The Turán number of Berge paths

The T urán n um b er of Berge paths Xin Cheng ∗ , Dániel Gerbner † , Hilal Hama Karim † , ‡ , Sh ujing Miao § , Junp eng Zhou ¶ ‖ Abstract A Berge path of length k in an r -uniform h yp ergraph is a collection of k h yp eredges h 1 , . . . , h k and k + 1 v ertices v 1 , . . . , v k +1 suc h that v i , v i +1 ∈ h i for eac h 1 ≤ i ≤ k . Gy őri, Katona and Lemons [ Eur op e an J. Combin. 58 (2016) 238–246 ] generalized the Erdős-Gallai theorem to Berge paths and established b ounds for the T urán n umber of Berge paths. Ho w ever, these b ounds are sharp only when some divisibility conditions hold. Gy őri, Lemons, Salia and Zamora [ J. Combin. The ory Ser. B 148 (2021) 239– 250 ] determined the exact v alue of the T urán num b er of Berge paths in the case k ≤ r . In this pap er, w e settle the nal open case k > r , thereb y completing the determination of the T urán num b er of Berge paths. Keyw ords : T urán num b er, hypergraph, Berge path AMS sub ject classications: 05C35, 05C65 1 In tro duction A hyp er gr aph H = ( V ( H ) , E ( H )) consists of a vertex set V ( H ) and a hyperedge set E ( H ) , where eac h h yp eredge in E ( H ) is a nonempty subset of V ( H ) . If | e | = r for ev ery e ∈ E ( H ) , then H is called an r -uniform hyp er gr aph ( r -graph for short). F or simplicit y , let e ( H ) := ∗ Sc ho ol of Mathematics and Statistics, North western P olytechnical Univ ersit y and Xi’an- Budap est Joint Research Center for Com binatorics, Xi’an 710129, Shaanxi, P .R. China. Email: xincheng@mail.nwpu.edu.cn . † Alfréd Rén yi Institute of Mathematics. Email: gerbner.daniel@renyi.hu . ‡ Departmen t of Computer Science and Information Theory , F aculty of Electrical Engineering and Infor- matics, Budap est Universit y of T echnology and Economics, Műegyetem rkp. 3., H-1111 Budap est, Hungary . E-mail: hilal.hamakarim@edu.bme.hu . § Sc ho ol of Mathematics and Statistics, and Hub ei Key Lab–Math. Sci., Central China Normal Univ ersity , W uhan 430079, China. Email: sjmiao2020@sina.com . ¶ Corr esp onding author. Departmen t of Mathematics, Shanghai Universit y , Shanghai 200444, P .R. China. Email: junpengzhou@shu.edu.cn . ‖ Newtouc h Center for Mathematics of Shanghai Universit y , Shanghai 200444, P .R. China. 1 | E ( H ) | . The de gr e e d H ( v ) of a vertex v is the num b er of h yp eredges con taining v in H . Let H 1 ∪ H 2 denote the disjoint union of hypergraphs H 1 and H 2 , and k H denote the disjoin t union of k h yp ergraphs H . Let F b e a family of r -graphs. An r -graph H is called F -fr e e if H do es not contain an y mem b er in F as a subhypergraph. The T ur án numb er ex r ( n, F ) of F is the maxim um n umber of h yp eredges in an F -free r -graph on n v ertices. If F = {G } , then we write ex r ( n, G ) instead of ex r ( n, {G } ) . When r = 2 , w e write ex( n, F ) instead of ex 2 ( n, F ) . The study of this function is a central theme in extremal combinatorics. T urán’s theorem [19] determines its exact v alue when G is a complete graph K k , and the Erdős-Stone-Simonovits theorem [5, 6] settles asymptotically the case when a graph G has chromatic n um b er at least 3. The Erdős-Gallai theorem states that a graph of order n containing no P k as a subgraph con tains at most ( k − 1) n 2 edges, where P k denote a path with k edges. In 2016, Győri, Katona and Lemons [13] dened the concept of Berge paths and generalized the Erdős-Gallai theorem to Berge paths. Denition 1.1 (Gy őri, Katona and Lemons [13]) . A Ber ge p ath of length k in an r -uniform hyp er gr aph is a c ol le ction of k hyp er e dges h 1 , . . . , h k and k + 1 vertic es v 1 , . . . , v k +1 such that for e ach 1 ≤ i ≤ k we have v i , v i +1 ∈ h i . W e refer to { v 1 , . . . , v k +1 } as the set of dening vertic es and to { h 1 , . . . h k } as the set of dening hyp er e dges . Note that for a xed path P k there are man y h yp ergraphs that are a Berge- P k . F or con v enience, we refer to this collection of h yp ergraphs as “Berge- P k ” . The T urán num b er of Berge paths of length k in r -graphs was studied by Győri, Katona and Lemons [13], who prov ed sharp b ounds for almost ev ery pair k , r . The missing case when k = r + 1 > 2 was settled by Da voo di, Gy őri, Meth uku and T ompkins [3]. Theorem 1.2 (Gy őri, Katona, Lemons [13], Dav o odi, Győri, Methuku, T ompkins [3]) . (i) If k ≥ r + 1 > 3 , then ex r ( n, Berge - P k ) ≤ n k  k r  . F urthermor e, this b ound is sharp whenever k divides n . (ii) If r ≥ k > 2 , then ex r ( n, Berge - P k ) ≤ n ( k − 1) r +1 . F urthermor e, this b ound is sharp whenever r + 1 divides n . Both upp er b ounds in Theorem 1.2 are sharp as shown b y the follo wing examples. In the case when k > r , supp ose that k divides n and partition the n vertices into sets of size k . In eac h k -set, tak e all p ossible subsets of size r as h yp eredges of the hypergraph. The resulting r -graph has exactly n k  k r  h yp eredges and clearly contains no copy of any Berge- P k . In the case when k ≤ r and r + 1 divides n , we partition the n vertices in to sets of size r + 1 , and from each ( r + 1) -set, select exactly k − 1 of its subsets of size r as h yp eredges of the hypergraph. The resulting r -graph has exactly k − 1 r +1 n hyperedges. Since each comp onen t con tains exactly k − 1 hyperedges, it is clear that there are no Berge- P k . Observ e how ever that these b ounds are sharp only in the case the ab o ve divisibility conditions hold. Gy őri, Lemons, Salia and Zamora [14] sho w ed that ex r ( n, Berge - P k ) =  n r +1  ( k − 1) + 1 r +1 | n +1 if 3 ≤ k ≤ r , where 1 r +1 | n +1 = 1 if r + 1 | n + 1 , and 1 r +1 | n +1 = 0 otherwise. Note that if k = 2 , then ex r ( n, Berge - P 2 ) =  n r  . 2 Here, we complete the study of the T urán num b er of Berge paths by the following theo- rem. Theorem 1.3. L et k ≥ r + 1 and n = pk + q with q < k . Then ex r ( n, Berge- P k ) = p  k r  +  q r  . This also strengthens a result of Chakrab orti and Chen [2]. Let N ( G, H ) denote the n umber of copies of G in the graph H . Theorem 1.4 (Chakrab orti and Chen [2]) . F or any p ositive inte gers n and 3 ≤ r ≤ k , if G is a P k -fr e e gr aph on n vertic es, then N ( K r , G ) ≤ N ( K r , pK k ∪ K q ) , wher e n = pk + q , 0 ≤ q ≤ k − 1 . Giv en graphs G and F , the gener alize d T ur án numb er ex( n, G, F ) is the largest num b er of copies of G in F -free n -vertex graphs. F or a v ery recen t survey on generalized T urán problems, one ma y refer to the work of Gerbner and P almer [12]. Berge [1] dened the Berge cycle, and a Ber ge cycle of length k in an r -uniform h yp er- graph is an alternating sequence of distinct vertices and h yp eredges of the form v 1 , h 1 , v 2 , h 2 , . . . , v k , h k , v 1 where v i , v i +1 ∈ h i for each i ∈ { 1 , 2 , . . . , k − 1 } and v k , v 1 ∈ h k . Similarly , we refer to { v 1 , . . . , v k } as the set of dening vertic es and to { h 1 , . . . h k } as the set of dening hyp er e dges . Gerbner and Palmer [10] generalized the established concepts of Berge cycle and Berge path to general graphs. Let F b e a graph. An r -graph H is a Ber ge- F if there is a bijection ϕ : E ( F ) → E ( H ) such that e ⊆ ϕ ( e ) for each e ∈ E ( F ) . F or a xed graph F there are many hypergraphs that are a Berge- F . Similarly , we refer to this collection of h yp ergraphs as “Berge- F ” . A simple connection betw een generalized T urán problems and Berge hypergraphs is ex( n, K r , F ) ≤ ex r ( n, Berge- F ) . Indeed, if w e tak e an n -v ertex F -free graph with ex( n, K r , F ) copies of K r and add a hyperedge on each copy of K r , then the resulting r -graph is clearly Berge- F -free. Therefore, Theorem 1.3 is indeed a strengthening of Theorem 1.4. Note that Chakrab orti and Chen [2] also determined the cases where equalit y is achiev ed. The rest of this pap er is organized as follo ws. Similarly to Theorem 1.2, the case k = r + 1 needs a dieren t pro of, which we give in Section 2. In Section 3, we prov e Theorem 1.3 in the case k > r + 1 . 