MXFP4 Block Quantization Error

Problem

Estimate the RMS error of MXFP4 block quantization. Elements use the E2M1 value set {0, 0.5, 1, 1.5, 2, 3, 4, 6} with a shared power-of-two scale 2^(floor(log2(amax)) - 2). Round each element to the nearest level (ties to the smaller), dequantize with the sign, and return the root-mean-square error; an all-zero block returns 0.0.

The signature is mxfp4_block_error(values).

Hint

The shared exponent is chosen from the block amax; the largest E2M1 level is 6.

Test cases

case 1

in: [[1,-1,0.5,-0.5]]

expected: 0

case 2

in: [[1.5,-1.5,0,3]]

expected: 0

case 3

in: [[1.7,0]]

expected: 0.14142135623730948

case 4

in: [[0,0]]

expected: 0

case 5

in: [[5]]

expected: 1

Starter codePython

import math
def mxfp4_block_error(values):
    # Your code here
    pass

5 test cases · run your code against them in the editor