ltdt- bai 02 - tong quan ltdt

Bi ging L thuyt th

th (graph) l mt cu trc ri rc gm cc nh v cc cnh ni cc nh .

th c k hiu l G = (V, E), trong :

V l tp nh (vertex),

E V V l tp hp cc cnh (edge).

HCMUS 2010 Bi ging L thuyt th ng Nguyn c Tin 2

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2

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CC LOI TH


HCMUS 2010 Bi ging L thuyt th ng Nguyn c Tin

Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

4

Mt n th (simple graph) G = (V, E) gm mt tp khng rng V v mt tp cnh E l cc cnh khng sp th t ca cc nh phn bit.


1

2

3


Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

6

Mt a th (multigraph) G = (V, E) gm mt tp cc nh V, mt tp cc cnh E v mt hm f t E ti {{u, v} | u, v V, u v}. Cc cnh e1, e2 c gi l cnh song song (parallel) (hay cnh bi (multiple)) nu f(e1) = f(e2).



Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

8

Mt gi th (pseudo graph) G = (V, E) gm mt tp nh V, mt tp cc cnh E v mt hm f t E ti {{u, v} | u, v V}.

Mt cnh l khuyn (loop) nu f(e) = {u, u} = {u} vi mt nh u no .


9


Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

10

Mt th c hng (directed graph hoc digraph) G = (V, E) gm tp cc nh V v tp cc cnh E l cc cp c th t ca cc phn t thuc V. Cc cnh y cn c gi l cung (arc).



Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

12

Mt a th c hng (directed multigraph) G = (V, E) gm mt tp cc nh V, tp cc cnh E v mt hm f t E ti {(u, v) | u, v V}.

Cc cnh e1 v e2 l cc cnh bi nu f(e1) = f(e2).


Loi Cnh C cnh bi? C khuyn?

n th V hng Khng Khng

a th V hng C Khng

Gi th V hng C C

th c hng C hng Khng C

a th c hng C hng C C


CC M HNH TH



A

B

C

D

AB AC AD

BC BD

A

B

C

D

A

B

C

D

DB DC

AB AC AD

BC BD

DB DC

18


A

B

C

D

AB BA

AD

BC

BD

A

B

C

D

A

B

C

D

CB

CD

A

B

C

D

AC

CA

DB

DC

DA

19


Tin

Nam

Minh Vn

Th

c

th quen bit trn tri t c hn 6 t nh v c th hn t t cnh!

20


Linda

Brian

Raffa Andy

John

Steve

21

nh: tc gi

Cnh: 2 ngi vit chung mt bi bo

th cng tc (2001) c hn 337000 nh v 496200 cnh.



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4 5

6

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1. A = 0

2. B = 1

3. C = A + 1

4. D = B + A

5. E = D + 1

6. E = C + D


S1 S2

S3 S4

S5 S6

24

1. Xc nh loi th ca cc th sau:


a

b

c

d

25

2. Xy dng th nh hng cho cc thnh vin lnh o ca mt cng ty nu:

Ch tch c nh hng ln gim c nghin cu & pht trin, gim c marketing, gim c iu hnh;

Gim c nghin cu & pht trin c nh hng ln gim c iu hnh;

Gim c Marketing nh hng ln Gim c iu hnh;

Khng ai c th nh hng ln trng phng ti chnh v Trng phng ti chnh khng nh hng ln bt c ai.


3. Hy xy dng th u tin cho chng trnh sau:

1. x = 0;

2. x = x + 1;

3. y = 2;

4. z = y;

5. x = x + 2;

6. y = x + z;

7. z = 4;


CNH K, NH K, BC


Hai nh u v v trong mt th v hng G c gi l k hay lin k (adjacent) (hay lng ging (neibor)) nu {u, v} l mt cnh ca G.

Nu e = {u, v} th e c gi l cnh lin thuc (incident) vi cc nh u v v. Cnh e cng c gi l cnh ni (connect) cc nh u v v.