2 Pro of of Theorem 1.3 for k = r + 1 Lemma 2.1. In a Ber ge- P r +1 -fr e e, Ber ge- C r +1 -fr e e r -gr aph H , the endp oints of a Ber ge- P r ar e c ontaine d only in hyp er e dges that ar e dening hyp er e dges of the Ber ge- P r . Pro of. Consider the Berge path P of length r in H . Observe that an y non-dening hyper- edge that con tains an endp oin t of P and a v ertex outside the set of dening vertices of P w ould create a Berge path of length r + 1 , and any non-dening h yp eredge that con tains both endp oin ts of P w ould create a Berge cycle of length r + 1 . The only possible non-dening h yp eredge h is the one that con tains all the dening v ertices of P except for an endp oin t, sa y v 1 . But then for the edge on the other end, sa y v r v r +1 , w e can c hange the dening h yp eredge 3 to h , and then we hav e a non-dening hyperedge con taining an endp oin t v r +1 . Based on the ab o v e analysis, this leads to a multiple hyperedge, and th us a contradiction. ■ Lemma 2.2. If a Ber ge- P r +1 -fr e e, Ber ge- C r +1 -fr e e r -gr aph H c ontains a Ber ge- C r , then it c ontains r + 1 vertic es that ar e c ontaine d in only r hyp er e dges. Pro of. If there is a Berge cycle C of length r , then all but at most one of the dening h yp eredges of C con tain a v ertex outside the set of dening vertices of C . F or the dening h yp eredge of the edge v i v i +1 , let x i b e this outside v ertex. Then w e can go from v i to x i instead of v i +1 , th us w e found a Berge path with the same dening hyperedges and endpoints v i , x i . F or all such dening vertices v i of the cycle and all such vertices x i , we nd such a Berge path of length r that has x i as an endp oin t. Thus, these at least r + 1 v ertices are con tained only in the r dening hyperedges of C by Lemma 2.1. ■ F or a set S of vertices, let N H ( S ) denote the set of hyperedges that contain at least one v ertex of S . Lemma 2.3. L et H b e an r -gr aph with no Ber ge cycles of length at le ast r such that the longest Ber ge p ath has length r . Then at le ast one of the fol lowing holds. (i) Ther e is a set S of size r − 1 with | N H ( S ) |≤ 1 . (ii) Ther e is a set S of size r + 1 with | N H ( S ) |≤ r + 1 . This is a v ariant of Lemma 1 in [14]. There the assumption is that there is no Berge cycle of length at least k (where k ≤ r , for them) and it also works for m ulti-h yp ergraphs. Here we consider the case k = r + 1 and simple h yp ergraphs, and also assume the condition on Berge paths. There are three p ossible consequences in [14]. With our restrictions, (i) in [14] is identical to the corresp onding statemen t here, while (ii) in [14] states that there is a set S of size r with | N H ( S ) |≤ r − 1 . There is a third p ossibility in [14] that w e do not describ e here. W e remark that simply applying Lemma 1 from [14] is almost enough for us. The statemen ts (i) and (iii) there could b e applied in our pro of. Ho w ev er, nding a set S of size r with | N H ( S ) |≤ r − 1 is not suitable for our purp oses, but we can use a set S of size r + 1 with | N H ( S ) |≤ r + 1 . W e follow the line of though t of the pro of in [14] b elo w. Pro of. Let us c ho ose a Berge path v 1 , e 1 , v 2 , . . . , e r , v r +1 of length r . Let F = { e 1 , . . . , e r − 1 } , U = { v 2 , . . . , v r } and U ′ = U ∪ { v r +1 } . Observ e that e 1 do es not con tain v r +1 , since then v r +1 , e 1 , v 2 , . . . , e r , v r +1 w ould form a Berge- C r . Analogously , v 1 is not in e r , hence b y Lemma 2.1 v 1 is con tained only by hyperedges in F . Claim 2.4. (i) N H ( e 1 \ U ) ⊆ F . (ii) If for some 2 ≤ i ≤ r we have v i ∈ e 1 , then N H ( e i − 1 \ U ) ⊆ F . (iii) If ther e ar e two vertic es v i , v j ∈ e 1 with i > j such that ( e i − 1 ∩ e j − 1 ) \ U ′  = ∅ , then N H ( v i − 1 ) ⊆ F and N H ( v j ) ⊆ F . (iv) If for some i we have v 1 ∈ e i − 1 , then N H ( v i − 1 ) ⊆ F . 4 Note that (i) , (ii) and (iii) are equiv alent to statements pro v ed in [14], but for the sak e of completeness, w e prov e them. Pro of of Claim. Any v ertex of e 1 \ U can replace v 1 to b e the endvertex of a Berge path of length r with dening hyperedges F ∪ { e r } , thus w e can apply the argument ab o v e the claim to sho w that they are contained only in h yp eredges of F . This prov es (i) . F or w ∈ e i − 1 \ U , consider w , e i − 1 , v i − 1 , . . . , v 2 , e 1 , v i , e i , . . . , e r , v r +1 . If w = v r +1 , this is a Berge- C r , a contradiction. Otherwise, this is a Berge- P r with w as an endp oin t and F ∪ { e r } as dening h yp eredges. If e r con tains w , we clearly obtain a Berge- C r , completing the pro of of (ii) . Fix u ∈ ( e i − 1 ∩ e j − 1 ) \ U ′ , and consider the Berge paths v i − 1 , e i − 2 , v i − 2 . . . , e j +1 , v j , e 1 , v 2 , e 2 , . . . , e j − 1 , u, e i − 1 , v i , e i , . . . , e r , v r +1 and v j , e j , v j +1 , . . . , v i − 1 , e i − 1 , u, e j − 1 , v j − 1 , . . . , v 2 , e 1 , v i , e i , . . . , e r , v r +1 . Both v i − 1 and v j are endp oints of Berge- P r with dening h yp eredges in F ∪ { e r } , th us Lemma 2.1 completes the pro of of (iii) . Finally , to pro ve (iv) , the Berge path v i − 1 , e i − 2 , . . . , v 2 , e 1 , v 1 , e i − 1 , v i , e i , . . . , e r , v r +1 with Lemma 2.1 completes the pro of. ■ Let us return to the pro of of the lemma. Claim 2.5. If e 1 ∩ U = { v 2 } , then either (i) or (ii) of the L emma holds. Pro of of Claim. By Claim 2.4 (i) , N H ( e 1 \ U ) ⊆ F . If there is no w ∈ e 1 \ U that is con tained in some e i with i > 1 , then (i) of the Lemma holds. Otherwise, the Berge- path v i , e i − 1 , . . . , v 2 , e 1 , w , e i , v i +1 , e i +1 , . . . , e r , v r +1 sho ws that N H ( v i ) ⊆ F ∪ { e r } . Finally , N H ( v r +1 ) ⊆ F ∪ { e r } by Lemma 2.1, th us N H (( e 1 \ U ) ∪ { v i , v r +1 } ) ⊆ F ∪ { e r } , hence (ii) of the Lemma holds. ■ Let us return to the pro of of the Lemma. Let e 1 ∩ U = { v i 0 , v i 1 , . . . , v i s } , then v i 0 = v 2 . W e dene recursively the sets Q j the follo wing wa y . Let Q 1 = e 1 \ U and for 1 ≤ j ≤ s , if ( e i j − 1 \ U ′ ) ∩ Q j = ∅ , and e i j − 1  = U ′ then let Q j +1 = Q j ∪ ( e i j − 1 \ U ′ ) . Otherwise let Q j +1 = Q j ∪ ( e i j − 1 \ U ′ ) ∪ { v i j − 1 } . W e claim that | Q j +1 | > | Q j | . Indeed, when we add v i j − 1 , it is a new vertex, since w e only added vertices outside U and v ertices v i ℓ − 1 for some ℓ < j . Otherwise, we add a set e i j − 1 \ U ′ that con tains an element not in Q j , unless e i j − 1 \ U ′ = ∅ . Therefore, | Q s +1 |≥ | Q 1 | + s ≥ r − 1 . Observe that by Claim 2.4, N H ( Q s +1 ) ⊆ F . W e ha ve N H ( Q s +1 ∪ { v r +1 } ) ⊆ F ∪ { e r } , thus we are done unless | Q s +1 | = r − 1 . No w, we rep eat the ab o v e pro cedure for the same path, but starting from v r +1 , and going to the other direction. This wa y w e obtain R t +1 with | R t +1 |≥ r − 1 and N H ( R t +1 ∪ { v 1 } ) ⊆ F ∪ { e r } . W e are done if | Q s +1 ∪ R t +1 |≥ r + 1 . Otherwise, since v r +1 ∈ Q s +1 and v 1 ∈ R t +1 , we ha ve that Q s +1 \ { v 1 } = R t +1 \ { v r +1 } . In particular, since e 1 \ U ⊂ Q s +1 but it cannot b e in R t +1 (otherwise a Berge- C r w ould b e found), we hav e that e 1 \ U = { v 1 } , and thus e 1 = { v 1 , v 2 , . . . , v r } and e r = { v 2 , v 3 , . . . , v r +1 } . If v 1 is con tained in e i for some 1 < i < r , then the Berge path v i , e i − 1 , . . . , v 2 , e 1 , v 1 , e i , v i +1 , . . . , e r , v r +1 also has that the rst edge consists of the rst r v ertices, th us e i = e 1 , a contra- diction. 5 Therefore, eac h endp oin t of a Berge- P r has degree 1. Then eac h e i con tains a non-dening v ertex w i , thus w i , e i , v i , e i − 1 , . . . , v 2 , e 1 , v i +1 , e i +1 , . . . , v r , e r , v r +1 sho ws that w i is contained only in e i . Thus, the r v ertices v 1 , w 2 , . . . , w r − 1 , v r +1 eac h ha v e degree 1 . Consider now some v i ( i = 2 , . . . , r ). If there is no non-dening h yp eredge that contains v i , then (ii) holds for { v 1 , w 2 , . . . , w r − 1 , v r +1 , v i } . Assume that there exists a non-dening h yp eredge h con taining v i . If h contains only dening v ertices, then h ⊆ { v 1 , . . . , v r +1 } and h  = e 1 , e r . This implies that v 1 , v r +1 ∈ h , and th us h forms a Berge- C r +1 with v 1 , e 1 , v 2 , . . . , e r , v r +1 , a contradiction. Therefore, h con- tains a non-dening vertex v  = w i as d ( w i ) = 1 . The Berge path v , h, v i , e i − 1 , . . . , v 2 , e 1 , v i +1 , e i +1 , . . . , v r , e r , v r +1 implies that v is con tained only in h . Th us, (ii) holds for { v 1 , w 2 , . . . , w r − 1 , v r +1 , v } , and we are done. ■ Let H b e an r -graph and U ⊆ V ( H ) b e a nonempty subset. Let H − U denote the r -graph obtained from H by remo ving the vertices in U and the h yp eredges inciden t to them. No w we are ready to prov e our main theorem for k = r + 1 . Recall that it states that if n = p ( r + 1) + q with q < r + 1 , then ex r ( n, Berge- P r +1 ) = p ( r + 1) +  q r  . Pro of of the case k = r + 1 of Theorem 1.3. Let H b e a counterexample with the small- est n umber of v ertices. Since H is Berge- P r +1 -free, it con tains no Berge cycles of length at least r + 2 . Suppose H con tains a Berge- C r +1 , say v 1 , e 1 , . . . , v r +1 , e r +1 . Assume that a dening h yp eredge, sa y e 1 con tains a non-dening v ertex u . Then u, e 1 , . . . , v r +1 , e r +1 , v 1 is a Berge- P r +1 , a contradiction. Therefore, these h yp eredges form a clique, and an y further h yp eredge in tersecting this clique would also create a Berge- P r +1 . Deleting this clique w e obtain another coun terexample on n − r − 1 vertices, a contradiction. No w w e can apply Lemma 2.2. If there are r + 1 v ertices inciden t to at most r hyperedges, then deleting those vertices we get a smaller coun terexample. Otherwise, b y Lemma 2.2 there is no Berge- C r in H . Note that e ( H ) > p ( r + 1) +  q r  > p ( r − 1) + 1 =  n r + 1  ( r − 1) + 1 ≥ ex r ( n, Berge - P r ) . This implies that the longest Berge path in H has length r , and thus w e can apply Lemma 2.3. If there is a set S of size r − 1 with | N H ( S ) |≤ 1 and 0  = q  = r − 1 , w e delete S and nd a smaller counterexample. If q = 0 , the statemen t already follows from the result b y Da voo di, Gy őri, Methuku and T ompkins (Theorem 3 in [3]). No w we consider the case q = r − 1 . If | N H ( S ) | = 0 , then we delete S and nd a smaller coun terexample. If | N H ( S ) | = 1 , then we consider the subhypergraph H − S . Note that | V ( H − S ) | = p ( r + 1) and e ( H − S ) ≥ p ( r + 1) . Recall that H con tains no Berge cycles of length at least r . Then H − S con tains no Berge cycles of length at least r . Th us, b y 6 Theorem 10 in [14], we hav e e ( H − S ) ≤ max  | V ( H − S ) |− 1 r  ( r − 1) , | V ( H − S ) |− r + 1  = max  p ( r + 1) − 1 r  ( r − 1) , p ( r + 1) − r + 1  < p ( r + 1) , whic h contradicts the fact that e ( H − S ) ≥ p ( r + 1) . Finally , if there is a set S of size r + 1 with | N H ( S ) |≤ r + 1 , we delete S and nd a smaller coun terexample. This completes the pro of. ■ 3 Pro of of Theorem 1.3 for k > r + 1 W e hav e mentioned the b ound ex( n, K r , F ) ≤ ex r ( n, Berge- F ) . There is also a b ound ex r ( n, Berge- F ) ≤ ex( n, K r , F ) + ex( n, F ) [11]. This was strengthened b y Füredi, Kostochka and Luo [7] and by Gerbner, Methuku and Palmer [8] indep enden tly in the follo wing wa y . A r e d-blue gr aph is a graph with ev ery edge colored red or blue. W e say that a red-blue graph is F -fr e e if it is F -free without considering the colors. Giv en a red-blue graph G , w e denote b y G r the subgraph consisting of the red edges and by G b the subgraph consisting of the blue edges. Let g ( G ) := N ( K r , G b ) + e ( G r ) . Lemma 3.1 (Gerbner, Meth uku, and P almer [8]) . Given a Ber ge- F -fr e e n -vertex r -gr aph H , ther e is an F -fr e e r e d-blue n -vertex gr aph G such that H has at most g ( G ) hyp er e dges. This lemma w as used in [8] to obtain sev eral b ounds on Berge-T urán problems, in par- ticular it can b e used to pro v e Theorem 1.2. Ho wev er, it cannot prov e Theorem 1.3, since an n -v ertex P k -free graph may hav e g ( G ) larger than the claimed b ound. Indeed, if q < r , then a q -clique do es not contain any K r , but ma y contain red edges. F urthermore, if q ≤ r + 1 , then K q con tains more edges than r -cliques, thus it makes more sense to color it red. W e determine the largest v alue of g ( G ) among n -v ertex P k -free red-blue graphs. In particular, this pro ves Theorem 1.3 in the case q ≥ r + 2 . W e will use the follo wing simple inequalities. Lemma 3.2. (i) If k ≥ r + 2 , then ( k − 1) / 2 ≤  k − 1 r − 1  /r . (ii) If 3 ≤ r ≤ ℓ ≤ k − 2 , then ( k − ℓ − 1) / 2 ≤ ( k − 1 r − 1 ) − ( ℓ r − 1 ) r . Pro of. Observ e that (i) holds with equality if k = r + 2 and increasing k increases the righ t-hand side by more than the left-hand side. T o prov e (ii) , we use induction on k − ℓ − 1 . In the case k − ℓ − 1 = 1 , w e hav e that 1 2 ≤ r ( r − 1) 2 r ≤ ( k − 2)( r − 1) 2 r ≤ ( k − 2)( k − 3) · ( k − 4) · · · ( k − r + 1) · 2 2 r · ( r − 2) · · · 3 · 2 =  k − 2 r − 2  r =  k − 1 r − 1  −  k − 2 r − 1  r . 7 If ℓ decreases by 1, then the left-hand side increases by 1 2 , and the right-hand side increases b y ( ℓ r − 1 ) − ( ℓ − 1 r − 1 ) r = ( ℓ − 1 r − 2 ) r ≥ r − 1 r ≥ 1 2 since ℓ ≥ r . ■ W e will follow the strategy of the pro of of Theorem 1.4 by Chakrab orti and Chen [2]. In particular, w e will use Karamata’s inequality [17]. Lemma 3.3 (Karamata’s inequality [17]) . L et f b e a r e al value d c onvex function dene d on N . If x 1 , x 2 , . . . , x N and y 1 , y 2 , . . . , y N ar e inte gers such that x 1 ≥ x 2 ≥ . . . ≥ x N and y 1 ≥ y 2 ≥ . . . ≥ y N , x 1 + x 2 + · · · + x i ≥ y 1 + y 2 + · · · + y i for al l i < N , and x 1 + x 2 + · · · + x N = y 1 + y 2 + · · · + y N , then f ( x 1 ) + f ( x 2 ) + · · · + f ( x N ) ≥ f ( y 1 ) + f ( y 2 ) + · · · + f ( y N ) . Theorem 3.4. L et n = pk + q with q < k . Then for any P k -fr e e r e d-blue n -vertex gr aph G we have g ( G ) ≤ p  k r  +   q r  if q ≥ r + 2 ,  q 2  if q ≤ r + 2 . F urthermor e, this upp er b ound c an b e achieve d. Pro of. The lo wer b ounds are giv en by the graph consisting of p mono-blue copies of K k and a cop y of K q that is mono-blue if q ≥ r + 2 and mono-red otherwise. T o pro ve the upp er b ound, w e follo w the strategy of Chakraborti and Chen [ 2]. F or a v ertex v , let d b ( v ) denote the num b er of blue edges inciden t to v and d r ( v ) denote the num b er of red edges inciden t to v . Let k ( v ) denote the num b er of blue copies of K r con taining v . The c ontribution of v is c ( v ) := k ( v ) r + d r ( v ) 2 . Clearly ,  v ∈ V ( G ) c ( v ) = g ( G ) . F or any v ertex v with d b ( v ) = a ( v )( k − 1) + b ( v ) w e ha ve that k ( v ) ≤ a ( v )  k − 1 r − 1  +  b ( v ) r − 1  using Theorem 1.4, since the neighborho o d of v is P k − 1 -free. Therefore, g ( G ) ≤  v ∈ V ( G )  a ( v )  k − 1 r − 1  r +  b ( v ) r − 1  r + d r ( v ) 2  . (1) The pro of of Theorem 1.4 in [2] go es the following w ay . If d r ( v ) = 0 , then the right- hand side of the ab o v e equation corresponds to a sequence { y i } N i =1 consisting of  v ∈ V ( G ) a ( v ) copies of k − 1 and b ( v ) for eac h v . They construct a sequence x i suc h that x i = k − 1 for i ≤ pk , x i = q − 1 for all pk < i ≤ pk + q , and x i = 0 for i > pk + q . Then they sho w that the conditions of Karamata’s inequality hold for x i and y i , whic h completes their pro of (using that  x r − 1  /r is a conv ex function). Let y :=  v ∈ V ( G ) d r ( v ) . W e will now increase the righ t-hand side of (1) the following w ay . W e will obtain a quan tity  m i =1 ( z i r − 1 ) r + z / 2 for some m such that the following three conditions hold for the sequence z , z 1 , z 2 , . . . , z m . 1, k − 1 ≥ z 1 ≥ z 2 ≥ . . . ≥ z m , 2,  pk +1 i =1  z i r − 1  ≤ pk  k − 1 r − 1  +  q − 1 r − 1  and 3, z +  m i =1 z i = 2 e ( G ) . First w e sho w that these conditions hold originally , i.e., for the sequence y , y 1 , . . . , y N . Observ e that  v ∈ V ( G ) [ a ( v )( k − 1) + b ( v ) + d r ( v )] =  v ∈ V ( G ) d ( v ) = 2 e ( G ) , thus the third 8 condition holds for y i . The rst condition holds for y i b y denition, while the second condition holds b ecause  pk +1 i =1  y i r − 1  ≤  pk +1 i =1  x i r − 1  = pk  k − 1 r − 1  +  q − 1 r − 1  . Recall that the conditions of Karamata’s inequalit y hold for x i and y i ; in particular, there are at most pk elements in { y i } N i =1 equal to k − 1 . If there are few er than pk such elements, consider the next t wo elemen ts b ( v 1 ) , b ( v 2 ) . If b oth are larger than 0, we com bine them in the following sense. If b ( v 1 ) + b ( v 2 ) ≤ k − 1 , w e replace them b y b ( v 1 ) + b ( v 2 ) and 0. If b ( v 1 ) + b ( v 2 ) > k − 1 , we replace them by k − 1 and b ( v 1 ) + b ( v 2 ) − k + 1 . After reorganizing the sequence, the rst and third conditions ob viously still hold. The second condition holds since w e c hanged y i and y i +1 for some i ≤ pk , and then only the i th elemen t can increase. W e rep eat this pro cedure as long as there are fewer than pk elemen ts in our sequence that are equal to k − 1 . Then the required conditions hold at the end. This procedure can end in tw o w ays: either only at most one b ( v ) > 0 remains, or w e ha ve pk elemen ts equal to k − 1 . W e let B :=  m i =1 ( z i r − 1 ) r and R := z / 2 . Note that B is an upp er b ound on