ ‣ 10B, ➎ ‣ 10B, and ➏ ‣ 10B imply ➋ ‣ 10B; indeed
[TABLE]
∎
10.7. Exercise. * Suppose that is a toroidal graph with girth . Show that*
[TABLE]
where and denote number of vertices and edges in .
10.8. Exercise. * Is there a drawing of on the torus with only square regions? If ‘‘yes’’, then draw a square diagram; if ‘‘no’’, explain why.*
C The seven-color theorem
10.9. Theorem. * The chromatic number of any toroidal graph cannot exceed .*
Recall that is toroidal; see diagram on page 10. Therefore, the theorem gives an optimal bound. In the proof, we will use the following lemma, which is a simple corollary of Euler’s formula (10B).
10.10. Lemma. * Let be a toroidal graph with vertices and edges. Then*
[TABLE]
Proof. Choose a drawing of on a torus. By Euler’s inequality we have
[TABLE]
where denotes the number of regions in the drawing.
Note that each region in the drawing has at least sides and each edge of appears twice as a side of a region. Therefore,
[TABLE]
These two inequalities imply the corollary. ∎
Proof of 10C. Suppose that there is a toroidal graph that requires colors. Choose a critical subgraph in with chromatic number 8. Denote by and the number of vertices and edges in .
By [21, Theorem 2.1.3], each vertex in has a degree of at least . By the handshake lemma [21, Theorem 1.1.1], we have
[TABLE]
On the other hand, by Lemma 10C, we have
[TABLE]
These two inequalities contradict each other. ∎
D A remark about forbidden minors
It is straightforward to see that if is toroidal, then any graph obtained from by the deletion of a vertex or an edge or by contracting an edge is also toroidal.
A graph that can be obtained from a given graph by applying a sequence of such operations is called a minor of . The pseudograph is considered to be a minor of itself. The other minors require at least one deletion or contraction; they are called proper minors.
Note that the statement above implies that any minor of a toroidal graph is toroidal, or in other words, toridality is inherited by minors.
The following deep result was proved by Neil Robertson and Paul Seymour [38].
10.11. Theorem. * Any property of pseudographs that is inherited by minors can be described by a finite set of forbidden minors; that is, a pseudograph meets the property if and only if none of its minors are forbidden.*
For example, note that deletion and contraction do not create cycles; that is, any minor of a forest is a forest. Forests can be described by one forbidden minor — a pseudograph formed by one loop. Indeed, if a pseudograph has a cycle, then by a sequence of deletions and contractions, one can get a single loop from it.
10.12. Exercise. * Describe the following classes of graphs by a single forbidden minor.*
- (a)
Graphs that do not contain trees with 5 end vertices. 2. (b)
Graphs in which any two cycles have at most one common vertex.
A more complicated example is given by the Pontryagin–Kuratowski theorem; it states that planar graphs are characterized by two forbidden minors: and .
Since toroidality is inherited by its minors, it can be described by a set of forbidden minors. The complete list of forbidden minors for this problem has to be huge and is not yet known.
10.13. Advanced exercise. * Show that the graph on the diagram is not toroidal, but every proper minor of this graph is toroidal.*
E Other surfaces
One may draw graphs on the so-called the surfaces of genus . These surfaces can be obtained by attaching toruses to each other;
a surface of genus is shown in the picture. Namely, to construct a surface of genus , start with a surface of genus and a torus , drill a hole in each and reconnect them to each other as shown.333The described construction is called connected sum; so we can say that connected sum of a surface of genus and a torus is a surface of genus .
It is natural to assume that the sphere has genus [math] and the torus has genus . (In general, genus tells how many disjoint closed curves one could draw on the surface so that they do not cut the surface into pieces.)
The simple regions of a drawing on a surface of genus can be defined the same way as on the torus. Euler’s formula given in Theorem 10B admits the following straightforward generalization
10.14. Theorem. * For any drawing of a pseudograph on the surface of genus we have*
[TABLE]
where , , and denote the number of vertices, edges and regions of .
Moreover, equality holds if all regions of are simple.
The seven-color theorem also admits the following straightforward generalization; it was proved by Percy John Heawood [22].
10.15. Theorem. * If a graph admits a drawing on a surface of genus , then its chromatic number cannot exceed*
[TABLE]
Note that for a sphere (that is, for ) the formula gives , which is the right bound for the chromatic number for planar graphs. However, this is just a coincidence; the proof of Heawood works only for .
The estimate in the last theorem is sharp. The latter was proved by constructing a drawing on the surface of genus of the complete graphs for any . The final step in this construction was made by Gerhhard Ringel and Ted Youngs [37]. The solution uses the so-called rotations of graphs; which is a combinatoric way to encode a drawing of a graph on a surface. This is the subject of [21, Chapter 10].