Cc nh u v v gi l cc im u mt (endpoint) ca cnh {u, v}.


Bc (degree) ca mt nh trn th v hng l s cc cnh lin thuc vi n, ring khuyn ti mt nh c tnh hai ln cho bc ca n. Ngi ta k hiu bc ca nh v l deg(v).


30


Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

3

6

5

3

5

31

Cho G = (V, E) l mt th v hng c e cnh. Khi :

C bao nhiu cnh trong th c 10 nh, mi nh c bc bng 7?


Vv

ve )deg(2

32

nh l: Mt th v hng c s lng nh bc l l mt s chn.


Khi (u, v) l cnh ca th c hng G, th u c gi l ni ti v v v c gi l ni t u. nh u c gi l nh u (initial vertex), nh v gi l nh cui (terminal hoc end vertex) ca cnh (u, v).

Cnh e = (u, v) c gi l i t nh u ti nh v hoc i ra nh u vo nh v.


Trong th c hng, bc vo (in-degree) ca nh v, k hiu l deg-(v) l s cc cnh c nh cui l v.

Bc ra (out-degree) ca nh v, k hiu l deg+(v) l s cc cnh c nh u l v. (Mt khuyn s gp thm 1 n v vo bc vo v 1 n v vo bc ra ca nh cha khuyn)



Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

d- = 2, d+ = 1

d- = 4, d+ = 1

d- = 2, d+ = 3

d- = 2, d+ = 4

d- = 1, d+ = 2

36

Cho G = (V, E) l mt th c hng. Khi :


VvVv

Evv )(deg)(deg

37

nh treo (pendant vertex) l nh c bc bng 1.

nh c lp (isolated vertex) l nh c bc bng 0.



Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

Ph Quc

39

MT S N TH C BIT


th y (complete graph) n nh, k hiu l Kn, l mt n th cha ng mt cnh ni mi cp nh phn bit.


th chnh quy (regular graph) l n th m bc ca mi nh u bng nhau.

Nu bc ca cc nh l n, th th ny c gi l n chnh quy (n regular).



2 chnh quy

3 chnh quy

4 chnh quy

43

th vng (cycle) Cn, n 3 l mt th c n nh v1, v2 vn v cc cnh {v1, v2}, {v2, v3}, {vn-1, vn} v {vn, v1}.


Mt n th G c gi l th phn i (bipartite graph) nu tp cc nh V c th phn lm hai tp con khng rng, ri nhau V1 v V2 sao cho mi cnh ca th ni mt nh ca V1 vi mt nh ca V2.


th C6 v K3 c phi l cc th phn i? Gii thch.


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3

4 5

6

1

2

3

47

th phn i y (complete bipartite graph) Km,n l th c tp nh c phn thnh hai tp con tng ng c m nh v n nh v c mt cnh gia 2 nh nu v ch nu mt nh thuc tp con ny v nh th hai thuc tp con kia.


TH CON

TH B PHN


th con (subgraph) ca th G = (V, E) l th H = (W, F) trong W V v F E.

th H l con ca th G c gi l th b phn (spanning subgraph) ca G khi W = V.



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4 5

6

1

3

4 5

6

1 2

3

4 5

6

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Th no l 2 nh k nhau?

Bc ca nh l g?

Mi lin quan gia s cnh v bc?

S lng nh bc l ca mt th?

nh treo? nh c lp?

th y ?

th vng?

th phn i?

th phn i y ?

th con? th b phn


1. Cho G l mt th n, v hng c s nh n > 3.

Chng minh G c cha 2 nh cng bc.

Hng dn:

Dng nguyn tc Dirichlet


2. C th tn ti th n c 15 nh, mi nh c bc

bng 5 hay khng?

Hng dn

Dng nh l bt tay.


