1// I've called the primary data structure in this module a "range trie." As far
2// as I can tell, there is no prior art on a data structure like this, however,
3// it's likely someone somewhere has built something like it. Searching for
4// "range trie" turns up the paper "Range Tries for Scalable Address Lookup,"
5// but it does not appear relevant.
6//
7// The range trie is just like a trie in that it is a special case of a
8// deterministic finite state machine. It has states and each state has a set
9// of transitions to other states. It is acyclic, and, like a normal trie,
10// it makes no attempt to reuse common suffixes among its elements. The key
11// difference between a normal trie and a range trie below is that a range trie
12// operates on *contiguous sequences* of bytes instead of singleton bytes.
13// One could say say that our alphabet is ranges of bytes instead of bytes
14// themselves, except a key part of range trie construction is splitting ranges
15// apart to ensure there is at most one transition that can be taken for any
16// byte in a given state.
17//
18// I've tried to explain the details of how the range trie works below, so
19// for now, we are left with trying to understand what problem we're trying to
20// solve. Which is itself fairly involved!
21//
22// At the highest level, here's what we want to do. We want to convert a
23// sequence of Unicode codepoints into a finite state machine whose transitions
24// are over *bytes* and *not* Unicode codepoints. We want this because it makes
25// said finite state machines much smaller and much faster to execute. As a
26// simple example, consider a byte oriented automaton for all Unicode scalar
27// values (0x00 through 0x10FFFF, not including surrogate codepoints):
28//
29// [00-7F]
30// [C2-DF][80-BF]
31// [E0-E0][A0-BF][80-BF]
32// [E1-EC][80-BF][80-BF]
33// [ED-ED][80-9F][80-BF]
34// [EE-EF][80-BF][80-BF]
35// [F0-F0][90-BF][80-BF][80-BF]
36// [F1-F3][80-BF][80-BF][80-BF]
37// [F4-F4][80-8F][80-BF][80-BF]
38//
39// (These byte ranges are generated via the regex-syntax::utf8 module, which
40// was based on Russ Cox's code in RE2, which was in turn based on Ken
41// Thompson's implementation of the same idea in his Plan9 implementation of
42// grep.)
43//
44// It should be fairly straight-forward to see how one could compile this into
45// a DFA. The sequences are sorted and non-overlapping. Essentially, you could
46// build a trie from this fairly easy. The problem comes when your initial
47// range (in this case, 0x00-0x10FFFF) isn't so nice. For example, the class
48// represented by '\w' contains only a tenth of the codepoints that
49// 0x00-0x10FFFF contains, but if we were to write out the byte based ranges
50// as we did above, the list would stretch to 892 entries! This turns into
51// quite a large NFA with a few thousand states. Turning this beast into a DFA
52// takes quite a bit of time. We are thus left with trying to trim down the
53// number of states we produce as early as possible.
54//
55// One approach (used by RE2 and still by the regex crate, at time of writing)
56// is to try to find common suffixes while building NFA states for the above
57// and reuse them. This is very cheap to do and one can control precisely how
58// much extra memory you want to use for the cache.
59//
60// Another approach, however, is to reuse an algorithm for constructing a
61// *minimal* DFA from a sorted sequence of inputs. I don't want to go into
62// the full details here, but I explain it in more depth in my blog post on
63// FSTs[1]. Note that the algorithm not invented by me, but was published
64// in paper by Daciuk et al. in 2000 called "Incremental Construction of
65// MinimalAcyclic Finite-State Automata." Like the suffix cache approach above,
66// it is also possible to control the amount of extra memory one uses, although
67// this usually comes with the cost of sacrificing true minimality. (But it's
68// typically close enough with a reasonably sized cache of states.)
69//
70// The catch is that Daciuk's algorithm only works if you add your keys in
71// lexicographic ascending order. In our case, since we're dealing with ranges,
72// we also need the additional requirement that ranges are either equivalent
73// or do not overlap at all. For example, if one were given the following byte
74// ranges:
75//
76// [BC-BF][80-BF]
77// [BC-BF][90-BF]
78//
79// Then Daciuk's algorithm also would not work, since there is nothing to
80// handle the fact that the ranges overlap. They would need to be split apart.
81// Thankfully, Thompson's algorithm for producing byte ranges for Unicode
82// codepoint ranges meets both of our requirements.
