1. The line graph of a graph.
Let G = (V, E) be a graph with E 6= ∅. The line graph of G is the graph
LG such that V (LG ) = E and
E(LG ) = {ef : ∀e, f ∈ E : e and f are adjacent}.
Examples 1.1.
(1) If G is a path of length n, then LG is a path of
length n − 1.
(2) If G is a cycle of length n, then LG ∼
= G.
(3) If G is a star with n vertices i.e. |G| ≥ 3 and G has a vertex u such
that E = {vu, ∀v ∈ V r {u}}, then LG ∼
= K n−1 .
The converses also hold, see Exercise 1.6.
For an edge uv ∈ V we write (uv) ∈ V (LG ) for the corresponding vertex
of the line graph. But we abuse notation using the same symbol for edges,
or subsets of edges, in G and vertices, or subsets of vertices, in LG .
Lemma 1.2. Let G be a graph with E(G) 6= ∅. Then
(1)
X d(v)
|E(LG )| =
;
2
v∈V
(2) for any F ⊂ E(G) we have LG−F = LG − F ;
(3) for any v ∈ V (G) we have LG−v = LG − E(v).
Proof. For any v ∈ V (G), every pair of edges in E(v) corresponds to an
edge in LG , and all the edges of LG are obtained in this way. Hence the
first formula follows. The remaining two identitities follow trivially from the
definition.
Let
P = e0 . . . em ⊂ LG
be a path in LG ; we can associated to P the following alternating sequence
of vertices and edges in G:
w0 , e0 , w1 , e1 , . . . wm , em , wm+1 ,
where wi ∈ V (G) are such that e0 = w0 w1 , em = wm wm+1 and wi is the
vertex adjacent to ei−1 and ei for i = 1, . . . , m. By definition w0 6= w1
and wm 6= wm+1 , while we may have wi = wj for the remaining cases; see
Example 1.4. From the above sequence we extract the following sequence of
(not necessarily distinct) vertices in G
(1.1)
ω(P ) := (w0 , w1 , . . . , wm , wm+1 ).
We shall use this notation in the next basic Lemma.
Lemma - Definition 1.3. Let G be a graph and LG its line graph.
(1) Let Q = v0 . . . vn ⊂ G be a path. LQ is a path in LG .
(2) Let P ⊂ LG be a path such that ω(P ) consists of distinct vertices.
Then there exists a path QP ⊂ G such that P = LQP . We have
V (QP ) = ω(P )
and
length(QP ) = length(P ) + 1
1
2
(3) Fix e, f ∈ E(G) and let P ⊂ LG be an ef -path of minimal length.
Then P satisfies the assumption of (2).
(4) Let P ⊂ LG be an ef -path. Then there exists an ef -path P ⊂ LG
such that ω(P ) ⊂ ω(P ) and such that P satisfies the assumption of
(2).
Proof. (1). Let Q = v0 . . . vn be a path in G, we already noticed that LQ is
a path. Now vi vi+1 ∈ E and hence (vi vi+1 ) is a vertex of LG ; as the vi are
all distinct, the (vi vi+1 ) are all distinct. Now, the edges vi vi+1 and vi−1 vi
are adjacent for all i ≥ 1, hence (vi−1 vi )(vi vi+1 ) is an edge of LG . Therefore
LQ ⊂ LG .
(2). Keeping the above notation, assume w0 , . . . , wm+1 are all distinct.
By definition, G has the edge wi wi+1 = ei for all i ≤ m, hence Q =
w0 w1 . . . wm+1 is a path in G. Finally, LQ = P since, as we just noticed,
the vertex (wi wi+1 ) equals the vertex ei in LG . We set Q = QP and we are
done.
(3). We need to prove that if P = e0 . . . em has minimal length between
e = e0 and f = em , then the elements in ω(P ) are all distinct. By contradiction, let h be minimum such that there exists k > h with wh = wk . If
h > 0 then eh−1 is adjacent to ek , hence
(1.2)
P1 = e0 . . . eh−1 ek . . . em
is a path from e to f shorter than P , which is not possible. If h = 0 then
k ≥ 2, hence the following
(1.3)
P2 = e0 ek . . . em
is too short a path from e to f , a contradiction.
