Math 4575 : HW #7
Chapter 11: #1, 2, 3, 4, 8, 9, 10, 20, 39, 44, 56, 62
• #1
How many nonisomorphic graphs of order 1 are there? of order 2? of order
3? Explain why the answer to each of the preceding questions is ∞ for general
graphs.
There is one nonisomorphic graph of order 1, the edge empty graph. There are
two graphs of order 2—either this is an edge, or there isn’t. There are four graphs
of order 3—in this case, again the isomorphism type is determined by the number
of edges, which is either 0, 1, 2, or 3.
The answer is ∞ in the case of general graphs (i.e. multigraphs), since there is
no bound on the number of edges, even for a graph of order 1. A single vertex
may have arbitrarily many loops.
• #2
Determine each of the 11 nonisomorphic graphs of order 4, and give a planar
representation of each.
• #3
Does there exist a graph of order 5 whose degree sequence equals (4, 4, 3, 2, 2)?
No, since the number of vertices of odd degree is always even. There is not even
a multigraph with this degree sequence.
• #4
Does there exist a graph of order 5 whose degree sequence equals (4, 4, 4, 2, 2)?
a multigraph?
There is not a graph of order 5 with this degree sequence. Indeed, if vertices 1,
2, and 3 each have degree 4 then they are each connected to every other vertex,
so vertices 4 and 5 must each have degree at least 3.
On the other hand, there is a multigraph with this degree sequence. Consider
the graph that is the union of a triangle 123 and an edge 45. Then the degree
sequence is (2, 2, 2, 1, 1), so replacing each edge by a “double edge” results in
1
degree sequence (4, 4, 4, 2, 2).
• #8
Let G be a graph with degree sequence (d1 , d2 , . . . , dn ). Prove that, for each k
with 0 < k < n,
(∗)
k
X
di ≤ k(k − 1) +
i=1
n
X
min{k, di }.
i=k+1
We interpret the left side as counting incident pairs (v, e), here v is a vertex in
the first k vertices of G, and e is an edge containing v.
Now we try to count incident pairs (v, e), but by summing over edges rather than
over vertices. Every edge counted in an incident pair is in one of two categories:
either both
its ends are in the first k vertices, or only one end. There are at
most k2 edges with both ends on first k vertices, and each such edge would get
counted twice, once for each end of the edge, so these edges contribute at most
k(k − 1) to the sum.
Every other edge has one end on one of the last n − k vertices, and the other end
on one of the first k vertices. For i = k + 1, . . . , n, the contribution of vertex i to
the count of incident pairs can be at most k, since each edge must have its other
end in the first k vertices. On the other hand we also note that its contribution
can be at most di , since that is its total degree. Since both these bounds hold, we
have that the contribution from vertex i is at most min{k, di }.
Putting it all together, we have the desired inequality.
Comment: the Erdős–Gallai theorem is a famous theorem in graph theory, that
a necessary and sufficient condition for d1 ≥ d2 ≥ . . . dn ≥ 0 to be the degree
sequence of graph is that (1) the sum d1 + d2 + · · · + dn be even, and (2) the
inequality (∗) be satisfied for every k. We have shown in this exercise that the
condition is necessary, but showing that it is sufficient is harder.
• #9
Draw a connected graph whose degree sequence equals (5, 4, 3, 3, 3, 3, 3, 2, 2).
2
• #10
Prove that any two connected graphs of order n with degree sequence (2, 2, . . . , 2)
are isomorphic.
We know that a connected graph of order n with this degree sequence admits a
closed Eulerian trail passing through every edge. Since it passes through every
edge, it passes through every vertex. Because every degree is 2, the trail must
enter and exit every vertex exactly once along the trail. In other words, we may
label the vertices v1 , v2 , . . . , vn such that every vertex vi is adjacent to vi−1 and
vi+1 . (The indices are read modulo n.) So such a graph must be isomorphic to
the cycle Cn .
• #20
Prove that a graph of order n with at least
(n − 1)(n − 2)
+1
2
edges must be connected. Give an example of a disconnected graph of order n
with one fewer edge.
If G = (V, E) is disconnected, then it has `
at least two connected components,
and it’s vertices may be partitioned V = A B so that A and B are nonempty,
and there are no edges between A and B. Suppose that A = k and B = n − k.
Then the maximum number of edges in such a graph is
n
− k(n − k).
2
Since n is given, maximizing this function is equivalent to minimizing k(n − k).
It is straightforward to check that the minimum of the function f (k) = k(n − k)
over the interval k ∈ [1, n − 1] are obtained at the endpoints, when k = 1 or
k = n − 1. (One could use calculus, for example, to see that there is only one
critical point inside the interval, a local maximum at k = n/2.) In either case,
k(n − k) = n − 1 and the number of edges in the graph is
n
n−1
− (n − 1) =
.
2
2
The only example of a disconnected graph with n−1
edges is a clique of order
2
n − 1 together with an isolated vertex. Our argument shows that a graph with
more edges than this must be connected, which is the desired result.
• #39
Call a graph cubic if each vertex has degree equal to 3. The complete graph K4
is the smallest example of a cubic graph. Find an example of a connected, cubic
3
graph that does not have a Hamilton path.
• #44
Which complete bipartite graphs Km,n have Hamilton cycles? Which have
Hamilton paths?
Suppose first that v1 − v2 − · · · − vm+n − v1 is a Hamiltonian cycle. Then first
of all, m + n is even, since Km,n is bipartite and only contains cycles of even
length. Moreover, every odd labelled vertex is in part A and every even labelled
vertex in part B. Since every vertex is in the cycle, this gives that |A| = |B|, or
m = n. Finally, if there is a cycle at all then m and n must be at least 2. So
Km,n has a Hamiltonian cycle if and only if m = n and m, n ≥ 2.
By essentially the same reasoning, Km,n has a Hamiltonian path if and only if
|m − n| ≤ 1 and m, n ≥ 1.
• #56
Grow all the nonisomorphic trees of order 7.
• #62
Prove that, if a tree has a vertex of degree p, then it has at least p pendent vertices.
Suppose that the tree T has n vertices, with degree sequence d1 , d2 , . . . , dn .
Since T has n − 1 edges, we have
d1 + d2 + · · · + dn = 2(n − 1) = 2n − 2.
4
Every vertex has degree at least one. Suppose without loss of generality that
d1 = p. Then
d2 + d3 + · · · + dn = 2n − 2 − p.
Suppose that m of the remaining vertices are pendent, i.e. have degree 1, and the
remaining n − 1 − m vertices in this sum all have degree ≥ 2. Then the total of
these vertex degrees is at least m + 2(n − 1 − m), so we have
m + 2(n − 1 − m) ≤ 2n − 2 − p.
After some cancellation, it follows that m ≥ p, as desired.
5