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Graph Theory
Po-Shen Loh
24 June 2008
At first, graph theory may seem to be an ad hoc subject, and in fact the elementary results have proofs of that nature. The methods recur, however, and the way to learn them is to work on problems. Later, when you see an Olympiad graph theory problem, hopefully you will be sufficiently familiar with graph-theoretic arguments that you can rely on your own ingenuity to invent an approach.
1 Warm-up
- Graph the function f (x) = x x−+3^2.
- There are 21 friends at a party. Each person counts how many of the other people are he/she has met before. Is it possible for the sum of these counts to be an odd number?
2 Terminology
Definition. A graph G = (V, E) is a collection V of vertices and E ⊂ V × V of edges.
Remark. Informally, we think of the edges as linking the pairs of vertices that they correspond to, and typically represents graphs by drawings in which we connect the endpoints by a curve.
Another remark. If every edge links a unique pair of distinct vertices, then we say that the graph is simple. Most of our work will be with simple graphs, so we usually will not point this out.
Glossary of terms.
- Two vertices are adjacent if there is an edge that has them as endpoints.
- A graph is bipartite if the vertex set can be partitioned into two sets V 1 ∪ V 2 such that edges only run between V 1 and V 2.
- The chromatic number of a graph is the minimum number of colors needed to color the vertices without giving the same color to any two adjacent vertices.
- A clique on n vertices, denoted Kn, is the n-vertex graph with all
(n 2
possible edges.
- A complete graph on n vertices, denoted Kn, is the n-vertex graph with all
(n 2
possible edges.
- A graph is connected if there is a path between every pair of distinct vertices.
- A cycle is a path for which the first and last vertices are actually adjacent.
- The degree d(v) of a vertex v is the number of edges that are incident to v.
- An Eulerian circuit is a walk that traverses every edge exactly once, and returns to its starting point.
- A forest is a not-necessarily-connected graph with no cycles. See also the definition of tree below.
- If U is a subset of the vertices, then the induced subgraph G[U ] is the graph obtained by deleting all vertices outside U , keeping only edges with both endpoints in U.
- A Hamiltonian path is a path that includes every vertex. A Hamiltonian cycle is a cycle that includes every vertex.
- We say that an edge e is incident to a vertex v if v is an endpoint of e.
- A path is a sequence of distinct, pairwise-adjacent vertices. See also the definition of walk below.
- A graph is planar if it is possible to draw it in the plane without any crossing edges.
- A tree is a connected graph with no cycles. See also the definition of forest above.
- A walk is a sequence of not-necessarily-distinct, pairwise-adjacent vertices. See also the definition of path above.
3 Tools
3.1 Bare-hands
Each of these well-known results can be proved without any fancy theorems. Prove them.
- The sum of all of the degrees is equal to twice the number of edges. Deduce that the number of odd-degree vertices is always an even number. Solution: By counting in two ways, we see that the sum of all degrees equals twice the number of edges.
- Every tree contains a vertex of degree exactly 1, which is called a leaf. Solution: If all vertices had degree ≥ 2, then we could take a walk around the graph (with the rule that we are not allowed to go back over the edge that we took in the previous step), and then we will eventually intersect our path again. This forms a cycle.
- Every tree can be constructed by starting with a single vertex, and adding one new leaf at a time. Solution: Since every tree has a leaf, we can “destruct” any tree by pulling off a leaf. Reversing these steps gives a construction.
- Every connected graph contains a spanning tree. This is a subgraph which is a tree, that includes all of the original vertices. Solution: Keep deleting one edge from a remaining cycle until all cycles are gone. This cannot hurt connectedness, so we end up with a tree. Alternatively, consider the breadth-first-search or depth-first- search.
- A connected graph is a tree if and only if it has exactly V − 1 edges. Solution: Forward implication follows from the construction of a tree by adding one leaf at a time. Reverse implication follows by considering a spanning tree inside the connected graph, and noting that it already consumes all of the edges.
- The vertices can be partitioned into sets V 1 ∪.. .∪Vr , where each induced subgraph G[Vi] is a connected graph, and there are no edges between any pair of distinct {Vi, Vj }. These are called the connected components of G. Solution: Trivial observation.
