basic fsst+ beginnings

Signed-off-by: Andrew Duffy <[email protected]>

Author: Andrew Duffy

encodings/fsst/src/lib.rs

@@ -19,6 +19,7 @@ mod ops;
 mod serde;
 #[cfg(test)]
 mod tests;
+mod prefix;
 
 pub use array::*;
 pub use compress::*;

encodings/fsst/src/prefix.rs

@@ -0,0 +1,188 @@
+//! An implementation of the FSST+ extension, original at https://github.com/cwida/fsst_plus/.
+//!
+//! FSST+ augments FSST with the addition.
+
+#![allow(unused)]
+
+#[derive(Debug)]
+pub struct SimilarityBlock {
+    start_idx: usize,
+    prefix_len: usize,
+}
+
+/// Maximum shared prefix length.
+pub const MAX_PREFIX: usize = 128;
+
+/// Find the longest-common-prefix between adjacent encoded texts.
+pub fn longest_common_prefix(codes: &[&[u8]]) -> Vec<usize> {
+    // LCP for each pair of successive strings.
+    // LCP[i] is the length of longest common prefix between the i and i+1 string.
+    // For example, lcp(["abc", "abcd", "ab"]) -> [3, 2]
+    let mut longest_common_prefix = Vec::new();
+
+    // Calculate the LCP of consecutive strings in the input.
+    for w in codes.windows(2) {
+        let s1 = w[0];
+        let s2 = w[1];
+
+        // Consecutive strings to evaluate LCP
+        let mut lcp = 0;
+        for (&a, &b) in s1.iter().zip(s2.iter()) {
+            if a == b {
+                lcp += 1;
+            } else {
+                break;
+            }
+        }
+
+        longest_common_prefix.push(lcp);
+    }
+
+    longest_common_prefix
+}
+
+/// Input: a vector of FSST-encoded strings.
+/// Output: a vector of "similarity blocks".
+///
+/// We first calculate the LCP between adjacent strings, then once that has been completed, we find
+/// the optimal split points to create "similarity" blocks that all share a maximal prefix.
+///
+///
+/// The simple recursive solution structure looks something like this:
+///
+/// ```python
+/// def cost(i, j, min_lcp, lcp):
+///     # find the min prefix between i and j
+///     min_prefix = min_lcp[j][i]
+///     block_assignment = []
+///
+///     # figure out cost of associating i with i-1
+///
+///     # determine if grouping strings i/j is advantageous.
+///     for prefix_len in [0, min_prefix]:
+///         n_strings = i - j
+///         # if the strings are
+/// ```
+#[allow(clippy::needless_range_loop)]
+pub fn chunk_by_similarity(strings: &[&[u8]]) -> Vec<SimilarityBlock> {
+    // Calculate LCP between all items first
+    let lcp = longest_common_prefix(strings);
+
+    // min_lcp[i][j] = min(LCP[i], LCP[i+1], ..., LCP[j])
+    let mut min_lcp = vec![vec![0; strings.len()]; strings.len()];
+
+    // ... the diagonals are just the LCP[i]'s
+    for i in 0..strings.len() {
+        // up to 128 is longest prefix we allow.
+        min_lcp[i][i] = strings[i].len().min(MAX_PREFIX);
+        for j in (i + 1)..lcp.len() {
+            min_lcp[i][j] = min_lcp[i][j - 1].min(lcp[j - 1]);
+        }
+    }
+
+
+    // Cost is the total cost of the block split.
+    let mut cost = vec![usize::MAX; 1+strings.len()];
+    cost[0] = 0;
+
+    let mut block = vec![0; 1+strings.len()];
+    let mut prefix = vec![0; 1+strings.len()];
+
+    // Estimate the cost for all strings instead here.
+    let estimate_cost = |i: u32, j: u32| min_lcp[i as usize][j as usize];
+
+    // cum_len[k] = sum(strings[i].len() for i in 0..k)
+    let mut length_prefix_sum =  Vec::new();
+    length_prefix_sum.push(0);
+    for string in strings {
+        length_prefix_sum.push(length_prefix_sum.last().unwrap() + string.len());
+    }
+
+    // cost[j] holds the cost up to j.
+    // block[j] holds the start of the block that j is a part of
+    // prefix[j] is the prefix length for the j-th string
+    for end in 1..=strings.len() {
+        for start in 0..end {
+            dbg!((start, end));
+            let min_prefix = min_lcp[start][end - 1];
+            dbg!(min_prefix);
+            for prefix_len in [0, min_prefix] {
+                dbg!(prefix_len);
+                let n = end - start;
+                dbg!(n);
+                let per_string_overhead = if prefix_len == 0 { 1 } else { 3 };
+                dbg!(per_string_overhead);
+                let overhead = n * per_string_overhead;
+                dbg!(overhead);
+                let sum_len = length_prefix_sum[end] - length_prefix_sum[start];
+                dbg!(sum_len);
+                let total_cost = cost[start] + overhead + sum_len - ((n - 1) * prefix_len as usize);
+                dbg!(total_cost);
+
+                if total_cost < cost[end] {
+                    cost[end] = total_cost;
+                    block[end] = start;
+                    prefix[end] = prefix_len;
+                }
+            }
+        }
+    }
+
+    // Traverse the blocks from end to start.
+    let mut blocks = Vec::new();
+
+    let mut idx = strings.len();
+    while idx > 0 {
+        let start_idx = block[idx];
+        let prefix_len = prefix[idx];
+
+        blocks.push(SimilarityBlock {
+            start_idx, prefix_len,
+        });
+
+        // Advance backward.
+        idx = start_idx;
+    }
+
+    // Reverse the list of blocks so they're in order now.
+    blocks.reverse();
+
+    blocks
+}
+
+#[cfg(test)]
+mod tests {
+    use fsst::Compressor;
+
+    use crate::prefix::{chunk_by_similarity, longest_common_prefix};
+
+    #[test]
+    fn test_urls() {
+        let strings = vec![
+            "reddit.com".as_bytes(),
+            "reddit.com/a".as_bytes(),
+            "reddit.com/a/b".as_bytes(),
+            "reddit.com/c".as_bytes(),
+            "reddit.com/c/d/d".as_bytes(),
+            "reddit.com/c/e".as_bytes(),
+            "google.com".as_bytes(),
+            "google.com/search?q=beans".as_bytes(),
+            "google.com/search?q=black+beans".as_bytes(),
+            "google.com/search?q=lima+beans".as_bytes(),
+            "google.com/search?q=taylor+swift".as_bytes(),
+        ];
+
+        let compressor = Compressor::train(&strings);
+        let result = compressor.compress_bulk(&strings);
+
+        let lcps = longest_common_prefix(strings.as_slice());
+        assert_eq!(lcps, vec![10, 12, 11, 12, 13, 0, 10, 21, 20, 20]);
+
+        let chunks = chunk_by_similarity(&strings);
+
+        dbg!(&chunks);
+
+        // Once we have calculate the adjacent_lcp, expand into a new LCP with a bunch of strings
+        // in a single block.
+    }
+}