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Algoritmica Grafurilor - Cursul 7
13 noiembrie 2020
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Cuprins
1
Cuplaje
Formularea analitic a problemei cuplajului maxim Cuplaje perfecte - Teorema lui Tutte
Cuplaje de cardinal maxim - Teorema lui Berge
Cuplaje de cardinal maxim - Algoritmul Hopcroft-Karp
2
Fluxuri în reµele Reµele de transport
Flux, valoarea uxului, problema uxului maxim
3
Exerciµii pentru seminaru urm tor
4
Exerciµii rezolvate (parµial)
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Formularea analitic problemei cuplajului maxim
Fie G = (V ; E) un graf ³i M
Gfamilia cuplajelor sale.
Problem cuplajului maximum.
P
1dat un graf G = (V ; E), determinaµi M
2 M
Gastfel încât jM
j = max
M 2MG
jM j:
Fie B = (b
ij)
nmmatricea de incidenµ a lui G (se consider câte o ordonare xat a celor n noduri al lui G ³i a celor m muchii ale lui G) cu
b
ij=
® 1; dac muchia j este incident cu nodul i 0; altfel
Dac 1
peste vectorul p-dimensional cu toate componentele 1, atunci
problema P
1poate descris echivalent astfel:
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Formularea analitic a problemei cuplajului maxim
P
01max ¶
1
Tmx : Bx 6 1
n; x > 0; x
i2 f0; 1g; i = 1; m © :
P
01este o problem ILP (Integer Linear Programming), care este în gen- eral NP-hard. Vom încerca s rezolv m P
01, folosind problema (relaxat ) asociat LP (Linear Programming)
LP
01max ¶
1
Tmx : Bx 6 1
n; x > 0 © :
Soluµiile optime ale lui LP
01pot avea componente ne-întregi ³i de asemeni
valorile lor optime pot mai mari decât (G). De exemplu, dac G =
C
2n+1, atunci (G) = n, dar soluµia x
i= 1=2, 81 6 i 6 2n + 1 este
optim pentru problema corespunz toare LP
01cu valoarea optim n +
1=2 > n.
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Formularea analitic a problemei cuplajului maxim
b b
bb b
1/2 1/2
1/2 1/2
1/2
Urmeaz c dac graful G are circuite impare, problema P
1nu poate rezolvat folosind LP
01. Mai precis,
Teorema 1
(Balinski, 1971) Punctele extreme ale politopului Bx 6 1
n, x > 0, x 2 R
m, au componentele din f0; 1=2; 1g. Componentele 1=2 apar dac
³i numai dac G are circuite impare.
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Formularea analitic a problemei cuplajului maxim
Astfel, problema P
1este u³oar dac graful este bipartit: se rezolv problema LP
01³i soluµia (întreag ) optim g sit este o soluµie op- tim ³i pentru P
1.
Adapt ri combinatoriale ale algoritmului simplex pentru rezolvarea problemei de programare liniar au condus la a³a numita metod ungar pentru rezolvarea P
1în grafuri bipartite.
Vom discuta o soluµie mai rapid g sit de Hopcroft ³i Karp (1973).
Cu toate acestea, teorema de de dualitate din programarea liniar ³i
integralitatea soluµiei optime pot folosite pentru a obµine ³i explica
rezultatele (deja dovedite) despre cuplaje în grafuri bipartite:
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Cuplaje perfecte - Teorema lui Tutte
Teorema 2
(Hall, 1935) Fie G = (S; T; E) un graf bipartit. Exist un cuplaj în G care satureaz toate nodurile din S dac ³i numai dac
jN
G(A)j > jAj; 8A S:
Teorema 3
(Konig, 1930) Fie G = (S; T; E) un graf bipartit. Cardinalul maxim
al unui cuplaj în G este egal cu cardinalul minim al unei acoperiri cu
noduri a lui G, (G) = n (G).
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Cuplaje perfecte - Teorema lui Tutte
Fie G be un graf. Evident, jM j 6 jGj=2, 8M 2 M
G. Un cuplaj perfect (sau 1-factor) în G este un cuplaj M cu proprietatea jM j = jGj=2 (i.
e., S(M ) = V (G)).
