WordPiece Tokenizer
Implement a greedy longest-match WordPiece tokenizer.
Given a vocabulary (set of subword tokens) and a word, tokenize the word using greedy longest-match from left to right. Subword tokens after the first piece are prefixed with "##".
Rules:
- Start from the beginning of the word
- Find the longest prefix that exists in the vocabulary
- If found, add it as a token; continue with the remainder (prefix "##" to remaining pieces)
- If no match is found for a character, output [UNK] for the entire word
Input:
- Line 1: space-separated vocabulary tokens
- Line 2: word to tokenize
Output: space-separated tokens
Example:
un ##able ##ing play ##s playing playing
play ##ing
- The vocabulary tokens are
un,##able,##ing,play,##s, andplaying, and the word to tokenize isplaying. - Starting from the beginning of the word, the longest prefix that exists in the vocabulary is
play. - The remainder of the word is
ing, which is found in the vocabulary as##ing, so it is added as the next token. - The final output is the combination of these two tokens:
play ##ing.
Constraints:
- Vocabulary contains lowercase subword tokens (some prefixed with ##)
- Word is a single lowercase string with no spaces
- Output [UNK] if any portion can't be matched
Background Knowledge
The WordPiece Tokenizer is a type of tokenization algorithm used in natural language processing (NLP) to split words into subwords or word pieces. This is particularly useful for handling out-of-vocabulary (OOV) words, where the word is not present in the training vocabulary. The WordPiece Tokenizer uses a greedy approach to find the longest matching subword from a given vocabulary.
The concept of subword tokenization is essential in modern NLP models, especially in language models. It allows the model to represent rare or unseen words as a combination of subwords, enabling better handling of OOV words. The vocabulary in this context refers to a set of predefined subword tokens that the tokenizer can use to split the input words. The tokenizer's goal is to find the most suitable subword tokens that represent the input word.
In the context of this problem, the greedy longest-match approach means that the tokenizer will always choose the longest prefix of the input word that matches a subword token in the vocabulary. If a match is found, the tokenizer will add the matched subword token to the output and continue with the remaining part of the word. If no match is found, the tokenizer will output [UNK] for the entire word, indicating that the word is unknown.
Algorithm/Approach
The algorithm for this problem involves a simple iterative approach:
- Initialize the input word and the vocabulary
- Iterate through the word from left to right, finding the longest matching prefix in the vocabulary
- If a match is found, add the matched token to the output and continue with the remaining part of the word
- If no match is found for a character, output [UNK] for the entire word
This approach is a classic example of a greedy algorithm, where the tokenizer makes the locally optimal choice at each step, hoping to find a global optimum solution.
Step-by-Step Strategy
To implement the solution, follow these steps:
- Read the input vocabulary and store it in a suitable data structure (e.g., a set or list)
- Read the input word and initialize an empty list to store the output tokens
- Iterate through the word from left to right, using a sliding window approach to find the longest matching prefix in the vocabulary
- If a match is found, add the matched token to the output list and update the remaining part of the word
- If no match is found for a character, output [UNK] for the entire word and stop the iteration
- Post-process the output list to prefix subword tokens (except the first one) with "##"
Common Pitfalls
Some common pitfalls to watch out for:
- Not handling the case where no match is found for a character, leading to incorrect output
- Not prefixing subword tokens with "##" correctly
- Not using an efficient data structure to store the vocabulary, leading to slow lookup times
Time & Space Complexity
The expected time complexity for this problem is O(nâ‹…m), where n is the length of the input word and m is the size of the vocabulary. The space complexity is O(n+m), where n is the length of the input word and m is the size of the vocabulary. The actual complexity may vary depending on the implementation details and the choice of data structures.