TANS with separate probability

Sorry for the delay. I'm not sure I understand the question. Of course you can use the same probabilistic model to independently compress/decompress several messages. And yes, in this case you have to keep the model in memory only once, since compression and decompression don't consume the model, they only need a reference to it. Admittedly, this is a bit obscured by the generic nature of the API; for example, the method AnsCoder::encode_iid_symbols_reverse takes a generic argument model whose type has to implement EncoderModel, so it may indeed seem like you'd have to provide a fresh entropy model every time. But there's a blanket implementation of EncoderModel for any reference &M where M implements EncoderModel, so you only need a single owned EncoderModel and can hand out as many shared references to it as you like (some small entropy models also implement Copy; for those, it's usually more performant to pass them by value).

I'm attaching an example of a full compression/decompression round trip below. But in brief, if I understand correctly what you're trying to achieve, then your struct for the compressed representation of Index should probably look something like this:

struct CompressedIndex {
    doc: Vec<Vec<u32>>, // Note that constriction represents bit strings in 32-bit chunks by default for performance reasons.
    probs: DefaultContiguousCategoricalEntropyModel, // (for example; you can use any entropy model in `constriction::stream::model`)
    alphabet: Vec<char>, // List of all distinct characters that can appear in a message (see full example below).
}

And there's nothing that holds you back from encoding or decoding each entry of doc independently, using the shared entropy model probs and the shared alphabet (see full round-trip example below).

From what I’ve seen, it seems like we need to provide the probabilities for the symbol we're currently compressing.

I'm not sure I understand. Of course you have to provide the probabilities anytime you encode or decode a symbol (in fact, you have to provide the entire entropy model, not just the probability of the specific symbol you're currently encoding or decoding). That's not a limitation of constriction, it's a fundamental theoretical limitation of source coding: one cannot (losslessly) compress data without a probabilistic model of the data source ("source coding theorem").

Full Example

use std::collections::HashMap;

use constriction::{
    backends::Cursor,
    stream::{
        model::DefaultContiguousCategoricalEntropyModel, stack::DefaultAnsCoder, Decode, Encode,
    },
    UnwrapInfallible,
};

#[derive(Debug, PartialEq, Eq)]
struct UncompressedIndex {
    doc: Vec<String>,
}

#[derive(Debug)]
struct CompressedIndex {
    doc: Vec<Vec<u32>>, // Note that constriction represents bit strings in 32-bit chunks by default for performance reasons.
    probs: DefaultContiguousCategoricalEntropyModel, // (for example; you can use any entropy model in `constriction::stream::model`)
    alphabet: Vec<char>, // List of all distinct characters that can appear in a message.
}

impl UncompressedIndex {
    fn compress(
        &self,
        probs: DefaultContiguousCategoricalEntropyModel,
        alphabet: Vec<char>,
    ) -> CompressedIndex {
        let inverse_alphabet = alphabet
            .iter()
            .enumerate()
            .map(|(index, &character)| (character, index))
            .collect::<HashMap<_, _>>();

        let doc = self
            .doc
            .iter()
            .map(|message| {
                let mut coder = DefaultAnsCoder::new();

                // Start with a special EOF symbol so that `CompressedIndex::decompress` knows when to terminate:
                coder.encode_symbol(alphabet.len(), &probs).unwrap();

                // Then encode the message, character by character, in reverse order:
                for character in message.chars().rev() {
                    let char_index = *inverse_alphabet.get(&character).unwrap();
                    coder.encode_symbol(char_index, &probs).unwrap();
                }

                coder.into_compressed().unwrap_infallible()
            })
            .collect();

        CompressedIndex {
            doc,
            probs,
            alphabet,
        }
    }
}

impl CompressedIndex {
    fn decompress(&self) -> UncompressedIndex {
        let doc = self
            .doc
            .iter()
            .map(|data| {
                let mut coder =
                    DefaultAnsCoder::from_compressed(Cursor::new_at_write_end(&data[..])).unwrap();
                core::iter::from_fn(|| {
                    let symbol_id = coder.decode_symbol(&self.probs).unwrap();
                    self.alphabet.get(symbol_id).copied() // Returns `None` if `symbol_id` is the EOF token, which terminates the iterator.
                })
                .collect()
            })
            .collect();

        UncompressedIndex { doc }
    }
}

#[test]
fn round_trip() {
    let uncompressed = UncompressedIndex {
        doc: vec!["Hello, World!".to_string(), "Goodbye.".to_string()],
    };

    let alphabet = vec![
        'H', 'e', 'l', 'o', ',', ' ', 'W', 'r', 'd', '!', 'G', 'b', 'y', '.',
    ];
    let counts = [1., 2., 3., 4., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 2.]; // The last entry is for the EOF token.
    let probs =
        DefaultContiguousCategoricalEntropyModel::from_floating_point_probabilities(&counts)
            .unwrap();

    let compressed = uncompressed.compress(probs, alphabet);
    let reconstructed = compressed.decompress();
    assert_eq!(uncompressed, reconstructed);
}

bamler-lab / constriction

TANS with separate probability #52

Full Example