O component conex a grafului G este par (impar ) dac num rul nodurilor sale este par (impar). Not m cu q(G) num rul componentelor conexe impare ale lui G.
Remarc
Fie G un graf care are un cuplaj perfect. Evident, ecare component conex a lui G este par . Astfel, q(G) = 0. Mai mult, dac S V (G) atunci pentru ecare component conex impar din graful G S trebuie s existe o muchie în cuplajul perfect al grafului cu o extremitate în aceast component conex ³i cealalt în S. Deoarece extremit µile muchiilor diferite sunt distincte, urmeaz c jSj > q(G S).
Pentru S = ? în (T), obµinem q(G ?) 6 0, deci q(G) = 0.
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Cuplaje perfecte - Teorema lui Tutte
bbbbbb b
b bb
b b
b
bb b
S even
components odd
components
Teorema 4
(Tutte, 1947) Un graf G admite cuplaj perfect dac ³i numai dac
(T) q(G S) 6 jSj; 8S V (G):
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Cuplaje perfecte - Teorema lui Tutte
Demonstraµie. Necesitatea condiµiei (T) a fost dovedit în discuµia de mai sus. Demonstr m prin inducµie dup n = jGj c dac un graf G = (V ; E) satisface (T), atunci G are un cuplaj perfect.
Pentru n = 1; 2 teorema are loc evident. În pasul inductiv, e G un graf cu n > 3 noduri care satisface (T) ³i s presupunem c orice graf G
0cu jG
0j < n ³i care satisface (T) are un cuplaj perfect.
Fie S
0V (G) astfel încât q(G S
0) = jS
0j ³i maximal cu proprietatea c avem egalitate în (T) (i. e., pentru orice supramulµime S a lui S
0avem q(G S) < jSj).
Observ m c familia de submulµimi ale lui V (G) pentru care (T) este satisf cut cu egalitate este nevid , ³i, deci, exist un S
0(pentru ecare nod v
0care nu este punct de articulaµie în componetnta lui din G, avem q(G v
0) = 1 = jfv
0gj, deoarece ecare component conex este par
³i în ecare component conex cu cel puµin un nod, exist un astfel de
nod v
0).
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Cuplaje perfecte - Teorema lui Tutte
Fie m = jS
0j > 0, C
1; C
2; : : : ; C
mcomponentele impare ale lui G S
0³i D
1; D
2; : : : ; D
kcomponentele pare ale lui G S
0(k > 0):
b
b
bb
b
b b
bb bb
bbb
b b
S0
b b b
b
b
b C1
C2
C3
Ci
Cm D1D2Dk s1s2
sm si v1
v2
v3
vi
vm
Vom construi un cuplaj perfect compus din
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Cuplaje perfecte - Teorema lui Tutte
a) câte un cuplaj perfect în ecare component conex par D
i; b) un cuplaj cu m muchii, fe
1; : : : ; e
mg, muchia e
iavând un cap t
s
i2 S ³i cel lalt v
i2 C
i(i = 1; 1; m);
c) cuplaj perfect in ecare subgraph C
iv
i(i = 1; 1; m).
a) Pentru orice 1 6 i 6 k graful [D
i]
Gadmite cuplaj perfect. Într- adev r, deoarece m > 0, urmeaz c jD
ij < n ³i din ipoteza inductiv este sucient s ar t m c G
0= [D
i]
Gsatisface (T).
Fie S
0D
i. Dac q(G
0S
0) > jS
0j, atunci obµinem urm toarea contradicµie:
q(G (S
0[S
0)) = q(G S
0)+q(G
0S
0) = jS
0j+q(G
0S
0) > jS
0[S
0j:
Astfel, q(G
0S
0) 6 jS
0j, 8S
0D
i, i. e., G
0satisface (T).
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Cuplaje perfecte - Teorema lui Tutte
b) Fie H = (S
0; fC
1; : : : ; C
mg; E
0) graful bipartit cu o clas a bipartiµiei S
0, cealalt clas mulµimea componentelor conexe impare ale lui G S
0,
³i cu muchiile de forma fs; C
ig, unde s 2 S
0astfel c exist v 2 C
icu sv 2 E(G).