(4). Let P = e0 . . . em be a path from e = e0 to f = em and
ω(P ) = (w0 , w1 , . . . , wm+1 )
the associated sequence of vertices. If ω(P ) has no multiple vertices (i.e. if
wi 6= wj for all i 6= j) then, by (2), we can take P = P and we are done.
We shall prove (4) by induction on the number of multiple vertices in ω(P ),
i.e. the number
MP = |{i = 0, . . . , m + 1 : ∃j 6= i : wi = wj }|.
The induction base, MP = 0, has already been treated.
Let h be minimum such that there exists k > h with wh = wk . Then,
arguing exactly as in the proof of (3), we construct a shorter path from e to
f . More precisely, if h > 0, then the path is P1 , given in (1.2), and
ω(P1 ) = (w0 , w1 , . . . , wh , wk+1 , . . . wm+1 ) ( ω(P )
with h < k; thus the “multiple” vertex wk is not in ω(P1 ), hence MP1 ≤
MP − 1. The iduction hypotesis gives an ef -path P1 satisfying (2) and such
that ω(P1 ) ⊂ ω(P1 ) ⊂ ω(P ). So we are done.
In the case where the path is P2 , given in (1.3), we have
ω(P2 ) = (w1 , w0 , wk+1 , . . . wm+1 ) ( ω(P )
with k ≥ 2, so the same argument as before applies.
The next example illustrates parts of the Lemma and of its proof.
3
Example 1.4. In the picture below we have a star graph, G, and its line
graph, where we highlighted the path P described in the sequel. In G we
set ei = vi v0 for all i = 1, . . . , 5. Next, consider the following path in LG
joining e1 and e5
P = e1 e3 e4 e2 e5 ⊂ LG
hence
ω(P ) = (v1 , v0 , v0 , v0 , v0 , v5 ).
The proof yields
ω(P ) = (v1 , v0 , v5 )
and
P = e1 e5 .
Finally,
QP = v1 v0 v5 ⊂ G
and, of course,
Lv1 v0 v5 = (v1 v0 )(v0 v5 ) = e1 e5 .
•9v91
G=
v5
•
e1
v2
•
99 9v0
•
e2
•RRRRlll•
v3
•
• v4
LG =
lR
lll RRRRR e
e5llll
R3
•
s•
ssss
sss
• e4
Figure 1
Lemma 1.5. Let G = (V, E) be a graph and LG its line graph.
G is connected if and only if LG is connected and δ(G) ≥ 1.
Proof. Suppose G connected, let e = uw and f = xy be two edges of G; we
need to find a path in LG between e and f . Let Q = uv1 . . . vn x be a ux-path
in G of minimal length. Consider the path LQ ⊂ LG . If LQ contains e and
f we are done. If LQ does not contain e, then, by the minimality of Q, the
vertex w is not in Q. Hence eLQ is a path in LG containing e. If this path
does not contain f , arguing as above we have the path eLQ f contains e and
f . So we are done.
Conversely, assume LG connected and G free from isolated vertices. Fix
x, y ∈ V (G); we need to find a path joining x and y. Pick two edges, ex and
ey , adjacent to each of them (if ex = ey then we are done). Let P = e0 . . . en
be a path of minimal length in LG such that ex = e0 and ey = en . By
Lemma 1.3 there exists a path QP ⊂ G such that LQP = P ; by part (2) of
the Lemma the vertices of the path QP are those in ω(P ), which contains x
and y; so QP contains x and y, and we are done.
Exercise 1.6. Let G be such that δ(G) > 0. Prove the following.
(1) If LG is a path of length n, then G is a path of length n + 1.
(2) If LG is a cycle, then G ∼
= LG .
(3) If LG = K n , then G is a star with n edges.
4
Exercise 1.7. Let G = (V, E) and e, f ∈ E such that e = xy and f = uv.
Set M = min{d(x, u), d(x, v), d(y, u), d(y, v)}. Show that dLG (e, f ) = M +1.