One may consider nonoriented surfaces; for example, the so-called Möbius strip shown on the diagram. It turns out that it is possible to generalize Euler’s inequality and get an exact upper bound for the chromatic numbers for graphs that can be drawn on such surfaces. (The problem is slightly harder for the so-called Klein bottle, but still, its difficulty is not comparable with the four-color theorem.)
10.16. Exercise. * Draw the complete graph on a Möbius strip (assume it is made from a transparent material).*
Chapter 11 Rado graph
In this chapter, we consider one graph with many surprising properties. Unlike most of the graphs we have considered so far, this graph has an infinite set of vertices.
A Definition
Recall that a set is countable if it can be enumerated by natural numbers ; it might be infinite or finite.
A countable graph is a graph with a countable set of vertices; the set of vertices can be infinite or finite, but it cannot be empty since we always assume that a graph has a nonempty set of vertices.
11.1. Definition. * A Rado graph is a countable graph satisfying the following property:*
Given two finite disjoint sets of vertices and , there exists a vertex that is adjacent to any vertex in and nonadjacent to any vertex in .
The property in the definition will be called the Rado property; so we can say that for the sets of vertices and in a graph, the Rado property holds or does not hold.
11.2. Exercise. * Show that any Rado graph has an infinite number of vertices.*
11.3. Exercise. * Show that any Rado graph has diameter 2.*
B Stability
The following exercises show that the Rado property is very stable — small changes cannot destroy it.
11.4. Exercise. * Assume the set of vertices of a Rado graph is partitioned into two subsets. Show that the subgraph induced by one of these subsets is Rado.*
11.5. Exercise. * Let be a countable graph.*
- (a)
Assume is an edge in . Show that is a Rado graph if and only if so is . 2. (b)
Assume is a vertex in . Show that is a Rado graph if so is . 3. (c)
Assume is a vertex in . Consider the graph obtained from by replacing each edge from by a non-edge, and each non-edge from by an edge (leaving the rest unchanged). Show that is a Rado graph if and only if so is .
11.6. Exercise. * Let be a Rado graph. Assume that is the set of all vertices in adjacent to a given vertex . Show that the subgraph induced by is Rado.*
C Existence
11.7. Theorem. * There exists a Rado graph.*
Proof. Let be a finite graph. Denote by the graph obtained from according to the following rule: for each subset of vertices in add a vertex and connect it to all the vertices in .
Note that if has vertices, then has vertices — it has vertices of and additional vertices — one for each of subsets of the -element set (including the empty set).
The original graph is an induced subgraph in . Note also that is finite — it has vertices.
By construction, the Rado property holds in for any two sets and of vertices in — the required vertex is the vertex in that corresponds to the subset .
Let be a graph with one vertex.
By repeating the construction, we get a sequence of graphs , such that for any . The graphs are shown on the diagram.111It would be hard to draw since it contains vertices, and it is impossible to draw — it has vertices which exceeds by many orders the number of particles in the observable universe.
Since is a subgraph of for any , we can consider the union of the graphs in the sequence ; denote it by . By construction, each graph is a subgraph of induced by finitely many vertices. Moreover, any vertex or edge of belongs to any with a sufficiently large .
It remains to show that is Rado. Indeed, any two finite sets of vertices and belong to for some . From above, the Rado property holds for and in , and therefore in . ∎
Another construction. One could also construct a Rado graph by directly specifying which vertices are adjacent. Namely, consider the graph as on the diagram with vertices
such that is adjacent to for some if the -th bit of the binary representation of is 1.
For instance, vertex is adjacent to all with odd , because the numbers whose 0-th bit is nonzero are exactly the odd numbers. Vertex is adjacent to (since 1 is odd) and to all with or ; and so on.
11.8. Exercise. * Show that the described graph is Rado.*
D Uniqueness
In this section, we will prove that any two Rado graphs are isomorphic, so essentially, there is only one Rado graph. First, let us prove a simpler statement.
11.9. Theorem. * Let be a Rado graph. Then any countable graph (finite or infinite) is isomorphic to an induced subgraph of .*
Proof. Enumerate the vertices of as (the sequence might be finite or infinite).
It is sufficient to construct a sequence of vertices in such that is adjacent to if and only if is adjacent to . In this case, the graph is isomorphic to the subgraph of induced by .
We may choose any vertex of as . Suppose that the sequence is constructed. If has vertices, then the required sequence is already constructed. Otherwise, note that the Rado property implies that there is a vertex in that is adjacent to for if and only if is adjacent to .
Clearly, the new vertex meets all the required properties. Repeating this procedure infinitely many times, or until the sequence terminates, produces the required sequence . ∎
11.10. Exercise. * Show that any two vertices in a Rado graph can be connected by a path of length 10.*
11.11. Theorem. * Any two Rado graphs and are isomorphic.*
Moreover any isomorphism between finite induced subgraphs in and can be extended to an isomorphism .
Note that Theorem 11D implies that is isomorphic to an induced subgraph in and the other way around — is isomorphic to an induced subgraph in . For finite graphs these two properties would imply that the graphs are isomorphic; see Exercise 11D. As the following example shows, it does not hold for infinite graphs. It is instructive to understand this example before going into the proof.
