chuong2. duyet_dequi

7/23/2019 Chuong2. Duyet_Dequi

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CHNG 2. DUYT V QUI

NI DUNG:2.1. Thut ton duyt

2.2. Thut ton sinh

2.3. Thut ton qui

2.4. Thut ton quay lui

2.5. Kh qui

2.6. CASE STUDY

An evaluation version of novaPDFwas used to create this PDF file.

Purchase a license to generate PDF files without this notice.
http://www.novapdf.com/http://www.novapdf.com/


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2.1. Thut ton duyt

Phng php gii quyt bi ton (Problem):

S dng cc cng c ton hc: S dng cc nh l, mnh , lp lun v suy logic ca trong ton hc tm ra nghim ca bi ton.

u im:d dng my tnh ha cc bi ton c thut gii (MathLab).

Nhc im: ch thc hin c trn lp cc bi ton c thut gii.

Lp bi ton ny rt nh so vi lp cc bi ton thc t.

S dng my tnh v cc cng c tnh ton:

Gii c mi bi ton c thut gii bng my tnh.

i vi mt s bi ton cha c thut gii, ta c th s dng my tnh xem xt tt c cc kh nng c th t a ra nghim ca bi ton.

Mt thut ton duyt cn tha mn hai iu kin:

Khng c lp li bt k kh nng no.

Khng c b st bt k cu hnh no.




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V d 1. Cho hnh vung gm 25 hnh vung n v. Hy in cc s t 0 n 9 vomi hnh vung n v sao cho nhng iu kin sau c tha mn.

c t tri sang phi theo hng ta nhn c 5 s nguyn t c 5 ch s;

c t trn xung di theo ct ta nhn c 5 s nguyn t c 5 ch s;

c theo hai ng cho chnh ta nhn c 2 s nguyn t c 5 ch s;

Tng cc ch s trong mi s nguyn t u l S cho trc.

V d hnh vung di y vi S = 11.

3 5 1 1 1

5 0 0 3 3

1 0 3 4 3

1 3 4 2 1

1 3 3 1 3




4/39

Thut gii duyt. Chia bi ton thnh hai bi toncon nh sau:Tm X ={ x[10001,...,99999] | x l nguyn t v tng

cc ch s l S.Chin lc vt cn c thc hin nh sau: Ly xX t vo hng 1(H1): ta in c vung 1, 2, 3, 4, 5.

Ly xX c s u tin trng vi s 1 t voct 1 (C1): ta in c vung 6, 7, 8, 9.Ly xX c s u tin trng vi s 9, s cuicng trng vi s 5 t vo ng cho chnh 2(D2): ta in c vung 10, 11, 12.Ly xX c s th nht v s th 4 trng vi s 6 v 12 t vo hng 2 (H2): ta in c

vung 13, 14, 15.Ly xX c s th nht, th hai, th 4 trng vi s 2, 13, 10 t vo ct 2 (C2): ta in c vung 16, 17.

Lm tng t nh vy ta in vo hng 5 s25.

Cui cng ta ch cn kim tra D1X v C5 X?

1 2 3 4 5

6 13 14 12 15

7 16 11 18 19

8 10 20 22 239 17 21 24 25

3 5 1 1 1

5 0 0 3 3

1 0 3 4 3

1 3 4 2 1

1 3 3 1 3




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2.2. Thut ton sinh (Generation)

Thut ton sinh c dng gii lp cc bi ton tha mn hai iu kin:

Xc nh c mt th ttrn tpcc cuhnh cn litk cabi ton. Bit

c cuhnh utin, bit c cuhnh cuicng.T mt cu hnh cuicng, ta xy dng c thutton sinh ra cuhnhngngay sau n theo th t.

Thut ton:

Thut ton Generation:

Bc1 (Khi to):;

Bc 2 (Bc lp):

while ()do

;

;

endwhile;

End .




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V d1. Cho s t nhin N (N100). Hy li k tt c cc cchchia s t nhin N thnh tng ca cc s t nhin nh hn N.Cc cch chia l hon v ca nhau ch c tnh l mt cch.

V d vi N= 5 ta c 7 cch chia nh sau:5

4 13 2

3 1 12 2 12 1 1 11 1 1 1 1




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Thut ton:Input:

S t nhin n.Output:

Cc cch chia n thnh tng cc s nh hn n.Actions:Bc 1(Khi to):

k =1; X[k] = n; OK = TRUE; //a ra cch chia u tin.Bc 2(Lp):

while (OK) { //Lp n khi no cch chia khng phi l n s 1Result(); //a ra phng n hin ti.