the total con tribution of the blue cliques and R is an upp er bound on the total contribution of the red edges, but they do not corresp ond to blue cliques or red edges. CASE 1. W e arrived at a sequence { z i } that has pk − t elements equal to k − 1 and at most one elemen t equal to k − s for some in tegers t > 0 and s > 1 . Then w e ha ve that B is at most ( pk − t )  k − 1 r − 1  /r +  k − s r − 1  /r . Note that the underlying graph of G is P k -free. By the third condition, z = 2 e ( G ) −  m i =1 z i ≤ 2 · ex( n, P k ) −  m i =1 z i = ( t − 1)( k − 1) + s − 1 + q ( q − 1) , hence R is at most ( t − 1)( k − 1) / 2 + ( s − 1) / 2 + q ( q − 1) / 2 . W e hav e that ( t − 1)( k − 1) / 2 ≤ ( t − 1)  k − 1 r − 1  /r and ( s − 1) / 2 ≤ ( k − 1 r − 1 ) − ( k − s r − 1 ) r b y Lemma 3.2. Therefore, our upp er bound on the sum of con tributions is at most pk  k − 1 r − 1  /r +  q 2  , whic h completes the pro of. CASE 2. W e arrived at a sequence { z i } that has pk elemen ts equal to k − 1 and some other elemen ts less than q . W e ha ve that z ′ :=  m i = pk +1 z i = 2 e ( G ) − pk ( k − 1) − z . Let z ′ = a ( q − 1) + b with b < q − 1 , then we further increase the righ t-hand side of (1) if we replace the z i ’s with i > pk by a elements equal to q − 1 and an element equal to b . W e ha v e that z ≤ q ( q − 1) − z ′ . If q ≥ r + 2 and b ≥ r − 1 , then a ( q − 1 r − 1 ) + ( b r − 1 ) r + z 2 ≤ ( a +1) ( q − 1 r − 1 ) r + z − q +1+ b 2 ≤ ( a +1) ( q − 1 r − 1 ) r + ( q − a − 1)( q − 1) 2 ≤ q ( q − 1 r − 1 ) r . Here, the rst inequality holds b ecause b y decreasing z and increasing b by 1, the v alue of the expression increases by at least  b r − 2  /r − 1 / 2 . The second inequality holds b ecause z ≤ q ( q − 1) − z ′ , and the third inequality holds because the expression increases b y ( q − a − 1)( ( q − 1 r − 1 ) r − ( q − 1) / 2) , which is non-negativ e by (i) of Lemma 3.2. If q ≥ r + 2 and b < r − 1 , then observe that z ≤ ( q − a )( q − 1) , th us z / 2 ≤ ( q − a )( q − 1) / 2 ≤ ( q − a )  q − 1 r − 1  /r . Therefore, a ( q − 1 r − 1 ) + ( b r − 1 ) r + z 2 ≤ q ( q − 1 r − 1 ) r , using that  b r − 1  = 0 . This implies that our upper bound on g ( G ) is at most pk  k − 1 r − 1  /r +  q r  , whic h completes the pro of. If q ≤ r + 1 , then b is at most r − 1 , th us  b r − 1  is either 0 or 1. W e increase a ( q − 1 r − 1 ) r to a ( q − 1) / 2 and if b = r − 1 , then we increase ( b r − 1 ) r to b/ 2 . This wa y we increased a ( q − 1 r − 1 ) + ( b r − 1 ) r + z 2 to ( z + z ′ ) / 2 ≤ q ( q − 1) / 2 . Therefore, our upp er bound on the sum of contributions is at 9 most pk  k − 1 r − 1  /r +  q 2  , whic h completes the pro of. ■ W e sa y an r -graph H is hamiltonian-c onne cte d if it con tains a hamiltonian Berge path b et w een any pair of its vertices. Let δ ( H ) denote the minimum degree of v ertices in H . F or S ⊆ V ( H ) , let H [ S ] denote the subh yp ergraph of H induced by S . Theorem 3.5 (K osto c hka, Luo and McCourt [ 18]) . L et n ≥ r ≥ 3 . Supp ose H is an n - vertex r -gr aph such that (i) r ≤ n 2 and δ ( H ) ≥  ⌊ n/ 2 ⌋ r − 1  + 1 , or (ii) n − 1 ≥ r > n 2 ≥ 3 and δ ( H ) ≥ r − 1 , or (iii) r = 3 , n = 5 and δ ( H ) ≥ 3 . Then H is hamiltonian-c onne cte d. Giv en an r -graph H with the longest Berge path of length ℓ > r , we say that a set S ⊆ V ( H ) is go o d if | N H ( S ) |≤ | S |  ℓ r − 1  /r , and a 2-set S is very go o d if | N H ( S ) |≤  ℓ r − 1  /r +  ℓ − 1 r − 1  /r . The pro of of the b ound ex r ( n, Berge - P k ) ≤ n k  k r  in [13] go es b y sho wing a go od set and applying induction. W e need a go o d set of order exactly k to apply induction. W e will use the follo wing lemma. Lemma 3.6. A ssume that the longest Ber ge p ath in an r -gr aph H has length ℓ > r . Then at le ast one of the fol lowing holds. • Ther e is a go o d set of size 1 and a go o d set of size 2. • Ther e is a go o d set of size 2 and a go o d set of size 3. • The dening vertic es of a Ber ge p ath of length ℓ form a c onne cte d c omp onent that either c ontains a Ber ge cycle of length ℓ + 1 , or c ontains a Ber ge cycle of length ℓ and the r emaining vertex of the Ber ge p ath has de gr e e 1 in H . Mor e over, if we found a go o d set S of size two this way, and ther e ar e at most ℓ − 1 vertic es in addition to S in H and ℓ > r + 1 , then the numb er of hyp er e dges is at most  ℓ − 1 r  +  ℓ − 1 r − 1  /r +  ℓ r − 1  /r . Note that the moreov er part follows if w e nd a very go o d set of size 2. Note also that in [13], a go o d set of unsp ecied size was found. W e b orro w a lot from the pro of in [13], but there are tw o small mistakes in the argument there that w e also hav e to correct. W e remark that in the case k ≥ r + 3 , one can nd a go od set of size 1 with a little bit more w ork. That w ould b e enough for our purp oses, but w e also hav e to deal with the case k = r + 2 . Pro of. Consider a Berge path v 1 , e 1 , v 2 , . . . , e ℓ , v ℓ +1 of length ℓ . First, assume there is a Berge cycle with dening vertices v 1 , . . . , v ℓ +1 . Without loss of generality , the dening hyperedges are e 1 , . . . , e ℓ and a h yp eredge e ℓ +1 con taining v ℓ +1 and v 1 . If there is a hyperedge h that is not a dening h yp eredge and contains v i and a vertex v that is not a dening vertex, then v , h, v i , e i , v i +1 , . . . , v i − 1 is a Berge path of length ℓ + 1 , a con tradiction. If a dening h yp eredge e i con tains a non-dening v ertex v , then v , e i , v i +1 , . . . , v i − 1 , e i − 1 , v i is a Berge path of length ℓ + 1 , a contradiction. Therefore, the third bullet p oin t holds for this cycle. Hence, w e are done unless there is no Berge- C ℓ +1 in H on the vertices v 1 , v 2 , . . . , v ℓ +1 . 10 Let H ′ denote the set of hyperedges in H that are not the dening hyperedges of our Berge path. Note that the moreov er part follo ws if the go o d set of size 2 is v ery go o d. CASE 1. No h yp eredge of H ′ con tains v 1 or v ℓ +1 . Then N H ( { v 1 , v ℓ +1 } ) ≤ ℓ . Observ e that since ℓ > r , we hav e  ℓ r − 1  /r ≥  ℓ ℓ − 2  / ( ℓ − 1) = ℓ/ 2 . Therefore, { v 1 , v ℓ +1 } is a go od set. Moreo ver, { v 1 , v ℓ +1 } is a very go o d set since  ℓ r − 1  /r +  ℓ − 1 r − 1  /r ≥ ℓ when ℓ > r + 1 . W e hav e found a go o d set of size 2 . Next, we will nd a go o d set of size 1 or 3 . If any of v 1 and v ℓ +1 is contained in at most ⌊ ℓ 2 ⌋ dening h yp eredges, then it will b e a go o d set of size 1, and w e are done. Now assume that eac h of v 1 and v ℓ +1 is con- tained in at least ⌊ ℓ 2 ⌋ + 1 dening h yp eredges. T ak e a h yp eredge e i that contains v ℓ +1 . Then v 1 , e 1 , v 2 , . . . , v i , e i , v ℓ +1 , e ℓ , v ℓ , . . . , v i +1 is a Berge- P ℓ with the same dening hyper- edges. If v i +1 is contained only in the dening hyperedges, then { v 1 , v ℓ +1 , v i +1 } is a go od set of size three, and we are done. W e claim that an y non-dening hyperedge that con- tains v i +1 m ust only contain dening vertices, otherwise we can get a Berge- P ℓ +1 . Let A i +1 b e the set of dening vertices that are in a non-dening hyperedge with v i +1 . Note that if some v j is in A i +1 , then v 1 cannot b e in e j , otherwise we get a Berge- C ℓ +1 . Th us, | A i +1 |≤ ℓ − ( ⌊ ℓ 2 ⌋ + 1) = ⌈ ℓ 2 ⌉ − 1 . So, there are at most  ⌈ ℓ 2 ⌉− 1 r − 1  non-dening hyperedges con taining v i +1 . Therefore, { v 1 , v ℓ +1 , v i +1 } is inciden t to at most ℓ +  ⌈ ℓ 2 ⌉− 1 r − 1  ≤ ℓ +  ℓ r − 1  /r h yp eredges, hence forms a go od set of size 3. CASE 2. F or eac h Berge path of length ℓ , at least one of the endvertices is con tained in a non-dening hyperedge. Note that suc h non-dening hyperedges con tain only the dening v ertices, otherwise we get a Berge- P ℓ +1 . Let U b e the dening v ertex set of the Berge path v 1 , e 1 , . . . , v l , e ℓ , v ℓ +1 . Let A 1 denote the set of vertices in U \ { v 1 } that are con tained in a h yp eredge of H ′ together with v 1 and A ℓ +1 denote the set of v ertices in U \ { v ℓ +1 } that are con tained in a hyperedge of H ′ together with v ℓ +1 . A hyperedge in H ′ con taining v 1 and v ℓ +1 w ould create a Berge cycle of length ℓ + 1 . Meanwhile, if v i ∈ A ℓ +1 and v i +1 ∈ A 1 , then v i , e i − 1 , v i − 1 , . . . , e 1 , v 1 , h 1 , v i +1 , e i +1 , . . . , e ℓ , v ℓ +1 , h 2 , v i is a Berge cycle of length ℓ + 1 , where v 1 , v i +1 ∈ h 1 ∈ H ′ and v i , v ℓ +1 ∈ h 2 ∈ H ′ . Therefore, | A 1 | dening vertices are forbidden from A ℓ +1 . Note that v 1 and v ℓ +1 are also not in A ℓ +1 . The pro of in [13] concludes that | A 1 | + | A