The first graph on the diagram has an infinite number of vertices, non of which has degree 3. The second graph has exactly one vertex of degree 3. Therefore, these two graphs are not isomorphic.
Deleting the marked vertices from one graph produces the other one. Therefore, is isomorphic to a subgraph of and the other way around.
The proof below uses the same construction as in the proof of Theorem 11D, but it is applied back and forth to ensure that the constructed subgraphs contain all the vertices of the original graph.
Proof. Once we have proved the second statement, the first statement will follow if you apply it to single-vertex subgraphs and .
Since the graphs are countable, we can enumerate the vertices of and , as and , respectively. We will construct a sequence of induced subgraphs in and in with a sequence isomorphisms .
Suppose that an isomorphism is constructed.
If is even, set to be the smallest index such that not in . The Rado property guarantees that there is a vertex such that for any vertex in , is adjacent to if and only if is adjacent to . Set to be the graph induced by vertices of and ; further set to be the graph induced by vertices of and . The isomorphism can be extended to the isomorphism by setting .
If is odd, we do the same, but backwards. Let be the smallest index such that not in . The Rado property guarantees that there is a vertex which is adjacent to a vertex in if and only if is adjacent to . Set to be the graph induced by vertices of and ; further set to be the graph induced by vertices of and . The isomorphism can be extended to the isomorphism by setting .
Note that if , then for all . Therefore, we can define if for some .
By construction we get that
is defined for any . Therefore, is defined at any vertex of .
lies in the range of for any . Therefore, the range of contains all the vertices of .
is adjacent to if and only if is adjacent to .
Therefore, is an isomorphism. ∎
11.12. Exercise. * Explain how to modify the proof of 11D to prove the following theorem.*
11.13. Theorem. * Let be a Rado graph. A countable graph is isomorphic to a spanning subgraph of if and only if, given any finite set of vertices of , there is a vertex that is not adjacent to any vertex in .*
11.14. Exercise. * Let and be two vertices in a Rado graph . Show that there is an isomorphism from to itself that sends to .*
11.15. Exercise. * Let and be two finite graphs. Assume is isomorphic to a subgraph of and the other way around — is isomorphic to a subgraph of . Show that is isomorphic to .*
E The random graph
The following theorem explains why a Rado graph is also named the random graph.
11.16. Theorem. * Suppose is an infinite countable set of vertices. Let us connect each pair in by an edge independently and with probability . Then, with probability 1, we obtain a Rado graph.*
Proof. It is sufficient to show that for two given finite sets of vertices and , the Rado property fails with probability 0.
Assume ; that is, is the total number of vertices in and . The probability that a given vertex outside of and satisfies the Rado property for and is . Therefore, probability that a given vertex does not satisfy this property is .
Note that events that a given vertex does not satisfy the property are independent. Therefore, the probability that different vertices outside of and do not satisfy the Rado property for and is
[TABLE]
This value tends to 0 as ; therefore the event that no vertex is correctly joined has probability 0. ∎
11.17. Exercise. * Let . Assume an infinite countable graph is chosen at random by selecting edges independently with probability from the set of 2-element subsets of the vertex set. Show that with probability 1, the resulting graph is a Rado graph.*
F Remarks
The Rado graph is also called the Erdős–Rényi graph or random graph; it was first discovered by Wilhelm Ackermann, rediscovered later by Paul Erdős and Alfréd Rényi, and yet again by Richard Rado. An analogous object in metric geometry was discovered much earlier by Pavel Urysohn.
Theorem 11E was proved by Paul Erdős and Alfréd Rényi. A good survey on the subject is written by Peter Cameron [4].
Chapter 12 Rewriting systems
Suppose a certain set of objects and a set of elementary transformations for these objects are given, and we are interested in the following problem.
Under what conditions do these transformations reduce each object to a single form?
As we will see, this question is equivalent to an elegant statement about graphs (Lemma 12F).
For a precise formulation, we need to introduce rewriting systems. They are understood in two senses: narrow and broad. In the narrow sense, certain word transformations are considered. These are the so-called string rewriting systems, which will be defined in the next section. In the broad sense, we talk about any objects and transformations; these are called abstract rewriting systems.
A String rewriting systems
Let us choose a finite set of symbols and call it an alphabet. Any finite strings (including the empty string) of symbols in the alphabet will be called a word. If the alphabet has only two symbols , then the words are , , , , , , , , and so on.
Suppose a list of pairs of words, , is given; we are allowed to change any occurrence of to . More precisely, a word can be exchanged for the word ; here and are arbitrary words. In this case we write .
For instance, we might be allowed to change to and to anywhere in the word. All possible words that can be reached from are shown on the diagram.
We have just described the so-called string rewriting systems. To describe the given example, we will use the notation
[TABLE]
So, after , we list the symbols of the alphabet, and after , we list all allowed replacements, closing it with .