- (BAMO 2004/3) NASA has proposed populating Mars with 2,004 settlements. The only way to get from one settlement to another will be by a connecting tunnel. A bored bureaucrat draws on a map of Mars, randomly placing N tunnels connecting the settlements in such a way that no two settlements have more than one tunnel connecting them. What is the smallest value of N that guarantees that, no matter how the tunnels are drawn, it will be possible to travel between any two settlements? Solution: This is asking for the max number of edges in a disconnected graph, after which we add 1. Suppoose we have a maximal disconnected graph. Then each connected component must be a clique. By convexity, the answer corresponds to a single K 2003 plus an isolated vertex.
- (BAMO 2005/4) There are 1000 cities in the country of Euleria, and some pairs of cities are linked by dirt roads. It is possible to get from any city to any other city by traveling along these roads. Prove that the government of Euleria may pave some of the roads so that every city will have an odd number of paved roads leading out of it. Solution: The key is that 1000 is even. Reduce to the case when the graph is a spanning tree. Take a leaf and its neighbor, pave the edge between them, and then delete both. Repeating 500 times, we will find a 1-factor.
- (St. Petersburg 1996/4) In a group of several people, some are acquainted with each other and some are not. Every evening, one person invites all of his acquaintances to a party and introduces them to each other. Suppose that after each person has arranged at least one party, some two people are still unacquainted. Prove that they will not be introduced at the next party. Solution: We prove the stronger statement that at the end of this process, all connected components are cliques. For this, it suffices to show that if the initial graph was connected, then the final graph is a clique. Use induction: let v be the last guy, and let G′^ be the acquaintance graph right before his party. Then G′^ − v is a disjoint union of cliques, and each clique is connected back to v. So when v holds his party, everything connects into a big clique.
- (Czech-Slovak Match 1997/2) In a community of more than six people, each member exchanges letters with precisely three other members of the community. Prove that the community can be divided into two nonempty groups so that each member exchanges letters with at least two members of the group he belongs to. Solution: Consider a shortest cycle, and let that be one of the groups. Put everybody else in the other group, and then do small alterations if necessary.
3.2 Extremal graph theory
Extremal graph theory, in its strictest sense, is a branch of graph theory developed and loved by Hungarians. (The opening sentence in Extremal Graph Theory, by B´ela Bollob´as.)
This very interesting field happens to be the subject of my own research, as well as one of the most common sources of advanced graph theory problems in Olympiads. The most famous theorems concern what substructures can be forced to exist in a graph simply by controlling the total number of edges. The classical starting point is Tur´an’s theorem, which proves the extremality of the following graph: let Tr (n) be the complete r-partite graph with its n vertices distributed among its r parts as evenly as possible (because rounding errors may occur).
Theorem. (Tur´an) For r ≥ 3 , the Tur´an graph Tr− 1 (n) is the unique n-vertex graph with the maximum number of edges subject to having no Kr subgraphs.
An excellent proof of Tur´an’s theorem can be found on page 167 of the book Graph Theory, by Reinhard Diestel. This is a well-written book which has an electronic edition freely available on the author’s website!
Another fundamental question is to ask how many subgraphs of a certain type must be forced by a given number of edges. The following results can be proven by simply letting di denote the degree of the i-th
vertex, and using the inequality^1 that
∑n i=
(di r
≥ n ·
(d r
, where d is the average value of di.
- Any n-vertex graph with m edges must have ≥ n ·
( 2 m/n 2
subgraphs isomorphic to the 3-vertex “V- shape” K 1 , 2. Here, we extend the
(x r
notation to allow non-integral values of x, by defining it to be 1 r! ·^ x(x^ −^ 1)^ · · ·^ (x^ −^ r^ + 1).
- More generally, for any integer r ≥ 2, any n-vertex graph with m edges must have ≥ n ·
( 2 m/n r
subgraphs isomorphic to the r-pointed “star” K 1 ,r.
Now try these problems.
- Prove that the number of edges in the Tur´an graph Tr (n) is less than or equal to
1 − (^1) r
) (^) n 2 2 , with equality when r divides n. Solution: Let xi be the part sizes. Then
xi = n, and the number of edges is
i<j xixj^ = 1 2
[( ∑
xi
x^2 i
]
. By convexity,
x^2 i ≥ r ·
( (^) n r
= n
2 r , and we just plug that in.
- Prove that the Tur´an graph Tr (n) is the r-colorable graph with the maximum number of edges. Solution: Clearly Tur´an graph is r-colorable. But any graph with more edges contains a Kr+1, so that portion cannot be r-colored, hence the entire graph cannot be r-colored.
