2. Why Do We Need Compression?
To save
⢠Memory
⢠Bandwidth
⢠Cost
3. How Can We Compress?
⢠Coding redundancy
â Neighboring pixels are not independent but
correlated
⢠Interpixel redundancy
⢠Psychovisual redundancy
4. Information vs Data
REDUNDANTDAT
A
INFORMATION
DATA = INFORMATION + REDUNDANT DATA
6. Overview of SVD
⢠The purpose of (SVD) is to factor matrix A into
T
USV .
⢠U and V are orthonormal matrices.
⢠S is a diagonal matrix
⢠. The singular values Ď1 > ¡ ¡ ¡ > Ďn > 0 appear in
descending order along the main diagonal of S.
The numbers Ď12¡ ¡ ¡ > Ďn2 are the eigenvalues of
T T
AA and A A.
T
A= USV
7. Procedure to find SVD
⢠Step 1:Calculate AAT and ATA.
⢠Step 2: Eigenvalues and S.
⢠Step 3: Finding U.
⢠Step 4: Finding V.
⢠Step 5: The complete SVD.
15. SVD Compression
How SVD can compress any form of data.
⢠SVD takes a matrix, square or non-
square, and divides it into two
orthogonal matrices and a diagonal
matrix.
⢠This allows us to rewrite our original
matrix as a sum of much simpler rank
one matrices.
16.
17. ⢠Since Ď1 > ¡ ¡ ¡ > Ďn > 0 , the first term of
this series will have the largest impact on
the total sum, followed by the second
term, then the third term, etc.
⢠This means we can approximate the
matrix A by adding only the first few
terms of the series!
⢠As k increases, the image quality
increases, but so too does the amount of
memory needed to store the image. This
means smaller ranked SVD
approximations are preferable.
18. If we are going to increase the rank then we can improve the quality of the image
and also the memory used is also high
19. SVD vs Memory
⢠Non-compressed image, I, requires
With rank k approximation of I,
⢠Originally U is an mĂm matrix, but we
only want the first k columns. Then UM
= mk.
⢠similarly VM = nk.
AM = UM+ VM+â M
AM = mk + nk + k
AM = k(m + n + 1)
20. Limitations
⢠There are important limits on k for which
SVD actually saves memory.
AM â¤IM
k(m + n + 1) < mn
k <mn/(m+n+1)
⢠The same rule for k applies to color images.
⢠In the case of color IM =3mn. While
AM =3k(m+n+1)
AM â¤IM
â 3k(m+n+1) < 3mn
Thus, k <mn/(m+n+1)