ℓ +1 |≤ ℓ − 1 , but this is incorrect, since v 1 ma y b e one of the | A 1 | forbidden v ertices, th us we only hav e | A 1 | + | A ℓ +1 |≤ ℓ . Clearly , the ab o v e imply that at least one of | A 1 |≥ r − 1 and | A ℓ +1 |≥ r − 1 holds, and there are at most  | A 1 | r − 1  h yp eredges in H ′ that contain v 1 and analogously at most  | A ℓ +1 | r − 1  h yp eredges in H ′ that contain v ℓ +1 . Observ e that if v i ∈ A 1 , and e i − 1 con tains v ℓ +1 , then v i , e, v 1 , e 1 , . . . , v i − 1 , e i − 1 , v ℓ +1 , e ℓ , v ℓ , . . . , v i is a Berge cycle of length ℓ + 1 , where { v 1 , v i } ⊆ e ∈ H ′ . Therefore, v ℓ +1 is contained in at most ℓ − | A 1 | dening hyperedges and analogously v 1 is con tained in at most ℓ − | A ℓ +1 | dening h yp eredges. Thus, if the smaller of A 1 and A ℓ +1 con tains i elements, then the degree of an endp oin t is at most  i r − 1  + ℓ − i . If i ≥ r − 1 , then increasing i increases this n umber, thus its maximum is  ⌊ ℓ/ 2 ⌋ r − 1  + ⌈ ℓ/ 2 ⌉ . If i < r − 1 , then i = 0 , say , A ℓ +1 = ∅ . Then | A 1 |≥ r − 1 , th us the degree of v ℓ +1 is at most ℓ − r + 1 . Moreo ver, if i ≥ r − 1 , then ℓ ≥ 2 r − 2 . W e hav e obtained that there is a v ertex, say v ℓ +1 of degree at most x , where x = ℓ − r + 1 11 if ℓ < 2 r − 2 and x = max {  ⌊ ℓ/ 2 ⌋ r − 1  + ⌈ ℓ/ 2 ⌉ , ℓ − r + 1 } if ℓ ≥ 2 r − 2 . W e need to show that x ≤ y :=  ℓ r − 1  /r . Let us remark that [13] arriv ed at a similar inequalit y and left chec king the details to the reader, but their inequalit y do es not hold for r = 3 , k = 6 . W e claim that if r ≥ 4 and ℓ increases by 1, then y − x do es not decrease. Clearly , y increases b y  ℓ r − 2  /r , while x increases b y at most  ⌊ ℓ/ 2 ⌋ r − 2  + 1 . If ℓ < 2 r − 2 , then x increases b y 1 and w e are done. Otherwise, w e hav e  ⌊ ℓ/ 2 ⌋ r − 2  ≤  ℓ/ 2 r − 2  = ℓ ( ℓ − 2) ... ( ℓ − 2 r +6) 2 r − 2 ( r − 2)! <  ℓ r − 2  /r . Therefore, it is enough to chec k that x ≤ y holds for the smallest possible ℓ , whic h is ℓ = 2 r − 2 . In that case, w e hav e to sho w that r ≤  2 r − 2 r − 1  /r . This can b e easily chec k ed. It is left to deal with the case r = 3 . In that case, w e ha ve that y = ℓ ( ℓ − 1) / 6 (in which case ℓ ≥ 4 ), while for o dd ℓ , either x ≤ ( ℓ − 1)( ℓ − 3) / 8 + ( ℓ + 1) / 2 , or x = ℓ − 2 . Then one can easily c heck that x ≤ y if ℓ ≥ 7 . If ℓ = 5 , then y = 10 / 3 , while it is p ossible that x = 4 . Ho wev er, in this case w e hav e that | A 1 | , | A 6 |≥ 2 . Recall that if v i ∈ A 1 , then v i − 1 ∈ A 6 and e i − 1 do es not con tain v 6 . Analogously , if v i ∈ A 6 , then v i +1 ∈ A 1 and e i do es not contain v 1 . First, assume that v 2 ∈ A 1 . Then we either ha v e that v 2 ∈ A 6 (in which case A 1 = { v 4 , v 5 } ) or v 2 ∈ A 6 . In the rst case, w e hav e that A 6 = { v 2 , v 5 } , v 1 is contained in e 1 , e 3 , e 4 and v 6 is con tained in e 1 , e 2 , e 5 . Then v 1 v 4 v 5 is a dening h yp eredge, th us v 1 is not con tained in an y non-dening h yp eredge, hence v 1 is con tained in 3 h yp eredges, a con tradiction. In the second case, w e ha ve that A 1 ∪ A 6 ⊆ { v 3 , v 4 , v 5 } , which is p ossible only if A 1 = A 6 = { v 3 , v 5 } (and v 1 is contained in e 1 , e 2 , e 4 , v 6 is contained in e 1 , e 3 , e 5 ) or A 1 = { v 3 , v 4 } and A 6 = { v 4 , v 5 } (and v 1 is con tained in e 1 , e 2 , e 3 , v 6 is con tained in e 1 , e 4 , e 5 ). In the last case, v 1 v 3 v 4 is a dening h yp eredge, thus v 1 is not contained in any non-dening hyperedge, a contradiction as abov e. If A 1 = A 6 = { v 3 , v 5 } , then the hyperedges are v 1 v 2 v 6 ( e 1 ) , v 1 v 2 v 3 ( e 2 ) , v 3 v 4 v 6 ( e 3 ) , v 1 v 4 v 5 ( e 4 ) , v 5 v 6 x ( e 5 ) , v 1 v 3 v 5 , v 3 v 5 v 6 , for some x . Then x m ust b e a dening v ertex, otherwise w e use v 3 v 5 v 6 for the edge v 5 v 6 in the original Berge path, and then con tinue with v 6 xv 5 to x , obtaining a Berge path of length 6, a contradiction. Then x ∈ { v 1 , v 2 , v 4 } . If x ∈ { v 1 , v 2 } , then again, we use v 3 v 5 v 6 for the edge v 5 v 6 in the original Berge path, and then the non- dening hyperedge containing v 6 either contains v 1 , forming a Berge cycle of length 6, a con tradiction, or con tains v 2 , arriving to the previous case. If x = v 4 , then w e can nd a Berge- C 6 ( v 5 v 3 v 5 v 6 − − − → v 6 e 5 − → v 4 e 3 − → v 3 e 2 − → v 2 e 1 − → v 1 v 1 v 3 v 5 − − − → v 5 ), a contradiction. W e hav e obtained that v 2 ∈ A 1 , and then analogously v 5 ∈ A 6 . W e can replace e 1 b y the non-dening hyperedge h containing v 1 and v 2 . Then the third v ertex of e 1 m ust b e a dening vertex v i . Let A ′ 1 = A 1 ∪ { v i } . If v i ∈ A 1 , then | A ′ 1 |≥ 3 . If v i ∈ A 1 , then there is a non-dening hyperedge { v 1 , v i , x } , where x  = v 2 since { v 1 , v i , x }  = e 1 . Therefore, | A ′ 1 |≥ 3 in this case as w ell. Analogously , let A ′ 6 = A 6 ∪ { v j } , where v j is the v ertex of e 5 dieren t from v 5 and v 6 . Then we hav e that | A ′ 6 |≥ 3 . Observe that v q ∈ A ′ 6 and v q +1 ∈ A ′ 1 is imp ossible, b y the same argument that works for A 6 and A 1 . This shows that | A ′ 1 | + | A ′ 6 |≤ 5 , a con tradiction. F or r = 3 and ev en ℓ , w e claim that there is a v ertex of degree at most ℓ ( ℓ − 2) / 8 + ℓ/ 2 − 1 . This clearly holds if x = ℓ − 2 , thus w e can assume that x = ℓ ( ℓ − 2) / 8 + ℓ/ 2 . W e are done unless v ℓ +1 has degree exactly x . In that case, w e hav e | A 1 | = | A ℓ +1 | = ℓ/ 2 and eac h 12 dening h yp eredge con tains exactly one of v 1 and v ℓ +1 . Indeed, this second condition has b een established for e 1 and e ℓ , the other dening hyperedges cannot con tain b oth v 1 and v ℓ +1 since r = 3 , and hence eac h of v 1 and v ℓ +1 is con tained in ℓ/ 2 dening h yp eredges. Moreo ver, eac h subset of A 1 of size r − 1 = 2 forms a non-dening h yp eredge when extended with v 1 . W e know that v i ∈ A 1 for some i > 2 . Then, as we hav e observed, e i − 1 cannot con tain v ℓ +1 , thus con tains v 1 . Consider now the Berge path v 2 , e 2 , v 3 , . . . , v i − 1 , e i − 1 , v 1 , h, v i , e i , . . . , v ℓ +1 , where h is a non-dening hyperedge of our Berge path that contains v 1 and v i . This Berge path has length ℓ , thus we could start with this Berge path instead of our Berge path. If w e do not nd a v ertex of degree at most ℓ ( ℓ − 2) / 8 + ℓ/ 2 − 1 there, then eac h dening h yp eredge of this Berge path contains one of the endp oin ts v 2 or v ℓ +1 . More precisely , ℓ/ 2 dening h yp eredges of the new path con tain v 2 , but we also kno w that only e 2 and h may con tain v 2 , th us w e are done if ℓ > 4 . If ℓ = 4 , we are done unless h consists of v 1 , v 2 , v i . In that case A 1 = { v 2 , v i } . If i = 3 , then h = { v 1 , v 2 , v 3 } , and hence A 1 = { v 2 , v 3 } . Since v 3 ∈ A 1 , v l +1 / ∈ e 2 , but then e 2 m ust contain v 1 . Thus, e 2 = { v 1 , v 2 , v 3 } = h , a contradiction. If i = 4 , then A ℓ +1 = A 1 = { v 2 , v 4 } . W e claim that e 3 cannot contain any of v 1 and v ℓ +1 . Otherwise, w e can nd a Berge- C ℓ +1 ( v 3 → v 2 → v 5 → v 4 → v 1 → v 3 ) if v 1 ∈ e 3 , and a Berge- C ℓ +1 ( v 4 → v 5 → v 3 → v 2 → v 1 → v 4 ) if v 5 ∈ e 3 , a con tradiction. Recall that e 3 do es not con tain v ℓ +1 but contains v 1 (since v 4 ∈ A 1 ), whic h is a con tradiction. If i = 5 , it is easy to see that there is a Berge- C ℓ +1 , a contradiction. Thus, we are done, since ℓ ( ℓ − 2) / 8 + ℓ/ 2 − 1 ≤ y for even ℓ . W e ha ve found a go o d set of size 1. More precisely , we show ed that at least one end v ertex of any Berge path of length ℓ forms a go o d set. Without loss of generality , supp ose that in the Berge path v 1 , e 1 , v 2 , . . . , e ℓ , v ℓ +1 , the vertex v ℓ +1 forms a go od set of size 1. Next, w e will nd a go o d set of size 2. Assume now that v ℓ +1 is in e i for some i < ℓ . Then consider the Berge path v 1 , e 1 , . . . , v i , e i , v ℓ +1 , e ℓ , v ℓ , . . . , e i +1 , v i +1 . It has length ℓ , th us at least one of v 1 and v i +1 has degree at most  ℓ r − 1  /r and forms a goo d set together with v ℓ +1 . If v ℓ +1 is con tained in a non-dening h yp er- edge h , then that con tains only dening vertices. Let v i ∈ h , then v 1 , e 1 , . . . , e i − 1 , v i , h, v ℓ +1 , e ℓ , v ℓ , . . . , v i +1 is a Berge path of length ℓ , thus at least one of its endp oin ts has degree at most  ℓ r − 1  /r and forms a go o d set together with v ℓ +1 . Assume now that v ℓ +1 is contained in only one hyperedge, i.e., e ℓ . If e ℓ con tains a non- dening v ertex, then that can replace v ℓ +1 , and b y the ab o