2. Connectivity via paths
The definition of a connected graph is based on the notion of path. Indeed
let G = (V, E) be a graph. Then, by definition,
G is 1-connected if and only if for every x, y ∈ V there exists an xy-path
P ⊂ G.
Next, already know that
G is 2-connected if and only if for every x, y ∈ V there exist two independent xy-paths P1 , P2 ⊂ G
(recall that to say that P1 and P2 are independent is to say that i.e.
P1 ∩ P2 = {x, y}).
We shall prove in Theorem 2.7 that the analog holds for every k. For now
we prove the following result, which is the “easy” implication of Theorem 2.7.
Proposition 2.1. Let G be a graph containing k independent xy-paths for
every pair of vertices x, y. Then G is k-connected.
Proof. To prove that G is k-connected pick a pair of vertices x, y any a set
S ⊂ V r {x, y} separating x from y; we need to show that |S| ≥ k. Now,
let P1 , . . . , Pk be k independent xy-paths in G; then S each of them must
intersect S, so let si ∈ S ∩ V (Pi ); since the Pi are independent, we have
si 6= sj for all i 6= j; hence |S| ≥ k.
The proof of the converse is non-trivial, and it is based on an important
auxiliary result, interesting in its own right. To begin with, we introduce
the following notation: for any graph G = (V, E) and any A, B ⊂ V (not
necessarily disjoint) we set
(2.1)
κG (A, B) = min{|S| : ∀S ⊂ V (G) : S separates A from B}
and
(2.2)
πG (A, B) = maximum number of disjoint AB-paths in G.
Example 2.2. Assume G connected. If A = {x} and B = {y} with x 6= y
we can separate x from y with {x}; moreover it is clear that two xy-paths
will never be disjoint, hence
κG (x, y) = πG (x, y) = 1.
Example 2.3. Suppose B = V . Then an AB-path P must be of type P = v
with v ∈ A (for otherwise P would intersect B in two points, which is not
possible). Hence
πG (A, V ) = |A|;
now a set S intersecting every AB-path must contain A, therefore
κG (A, V ) = |A|
Lemma 2.4. Let G = (V, E) be k-connected and let x, y ∈ V be distinct
and not adjacent. Then kG (N (x), N (y)) ≥ k.
5
Proof. Set A = N (x) and B = N (y); to every xy-path we can associate an
AB-path as follows
{xy − paths} −→ {AB − paths}
xv1 . . . vn y 7→
v1 . . . vn
which is well defined, as xy 6∈ E. We denote by P be the image of the above
map; observe that P is the set of AB-paths not containing x or y, and it is
not empty, as G is connected.
Let now S ⊂ V be a set separating A and B. Then S intersects every path
in P, hence S r {x, y} intersects every path in P, hence S r {x, y} intersects
every xy-path, hence S r {x, y} separates x and y. As G is k-connected,
|S r {x, y}| ≥ k and hence, a fortiori, |S| ≥ k. Therefore kG (A, B) ≥ k as
wanted.
Now, since every S ⊂ V separating A and B must contain a vertex from
every AB-path is clear that
πG (A, B) ≤ κG (A, B).
It is quite remarkable that, just as in the previous example, equality holds
in general, as the following theorem states.
Theorem 2.5. Fix a graph G = (V, E) and two subsets A, B of V . Then
the minimum number of vertices separating A and B equals the maximum
number of disjoint AB-paths. In symbols
πG (A, B) = κG (A, B).
Proof. By what we observed above, it suffices to show that G has at least
κG (A, B) disjoint AB-paths. We shall do that by proving the following
claim. Let P be a set of disjoint AB-paths with |P| < κG (A, B); then there
e of disjoint AB-paths such that |P|
e = |P| + 1 and such that
exists a set P
e contains the set of endpoints of
the set of endpoints in B of the paths in P
the paths in P.
We shall prove the claim by descending induction on |B|; observe that
the claim is stronger than the theorem, but the additional requirement on
the endpoints is needed for the proof to work.
The basis of the induction is the case B = V , which has been treated in
Example 2.3 above. We set k = κG (A, B).
