1. Integer transform and Triangular matrix scheme
Presenter : Shiang-Chih Hua
Advisor : Prof. Jian-jiun Ding

2. Outline Introduction Truncation Lifting scheme
Triangular Matrix Scheme
Improvements of Triangular Matrix Scheme

3. integer & non-integer transform

4. “integer” inputs ←→ “integer” outputs
integers: 1, 2, 100, -3, 1/8 → C x 2^(-b)
𝐴= 𝑎 11 ⋯ 𝑎 1𝑁 ⋮ ⋱ ⋮ 𝑎 𝑁1 ⋯ 𝑎 𝑁𝑁 𝑁×𝑁
𝑎 𝑚𝑛 = 𝐶 2 𝑏 , 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑏,𝐶∈ℤ, ∀ 1≤𝑚,𝑛≤𝑁

5. Advantages of integer transform
Simple
Fast
Accuracy

6. Convert non-integer to integer
JPEG-2000
𝑌= 𝑅+2𝐺+𝐵 4 , 𝐶 𝑏 =𝐵−𝐺, 𝐶 𝑟 =𝑅−𝐺
𝑌 𝐶 𝑏 𝐶 𝑟 = −1 1 1 − 𝑅 𝐺 𝐵

7. Convert non-integer to integer
Walsh transform & DCT
𝐴= −1 −1 1 −1 −1 1 1 −1 1 −1

8. Truncation Rounding function Q, e.g. 𝑥 , 𝑥 , 𝑥+0.5
𝑄 𝑛+𝛼 =𝑛+𝑄 𝛼 ∀𝑛∈ℤ, 𝛼∈ℝ
Forward integer transform 𝑦=𝐴𝑥 → 𝑦=𝑄 𝐴𝑥
Inverse integer transform 𝑧= 𝐴 −1 𝑦 → 𝑧=𝑄 𝐴 −1 𝑦
𝑧=𝑥 ?

9. Truncation 𝑦=𝑄 𝐴𝑥 =𝐴𝑥+𝑒, 𝑒 ∞ ≤0.5
𝑦=𝑄 𝐴𝑥 =𝐴𝑥+𝑒, 𝑒 ∞ ≤0.5
𝑧=𝑄 𝐴 −1 𝑦 =𝑄 𝐴 −1 𝐴𝑥+𝑒 =𝑄 𝑥+ 𝐴 −1 𝑒 =𝑥+𝑄 𝐴 −1 𝑒 =𝑥 ?
∀𝑒, 𝑒 ∞ ≤0.5⇒ 𝐴 −1 𝑒 ∞ ≤0.5
𝐴 −1 ∞ ≜ 𝑚𝑎𝑥 𝐴 −1 𝑒 ∞ 𝑒 ∞ ≤1
𝑧= 𝐴 −1 𝑦+ 𝑒 ⇒ 𝐴 ∞ ≤1

10. Truncation 𝑒= 𝑒 1 𝑒 2 𝑒 3 𝑇 , 𝑒 ∞ = 𝑚𝑎𝑥 𝑒 1 , 𝑒 2 , 𝑒 3
𝑒= 𝑒 1 𝑒 2 𝑒 3 𝑇 , 𝑒 ∞ = 𝑚𝑎𝑥 𝑒 1 , 𝑒 2 , 𝑒 3
𝐴 −1 = 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 ,
𝐴 −1 ∞ = 𝑚𝑎𝑥 𝑎 𝑎 𝑎 13 , 𝑎 𝑎 𝑎 23 , 𝑎 𝑎 𝑎 33

11. Truncation 𝐴𝐵 ∞ = 𝑚𝑎𝑥 𝐴𝐵𝑒 ∞ 𝑒 ∞ = 𝑚𝑎𝑥 𝐴 𝐵𝑒 ∞ 𝐵𝑒 ∞ 𝐵𝑒 ∞ 𝑒 ∞
𝐴𝐵 ∞ = 𝑚𝑎𝑥 𝐴𝐵𝑒 ∞ 𝑒 ∞ = 𝑚𝑎𝑥 𝐴 𝐵𝑒 ∞ 𝐵𝑒 ∞ 𝐵𝑒 ∞ 𝑒 ∞
≤ 𝑚𝑎𝑥 𝐴 𝐵𝑒 ∞ 𝐵𝑒 ∞ 𝑚𝑎𝑥 𝐵𝑒 ∞ 𝑒 ∞ = 𝐴 ∞ 𝐵 ∞
𝐴 ∞ 𝐴 −1 ∞ ≥ 𝐴 𝐴 −1 ∞ =1