- (approximation to Erd˝os-S´os conjecture, does not use Tur´an’s Theorem) If a graph G has average degree d, then it has a subgraph H ⊂ G whose minimum degree is ≥ d/2. Corollary: for any tree T , if a graph G has average degree ≥ 2 |T |, then G contains T as a subgraph. Solution: The condition that average degree ≥ d is equivalent to the number of edges being ≥ nd/2. Now keep deleting vertices of degree < d/2. This will preserve the above condition, but must terminate because the average degree never goes below d, so in particular the number of vertices cannot go below d. Thus, we will stop at a graph with all vertices of degree ≥ d/2. Then embed the tree greedily.
- (Japan 1998/2) A country has 1998 airports connected by some direct flights. For any three airports, some two are not connected by a direct flight. What is the maximum number of direct flights that can be offered? Solution: By Tur´an, the largest triangle-free subgraph of K 1998 is bipartite with sides of size 999, so 999^2 is the answer.
- (Japan 1997/3) Let G be a graph with 9 vertices. Suppose given any five points of G, there exist at least 2 edges with both endpoints among the five points. What is the minimum possible number of edges in G? Solution: The optimal configuration is a union of 3 disjoint triangles. But more work is required to prove that it is actually optimal.
3.3 Matching
Consider a bipartite graph G = (V, E) with partition V = A ∪ B. A matching is a collection of edges which have no endpoints in common. We say that A has a perfect matching to B if there is a matching which hits every vertex in A.
(^1) This inequality will be proved in my Convexity lecture.
- (Diestel 2.8) Let A be a finite set with subsets A 1 ,... , An, and let d 1 ,... , dn ∈ N. Show that there are disjoint subsets Dk ⊂ Ak with |Dk| = dk for all k, if and only if ∣ ∣ ∣ ∣ ∣
i∈I
Ai
i∈I
di,
for all I ⊂ { 1 ,... , n}.
3.4 Ramsey theory
Complete disorder is impossible. — T. S. Motzkin, on the theme of Ramsey Theory.
Let s and t be positive integers. We define the Ramsey Number R(s, t) to be the minimum integer n for which every red-blue coloring of the edges of Kn contains either a completely red Ks or a completely blue Kt. Ramsey’s Theorem states that R(s, t) is always finite, and we will prove this in the first exercise below. The interesting question in this field is to find upper and lower bounds for these numbers, as well as for quantities defined in a similar spirit.
Now try these problems.
- Prove by induction that R(s, t) ≤
(s+t− 2 s− 1
. Note that in particular, R(3, 3) ≤ 6. Solution: Observe that R(s, t) ≤ R(s− 1 , t)+R(s, t−1), because if we have that many vertices, then if we select one vertex, then it cannot simultaneously have < R(s − 1 , t) red neighbors and < R(s, t − 1) blue neighbors, so we can inductively build either a red Ks or a blue Kt. But ( (s − 1) + t − 2 (s − 2)
s + (t − 1) − 2 s − 1
s + t − 2 s − 1
because in Pascal’s Triangle the sum of two adjacent guys in a row equals the guy directly below them in the next row.
- (IMO 1964/4) Seventeen people correspond by mail with one another-each one with all the rest. In their letters only three different topics are discussed. each pair of correspondents deals with only one of these topics. Prove that there are at least three people who write to each other about the same topic. Solution: This is asking us to prove that the 3-color Ramsey Number R(3, 3 , 3) is ≤ 17. By the same observation as in the previous problem, R(a, b, c) ≤ R(a− 1 , b, c)+R(a, b − 1 , c)+R(a, b, c−1)−1. Then using symmetry, R(3, 3 , 3) ≤ 3 R(3, 3 , 2) − 1. It suffices to show that R(3, 3 , 2) ≤ 6. But this is immediate, because if we have 6 vertices, if we even use the 3rd color on a single edge, we already get a K 2. So we cannot use the 3rd color. But then from above, we know R(3, 3) ≤ 6, so we are done.
- (South Africa 1997/5) Six points are joined pairwise by red or blue segments. Must there exist a closed path consisting of four of the segments, all of the same color? Solution: Yes. Proof: assume not. Let the vertices be a, b, c, d, e, f. Use Ramsey to get a monochro- matic triangle first, and suppose it is a, b, c. WLOG it is blue. Now d cannot have 2 blue edges into a, b, c, or else we get blue C 4. Same for e and f. Also, if, say, a, b both have red edges to each of say, d, e, then we get C 4. Therefore, the only possible configuration is to have a blue matching between {a, b, c} and {d, e, f }, and all other edges between those sets are red. WLOG the blue matching is ad, be, cf. But then edges de and ef are both forced red, or else we have blue C 4 , say abed. And then there is a red C 4 : bf ed.
