unigram 0.1.0

Bijective codec between bytes and single-token words, for moving identifiers through a language model
Documentation

unigram

A bijective codec between bytes and words that cost exactly one LLM token.

let words = unigram::encode(&[0x3d, 0x9a, 0x00, 0xff]);
// "check music access world"

assert_eq!(unigram::decode(&words)?, vec![0x3d, 0x9a, 0x00, 0xff]);

Why

Machine identifiers are routinely handed to a language model and asked back: an acknowledgement token, a digest, a correlation id. Hexadecimal is the worst possible carrier for that trip. It is expensive, because a hex run shreds into a fragment every character or two under every tokenizer; and it is undetectably fragile, because every character is drawn from the same sixteen, so a corrupted one still looks like a valid digest.

unigram carries the same bytes as words drawn from a fixed alphabet of 256. Two properties follow from that size, and they are the whole design.

One word is exactly one byte. Encoding is a table lookup per byte — no bit-packing, no padding, no length convention. Every byte string has exactly one encoding, and every sequence of alphabet words decodes.

Every word is exactly one token. An encoded value costs one token per byte, and the same for every value. Against hex of the same payload, under Claude:

payload hex (mean / worst) unigram
4 bytes 6.0 / 8 4
16 bytes 21.5 / 25 16
32 bytes 42.2 / 49 32

Roughly a quarter cheaper on average — but the flat cost matters more than the mean. Hex swings with the value, so a token budget built on it has to assume the worst case. This one is known before the value is minted.

Corruption becomes visible. The alphabet is 256 words out of every string that could be written, and no two entries are within one character edit of each other, so a mangled word is overwhelmingly likely to be no word at all. decode says so, and names the word:

unigram::decode("check musix access")?;
// Err(UnknownWord { position: 1, word: "musix" })

Hex cannot do this. Every single-character corruption of a hex digest is another valid hex digest.

Surviving the round trip

decode is liberal in what it accepts. Any run of characters that is not an ASCII letter separates words, and case is ignored — so a value that came back hyphenated, re-wrapped across lines, comma-joined, quoted, or shouted still decodes to the bytes that were sent.

unigram::mint(4);                        // 32 fresh bits, 4 tokens
unigram::matches(issued, presented);     // comparison that forgives the damage

matches compares decoded bytes when both sides are encoded values, and normalized strings otherwise — so values issued in some older format keep matching themselves without a migration.

Choosing a length

One word is one byte and one token, so a value's length is its entropy budget and its token budget at once — the two cannot drift apart, which is most of why this is easier to reason about than hex.

words bits distinct values values before a 1-in-a-million collision
2 16 65,536 fewer than 1
3 24 16.8 million 5
4 32 4.3 billion 92
6 48 281 trillion 23,700
8 64 1.8 × 10¹⁹ 6 million
16 128 3.4 × 10³⁸ 2.6 × 10¹⁶
32 256 1.2 × 10⁷⁷ 4.8 × 10³⁵

The right column is the birthday bound, k ≈ √(2·N·p), and it is the column to size against: collisions arrive at the square root of the space, not at the space. Sixteen words is a UUID's width, thirty-two a SHA-256's.

Two questions hide in that table and it answers only one. Collision is the right column — how many values may be outstanding before two coincide. Guessing is separate: mint draws from the OS CSPRNG, so every bit is unpredictable, but four words is 4.3 billion candidates, which is an afternoon for anything that can ask freely. Four words suits a value that is scoped, short-lived, and rate-limited — an acknowledgement nonce, a correlation id. A value a stranger can grind at wants eight or more, and at equal entropy the words are still cheaper than the hex: 64 bits costs 8 tokens here against a mean of 11.2 and a worst case of 14.

The join is a space, deliberately

Tokenizer vocabularies hold their canonical word entries space-prefixed, so the space between two words is absorbed into the word that follows it and costs nothing. No other separator is free. Measured across all five families, an eight-byte value:

separator GPT-4o GPT-3.5/4 GPT-3 GPT-2 Llama Claude
space 8 8 8 8 8 8
_ . 8 8 15 15 15 15
- 11 9 15 15 15 15
, \n 13–15 12–15 15 15 15 15

The join would cost almost as much as the payload. Encoded values travel inside quoted strings in practice, where embedded spaces are free.

The alphabet

256 entries of lowercase ASCII English, 4 to 11 characters, chosen under four constraints:

  • One token under Claude, GPT-2/3 (r50k, p50k), GPT-3.5/4 (cl100k), GPT-4o (o200k), and Llama's SentencePiece — spanning both the BPE and SentencePiece families.
  • No two entries within one character edit of each other, which is what makes a single-character slip land outside the alphabet instead of on a different valid word.
  • Nothing charged — no death, violence, race, gender, religion, or politics. These strings surface unbidden in transcripts, logs, and user-facing errors.
  • No entry is an inflection of another, so a dropped plural cannot silently decode to a different byte.

The table is indexed by the byte each word encodes, so it is appended to, never rearranged: reordering an entry changes what every previously issued value decodes to.

Verifying it

The crate depends on nothing but the OS CSPRNG, at runtime or under test, and never tokenizes. cargo test covers the codec and the table's structure; it says nothing about cost.

Every cost claim above is checked by verify-alphabet.py, which reads the alphabet straight out of src/lib.rs and re-measures it against all five families:

uv run verify-alphabet.py

Run it after any edit to the table. A green test suite alone establishes none of what this crate is named for.

License

MIT.