Cay Dinh Danh

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Contents

Tng quan v khoa hc tr tu nhn to1

Cc phng php gii quyt vn c bn2

Tri thc v cc phng php biu din tri thc3

My hc4

Mng Nron5


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Chng 4

MY HC

4.1

4.2

TH NO L MY HC

HC BNG CCH XY DNG CY NH DANH


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4.1 TH NO L MY HC

Thutng"hc" l tip thu tri thc.

Qu trnh hcdin ra dinhiu hnh thcnh:

hcthuc lng (hcvt),

hc theo kinh nghim(hcda theo trnghp),

hc theo kiu nghe nhn,...


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4.1 TH NO L MY HC

Kho st phng php hcda theo trnghp.

h thngc cung cpmtscc trnghp"mu",

da trn tp mu h thng s tin hnh phn

tch v rt ra cc quy lut.

Sau ,hthngda trn cc lut ny "nhgi" cc trnghp khc.


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4.1 TH NO L MY HC

C th khi qut qu trnh hc theo trnghp di dng hnh thc nh sau:

f : P R

p | r


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4.1 TH NO L MY HC

Tuy nhin, tp P nh (hu hn) so vi tp tt ccc trng hp cn quan tm P (P P).

Mc tiu l xy dng nh x f sao cho c thng mi trng hp p trong tp P vi mt "lp"r trong tp R.

Hn na, f phi bo ton f, ngha l :

Vi mi p P th f(p) f (p)


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4.1 TH NO L MY HC

Hc theo trng hp l tm cch xy dng nh x fda theo nh x f. f c gi l tp mu.


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4.2.1

4.2.2

4.2.3

4.2.4

m chiPhng n chn thuc tnh phn hochPht sinh tp lut

Ti u tp lut

4.2 HC BNG CCH XY DNG CY NH DANH


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Xt mt v d: Cn xy dng cc quy lutktlunmtnginhthno khi

itmbin th b chy nng.

Ta gi tnh cht chy nng hay khng chy nng l thuc tnhquan tm (thuc tnh mc tiu).

Trong trnghp ny,

tp Rgm c hai phnt{"chy nng", "bnh thng"}.

tp P l ttcnhngngiclit k trong bng (8 ngi)

Hintng chy nngda trn 4 thuc tnh sau:

chiu cao (cao, trung bnh, thp),

mu tc (vng, nu, ) cn nng(nh, TB, nng),

dng kem (c, khng).

Ta gi cc thuc tnh ny l thuc tnh dnxut.


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Tn Tc Ch.Cao Cn Nng Dng kem Kt quSarah Vng T.Bnh Nh Khng Chy

Dana Vng Cao T.Bnh C Khng

Alex Nu Thp T.Bnh C KhngAnnie Vng Thp T.Bnh Khng Chy

Emilie T.Bnh Nng Khng Chy

Peter Nu Cao Nng Khng Khng

John Nu T.Bnh Nng Khng Khng

Kartie Vng Thp Nh C Khng


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tngPhn hoch tp P thnh cc tp Pi sao cho tt c

cc phn t trong cc tp Piu c chung thuctnh mc tiu.

P = P1 P2 ... Pn v (i,j) ij: th (Pi Pj = )

v

i, k,l: pk Pi v pl Pj th f(pk) = f(pl)


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Cc phn hoch Pic c trng bi thuc tnhch ri (ri R),

ng vi mi phn hoch Pita xy dng lut Li:GTi ritrong cc GTil mnh c hnhthnh bng cch kt hp cc thuc tnh dnxut.


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C hai cch phn hochhin nhin: Cch u tin l cho mi ngi vo mt phn

hoch ring (P1 = {Sarah}, P2 = {Dana}, tngcngs c 8 phn hoch cho 8 ngi).

Cch th hai l phn hoch thnh hai tp, mttp gm tt c nhng ngi chy nng v tp

cn ligmttcnhngngi khng chy nng.


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Phng php khc. quan st thuc tnh mutc. c 3 phn hoch:

Pvng = { Sarah, Dana, Annie, Kartie }

Pnu = { Alex, Peter, John }

P = { Emmile }

4.2.1. m chi


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Dng s cy m t phn hoch:

4.2.1. m chi


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Nhn xt:

Tp Pnu cha ton ngi khng chy nng,

Tp P cha ton ngi chy nng,

Tp Pvng cha ln ln ngi chy nng v khngchy nng.

