. Démonstration, Considérons une solution obtenue avec l'algorithme glouton Notons a 1 , a k les lettres où il y a un changement : ces a i sont tels que pour w d(i) = w f (i) = a i , d(i) < f (i), w f (i)?1 et w f (i) sont de couleurs diérentes (d(i) comme début, f (i) comme n) On indexe les a i de manière à ce que les f (i) soient rangés dans l'ordre croissant

. Montrons and ?. Dans, est pas un sous-mot de w, alors les intervalles I i := {d(i), d(i) + 1, . . . , f (i) ? 1} sont disjoints. La proposition sera alors démontrée puisque toute solution optimale est supérieure ou

I. Il-sut-de-montrer-que, ?. I. , =. Pour-tout-i, and ?. , Comme les f (i) sont croissants, cela impliquera automatiquement que tous les I i sont disjoints. Par dénition de f (i) et f (i+1), le premier changement de couleur après w f (i) a lieu entre w f (i+1)?1 et w f (i+1) Donc les couleurs de w f (i) et w f (i+1) sont diérentes, de même donc que celles de w d(i) et w d(i+1) Il n'y a pas de sous

G. Même-dans-le-cas, planaire, la solution n'est en général pas gloutonne : par exemple pour w = ABACDCBD. G(w) est alors planaire puisqu'il a quatre sommets Cela dit, la solution optimale est 10000111, i.e. 2 changements de couleurs (on note 1 et 0 les deux couleurs). L'algorithme glouton donne : 11000101, i.e. 4 changements. Même dans le cas couvert par la Proposition 5.3, l'algorithme glouton ne fonctionne pas : w = ABACCB, la solution optimale est 011100

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