3. Trong mt bui chiu i, mi ngi bt tay nhau.

Chng t rng tng s ngi c bt tay vi mt s

l ngi khc l mt s chn. Gi s khng ai t bt

tay mnh

Hng dn

Chng minh tng t nh l s nh bc l ca th


4. V cc th:

a. K7

b. K1,8

c. K4,4


5. Cc th sau y c bao nhiu cnh?

a. Kn

b. Km,n

Hng dn

Kn c s cnh l C(2, n) = n(n-1)/2.

Km,n c s cnh l mn.


6. th s c bao nhiu cnh nu n c cc nh bc 4, 3, 3, 2,

2? V mt th nh vy.

Hng dn

Tng s bc nh ca th l 4 + 3 + 3 + 2 + 2 = 14, vy th ny c 7 cnh (nu tn ti th).


c a

e d

b

59

7. C tn ti th n cha nm nh vi cc bc sau

y? Nu c hy v th .

a. 3, 3, 3, 3, 2

b. 1, 2, 3, 4 ,5

c. 1, 1, 1, 1, 1

d. 1, 2, 1, 2, 1

e. 2, 1, 0, 2, 1

Hng dn

a, e). C.

b, c, d). Khng th (v c s nh bc l l mt s l).


c a

e d

b

c a

e d

b

60

9. Cc th sau y c phi l th phn i khng?

Hng dn

th bn tri: c.

th bn phi: khng.


TH NG CU


Cc n th G1 = (V1, E1) v G2 = (V2, E2) l ng cu (isomorphic) nu c hm song nh f t V1 ln V2 sao cho cc nh u v v l lin k trong G1 nu v ch nu f(u) v f(v) l lin k trong G2 vi mi u, v trong V1. Hm f nh vy c gi l mt ng cu.



th G1 v G2 l ng cu vi nhau:

f(1) = a, f(2) = b, f(3) = c, f(4) = d

e1 = E1, e2 = E2

1

3

2

4

a

b c

d

e1

e2

e3

e4 e5

e6

E1 E2

E3

E4

E5

E6

64

xc nh xem cc th l ng cu hay khng l rt kh khn!

chng minh 2 th l ng cu, cn a ra mt quan h tng ng (ng cu) gia 2 th ny.

chng minh 2 th khng ng cu, ch ra chng khng c chung mt tnh cht m cc th ng cu phi c.



c a

e d

b

z x

v t

y

70


b a

d c

e f

h g

t s

v u

w x

z y

71

BIU DIN TH


nh Cc nh k

a b, e

b a, c

c b, d, e

d c, e

e a, d


c a

e d

b

73

nh u nh cui

1 2, 3, 4, 6

2 3

3

4 3

5

6 3


1 2

3

4 5

6

74

Gi s G = (V, E) trong V = {v1, v2, }, |V| = n.

Ma trn k (Adjacency Matrix) A (hay AG) ca G l mt ma trn 0-1 cp nxn c phn t aij ti dng i, ct j bng 1 nu vi v vj k nhau v bng 0 nu vi v vj khng k nhau.


a b c d e

a 0 1 0 0 1

b 1 0 1 0 0

c 0 1 0 1 1

d 0 0 1 0 1

e 1 0 1 1 0


c a

e d

b

76


1 2

3

4 5

6

1 2 3 4 5 6

1 0 1 1 1 0 1

2 0 0 1 0 0 0

3 0 0 0 0 0 0

4 0 0 1 0 0 0

5 0 0 0 0 0 0

6 0 0 1 0 0 0

77

1 2 3 4

1 1 1 0 1

2 1 1 2 0

3 0 2 1 2

4 1 0 2 0


2 1

4 3

78

Gi s G = (V, E) trong :

V = {v1, v2, }, |V| = n.

E = {e1, e2, }, |E| = e.

Ma trn lin thuc (incidence matrix) M ca G l mt ma trn 0-1 kch thc nxe c phn t aij ti dng i, ct j bng 1 nu cnh ej ni vi nh vi v bng 0 nu cnh ej khng ni vi nh vi.


