# Homework MA 725 Spring, 2012 C. Huneke SELECTED ANSWERS ```Homework
MA 725
Spring, 2012
C. Huneke
1.1.25 Prove that the Petersen graph has no cycle of length 7.
Solution: There are 10 vertices in the Petersen graph G. Assume there is a
cycle C of length 7. It leaves out exactly 3 vertices. Since each vertex is a pair of
numbers from 1 to 5, at least two of these 3 vertices share a common number, say
i. That leaves only two vertices in C which contain i. In particular, there must be
a path P = P3 in the cycle C where no vertex in P contains the number i. Fix the
middle vertex of P , say {j, k}. But there is then a unique vertex adjacent to {j, k}
which does not contain i, namely the two element subset \{i, j, k}, contradicting
that there are two vertices adjacent to {j, k}.
Second Solution: (Lucas Chaffee, similar to several others) Proceed by way of
contradiction. Assume that G has a 7-cycle, C. Since G is 3-regular, each vertex
in C is connected to at least one vertex not in C, else there would be a 3-cycle
or 4-cycle in G. But we know the girth of G is 5. Fix a vertex v in C, and let w
and u be two vertices in C such that wu is an edge in C across from v. There is a
common neighbor x of v and u and a common neighbor y of v and w with x and y
not vertices in C. Since the degree of v is 3, it follows that x = y. But this makes
1.1.26 Let G be a graph with girth 4. Assume that every vertex of G has degree k. Prove
that G has at least 2k vertices. Determine all graphs with exactly 2k vertices.
Solution (by several people): Fix two vertices x and y which are neighbors. Then
N (x) ∩ N (y) = ∅, since otherwise G would have girth at most 3. Hence G must
have at least 2k vertices, since each of N (x) and N (y) have at least k vertices. If
G has exactly 2k vertices, then these are all the vertices. But every neighbor of
x can only have neighbors in N (y) (otherwise there is a C3 in G), and similarly
every neighbor of y can only have neighbors in N (x). It follows that every vertex
in N (x) is connected by an edge to every vertex in N (y), so that G is isomorphic
to Kk,k .
1.1.27 Let G be a graph with girth 5. Assume that every vertex of G has degree at least
k. Prove that G has at least k 2 + 1 vertices. For k = 2 and k = 3, find one such
graph with exactly k 2 + 1 vertices.
Solution: For k = 2, the 5-cycle is an example, and for k = 3, the Petersen
graph is an example.
To prove the main statement, fix a vertex v of G. Then |N (v)| ≥ k by assumption. For each neighbor u of v, N (u) has at least k − 1 members other than
v. Moreover the sets N (u) for u ∈ N (v) are all disjoint. For if a neighbor of u
is a neighbor of v, G would have a 3-cycle, while if two neighbors of v share a
1
2
common neighbor other than v, G would have a 4-cycle. Hence there are at least
1 + k + k(k − 1) = k 2 + 1 vertices.
1.1.28 Prove that the girth of the odd graph Ok is 6 if k ≥ 3.
Solution: We show that there is no 5-cycle. Proving there are no 4 or 3 cycles
is easier. Assume by way of contradiction that there is a 5-cycle, say with vertices
a, b, c, d, e and edges ab, bc, cd, de, ea. Without loss of generality, a = {k + 2, ..., 2k +
1}. Then both b and e must be k-element subsets of {1, ..., k + 1}. In particular,
they must share k − 1 elements. Since a is arbitrary, this argument proves that
any two vertices which share a common neighbor in the 5-cycle must have k − 1
elements in common. In particular a and d share k − 1 elements and a and c share
k −1 elements. Since a is a k-element set, it follows that d and c must share at least
k − 2 elements. But they are neighbors, hence disjoint. It follows that k − 2 ≤ 0.
This contradicts the assumption that k ≥ 3.
To prove there is a cycle of length 6, do this as I did in class. First write an
explicit 6-cycle when k = 3. Then divide the remaining 2k + 1 − 7 = 2k − 6 into two
disjoint sets S and T , and attach them to the 6-cycle written down for k = 3 by
alternating them among the vertices. This gives an example of a 6-cycle in general.
Section 1.2
1.2.20 Let v be a cut-vertex of a simple graph G. Prove that G − v is connected.
Solution: This is done as we discussed in class. Let x and y be vertices of G − v.
