Hog processing

Post on 05-Jul-2015

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Dx = (−1 , 0 , 1)

Dy = (−1 , 0 , 1)T

I

Ix = I ∗Dx

Iy = I ∗Dy

| G |=�

I2x

+ I2y

θ = arctan( Iy

Ix)

[0 , 40 )

[40 , 80 )

[80 , 120 )

v � v �k ε

v → v/(� v �1 +ε)

v → v/

�� v �2

2 +ε2

(4× 8)× (2× 2)× 9 = 1152

{xk , yk} ∈ χ×{−1 , 1} xk

yk

xk φ

f(x) = w ·φ(x)+ b f(x)

φ(xi) x

f(x)

yi, i = 1, . . . , n

Hi

f̂(y) =1

n(2π)3/2

n�

i=1

| Hi |−1/2t(wi) exp(−D

2 [y, yi, Hi]

2)

D2 [y, yi, Hi] = (y − yi)

�H−1i

(y − yi)

y yi t(wi)

∇f̂(y) =1

n(2π)3/2

n�

i=1

| Hi |−1/2H−1i

(yi − y)t(wi) exp(−D2 [y, yi, Hi]

2) =

1

n(2π)3/2

�n�

i=1

| Hi |−1/2H−1i

yit(wi) exp(−D2 [y, yi, Hi]

2)

1

n(2π)3/2

��n�

i=1

| Hi |−1/2H−1i

t(wi) exp(−D2 [y, yi, Hi]

2)

�y

�i

�i(y) =| Hi |−1/2

t(wi) exp(−D2[y,yi,Hi]

2 )n�

i=1| Hi |−1/2 t(wi) exp(−D2[y,yi,Hi]

2 )

n�i=1

�i = 1

∇f̂(y)

f̂(y)=

n�

i=1

�i(y)H−1i

yi −�

n�

i=1

�i(y)H−1i

�y

H−1h

(y) =n�

i=1

�i(y)H−1i

Hi

m(y) = Hh

∇f̂(y)

f̂(y)≡ Hh(y)

�n�

i=1

�i(y)H−1i

yi

�− y

∇f̂(y) = 0 m(y) = 0

ym = Hh(ym)

�n�

i=1

�i(y)H−1i

yi

yi ym

yi, i =

1, . . . , n ym

Hi Hi diag [Hi]

diag [Hi] =�(exp (si) σx)

2, (exp (si) σy) , (σs)

2�

σx, σy, σs

MissRate = #falseNegatives

#truePositives+#falseNegatives

Precision = #truePositives

#truePositives+#falsePositives

#falseNegatives

#truePositives

#falsePositives

ao bp

bgt

ao =area (bp ∩ bgt)

area (bp ∪ bgt)

ao

area (bp ∩ bgt) = (396− 305)× (411− 127) = 91× 284 = 25844

area (bp ∪ bgt) = (443− 259)× (458− 90) = 184× 368 = 67712

ao = 2584467712 = 0.38

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