1 2 3 4 5 6

a 1 0 1 0 0 0

b 1 1 0 0 0 0

c 0 1 0 1 0 1

d 0 0 0 1 1 0

e 0 0 1 0 1 1

(a, b) (b, c) (a, e) (c, d) (d, e) (c, e)


c a

e d

b e1 e2

e3 e4

e5

e6

80

1 2 3 4 5 6 7 8 9

1 1 1 0 0 0 0 0 0 1

2 0 1 1 1 1 0 0 0 0

3 0 0 0 1 1 1 1 1 0

4 0 0 0 0 0 0 1 1 1

(1) (1, 2) (2) (2, 3) (2, 3) (3) (3, 4) (3, 4) (1, 4)


2 1

4 3

e1

e2 e3

e5 e4

e6

e7

e8

e9

81

Danh sch k

Cch 1:

int ke[MAX][MAX];

int soDinhKe[MAX];

Cch 2:

list ke[MAX];


Ma trn k & Ma trn lin thuc

int a[MAX][MAX];


Chn la cch biu din no l ph hp?


1. Hy biu din cc th sau y bng 3 cch biu din hc. Liu c th biu din c?


2 1

4 3

2 1

4 3

5

85

2. Hy biu din cc th sau bng cc cch biu din hc

a) K4

b) K1,4

c) C4

d) K2, 3


3. Hy v cc th v hng cho bi ma trn k sau:


1 3 2

3 0 4

2 4 0

1 2 0 1

2 0 3 0

0 3 1 1

1 0 1 0

0 1 3 0 4

1 2 1 3 0

3 1 1 0 1

0 3 0 0 2

4 0 1 2 3

87

4. Hy m t hng v ct ca ma trn k ca th tng ng vi nh c lp.

5. Cc n th vi ma trn k sau y c ng cu khng?


0 0 1

0 0 1

1 1 0

0 1 1

1 0 0

1 0 0

0 1 0 1

1 0 0 1

0 0 0 1

1 1 1 0

0 1 1 1

1 0 0 1

1 0 0 1

1 1 1 0

88

6. Cc th sau y c ng cu?


a)

b)

89

7. Cc th sau y c ng cu?


a)

b)

90

NG I, CHU TRNH


ng i (path) ( di n) t u ti v trong mt th v hng l mt dy cc cnh e1, e2 en ca th sao cho f(e1) = {x0, x1}, f(e2) = {x1, x2} f(en) = {xn-1, xn}, vi x0 = u v xn = v.

Khi th l n ta k hiu ng i ny bng dy cc nh x0, x1, xn.


ng i c gi l chu trnh (cycle/circuit) nu n bt u v kt thc ti mt nh (ngha l u = v).

ng i hay chu trnh gi l n nu n khng i qua cng mt cnh qu mt ln.


{a, b, c, e, d} l mt ng i di 4.

{a, b, e, d} khng l ng i.

{a, b, c, e, a} l mt chu trnh.

{c, e, d, e, c} khng phi l mt ng i n.


c a

e d

b

94

Nu 2 th ng cu vi nhau, n s c cc chu trnh c cng di k vi nhau, trong k > 2.

(iu ny suy ra Nu iu kin trn khng tha, ngha l 2 th khng ng cu vi nhau)



a

f b

e c

d

1

6 2

5 3

4 a

f b

e c

1

5 2

4 3

97

LIN THNG


Mt th v hng c gi l lin thng (connected) nu c ng i gia mi cp nh phn bit ca th.

Ngc li, th ny c gi l khng lin thng (disconnected).


Mt th khng lin thng s bao gm nhiu th con lin thng, cc th con ny c gi l cc thnh phn lin thng (connected component).



a b

c

k l

g

d e

h

i j

f

102

nh l: Nu mt th G (khng quan tm lin thng hay khng) c ng 2 nh bc l, chc chn s c mt ng i ni 2 nh ny.

