Examen d` Optimisation dans les Graphes

5/3T
G= (V, E)T⊆V T
k:E→Z+
v∈V
X
e∈δ(v)
k(e)⇐⇒ v∈T.
(V, B)B={e∈E:k(e)≥1}
v B
T T
T
G w :E→R+T k
Pe∈Ek(e)w(e)T
T
G= (V, E)T⊆V
w:E→R+
T
T

T
≤αOPT OPT
α
T
k{0,1,2}
(V, F )G
(V, F )T
T F (T)TF
v∈VdegF(v)
F\F(T)T4TF(V, F )
k∗T
{0,1,2}G∗= (V, E∗)
G e k∗(e)
w(E∗) = Pe∈Ek∗(e)w(e)
G∗TF
F0TF
E∗\F0T4TFG∗
F0
E∗
F0TFG
T4TFG
w≤2
3w(E∗)
w(F\F(T)) ≤2
3w(F)w(E∗\
F0)≤2
3w(E∗)
F(T)4F0T4TFG

e
F T 4TFG e ∈E
e
k(e) = 
2e∈F∩e F
1e∈F4e F
0
e
k T 5/3T
n Kn=
(V, E)s t w :E→R+
s t