12. Truncation
𝑑𝑒𝑡 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 = 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 32
− 𝑎 𝑎 𝑎 32 − 𝑎 𝑎 𝑎 33 − 𝑎 𝑎 𝑎 31
𝑑𝑒𝑡 𝐴 ≤ 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 ≤1
only permutation & -1

13. Truncation Triangular matrices are always reversible
𝐿= 𝑎 1 0 𝑏 𝑐 1 , 𝑦 1 𝑦 2 𝑦 3 = 𝑥 1 𝑄 𝑎 𝑥 1 + 𝑥 2 𝑄 𝑏 𝑥 1 +𝑐 𝑥 2 + 𝑥 3 , 𝑥 1 𝑥 2 𝑥 3 = 𝑦 1 −𝑄 𝑎 𝑥 1 + 𝑦 2 −𝑄 𝑏 𝑥 1 +𝑐 𝑥 2 + 𝑦 3

14. Goal
Given a square matrix A with |det(A)| = 1, find a reversible integer mapping T, such that the error of T(x) – Ax is small
e.g. 𝑇 𝑥 =𝑄 𝑇 𝑚 𝑄 𝑇 𝑚−1 𝑄 …𝑄 𝑇 1 𝑥 …
𝑇 1 , 𝑇 2 ,…, 𝑇 𝑚 are reversible integer matrix

15. Goal Integerization 𝑡 𝑚𝑛 = 𝐶 2 𝑏 , 𝑡 𝑚𝑛 = 𝐶 2 𝑏
Reversibility 𝑇 𝑇 −1 =𝑖𝑑𝑒𝑛𝑡𝑖𝑡𝑦
Accuracy 𝑇𝑥≈𝜎𝐴𝑥 𝐴 −1 ∞ ≤𝜎
Bit Constraint b should be small
Less Complexity

16. Lifting Scheme For 2x2 matrix
Degrees of freedom = 4 entries – 1 constraint (unit det.) = 3
𝐴= 𝑎 𝑏 𝑐 𝑑 = 𝑑−1 𝑏 𝑏 𝑎−1 𝑏 1
𝑡=𝑄 𝑎−1 𝑏 𝑥 1 + 𝑥 2 , 𝑦 1 = 𝑥 1 +𝑄 𝑏𝑡 , 𝑦 2 =𝑄 𝑑−1 𝑏 𝑦 1 +𝑡
𝑥 1
𝑦 1
𝑥 2
𝑦 2
𝑎−1 𝑏
𝑑−1 𝑏 𝑏 𝑡

17. Triangular Matrix Scheme
For NxN matrix
Degrees of freedom = 𝑁 2 −1= 1 2 𝑁 𝑁− 𝑁 𝑁−1 + 𝑁−1
A = LUS
L: lower triangular matrix
U: upper triangular matrix
S: one row lower triangular matrix

18. Triangular Matrix Scheme
𝐴= 𝑎 1,1 𝑎 1,2 ⋯ 𝑎 1,𝑁−1 𝑎 1,𝑁 𝑎 2,1 𝑎 2,2 ⋯ 𝑎 1,𝑁−1 𝑎 2,𝑁 ⋮ ⋮ ⋱ ⋮ ⋮ 𝑎 𝑁−1,1 𝑎 𝑁−1,2 ⋯ 𝑎 𝑁−1,𝑁−1 𝑎 𝑁,𝑁−1 𝑎 𝑁,1 𝑎 𝑁,2 ⋯ 𝑎 𝑁,𝑁−1 𝑎 𝑁,𝑁 , 𝐿= 1 0 ⋯ 0 0 𝑙 2,1 1 ⋯ 0 0 ⋮ ⋮ ⋱ ⋮ ⋮ 𝑙 𝑁−1,1 𝑙 𝑁−1,2 ⋯ 1 0 𝑙 𝑁,1 𝑙 𝑁,2 ⋯ 𝑙 𝑁,𝑁−1 1
𝑈= 1 𝑢 1,2 ⋯ 𝑢 1,𝑁−1 𝑢 1,𝑁 0 1 ⋯ 𝑢 1,𝑁−1 𝑢 2,𝑁 ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 ⋯ 1 𝑢 𝑁,𝑁− ⋯ 0 1 , 𝑆= 1 0 ⋯ ⋯ 0 0 ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 ⋯ 1 0 𝑠 𝑁,1 𝑠 𝑁,2 ⋯ 𝑠 𝑁,𝑁−1 1