83//
84// ... however, we would also like to be able to compile UTF-8 automata in
85// reverse. We want this because in order to find the starting location of a
86// match using a DFA, we need to run a second DFA---a reversed version of the
87// forward DFA---backwards to discover the match location. Unfortunately, if
88// we reverse our byte sequences for 0x00-0x10FFFF, we get sequences that are
89// can overlap, even if they are sorted:
90//
91// [00-7F]
92// [80-BF][80-9F][ED-ED]
93// [80-BF][80-BF][80-8F][F4-F4]
94// [80-BF][80-BF][80-BF][F1-F3]
95// [80-BF][80-BF][90-BF][F0-F0]
96// [80-BF][80-BF][E1-EC]
97// [80-BF][80-BF][EE-EF]
98// [80-BF][A0-BF][E0-E0]
99// [80-BF][C2-DF]
100//
101// For example, '[80-BF][80-BF][EE-EF]' and '[80-BF][A0-BF][E0-E0]' have
102// overlapping ranges between '[80-BF]' and '[A0-BF]'. Thus, there is no
103// simple way to apply Daciuk's algorithm.
104//
105// And thus, the range trie was born. The range trie's only purpose is to take
106// sequences of byte ranges like the ones above, collect them into a trie and
107// then spit them in a sorted fashion with no overlapping ranges. For example,
108// 0x00-0x10FFFF gets translated to:
109//
110// [0-7F]
111// [80-BF][80-9F][80-8F][F1-F3]
112// [80-BF][80-9F][80-8F][F4]
113// [80-BF][80-9F][90-BF][F0]
114// [80-BF][80-9F][90-BF][F1-F3]
115// [80-BF][80-9F][E1-EC]
116// [80-BF][80-9F][ED]
117// [80-BF][80-9F][EE-EF]
118// [80-BF][A0-BF][80-8F][F1-F3]
119// [80-BF][A0-BF][80-8F][F4]
120// [80-BF][A0-BF][90-BF][F0]
121// [80-BF][A0-BF][90-BF][F1-F3]
122// [80-BF][A0-BF][E0]
123// [80-BF][A0-BF][E1-EC]
124// [80-BF][A0-BF][EE-EF]
125// [80-BF][C2-DF]
126//
127// We've thus satisfied our requirements for running Daciuk's algorithm. All
128// sequences of ranges are sorted, and any corresponding ranges are either
129// exactly equivalent or non-overlapping.
130//
131// In effect, a range trie is building a DFA from a sequence of arbitrary
132// byte ranges. But it uses an algoritm custom tailored to its input, so it
133// is not as costly as traditional DFA construction. While it is still quite
134// a bit more costly than the forward's case (which only needs Daciuk's
135// algorithm), it winds up saving a substantial amount of time if one is doing
136// a full DFA powerset construction later by virtue of producing a much much
137// smaller NFA.
138//
139// [1] - https://blog.burntsushi.net/transducers/
140// [2] - https://www.mitpressjournals.org/doi/pdfplus/10.1162/089120100561601
141
142use std::cell::RefCell;
143use std::fmt;
144use std::mem;
145use std::ops::RangeInclusive;
146use std::u32;
147
148use regex_syntax::utf8::Utf8Range;
149
150/// A smaller state ID means more effective use of the CPU cache and less
151/// time spent copying. The implementation below will panic if the state ID
152/// space is exhausted, but in order for that to happen, the range trie itself
153/// would use well over 100GB of memory. Moreover, it's likely impossible
154/// for the state ID space to get that big. In fact, it's likely that even a
155/// u16 would be good enough here. But it's not quite clear how to prove this.
156type StateID = u32;
157
158/// There is only one final state in this trie. Every sequence of byte ranges
159/// added shares the same final state.
160const FINAL: StateID = 0;
161
162/// The root state of the trie.
163const ROOT: StateID = 1;
164
165/// A range trie represents an ordered set of sequences of bytes.
166///
167/// A range trie accepts as input a sequence of byte ranges and merges
168/// them into the existing set such that the trie can produce a sorted
169/// non-overlapping sequence of byte ranges. The sequence emitted corresponds
170/// precisely to the sequence of bytes matched by the given keys, although the
171/// byte ranges themselves may be split at different boundaries.
172///
173/// The order complexity of this data structure seems difficult to analyze.
174/// If the size of a byte is held as a constant, then insertion is clearly
175/// O(n) where n is the number of byte ranges in the input key. However, if
176/// k=256 is our alphabet size, then insertion could be O(k^2 * n). In
177/// particular it seems possible for pathological inputs to cause insertion
178/// to do a lot of work. However, for what we use this data structure for,
179/// there should be no pathological inputs since the ultimate source is always
180/// a sorted set of Unicode scalar value ranges.
