LTDT- Bai 02 - Tong Quan LTDT Sep19 2011 Slide

15/09/2011

1

Bi ging L thuyt th(cp nht thng 01/2010)

ng Nguyn c Tin

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

CC LOI TH

HCMUS 2010 Bi ging L thuyt th ng Nguyn c Tin 3 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.


Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

6

15/09/2011

2

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 .

HCMUS 2010 Bi ging L thuyt th ng Nguyn c Tin

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

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

15/09/2011

3

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


Ch lm vic vi th c tp nh v tp cnh hu hn. Thut ng th c ngm hiu l th hu hn.


CC M HNH TH



Gu trc

Diu hu

Sc Qu

C

Chim g kin

Chut

Chut ch

Th c ti


Tin

Nam

Minh Vn

Th

c

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

18

15/09/2011

4


Linda

Brian

RaffaAndy

John

Steve

19

nh: tc gi

Cnh: 2 ngi vit chung mt bi bo

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


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

S5S6

21

1. Xc nh loi th ca cc th sau:


a

b

c

d

22

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;


15/09/2011

5

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).



Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

3

6

5

3

5

28

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

Chng minh:

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


Vv

ve )deg(2

29

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

Chng minh:


15/09/2011

6

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

33

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


VvVv

Evv )(deg)(deg

34

nh treo (pendant vertex, end vertex) l nh c bcbng 1.

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


Tin Giang

Tr Vinh

Vnh Long

ng Thp

Cn Th

Ph Quc

36

15/09/2011

7

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

40

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.


15/09/2011

8


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


1 2

3

45

6

1

2

3

44

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.

HCMUS 2010 Bi ging L thuyt th ng Nguyn c Tin 45 HCMUS 2010 Bi ging L thuyt th ng Nguyn c Tin 46

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

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


1 2

3

45

6

1

3

45

6

1 2

3

45

6

48

15/09/2011

9

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.


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

bng 5 hay khng?


3. Trong mt bui chiu i, mi ngi u 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


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


15/09/2011

10

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

2? V mt th nh vy.


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


8. Mt bui tic c 6 ngi tham d. Chng minh rngc 3 ngi tng cp quen nhau hoc 3 ngikhng quen nhau.


9. Cc th sau y c phi l th phn ikhng?


TH NG CU


Cc n th G1 = (V1, E1) v G2 = (V2, E2) l ngcu (isomorphic) nu c hm song nh f t V1 ln V2sao cho cc nh u v v l lin k trong G1 nu vch 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.


15/09/2011

11


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

e4e5

e6

E1 E2

E3

E4

E5

E6

61

xc nh xem mt 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.



ca

e d

b

zx

v t

y


ba

d c

e f

h g

ts

v u

w x

z y

64

BIU DIN TH


nh Cc nh k

a b, e

b a, c

c b, d, e

d c, e

e a, d


ca

e d

b

66

15/09/2011

12

nh u nh cui

1 2, 3, 4, 6

2 3

3

4 3

5

6 3


1 2

3

45

6

67

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 vjkhng 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


ca

e d

b


1 2

3

45

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

70

1 2 3 4

1 1 1 0 1

2 1 1 2 0

3 0 2 1 2

4 1 0 2 0


21

4 3

71

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.


15/09/2011

13

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)


ca

e d

be1 e2

e3e4

e5

e6

73

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)


21

4 3

e1

e2e3

e5e4

e6

e7

e8

e9

74

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?


Hai nh l sau y s gip chng ta xc nh s ng cu ca hai th ( gn nhn):

nh l 1: Hai th l ng cu vi nhau khi v ch khi cc nh ca n c th c gn nhn sao cho ma trn k tng ng l ging nhau.

nh l 2: Hai th c gn nhn G1 v G2 vi hai ma trn k tng ng l A1 v A2 ng cu vi nhau khi v ch khi tn ti ma trn hon v P sao cho

P A1 PT = A2


Ghi ch: Ma trn hon v l mt ma trn c c bng cch hon v cc hng v/hoc ctca mt ma trn n v n x n. Nh vy ma trn hon v l mt ma trn vung m mi hng

v ct ch c 1 phn t c gi tr '1', cc phn t cn li c gi tr '0'.

