Math 240: Discrete Structures I
Assignment 6: Solutions
1. Cycles and Circuits.
(a) Let G be a graph on 8 vertices with 25 edges. By the handshaking lemma,
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we have u∈V deg(u) = 2m = 50. Since the graph is simple, the maximum
degree is 7. Since 8 · 6 = 48, there must be at least 2 nodes of degree 7,
contradicting the fact that the graph is Eulerian.
If G had 24 edges, the graph can in fact be Eulerian: consider K8 , the
complete graph on 8 vertices, and label the vertices v1 , v2 , . . . , v8 . Remove
the edges {v1 , v2 }, {v3 , v4 }, {v5 , v6 } and {v7 , v8 }, to get a graph where all
degrees are 6, and is therefore Eulerian.
(b) Let G be a graph on 21 vertices with at least 200 edges. The complete
graph would have had 21
= 210 edges, so this is a complete graph minus
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at most 10 edges. Note that the vertices in the complete graph have degree
20.
If there is no single vertex incident to all 10 removed edges, then the degree
, and thus, by Dirac’s theorem,
of every vertex must be at least 11 > 21
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the graph is Hamiltonian.
If there is one vertex to which all the removed edges are incident, then the
graph is complete on the 20 other vertices. Construct some Hamiltonian
walk on these vertices, and simply add-in the remaining vertex. Thus, in
either case, the graph is Hamiltonian.
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2. Trees.
(a) Recall that in a tree T , there is a unique path between any two given
vertices. Let Pu,v be the path from u to v. Let a be the vertex where Pu,v
meets the path Pw,v . Thus, we have Pu,v = Pu,a ∪Pa,v and Pw,v = Pw,a ∪Pa,v ,
where some of these paths may be empty (e.g. if w lies on Pu,v , then a = w
and Pw,a is empty.)
Since |Pu,v | and |Pw,v | are both even, we have that |Pu,a | and |Pa,v | must
have the same parity, and |Pw,a | and |Pa,v | must have the same parity. But
this implies |Pu,a | and |Pw,a | = |Pa,w | have the same parity, so Pu,w must
be of even length.
(b) For the counterexample in the general case, consider the following graph:
there exist even paths from u to v and from w to v, but there are no even
paths from u to w, since a path cannot cross the same vertex twice, and
so must stay on the bottom path.
v
u
w
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3. Joyal Codes.
(a) For the Joyal code 379728976, we create a directed graph by consideration
of
123456789
379728976
to obtain
1
3
9
6
8
7
2
5
4
And so the set of vertices which belong to a cycle is {6, 7, 8, 9} in sorted
(basic) order. This gives an f -order via the mapping 68 79 87 96 . Thus, the
‘left-right path of the tree is 8 → 9 → 7 → 6, which gives us the following
tree:
left 8
9
7
3
2
1
5
6 right
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(b) The set of vertices in cycles is {4, 8, 3, 2} as they lie on the path P between
the left and right vertices. This is their f -order. Adding their sorted (basic)
order gives
2348
2384
For the remaining nodes, we copy their edges in the tree and direct them
towards the path P ¿ Thus, we have 1 → 3, 5 → 6 → 2, 7 → 4 and 9 → 2,
Together we get the mapping
123456789
323862442
which gives the code 323862442.
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4. Connectivity.
(a) G is connected so it contains a spanning tree T . Construct the list of
vertices as follows: take a leaf of T and call it v1 . Then take a leaf of T −v1
and call it v2 . Proceed like this, taking a leaf of T − {v1 , v2 , . . . , vi−1 } and
calling it vi . Removing the leaves one at a time leaves the spanning tree
connected. Since the spanning tree is a subset of G, then this also leaves
G connected, as desired.
(b) We wish to show either G or its complement G is connected (or both). If
G is connected, we are done. If not, take any pair of vertices u and v. We
have two cases:
(i) u and v are in different components of G. In particular, G must not
contain the edge (u, v). Therefore G does contain the edge (u, v), and there
is a (trivial) walk from u to v in G (ii) u and v are in the same component
of G. Since G is disconnected it has at least two components. So take
a vertex w is in a different component. Then G contains both the edges
(u, w) and (v, w). Thus, there is a path {u, w, v} between any u and v in
G.
Since there is a path from u to v in G for every pair of vertices we have
that G is connected.
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5. Planar Graphs.
(a) We begin by noting that, if infinite graphs are allowed, then this is possible
for every choice of k and k 0 , but the construction is a little difficult.