B Abstract rewriting systems
An abstract rewriting system is a digraph; that is, a pseudograph (typically infinite) on each edge of which one of two directions is chosen. The vertices of the digraph are often referred to as objects, and directed edges as transformations. For two objects and , we write if there is a directed edge that starts at vertex and ends at vertex ; it defines a relation on the set of objects.
We will be interested in the relation on the set of vertices. Of course, if we know the digraph, then we know the relation .
A string rewriting system can be interpreted as an abstract rewriting system with the set of vertices formed by all words in the alphabet.
Let us introduce two new binary relations on the set of vertices of the digraph that will be denoted by and .
The first relation means that following the arrows in the digraph, one can get from to ; it includes the case . More precisely, there is a sequence of objects for some integer .
The expression says that one can walk along edges in the digraph from to ; it is allowed to go in both directions of the edges. In other words, means that and lie in the same connected component of the undirected pseudograph.
C Terminating systems
A system is called terminating if there are no infinite sequences In other words, one cannot travel around the digraph indefinitely, following the directions of the arrows.
An object (a vertex) is said to be irreducible if there is no object such that . In other words, there is no way out of .
It is clear that in a terminating rewriting system, starting at vertex and traveling along the arrows, we will arrive at some irreducible vertex, say . In the latter case, we will say that is a normal form of .
Any object in the terminating system has at least one normal form, and there can be many of them. If has a unique normal form, then it will be denoted by .
12.1. Exercise. * Give an example of*
- (a)
a finite nonterminating system. 2. (b)
a terminating system such that a vertex has an arbitrary long sequence . 3. (c)
an infinite nonterminating system such that any object has a unique normal form.
D Confluent systems
A system is called confluent if the conditions and for any three objects , , imply that there is an object such that and . Informally, it means that if two people start at and follow the arrows for some time, then, going further along the arrows, they can always converge at some vertex .
12.2. Lemma. * Suppose two objects and in a confluent system have normal forms and , respectively. If , then .*
Proof. Since and , we get and for some object . Since is irreducible, we get ; that is .
Since and , we get and for some object . Since and are irreducible, we get . ∎
A terminating confluent system is called convergent.
12.3. Theorem. * In a convergent system, any object has a unique normal form. Moreover, for any two objects and in the system we have*
[TABLE]
Proof. Since the system is terminating, any object has a normal form.
Let and be normal forms of ; in particular and . Since the system is confluent, and for some object . However, since both objects and are irreducible, we have and ; that is, any object has a unique normal form, which proves the main statement.
Suppose . Then and . Therefore, ; we proved the if part of the second statement.
Now suppose ; so there is a sequence such that or for each . By 12D,
[TABLE]
Hence the only-if part follows. ∎
E Proving termination
So, how do we prove that a system is terminating? Typically, we construct a relation ‘‘’’ on the set of vertices such that implies . If any decreasing sequence with respect to is terminating, then our original system is terminating as well.
More precisely, ‘‘’’ is a strict partial order; that is, ‘‘’’ is a relation that satisfies the following conditions for all objects , , and :
.
If , then .
If and then .
If there is no infinite sequence , and implies , then our system is terminating.
12.4. Exercise. * Let be an abstract rewriting system with a set of objects formed by positive integers and the transformations if is even and if is odd. Show that is terminating.*
For string rewriting systems, the words can be ordered first by length, and if the lengths are equal, lexicographically. This is the so-called shortlex order. Note that for this order, there is no infinite decreasing sequence. In this case, to ensure that the system is terminating, it is sufficient that when moving from one word to another along the arrow, either the length of the word decreases, or it does not change, but the word decreases relative to the order of words in the dictionary.
12.5. Exercise. * Show that the following process always terminates. There is a box with a finite number of black and white balls. Each step consists of removing an arbitrary ball from the box. If it happens to be a black ball, one also adds an arbitrary (but finite) number of white balls to the box.*
12.6. Exercise. * Show that the following string rewriting system*
[TABLE]
is terminating.
F Proving confluence
Checking confluence is a more difficult problem. One needs to keep track of all pairs of paths leaving the same vertex. It turns out that this is not necessary.
An abstract rewriting system is called locally confluent if for any objects , , , from the conditions , , follows the existence of an object such that , . This means that if two people started from the same vertex and each walked along one directed edge, then they can converge somewhere, following the arrows.
It is clear that every confluent system is locally confluent.
12.7. Exercise. * Construct a system that is locally confluent but not confluent.*
12.8. Diamond lemma. * A terminating rewriting system is confluent if and only if it is locally confluent.*
Let us describe an induction-type argument that will be used in the proof.
Suppose we want to prove a statement for every object of the terminating system. Then it is sufficient to prove the induction step: * holds under the assumption that holds for all downstream objects *. In other words, we need to show that if holds for any object such that , then holds as well.
