c1(1,1)

c11(1.5,1.5)

c12(1.5,0.5)c13(0.5,0.5)

c14(0.5,1.5)

c2(1,-1)

c21(1.5,-0.5)

c22(1.5,-1.5)c23(0.5,-1.5)

c24(0.5,-0.5)

c3(-1,-1)

c31(-0.5,-0.5)

c32(-0.5,-1.5)c33(-1.5,-1.5)

c34(-1.5,-0.5)

c4(-1,1)

c41(-0.5,1.5)

c42(-0.5,0.5)c43(-1.5,0.5)

c44(-1.5,1.5)

c1 c2 c3 c4

c11 c12 c13 c14c21 c22 c23 c24

c31 c32 c33 c34c41 c42 c43 c44

root

I, ={ C-

of,i. s )

,

(

1.5¥)

}We are given two ' map Using the vocabulary

±,

.ES#,c.o.5,

- i. ⇒ I,

and Izthe ' map-

tree find The

# T.TL#IpdtIsiLaIL?|normalized difference

✓ - between I,

and -7-

@A←=a

a {Thing@ A-

t '

-

t 'IIB} {...$¢

D-

- - -

+1

+ ,Lt's

C)@TO find the normalized differencewe

represent each I,

and I-

using zi dimensional vectors oh.

and oti s(q, ,gD= ) q ,

- of,)=

666

of ,= ( 2

,0,1,

71,999*9999994%9? )

We ignorethenormalisation

on = ( z,

.it#9E9oiM9'' 999499 t.ITT.TL?sItetEEdefs.

.

Vocabulary tree represents a

hierarchical clustering d the

descriptors . The cluster✓ centers

have the Same dimension as The

descriptors .

Given a descriptor how do youparse the tree

.

Let d.= ( a ,b ) and dz=Gd )

The Similarity between d. and d-

Can be computed usingthe angle

between The vectors

Cos O =d

,

.dz-

If d,

and de✓ ¥2 Jade

arethe

Same = a .cc +td

then theyhave -

thehshett ftp.d

\ = varies from -1

Similarity I

Ifyou

See the Same vector

as a cluster Center Then it

has the man .m -m

Similarity