Partial solutions to P

MATH 4707 PROBLEM SET 6 SOLUTIONS
JOEL LEWIS
SOME SOLUTIONS COURTESY OF JED YANG AND MAX GLICK
13.4.2 Let G be a connected graph such that all vertices have degree at most d, and there exists
a vertex with degree strictly less than d. Prove that G is d-colourable.
Proof. We prove by induction on the number of vertices. Base case is trivial. Let G be a connected
graph, and v a vertex with degree strictly less than d. Let G0 = G − v be the graph obtained by
deleting v from G. Note that all vertices of G0 have degree at most d. As G is connected, each
connected component of G0 has a vertex that is adjacent to v in G. This vertex has degree strictly
less than d in G0 , so the induction hypothesis is satisfied for each connected component. Fix a
set of d colours and properly colour each of the connected components of G0 (here we use strong
induction). Since v has fewer than d neighbours, there is still some colour available for v, as desired.
Alternatively, recall that we proved in class that if a graph G has the property that all of
its subsets have a vertex of degree less than d, then G is d-colorable. (The argument was by an
induction similar to the previous solution.) We will show that this is true of our graph G. Let G
and v be as above and let H be an arbitrary subgraph of G. If H contains v, then H contains a
vertex of degree less than d, as desired. Suppose H does not contain v. Let A be the set of vertices
of H and let B be the (nonempty) set of vertices not in H. Since G is connected, there is some
edge {u, w} connecting a vertex in u ∈ A to a vertex w ∈ B. Since the degree of u in G is at most
d, its degree in H must be strictly less than d (because one of its neighbors is outside H). So again
H contains a vertex of degree less than d. Thus every subgraph of G contains a vertex of degree
less than d. Apply the theorem from class to declare victory.
13.4.4 Since K5 has 10 edges, G = (V, E) has 10 vertices, namely
V = {12, 13, 14, 15, 23, 24, 25, 34, 35, 45}.
Two of the vertices are adjacent if and only if they have a common number (e.g. 13 and 34 are
adjacent but 13 and 24 are not). We must give colors to these 10 vertices so that vertices of the
same color are never adjacent. One possible 5-coloring is
Color Vertices
Red
12, 35
Yellow 23, 14
Green 34, 25
Blue
45, 13
Purple 15, 24
(draw a picture!). Suppose for the sake of contradiction that a 4-coloring existed. There are 10
total vertices so by the pigeonhole principle there would have to be 3 vertices of the same color,
say red. Each vertex consists of 2 numbers from 1 to 5, so by pigeonhole again some 2 of the red
vertices would need to have a number in common. This contradicts the definition of a coloring.
A 4-coloring is not possible so the chromatic number is 5.
(The graph G is called the line graph of K5 , and the chromatic number of the line graph is called
the chromatic index. Vizing’s Theorem asserts that the chromatic index of every graph G is either
equal to the maximum degree of G or one more than the maximum degree, so in particular coloring
line graphs is in some sense less complicated than coloring general graphs.)
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JOEL LEWIS SOME SOLUTIONS COURTESY OF JED YANG AND MAX GLICK
(1)
Proof. a) Suppose that G is a planar graph of girth g, and consider a drawing of G in the plane.
Every face of the resulting drawing is bounded by a closed walk, which includes a cycle. Since every
cycle of G has at least g edges, every face in the drawing of G has at least g boundary edges. For
a plane drawing of a graph G, we have
X
2E =
#{boundary edges of f },
f a face
and thus in our case
2E ≥ g · F.
Applying Euler’s formula, we get that
2=V −E+F ≤V −E+
2E
g−2
=V −
· E,
g
g
and rearranging gives the desired result.
b) The graph K3,3 has V = 6, E = 9 and g = 4. Thus, if it were planar we would have by part
a that 2 ≤ 6 − 24 · 9 = 1.5, a contradiction.
(2) Recall that we defined a contracted graph as follows: given a graph G with vertices a, b
connected by an edge e = {a, b}, the contraction G/e has vertex set (V (G) r {a, b}) ∪ {ab}
(i.e., all the vertices of G except a and b, with a new vertex added called “ab”) and edge set
{{c, d} : {c, d} ∈ E(G) and c, d 6= a, b} ∪ {{c, ab} : {c, a} ∈ E(G) or {c, b} ∈ E(G)} (i.e., leave
edges not touching a, b as they were, and connect old vertex c to the new vertex ab if and only
if c was connected to at least one of a, b). For example, if G has vertices {w, x, y, z, c} and edges
{{w, x}, {x, y}, {y, z}, {z, w}, {c, w}, {c, x}, {c, y}, {c, z}} (draw it!) then the contraction G/{c, w}
has the four vertices {cw, x, y, z} and the five edges {{x, y}, {y, z}, {cw, x}, {cw, y}, {cw, z}}. Define
T (G) to be the number of spanning trees of a graph G.
a) Suppose that G is a graph and {a, b} is an edge of G. Show that the number of spanning
trees of G that do not contain {a, b} is equal to T (G r {a, b}) (where G r {a, b} is the result
of deleting edge {a, b} from G).
b) Suppose that G is a graph and {a, b} is an edge of G and no vertex is connected to both a, b.
