By Yves André (auth.), Umberto Zannier (eds.)

Since 2001 the Scuola Normale Superiore di Pisa has prepared the "Colloquio De Giorgi", a sequence of colloquium talks named after Ennio De Giorgi. The Colloquio is addressed to a common mathematical viewers, and particularly intended to draw graduate scholars and complex undergraduate scholars. The lectures are meant to be no longer too technical, in fields of large curiosity. they need to offer an summary of the overall subject, in all probability in a ancient viewpoint, including an outline of more moderen development. the belief of amassing the fabrics from those lectures and publishing them in annual volumes got here out lately, as a attractiveness in their intrinsic mathematical curiosity, and likewise with the purpose of holding reminiscence of those events.

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**Example text**

This is in Figure 16. The proof is much more difficult than the approximate version. ) 1 + ε n. 2 25 Is laziness paying off? (“Absorbing” method) For a k-uniform hypergraph H and an integer m by an m cycle of length l we mean a k-uniform hypergraph whose vertices can be ordered cyclically x1 , x2 , . . , xl in such a way that (x1 , x2 , . . , xk ) ∈ H, (xk−m−1 , xk−m+1 , xk , xk+1 , . . , x2k−m ) ∈ H and if (x j , x j+1 , . . , x j+k−1 ) ∈ H, then (x j+k−1−m−1 , . . , x j+2k−1−m ) ∈ H. Figure 17.

V. R¨odl, A. Ruci´nski, E. Szemer´edi [24] proved that if δ2 (H) > 1 + ε n, 2 then H contains a Hamiltonian cycle. 1 ∗ (H) + ε n, then H contains Later in [25] they proved that if δk−1 2 a Hamiltonian cycle. In [26] they proved the exact result. They proved that if δ2 (H) 1 n, then H contains a Hamiltonian cycle. Moreover, there is only one 2 example where the result is tight. This is in Figure 16. The proof is much more difficult than the approximate version. ) 1 + ε n. 2 25 Is laziness paying off?

Since n 0 was chosen large enough n n ≤ γ k−1 edges which contain v1 and v j there are at most (k − 1) k−2 for some j ∈ {2, . . , k}. Due to the minimum degree of H there are at n /2 edges containing v1 but none of the vertices v2 , . . , vk . We least k−1 fix one such edge {v1 , u 2 , . . , u k } and set U1 = {u 2 , . . , u k }. For each i = 2, 3, . . , k and each pair u i , vi suppose we succeed to choose a set Ui such that Ui is disjoint to Wi−1 = j∈[i−1] U j ∪ T and both Ui ∪ {u i } and Ui ∪ {vi } are edges in H.