- (IMO 1978/6) The members of an international society belong to 6 different countries (each to only one country). The list of the members contain 1978 names numbered 1, 2,... , 1978. Prove that there exists at least one member whose number is the sum of the numbers of two of his compatriots, not necessarily distinct.
(The IMO problems were listed in the article Ramsey Theory and the IMO, by Cambridge professor Ben Green.)
3.5 Planarity
When we represent graphs by drawing them in the plane, we draw edges as curves, permitting intersections. If a graph has the property that it can be drawn in the plane without any intersecting edges, then it is called planar. Here is the tip of the iceberg. Perhaps the most useful planarity theorem in Olympiad problems is the Euler Formula. (The second result is less useful for Olympiads, but is too famous to omit.)
- (Euler Formula) Every connected planar graph satisfies V − E + F = 2, where V is the number of vertices, E is the number of edges, and F is the number of faces.
- (Four-color theorem) Every planar graph is 4-colorable.
Now use the Euler Formula to solve these problems.
- If G is a connected, planar, simple graph, then E ≤ 3 V − 6. Solution: For each face, calculate its perimeter, and add all of these up. This double-counts each edge. Each face has perimeter ≥ 3, so we get 2E ≥ 3 F. Plugging in, we have
2 = V − E + F ≤ V − E +
E = V −
E.
- Show that K 5 is not planar. Solution: V = 5, E = 10, so we must have F = 2 − V + E = 7. But as in the previous solution, we need to have 2E ≥ 3 F , which is not the case.
- Show that K 3 , 3 is not planar. Solution: V = 6, E = 9, so we must have F = 2 − V + E = 5. But as in the previous solution, we need to have 2E ≥ 3 F. Actually, we need 2E ≥ 4 F , because K 3 , 3 has no triangles. But this stronger inequality is false.
4 Problems
- (IMO Shortlist 2004/C3) The following operation is allowed on a finite graph: Choose an arbitrary cycle of length 4 (if there is any), choose an arbitrary edge in that cycle, and delete it from the graph. For a fixed integer n ≥ 4, find the least number of edges of a graph that can be obtained by repeated applications of this operation from a complete graph on n vertices (where each pair of vertices are joined by an edge). Solution: First we show that we cannot end up with any graph with ≤ n − 1 edges. We are only breaking cycles, so we cannot destroy connectivity. Therefore, any final graph with ≤ n − 1 edges must have exactly n − 1 edges, and be a tree, hence bipartite. But if we consider the reverse process, observe that if we start from a bipartite graph and complete C 4 ’s, we will stay bipartite, and Kn is not bipartite! It remains to find an n-edge graph that we can reach. I think one such graph is a triangle plus a single path leading out of one of the vertices of the triangle. See if you can prove this.
- (List coloring conjecture) Let χ′(G) be the minimum number of colors required to color the edges of G such that incident edges receive different colors. Now, consider a slightly different notion of coloring, where instead of using the same pool of colors for each edge, we give each edge an individual list of colors (possibly drawn from a much larger pool) that it can pick from. The parameter ch′(G) is then defined as the minimum integer k such that no matter how we assign lists of size k to the edges of G, it is always possible to color the edges from the lists in such a way that incident edges receive different colors. Prove that χ′(G) = ch′(G) for all graphs G.
- (Hadwiger’s conjecture) If a graph has chromatic number at least r, then it contains a Kr -minor. That is, we can find disjoint subsets V 1 ,... , Vr of the vertex set such that: - every induced subgraph G[Vi] is a connected graph, and - for every i < j, we can find an edge that has one endpoint in Vi and one endpoint in Vj.
7 Really harder problems
Determine the exact value of the Ramsey number R(5, 5). Hint: it is known to be one of { 43 , 44 , 45 , 46 , 47 , 48 , 49 }. You may use as many supercomputers as you want.
Believe it or not, this is unknown. For a greater challenge, determine R(6, 6). If you succeed, Paul Erd˝os would have been proud.
Imagine an alien force, vastly more powerful than us landing on Earth and demanding the value of R(5, 5) or they will destroy our planet. In that case, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they asked for R(6, 6), we should attempt to destroy the aliens.
-Paul Erd˝os
