Abstract
AbstractGiven a hypergraph $${{\mathcal {H}}}$$
H
and a graph G, we say that $${{\mathcal {H}}}$$
H
is a Berge-G if there is a bijection between the hyperedges of $${{\mathcal {H}}}$$
H
and the edges of G such that each hyperedge contains its image. We denote by $$\textrm{ex}_k(n,Berge- F)$$
ex
k
(
n
,
B
e
r
g
e
-
F
)
the largest number of hyperedges in a k-uniform Berge-F-free graph. Let $$\textrm{ex}(n,H,F)$$
ex
(
n
,
H
,
F
)
denote the largest number of copies of H in n-vertex F-free graphs. It is known that $$\textrm{ex}(n,K_k,F)\le \textrm{ex}_k(n,Berge- F)\le \textrm{ex}(n,K_k,F)+\textrm{ex}(n,F)$$
ex
(
n
,
K
k
,
F
)
≤
ex
k
(
n
,
B
e
r
g
e
-
F
)
≤
ex
(
n
,
K
k
,
F
)
+
ex
(
n
,
F
)
, thus if $$\chi (F)>r$$
χ
(
F
)
>
r
, then $$\textrm{ex}_k(n,Berge- F)=(1+o(1)) \textrm{ex}(n,K_k,F)$$
ex
k
(
n
,
B
e
r
g
e
-
F
)
=
(
1
+
o
(
1
)
)
ex
(
n
,
K
k
,
F
)
. We conjecture that $$\textrm{ex}_k(n,Berge- F)=\textrm{ex}(n,K_k,F)$$
ex
k
(
n
,
B
e
r
g
e
-
F
)
=
ex
(
n
,
K
k
,
F
)
in this case. We prove this conjecture in several instances, including the cases $$k=3$$
k
=
3
and $$k=4$$
k
=
4
. We prove the general bound $$\textrm{ex}_k(n,Berge- F)= \textrm{ex}(n,K_k,F)+O(1)$$
ex
k
(
n
,
B
e
r
g
e
-
F
)
=
ex
(
n
,
K
k
,
F
)
+
O
(
1
)
.
Funder
Nemzeti Kutatási Fejlesztési és Innovációs Hivatal
Publisher
Springer Science and Business Media LLC