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Chng I: Gii thiu bi ton phn cm v khai ph d liuI.Gii thiu

II. Tch tuyn tnh hoc li1.Tch tuyn tnh.2 .Tch ton phng li.

III. Cm theo tm1.Trng hp .||),( 1 = =

n

j jjxaxad

2.Trng hp .||||),(1

2

= =n

j jjxaxad

3.Ti u n iu.IV. Cm theo siu phngV.Phn cm trn mt phng v mt cu

1. Biu Voronoi.2. bi ton k-median lin tc.3.Phn b k im trn hnh cu.

Chng II: tng thut tonI. Cch gii quytII. Trng hp gp phi II. Trng hp gp phi

Chng III: Phn tch v thit kI. Cc bc thc hin I. Cc bc thc hin

II.Xy dng lp.1.Cc thuc tnh ca i tng2.Cc phng thc ca i tng.

III. Gii thiu chng trnh

Chng IV: Kt lunI. Nhn xt v nh gi kt qu

1.S im n=1000; S cm 102.S im n=2000; S cm 203.S im n=3000; S cm 304.Nhn xt

II. Kt lun

Chng I: Gii thiu bi ton phn cm v khai ph d liu

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I.Gii thiuTrong nhiu tnh hung thc t ta gp vn sau y: ta c mt tp hp i tng no v mun chia tp ra thnh mt s cm (nhm), mi cm gm cc phn t gnnhau(ging nhau) tho mt ngha nht nh. Vn ny xut hin t lu, vo nhngnm 60 th k trc mi c ch v nghin cu bng phng php ton hc, bt u

hai lnh vc: phn loi trong sinh hc v nhn dng trong vic dy my. Ngy nayphm vi ng dng ca bi ton chia cm bao gm nhiu lnh vc, t sinh hc, tm l hc,y hc, k thut, cho n kinh t tip th, v.vVi d:1.(Y hc) Bnh nhn l cc d liu vt l v nhng bnh nhn b nghi ung thu v v tamun da v d liu phn cc i tng thnh hai nhm u lnh v u c.2.(Kinh t) i tng l cc d liu ti chnh v nhng doanh nghip ang hot ng vta mun da vo cc d liu y phn cc i tng thnh hai nhm: Doanh nghip trn tht bi v doanh nghip ang lm n bnh thng.C nhiu cch tip cn khc nhau i vi vn nu: thng k, my mc, ti u,trong cch tip cn ti u t ra c hiu qu hn c. Vi bi ton ny chng ta tip

cn bi ton dng ti u ton cc.II. Tch tuyn tnh hoc liBi ton t ra nh sau: cho hai tp A={ raaa ,,, 21 } v B={ rbbb ,,, 21 }, hy tm

mt hm )(xf thuc mt lp hm cho trc F tch hai tp A,B, ngha l sao cho

1)(

1)(

>

c[j].c-c[j].d))) j=i;

return j;}

//tm cnh max cm x sau khi to tm x v nx chia cm x thnh //hai cum tiu nht

void PhanCum::XDTam(int x)

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{

int p,q;

p=0;q=1;

for (int i=c[x].d;i

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for (int i=c[x].d;i

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double PhanCum::Tinh(int x)

{

double E=0;

for (int i=c[x].d;i

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x+=d[j].td.x;

y+=d[j].td.y;

}

if (k!=0){

c[i].td.x=int(x/k);

c[i].td.y=int(y/k);

}

}

}

//tnh li kt qu ti u cho cc cm t 1 dn nx

double PhanCum::Tinh()

{

double E=0;

for (int i=0;i

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DIEM temp=d[i];

d[i]=d[j];

d[j]=temp;

}i=0;j=0;

int k=0;

long x=0,y=0;

do

{

if (d[j+1].c==d[j].c) {

d[j].c=i;

x+=d[j].td.x;

y+=d[j].td.y;

}

else {

d[j].c=i;

x+=d[j].td.x;

y+=d[j].td.y;

c[i].td.x=int(x/(j-k+1));

c[i].td.y=int(y/(j-k+1));

x=y=0;

c[i].d=k;

c[i].c=j+1;

k=j+1;

i++;

}

j++;

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} while (j

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Chng IV: Kt lun

I. Nhn xt v nh gi kt quVi cc d liu ngu nhin:

1.S im n=1000; S cm 10

2.S im n=2000; S cm 20;

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3:S im n=3000; S cm 30;

4.Nhn xtVi s im b tr u th cc cm phn b rt u ging nh cc t ong.

II. Kt lunBi ton phn cm v khai ph d liu l mt bi ton hay v c nhiu cch gii

qut khc nhau, y l mt cch gii quyt mang tnh cht tham lam na quy nhiuhn mong a n mt kt qu ti u nht. Bi ton ngy cng ti u hn sau mibc lp nn n s dng li sau mt s hu han ln.

Cch gii quyt bi ton ny ch mang li kt qu ti u c b nhng n chy

c d liu rt ln. Chng trnh chy n nh. Vi cc d liu nh th n rt chnhxc.Chng ta so th ci tin mt s on chng trnh chng trnh chy nhanh

hn nh vic sp xp chng ta c th dng quicksort hay dng sp xp vun ng...Ta c th p dng bi ton ny vo mt s trng hp ti u v khong cch nh

vic b tr cc trng hc hay bnh vin sao cho dn di li l tin nht v khong cch.