If they lie in different connected components of G − v, then they are adjacent in
G − v. If not, since G − v has at least two components, choose a vertex z in a
different component than the one containing both x and y. In G − v, xz is an edge,
as well as yz. Hence there is a path from x to y in G − v.
1.2.24 Let G be a simple graph having no isolated vertex and no induced subgraph with
exactly two edges. Prove that G is a complete graph.
Solution: (several people): First observe that G must be connected; if not, then
since it has no isolated vertices the connected components must each contain at
least one edge. But two edges in two different connected components contradicts
the assumption that no induced subgraph has exactly two edges.
If G has two vertices that are not adjacent, then there is a shortest path connecting them of length at least two. Then any three consecutive vertices along this
path would induce a subgraph with exactly two edges, contradiction.
1.2.29 Let G be a connected simple graph not having P4 or C3 as an induced subgraph.
Prove that G is a biclique. (That is, isomorphic to a complete bipartite graph.)
Solution. We first prove that G is bipartite. It suffices to prove that G has no
odd cycles. Suppose G does have an odd cycle, say C2k+1 . Choose k smallest. By
assumption, k ≥ 2. In that case the cycle contains a P4 subgraph, which cannot
3
be an induced subgraph. It therefore has a chord, which will lead to a smaller odd
cycle, contradiction. Hence G is bipartite.
We prove that G is complete. Let X and Y denote the sets of vertices which
realize G as a bipartite graph, and let x ∈ X, y ∈ Y . If xy is not an edge, choose
the shortest path between them (G is connected.) Since G is bipartite, this path
must contain an induced P4 , contradiction.
1.2.30 Let G be a simple graph with vertices v1 , . . . , vn . Let A = (aij ) be the adjacency
matrix of G. Prove that (Ak )ij is the number of vi , vj walks of length K in G.
P
Solution: Induct on k. For k = 2, note that (A2 )ij = k ail alj , and a term in
this sum is nonzero if and only if ail 6= 0 and alj 6= 0, in which case both are one,
and their product is one. Thus the sum counts the number of nonzero terms in the
sum. But a term is nonzero if and only if vi vl vj is a walk in G. This proves the
case k = 2.
The general case follows by induction. Set Ak−1 = (bij ). By induction
bij is the
P
th
k
number of vi , vj walks of length k − 1. The ij entry of A is l bil alj . A term
is nonzero in this sum if and only if there is a walk of length k − 1 from vi to vl
and an edge vl vj , and in this case that term counts the number of walks of length
k from vi to vj whose next to last step is vl . Summing over all l then counts the
total number of walks from vi to vj of length k.
1.2.40 Let P and Q be paths of maximum length in a connected graph G. Prove that P
and Q have a common vertex.
Solution: (several people) Set the length of these paths to be m. If their vertices
are disjoint, then fixing any two vertices, one in P and one in Q, there is a path
from one to the other. Choose a vertix x ∈ P , and a vertex y ∈ Q such that the
path from x to y is the shortest among all such paths. The length of a path P 0
from x to the furthest endpoint of P is at least m
2 , and similarly the length of a
0
path Q from y to the furthest endpoint of Q is at least m
2 . Using the shortest path
0
0
from x to y and combining it with the paths P and Q gives a path of length more
Section 1.3
1.3.12 Prove that an even graph has no cut-edge. For each k ≥ 1, construct a (2k + 1)regular simple graph having a cut-edge.
Solution (Ilya Smirnov): We know that an even graph decomposes into cycles.
Moreover, we know a edge is a cut-edge if and only if it belongs to no cycle. Hence
even graphs have no cut-edges.
To construct the required graph, start with K2k+2 . Choose k pairs of vertices
from this graph, say x1 , ..., xk and y1 , ..., yk . Remove the edges xi yi from K2k+2 ,
for i = 1, ..., k. The resulting graph has 2k vertices of degree 2k, and 2 vertices of
degree 2k + 1. Finally add one new vertex z, and edges zxi and zyi . This gives a
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graph G with one vertex (namely z) of degree 2k, and all other vertices of degree
2k + 1. Now take two disjoint copies of G, say G and G0 , and add one edge joining
z and z 0 . This edge is a cut-edge, and the graph is (2k + 1)-regular.
1.3.14 Prove that every simple graph with at least two vertices has two vertices of equal
degree.
Solution: Suppose there are n vertices. The degrees of the vertices are integers
between 0 and n − 1. The Pigeon-Hole principle shows that two have the same
degree unless the degrees of the n vertices are exactly the integers between 0 and
n − 1. But this is impossible since a vertex of degree n − 1 is connected to every
vertex, so there could not be a vertex of degree 0 in this case.
