Divers Marc SAGE

(1)

Divers

Marc SAGE

Table des matières

1 Sur les permutations 2

1.1 1 . . . 2 1.2 2 . . . 2 1.3 3 . . . 2

2 Petit pb de combi 3

1

(2)

http ://en.wikipedia.org/wiki/Lubell-Yamamoto-Meshalkin_inequality http ://en.wikipedia.org/wiki/Sperner%27s_theorem

1 Sur les permutations

SoitS un ensemble de cardinalnetS_kl’ensemble des parties deSàkéléments.kétant …xé, peut-on injecter pour nassez gradS_k ,!S_k+1 de sorte que A (A)pour toutA2S_k?

Clairement, il fautn 2k+ 1par des considérations sur les cardinaux. Est-ce su¢ sant ?

1.1 1

En fait, by a theorem of de Bruijn, Tenbergen and Kruyswijk, les parties deSpeuvent s’écrire comme l’union disjointe de chaînes symétriques, où une chaîne symétrique consiste en un emboitement de parties

A1 A2 ::: Am

avecjA_i+1j=jA_ij+ 1 etjA₁j+jA_mj=n.

See e.g. chapter 3 of I. Anderson, "Combinatorics of Finite Sets", Oxford 1987, ISBN 0-19-853367-5 and 0-19-853379-9.

1.2 2

This is done with the marriage lemma. Viz. : ifR X Y is a binary relation between …nite setsX and Y, such that for every subsetEofX the cardinality of the projection inY ofR\(E Y)(i.e., the image ofE in Y) is at least equal to the cardinality ofE, then there exists an injetion ofX inY whose graph is included in R).

With the notation above, letX beS_k andY be S_k+1, and let Rbe the relation "A B" (forA inX and B inY). The marriage lemma shows that to obtain anf such as you ask, it is su¢ cient that, for any setE of m subsets ofS of sizek, the setE⁰ obtained by adding one element in every possible way to every element of E is of cardinality at leastm. But the relation Rrestricted toE E⁰ matches each element ofE with exactly n kelements ofE⁰ and each element of E⁰ with at mostk+ 1elements ofE; so as long asn k k+ 1, i.e.

n 2k+ 1, we have indeedjE⁰j jEj(by double counting).

1.3 3

It is su¢ cient indeed. LetS have2k+ 1elements and consider the poset (P(S), ) of subsets ofS, ordered by set-theoretic inclusion. A chain onP(S)is a collectionC of subsets ofS, such that every two elements ofC are comparable by . An antichain is a collection Aof such subsets, such that *no* two elements ofA are so comparable.

A well-known theorem of Sperner’s states that the largest antichain inP(S)has sizeN = 2k+ 1 k http ://en.wikipedia.org/wiki/Sperner_family

Next, one invokes Dilworth’s theorem, which says that, if a poset’s largest antichain has size k, then that poset is the union ofkchains.

http ://en.wikipedia.org/wiki/Dilworth’s_theorem

Combining these, we may decomposeP(S)into the union of N chains. Let D be the set of chains in this decomposition.

It is clear that any chain inP(S)has at most one element of each cardinality. Further, both Sk and Sk+1

have N elements, so each chain in D must contain exactly one element from each of these. Now build your function from that : givenA2Sk, there is a unique chain C(A)in Dsuch that A2C(A); de…nef(A)to be the unique element ofSk+1\C(A).

If you’re interested, I have a couple posts on my weblog about Dilworth’s theorem, with a visual proof : http ://pietrokc.wordpress.com/2008/03/06/dilworths-theorem-i/

http ://pietrokc.wordpress.com/2008/03/17/dilworths-theorem-ii/

2

(3)

2 Petit pb de combi

Soient k et mdeux entiers …xes, avec k 2,m 5 et m k+ 2:

Soient x1; :::; xm des reels (ou des complexes).

Pour I inclus dans f1; :::; mg de cardinal k, on considere la somme s(I)des xi, pour i appartenant à I.

Il est facile de voir que si s(I)est nul pour tout I (de cardinal k …xe), alors les x_i sont tous nuls.

Ma question est la suivante : quel est le plus petit entier utel que l’existence de uparties I (de cardinal k) telles que s(I) = 0implique que les x_i sont tous nuls ?

On a necessairementu 1 + m 1

k en prenantx₁ non nul et les autresx_i nuls ; on a en e¤et dans ce cas s(I) = 0des queIest contenu dansf2; ::; mg.

Hé bienu= 1 + m 1

k justement, non ?

En e¤et tu regardes des matrices àmcolonnes et ulignes distinctes telles qu’on ait k‘1’et m k‘0’dans chaque ligne, et tu cherches unuminimal tel que toute matrice de ce type ait un rangm.

On peut donc regarder unumaximal compatible avec un rangm 1. On a des lignes génératricesL1; :::; Lm 1: Est-il possible d’obtenir une ligne valide avec une combinaison linéaire dek+ 1Li? Si ce n’est pas le cas, le théorème est démontré.

12/11/98 19h02

3