181///
182/// Internally, this trie is setup like a finite state machine. Note though
183/// that it is acyclic.
184#[derive(Clone)]
185pub struct RangeTrie {
186 /// The states in this trie. The first is always the shared final state.
187 /// The second is always the root state. Otherwise, there is no
188 /// particular order.
189 states: Vec<State>,
190 /// A free-list of states. When a range trie is cleared, all of its states
191 /// are added to list. Creating a new state reuses states from this list
192 /// before allocating a new one.
193 free: Vec<State>,
194 /// A stack for traversing this trie to yield sequences of byte ranges in
195 /// lexicographic order.
196 iter_stack: RefCell<Vec<NextIter>>,
197 /// A bufer that stores the current sequence during iteration.
198 iter_ranges: RefCell<Vec<Utf8Range>>,
199 /// A stack used for traversing the trie in order to (deeply) duplicate
200 /// a state.
201 dupe_stack: Vec<NextDupe>,
202 /// A stack used for traversing the trie during insertion of a new
203 /// sequence of byte ranges.
204 insert_stack: Vec<NextInsert>,
205}
206
207/// A single state in this trie.
208#[derive(Clone)]
209struct State {
210 /// A sorted sequence of non-overlapping transitions to other states. Each
211 /// transition corresponds to a single range of bytes.
212 transitions: Vec<Transition>,
213}
214
215/// A transition is a single range of bytes. If a particular byte is in this
216/// range, then the corresponding machine may transition to the state pointed
217/// to by `next_id`.
218#[derive(Clone)]
219struct Transition {
220 /// The byte range.
221 range: Utf8Range,
222 /// The next state to transition to.
223 next_id: StateID,
224}
225
226impl RangeTrie {
227 /// Create a new empty range trie.
228 pub fn new() -> RangeTrie {
229 let mut trie = RangeTrie {
230 states: vec![],
231 free: vec![],
232 iter_stack: RefCell::new(vec![]),
233 iter_ranges: RefCell::new(vec![]),
234 dupe_stack: vec![],
235 insert_stack: vec![],
236 };
237 trie.clear();
238 trie
239 }
240
241 /// Clear this range trie such that it is empty. Clearing a range trie
242 /// and reusing it can beneficial because this may reuse allocations.
243 pub fn clear(&mut self) {
244 self.free.extend(self.states.drain(..));
245 self.add_empty(); // final
246 self.add_empty(); // root
247 }
248
249 /// Iterate over all of the sequences of byte ranges in this trie, and
250 /// call the provided function for each sequence. Iteration occurs in
251 /// lexicographic order.
252 pub fn iter<F: FnMut(&[Utf8Range])>(&self, mut f: F) {
253 let mut stack = self.iter_stack.borrow_mut();
254 stack.clear();
255 let mut ranges = self.iter_ranges.borrow_mut();
256 ranges.clear();
257
258 // We do iteration in a way that permits us to use a single buffer
259 // for our keys. We iterate in a depth first fashion, while being
260 // careful to expand our frontier as we move deeper in the trie.
261 stack.push(NextIter { state_id: ROOT, tidx: 0 });
262 while let Some(NextIter { mut state_id, mut tidx }) = stack.pop() {
263 // This could be implemented more simply without an inner loop
264 // here, but at the cost of more stack pushes.
265 loop {
266 let state = self.state(state_id);
267 // If we're visited all transitions in this state, then pop
268 // back to the parent state.
269 if tidx >= state.transitions.len() {
270 ranges.pop();
271 break;
272 }
273
274 let t = &state.transitions[tidx];
275 ranges.push(t.range);
276 if t.next_id == FINAL {
277 f(&ranges);
278 ranges.pop();
279 tidx += 1;
280 } else {
281 // Expand our frontier. Once we come back to this state
282 // via the stack, start in on the next transition.
283 stack.push(NextIter { state_id, tidx: tidx + 1 });
284 // Otherwise, move to the first transition of the next
285 // state.
286 state_id = t.next_id;
287 tidx = 0;
288 }
289 }
290 }
291 }
292
293 /// Inserts a new sequence of ranges into this trie.
294 ///
295 /// The sequence given must be non-empty and must not have a length
296 /// exceeding 4.
