Integer Transform
●
What Integer Transform
–
Definition and Property
–
Verifying Fixed-point Number
–
Integer Transform
●
Goals to Meet
●
Some Problems of Integer Transform
●
Prototype Method
●
Lifting Scheme
●
Triangular Scheme
Fixed-point number
y=∑k a k 2 where a k ∈{0,1 }, k ∈ℤ
k
For example:
5
4
3
1
0
−3
59.125=2 2 2 2 2 2
k
ak
5
1
4
1
3
1
2
0
1
1
0
1
-1
0
Radix point
-2
0
-3
1
Verifying fixed-point number
5
4
3
1
−3
0
59.125=2 2 2 2 2 2
Radix point
k
ak
8
1
7
1
k
ak
5
1
4
1
3
1
2
0
1
1
0
1
6
1
5
0
4
1
3
1
2
0
1
0
0
1
-1
0
-2
0
Radix point
3
8
7
6
4
3
0
473=2 ×59.125=2 2 2 2 2 2
k
a
2
∑k k
, where k ∈{0,1,2,. .. }, for some k 0
k
2
0
-3
1
Integer transform
●
●
The entries of the transform matrix Aij can be
represented by fixed-point number.
Property:
–
k
v
=
a
2
If the entries of the input j ∑k jk for some a jk ,
k
then Av=∑k a jk 2 for some a jk
[
∑k a k 2
11
∑k a k 2
21
k
k
∑k a k 2
12
∑k a k 2
22
k
k
][
][
∑k
∑k
=
21 k
21 k
x
2
y
∑k k
∑k k 2
11
xk 2
k
11
yk 2
k
]
Goals to meet
I. If x is fixed-point number, that y= A x is also
fixed-point number
−1
II.Reversibility A Ax= x
III.High accuracy
Some Problems of Integer
Transform
[ ] [ ]
2
1
If A=10, A = !
10
−1
If A=
1
2
1
2
−1
,A =
1
2
2
3
−2
3
k
det  A=2 ∧ Ais integer transform ,
−1
p
−1
⇔ det  A =2 ∧ A is also integer transform.
1
A =
adj  A
det  A
−1
−2
3
!
8
3
Prototype Method
J=
¿
Implementation
Lifting Scheme
Lifting Scheme
[ ] [ ][ ][ ] [ ] [
y0 1
=
y1 0
A1 1 0 x 0
1 A0 1 x 1
][
][ ]
x0
1 −A1 y 0
1
0
=
x 1 − A0 1 0
1 y1
Lifting Scheme
[
[
1 1
1 −1
]
1 1
1 −1
−1
]
[ ]
1
1
2 2
=
1
1
−
2
2
[