Next_Division(); //Sinh ra php chia k tip}

EndAction.Procedure Next_Division() { int i = k, j, R, S,D;

while (i > 0 && X[i]==1 ) i--; //Lp t bn phi khi no X[i] =1if (i>0 ) { X[i] = X[i] 1; D = k - i + 1;R = D / X[i]; S = D % X[i]; k= i;

if (R>0) { for ( j = i +1; j0 ){ k = k +1; X[k] = S; }

}else OK =0;

}An evaluation version of novaPDFwas used to create this PDF file.



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//Hy cho bit kt qu thc hin ?

#include #include

#define MAX 100#define TRUE 1#define FALSE 0int n, k, X[MAX], dem =0, OK =TRUE;void Init(void ){

coutn;k = 1; X[k] = n;

}void Result(void) {

cout


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2.3. Thut ton qui (Recursion)

Phngphp nh ngha bng qui: Mt i tng c nh ngha trc tiphoc gin tip thng qua chnh n c gi l php nh ngha bng qui.

Thutton qui : Thut ton gii bi ton P trc tip hoc gin tip thng qua biton P ging nh P c gi l thut ton qui. Mt hm c gi l qui nun c gi trc tip hoc gin tip n chnh n. Mt bi ton gii c bng qui nu n tha mn hai iu kin:

Phn tch c: C th gii cbi ton P bngbi ton P ging nhP vchkhc P d liu uvo. Vic giibi ton P cng c thc hintheo

cch phn tch ging nhP.iu kin dng: Dy cc bi ton P ging nh P l hu hn v s dng timt bi ton xc nh no .

Thut ton qui tng qut c th c m t nh sau:

Thut ton Recursion ( P ) {

1. Nu P tha mn iu kin dng:;

2. Nu P khng tha mn iu kin dng:

Recursion(P).

}




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V d: Tm tng ca n s t nhin bng phng php qui.

Li gii. Gi Sn l tng ca n s t nhin. Khi :

Bcphn tch: Sn = n + S(n-1), n>1.

iu kin dng: s 1 = 1 nu n=1;T ta c li gii ca bi ton nh sau:

int Tong (int i ) {

if (i ==1 ) return(1); //iu kin dng

else return(i + Tong(i-1)); //iu kinphn tch c

}

V d. Tm n!.

Li gii. Gi Sn l n!. Khi :

Bcphn tch: Sn = n*(n-1)! nu n>0;

iu kin dng: s0=1 nu n=0.T ta c li gii ca bi ton nh sau:

long Giaithua (int i ) {

if (i ==0 ) return(1); //iu kin dng

else return(i *Giaithua(i-1)); //iu kinphn tch c




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V d: Tm c s chung ln nht ca a v b bng phng php qui.

Li gii. Gi d =USCLN(a,b). Khi :

Bcphn tch:

d = USCLN(a-b, b) nu a>b. d = USCLN(a, b-a) nu a b) return(USCLN(a-b, b));

else return(USCLN(a, b-a));

}}




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V d.Hy cho bit kt qu thc hin chngtrnh di y?

#include

#include

int convert(int X, int b){

if (X == 0) return 0;

else return (X % b + 10 * convert(X / b));

}

int main(){

int X, b, bin;

printf("Nhap so X: "); scanf("%d", &X);

printf("Nhap so b: "); scanf("%d", &b);

bin = convert(X, b);

printf("Ket qua %d la %d.\n", X, bin);

getch(); return 0;

}




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Bi tp.Hy cho bit kt qu thc hin chngtrnh di y?

#include

#include

#include

void reverse(char str[], int index, int size){

char temp = str[index];

str[index] = str[size];

str[index] = temp;

if (index>= size) return;

reverse(str, index + 1, size-1);

}

int main(){

char X[] ="ABCDEF";int size = strlen(X);reverse(X, 0, size - 1);

printf("Ket qua is: %s\n", X);

getch(); return 0;

}




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Kch ccy nhphn (size of a tree). Ta nh ngha kch c ca mt cy l s node cthc trn cy. Bi ton t ra l cho trc mt cy hy tm kch c ca cy. V d: cydi y c kch c l 13

40

30 60

25 35 50 70

20 28 65 903832

int Size(struct node* T) {

if (T==NULL)return 0;

elsereturn(Size(T->left) + 1 + Size(T->right));

}




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Xc nhhai cy nhphn gingnhau. Ta nh ngha hai cy nh phn ging nhaunu chng c cng chung node v mi node c sp t ging nhau trn cy. Cho haicy nh phn bt k, hy xc nh xem hai cy c ging nhau hay khng?