297 pub fn insert(&mut self, ranges: &[Utf8Range]) {
298 assert!(!ranges.is_empty());
299 assert!(ranges.len() <= 4);
300
301 let mut stack = mem::replace(&mut self.insert_stack, vec![]);
302 stack.clear();
303
304 stack.push(NextInsert::new(ROOT, ranges));
305 while let Some(next) = stack.pop() {
306 let (state_id, ranges) = (next.state_id(), next.ranges());
307 assert!(!ranges.is_empty());
308
309 let (mut new, rest) = (ranges[0], &ranges[1..]);
310
311 // i corresponds to the position of the existing transition on
312 // which we are operating. Typically, the result is to remove the
313 // transition and replace it with two or more new transitions
314 // corresponding to the partitions generated by splitting the
315 // 'new' with the ith transition's range.
316 let mut i = self.state(state_id).find(new);
317
318 // In this case, there is no overlap *and* the new range is greater
319 // than all existing ranges. So we can just add it to the end.
320 if i == self.state(state_id).transitions.len() {
321 let next_id = NextInsert::push(self, &mut stack, rest);
322 self.add_transition(state_id, new, next_id);
323 continue;
324 }
325
326 // The need for this loop is a bit subtle, buf basically, after
327 // we've handled the partitions from our initial split, it's
328 // possible that there will be a partition leftover that overlaps
329 // with a subsequent transition. If so, then we have to repeat
330 // the split process again with the leftovers and that subsequent
331 // transition.
332 'OUTER: loop {
333 let old = self.state(state_id).transitions[i].clone();
334 let split = match Split::new(old.range, new) {
335 Some(split) => split,
336 None => {
337 let next_id = NextInsert::push(self, &mut stack, rest);
338 self.add_transition_at(i, state_id, new, next_id);
339 continue;
340 }
341 };
342 let splits = split.as_slice();
343 // If we only have one partition, then the ranges must be
344 // equivalent. There's nothing to do here for this state, so
345 // just move on to the next one.
346 if splits.len() == 1 {
347 // ... but only if we have anything left to do.
348 if !rest.is_empty() {
349 stack.push(NextInsert::new(old.next_id, rest));
350 }
351 break;
352 }
353 // At this point, we know that 'split' is non-empty and there
354 // must be some overlap AND that the two ranges are not
355 // equivalent. Therefore, the existing range MUST be removed
356 // and split up somehow. Instead of actually doing the removal
357 // and then a subsequent insertion---with all the memory
358 // shuffling that entails---we simply overwrite the transition
359 // at position `i` for the first new transition we want to
360 // insert. After that, we're forced to do expensive inserts.
361 let mut first = true;
362 let mut add_trans =
363 |trie: &mut RangeTrie, pos, from, range, to| {
364 if first {
365 trie.set_transition_at(pos, from, range, to);
366 first = false;
367 } else {
368 trie.add_transition_at(pos, from, range, to);
369 }
370 };
371 for (j, &srange) in splits.iter().enumerate() {
372 match srange {
373 SplitRange::Old(r) => {
374 // Deep clone the state pointed to by the ith
375 // transition. This is always necessary since 'old'
376 // is always coupled with at least a 'both'
377 // partition. We don't want any new changes made
378 // via the 'both' partition to impact the part of
379 // the transition that doesn't overlap with the
380 // new range.
381 let dup_id = self.duplicate(old.next_id);
382 add_trans(self, i, state_id, r, dup_id);
383 }
384 SplitRange::New(r) => {
385 // This is a bit subtle, but if this happens to be
386 // the last partition in our split, it is possible
387 // that this overlaps with a subsequent transition.
388 // If it does, then we must repeat the whole
389 // splitting process over again with `r` and the
390 // subsequent transition.
391 {
392 let trans = &self.state(state_id).transitions;
393 if j + 1 == splits.len()
394 && i < trans.len()
395 && intersects(r, trans[i].range)
396 {
397 new = r;
398 continue 'OUTER;
399 }
400 }
401
402 // ... otherwise, setup exploration for a new
403 // empty state and add a brand new transition for
404 // this new range.
405 let next_id =
406 NextInsert::push(self, &mut stack, rest);
407 add_trans(self, i, state_id, r, next_id);
408 }
409 SplitRange::Both(r) => {
410 // Continue adding the remaining ranges on this
411 // path and update the transition with the new
412 // range.
413 if !rest.is_empty() {
414 stack.push(NextInsert::new(old.next_id, rest));
415 }
416 add_trans(self, i, state_id, r, old.next_id);
417 }
418 }
419 i += 1;
420 }
421 // If we've reached this point, then we know that there are
422 // no subsequent transitions with any overlap. Therefore, we
423 // can stop processing this range and move on to the next one.