1.3.15 For each k ≥ 3, determine the smallest n such that
a) there is a simple k-regular graph with n vertices.
b) there exist nonisomorphic simple k-regular graphs with n vertices.
Solution, a): Clearly n ≥ k + 1 since G is simple; otherwise the maximum degree
of a vertex would be at most k − 1. On the other hand there is a k-regular simple
graph with k + 1 vertice, namely Kk+1 . Hence n = k + 1 is the smallest number of
vertices for which there exists such a graph.
Solution, b) (several people): It is simplest to think in terms of the complement
of such graphs. A simple graph G with n vertices is k-regular if and only if G is
n − 1 − k-regular. Moreover the complement of two simple graphs are isomorphic if
and only if the graphs are isomorphic. Consider then the complement. If n = k +1,
G is 0-regular, i.e., a set of isolated points. Hence the only k-regular graph having
k + 1 vertices is Kk+1 , the complement of isolated points. If n = k + 2, then the
complement is a 1-regular graph on n vertices. There are no such graphs if n is
odd by the degree-sum formula, while if n is even, the only such graph is a disjoint
union of edges. Since this is unique, there is at most one graph which is k-regular
with k + 2 edges. However, if we take n = k + 3, then there are at least two
non-isomorphic simple 2-regular graphs having k + 3 vertices, for example, Ck+3
and the disjoint union of C3 and Ck (note k ≥ 3). Thus n = k + 3 is the answer.
1.3.18 For k ≥ 2, prove that a k-regular bipartite graph has no cut edge.
Solution: Since every component of a k-regular bipartite graph is also k-regular
and bipartite, we may assume that G is connected without loss of generality. Suppose G has a cut edge e. Let G1 and G2 be the two connected components of G − e.
Consider one of them, G1 , which will contain one of the vertices incident to e, say
u. G1 is still bipartite. Denote the two disjoint vertex sets as X and Y , so that
every edge of G1 goes between vertices in X and vertices in Y . Let n(X) = p and
n(Y ) = q. Every vertex has degree k except for u, which has degree k − 1. We
count the number of edges in two ways; if u ∈ X, then there are k(p − 1) + (k − 1)
5
edges going from X to Y , which is exactly kp − 1. But counting the same edges as
going from Y to X, there must be qk. Since k ≥ 2, this is impossible.
1.3.26 Count the number of 6-cycles in Q3 . Prove that every 6-cycle in Qk (k ≥ 3) lies in
exactly one three-dimensional subcube. Use this to count the number of 6-cycles
in Qk .
Solution: By either brute force or clever counting, one finds there are 16 6-cycles
in Q3 . Notice that in a 6-cycle in Qk , edges correspond to changing the entry in
exactly one position. Since one has to return back to the start, each time a position
is changed, it must be eventually changed back. Hence there must be exactly three
positions which change, and the other k−3
remain fixed. Since these other positions
k
can be fixed arbitrarily, and there are 3 choices of the three dimensional subcube,
there are exactly 16 k3 2k−3 6-cycles.
Section 1.4
1.4.8 Prove that there is an n-vertex tournament with indegree equal to outdegree at
every vertex if and only if n is odd.
Solution (Nick Packauskas and others): Let G be an n-vertex tournament with
indegree equal to outdegree at every vertex. Then n cannot be even since for every
vertex v, d+ (v) + d− (v) = n − 1 is odd.
We prove the converse by induction. It is clear for n = 1. Let n &gt; 1 and
suppose there an (n − 1)-vertex tournament H with indegree equal to outdegree at
every vertex. Partition V (H) into two sets, X and Y , of cardinalities n and n − 1
respectively. Add two new vertices u, v to H. Add edges going from u to each
vertex in X, from each vertex of X to v, from v to each vertex in Y , and from each
vertex of Y to u. This does it.
1.4.10 Prove that a digraph is strongly connected iff for each partition of the vertex set
into nonempty sets S and T , there is an edge from S to T .
Solution (Lucas Chaffee, similar to several others): We first prove the forward
direction. For an arbitrary partition S and T , let u ∈ S and v ∈ T . By the strong
connectedness, there exists a directed path from u to v, and so at some point we
traverse an edge from S to T .
