CS654: Digital Image Analysis

Lecture 12: Separable Transforms

Recap of Lecture 11

• Image Transforms

• Source and target domain

• Unitary transform, 1-D

• Unitary transform, 2-D

• High computational complexity

Outline of Lecture 12

• Unitary transforms

• Separable functions

• Properties of unitary transforms

Image transforms

• Operation to change the default representation space of a digital image (source domain target domain)

• All the information present in the image is preserved in the transformed domain, but represented differently;

• The transform is reversible

• Source domain = spatial domain and target domain= frequency domain

Unitary transform

1}Nn0{u(n), 1-D input sequence

1N

0n1Nk0,n)u(n)a(k,v(k)ro,Auv

If Transformed sequence

1N

0k

*T* 1Nn0,n)v(k)(k,au(n)orvAu

2-D sequence

v(k,l) a (m,n) u(m,n) , 0 k,l N 1k,ln 0

N 1

m 0

N 1

=

[𝑎𝑘𝑙 (0 ,0 ) ⋯ 𝑎𝑘𝑙 (0 ,𝑛 ) ⋯ 𝑎𝑘𝑙 (0 ,𝑁−1 )

⋮ ⋯ ⋯ ⋯ ⋮𝑎𝑘𝑙 (𝑚 ,0 ) ⋯ 𝑎𝑘𝑙 (𝑚 ,𝑛) ⋯ 𝑎𝑘𝑙(𝑚 ,𝑁−1)

⋮ … … … ⋮𝑎𝑘𝑙(𝑁−1,0) … 𝑎𝑘𝑙(𝑁−1,𝑛) … 𝑎𝑘𝑙(𝑁−1 ,𝑁−1)

]×[

𝑢 (0 ,0 ) ⋯ 𝑢 (0 ,𝑛) ⋯ 𝑢 (0 ,𝑁−1 )⋮ ⋯ ⋯ ⋯ ⋮

𝑢 (𝑚 ,0 ) ⋯ 𝑢(𝑚 ,𝑛) … 𝑢 (𝑚 ,𝑁−1)⋮ … … … ⋮

𝑢(𝑁−1,0) … 𝑢(𝑁−1 ,𝑙) … 𝑢(𝑁−1 ,𝑁−1)]

High computational complexity

O(N4)

Separable Transformations

• We like to design a transformation such that

𝑎𝑘 ,𝑙 (𝑚 ,𝑛)=𝑎𝑘 (𝑚 )𝑏𝑙 (𝑛 )=𝑎 (𝑘 ,𝑚 )𝑏( 𝑙 ,𝑛)

{𝑎𝑘 (𝑚 ) ,𝑘=0 ,…,𝑁−1}

{𝑏𝑙 (𝑛) , 𝑙=0 ,…,𝑁−1 }

𝐴={𝑎(𝑘 ,𝑚)} 𝐵={𝑏 (𝑙 ,𝑛)}

Let there be two sets

1-D complete orthonormal basis vectors

Separable Transformations

𝐴 𝐴∗𝑇=𝐴∗𝑇 𝐴=𝐼

Assumption: the separable matrices be same, then

𝐵𝐵∗𝑇=𝐵∗𝑇 𝐵=𝐼

𝑣 (𝑘 , 𝑙 )=∑𝑚=0

𝑁−1

∑𝑛=0

𝑁− 1

𝑎 (𝑘 ,𝑚 )𝑢 (𝑚 ,𝑛)𝑎(𝑙 ,𝑛)

What would be v in matrix notation?

𝑽=𝑨𝑼 𝑨𝑻

Reverse transformations

1N

0k

1N

0l

** ron),(l,al)v(k,m)(k,an)u(m, *T* VAAU

For non-square matrices

V A UAM N

U A VAM*T

N*T

V AUA , V A[AU]T T T

=

[𝑎 (0 ,0 ) ⋯ 𝑎 (0 ,𝑚 ) ⋯ 𝑎 (0 ,𝑁−1 )

⋮ ⋯ ⋯ ⋯ ⋮𝑎 (𝑘 ,0 ) ⋯ 𝑎(𝑘 ,𝑚) ⋯ 𝑎(𝑘 ,𝑁−1)

⋮ … … … ⋮𝑎(𝑁−1,0) … 𝑎(𝑁−1 ,𝑚) … 𝑎(𝑁−1 ,𝑁−1)

]×[

𝑢 (0 ,0 ) ⋯ 𝑢 (0 ,𝑛) ⋯ 𝑢 (0 ,𝑁−1 )⋮ ⋯ ⋯ ⋯ ⋮

𝑢 (𝑚 ,0 ) ⋯ 𝑢(𝑚 ,𝑛) … 𝑢 (𝑚 ,𝑁−1)⋮ … … … ⋮

𝑢(𝑁−1,0) … 𝑢(𝑁−1 ,𝑙) … 𝑢(𝑁−1 ,𝑁−1)]×

[𝑎 (0 ,0 ) ⋯ 𝑎 (0 ,𝑛) ⋯ 𝑎 (0 ,𝑁−1 )

⋮ ⋯ ⋯ ⋯ ⋮𝑎 (𝑙 ,0 ) ⋯ 𝑎(𝑙 ,𝑛) ⋯ 𝑎(𝑙 ,𝑁−1)

⋮ … … … ⋮𝑎(𝑁−1,0) … 𝑎(𝑁−1 ,𝑛) … 𝑎(𝑁−1 ,𝑁−1)

]

Computational complexity

O(N3)

Example

𝐴=12 [√ 3 1−1 √3 ] 𝑈=[2 3

1 2]𝑉=𝐴𝑈 𝐴𝑇

𝑉=12 [√ 3 1−1 √3 ][2 3

1 2] 12 [√ 3 −11 √3 ]

𝑉=[2+√3 20 2−√ 3]

Inverse transforms

𝑉=[2+√3 20 2−√ 3] 𝐴∗𝑇=1

2 [√ 3 −11 √3 ]

𝑢 (1,0 )=∑𝑘=0

1

∑𝑙=0

1

𝑎∗ (𝑘 ,1 )𝑣 (𝑘 , 𝑙 )𝑎∗( 𝑙 ,0)

1N

0k

1N

0l

** ron),(l,al)v(k,m)(k,an)u(m, *T* VAAU

Kronecker Products

• Kronecker Product

Arbitrary 1-D transformation

This will be separable if𝒜=𝐴1⨂ 𝐴2

= Kronecker Product

It is a generalization of the outer product

Kronecker Products

𝐴⨂𝐵=[ 𝑎0 , 1𝐵 … 𝑎0 ,𝑀2𝐵

⋮ ⋱ ⋮𝑎𝑀1−1,0

𝐵 … 𝑎𝑀 1−1 ,𝑀 2−1𝐵]

If and are and matrices then Kronecker product of and is defined as

block matrix of dimension

If

Computational complexity?? Fast image transforms

Basis Images

column of

Outer product

Inner product

⟨ 𝐹 ,𝐺 ⟩=∑𝑚=0

𝑁 −1

∑𝑛=0

𝑁−1

𝑓 (𝑚 ,𝑛 )𝑔∗(𝑚 ,𝑛)

Basis Images

Imagine originala

=

=

V(1,3)

+ + +

+ + + + + +

+ + + + … +

V(1,5) V(1,7) V(1,9)

V(1,13) V(1,15) V(2,1) V(2,9) V(3,1) V(3,5)

V(5,1) V(5,2) V(5,6) V(5,8) V(16,15)

Imagine aproximata

Keeping only 50% of coefficients

Thank youNext Lecture: Discrete Fourier Transform