424 break;
425 }
426 }
427 self.insert_stack = stack;
428 }
429
430 pub fn add_empty(&mut self) -> StateID {
431 if self.states.len() as u64 > u32::MAX as u64 {
432 // This generally should not happen since a range trie is only
433 // ever used to compile a single sequence of Unicode scalar values.
434 // If we ever got to this point, we would, at *minimum*, be using
435 // 96GB in just the range trie alone.
436 panic!("too many sequences added to range trie");
437 }
438 let id = self.states.len() as StateID;
439 // If we have some free states available, then use them to avoid
440 // more allocations.
441 if let Some(mut state) = self.free.pop() {
442 state.clear();
443 self.states.push(state);
444 } else {
445 self.states.push(State { transitions: vec![] });
446 }
447 id
448 }
449
450 /// Performs a deep clone of the given state and returns the duplicate's
451 /// state ID.
452 ///
453 /// A "deep clone" in this context means that the state given along with
454 /// recursively all states that it points to are copied. Once complete,
455 /// the given state ID and the returned state ID share nothing.
456 ///
457 /// This is useful during range trie insertion when a new range overlaps
458 /// with an existing range that is bigger than the new one. The part of
459 /// the existing range that does *not* overlap with the new one is that
460 /// duplicated so that adding the new range to the overlap doesn't disturb
461 /// the non-overlapping portion.
462 ///
463 /// There's one exception: if old_id is the final state, then it is not
464 /// duplicated and the same final state is returned. This is because all
465 /// final states in this trie are equivalent.
466 fn duplicate(&mut self, old_id: StateID) -> StateID {
467 if old_id == FINAL {
468 return FINAL;
469 }
470
471 let mut stack = mem::replace(&mut self.dupe_stack, vec![]);
472 stack.clear();
473
474 let new_id = self.add_empty();
475 // old_id is the state we're cloning and new_id is the ID of the
476 // duplicated state for old_id.
477 stack.push(NextDupe { old_id, new_id });
478 while let Some(NextDupe { old_id, new_id }) = stack.pop() {
479 for i in 0..self.state(old_id).transitions.len() {
480 let t = self.state(old_id).transitions[i].clone();
481 if t.next_id == FINAL {
482 // All final states are the same, so there's no need to
483 // duplicate it.
484 self.add_transition(new_id, t.range, FINAL);
485 continue;
486 }
487
488 let new_child_id = self.add_empty();
489 self.add_transition(new_id, t.range, new_child_id);
490 stack.push(NextDupe {
491 old_id: t.next_id,
492 new_id: new_child_id,
493 });
494 }
495 }
496 self.dupe_stack = stack;
497 new_id
498 }
499
500 /// Adds the given transition to the given state.
501 ///
502 /// Callers must ensure that all previous transitions in this state
503 /// are lexicographically smaller than the given range.
504 fn add_transition(
505 &mut self,
506 from_id: StateID,
507 range: Utf8Range,
508 next_id: StateID,
509 ) {
510 self.state_mut(from_id)
511 .transitions
512 .push(Transition { range, next_id });
513 }
514
515 /// Like `add_transition`, except this inserts the transition just before
516 /// the ith transition.
517 fn add_transition_at(
518 &mut self,
519 i: usize,
520 from_id: StateID,
521 range: Utf8Range,
522 next_id: StateID,
523 ) {
524 self.state_mut(from_id)
525 .transitions
526 .insert(i, Transition { range, next_id });
527 }
528
529 /// Overwrites the transition at position i with the given transition.
530 fn set_transition_at(
531 &mut self,
532 i: usize,
533 from_id: StateID,
534 range: Utf8Range,
535 next_id: StateID,
536 ) {
537 self.state_mut(from_id).transitions[i] = Transition { range, next_id };
538 }
539
540 /// Return an immutable borrow for the state with the given ID.
541 fn state(&self, id: StateID) -> &State {
542 &self.states[id as usize]
543 }
544
545 /// Return a mutable borrow for the state with the given ID.
546 fn state_mut(&mut self, id: StateID) -> &mut State {
547 &mut self.states[id as usize]
548 }
549}
550
551impl State {
552 /// Find the position at which the given range should be inserted in this
553 /// state.