ve reasoning, we nd either a go od set of size 2, or a v ertex of degree 1, which forms a go o d set with v ℓ +1 . Next, supp ose that e ℓ con tains no non-dening vertices. Let us delete now v ℓ +1 and the single hyperedge e ℓ con taining it and rep eat the argumen t of the pro of so far. If there is in the resulting h yp ergraph H 0 a Berge- P ℓ and a Berge cycle of length ℓ + 1 containing it, w e are done. If there is a Berge path of length ℓ , and CASE 1 holds inside H 0 , then w e nd a go od set { v , v ′ } in H 0 . Note that v and v ′ are the tw o endp oin ts of this Berge path of length ℓ . This implies that v , v ′ / ∈ e ℓ , otherwise w e would obtain a Berge- P ℓ +1 in H together with the h yp eredge e ℓ and the vertex v ℓ +1 , a con tradiction. Hence, { v , v ′ } is also a goo d set in H , and we are done. If there is a Berge path of length ℓ , and CASE 2 holds inside H 0 , then 13 w e nd at least one endvertex v that forms a go od set. Clearly , { v ℓ +1 , v } is a go od set as  ℓ r − 1  /r + 1 ≤ 2  ℓ r − 1  /r . If there is no Berge path of length ℓ in H 0 and ℓ − 1 > r , then we rep eat this argument with the Berge path v 1 , e 1 , . . . , e ℓ − 1 , v ℓ in H 0 . If we get a Berge- C ℓ , then together with v ℓ +1 , the third bullet p oin t holds. If CASE 1 holds in H 0 , then w e nd the required go o d sets S with | N H 0 ( S ) |≤ | S |  ℓ − 1 r − 1  /r ≤ | S |  ℓ r − 1  /r − 1 , thus | N H ( S ) |≤ | S |  ℓ r − 1  /r . If CASE 2 holds in H 0 , then w e nd a go o d v ertex, whic h forms a very go od set with v ℓ +1 . If we nd that v 1 , . . . , v ℓ form a connected comp onen t in H 0 , then there are tw o p ossibilities. If v 1 , . . . , v ℓ form a connected comp onen t that contains a v ertex of degree 1, then that v ertex with v ℓ +1 forms a goo d set. If v 1 , . . . , v ℓ form a connected component in H 0 whic h con tains a Berge cycle with length ℓ , then with v ℓ +1 they form a connected comp onen t in H with a Berge cycle of length ℓ and a vertex of degree 1. In each of the ab o ve cases, w e found the desired conguration. Finally , assume that ℓ − 1 = r . If there is a Berge cycle of length ℓ , then the third bullet p oin t holds and w e are done. Otherwise, w e can apply Lemma 2.1 to nd a vertex v only in dening hyperedges. Then | N H ( { v , v ℓ +1 } ) |≤ ℓ and w e are done. It is left to pro ve the moreo ver part. Recall that in CASE 1, w e ha v e pro ved the moreo ver part (we found a very go od 2-set), th us we can assume we are in CASE 2, in particular, v ℓ +1 forms a go o d set. Assume that H has exactly ℓ + 1 v ertices and delete v ℓ +1 . Then, in the remaining h yp ergraph H 0 the longest path has length ℓ − 1 . If there is a Berge cycle of length ℓ on the ℓ v ertices in H 0 , then there are t w o cases. If v ℓ +1 has degree 1 in H , then w e do not need to show the moreo ver part, b ecause the third bullet p oin t of the statement holds. If v ℓ +1 is in at least t wo h yp eredges, and H 0 is hamiltonian-connected, then we pick t wo distinct v ertices v i from one of the h yp eredges containing v ℓ +1 and v j from the other. Then the hamiltonian Berge path connecting v i and v j in H 0 extends to a Berge cycle of length ℓ + 1 , a case we hav e dealt with in the rst paragraph of the pro of. If H 0 is not hamiltonian-connected, then it contains a vertex v of degree at most  ⌊ ℓ/ 2 ⌋ r − 1  or r − 2 or 2 by Theorem 3.5. Then in each case, { v , v ℓ +1 } is a very go o d set. Assume now that there is no cycle of length ℓ in H 0 . If in H 0 eac h Berge path of length ℓ − 1 has an endp oin t con tained in some non-dening hyperedges, then w e can nd a go o d v ertex b y rep eating the ab o ve argument. The go o d vertex there is contained in at most  ℓ − 1 r − 1  /r hyperedges. Recall that d H ( v ℓ +1 ) ≤  ℓ r − 1  /r . Then e ( H ) ≤  ℓ r − 1  /r +  ℓ − 1 r − 1  /r +  ℓ − 1 r  , and thus w e are done. If there is a Berge path of length ℓ − 1 in the remaining graph suc h that none of its end v ertices is contained in a non-dening h yp eredge, then there are at most ℓ − 1 hyperedges con taining those tw o v ertices and at most  ℓ − 2 r  h yp eredges av oiding those t wo v ertices in H 0 . Therefore, there are at most  ℓ − 2 r  + ℓ − 1 +  ℓ r − 1  /r hyperedges in H . This is at most the desired b ound if and only if ℓ − 1 ≤  ℓ − 2 r − 1  +  ℓ − 1 r − 1  /r . Since ℓ > r + 1 , w e ha ve ℓ − 2 > r − 1 , th us the rst term of the right-hand side is at least ℓ − 2 . The second term of the right-hand side is at least ( ℓ − 1) /r ≥ 1 , thus w e are done. ■ No w we are ready to prov e Theorem 1.3 for k > r + 1 . Recall that it states that if n = pk + q with q < k , then ex r ( n, Berge- P k ) = p  k r  +  q r  . Pro of of Theorem 1.3. F or the lo w er b ound, we partition the n vertices in to p sets of size 14 k and a single q -set. In each set, take all p ossible subsets of size r as hyperedges of the h yp ergraph. The resulting r -graph has exactly p  k r  +  q r  h yp eredges and clearly con tains no cop y of any Berge- P k . Let us contin ue with the upp er b ound. If q = 0 , we are done by Theorem 1.2. If q ≥ r + 2 or q = 1 , we are done b y Lemma 3.1 and Theorem 3.4. F rom no w on, we assume that 2 ≤ q ≤ r + 1 . Assume that the statemen t do es not hold and consider a coun terexample on the least n umber of v ertices. It is an n -v ertex Berge- P k -free r -graph H for some n = pk + q . The statemen t is trivial for n ≤ k , th us w e can assume n > k . W e can also assume that H is connected, otherwise at least one of its connected comp onen ts is a smaller coun terexample. If H con tains a go od k -set, w e delete the go o d k -set to obtain a h yp ergraph H ∗ , then H ∗ is not a coun terexample, th us has at most ( p − 1)  k r  +  q r  h yp eredges. W e ha ve deleted at most k  k − 1 r − 1  /r =  k r  h yp eredges, thus H contains at most p  k r  +  q r  h yp eredges, a contradiction. If H con tains go od sets of size 1 and 2, or go o d sets of size 2 and 3, then we delete such a go o d set with parity equal to the parity of k to obtain H 1 . Then if H 1 has a go o d set of size 2 that satises the moreo ver part of Lemma 3.6, we delete such a goo d set to obtain H 2 , and so on. If we can delete exactly k vertices this w ay , then we found a go od k -set, a con tradiction. Therefore, there are at most k − 2 v ertices that we can delete this wa y , let S denote their set and H ′ denote the remaining hypergraph. Claim 3.7. Each c omp onent of H ′ has or der at most k and c ontains a Ber ge p ath of length mor e than r , exc ept at most one isolate d vertex. Pro of of Claim. Let us assume indirectly that there is a comp onen t C of H ′ that is P r +1 - free. If there is a Berge- C r +1 in C , then no hyperedge of H ′ con tains a dening v ertex from the Berge- C r +1 and a vertex that is not a dening vertex. Therefore, C is of order r + 1 , hence clearly contains a go o d set of size 2, since C has at most  r +1 r  h yp eredges. If there is a Berge- C r in C , then Lemma 2.2 gives us a goo d set of size 2. If there is no Berge- C r +1 nor Berge- C r but there is a Berge path of length r in C , then w e can apply Lemma 2.3 to nd a go od set of size 2, a con tradiction. If there is no Berge path of length r in C , then consider a longest Berge path P . Note that an y non-dening h yp eredge con taining the endp oin ts of P can only contain the dening v ertices in P . If P has fewer than r dening v ertices, this is imp ossible, th us the endp oin ts are contained only in dening hyperedges. If P has r dening vertices, and the hyperedge e formed by those r v ertices is in H but not a dening hyperedge of this Berge path, then w e can nd a Berge- C r , a contradiction. Therefore, the endvertices of P are con tained only in the dening h yp eredges, th us they form a go o d set. W e hav e established that each comp onen t con tains a Berge path of length at least r , except a comp onen t that consists of only one v ertex, since in that case w e only nd one endp oin t of P . Clearly , tw o isolated vertices would form a go o d set. It is left to sho w that eac h comp onen t has at most k v ertices. W e pic k a longest Berge path, then the third bullet p oin t of Lemma 3.6 holds, th us the v ertices of that path form a comp onen t. In particular, 15 the comp onen t has at most k vertices (since the n umber of dening v ertices in the longest Berge path is at most k ), and the same holds for the rest. Finally , we need to show that the go o d sets found this wa y satisfy the moreo ver part of