Graph Theory
Spring 2013
Prof. János Pach
Assist. Filip Morić
Exercise sheet 3: Solutions
Caveat emptor: These are merely extended hints, rather than complete solutions.
1. Show that a tree T has a perfect matching if and only if for every
vertex v the graph T − v has exactly one odd component.
Solution. If there’s a perfect matching in T , then clearly for every v
the only odd component in T − v will be the component that contains
u, the unique neighbor of v in the matching.
To prove the converse, we argue by induction. Choose an arbitrary leaf
u and let v be its unique neighbor. By assumption, the components of
T −v are u and several trees T1 , . . . , Tk of even order. It is not difficult
to see that each Ti satisfies the condition that Ti − vi has exactly one
component for each vertex vi of Ti . Thus, by the inductive hypothesis
all these trees have a perfect matching, which easily provides a perfect
matching for T .
2. Prove that every bridgeless cubic graph has a 1-factor.
Solution. Corollary 2.2.2 in Diestel’s book.
3. (a) Prove that a connected graph has an Euler tour if and only if
every vertex has even degree.
(b) Let G be a connected 2d-regular graph with an even number of
edges. Prove that it contains a d-factor.
(c) Prove that every regular graph of positive even degree has a 2factor.
Solution. (a) Theorem 1.8.1 in Diestel’s book.
(b) Since G is connected and all degrees are even, there exists an Euler
tour e1 e2 . . . e2m . Then the subgraph generated by e1 , e3 , . . . , e2m−1 is
a d-factor.
(c) Corollary 2.1.5 in Diestel’s book.
4. If A1 , . . . , An are finite sets, show that we can find at least
such that the union of no two is a third.
√
n of them
Graph Theory
Spring 2013
Prof. János Pach
Assist. Filip Morić
Solution. A family of sets is a poset w.r.t. the inclusion relation.
By Dilworth’s theorem we can find either a chain or an antichain of
√
size at least n. Both a chain and an antichain satisfy the desired
condition.
5. (a) Let X be a set with n elements, and let A1 , . . . , Ak be a family
of subsets of X, such that Ai 6⊆ Aj for all i 6= j. Prove that
n
k ≤ bn/2c
.
(b) Let x1 , . . . , xn be real numbers with |xi | ≥ 1 for all i, and let
I ⊂ R bePan arbitrary interval of length 2. Prove that the number
of sums k=1 k xk , where the
k are ±1, which fall in the interior
n
of I does not exceed bn/2c
.
Solution. (a) By Dilworth’s theorem, it suffices
to show that the
n
X
poset (2 , ⊂) can be decomposed into bn/2c chains. To this end,
we notice that by a simple application of Hall’s theorem, for each
k = 1, . . . , bn/2c there is a matching from the family of (k − 1)-subsets
to the family of k-subsets, such that for each pair in the matching one
subset is contained in the other. This family of matchings can be used
to produce a decomposition as needed. There are many other proofs of
this statement, known as Sperner’s theorem.
(b) W.l.o.g. we can assume that xi ≥ 1 for all i (do you see why?).
Now to each choice (1 , . . . , n ) for which the corresponding sum falls
into I, assign the set of indices i such that i = 1. We get a Sperner
family of subsets of {1, 2, . . . , n} and the conclusion follows from (a).
6. Alice and Bob perform a card trick. Alice shuffles a deck of cards, then
peels off the top 5 cards, shows them to Bob and to the audience, and
gives them to a volunteer for additional mixing. Bob leaves the room,
and Alice then has the volunteer lay out the cards in a face-up row on
the table, stressing how totally random everything is. Next, Alice has
the cards turned face-down. Finally, Bob is invited into the room, to
view the displayed cards. Alice silently turns over 2 cards, and invokes
the spirit of the late great Paul Erdős. Bob picks up the vibes, and
unhesitatingly identifies the 3 remaining face-down cards correctly, in
the right order. How is this possible?
Solution. Just observe that by Erdős-Szekeres theorem, among any
five numbers there is either an increasing or a decreasing subsequence
of length 3. Alice will turn over 2 other cards, moreover in the order
indicating whether the remaining sequence is increasing or decreasing.
Graph Theory
Spring 2013
Prof. János Pach
Assist. Filip Morić
7. Baron Münchhausen is very proud: “I chose a huge set of positive
integers smaller than 2013, and yet among any four elements there are
two such that one is divisible by the other.” He went on boasting to
his stupefied audience: “I bet no one could ever find such a set with
more elements.” However, the infamous storyteller was exaggerating
again. The set he had found was rather small. Help him find a set
that could save his honor.
Solution. Consider the poset P = ({1, 2, . . . , 2012}, |). We are looking
for a largest subposet in which there is no antichain of size 4. Let S
be a subposet with no antichain of size 4. Then by Dilworth’s theorem,
S can be decomposed into at most 3 chains, say S = C1 ∪ C2 ∪ C3 ,
where the Ci ’s are disjoint chains, possibly empty. For an n ∈ N, let
f (n) be the number of prime divisors of n, counted with multiplicity.
For a non-empty chain Ci , we have that |Ci | ≤ f (ci ) + 1, where ci
is the largest element of Ci . Moreover, if 1 ∈
/ Ci , then we have that
|Ci | ≤ f (ci ). Since maxn∈P f (n) = 10, and the maximum is attained
only for n ∈ {210 , 3 · 29 }, we have that
|S| = |C1 | + |C2 | + |C3 | ≤ 10 + 10 + 9 + 1 = 30 .
On the other hand, there are examples with 30 elements:
S = {1, 2, . . . , 210 , 3, 3 · 2, . . . , 3 · 29 , 5, 5 · 2, . . . , 5 · 28 } .
If you spot any mistakes on this sheet, please drop an email to [email protected].