MediumNLPnlp-077

Greedy Decoding

Problem

Decode one token per probability row by taking the argmax, breaking ties toward the smallest vocabulary index.

Stop before appending eos when the argmax token is eos, and return the generated tokens. An empty probability list yields an empty result.

Hint

Ties keep the earlier vocabulary index because only strictly greater values replace the best.

case 1

in: [["a","b","</s>"],[[0.1,0.7,0.2],[0.6,0.3,0.1],[0.1,0.1,0.8]],"</s>"]

expected: ["b","a"]

case 2

in: [["a","b","</s>"],[[0.1,0.2,0.7]],"</s>"]

expected: []

case 3

in: [["a","b","</s>"],[[0.5,0.5,0],[0.2,0.6,0.2]],"</s>"]

expected: ["a","b"]

case 4

in: [["x","y"],[],"z"]

expected: []

Starter codePython

def greedy_decode(vocab, probs, eos):
    # Your code here
    pass

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