Li gii. Thut ton gii quyt bi ton c thc hin nh sau:

in identicalTrees(struct node* T1, struct node* T2) {

if (T1==NULL && T2==NULL) //nu chai cy T1 v T2 u rng

return 1; //r rng chng gingnhau

if (T1!=NULL && T2!=NULL){ //nu chai cy khc rngreturn (T1->data == T2->data &&

identicalTrees(T1->left, T2->left) &&

identicalTrees(T1->right, T2->right) );

}

return 0; //mttrong hai cy khc rng}




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Tm cao cacy. cao ca cy c nh ngha l ng i di nht t node gcn node l. Cho mt cy nh phn bt k, hy xc cao ca cy?

Li gii. Thut ton gii quyt bi ton c thc hin nh sau:

int maxDepth(struct node * T) {

if (T==NULL)

return 0;

else {

int lDepth = maxDepth(T->left);//Tm cao ca cy con triint rDepth = maxDepth(T->right); //Tm cao ca cy con phi

if (lDepth > rDepth)

return(lDepth);

else

return(rDepth);

}

}




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Cy phn chiu(Mirror Tree). Cy phn chiu ca cy nh phn T l cy nh phnM(T), trong cy con bn tri ca cy T tr thnh cy con bn phi ca M(T) v cycon bn phi ca T tr thnh cy con tri ca M(T). Bi ton c t ra l hy dchchuyn cy nh phn T cho trc thnh cy nh phn phn chiu ca T l cy M(T).

30

25 35

20 28 3832

30

35 25

38 32 2028

void mirror(struct node* node) {if (node==NULL) return; //Nu cy rng th khng phi lm gelse { //Nu cy khng rng

struct node* temp; //S dng node trung gian tempmirror(node->left); //Gi n cy phn chiu bn trimirror(node->right);//Gi n cy phn chiu bn phitemp = node->left; node->left = node->right;node->right = temp; //Tro i hai node tri v phi

}




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n tt ccc node l trn cy:

Input: Output:

430

25 35

20 28 3832

unsigned int getLeafCount(struct node* node) {if(node == NULL) //Nu cy rng

return 0; //S node l l 0

if(node->left == NULL && node->right==NULL) //Nu cy c mt nodereturn 1; //S node l l 1else //Nu cy c t nht mt cy con

return getLeafCount(node->left)+ getLeafCount(node->right);}




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Kimtra mtcy thamn iu kinnode trung gian bng tnghai node con triv phihay khng? Node khng c node l tri hoc phi cxem c gi tr0.

10

8 2

5 3 2

int isSumProperty(struct node* node) {int left_data = 0, right_data = 0;if(node == NULL || (node->left == NULL && node->right == NULL)) return 1;else {

if(node->left != NULL)left_data = node->left->data;

if(node->right != NULL)right_data = node->right->data;

if((node->data == left_data + right_data)&& isSumProperty(node->left) &&isSumProperty(node->right))

return 1;else return 0;

}




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2.4. Thut ton quay lui (Back track)

Gi s ta cn xc nh b X =(x1, x2,..,xn) tha mn mt s rng buc no . ngvi mi thnh phn xi ta c ni kh nng cn la chn. ng vi mi kh nngjnidnh cho thnh phn xi ta cn thc hin:

Kim tra xem kh nng j c c chp thun cho thnh phn xihay khng?Nu kh nng j c chp thun th nu i l thnh phn cui cng (i=n) ta ghinhn nghim ca bi ton. Nu i cha phi cui cng ta xc nh thnh phnth i +1.

Nu khng c kh nng j no c chp thun cho thnh phn xith ta quay

li bc trc (i-1) th li cc kh nng khc.Thut ton Back-Track ( int i ) {

For ( j =; j


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V d . Bi ton N qun hu. Trn bn c kch c NN, hy t N qun hu miqun trn 1 hng sao cho tt c cc qun hu u khng n c ln nhau.