Let P be a set of AB-paths as above; to simplify the terminology, by an
“endpoint of P” we shall mean an endpoint of a path in P. Let S be the set
of endpoints of P lying in B; since |S| < k, by hypothesis S cannot separate
A and B. Hence there exists an AB path, R, not contained in P. If R is
disjoint from the paths in P we are done. If not, let v be the last vertex of
R lying in a path, P , in P. Now consider the following set B 0 ⊂ V
B 0 = B ∪ V (vR) ∪ V (vP )
and the following set of disjoint AB 0 -paths
P 0 = P r {P } ∪ {P v}
(P 0 is gotten from P by replacing P with its truncation at v).
By construction B ( B 0 , hence kG (A, B 0 ) ≥ kG (A, B), and hence |P 0 | <
kG (A, B 0 ); therefore we can apply the induction hypothesis to B 0 and P 0 .
6
f0 , of |P| + 1 disjoint AB 0 -paths whose endpoints in B
This gives a set, P
f0 contains a path,
contain the endpoints of the paths in P 0 . In particular, P
0
0
P , from A to v and a path, Q , from A to a point y ∈ B 0 other than the
f0 r {P 0 , Q0 } is made of disjoint AB-paths; we will
endpoints of P 0 . Hence P
conclude the proof by showing how to replace P 0 and Q0 with AB-paths so
e satisfying the claim.
as to get a set P
There are two possibilities, y 6∈ vP or y ∈ vP .
If y 6∈ vP we replace P 0 with P ∗ = P 0 + vP and if y 6∈ B, we replace Q0
with Q∗ = Q0 + yR; if y ∈ B we leave Q∗ = Q0 .
If y ∈ vP we replace P 0 with P ∗ = P 0 + yR and we replace Q0 with
∗
Q = Q0 + yP .
f0 r {P 0 , Q0 } ∪ {P ∗ , Q∗ } is a set of |P| + 1 disjoint
e := P
Finally, the set P
AB-paths, as we wanted.
Corollary 2.6. Let G = (V, E) be k-connected; fix A ⊂ V with |A| ≥ k
and v ∈ V r A. Then G contains k independent xA-paths with different
endpoints in A.
Proof. Set B = N (x); we claim kG (A, B) ≥ k. Indeed, let U ⊂ V be such
that |U | ≤ k − 1; pick a ∈ A r U and b ∈ B r U ; as G is k-connected
G − U is connected, hence it contains an ab-path. This shows that U does
not separate A and B. By Theorem 2.5, G contains k disjoint BA-paths,
P1 , . . . , Pk . Now x 6∈ Pi , since x 6∈ A; therefore xP1 , . . . , xPk are independent
xA-paths with different endpoints in A.
The previous Theorem 2.5 is a key ingredient in the proof of the following.
Theorem 2.7 (Menger). A graph G is k-connected if and only if for every
pair of vertices x, y it contains k independent xy-paths.
Proof. We proved one implication in Proposition 2.1.
We shall now show that for any x, y ∈ V distinct, G contains k independent xy-paths.
We first assume x and y not adjacent. By Lemma 2.4 and Theorem 2.5
there exist k disjoint N (x)N (y)-paths in G, written P1 , . . . , Pk . Set Pi =
v0 v1 . . . vn ; we claim that Pi does not contain x or y. Indeed if, say, x = vh
then we have h 6= 0 (for x 6∈ N (x)) and h 6= n (for x 6∈ N (y)); hence vh−1
and vh+1 are in P ∩ N (x), which is not possible as P meets N (x) only in
v0 . The claim is proved.
As Pi does not contain x or y the following Pi := xv0 v1 . . . vn y is an xypath for i = 1, . . . , k. It is clear that P1 , . . . , Pk are independent xy-paths,
so we are done.
Now suppose x and y adjacent. By contradiction, assume G has at most
k − 1 independent xy-paths; hence G − e has at most k − 2 independent xypaths. Recall now that G − e is (k − 1)-connected, since G is k-connected.