Let us explain why it works. Suppose we proved the induction step for some statement in a terminating system, but does not hold for an object . Then we may choose another object such that does not hold and . Continuing this way, we get an infinite descending sequence of so that , , and so on. The latter contradicts that our system is terminating.
Note that applying the step of induction to the empty set of objects, we get that holds for any irreducible object, and we can continue further. It explains why this version of induction requires no base.
Proof. The only-if part is trivial; let us prove the if part.
Given an object , we need to show that if and , then there is such that and . Denote the latter statement by .
If or , then there is nothing to prove. Therefore, we can assume that and . Consider the first edges and of the paths defining and ; so we have , , , and .
Applying the local confluence condition to a pair of edges , , we find a vertex such that and .
Next, the vertices and are descending with respect to , and the same applies to vertex . Therefore, by the induction hypothesis, the statements , , and hold true.
Since and we can get such that and . Likewise, we get such that and .
Finally, we use . Since and , we can find such that and .
Finally, implies . Likewise, we get . ∎
G Applications
Suppose is a word in a string rewriting system from which two arrows emanate and . So, and are obtained from by two elementary transformations, replacing some of its occurrences and with and , respectively. If occurrences do not overlap, then we can apply these two transformations in any order and get the same result, say . In this case, local confluence holds.
If the occurrences overlap, then two cases are possible: they can either be contained in one another (a rarer case) or overlap by some non-empty word. Without loss of generality, we can pass to the part of our word that is the union of occurrences and . The following two cases are subject to verification.
Case A. Suppose and overlap by non-empty word ; we may assume that and , for some words and . That is, we have a word , from which we can go along the edge to both the word and the word . By making transformations in these words, we want to reduce them to one word; that is, we need to find a word such that and .
Case B. Suppose is contained in ; in other words for some words and . In this case, from the vertex we can move along the edge to both and . The goal is to find a word such that , .
12.9. Example. **
[TABLE]
Applying the shortlex order, we find that the system is terminating.
To check confluence, we need to examine cases A and B. There is only one word here, but it overlaps with itself in . By gluing together two such words that overlap in , we come to checking the word . Applying the transformation at the left and right ends, we get two irreducible words and . It follows that the system is not confluent and, therefore, not convergent.
In this example, the words and are equivalent to the original word . Therefore, if we add to the existing rules, then the word equivalence relation will remain unchanged. It leads to the following example.
12.10. Example. **
[TABLE]
Again, applying the shortlex order, we find that the system is terminating.
Now we need to go thru all the options for overlapping words. There are only two such options. One of them is old and does not need to be checked. Now the words and from the previous example are automatically converted to in the new system. But we have a new relationship, and therefore a new overlap has arisen: overlaps with by . (Here you must always be careful, since a new word can give several new overlaps; two words can overlap not in one, but in many ways — all this must be checked.)
Gluing the overlapping words together leads to and we can get and . Since , we get that the system is confluent, and therefore convergent.
12.11. Exercise. * Check if the following systems are terminating and conluent.*
- (a)
[TABLE] 2. (b)
[TABLE] 3. (c)
[TABLE]
H Comments
This chapter is inspired by a blog post by Victor Guba [18]. Several examples and exercises are taken from the book by Franz Baader and Tobias Nipkow [3], which is very good for further study.
The diamond lemma (12F) is also known as Newman’s lemma; it was proved by Max Newman [32, Theorem 3].
The replenishment of the system in 12G to the system in 12G is called the Knuth–Bendix procedure. It may produce a convergent system after several iterations, but this process may also run forever; in our example, just one step was needed.
Appendix A Hints
1B**.** Here is a part of the needed graph.
{4^{*}_{4}}{\parallel}{0_{0}}$${2_{4}}{\parallel}{2^{*}_{0}}$${3_{3}}{\parallel}{1^{*}_{1}}$${3^{*}_{4}}{\parallel}{1_{0}}$${3_{4}}{\parallel}{1^{*}_{0}}$$2_{0}$$1_{1}$$1_{0}$$0_{1}$$1_{0}
2A**.** Use the following observation: if there is no blue , then all edges are red.
2D**.** Assuming such a subgraph exists. Fix a vertex . Note that we can assume that the subgraph contains ; otherwise rotate it. In each case, draw the subgraph induced by the vertices connected to . (If uncertain, see definition of induced subgraph.)
3B**.** Show that has Hamiltonian cycles. Find the probability that that a given Hamiltonian cycle is monochromatic. Show that is the expected number of Hamiltonian cycles. Estimate ; you can use Stirling’s inequality
[TABLE]
3D**;** a. Use that the distance between two vertices is the number of different digits in their sequences.
b. Use that expected value of sum is the sum of expected values.
c. Show and use that .
3D**;** a. Identify the vertices of by -dimensional vectors with components. Observe that two vertices and are at a distance if and only if ; here denotes the scalar product (also known as dot product).
