Token trie implementation notes

The round nodes represent tokens, the square nodes do not have a corresponding token.

The number (num_parents) specifies how many parents do you need to pop to get to the parent of the node which comes after our children in DFS order.

We also keep the token_id and a subtree_size (which includes the node itself) in each node. A bogus token_id is used for nodes that do not have a corresponding token.

graph TD
  root[ε, 0] -- a --> a((a, 1))
  root -- b --> b((b, 1))
  root -- c --> c((c, 1))
  a -- x --> ax((ax, 1))
  a -- y --> ay[ay, 1]
  a -- z --> az((az, 2))
  az -- a --> azq((aza, 3))
  ay -- a --> ayq((aya, 1))
  ay -- b --> ayw((ayb, 2))

Traversal algorithm - computing the set of tokens allowed by a stack-based recognizer. The set is stored in logits array - entries with 0.0 are allowed.

let mut logits = vec![-100.0; VOCAB_SIZE + 1];

A simple version of traversal algorithm:

fn traverse(n) {
    // mark token as allowed; nodes without token use `token_id == VOCAB_SIZE`
    logits[n.token_id] = 0.0;
    for c in n.children {
        // for every child that starts with an allowed byte
        if byte_allowed(c.byte) {
            push_byte(c.byte);
            // traverse it
            traverse(c);
            pop_bytes(1);
        }
    }
}

Now, assume the tree is laid out in memory in DFS order:

fn traverse(mut p) {
    let endp = p + nodes[p].subtree_size;
    p += 1; // move to first child
    while p < endp {
        let n = nodes[p];
        if byte_allowed(n.byte) {
            push_byte(n.byte);
            logits[n.token_id] = 0.0;
            // p is moved by n.subtree_size
            p = traverse(p);
            pop_bytes(1);
        } else {
            p += n.subtree_size;
        }
    }
}

Now, we get rid of the recursion:

let mut p = 0;
while p < nodes.len() {
    let n = nodes[p];
    if byte_allowed(n.byte) {
        push_byte(n.byte);
        logits[n.token_id] = 0.0;
        // if the node is a leaf, we need to pop all the parents
        pop_bytes(if n.subtree_size == 1 { n.num_parents } else { 0 });
        // move to first child, or sibling if no children
        p += 1;
    } else {
        // skip the children, and go to the sibling node
        p += n.subtree_size;
        // regardless if the node is a leaf, we need to pop all the parents
        pop_bytes(n.num_parents - 1);
    }
}

Note that the only branch that gets mis-predicted here is the if byte_allowed(n.byte). The if in argument to pop_bytes is compiled to bit operations, so it is branchless.

Actual code

See add_bias_inner in toktree.rs.

• it uses try_push_byte() which combines byte_allowed() and push_byte()
• it calls pop_bytes() at the beginning with a variable stored in previous iteration

The following is a breakdown of all memory reads and writes, when used with llguidance, see try_push_byte() in parser.rs. This only considers the fast lexer path.

• pop_bytes() - only register update (stack length)
• fetch current TrieNode (8 bytes)
• try_push_byte() - 3 reads, 1 write, see below
• updating token bit-mask - 1 read, 1 write

The try_push_byte() function:

• fetch lexer state from the stack (1 read)
• compute next DFA state: 1 read for alphabet compression if enabled, 1 read for transition table
• push lexer state to the stack (1 write)

Together, this is 5 reads and 2 writes per node. Dependency chain lengths are difficult to estimate, given the possible speculation and out-of-order execution.

On an AMD EPYC 7V13 a single node is processed in around 13 cycles (at 4.2 instructions per cycle); this drops by 1 cycle if the alphabet compression is disabled (likely only 1 because lexer stack fetch and alphabet compression fetch can be done in parallel).

The 7V13 has 4 cycles L1 latency (32KB), 13 cycles L2 latency (512KB), and 46 cycles L3 latency (up to 32MB per core, but shared). It also has 6-wide uop dispatch. Sources: EPYC Milan, Zen3, Zen2 (shares L1/L2 specs).