19. Advantages Systematic 𝐴 𝑆 −1 =𝐿𝑈 Inverse transform is easy
𝐴=𝐿𝑈𝑆⇒ 𝐴 −1 = 𝑆 −1 𝑈 −1 𝐿 −1
Inverse of integer triangular matrix is still an integer triangular matrix
Fast computation
Time(L & U) = N + 1

20. y1 y2 y3 y4 x1 x2 x3 x4 Q 1

21. Improvements of Triangular Matrix Scheme
Decomposition A = LUS, (x1, x2, …, xN), (y1, y2, …, yN) are not symmetric
𝐴= 𝐷 1 𝑃 1 𝐿𝑈𝑆 𝑃 2 𝐷 2
D1, D2: diagonal matrix with entries = ± 2^k
P, Q: permutation matrix
𝐵= 𝑃 1 𝑇 𝐷 1 −1 𝐴 𝐷 2 −1 𝑃 2 𝑇 =𝐿𝑈𝑆
Adjust A and apply LUS decomposition

22. Improvements of Triangular Matrix Scheme
𝑦= 𝐷 1 𝑃 1 𝑄 𝐿𝑄 𝑈𝑄 𝑆 𝑃 2 𝐷 2 𝑥 = 𝐷 1 𝑃 1 𝐿 𝑈 𝑆 𝑃 2 𝐷 2 𝑥+ 𝑒 1 + 𝑒 2 + 𝑒 3
𝑦=𝐴𝑥+ 𝐷 1 𝑃 1 𝐿𝑈 𝑒 1 +𝐿 𝑒 2 + 𝑒 3
Minimize the norm of L & U
𝐴= 𝐷 1 𝑃 1 𝐿+ Δ 3 𝑈+ Δ 2 𝑆+ Δ 1 𝑃 2 𝐷 2
𝐴≈ 𝐷 1 𝑃 1 𝐿𝑈𝑆+ Δ 3 𝑈𝑆+𝐿 Δ 2 𝑆+𝐿𝑈 Δ 1 𝑃 2 𝐷 2

23. Further Improvements
𝐴= 𝑃 1 𝐿 1 𝑃 2 𝐿 2 𝑃 3 𝐿 3 𝑃 4
𝑁! 4 possibilities

24. Example
𝑌 𝐶 𝑏 𝐶 𝑟 = −0.169 − −0.419 − 𝑅 𝐺 𝐵
𝐴= − − − − − −2 0

25. Best solution?
-10 20 10 𝐴=
1,−5 , 1,−4 , …, 1, 4 → 10,0
No points → 1,0 , 2,0 , …, 9,0 , 11,0 , …, 19,0
0 10 − = 1 0 − −0.1 1
0 0.1 −10 0 = 1 0 − −10 1

26. Best solution?
𝑦 1 𝑦 2 = 𝑥 1 𝑥 2
𝑦 1 =10 𝑥 1 + 𝑥 2 −10𝑄 0.1 𝑥 2
𝑦 2 =𝑄 0.1 𝑥 2
𝑥 1 =𝑄 0.1 𝑦 1
𝑥 2 =10 𝑦 2 + 𝑦 1 −10𝑄 0.1 𝑦 1
Yes

27. Best solution?
𝐴= =
= 1 0 −
0 0.1 − = −10 1
10 10 − = 1 0 − −10 1 No

28. Reference
[1] M. D. Adams, F. Koosentini, R. K. Ward,“Generalized S Transform,” IEEE Trans. Signal Processing, vol. 50, no. 11, pp , Nov
[2] S. C. Pei, J. J. Ding,“Improved Reversible Integer Transform,” Circuits and Systems, ISCAS Proceedings., 2006 IEEE International Symposium on May 2006 Page(s):4 pp
[3] P. Hao, Q. Shi,“Matrix Factorizations for Reversible Integer Mapping,” IEEE Trans. Signal Processing, vol. 49, pp , Oct
[4] S. Oraintara, Y. J. Chen, T. Q. Nguyen,“Integer Fast Fourier Transform,” IEEE Trans. Signal Processing, pp , Mar