Converse: For an arbitrary x ∈ V (G), let S be all vertices reachable by x with a
directed path. If S were not all of V (G), then by hypothesis there’s an edge from
S to the complement of S in V (G), meaning x can reach it, a contradiction. Hence
S = V (G), and since x was arbitrary, G is strongly connected.
Section 2.1
2.1.6 Let T be a tree with average degree a. In terms of a, determine n(T ).
6
P
Solution: By definition, a = nd(v) , where n = n(T ), and the sum is over all
)
vertices. By the degree-sum formula we obtain that a = 2e(T
n . Since T is a tree,
2
e(T ) = n − 1. Hence a = 2(n−1)
. Solving for n in terms of a yields n = 2−a
.
n
2.1.26 For n ≥ 3, let G be an n-vertex graph such that every graph obtained by deleting
one vertex is a tree. Determine e(G), and use this to determine G itself.
Solution: (Nick Packauskas) We claim that G is isomorphic to Cn . Let Gi =
G − vi , where V (G) = {v1 , ..., vn }. Each Gi is a tree and thus has n − 2 edges.
As there are n such subgraphs, the number of total edges in all these subgraphs is
n(n − 2). Each edge in G has two endpoints, and is therefore counted in exactly
n − 2 of the edge count in the subgraphs. Thus e(G) = n. It follows that G has
a cycle. But none of the Gi have a cycle, so the cycle must include every vertex.
Thus G is an n-cycle.
2.1.29 Every tree is bipartite. Prove that every tree has a leaf in its larger partite set,
and in both if they have equal size.
Solution (Rajib Anwar and others): Let A and B be two partite sets of a tree T
such that |A| ≥ |B|. If there is no leaf in A, then since every vertex in A will have
degree at least two, e(T ) ≥ 2|A| ≥ |A| + |B| = n(T ) = e(T ) + 1, a contradiction. If
|A| = |B|, the same argument shows that both have leaves.
2.1.37 Let T and T 0 be two spanning trees of a connected graph G. For e ∈ E(T ) \ E(T 0 ),
prove there is an edge e0 ∈ E(T 0 ) \ E(T ) such that T − e + e0 and T 0 − e + e0 are
both spanning trees of G.
Solution (Khaled Alhazmy and others): Since T is a tree, we know that e is a
cut-edge of T . Thus T − e has two connected components, say T1 and T2 , both of
which are trees as well. Let e = xy, with x ∈ T1 and y ∈ T2 . There is a unique
x − y path in T 0 (either by (2.1.4D) or the fact that T 0 + e contains a unique cycle),
/ T,
and hence this path contains an edge e0 which connects T1 to T2 . Clearly e0 ∈
0
else T contains a cycle. Now T − e + e is connected with n(G) − 1 edges and is
therefore a spanning tree. Likewise T 0 − e0 + e has n(G) − 1 edges and no cycles,
and is therefore a tree.
Section 2.2
2.2.1 Determine which trees have Prüfer codes that (a) contain only one value, (b) contain exactly two values, or (c) have distinct values in all positions.
Solution: (a) This means that one vertex is adjacent to all other vertices, so the
graph is K1,n for some n.
(b) With two values in the code, we know there are exactly two vertices which
are not leaves. Therefore the graph has two vertices with leaves coming out from
each of them.
(c) With distinct values in all positions, there are only two leaves in the tree.
Therefore the tree must be a path.
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2.2.10 Compute τ (K2,m ). Also compute the number of isomorphism classes of spanning
trees of K2, m.
Solution 1 (Farhana Abedin, similar to several): Let X be the vertex set on two
elements, and Y the other vertex set on m-elements. Each spanning tree of K2,m
has a unique vertex in Y which is a common neighbor of the vertices in X, and
this common neighbor can be chosen in m ways. The remaining vertices in Y form
leaves. For each leaf we can choose its neighbor in X in one of two ways. Hence
the number of spanning trees is exactly m2m−1 .
Vertices in X possess one common neighbor in Y . The leaves are distributed
among the vertices in X to determine the isomorphism classes. We can connect
z leaves to one vertex and m − 1 − z leaves to the other vertex, where 0 ≤ z ≤
b(m − 1)/2c. Hence there will be b(m + 1)/2c total isomorphism classes.
2.2.25 Prove that if a graph G is graceful and Eulerian, then e(G) is congruent to 0 or 3
mod 4.