78

15/09/2011

14

Xt hai th G1 v G2 c gn nhn nh bn di v hai ma trn k tng ng:


21

4 3

da

b c

1 2 3 4

1 0 1 0 1

2 1 0 1 0

3 0 1 0 1

4 1 0 1 0

a b c d

a 0 0 1 1

b 0 0 1 1

c 1 1 0 0

d 1 1 0 0

79

Xt ma trn hon v

Ta c:

Vy G1 v G2 ng cu vi nhau theo nh x: 1 a, 2 c, 3 b v 4 d.


1000

0010

0100

0001

P

21

0011

0011

1100

1100

1000

0010

0100

0001

0101

1010

0101

1010

1000

0010

0100

0001

APPA T

80

chng minh 2 th khng ng cu, ta ch cn a ra v d phn chng.

chng minh 2 th l ng cu, cn phi ch ra ma trn hon v P.

(C n n! hon v khc nhau vi th n nh, do vy, bi ton xc nh 2 th ng cu c phc tp rt ln!)


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


21

4 3

21

4 3

5

82

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

84

15/09/2011

15

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

85

6. Cc th sau y c ng cu?


a)

b)

86

7. Cc th sau y c ng cu?


a)

b)

87

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 v 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.


15/09/2011

16

{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.


ca

e d

b

91

Nhiu nh khoa hc x hi cho rng hu ht mi cp ngi trn th gii c kt ni bi mt chui kh nh, c th ch gm 5 ngi hoc t hn.


Tin

Nam

Minh Vnc

Th

92

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

4a

f b

e c

1

5 2

4 3

94

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).


15/09/2011

17


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

99

nh l: Nu mt th G (khng quan tm linthng hay khng) c ng 2 nh bc l, chc chns c mt ng i ni 2 nh ny.


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


2

)1)(( knkn

101

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 (underlying graph).

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.


15/09/2011

18


1 2

3

4

5

1 2

3

4

5

103

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.



Kh nng lin thng (connectivity) hay kh nng lin thng nh (vertex connectivity) (G) ca mt th l tp c lc lng nh nht cc nh m khi xa cc nh ny i th khng cn lin thng na.

Mt th G c gi l k lin thng (kconnectivity) hay k lin thng nh (k-vertex-connectivity) khi (G) k.

Nh vy, mt th y Kn, theo nh ngha s c (G) = n 1.

th song lin thng chnh l th 2 lin thng.

Ghi ch: c l kappa.


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


ca b

d e

107

2. Cc thnh phn lin thng trong th quen bit biu din ci g?

3. Hy tm thnh phn lin thng mnh ca cc th c hng sau:


ca b

f e d

ca b

h g f

d

e

108

15/09/2011

19

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

a) K3b) K4c) 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.

tng: nh thm cng mun s cng sm tr thnh nh duyt xong.


a b

c

g

d e

h f

1 2

3

4

5 6

78

112

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

78

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 tiF: nh chn c bc k tip.

114

15/09/2011

20

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 thm cng sm s cng sm thnh nh duyt xong.



a b

c

g

d e

h f

1 2

3

4

5 7

86

117

procedure BFS(v)

begin

QUEUE = ;

QUEUE v; // Kt np v vo QUEUEchaXt[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 tiF: nh chn c bc k tip.

a b

c

g

d e

h f

1 2

3

4

5 7

86

119

V D MINH HA


15/09/2011

21

1 2

3

4

Ln cn 4 Ln cn 8


Yu cu

Khi kho st cc o mt vng bin, ngi ta ghi kt qu kho st li thnh mt bn nh phn trong s 0 cho bit bin v s 1 l t lin. Mt o chnh l mt min lin thng 4 cc s 1 trn bn . T mt bn cho trc, hy m s lng o ca vng bin.


T mt th v hng, lin thng cho trc, liu c th nh hng li thnh th c hng lin thng mnh?

(nh l: th v hng lin thng l nh hng c khi v ch khi mi cnh ca n nm trn t nht mt chu trnh)


Hy tm ng i gia 2 nh bt k ca mt n th v hng lin thng.



LTDT- Bai 02 - Tong Quan LTDT Sep19 2011 Slide

Documents

Transcript of LTDT- Bai 02 - Tong Quan LTDT Sep19 2011 Slide