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Hi HN! I was looking for a tokenizer that would accurately(!) count tokens in browser, and I couldn't find one. So I thought "how hard can it be", and here we a

Hi HN! I was looking for a tokenizer that would accurately(!) count tokens in browser, and I couldn't find one. So I thought "how hard can it be", and here we are 2 weeks later...

zerojames 3y ago
I love this sentiment! Amazing work!
Solvency 3y ago
For those who would also think the same thing, what're some of the the tldr bulletpoints on why this is more complicated than it'd seem?
- belladoreai 3y ago
  I'll answer with an example. Consider the input string " grabbed". If we wanted to map this string to tokens by greedily going from left to right and choosing tokens from the vocabulary with the strategy of minimizing the number of tokens, our algorithm would be very simple. We would end up with the following tokenization: [17229, 2580] == [" grab", "bed"] Surprisingly, the LLaMA tokenizer does not work this way. It actually finds a "worse" tokenization for this input string: [2646, 1327, 287] == [" gra", "bb", "ed"] The tokenizer arrives at this 3 token output by applying "merges" in a priority order. For example, this is a merge: [" g", "r"] -> " gr". The trained data contains tens of thousands of these merges. When we apply the merges in the priority order, we end up with 3 tokens. Now you might be thinking, that's easy, we'll just iterate the list of merges and see if any of them apply. Only problem with that approach is that applying a merge can open up a new opportunity to merge something else that wasn't possible before. This right here is the key thing that makes this problem complicated. We can solve this problem by iterating all possible merges from the beginning after every time we apply a merge. This would produce the correct solution. Only problem is: our algorithm is now very slow and takes minutes to run...
  - pbhjpbhj 3y ago
    >Only problem is: // As a non-programmer this suggests use of a trie? But could you very roughly sketch what you did, or at least give me a keyword or two to lookup? Grazie mille.
    - belladoreai 3y ago
      That's quite a question coming from a self-proclaimed non-programmer (are you sure you're not one?) I didn't use a Trie. The main data structures used are Linked List and Priority Queue. The tokenization begins by transforming each character to a token. After that merges are applied in priority order as long as merges are possible. The current state of tokenization is represented as a linked list where each node (token) points to the previous node and next node. Nodes of the Linked List are added to a Priority Queue _if_ a merge to their corresponding "next node" is possible according to vocabulary (we use a Hash Map to perform these checks fast). Even though we are technically adding nodes of a linked list into the Priority Queue, you can think of the Priority Queue as holding "potential merges" in the order of their priority (priority according to the trained tokenizer merge data). When we poll a node from the priority queue, we first check that the merge is still possible (because the situation may have changed between the time the node was added to the priority queue, and the time when it was polled from the queue). If the merge is possible, then this is guaranteed to be the most highest-priority merge, so we apply the merge. At this point we need to do several things: we need to mark the previous nodes of the merge as "deleted", we need to create a new node representing the merged token, we need to update the pointers of the adjacent nodes in the linked list, and we need to check if new merges have become possible, and if they have, we need to add the corresponding nodes to the priority queue.
  - londons_explore 3y ago
    I take it if you do greedily go from left to right, you end up with different tokenization and the model doesn't work properly because that isn't what it was trained with? Did the model designers choose this 'difficult' tokenization because it performs better?
    - 3y ago
YetAnotherNick 3y ago
Great repo, but there was this for openAI which was bit hard to find: https://github.com/cogentapps/chat-with-gpt/blob/main/app/src/core/tokenizer/bpe.ts https://github.com/cogentapps/chat-with-gpt/blob/main/app/sr.... Not completely sure, but I think it will likely work as it is for llama as both are BPE following same rules.
- belladoreai 3y ago
  I think it would take quite a bit of work for someone to grab that BPE implementation and make it work for LLaMA. Less work than rewriting the whole tokenizer from scratch, for sure, but a non trivial amount of work anyway.

Hi HN! I was looking for a tokenizer that would accurately(!) count tokens in browser, and I couldn't find one. So I thought "how hard can it be", and here we are 2 weeks later...

zerojames 3y ago

I love this sentiment! Amazing work!

Solvency 3y ago

For those who would also think the same thing, what're some of the the tldr bulletpoints on why this is more complicated than it'd seem?

belladoreai 3y ago
I'll answer with an example. Consider the input string " grabbed". If we wanted to map this string to tokens by greedily going from left to right and choosing tokens from the vocabulary with the strategy of minimizing the number of tokens, our algorithm would be very simple. We would end up with the following tokenization: [17229, 2580] == [" grab", "bed"] Surprisingly, the LLaMA tokenizer does not work this way. It actually finds a "worse" tokenization for this input string: [2646, 1327, 287] == [" gra", "bb", "ed"] The tokenizer arrives at this 3 token output by applying "merges" in a priority order. For example, this is a merge: [" g", "r"] -> " gr". The trained data contains tens of thousands of these merges. When we apply the merges in the priority order, we end up with 3 tokens. Now you might be thinking, that's easy, we'll just iterate the list of merges and see if any of them apply. Only problem with that approach is that applying a merge can open up a new opportunity to merge something else that wasn't possible before. This right here is the key thing that makes this problem complicated. We can solve this problem by iterating all possible merges from the beginning after every time we apply a merge. This would produce the correct solution. Only problem is: our algorithm is now very slow and takes minutes to run...
- pbhjpbhj 3y ago
  >Only problem is: // As a non-programmer this suggests use of a trie? But could you very roughly sketch what you did, or at least give me a keyword or two to lookup? Grazie mille.
  - belladoreai 3y ago
    That's quite a question coming from a self-proclaimed non-programmer (are you sure you're not one?) I didn't use a Trie. The main data structures used are Linked List and Priority Queue. The tokenization begins by transforming each character to a token. After that merges are applied in priority order as long as merges are possible. The current state of tokenization is represented as a linked list where each node (token) points to the previous node and next node. Nodes of the Linked List are added to a Priority Queue _if_ a merge to their corresponding "next node" is possible according to vocabulary (we use a Hash Map to perform these checks fast). Even though we are technically adding nodes of a linked list into the Priority Queue, you can think of the Priority Queue as holding "potential merges" in the order of their priority (priority according to the trained tokenizer merge data). When we poll a node from the priority queue, we first check that the merge is still possible (because the situation may have changed between the time the node was added to the priority queue, and the time when it was polled from the queue). If the merge is possible, then this is guaranteed to be the most highest-priority merge, so we apply the merge. At this point we need to do several things: we need to mark the previous nodes of the merge as "deleted", we need to create a new node representing the merged token, we need to update the pointers of the adjacent nodes in the linked list, and we need to check if new merges have become possible, and if they have, we need to add the corresponding nodes to the priority queue.
- londons_explore 3y ago
  I take it if you do greedily go from left to right, you end up with different tokenization and the model doesn't work properly because that isn't what it was trained with? Did the model designers choose this 'difficult' tokenization because it performs better?
  - 3y ago

YetAnotherNick 3y ago

Great repo, but there was this for openAI which was bit hard to find: https://github.com/cogentapps/chat-with-gpt/blob/main/app/src/core/tokenizer/bpe.ts https://github.com/cogentapps/chat-with-gpt/blob/main/app/sr.... Not completely sure, but I think it will likely work as it is for llama as both are BPE following same rules.

belladoreai 3y ago
I think it would take quite a bit of work for someone to grab that BPE implementation and make it work for LLaMA. Less work than rewriting the whole tokenizer from scratch, for sure, but a non trivial amount of work anyway.