4

4

4
−cos   sin  

sin  4 
cos 
]
[
cos 
3
8
3
8
−sin 

sin 
 cos 
3
8
3
8


]
Lifting Scheme
Case 1
][
[
cos  −sin  1
T =
=
sin  cos
0
][
][
cos−1
1
0 1
sin 
sin  1
1
0
Case 2
[
][
][ ]
cos  sin 
cos  −sin  1 0
T =
=
sin  −cos sin  cos 0 −1
[
=
1
0
][
][
cos −1
−cos−1
1
0 1
sin 
sin 
sin  1
1
0
−1
]
cos −1
sin 
1
]
Lifting Scheme
Lifting Scheme
Lifting Scheme
abi cdi =acbd i
abi ×cdi =ac−bd ad bci
[
cos  −sin 
sin  cos
]
[
cos  sin 
−sin  cos
]
Lifting Scheme
Triangular Scheme
[
[
[
1 a b 1 0 0
0 1 c 0 1 0
0 0 1 0 0 1
]
1 a 0 1 0 −b
0 1 0 0 1 −c
0 0 1 0 0 1
]
1 0 0 1 −a ac−b
0 1 0 0 1
−c
0 0 1 0 0
1
]
[
[
[
1 0 0 1 0 0
a 1 0 0 1 0
b c 1 0 0 1
]
1 0 0 1 0 0
0 1 0 −a 1 0
0 c 1 −b 0 1
]
1 0 0
1
0 0
0 1 0 −a
1 0
0 0 1 ac−b −c 1
]
Triangular Scheme
Assume∣det  A∣=1, choose permutation P∧Q such that R=PAQ
[
][
][
][
]
r 11 r 12 r 13 1 s12 s13 1 0 0 1 u12 u 13
R= r 21 r 22 r 23 = 0 1 0 l 21 1 0 0 1 u 23 =T 1 T 2 T 3
r 31 r 32 r 33 0 0 1 l 31 l 32 1 0 0 K
K =det  A
Triangular Scheme
Algorithm I
A
b1
e1
h1
[ ][
[ ][
[ ][
[ ][
a1
d1
g1
c1
f1
i1
S1
1 b1
1 0 0
0 1 0 = j1 e1
s1 0 1
k 1 h1
L1
1 0 0 1 b1
l 1 1 0 j 1 e1
l 2 0 1 k 1 h1
1 b1
0 m1
0 o1
c1
n1
p1
][
][ ]
][ ]
][ ]
c1
1 b1 c 1
f 1 = 0 m 1 n1
i1
0 o1 p 1
S2
1 0 0 1 b1
0 1 0=0 1
0 s2 1
0 q1
L2
1 0 0 1 b1
0 1 0 0 1
0 l 3 1 0 q1
c1
f1
i1
c1
n1
p1
U
c1
1 b1 c 1
n1 = 0 1 n1
p1 0 0 r 1
]
s1=
1−a 1
c1
l 1=− j 1, l 2 =−k 1
L 2 L 1 A S 1 S 2 =U
[ ]
[ ]
1
0 0
Let L = L2 L 1=
l1
1 0
l 2 l 1 l 3 l 3 1
−1
1 0 0
S =S 1 S 2 = 0 1 0
s1 s2 1
−1
Then A= LUS
Triangular Scheme
Algorithm II
a 11
a 21
a 31
A
U
a 12 a 13 1 u 12 u 13
h11 h 12 h 13
a 22 a 23 0 1 u 23 = h21 1
0
a 32 a 33 0 0
1
h31 h 32 1
s11
0
0
L
S
h11 h 12 h 13
s12 s13 1
0 0
0
1 0 h21 1 0 = h21 1
0 1 h31 h 32 1
h31 h 32 1
a 11
a 21
a 31
−1
L
U
A
S
−1
a 12 a 13
s11 s 12 s13 1
0 0 1 u 12 u13
a 22 a 23 = 0 1 0 h 21 1 0 0 1 u 23
0 0 1 h 31 h 32 1 0 0
a 32 a 33
1
[
[
[
][ ] [
][ ] [
] [ ][
]
]
][
a 21 u 12 a 22 =1
a 21 u 13 a 22 u 23 a 23 =0
a 31 u13a 32 u 23 a 33 =1
s13=h13
s12 s13 h 32=h12
s11s12 h 21 s13 h31=h 11
]
Rounding
1
y= x
4
[ ][
][ ]
1 1
y1
x1
= 16 2
y2
x2
0 1
1
1
y 1 = x 1 x 2
16
2
If Q  xa= xQ a∀ x∈ℤ , we call Q is integer−shift invariant
Floor, Ceiling
Rounding
[ ][
][ ]
y 1 1 u12 u 13 x1
y 2 = 0 1 u 23 x2
y3 0 0
1 x3
Forward Transform
y1 =Q  x1u 12 x2 u 13 x3 = x1Q u12 x 2u 13 x3 
y 2=Q  x 2u 23 x3 = x 2Q u 23 x 3 
y 3=Q  x3 = x3
Inverse Transform
x 3= y 3
x 2= y2 −Q u 23 x3 
x 1= y1 −Q u 12 x2 u 13 x3 
Error