Polar SC Decode Step

Problem

Perform one 2-bit successive-cancellation decision for a polar code. Decide u0 = 1 when llr0 + llr1 < 0, otherwise 0. Take the hard decision of llr1 and set u1 = hard(llr1) XOR u0. Return [u0, u1].

Hint

u0 is decided first from the combined LLR, then it corrects the hard decision of u1.

Test cases

case 1

in: [1,1]

expected: [0,0]

case 2

in: [-1,-1]

expected: [1,0]

case 3

in: [-1,1]

expected: [0,0]

case 4

in: [1,-1]

expected: [0,1]

case 5

in: [-2,-0.5]

expected: [1,0]

Starter codePython

def polar_sc_step(llr0, llr1):
    # Your code here
    pass

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