H are un cuplaj perfect. Într-adev r, ar t m c H satisface condiµia din teorema lui Hall pentru existenµa unui cuplaj M
0care satureaz fC
1; : : : ; C
mg.
Fie A fC
1; : : : ; C
mg. Atunci B = N
H(A) S
0, ³i din construcµia
lui H , în graful G nu avem nicio muchie de la un nod v 2 S
0B la
un nod w 2 C
i2 A. Astfel componentele conexe impare din A r mân
componente conexe impare ³i în G B; astfel q(G B) > jAj. Deoarece
G satisface condiµia lui Tutte (T), avem jBj > q(G B). Am obµinut
c jN
H(A)j = jBj > jAj.
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Cuplaje perfecte - Teorema lui Tutte
Din teorema lui Hall H are un cuplaj M
0, care satureaz fC
1; : : : ; C
mg;
deoarece jS
0j = m, M
0este perfect.
M
0= fs
1v
1; s
2v
2; : : : ; s
mv
mg; S
0= fs
1; : : : ; s
mg; v
i2 C
i; 8i = 1; m:
c) 8i 2 f1; : : : ; mg graful G
0= [C
iv
i]
Gadmite cuplaj perfect. Folosind ipoteza inductiv , este sucient s dovedim c G
0satisface (T).
Fie S
0C
iv
i. Dac q(G
0S
0) > jS
0j, atunci, deoarece q(G
0S
0) + jS
0j 0 (mod 2) (pentru c jG
0j este par), urmeaz c q(G
0S
0) >
jS
0j + 2.
Dac S
00= S
0[ fv
ig [ S
0, avem
jS
00j > q(G S
00) = q(G S
0) 1+q(G
0S
0) = jS
0j 1+q(G
0S
0) >
> jS
0j 1 + jS
0j + 2 = jS
00j;
i. e., q(G S
00) = jS
00j, în contradicµie cu alegerea lui S
0(deoarece
S
0( S
00).
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Cuplaje perfecte - Teorema lui Tutte
Astfel 8S
0C
iv
i, q(G
0S
0) 6 jS
0j ³i G
0are cuplaj perfect (din ipoteza inductiv ).
Evident cuplajul lui G obµinut reunind cuplajele de la a), b), ³i c) de
mai sus satureaz toate nodurile lui G, ³i demonstraµia prin inducµie a
teoremei se încheie
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Cuplaje de cardinal maxim - Teorema lui Berge
Fie G = (V ; E) un graf ³i M 2 M
Gun cuplaj al lui G.
Deniµie
Un drum alternat în G relativ la cuplajul M este un drum P : v
0; v
0v
1; v
1; : : : ; v
k 1; v
k 1v
k; v
kastfel încât fv
i 1v
i; v
iv
i+1g \ M 6= ?, 8i = 1; k 1.
Observ m c , deoarece M este un cuplaj, dac P este un drum alternat relativ la M , atunci dintre orice dou muchii consecutive ale lui P exact una aparµine lui M (muchiile aparµin alternativ lui M ³i E n M ).
În cele ce urmeaz , când ne vom referi la un drum P vom înµelege
mulµimea muchiilor sale.
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Cuplaje de cardinal maxim - Teorema lui Berge
Deniµie
Un drum de cre³tere al lui G relativ la cuplajul M este un drum alternat care une³te dou noduri distincte expuse relativ la M .
Observ m c , din deniµia de mai sus, urmeaz c dac P este un drum de cre³tere relativ la M , atunci jP n M j = jP \ M j + 1.
b b b
b b b b bb
a
b
c d e
f g
h i
j
a, b, c, d- alternating even path f- alternating odd path
j- augmenting path g, f, d- alternating odd path a, b, c, d, e- closed alternating path a, b, c, d, f, g, h- augmenting path
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Cuplaje de cardinal maxim - Teorema lui Berge Teorema 5
(Berge, 1959) M este un cuplaj de cardinal maxim in graful G dac ³i numai dac nu exist drum de cre³tere în G relativ la M .
Demonstraµie. ")" Fie M un cuplaj de cardinal maxim în G. S presupunem c P este un drum de cre³tere în G relativ la M .
