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MATH 5707 HOMEWORK 1 SOLUTIONS
˙
CIHAN
BAHRAN
1. Show that a finite simple graph with more than one vertex has at least two
vertices with the same degree.
Let V be the set of vertices of the graph where n = |V |. Let d be the function which
assigns every vertex its degree. Note that because the graph is simple, we have d(v) ≤
n − 1 for every v ∈ V . So we can consider d as a function from V to {0, 1, · · · , n − 1}.
Suppose no two vertices have the same degree. This means that the function d is
injective. But
|V | = n = |{0, 1 . . . , n − 1}|
hence d must be a bijection. In particular there exist v, w ∈ V such that d(v) = 0 and
d(w) = n − 1. Now d(w) = n − 1 means that w is connected to every vertex in V r {w},
but since d(v) = 0 we must have v ∈
/ V r {w}. Thus v = w, so 0 = d(v) = d(w) = n − 1;
thus n = 1.
So we showed that d being injective implies |V | = 1, which is the contrapositive of what
we wanted.
2. Prove that a sequence of positive integers d1 , . . . , dn is a degree sequence of a
P
tree if and only if ni=1 di = 2(n − 1).
We observe that every (finite) tree has at least one leaf, that is, a vertex v with degree
1. The “only if” part follows at once from the following proposition:
Proposition 1. A tree with n vertices has exactly n − 1 edges.
Proof. Employ induction on n. For n = 1, this is trivial. Suppose the claim holds for
n − 1 and let T be a tree with n vertices where n ≥ 2. Let v be a leaf of T and let e be
the unique edge containing v. Then note that the graph T 0 obtained by deleting v and
e from T remains connected and clearly has no cycles as T has none. Thus T 0 is also a
tree. Now the inductive hypothesis kicks in and says that T 0 has (n − 1) − 1 = n − 2
edges. Since T 0 contains all the edges of T except e, we see that T has n − 1 edges. For the “if” part, we also employ induction. The statement is not quite true for n = 1
(the degree sequence will not have positive integers in it), so we deal with n = 2 as the
base case. Here the only sequence of positive integers d1 , d2 with d1 + d2 = 2 is 1, 1
and yes, there is a tree with that degree sequence, namely a single edge connecting two
vertices.
Now assume n ≥ 3 and let d1 , . . . , dn be a sequence of positive integers such that
Pn
i=1 di = 2(n − 1). We claim that dj = 1 for some j. Otherwise, being positive integers
every di is larger than 2 and we get
2(n − 1) =
n
X
di ≥
i=1
n
X
i=1
1
2 = 2n ,
MATH 5707 HOMEWORK 1 SOLUTIONS
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a contradiction. We may assume j = 1 so d1 = 1, by relabeling the vertices.
Similarly, we observe that dk ≥ 2 for some k: Otherwise 2(n − 1) = ni=1 di = n, so
n = 2, contradicting n ≥ 3. Again by relabeling, we may assume d2 ≥ 2.
P
Now since
(d2 − 1) +
n
X
i=3
di =
n
X
!
di − 2 = 2(n − 1) − 2 = 2(n − 2) ,
i=1
by the induction hypothesis there exists a tree whose degree sequence is the sequence
of positive integers d2 − 1, d3 , . . . , dn . Attaching a single edge to the vertex with degree
d2 − 1, we get a tree with a degree sequence 1, d2 , . . . , dn .
3. An Eulerian walk is a walk that passes through every edge of a graph exactly
once. Eulerian walk may end in a different vertex from where it started. Show that
a graph without isolated vertices has an Eulerian walk if and only if it is connected
and all vertices except at most two have even degree.
Recall the standard (hopefully) theorem of Euler which says that a graph has an Eulerian cycle if and only if it is connected and every vertex has even degree.
Let G be a graph with no isolated vertices.
Suppose G has an Eulerian walk π. Since π passes through every edge, and every
vertex has an edge incident to it (thanks to the no isolated vertex assumption), the
walk passes through every vertex as well, hence G is connected. If π is a cycle, then
by Euler’s theorem every vertex of G has even degree; so we are done. So suppose π
starts at v and ends at w, where v 6= w. Now insert an edge e that connects w back to
v, and call the resulting graph G0 = G ∪ {e}. Since the only edge that is in G0 but not
in G is e, πe is an Eulerian cycle of G0 . So by Euler’s theorem every vertex of G0 has
even degree. Writing dG and dG0 for the degree function for G and G0 respectively, we
have dG (x) = dG0 (x) if x 6= v, w and dG (v) = dG0 (v) − 1, dG (w) = dG0 (w) − 1. Thus v
and w have odd degree in G and the other vertices have even degree.
Conversely, assume G is connected and all vertices except at most two have even degree.
Because the sum of degrees is even, either no vertex has odd degree or exactly two
vertices have odd degree. In the former case, again by Euler’s theorem we conclude
that G has an Eulerian cycle. In the latter case, let v and w be the two vertices with odd
degree and add an edge e between them as above. Now in G0 := G ∪ {e}, every vertex
has even degree, so by Euler’s theorem G0 has an Eulerian cycle σ. Now removing e
from σ yields an Eulerian walk in G.
4. Call a graph on n vertices a dendroid if it has n edges and is connected. Characterize degree sequences of dendroids.
If n = 1, there is only one dendroid: a single vertex with a loop; so the degree sequence
is 2.