Suppose are -dimensional vectors such that if . Denote by the projection of to the hyperplane perpendicular to . Show that if . Apply it with induction by .
b Choose a hyperplane thru the center of the cube that does not pass thru any vertices of the cube. It subdivides the vertices in our set into two collections.
Project one of these collection to and argue as in a to show that it has at most vertices.
Remark. The equality in b holds only if is a power of 2. Try to prove it.
3E**.** Identify vertices of and . Calculate probability that ; it gives the first inequality. Note that there are ways to identify vertices of and ; it implies the second inequality.
The first inequality becomes equality if permuting vertices of gives the same graph. There only two graphs of that type; try to find both.
The second inequality becomes equality if has no symmetries; that is, there is no nontrivial permutation of vertices that induces an isomorphism from to itself. Try to construct such a graph with 10 vertices.
3E**.** Let be the probability that two vertices and in cannot be connected by a path of length .
Find . Show that is the expected number of such pairs. Apply Markov’s inequality. You should get that a typical graph has diameter at most 2.
Finally, find probability that has diameter . Make a conclusion.
3E**.** Denote by the probability that the subgraph induced by vertices in is ; note that .
Choose disjoint 100-element subsets of vertices in . We may assume that ; in particular, as .
Show that probability that at least one of induces a subgraph isomorphic to is at least
[TABLE]
Make a conclusion.
4B**.** Show and use that any spanning tree of contains the bridge.
4C**.** Use induction on and/or mimic the proof of 4C with the following analogs of :
4C**.** To construct the recursive relation, in addition to the ladders , you will need two of its analogs — and , shown on the diagram.
4C**.** Start with a spoke of the wheel. The diagram will contain wheels, and several analogs of fans.
5A**.** Apply induction on .
5B**.** Use property 3 on page 3.
5B**.** Revert the steps of the construction of the Kirchhoff minor. (Since the Kirchhoff minor is , the graph should have 6 vertices.)
5B**.** Apply 5B.
5C**.** Apply the construction of Kirchhoff minor.
5D**.** Apply property 3 twice and property 2 (see page 3).
5E**.** We can assume that connects the last two vertices of . In this case
[TABLE]
is the Kirchhoff minor of . It remains to find .
Follow the calculations in the proof of the Cayley formula.
5E**.** Note that
[TABLE]
is a Kirchhoff minor of .
First, show that , and then generalize the argument to .
6A**.** Use that the subgrahs and can be colored independently.
6A**.** Let be the center of . Suppose we have choices for color of ; once this choice is made, the rest of has to be colored in the remaining colors. Show and use that each such coloring corresponds to a coloring of in colors.
6A**.** By 6A, is a monic polynomial of degree . Show that for any connected graph . Use 6A to show that for any the coefficient of in front of vanish.
To show that , use the deletion-minus-contraction formula in an induction on .
6A**.** The only-if part follows from 6Aa. For the if part, apply 6A to show that is connected, has vertices and edges. Make a conclusion.
6A**.** Apply the induction on .
6B**.** Use that the matchings in and can be chosen independently.
6B**.** Use the definition of .
6B**;** a*.* Show and use that
[TABLE]
b*.* Apply it times to all edges at one vertex of .
6C**.** Use the definition of .
6C**.** Calculate using the expression for given in Theorem 6C and determine how much a tree with degree at the first vertex contributes to the value .
6C**.** Modify the proof of Theorem 6C.
7A**+7A.** Apply the definition of exponential generating function.
7B**.** For a, apply the definitions of Fibonacci numbers and exponential generating function.
For b, solve the differential equation in a and use that .
For c, use the Taylor expansion
[TABLE]
7D**.** Apply 6Bb.
7E**.** Apply 6Bb for .
7F**;** a*.* Enumerate the vertices of . Choose a cycle and list the numbers of the vertices in the order they appear on the cycle after . Count all possible orders. Show and use that we have exactly two different orders for one cycle.
8B**.** Suppose is a minimal connected graph with two cycles; that is, removing an edge or a vertex from makes it disconnected or reduces the number of cycles.
Show that contains two spanning trees that are not neighbors. Use it to prove the general case.
8B**.** Denote the two weights by and . Suppose is a spanning tree and is its neighbor. Show that
[TABLE]
Apply 8B.
8D**.** Apply 8B.
8D**.** Compare the number of edges that have to be checked in each algorithm. Note that the organization of data might give an essential difference. Think which steps could be done parallelly (say, on different computers).
8D**.** Part a should be evident. For part b, use 8B.
c*.* Start to walk from an arbitrary vertex without coming back. Stop at the end vertex or at the first vertex that was visited twice. Decide what to do in each case and start over.
9A**.** Show and use that has odd length.
9B**;** a*.* Note that a 1-factor is a matching. Apply 9B.
b*.* Remove the 1-factor provided by a. Apply a again and again.
Remark. If for an integer , then has an Euler’s circuit. Note that the total number of edges in is even, so we can delete all odd edges from the circuit. The obtained graph is regular with degree . Repeating the described procedure recursively times, we will end up with a 1-factor of .