But then, by the previous part of the proof, G − e has k − 1 independent
xy-paths, as x and y are not adjacent in G − e. This is a contradiction. Exercise 2.8. Set k ≥ 3; show that in a k-connected graphs any k vertices
lie in a common cycle. (The case k = 2 is already known.)
7
Exercise 2.9. Let G be a k-connected graph with at least 2k-vertices. Then
G contains a cycle of length at least 2k.
3. Edge-connectivity
Definition 3.1. Let k ≥ 0. The graph G = (V, E) is k-edge-connected if
|V | > 1 and for any F ⊂ E with |F | ≤ k − 1 the graph G − F is connected.
Remark 3.2.
(1) If G is k-connected, G is k-edge-connnected.
(2) If G is k-edge-connnected, then δ(G) ≥ k.
(3) G is 1-edge-connected if and only if it is 1-connected if and only if
it is connected and not K 1 .
(4) G is 2-connected if and only if it is connected and has no bridges.
Recall that for a graph G its line graph, introduced in Section 1, is denoted
by LG .
Proposition 3.3. A graph G is k-edge-connected if and only if LG is kconnected and δ(G) ≥ k.
Proof. The first statement follows directly from the definition of line graph.
For the second, we can assume k ≥ 2 by Lemma 1.5.
Assume G is k-edge-connected, then δ(G) ≥ k and hence |V (LG )| =
|E(G)| > k (here we use k ≥ 2, exercise). Let F ⊂ V (LG ) be such that the
graph LG − F is disconnected, hence, by (2), LG−F is disconnected. Now
Lemma 1.5 yields that G − F is disconnected, hence |F | ≥ k; this proves
that LG is k-connected.
Conversely, assume LG is k-connected and δ(G) ≥ k; hence V (G) > k.
For any F ⊂ E(G) such that |F | ≤ k − 1 we have δ(G − F ) ≥ 1; therefore,
by Lemma 1.5 and (2), if G − F is disconnected so is LG − F ; now this
is not possible because LG is k-connected. This proves that G is k-edgeconnected.
The assumption on δ(G) in the second part of the Proposition is really
needed: in Example 1.4 we have a graph G with δ(G) = 1, hence not 2edge-connected, whereas LG is 2-connected (in fact 4-connected!).
The following is a consequence of the previous result.
Theorem 3.4 (Menger). A graph G is k-edge-connected if and only if every
pair of vertices is joined by k edge disjoint paths.
Proof. Of course, we can assume k ≥ 2.
One implication is easy. If G contains k edge-disjoint paths between any
two vertices, any set of edges F ⊂ E such that G − F is disconnected must
contain at least an edge for each of those paths, hence |F | ≥ k, and hence
G is k-edge-connected.
Conversely, suppose G is k-edge connected; then, by the previous proposition, its line graph, LG , is k-connected. Consider the sets of edges adjacent
to them, E(x) and E(y), and the denote the corresponding sets of vertices
in the line graph by A = E(x) ⊂ V (LG ) and B = E(y) ⊂ V (LG ). As LG
is k-connected, the minimum number of vertices separating A and B is at
least k; hence, by Theorem 2.5, there exist k disjoint AB-paths, P1 , . . . , Pk ,
in LG . Now we apply Lemma 1.3; we replace each Pi by the path Pi defined
8
in (4) of that Lemma; then Pi is again an AB-path, and since Pi and Pj are
disjoint, one easily checks that Pi and Pj are disjoint. Moreover, again by
the Lemma, for every i = 1, . . . , k there is a path Qi ⊂ G such that LQi = Pi .
By construction the paths Q1 , . . . , Qk are edge-disjoint; moreover, as Pi is
an AB-path, Qi is an xy-path. The proof is complete.
Exercise 3.5. Let G be a graph. Show that G is k-edge connected if and
only if |G| > 1 and for every x, y ∈ V , any set F ⊂ E separating x and y
satisfies |F | ≥ k.
Exercise 3.6. Let G be k-edge-connected with k ≥ 1; show that G − e is
(k − 1)-edge-connected for any e ∈ E.
Give an example of a k-edge-connected graph such that G − x is not
(k − 1)-edge-connected for some x ∈ V .