Lemma 3.6. Observe that we need to deal only with the case where there is a Berge path of length more than r + 1 in H ′ , while here w e found go od sets either when assuming that the longest Berge path has length at most r , or applying Lemma 3.6 itself. ■ Therefore, we can assume that eac h comp onen t of H ′ , we sa y W 1 , . . . , W t , has order at most k . Moreov er, w e can apply Lemma 3.6, and the third bullet p oin t holds. This implies that for each comp onen t W i and each vertex of W i , there is a Berge path of length at least | W i |− 2 starting at that v ertex. Observe that if a comp onen t has order less than x , where x is the smallest in teger such that  x − 1 r − 1  >  k − 1 r − 1  /r , then its v ertices hav e degree at most  k − 1 r − 1  /r . In particular, x ≥ ⌈ k +1 2 ⌉ + 1 , since x = k 2 + 1 do es not satisfy r ( x − 1) · · · ( x − r + 1) > ( k − 1) · · · ( k − r + 1) . Moreov er, if there is a comp onen t of size at least 2 and at most r + 1 , then since any set S of size at least tw o in a comp onen t of size r + 1 has | N H ( S ) |≤ r + 1 ≤ | S |  k − 1 r − 1  /r , we nd a go od set of size 2, a contradiction. So w e hav e that eac h component of H ′ , except for the at most one isolated vertex has order at least x and at least r + 2 . Let S ′ denote the union of the comp onen t of order 1 (if exists) with S , then any subset of S ′ of size k that con tains S is a go o d set, th us | S ′ | < k . Let us delete those comp onen ts to o, i.e., we delete S ′ from H to obtain H ′′ with comp onen ts W 1 , . . . , W t ′ ( t ′ ≤ t ). Claim 3.8. The c omp onents of H ′′ ar e in dier ent c onne cte d c omp onents in H . Pro of of Claim. Assume otherwise and consider a shortest Berge path in H connecting u ∈ W i and v ∈ W j for some i  = j . If an in ternal vertex w of this Berge path is in H ′′ , then there is a shorter Berge path b et ween u and w and b et ween w and v . A t least one of these Berge paths is b et w een v ertices of distinct comp onen ts of H ′′ , a contradiction. If a h yp eredge of this Berge path is in H ′′ , then one of its dening vertices w is neither u , nor v but is in H ′′ , th us this is also imp ossible. Recall that there is inside W i in H ′′ a Berge path of length at least ⌈ ( k + 1) / 2 ⌉ − 1 ending in u . Similarly , there is a Berge path of length at least ⌈ ( k + 1) / 2 ⌉ − 1 in H ′′ starting in v inside W j . No h yp eredge or v ertex is used in b oth of these Berge paths, since W i and W j are distinct comp onen ts of H ′′ . Moreov er, no hyperedge or v ertex distinct from u and v is used in these Berge paths or the path b et ween u and v , since these tw o Berge paths use v ertices and h yp eredges in H ′′ , while the other path uses vertices and hyperedges not in H ′′ . Com bining these with the path b et w een u and v , w e obtain a Berge path of length at least k , a contradiction. ■ Recall that if H is disconnected, then one of its components also must b e a counterex- ample, a con tradiction to the minimality of H . Therefore, there is only one comp onen t, H consists of W 1 with r + 2 ≤ | W 1 |≤ k and S ′ with | S ′ |≤ k − 1 . In particular, n ≤ 2 k − 1 . Moreo ver, if | W 1 | = k , then by the third bullet p oint of Lemma 3.6, either W 1 forms a connected comp onen t of H (th us n = k and we are done), or W 1 16 con tains a v ertex of degree 1 (in whic h case there was no isolated vertex in H ′ b ecause these t wo vertices w ould form a go od set). W e mo ve that vertex to S ′ . This wa y w e obtain that r + 1 ≤ | W 1 | < k and | S ′ |≤ k − 1 , thus n ≤ 2 k − 2 . Let us order the vertices by placing the v ertices of W 1 rst, then the vertices of S ′ \ S and then the v ertices of S , in reversed order of their remov al in our earlier pro cedure. Let u i denote the i th vertex in this order. W e add the vertices one b y one in this order, except that w e add at the same time the vertices of S that w ere remo ved together. W e claim that when w e add u k − 1 , then w e add at most  k − 2 r − 1  /r hyperedges, and when w e add u i alone with i ≥ k , then we add at most  k − 1 r − 1  /r hyperedges. When we add u i and u i +1 together and i ≥ k , then we add at most 2  k − 1 r − 1  /r h yp eredges, and when we add u i , u i +1 and u i +2 together and i ≥ k , then w e add at most 3  k − 1 r − 1  /r h yp eredges. If i ≥ k , then this holds since eac h vertex in S ′ has degree at most  k − 1 r − 1  /r in H . If w e deleted u i together with u i +1 or together with u i +1 and u i +2 and i ≥ k , then this holds since w e deleted a go o d set. If u i ∈ S ′ \ S and i < k , then we add no hyperedges by the denition of S ′ . Also, later, w e know that w e only deleted go o d 2-sets together except p ossibly the rst time which may b e just one or three vertices at a time. How ev er, we reac h to this only if n = k + 1 . If u i ∈ S and i < k , then the Berge path containing u i has length at most i − 1 , th us by Lemma 3.6, u i has degree at most  i − 1 r − 1  /r . If we add u k − 1 together with tw o other v ertices, then we deleted a goo d set of size 3. This can happ en only at the very rst step, th us w e had at most k + 1 vertices altogether. This implies that n = k + 1 (i.e., q = 1 ), a case w e ha ve already dealt with. If w e add u k − 1 together with one other v ertex, thus there are at most  k − 2 r  +  k − 2 r − 1  /r h yp eredges inside the set of the rst k − 1 vertices. Therefore, the num b er of hyperedges in H is at most  k − 2 r  +  k − 2 r − 1  /r + ( q + 1)  k − 1 r − 1  /r ≤  k − 1 r  + ( q + 1)  k − 1 r − 1  /r =  k r  + ( q + 1 − r )  k − 1 r − 1  /r . If q ≤ r − 1 , then we are done. Assume now that r ≤ q ≤ r + 1 . CASE 1. | W 1 |≤ k − 2 and r ≥ 4 or r = q = 3 . Then the num b er of h yp eredges is at most  | W 1 | r  +  k − 2 i = | W 1 |  i r − 1  /r + ( q + 1)  k − 1 r − 1  /r ≤  k − 2 r  +  k − 2 r − 1  /r + ( q + 1)  k − 1 r − 1  /r ≤  k − 2 r  +  k − 2 r − 1  /r +  k − 1 r − 1  + ( q − r + 1)  k − 1 r − 1  /r . Recall that r + 1 ≤ | W 1 |≤ k − 2 , thus k ≥ r + 3 . CASE 1.1. r = q . W e hav e  k − 1 r − 1  = ( k − 1)  k − 2 r − 1  / ( k − r ) , hence  k − 2 r − 1  /r + ( q − r + 1)  k − 1 r − 1  /r = (1 + ( k − 1) / ( k − r ))  k − 2 r − 1  /r = (2 k − r − 1)  k − 2 r − 1  / ( r ( k − r )) ≤  k − 2 r − 1  . Therefore, the n umber of hyperedges in H is at most  k r  , completing the pro of. CASE 1.2. q = r + 1 and r ≥ 4 . W e hav e  k − 1 r − 1  = ( k − 1)  k − 2 r − 1  / ( k − r ) , hence  k − 2 r − 1  /r + ( q − r + 1)  k − 1 r − 1  /r = (1 + (2 k − 2) / ( k − r ))  k − 2 r − 1  /r = (3 k − r − 2)  k − 2 r − 1  / ( r ( k − r )) ≤  k − 2 r − 1  + r + 1 . Indeed, the inequality holds without the r + 1 term if r ≥ 5 or if r = 4 and k ≥ 10 . It is 17 straigh tforward to chec k the inequalit y if r = 4 and 8 ≤ k ≤ 9 . If r = 4 and 6 ≤ k ≤ 7 , we use that the num b er of h yp eredges is at most  k − 2 r  +  k − 2 r − 1  /r + ( q + 1)  k − 1 r − 1  /r . It is easy to see that in b oth cases this is at most  k r  +  q r  . Therefore, the num b er of hyperedges in H is at most  k r  +  q r  , completing the pro of. CASE 2. | W 1 | = k − 1 and r ≥ 4 or r = q = 3 . Note that either the vertices of W 1 are the dening vertices of a Berge cycle of length k − 1 with vertices v 1 , . . . , v k − 1 in this order, suc h that the dening hyperedges of this cycle are in H ′ , or W 1 con tains a single v ertex v of degree 1 in H ′ . In the latter case, w e mov e v to S ′ and pro ceed with CASE 1. In the follo wing part of the pro of, the only prop ert y of W 1 that we use is that it contains a Berge cycle of length k − 1 . Since H is connected, for every u ∈ S ′ there is a shortest Berge path from u to some v ∈ W 1 . Since a Berge path of length k − 2 inside W 1 starts from v , the Berge path from u to v m ust hav e length 1 (using that this Berge path cannot ha ve internal dening vertices from W 1 , th us cannot hav e h yp eredges from H ′ ). If a hyperedge h con tains t wo vertices v , v ′ from S ′ , then no other hyperedge contains v , nor v ′ . In this case, we can delete h, v , v ′ from H to obtain a smaller counterexample to the statement of the theorem, a con tradiction. W e obtained that each vertex v ∈ S ′ is contained in a Berge- P k − 1 in H , where the other dening v ertices are the v ertices of W 1 . If v i is in a hyperedge together with v , then v i +1 cannot b e in a h yp eredge together with v ′ ∈ S ′ if v ′  = v . Let B denote the set of vertices v i suc h that v i and v i − 1 or v i +1 are b oth in some h yp eredge with some v ∈ S ′ . Then they b elong to at most one suc h v . Let b v denote the num b er of v ertices in B that form at least one h yp eredge with v , then  v ∈ S ′ b v = | B | . F or the other vertices of W 1 , t wo consecutive ones cannot b e in h yp eredges with v ertices of S ′ . Therefore, each vertex v ∈ S ′ forms hyperedges with at most ⌊ k − 1 −| B | 2 ⌋ + b v v ertices. The num b er of h yp eredges con taining some v ertex from S ′ is at most  v ∈ S ′  ⌊ k − 1 −| B | 2 ⌋ + b v r − 1  . Clearly , this is maximized if for some v we hav e b v = | B | , th us this is at most q  ⌊ k − 1 −| B | 2 ⌋ r − 1  +  ⌊ k − 1+ | B | 2 ⌋ r − 1  . By the con v exity of the binomial function, this is maximized either at | B | = 0 or at | B | = k − 1 . In the second case, w e ha ve that the total n umber of hyperedges is at most  k − 1 r  +  k − 1 r − 1  =  k r  and w e are done. In the rst case, the num b er of h yp eredges is at most  k − 1 r  + ( q + 1)  ⌊ ( k − 1) / 2 ⌋ r − 1  . Note that compared to Lemma 3.6, the improv emen t is that the dening h yp eredges of the Berge path do not contain v , since they are in H ′ . When q = r , w e ha ve that the n umber of h yp eredges is at most  k − 1 r  + ( q + 1)  ⌊ ( k − 1) / 2 ⌋ r − 1  =  k − 1 r  + ( r + 1)  ⌊ ( k − 1) / 2 ⌋ r − 1  . W e ha v e that  ⌊ ( k − 1) / 2 ⌋ r − 1  ≤  k − 1 r − 1  / 2 r − 1 ≤  k − 1 r − 1  / ( r + 1) for r ≥ 3 , th us w e are done in this case. When q = r + 1 , w e ha ve that the n umber of h yp eredges is at most  k − 1 r  + ( q + 1)  ⌊ ( k − 1) / 2 ⌋ r − 1  =  k − 1 r  + ( r + 2)  ⌊ ( k − 1) / 2 ⌋ r − 1  . W e ha ve that  ⌊ ( k − 1) / 2 ⌋ r − 1  ≤  k − 1 r − 1  / 2 r − 1 ≤  k − 1 r − 1  / ( r + 2) if r ≥ 4 , th us we are done in this case. CASE 3. r = 3 , q = 4 . CASE 3.1. | W 1 | = k − 1 . Then, we can use the following result from Case 2: every v ertex in S ′ is contained b y at most  ⌊ ( k − 1) / 2 ⌋ r − 1  h yp eredges in H . This is the only prop ert y of W 1 that w e use. 18 W e claim that H [ W 1 ] is not hamiltonian-connected. Otherwise, if a h yp eredge h con tains t wo vertices v , v ′ not in W 1 , then this leads to a con tradiction as in CASE 2. Assume rst that there are tw o vertices v 1 , v 2 outside W 1 with hyperedges in tersecting W 1 . Then there exist tw o v ertices u 1 , u 2 ∈ W 1 and tw o h yp eredges h 1 , h 2 ∈ E 1 suc h that v 1 , u 1 ∈ h 1 and v 2 , u 2 ∈ h 2 . If H [ W 1 ] is hamiltonian-connected, w e can nd a Berge- P k − 2 from u 1 to u 2 in H [ A 1 ] . Then the Berge- P k − 2 and h 1 , h 2 form a Berge- P k , a con tradiction. Assume no w that there is only one v ertex v outside W 1 with a hyperedge h intersecting W 1 . Then there is a hyperedge h ′ outside W 1 that contains v , and it is easy to see that a Berge path of length k − 2 inside W 1 is extended b y h and h ′ to a Berge path of length k , a con tradiction. W e obtained that H [ W 1 ] is not hamiltonian-connected. Thus, by Theorem 3.5, there exists a v ertex w ∈ W 1 with d H [ W 1 ] ( w ) ≤  ( k − 1) / 2 2  . Note that d H [ W 1 ] ( v ) ≤  k − 2 2  for an y v ∈ W 1 \{ w } and recall that d H ( v ) ≤  ⌊ ( k − 1) / 2 ⌋ 2  for an y v ∈ S . Then the n um b er of h yp eredges of H is  v ∈ S d H ( v ) +  v ∈ W 1 d H [ W 1 ] ( v ) 3 ≤ 5  ⌊ ( k − 1) / 2 ⌋ 2  +  ( k − 1) / 2 2  + ( k − 2)  k − 2 2  3 ≤  k 3  +  4 3  , th us we are done in this case. CASE 3.2. | W 1 |≤ k − 3 . Then we can use from CASE 1 the upp er b ound  | W 1 | r  +  k − 2 i = | W 1 |  i r − 1  /r + ( q + 1)  k − 1 r − 1  /r on the num b er of hyperedges. This is at most  k − 3 3  +  k − 3 2  / 3 +  k − 2 2  / 3 + 5  k − 1 2  / 3 ≤  k 3  +  4 3  and w e are done. CASE 3.3. | W 1 | = k − 2 . If S ′  = S , then u k − 1 w as an isolated comp onen t after removing S , thus the num b er of hyperedges is at most  k − 2 3  + 0 + 5  k − 1 2  / 3 ≤  k 3  +  4 3  and we are done. If S ′ = S , then in the deletion pro cess deleted the t w o vertices u k , u k − 1 together at the last step. If there are at most k hyperedges inciden t to u k and u k − 1 when w e delete them, then our upp er b ound on the n um b er of h yp eredges turns to  k − 2 3  + k + 4  k − 1 2  / 3 ≤  k 3  +  4 3  and we are done. If there are at least k + 1 suc h h yp eredges, then at most k − 2 of them con tains b oth u k and u k − 1 , since the third vertex of such a h yp eredge is in W 1 . Out of the at least three other such hyperedges (each containing exactly one of u k and u k − 1 ), at least t wo contain the same vertex of { u k − 1 , u k } . Without loss of generality , h 1 and h 2 con tain u k − 1 , such that for b oth h 1 and h 2 , the other tw o vertices are in W 1 . Let x  = y b e v ertices with x ∈ h 1 , y ∈ h 2 , and let W ′ 1 = W 1 ∪ { u k − 1 } . If W 1 is hamiltonian-connected, then a hamiltonian path connecting x and y in W 1 , together with h 1 and h 2 forms a Berge cycle of length k − 1 . Then we con tinue as in CASE 3.1. Recall that the only prop ert y of W 1 used there is a degree condition obtained in CASE 2, and in CASE 2 we used only the existence of a Berge cycle of length k − 1 . Finally , if W 1 is not hamiltonian-connected, then b y Theorem 3.5 there is a v ertex of degree at most  ( k − 2) / 2 2  in W 1 . This impro ves our upper b ound on the n umber of h yp eredges to  ( k − 2) / 2 2  +  k − 3 3  +  k − 2 2  / 3 + 5  k − 1 2  / 3 ≤  k 3  +  4 3  and w e are done. ■ F unding : The research of Cheng is supp orted by the National Natural Science F ounda- tion of China (Nos. 12131013 and 12471334), the Shaanxi F undamental Science Research 19 Pro ject for Mathematics and Ph ysics (No. 22JSZ009) and the China Scholarship Council (No. 202406290241). The research of Gerbner is supp orted by the National Research, Developmen t and Inno- v ation Oce - NKFIH under the grant KKP-133819 and b y the János Bolyai scholarship. The research of Miao is supported b y the China Sc holarship Council (No. 202406770056). The researc h of Zhou is supp orted b y the National Natural Science F oundation of China (Nos. 12271337 and 12371347) and the China Scholarship Council (No. 202406890088). References [1] C. Berge, Hypergraphs: Combinatorics of Finite Sets, North-Hol land, A mster dam , 1989. [2] D. Chakrab orti, D. Q. Chen. Exact results on generalized Erdős-Gallai problems, Eu- r op e an J. Combin. 120 (2024) 103955. [3] A. Da voo di, E. Győri, A. Meth uku, C. T ompkins, An Erdős-Gallai t yp e theorem for uniform h yp ergraphs, Eur op e an J. Combin . 69 (2018) 159–162. [4] P . Erdős, T. Gallai, On maximal paths and circuits of graphs, A cta Math. A c ad. Sci. Hung . 10 (1959) 337–356. [5] P . Erdős, M. Simono vits, A limit theorem in graph theory , Studia Sci. Math. Hungar. 1 (1966) 51–57. [6] P . Erdős, A. Stone, On the structure of linear graphs, Bul l. A mer. Math. So c. 52 (1946) 1087–1091. [7] Z. F üredi, A. K osto c hka, R. Luo, A v oiding long Berge cycles, J. Comb. The ory Ser. B 137 (2019) 5 5–64. [8] D. Gerbner, A. Meth uku, C. Palmer, General lemmas for Berge-T urán h yp ergraph prob- lems, Eur op e an J. Combin . 86 (2020) 103082. [9] D. Gerbner, D. Nagy , B. P atkós, N. Salia, M. Vizer, Stabilit y of extremal connected h yp ergraphs a voiding Berge-paths, Eur op e an J. Combin . 118 (2024) 103930. [10] D. Gerbner, C. Palmer, Extremal results for Berge hypergraphs, SIAM J. Discr ete Math . 31 (4) (2017) 2314–2327. [11] D. Gerbner, C. P almer, Counting copies of a xed subgraph in F -free graphs, Eur op e an J. Combin . 82 (2019) 103001. [12] D. Gerbner, C. Palmer, Survey of generalized T urán problems—counting subgraphs, Ele ctr onic J. Combin. 33 (1) (2026), #P1.23 20 [13] E. Gy őri, G. Katona, N. Lemons, Hyp ergraph extensions of the Erdős-Gallai theorem, Eur op e an J. Combin . 58 (2016) 238–246. [14] E. Győri, N. Lemons, N. Salia, O. Zamora, The structure of hypergraphs without long Berge cycles, J. Combin. The ory Ser. B 148 (2021) 239–250. [15] E. Győri, A. Methuku, N. Salia, C. T ompkins, M. Vizer, On the maximum size of connected h yp ergraphs without a path of giv en length, Discr ete Math . 341 (9) (2018) 2602–2605. [16] E. Gy őri, N. Salia, O. Zamora, Connected hypergraphs without long Berge-paths, Eu- r op e an J. Combin . 96 (2021) 103353. [17] J. Karamata, Sur une inégalité relativ e aux fonctions con vexes, Publ. Math. Univ. Bel- gr ade , 1 (1932) 145–148. [18] A. Kostochka, R. Luo, G. McCourt, Minim um degree ensuring that a hypergraph is hamiltonian-connected, Eur op e an J. Combin. 114 (2023) 103782. [19] P . T urán, Egy gráfelméleti szélső értékfeladatról, Mat. Fiz. L ap ok , 48 (1941) 436–452. 21