We claim that a sequence of positive integers d1 , . . . , dn is the degree sequence of a
P
dendroid if and only if ni=1 di = 2n. The forward implication is clear as a dendroid
has n edges. Now assume d1 , . . . , dn is a sequence of positive integers whose sum is 2n,
we may assume d1 ≥ d2 ≥ · · · ≥ dn by relabeling. The n = 1 case is true by the above,
so we may also assume n ≥ 2.
MATH 5707 HOMEWORK 1 SOLUTIONS
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Now we claim that d1 ≥ 2, because otherwise d1 = · · · dn = 1 which contradicts
P
di = 2n. We consider two cases:
• d1 ≥ 3. Then d1 − 2, d2 , · · · , dn is a sequence of positive integers whose sum is
2n − 2. By problem 2, there is a tree T with this degree sequence. So T has
n vertices and n − 1 edges by Proposition 1. Adding a loop to the first vertex
yields a dendroid with the degree sequence d1 , . . . , dn .
• d1 = 2. Then we claim that d2 = 2, because otherwise d2 = · · · = dn = 1 so
n
X
di = 2 + n − 1 = n + 1 < 2n
i=1
because n ≥ 2. So now d1 − 1, d2 − 1, · · · , dn is a sequence of positive integers
whose sum is 2n − 2. By problem 2, there is a tree T with this degree sequence.
Adding an extra edge between the first and second vertices yields a dendroid
with the degree sequence d1 , d2 , · · · , dn .
5. Show that every graph with 2` ≥ 2 vertices of odd degrees is the edge-disjoint
union of ` paths.
This is true only for connected graphs. For example the disjoint union of K2 and K3
has only two vertices of odd degree, yet it is disconnected, so a single path cannot cover
it.
We employ induction on `. For ` = 1, let G be a connected graph with 2 vertices, say
u and v of odd degree. Since G is connected and it has at least two vertices, it has no
isolated vertices. Thus by Problem 3, G has an Eulerian walk π.
Now assume that the claim holds for every k < `, where ` ≥ 2. Let G be a graph
with 2` vertices of odd degrees. Let u and v be two vertices of odd degree in G. Insert
a new edge e between u and v and call the resulting graph G0 . Now G0 has precisely
2` − 2 = 2(` − 1) vertices of odd degrees. So G0 is the edge-disjoint union of ` − 1 paths,
say π1 , · · · , π`−1 ; here we may assume e lies in π`−1 . Removing e from π`−1 breaks it
into two paths, say σ`−1 and σ` . Because the only extra thing in G0 different from G was
the edge e, we see that G is the edge-disjoint union of the paths π1 , · · · , π`−2 , σ`−1 , σ` .
6. Prove that either a graph or its complement is connected.
Assuming G is a disconnected graph, we show that its complement Gc is connected.
Since G is disconnected, it can be written as the disjoint union of two subgraphs:
G = H t K, where H, K are nonempty. Since there are no edges in between H and
K in G, for every vertex u ∈ H and v ∈ K, there is an edge in Gc from between u to
v. In particular, this shows that vertices in H are connected with the vertices in K in
Gc . Let us show that if u, u0 ∈ H, then u and u0 are connected in Gc . For this, since
K is nonempty, pick v ∈ K. Then there exists an edge e between u and v, and edge
e0 between u0 and v. Now we see that the path ee0 connects u and u0 . Similarly, two
vertices in K are connected with each other in Gc (go to H, and come back). Thus Gc
is connected.
7. Show that vertices of every graph G can be partitioned V = U ∪ W so that the
number of edges between U and W is at least half of the total number of edges in
G.
MATH 5707 HOMEWORK 1 SOLUTIONS
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The statement is false for the graph with a single vertex and a loop, so we assume G
has no loops.
Employ induction on the number of vertices |V | of G. If |V | = 0, there are no edges so
there is nothing to show.
Let n ≥ 1, let G = (V, E) be a graph with |V | = n and assume the claim holds when
the number of vertices of a graph is less than n (inductive hypothesis). Pick a vertex
v ∈ V and let G0 = (V 0 , E 0 ) be the subgraph of G obtained by deleting v and all of the
edges incident to v.
Given subsets X and Y of vertices of G, let us write eG (X, Y ) for the number of edges
in G that are between X and Y .
Note that V 0 = V − {v}. So by the induction hypothesis, V − {v} can be partitioned
V − {v} = U 0 ∪ W 0 so that
|E 0 |
|E| − d(v)
eG0 (U 0 , W 0 ) ≥
=
2
2
where d(v) is the degree of v in G. Since G has no loops, we have
d(v) = eG (v, V − {v}) = eG (v, U 0 ) + eG (v, W 0 ) .
Therefore, either eG (v, U 0 ) ≥ d(v)
or eG (v, V 0 ) ≥ d(v)
. By relabeling if necessary, we
2
2
d(v)
0
0
may assume eG (v, U ) ≥ 2 . Letting U = U and W = W 0 ∪ {v}, we see that V is
partitioned V = U ∪ W such that
eG (U, W ) = eG (U, W 0 ) + eG (U, v)
= eG0 (U 0 , W 0 ) + eG (v, U )
≥
as desired.
|E| − d(v) d(v)
+
= |E| ,
2
2