554 ///
555 /// The position returned is always in the inclusive range
556 /// [0, transitions.len()]. If 'transitions.len()' is returned, then the
557 /// given range overlaps with no other range in this state *and* is greater
558 /// than all of them.
559 ///
560 /// For all other possible positions, the given range either overlaps
561 /// with the transition at that position or is otherwise less than it
562 /// with no overlap (and is greater than the previous transition). In the
563 /// former case, careful attention must be paid to inserting this range
564 /// as a new transition. In the latter case, the range can be inserted as
565 /// a new transition at the given position without disrupting any other
566 /// transitions.
567 fn find(&self, range: Utf8Range) -> usize {
568 /// Returns the position `i` at which `pred(xs[i])` first returns true
569 /// such that for all `j >= i`, `pred(xs[j]) == true`. If `pred` never
570 /// returns true, then `xs.len()` is returned.
571 ///
572 /// We roll our own binary search because it doesn't seem like the
573 /// standard library's binary search can be used here. Namely, if
574 /// there is an overlapping range, then we want to find the first such
575 /// occurrence, but there may be many. Or at least, it's not quite
576 /// clear to me how to do it.
577 fn binary_search<T, F>(xs: &[T], mut pred: F) -> usize
578 where
579 F: FnMut(&T) -> bool,
580 {
581 let (mut left, mut right) = (0, xs.len());
582 while left < right {
583 // Overflow is impossible because xs.len() <= 256.
584 let mid = (left + right) / 2;
585 if pred(&xs[mid]) {
586 right = mid;
587 } else {
588 left = mid + 1;
589 }
590 }
591 left
592 }
593
594 // Benchmarks suggest that binary search is just a bit faster than
595 // straight linear search. Specifically when using the debug tool:
596 //
597 // hyperfine "regex-automata-debug debug -acqr '\w{40} ecurB'"
598 binary_search(&self.transitions, |t| range.start <= t.range.end)
599 }
600
601 /// Clear this state such that it has zero transitions.
602 fn clear(&mut self) {
603 self.transitions.clear();
604 }
605}
606
607/// The next state to process during duplication.
608#[derive(Clone, Debug)]
609struct NextDupe {
610 /// The state we want to duplicate.
611 old_id: StateID,
612 /// The ID of the new state that is a duplicate of old_id.
613 new_id: StateID,
614}
615
616/// The next state (and its corresponding transition) that we want to visit
617/// during iteration in lexicographic order.
618#[derive(Clone, Debug)]
619struct NextIter {
620 state_id: StateID,
621 tidx: usize,
622}
623
624/// The next state to process during insertion and any remaining ranges that we
625/// want to add for a partcular sequence of ranges. The first such instance
626/// is always the root state along with all ranges given.
627#[derive(Clone, Debug)]
628struct NextInsert {
629 /// The next state to begin inserting ranges. This state should be the
630 /// state at which `ranges[0]` should be inserted.
631 state_id: StateID,
632 /// The ranges to insert. We used a fixed-size array here to avoid an
633 /// allocation.
634 ranges: [Utf8Range; 4],
635 /// The number of valid ranges in the above array.
636 len: u8,
637}
638
639impl NextInsert {
640 /// Create the next item to visit. The given state ID should correspond
641 /// to the state at which the first range in the given slice should be
642 /// inserted. The slice given must not be empty and it must be no longer
643 /// than 4.
644 fn new(state_id: StateID, ranges: &[Utf8Range]) -> NextInsert {
645 let len = ranges.len();
646 assert!(len > 0);
647 assert!(len <= 4);
648
649 let mut tmp = [Utf8Range { start: 0, end: 0 }; 4];
650 tmp[..len].copy_from_slice(ranges);
651 NextInsert { state_id, ranges: tmp, len: len as u8 }
652 }
653
654 /// Push a new empty state to visit along with any remaining ranges that
655 /// still need to be inserted. The ID of the new empty state is returned.
656 ///
657 /// If ranges is empty, then no new state is created and FINAL is returned.
658 fn push(
659 trie: &mut RangeTrie,
660 stack: &mut Vec<NextInsert>,
661 ranges: &[Utf8Range],
662 ) -> StateID {
663 if ranges.is_empty() {
664 FINAL
665 } else {
666 let next_id = trie.add_empty();
667 stack.push(NextInsert::new(next_id, ranges));
668 next_id
669 }
670 }
671
672 /// Return the ID of the state to visit.
673 fn state_id(&self) -> StateID {
674 self.state_id
675 }
676
677 /// Return the remaining ranges to insert.