Tip tc phn hoch tp Pvng thnh 3 tp con:

PVng, Thp = {Annie, Kartie}, PVng, T.Bnh= {Sarah} v

PVng,Cao= { Dana }

4.2.1. m chi


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4.2.1. m chi


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Qu trnh ny c th tip tc cho n khi tt ccc nt l ca cy khng cn ln ln gia chynng v khng chy nng.

qu trnh ny c gi l qu trnh "m chi".Cy m chng ta ang xy dng c gi l cynh danh.

Nu ban u ta khng chn thuc tnh mu tc phn hoch m chn thuc tnh khc nhchiu cao chng hn phn hoch th sao?Cui cng th cch phn hoch no s tt hn?

4.2.1. m chi


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4.2.1

4.2.2

4.2.3

4.2.4


Ti u tp lut



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"ng trc mt ng r, ta cn phi i vohng no?".

Hai phng php nh gi diys gip tachnc thuc tnh phn hoch timibc

xy dng cy nh danh.

4.2.2. Phng n chn thuc tnh phn hoch


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a.Quinlan

Quinlan quyt nh thuc tnh phn hoch bngcch xy dng cc vector c trng cho mi gi

tr ca tng thuc tnh dn xut v thuc tnhmc tiu. c th nh sau:

Vi mi thuc tnh dn xut A cn c th s dng phn hoch, tnh:

VA(j) = ( T(j , r1), T(j , r2) , , T(j , rn))



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T(j, ri) = (tng s phn t trong phn hoch cgi tr thuc tnh dn xut A l j v c gi trthuc tnh mc tiu l ri) / (tng s phn ttrong phn hoch c gi tr thuc tnh dn xutA l j)

Trong r1, r2, , rnl cc gi tr ca thuctnh mc tiu



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mt thuc tnh A c th nhn mt trong 5 gi trkhc nhau th n s c 5 vector c trng.

Mt vector V(Aj) c gi l vector n v nu

n ch c duy nht mt thnh phn c gi tr 1v nhng thnh phn khc c gi tr 0.

Thuc tnh c chn phn hoch l thuc

tnh c nhiu vector n v nht.



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Thuc tnh mu tc c 3 gi tr khc nhau (vng, , nu)nn s c 3 vector c trng tng ng l: VTc(vng)= (T(vng, chy nng), T(vng, khng chy nng))

S ngi tc vng l: 4

S ngi tc vng v chy nng l: 2

S ngi tc vng v khng chy nng l: 2

Do : VTc(vng) = (2/4 , 2/4) = (0.5, 0.5)

Tng t

VTc(nu)= (0/3, 3/3) = (0,1) (vector n v)

S ngi tc nu l: 3 S ngi tc nu v chy nng l: 0

S ngi tc nu v khng chy nng l: 3

VTc()= (1/1, 0/1) = (1,0) (vector n v)

Tng s vector n v ca thuc tnh tc vng l 2



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Cc thuc tnh khc c tnh tng t, kt qu nh sau: VC.Cao(Cao) = (0/2,2/2) = (0,1)

VC.Cao(T.B) = (2/3,1/3)

VC.Cao(Thp) = (1/3,2/3)

VC.Nng (Nh) = (1/2,1/2)

VC.Nng (T.B) = (1/3,2/3)

VC.Nng (Nng) = (1/3,2/3)

VKem (C) = (3/3,0/3) = (1,0) VKem (Khng) = (3/5,2/5)

thuc tnh mu tc c s vector n v nhiu nht nnc chn phn hoch.



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tiptc phn hochtp Pvng.

tnh vector ctrngivi cc thuc tnh cnli(chiu cao, cn nng, dng kem). tpdliu

cn li l:



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Tn Ch.Cao Cn Nng Dng kem Kt qu

Sarah T.Bnh Nh Khng Chy

Dana Cao T.Bnh C Khng

Annie Thp T.Bnh Khng Chy

Kartie

Thp

Nh

C

Khng


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VC.Cao(Cao) = (0/1,1/1) = (0,1)

VC.Cao(T.B) = (1/1,0/1) = (1,0)

VC.Cao(Thp) = (1/2,1/2)

VC.Nng (Nh) = (1/2,1/2)

VC.Nng (T.B) = (1/2,1/2)

VC.Nng (Nng) = (0,0)

VKem (C) = (0/2,2/2) = (0,1)

VKem (Khng) = (2/2,0/2) = (1,0)



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2 thuc tnh dng kem v chiu cao u c 2vector nv.