Hng dn:

Dng nh l s nh bc l trong 1 th


S cnh ti a ca mt n th khng lin thng G gm n nh v k thnh phn l:


2

)1)(( knkn

104

th c hng gi l lin thng mnh (strongly connected) nu c ng i t a ti b v t b ti a vi mi cp nh a v b ca th.

th c hng gi l lin thng yu (weakly connected) nu c ng i gia 2 nh bt k ca th v hng nn.

th c hng c gi l lin thng mt phn (unilaterally connected) nu vi mi cp nh a, b bt k, c t nht mt nh n c nh cn li.



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3

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5

1 2

3

4

5

106

nh khp (cut vertex/ articulation point) ca mt th v hng l nh m nu xa nh ny khi th v cc cnh ni n n th s thnh phn lin thng ca th s tng thm.

Cnh cu (bridge) ca mt th v hng l cnh m nu xa i khi th th s thnh phn lin thng ca th s tng thm.

th song lin thng (biconnectivity) l th khng cha nh khp.


1. Mi danh sch cc nh sau y c to nn ng i trong th cho hay khng? ng i no l ng i n? ng i no l chu trnh? di ca cc ng i? a) a, e, b, c, d

b) a, e, a, d, b, c, a

c) e, b, a, d, b, e

d) c, b, d, a, e, c


c a b

d e

110

3. Cc th sau y c phi l th song lin thng?

a) K3

b) K4

c) K2,3

d) C5


CC PHP DUYT TH


Depth First Search

Nguyn l: Khi t mt nh, i theo cc cnh (cung) xa nht c th. Tr li nh sau ca cnh xa nht, tip tc duyt nh trc cho n nh cui cng.



a b

c

g

d e

h f

115


a b

c

g

d e

h f

1 2

3

4

5 6

7 8

116

procedure DFS(v)

begin

Thmnh(v);

chaXt[v] = false;

for u K(V) do

if chaXt[u] then

DFS(u);

end



a b

c

g

d e

h f

1 2

3

4

5 6

7 8

ChaXt[ ] a b c d e f g H

Khi to T T T T T T T T

Thm ln 1 F T T T T T T T

Thm ln 2 F F T T T T T T

Thm ln 3 F F F T T T T T

Thm ln 4 F F F T T T F T

Thm ln 5 F F F F T T F T

Thm ln 6 F F F F F T F T

Thm ln 7 F F F F F F F T

Thm ln 8 F F F F F F F F

F: nh hin ti F: nh chn c bc k tip.

118

Chng trnh chnh

BEGIN

for v V do

chaXt[v] = true;

for v V do

if chaXt[v] then

DFS(v);//BFS(v)

END.


Breadth First Search

tng: nh gn u tin thm trc



a b

c

g

d e

h f

121


a b

c

g

d e

h f

1 2

3

4

5 7

8 6

122

procedure BFS(v)

begin

QUEUE = ;

QUEUE v; // Kt np v vo QUEUE chaXt[v] = false;

while QUEUE do

p QUEUE; Thmnh(p);

for u K(p) do

if chaXt[u] then

QUEUE u; chaXt[u] = false;

end;



ChaXt[ ] a b c d e f g H

Khi to T T T T T T T T

Thm ln 1 F T T T T T T T

Thm ln 2 F F T T T T T T

Thm ln 3 F F F T T T T T

Thm ln 4 F F F T T T F T

Thm ln 5 F F F F T T F T

Thm ln 6 F F F F T T F F

Thm ln 7 F F F F F T F F

Thm ln 8 F F F F F F F F

F: nh hin ti F: nh chn c bc k tip.

a b

c

g

d e

h f

1 2

3

4

5 7

8 6

124

ltdt- bai 02 - tong quan ltdt

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