678 fn ranges(&self) -> &[Utf8Range] {
679 &self.ranges[..self.len as usize]
680 }
681}
682
683/// Split represents a partitioning of two ranges into one or more ranges. This
684/// is the secret sauce that makes a range trie work, as it's what tells us
685/// how to deal with two overlapping but unequal ranges during insertion.
686///
687/// Essentially, either two ranges overlap or they don't. If they don't, then
688/// handling insertion is easy: just insert the new range into its
689/// lexicographically correct position. Since it does not overlap with anything
690/// else, no other transitions are impacted by the new range.
691///
692/// If they do overlap though, there are generally three possible cases to
693/// handle:
694///
695/// 1. The part where the two ranges actually overlap. i.e., The intersection.
696/// 2. The part of the existing range that is not in the the new range.
697/// 3. The part of the new range that is not in the old range.
698///
699/// (1) is guaranteed to always occur since all overlapping ranges have a
700/// non-empty intersection. If the two ranges are not equivalent, then at
701/// least one of (2) or (3) is guaranteed to occur as well. In some cases,
702/// e.g., `[0-4]` and `[4-9]`, all three cases will occur.
703///
704/// This `Split` type is responsible for providing (1), (2) and (3) for any
705/// possible pair of byte ranges.
706///
707/// As for insertion, for the overlap in (1), the remaining ranges to insert
708/// should be added by following the corresponding transition. However, this
709/// should only be done for the overlapping parts of the range. If there was
710/// a part of the existing range that was not in the new range, then that
711/// existing part must be split off from the transition and duplicated. The
712/// remaining parts of the overlap can then be added to using the new ranges
713/// without disturbing the existing range.
714///
715/// Handling the case for the part of a new range that is not in an existing
716/// range is seemingly easy. Just treat it as if it were a non-overlapping
717/// range. The problem here is that if this new non-overlapping range occurs
718/// after both (1) and (2), then it's possible that it can overlap with the
719/// next transition in the current state. If it does, then the whole process
720/// must be repeated!
721///
722/// # Details of the 3 cases
723///
724/// The following details the various cases that are implemented in code
725/// below. It's plausible that the number of cases is not actually minimal,
726/// but it's important for this code to remain at least somewhat readable.
727///
728/// Given [a,b] and [x,y], where a <= b, x <= y, b < 256 and y < 256, we define
729/// the follow distinct relationships where at least one must apply. The order
730/// of these matters, since multiple can match. The first to match applies.
731///
732/// 1. b < x <=> [a,b] < [x,y]
733/// 2. y < a <=> [x,y] < [a,b]
734///
735/// In the case of (1) and (2), these are the only cases where there is no
736/// overlap. Or otherwise, the intersection of [a,b] and [x,y] is empty. In
737/// order to compute the intersection, one can do [max(a,x), min(b,y)]. The
738/// intersection in all of the following cases is non-empty.
739///
740/// 3. a = x && b = y <=> [a,b] == [x,y]
741/// 4. a = x && b < y <=> [x,y] right-extends [a,b]
742/// 5. b = y && a > x <=> [x,y] left-extends [a,b]
743/// 6. x = a && y < b <=> [a,b] right-extends [x,y]
744/// 7. y = b && x > a <=> [a,b] left-extends [x,y]
745/// 8. a > x && b < y <=> [x,y] covers [a,b]
746/// 9. x > a && y < b <=> [a,b] covers [x,y]
747/// 10. b = x && a < y <=> [a,b] is left-adjacent to [x,y]
748/// 11. y = a && x < b <=> [x,y] is left-adjacent to [a,b]
749/// 12. b > x && b < y <=> [a,b] left-overlaps [x,y]
750/// 13. y > a && y < b <=> [x,y] left-overlaps [a,b]
751///
752/// In cases 3-13, we can form rules that partition the ranges into a
753/// non-overlapping ordered sequence of ranges:
754///
755/// 3. [a,b]
756/// 4. [a,b], [b+1,y]
757/// 5. [x,a-1], [a,b]
758/// 6. [x,y], [y+1,b]
759/// 7. [a,x-1], [x,y]
760/// 8. [x,a-1], [a,b], [b+1,y]
761/// 9. [a,x-1], [x,y], [y+1,b]
762/// 10. [a,b-1], [b,b], [b+1,y]
763/// 11. [x,y-1], [y,y], [y+1,b]
764/// 12. [a,x-1], [x,b], [b+1,y]
765/// 13. [x,a-1], [a,y], [y+1,b]
766///
767/// In the code below, we go a step further and identify each of the above
768/// outputs as belonging either to the overlap of the two ranges or to one
769/// of [a,b] or [x,y] exclusively.