Solution (Peidi Gu, similar to others): Let f be a graceful labelling of G which
we assume has m + 1 vertices. The parity of the sum of the labels of an edge is the
same as the parity of the absolute value of their difference, and therefore the sum
of the absoute values of the differences of the labels, which is m
( m + 1)/2 since the
P
labels are from 1 to m, is congruent mod 2 to v∈V (G) d(v)f (v). As G is Eulerian,
each d(v) is even, so the sum is even. It follows that 4 divides m(m + 1), and the
problem follows at once.
Section 2.3
2.3.1 Assign integer weights to the edge of Kn . Prove that the total weight on every
cycle is even if and only if the total weight on every triangle is even.
Solution (CJ Harries, similar to several): One direction is trivial: if every cycle
has total weight even, obviously every triangle has total weight even. For the
converse, use induction on the size m of the cycle. We can assume m ≥ 4. Choose
a path P3 within the cycle C, say with edges e and f , and endpoints u and v. By
induction the cycle formed by deleting e and f and replacing them with the edge
g = uv has even total weight. Moreover, w(e) + w(f ) = w(g) modulo 2. Letting
W be the sum of the weights of the the edges of C − e − f , we have that the total
weight of C is w(e) + w(f ) + W ≡ w(g) + W ≡ 0 modulo 2.
2.3.7 Let G be a weighted connected graph with distinct edge weights. Without using
Kruskal’s algorithm, prove that G has a unique minimum weight spanning tree.
Solution (Lucas Chaffee): Assume that there are two distinct minimal weight
spanning trees, T and T 0 . By problem 2.1.37, there are edges e of T and e0 of T 0
such that both T − e + e0 and T 0 − e0 + e are spanning trees. Since one of e or e0 has
strictly smaller weight than the other, one of these trees has smaller weight than
T or T 0 , contradiction.
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Section 3.1
3.1.5 Prove that α(G) ≥
n(G)
∆(G)+1
for every graph G.
Solution: Let X be an independence set of G of size α(G) = a. There are at most
∆(G) neighbors for each element of X, and since X is maximum, G ⊂ X ∪ NG (X).
Hence n ≤ a + ∆(G)a, implying the result.
3.1.8 Prove or disprove: Every tree has at most one perfect matching.
Solution (several people): Some did this by induction. But I think the nicest
solution is the following: Assume that M and M 0 are perfect matching of a tree,
and consider their symmetric difference F . Every vertex in F has degree 0 or 2,
which implies that every connected component of F is either an isolated vertex or
a cycle. Since T has no cycles, F is a collection of isolated vertices. This means
every edge in M is an edge in M 0 , and vice-versa, so M = M 0 .
Section 4.1
4.1.5 Let G be a connected graph with at least three vertices. Form G0 from G by adding
an edge with endpoints x, y whenever dG (x, y) = 2. Prove that G0 is 2-connected.
Solution: Since n(G) &gt; 2, it suffices to prove that G has no cut-vertex. Let
x ∈ G0 . If G0 − x is disconnected, then obviously so is G − x. Let u and v be in two
different components of G − x. Since G is connected, there is a path in G between
these two vertices which necessarily must go through x. Let a and b be the two
vertices nearest x on this path. Then dG (a, b) = 2, so that a and b are connected
in G0 . It follows that G0 − x is not disconnected, a contradiction.
4.2.4 Prove or disprove: If P is a u, v path in a 2-connected graph G, then there is a
u, v-path Q that is internally disjoint from P .
Solution: This is false. A counterexample is given by K4 minus any edge.
4.2.12 Use Menger’s theorem to prove that κ(G) = κ0 (G) when G is 3-regular.
Solution (L. Chaffee): Let S ⊂ V (G) be a set which separates X, Y ⊂ V (G) with
|S| = κ(G), and let F = [H, H 0 ] with |F | = κ0 (G). Note that either X ∩ H 6= ∅ and
Y ∩ H 0 6= ∅ or X ∩ H 0 6= ∅ and Y ∩ H 6= ∅, or both. Without loss of generality, let
x ∈ X ∩ H and y ∈ Y ∩ H 0 . By Menger’s theorem 4.2.19, we have that κ0 (G) is the
maximum number of edge disjoint x, y-paths. These edge disjoint paths must be
internally disjoint, else the common vertex would have degree at least four, and so
there are at least κ0 (G) internally disjoint x, y-paths as well. By the other version
of Menger’s theorem, 4.2.17, we have that the minimum size of a x, y disconnecting
set is at least κ0 (G), and therefore κ(G) ≥ κ0 (G). Since we always have the other
inequality, they must be equal.
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