Show that T (G/{a, b}) (this is a contraction) is the number of spanning trees of G that do
contain the edge {a, b}.
c) Is the result of part (b) true in general? (Hint: consider K3 , and be careful about the
definition of contraction: the result of contraction is always a (simple) graph.) Can you
explain how to fix it with multigraphs?
d) Use the preceding parts to give a recursive formula for T (G) in terms of smaller graphs (i.e.,
those with strictly fewer edges or with strictly fewer vertices).
Proof. Part (a) is trivial: a spanning tree of G r e is precisely the same thing as a spanning tree of
G that doesn’t include e.
Part (b) is trivial if G has only the two vertices a, b, so assume that this is not the case. We
argue bijectively, as follows: given a spanning tree T of G including the edge e = {a, b}, choose a
subset S of the edges of G/e as follows: if {x, y} is an edge of T not involving a, b, add {x, y} to
S. If {a, x} (or {b, y}) is an edge of T , add {ab, x} (or {ab, y}) to S. We claim that S is the set
of edges of a spanning tree of G/e. If G has n vertices then T has n − 1 edges and so S has n − 2
edges (because we get one edge for each edge of T other than e). Since G/e has n − 1 vertices, it
suffices to establish our claim to show that S is connected and spans the vertices of G/e.
That S spans is straightforward: every vertex x other than a, b was in some edge of T , and so
is in the corresponding edge of S. Because G has at least three vertices, at least one of a, b must
have been connected in T to some other vertex in V (G), and the corresponding edge in S connects
MATH 4707 PROBLEM SET 6 SOLUTIONS
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to ab. Similarly, we have that S must be connected: every x 6= a was connected to a by a path in
T ; the image of the edges in this path is a path from x to ab via S. Since S connects every vertex
of G/e to ab, it is a connected subgraph. Putting everything together, S is a spanning tree of G/e.
Similarly, given a spanning tree S of G/e, we may construct a spanning tree of G containing e as
follows: if {x, y} ∈ S does not involve ab then put {x, y} as an edge in T . If {x, ab} ∈ S then (by
definition of contraction) x must be connected to one of a, b in G. By hypothesis1, a and b have no
common neighbors, so in fact x is connected in G to exactly one of a, b. Add this edge to T . One
must check that the resulting graph really is a spanning tree of G, a step we omit.
These two operations are mutually inverse bijections between the spanning trees of G containing
e and the spanning trees of G/e. This gives the desired conclusion.
For the first part of (c), observe that in the graph K3 , the number of spanning trees containing
edge 12 is 2, but the contracted graph is just a single edge connecting vertices 12 and 3 and has
only a single spanning tree.
For the rest of part (c) and (d), we make a new definition of contraction, this time on multigraphs:
if e = {a, b} is an edge in a multigraph G, the contraction G/e should have one edge for every edge
of G not touching a, b, and also an edge {x, ab} for every edge {x, a} or {x, b}.2 Thus, for example,
with this definition we have that K3 /{1, 2} is a graph with two vertices {12, 3} joined by two edges
(one from the edge 13 and one from the edge 23). Now the argument of part (b) works without
needing the extra condition about no triangles. In the end, we conclude that with this alternate
definition we have the deletion-contraction relation
T (G) = T (G r e) + T (G/e)
for any multigraph G and edge e.
(3) Show that it is possible to draw K5 on a torus. What are the values of V , E, F for this
drawing? [Note: it may be helpful to think of a torus as a square with the opposite edges identified,
as in some old video games, so that moving off the left edge returns you on the right edge similarly
for the top and bottom. If you draw things like this then some edges and faces will “wrap around.”]
Proof. See the scanned pdf linked from the course webpage for drawings of K5 and K6 on a torus.
In fact, it is possible to draw K7 (but not K8 !) on a torus without crossings.
1Note
that this is the only place we use the assumption that a and b have no common neighbors, but it is absolutely
essential for this step to be well-defined!
2What to do about loops? We can either allow them, so that if there were multiple edges between a and b then
after contracting there are a bunch of loops at vertex ab, or we can throw them out. Since they won’t be part of any
spanning tree, it doesn’t matter.