Li gii. Gi X =(x1, x2,..,xn) l mt nghim ca bi ton. Khi , xi = j c hiu lqun hu hng th i t ct j. cc qun hu khc khng th n c, qun

hu th i cn khng c ly trng vi bt k ct no, khng c cng ngcho xui, khng c cng trn ng cho ngc. Ta c n ct A = (a1,..an), cXuoi[2*n-1] ng cho xui, Nguoc[2*n-1] ng cho ngc.

Nu xi =j th A[j] = True, Xuoi[i-j+n] = True, Nguoc[i + j -1] = True.

1

2

3

4

5

6

7

15 14 13 12 11 10 9 8

1

2

3

4

5

6

7

8 9 10 11 12 13 14 15

ng cho xui: Xuoi [i j + n] ng cho ngc: Nguoc [i + j -1]




22/39

Thut ton:

void Try (int i){

for(int j=1; j


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2. Sudoku. Cho mt hnh vung gm li cp 99 hnh vung n v trong

mt s hnh vung in cc con s t 1 n 9. Hy in cc s t 1

n 9 vo cc hnh vung cn li sao cho mi hng, mi ct v mi li

con kch c 33 cha ng ch mt trong cc s t 1 n 9.

V d vi li di y s cho ta kt qu sau:

3 1 6 5 7 8 4 9 2

5 2 9 1 3 4 7 6 8

4 8 7 6 2 9 5 3 1

2 6 3 4 1 5 9 8 7

9 7 4 8 6 3 1 2 5

8 5 1 7 9 2 6 4 3

1 3 8 9 4 7 2 5 6

6 9 2 3 5 1 8 7 4

7 4 5 2 8 6 3 1 9




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Thut ton:

bool Sudoku( int grid[N][N]){ // N = 9

int row, col; //hng, ct cn t

if (!FindUnassignedLocation(grid, row, col)) //nu in c tt creturn true; // thnh cng!

for (int num = 1; num


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3. Dy con c tng bngK (Set Sum). Cho tp cc s gm n s khcnhau. Hy tm tt c cc tp con ca tp cc s cho sao cho tng cc

phn t ng bng K. V d di y a ra cy quay lui ca bi tonvi n=4.




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4. Bi ton t mu th (Graph Coloring). Cho th v hngG= v s t nhin m. Hy tm cch t mu th vi nhiu nht m

mu sao cho khng c hai nh k nhau no c t bi cng mt mu.V d di y chra mt cch t vi 3 mu.




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2.5.Kh qui

Nguyn tc lptrnh: Khng nn lm dng gii thut qui trong khi xydng ng dng. Mi bi ton gii c bng qui u c th dch chuyn

v cc cu trc lnh lp kt hp vi cc cu trc d liu ph hp.

Nguyn tc kh qui: Da trn b xp chng ca chng trnh dch (stack). Da trn cc cu trc lnh lp biu din li thut ton.

Bi tp1. Kh qui gii thut di y?




}




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}

int Convert(int X, int b){

int S=0, i=0;

while(X!=0) {

int du = X%b; //ly phn dint k = (int) pow(10, i);

du = du* k;

S= S + du;

X =X / b; i++;

}

return(S);

}




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Bi tp 2. Kh qui php duyt cy theo th t trc (NLR).Algorithm NLR ( Tree *T) {

if (T!= NULL ) {;

NLR(T -> left); //duyt th t trcsang bn triNLR( T -> right); //duyt th t trcsang bnphi

}}V d : NLR(T) = 10, 8, 5, 6, 12, 3, 9, 2, 1, 4, 7.

10

8 9

5 3 2 7

1 46 12




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Kim nghim.

Bc Trng thi stack Cc node thm

1 10

2 9, 8 10

3 9, 3, 5 10, 8

4 9, 3, 12, 6 10, 8, 5

5 9, 3, 12 10, 8, 5, 66 9, 3 10, 8, 5, 6, 12

7 9 10, 8, 5, 6, 12, 3

8 7, 2 10, 8, 5, 6, 12, 3, 9

9 7, 4, 1 10, 8, 5, 6, 12, 3, 9, 2

10 7, 4 10, 8, 5, 6, 12, 3, 9, 2, 1

11 7 10, 8, 5, 6, 12, 3, 9, 2, 1, 4

12 10, 8, 5, 6, 12, 3, 9, 2, 1, 4, 7




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Bi tp 3. Kh qui php duyt cy theo th t gia (LNR).Thut ton LNR ( Tree *T) {

if (T!= ) { / /nuT khc rngLNR(T -> left; //Duyttheo th t giasang bn tri

;LNR( T -> right); //Duyt th t giasang bnphi

}}V d : LNR(T) =6, 5, 12, 8, 3, 10, 1, 2, 4, 9, 7.