Tuy nhin, sphn hochca thuc tnh dng

kem l t hn nn chn phn hoch theo thuctnh dng kem.

Cy nh danh cui cng nhsau:



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b. o hn lonng vi mi thuc tnh dn xut ta cn tnh

o hn lon v la chn thuc tnh no c ohn loi thp nht. Cng thc tnh nh sau:


TA =

trong :

bt: tng s phn t c trong phn hochbj: tng s phn t c thuc tnh dn xut A c gi tr j.

bri: tng s phn t c thuc tnh dn xut A c gi tr j vthuc tnh mc tiu c gi tr i.


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4.2.1

4.2.2

4.2.3

4.2.4


Ti u tp lut



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Nguyn tc pht sinh tp lut:ng vi mi nt l, i t nh cho n nt l

v pht sinh ra lut tng ng.

C th l t cy nh danh kt qu trn ta ccc lut sau (xt cc nt l t tri sang phi)

(Mu tc vng) v (c dng kem) khng chy nng

(Mu tc vng) v (khng dng kem) chy nng (Mu tc nu) khng chy nng

(Mu tc ) chy nng

4.2.3. Pht sinh tp lut


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4.2.1

4.2.2

4.2.3

4.2.4


Ti u tp lut



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a.Loi b mnh thaphng php loi b mnh tha da vo d

liu.Vi v d v tp lut c phn trc, hy

quan st lut sau: (Mu tc vng) v (c dng kem) khng chy nng

4.2.4. Ti u tp lut


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lpbng Contigency, thng k nhngngi cdng kem tng ng vi tc mu vng v bchy nng hay khng. Trong dliu cho, c3 ngi khng dng kem.

4.2.4. Ti u tp lut

Khng chy nng Chy nng

Mu vng 2 0

Mu khc 1 0


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Theo bng thng k: thuc tnh tc vng khng kt lunchy nng hay khng (c 3 ngi dng kem u khngchy nng) nn loi b thuc tnh tc vng ra khi tplut.

Sau khi loi b mnh tha, tp mnh trong v dtrn s cn:

(c dng kem) khng chy nng

(Mu tc vng) v (khng dng kem) chy nng

(Mu tc nu) khng chy nng

(Mu tc ) chy nng

4.2.4. Ti u tp lut


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gi s lut ca chng ta c n mnh :

A1 v A2v v An R

kim tra xem c th loi b mnh Ai hay

khng, hy lp ra mt tp hp P bao gm ccphn t tha tt c mnh A1 , A2, Ai, Ai+1,, An(lu : khng cn xt l c tha Ai haykhng, ch cn tha cc mnh cn li lc)

4.2.4. Ti u tp lut


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Sau ,bn hy lpbng Contigency nhsau:

4.2.4. Ti u tp lut

R R

Ai E F

Ai G H

Trong :

E l s phn t trong P tha c Ai v R.

F l s phn t trong P tha Aiv khng tha RG l s phn t trong P khng tha Aiv tha R

H l s phn t trong P khng tha Aiv khng tha R

Nu tng F+H = 0 th c th loi b mnh Aira khi lut.


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b.Xy dng mnh mc nhmt vn t ra l:

khi gp phi mt trng hp m tt c cc lut

u khng tha th phi lm nh th no? Mtcch hnh ng l t ra mt lut mc nh iloi nh:

Nu khng c lut no tha chy nng (1)Hoc

Nu khng c lut no tha khng chy nng.(2)

4.2.4. Ti u tp lut


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Gi s ta chn lut mc nh l (2) th tp lutca chng ta s tr thnh:

(Mu tc vng) v (khng dng kem) chy nng

(Mu tc ) chy nng Nu khng c lut no tha khng chy nng. (2)

4.2.4. Ti u tp lut


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Mt s quy tc khi chn lut mc nh:

(1) Chn lut mc nh sao cho n c th thay th chonhiu lut nht.

(2) Chn lut mc nh c kt lun ph bin nht. (3) Chn lut mc nh sao cho tng s mnh ca

cc lut m n thay th l nhiu nht.

4.2.4. Ti u tp lut


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TRNG CAO NG CNTT HU NGH ViT - HN


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TRNG CAO NG CNTT HU NGH ViT HN

KHOA KHOA HC MY TNH-----------***-----------

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