There is a tricky way to make this idea work for arbitrary , not necessarily a power of ; it was discovered by Noga Alon [see \ncitealon and also \ncitekalai].
9B**.** Assume the graph, say , has a matching. Then the central vertex is matched with one of its three neighbors, say . Show and use that has even number of vertices in each connected component.
9B**.** Consider the bigraph with vertices labeled by rows and countries. Connect a row-vertex to a country-vertex if a kid from the corresponding county stands in the corresponding row. Try to apply 9B.
9B**.** Consider the bigraph with vertices labeled by 23 old parts and 24 new parts. Connect an old-part vertex to a new-part vertex if the corresponding parts overlap. Try to apply 9B.
9B**.** Consider the bigraph with vertices labeled by rows and columns. Connect a row vertex to a column vertex if a positive number stands in the common cell. Try to apply 9B.
9B**.** Add boys that everyone likes. Apply 9B and remove these boys.
9C**.** Assume that two -alternating paths starting from and can have a common vertex . Show and use that there is an -alternating paths starting from to .
9C**.** Consider the bigraph with a vertex for each column and row. Connect a row vertex to a column vertex if their common cell is marked. Apply 9C.
9E**.** Let be the minimal number of edges one can remove from so that there will be no directed path from to . Show that the assumptions in 9B imply that . Construct a bijection between directed paths from to in corresponds to an edge in to an edge in . Apply 9E.
10**.** Attach a small handle to the sphere so that one edge can go thru it over the crossing.
10**.** (a) Modify the diagram for .
(b)+(c) Start with the following square diagrams and draw the remaining edges.
10A**.** Suppose that is a nonsimple region in the drawing. Cut from the sphere and glue instead a disc or two to get a sphere.
10B**.** Show that and apply 10B.
10B**.** Assume it is possible. Show that the drawing has 5 squares. Try to glue 5 squares to each other side-to-side to obtain a torus.
10D**.** Show that the answers are and .
10D**.** Formally, one needs to show that deletion and contraction of every edge makes the graph toroidal. (Indeed, deleting a vertex produces a minor in a graph after deleting an edge.) But the graph has many symmetries. It reduces the number of cases to four: two deletions and two constructions.
10E**.** Make a Möbius strip; better use transparent paper. Redraw the diagram on the strip and try to draw the remaining 6 edges.
11A**.** Assume it is finite, then apply the definition including all the vertices in .
11A**.** Apply the definition for and a single vertex in . Conclude that the diameter is at least 2.
Choose two vertices and . Apply the definition for and . Conclude that there is a path of length 2 from to , and therefore the diameter is at most 2.
11B**.** Let and be the induced subgraphs in the Rado graph . Assume is not Rado; that is, there is a pair of finite vertex sets and in , such that any vertex in that meet the Rado property for and does not lie in (and therefore it lies in ). Use and to show that is Rado.
11B**+11B+11C.** Check the definition in each case. For 11C, use the binary numeral system.
11D**.** Argue as in 11D.
11D**.** Find how to weaken the conditions at the odd steps of the construction.
11D**.** Apply 11D.
11D**.** Show and use that the graphs have the same number of vertices and edges.
11E**.** Choose two finite subsets of vertices and , and a vertex . Show that satisfies the definition with a fixed positive probability . Conclude that the definition holds with probability 1.
Finally, use that there is only countably many of choices for and .
12E**.** Use the order
[TABLE]
Remark. Proving that a given rewriting system is terminating might be difficult. For example, it is unknown if the following system is terminating: the vertex set formed by all positive integers and the rules if is even and if is add. This is the so-called -problem.
12E**.** Consider the following order on pairs of integers: in two cases: if , or if and . (In fact it is the lexicographic or on pairs.) Apply this order assuming that the pair describes a box with white balls and black balls.
12E**.** Use the shortlex order.
12F**.** Try to construct such system with four objects, two irreducible and two reducible.
12G**.** a It is terminating but not confluent.
Remark. Note that the systems in 12G, 12G, and a have the same equivalence relation . The Knuth–Bendix procedure that led us from 12G to 12G lasts indefinitely, but all new relations that arise have the form
[TABLE]
All of them, taken at together with the relation , form a convergent system.
b Use lexicographic order to show that the system is terminating. For confluence, check the following diagram and show that this is the only thing to check.
c Note that each rule shortens the word. Use it to show that the system is terminating.
There are 6 possible overlaps here, but they are all of the same type. It is sufficient to consider one word .
Remark. А discerning reader could notice that this example describes a free group of rank 3; capital letters correspond to inverses.
Corrections and additions
Here we include corrections and additions to [21].
A Correction to 3.2.2
The proof of Theorem 3.2.2 about decomposition of a cubic graph with a bridge into 1-factors does not explain why ‘‘each bank has an odd number of vertices’’.