Atunci, M
0= M P = (P nM )[(M nP) 2 M
G. Într-adev r, M
0poate obµinut prin interschimbarea muchiilor din M cu cele din afara lui M de-a lungul drumului P:
b b
b b b b
a
b
c d e
f g
h i
j
bc bc
bc
a
b
c d e
f g
h i
bc j
P=a, b, c, d, f, g, h- augmenting path M∆P
b bb
b b b bb
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Cuplaje de cardinal maxim - Teorema lui Berge
Mai mult, jM
0j = jP \ M j + 1 + jM n Pj = jM j + 1, în contradicµie cu alegerea lui M .
"(" Fie M un cuplaj în G cu proprietatea c nu exist drumuri de cre³tere în G relativ la M .
Dac M
este un cuplaj de cardinal maxim în G, vom ar ta c jM
j = jM j. Fie G
0subgraful generat de M M
în G (G
0= (V ; M M
)).
S observ m c d
G0(v) 6 2, 8v 2 V ³i deci componentele conexe ale lui G
0pot noduri izolate, drumuri de lungime cel puµin unu, sau circuite.
Avem cinci posibilit µi (se v d mai jos; muchiile albastre sunt din M
, muchiile negre sunt din M ).
Cazul b) nu apare deoarece este un drum de cre³tere relativ la M
, care
este un cuplaj de cardinal maxim. Cazul c) nu apare deoarece este un
drum de cre³tere relativ la M .
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Cuplaje de cardinal maxim - Teorema lui Berge
bb bbbbbbb b
b bb
b
bbbb
b
b
b
bb b b bbbb
b b b b
b) c)
d) e) a)
Dac not m cu m
M(C ) num rul de muchii din M din componenta conex C a lui G
0³i cu m
M(C ) num rul de muchii din M
din aceea³i component conex a lui G
0, obµinem c m
M(C ) = m
M(C ). Astfel,
jM n M
j =
XC comp. a luiG0
m
M(C ) =
=
XC comp. a luiG0
m
M(C ) = jM
n M j
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Cuplaje de cardinal maxim - Teorema lui Berge
Deci jM j = jM
j.
Obµinem o strategie pentru a determina un cuplaj de cardinal maxim:
e M un cuplaj în G (e. g., M = ?);
while (9P drum de cre³tere relativ la M ) do M M P;
end while
La ecare iteraµie while cardinalul lui M cre³te cu 1, astfel, în cel mult
n=2 iteraµii obµinem un cuplaj f r drumuri de cre³tere, adic de cardinal
maxim. Neajunsul acestui algoritm este urm torul: condiµia din while
trebuie implementat în timp polinomial. Acest lucru a fost f cut pentru
prima oar de c tre Edmonds (1965).
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Cuplaje de cardinal maxim - Algoritmul Hopcroft-Karp
Lema 1
Fie M ; N 2 M
G, jM j = r, jN j = s ³i s > r. Atunci în M N exist cel puµin s r drumuri de cre³tere relativ la M disjuncte pe noduri.
Demonstraµie. Fie G
0= (V ; M N ) ³i C
i(i = 1; p) componentele conexe ale lui G
0. Pentru ecare 1 6 i 6 p, not m cu (C
i) diferenµa dintre num rul de muchii ale lui N din C
i³i num rul de muchii ale lui M din C
i:
(C
i) = jE(C
i) \ N j jE(C
i) \ M j:
Se observ c , deoarece M ³i N sunt cuplaje, C
isunt drumuri sau circuite. Astfel (C
i) 2 f 1; 0; 1g. (C
i) = 1 dac ³i numai dac C
ieste un drum de cre³tere relativ la M .
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Cuplaje de cardinal maxim - Algoritmul Hopcroft-Karp
Demonstraµie (continuare). Deoarece
Xpi=1
(C
i) = jN n M j jM n N j = s r;
urmeaz c exist cel puµin s r componente conexe ale lui G
0cu (C
i) = 1, adic , exist cel puµin s r drumuri de cre³tere disjuncte pe noduri conµinute în M N .
Lema 2
Dac (G) = s ³i M 2 M
Gcu jM j = r < s, atunci exist în G un drum de cre³tere relativ la M de lungime cel mult 2dr=(s r)e + 1.
Demonstraµie. Fie N 2 M
Gcu jN j = s = (G).
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