10

8 9

5 3 2 7

1 46 12




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Thutton LNR davo stack:

Thutton LNR (Tree *T) {Bc1 (Khi to): done = false; //trngthi duytban ul false

Stack =; // toStack rngNode = T; //Node hin ti tr nnode gcBc2 ( Lp):

while(! done) { //lp nkhi done l trueif (Node ) { / /nunode khc rng

Push(Stack, Node); /lanode vo stack

Node = Node -> left; //chuynNode sang bn tri}else { //nuNode rng

if (Stack ) { / /nuStack khc rngNode = Pop(Stack); //lynode ra khistack;

Node = Node ->right; //Node trsang bnphi}else //nuStack rng

done = true; //ktthc duyt}

}




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1 10

2 10, 8

3 10, 8, 5

4 10, 8, 5, 6

5 10, 8, 5 6

6 10, 8, 12 6, 5

7 10, 8 6, 5, 12

8 10,3 6, 5, 12, 8

9 10 6, 5, 12, 8, 3

10 9 6, 5, 12, 8, 3, 10

11 9, 2 6, 5, 12, 8, 3, 10

12 9, 2, 1 6, 5, 12, 8, 3, 10

13 9, 2 6, 5, 12, 8, 3, 10, 1

14 9, 4 6, 5, 12, 8, 3, 10, 1, 2

15 9 6, 5, 12, 8, 3, 10, 1, 2, 4

16 7 6, 5, 12, 8, 3, 10, 1, 2, 4, 9

17 6, 5, 12, 8, 3, 10, 1, 2, 4, 9, 7An evaluation version of novaPDFwas used to create this PDF file.



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Bi tp 4. Kh qui php duyt cy theo th t sau (LRN).Thut ton LRN ( Tree *T) {

if (T!= NULL ) {LRN (T -> left;

LRN ( T -> right);;

}}V d : LNR(T) =6, 12, 5, 3, 8, 1, 4, 2, 7, 9, 10.

10

8 9

5 3 2 7

1 46 12




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Thutton LRN davo stack:

Thutton LRN (Tree *T) {Bc1 (Khi to): if (T = ) return; //nu T l rng

Stack =

; Node = T; //tostack rngv Node tr n gcT

Bc2 ( Lp):do { //bt u lp

while(Node) { / /lp nkhi node l rngif (Node -> right ) //nunodephikhc rng

Push(Stack, Node->right); /lanodephivo stack

Push(Stack, Node); /lanode vo stackNode = Node ->left; //node trsang bn tri}Node = Pop(Stack); //anode ra khistack;if (Node->right && Top(Stack) == Noot->right){

Pop(stack); // loinode ny khistack

Push(Stack, Node); // anode vo stackNode = Node->right; // Trsang nodephi

}else { ; Node = NULL; }

} while (Stack));}




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1 9, 10, 3, 8, 12, 5,6

2 9, 10, 3, 8, 12, 5 6

3 9, 10, 3, 8, 12, 5 6

4 9, 10, 3, 8, 5, 12 6

5 9, 10, 3, 8, 5 6, 12

6 9, 10, 3, 8 6, 12, 5

7 9, 10, 8, 3 6, 12, 5

8 9, 10, 8 6, 12, 5, 3

9 9, 10 6, 12, 5, 3, 8

10 10, 9 6, 12, 5, 3, 8

11 10, 9, 4, 2,1 6, 12, 5, 3, 8

12 10, 9, 4, 2 6, 12, 5, 3, 8, 1

13 10, 9, 2, 4 6, 12, 5, 3, 8, 1

14 10, 9, 2 6, 12, 5, 3, 8, 1, 4

15 10, 9 6, 12, 5, 3, 8, 1, 4, 2

16 10, 9, 7 6, 12, 5, 3, 8, 1, 4, 2

17 10, 9 6, 12, 5, 3, 8, 1, 4, 2, 7

18 10 6, 12, 5, 3, 8, 1, 4, 2, 7, 9

19 6, 12, 5, 3, 8, 1, 4, 2, 7, 9, 10

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