However, if we restrict ourselves to finite graphs — which is the case in
this course — there are two possible answers, whether or not you consider
that the dual must be simple. If yes, there are only 5 finite connected
simple k-regular graphs whose dual is a finite connected simple k 0 -regular
graph, that is the platonic solid graphs. If no, then there are the same
5 graphs above, one 1-regular connected simple graph whose dual is a 2regular connected non-simple graph, and infinitely many 2-regular simple
connected graphs whose dual is a non-simple k 0 regular graph. We will
show all of these below.
We first begin by recalling Euler’s formula for finite connected (but not
necessarily simple) connected planar graphs: |V | − |E| + |F | = 2. Note
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that u∈V deg(u) = 2|E|, that the original graph has as many edges as its
dual, and that the number of faces of the original graph, |F |, is equal to
the number of vertices of the dual, |V 0 |. This gives us the equation
|V | − 12 · |V | · k + |V 0 | = 2 = |V 0 | − 21 · |V 0 | · k 0 + |V |
Cancelling terms, we get
0
0
k · |V | = k · |V |
⇐⇒
|V 0 |
|V |
=
k0
k
Now for k = 1, the only possible graph is an edge
a self-loop .
whose dual is simply
For k = 2, we need 2|V | = k 0 |V 0 |. This can be attained with k 0 = |V | and
|V 0 | = 2, by constructing a cycle. Note that the dual is not simple
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It remains to handle the case k ≥ 3 (and by symmetry k 0 ≥ 3). This was
done in class. Namely, by the handshaking lemma, we have 21 |V | · k = |E|,
so |V | = k2 |E|. Similarly, |V 0 | = k20 |E 0 | = k20 |E|. Thus,
2|E|
1
k
− 12 +
1
k0
=2
=⇒
1
k
+
1
k0
>
1
2
Hence, if k = 3, k 0 = 3, 4, 5, if k = 4, then k 0 = 3, (but (3, 4) was already
considered, by symmetry), and if k = 5, then k 0 = 3, which were already
considered. We note that (k, k 0 ) = (3, 3), (3, 4), (4, 3), (3, 5), (5, 3) are
simply the 5 platonic solid graphs.
Remark. We can also create infinite graphs. As then, in the limit, k1 + k10 >
1
can also be satisfied for (3, 6), (4, 4) and (6, 3). ((4, 4) is the infinite tiling
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of the plane with squares, (3, 6) with hexagons, and (6, 3) with triangles.
(b) We know G contains a vertex v1 of degree at most 5. Colour the edges
incident to v1 with (at most five) distinct colours. Now G−{v1 } is planar so
also contains a vertex v2 of degree at most five. Colour the edges incident
to v2 in G − {v1 } with (at most five) distinct colours. Repeat. Let Ei be
the set of edges of colour i. We claim Ei forms a forest Fi . Note that v1 is
a leaf in Fi (or v1 is not a vertex of the forest Fi ). The result then follows
by induction on G − {v1 }.
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6. Dual Graphs.
(⇒) Let G be any planar graph, and let F be the outside face. Let f be any
other face, it is clear that there must be a path from f to F in the dual, so the
dual must be connected. Now, if G is disconnected, then its dual is connected,
and so is the dual of its dual. Thus, G can not be the dual of its dual.
(⇐) If G is connected, then both G and its dual, call it G0 = (V 0 , E 0 ), satisfy
Euler’s formula, |V | − |E| + |F | = 2. Since |F | = |V 0 | and |E| = |E 0 | (there is
a dual vertex for every face in G, and a dual edge for every edge in G) then we
have |V 0 | = 2 + |E| − |V | and |V 00 | = 2 + |E 0 | − |V 0 |, where G00 = (V 00 , E 00 ) is the
dual of the dual. But this is simply
|V 00 | = 2 + |E 0 | − |V 0 |
= 2 + |E| − (2 + |E| − |V |)
=
|V |
Hence, |V | = |V 00 | and |E| = |E 0 | = |E 00 |. Now, let u and v be any two vertices
of G. The edge {u, v} must be crossed by some edge {f, g} in the dual, where
f and g are the two faces on either side of {u, v}. But now u and v are the
dual-faces on either side of {f, g} in the dual. This implies that we have a
mapping φ : V → V 00 such that {u, v} ∈ E implies {φ(u), φ(v)} ∈ E 00 . Since we
know that |E| = |E 00 |, we can conclude that G is isomorphic to G00 , as desired.
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