This is true since the 1-factor containing the bridge breaks all the vertices of each bank into pairs, except for the end vertex of the bridge.
B Correction to 8.4.1
There is an inaccuracy in the proof of Theorem 8.4.1 about stretchable planar graphs. Namely, in the planar drawing of , the region might be unbounded.
To fix this inaccuracy, one needs to prove a slightly stronger statement. Namely that any planar drawing of the maximal planar graph can be stretched. In other words, given a planar drawing of , there is a stretched drawing of and a bijection between the bounded (necessarily triangular) regions such that corresponding triangles have the same edges of as the sides.
The rest of the proof remains unchanged.
Index
- -factor, \hyperpage51
- (null graph), \hyperpage38
- (normal form), \hyperpage86
- abstract rewriting system, \hyperpage85
- adjacency matrix, \hyperpage27
- adjacent, \hyperpage6
- alphabet, \hyperpage84
- alternating path, \hyperpage60
- assignment problem, \hyperpage55, \hyperpage67
- augmenting path, \hyperpage60
- bank, \hyperpage21
- bigraph, \hyperpage60
- Binet’s formula, \hyperpage47
- binomial coefficient, \hyperpage9
- binomial expansion, \hyperpage9
- bipartite graph, \hyperpage60
- bridge, \hyperpage21
- chromatic polynomial, \hyperpage38
- complete graph, \hyperpage33
- confluent system, \hyperpage86
- connected sum, \hyperpage75
- constant-recursive sequences, \hyperpage24
- convergent, \hyperpage86
- countable graph, \hyperpage77
- countable set, \hyperpage77
- degree, \hyperpage5
- deletion-minus-contraction formula, \hyperpage36
- deletion-plus-contraction formula, \hyperpage22
- digraph, \hyperpage65, \hyperpage85
- directed graph, \hyperpage65
- distance, \hyperpage17
- double factorial, \hyperpage49
- drawing, \hyperpage68
- edge, \hyperpage5
- edge cover, \hyperpage64
- end vertex, \hyperpage5
- Erdős–Rényi graph, \hyperpage82
- expected value, \hyperpage13
- exponential generating function, \hyperpage44
- fan, \hyperpage23
- Fibonacci numbers, \hyperpage23
- forbidden minior, \hyperpage74
- forest, \hyperpage52
- free term, \hyperpage37
- graph, \hyperpage5
- greedy algorithm, \hyperpage55, \hyperpage57
- Hamiltonian cycle, \hyperpage51
- homogeneous polynomial, \hyperpage43
- incident, \hyperpage6
- independent random variables, \hyperpage13
- irreducible object, \hyperpage85
- isolated vertex, \hyperpage5
- Kirchhoff matrix, \hyperpage28
- Kirchhoff minor, \hyperpage28
- Knuth–Bendix procedure, \hyperpage91
- labeled graph, \hyperpage27
- ladder, \hyperpage25
- leading coefficient, \hyperpage37
- locally confluent, \hyperpage88
- loop, \hyperpage5
- Lucas numbers, \hyperpage26
- Markov’s inequality, \hyperpage14
- matching, \hyperpage39, \hyperpage60
- matching number, \hyperpage39
- matching polynomial, \hyperpage40
- minimal criminal, \hyperpage10, \hyperpage31
- minimum spanning tree proble, \hyperpage55
- minor, \hyperpage74
- monic polynomial, \hyperpage37
- multigraph, \hyperpage5
- nearest neighbor algorithm, \hyperpage54
- neighbor, \hyperpage55
- Newman’s lemma, \hyperpage91
- normal form, \hyperpage85
- null graph, \hyperpage38
- object, \hyperpage85
- orientation, \hyperpage65
- P=NP problem, \hyperpage54
- parallel edges, \hyperpage5
- perfect matching, \hyperpage49
- planar graph, \hyperpage68
- proper minor, \hyperpage74
- pseudograph, \hyperpage5
- Rado graph, \hyperpage77
- Rado property, \hyperpage77
- random graph, \hyperpage82
- random variable, \hyperpage13
- reverse-delete algorithm, \hyperpage59
- rotation, \hyperpage76
- set of neighbors, \hyperpage61
- shortest path problem, \hyperpage55
- shortlex order, \hyperpage87
- simple region, \hyperpage69
- spanning cycle, \hyperpage51
- spanning-tree polynomial, \hyperpage41
- square diagram, \hyperpage68
- squaring the square problem, \hyperpage26
- star, \hyperpage64
- string rewriting system, \hyperpage85
- surface of genus , \hyperpage74
- symmetric matrix, \hyperpage27
- symmetric polynomial, \hyperpage43
- terminating system, \hyperpage85
- toroidal graph, \hyperpage68
- transformation, \hyperpage85
- traveling salesman problem, \hyperpage54
- typical property, \hyperpage20
- vertex, \hyperpage5
- vertex cover, \hyperpage63
- weight, \hyperpage54
- word, \hyperpage84
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