The MUL unit of APOT #19
Description
Hi,
Do you have the specific design of the MUL (Multiplication) unit for APOT quantization?
We know that uniform(Int) quantization or POT quantization are friendly to hardware.
Assume that:
R = real number
S = Scale number
T = quantized number
R1 = S1 * T1
R2 = S2 * T2
Uniform quantization simply adopts the INT MUL unit:
T1 = m
T2 = n
So, we have:
R1 * R2 = (S1 * S2) * (m * n)
For POT:
T1 = 2^m
T2 = 2^n
So, we have:
R1 * R2 = (S1 * S2) * (2^m * 2^n)
= (S1 * S2) * 2^(m + n)
The POT is similar to the only-exponent float MUL.
However, for APOT, I have two questions about the MUL design. There are additive elements in the data.
Assume a 4-bit POT:
The first two bits decoder table:
| 00 | 01 | 10 | 11 |
|---|---|---|---|
| 2^0 | 2^-1 | 2^-3 | 2^-5 |
And the last two bits:
| 00 | 01 | 10 | 11 |
|---|---|---|---|
| 2^0 | 2^-2 | 2^-4 | 2^-6 |
For the first two bits, the decoder table is not continuous: 0, -1, -3, -5.
Q1: How do you efficiently decode the binary code to the APOT, especially in the MUL unit?
Aussume the two number in APOT:
0101: T1 = 2^-1 + 2^-2
1010: T2 = 2^-3 + 2^-4
T1 * T2 = (2^-1 + 2^-2) * (2^-3 + 2^-4)
= (2^-1 * 2^-3) + (2^-1 * 2^-4) + (2^-2 * 2^-3) + (2^-2 * 2^-4)
= 2^-4 + 2^-5 + 2^-5 + 2^-6
Obviously, the calculation has 4x (9x) add operations than POT in 4-bit (6-bit). And the result violates the definition of APOT, which won't have the same additive element in a number, such as 2^-5.
Q2: How do you deal with the complex computation and the subnormal number for APOT?
One direct solution is to convert a float with fake quantization. But is it a violation of the principle of quantization?