770#[derive(Clone, Debug, Eq, PartialEq)]
771struct Split {
772 partitions: [SplitRange; 3],
773 len: usize,
774}
775
776/// A tagged range indicating how it was derived from a pair of ranges.
777#[derive(Clone, Copy, Debug, Eq, PartialEq)]
778enum SplitRange {
779 Old(Utf8Range),
780 New(Utf8Range),
781 Both(Utf8Range),
782}
783
784impl Split {
785 /// Create a partitioning of the given ranges.
786 ///
787 /// If the given ranges have an empty intersection, then None is returned.
788 fn new(o: Utf8Range, n: Utf8Range) -> Option<Split> {
789 let range = |r: RangeInclusive<u8>| Utf8Range {
790 start: *r.start(),
791 end: *r.end(),
792 };
793 let old = |r| SplitRange::Old(range(r));
794 let new = |r| SplitRange::New(range(r));
795 let both = |r| SplitRange::Both(range(r));
796
797 // Use same names as the comment above to make it easier to compare.
798 let (a, b, x, y) = (o.start, o.end, n.start, n.end);
799
800 if b < x || y < a {
801 // case 1, case 2
802 None
803 } else if a == x && b == y {
804 // case 3
805 Some(Split::parts1(both(a..=b)))
806 } else if a == x && b < y {
807 // case 4
808 Some(Split::parts2(both(a..=b), new(b + 1..=y)))
809 } else if b == y && a > x {
810 // case 5
811 Some(Split::parts2(new(x..=a - 1), both(a..=b)))
812 } else if x == a && y < b {
813 // case 6
814 Some(Split::parts2(both(x..=y), old(y + 1..=b)))
815 } else if y == b && x > a {
816 // case 7
817 Some(Split::parts2(old(a..=x - 1), both(x..=y)))
818 } else if a > x && b < y {
819 // case 8
820 Some(Split::parts3(new(x..=a - 1), both(a..=b), new(b + 1..=y)))
821 } else if x > a && y < b {
822 // case 9
823 Some(Split::parts3(old(a..=x - 1), both(x..=y), old(y + 1..=b)))
824 } else if b == x && a < y {
825 // case 10
826 Some(Split::parts3(old(a..=b - 1), both(b..=b), new(b + 1..=y)))
827 } else if y == a && x < b {
828 // case 11
829 Some(Split::parts3(new(x..=y - 1), both(y..=y), old(y + 1..=b)))
830 } else if b > x && b < y {
831 // case 12
832 Some(Split::parts3(old(a..=x - 1), both(x..=b), new(b + 1..=y)))
833 } else if y > a && y < b {
834 // case 13
835 Some(Split::parts3(new(x..=a - 1), both(a..=y), old(y + 1..=b)))
836 } else {
837 unreachable!()
838 }
839 }
840
841 /// Create a new split with a single partition. This only occurs when two
842 /// ranges are equivalent.
843 fn parts1(r1: SplitRange) -> Split {
844 // This value doesn't matter since it is never accessed.
845 let nada = SplitRange::Old(Utf8Range { start: 0, end: 0 });
846 Split { partitions: [r1, nada, nada], len: 1 }
847 }
848
849 /// Create a new split with two partitions.
850 fn parts2(r1: SplitRange, r2: SplitRange) -> Split {
851 // This value doesn't matter since it is never accessed.
852 let nada = SplitRange::Old(Utf8Range { start: 0, end: 0 });
853 Split { partitions: [r1, r2, nada], len: 2 }
854 }
855
856 /// Create a new split with three partitions.
857 fn parts3(r1: SplitRange, r2: SplitRange, r3: SplitRange) -> Split {
858 Split { partitions: [r1, r2, r3], len: 3 }
859 }
860
861 /// Return the partitions in this split as a slice.
862 fn as_slice(&self) -> &[SplitRange] {
863 &self.partitions[..self.len]
864 }
865}
866
867impl fmt::Debug for RangeTrie {
868 fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
869 writeln!(f, "")?;
870 for (i, state) in self.states.iter().enumerate() {
871 let status = if i == FINAL as usize { '*' } else { ' ' };
872 writeln!(f, "{}{:06}: {:?}", status, i, state)?;
873 }
874 Ok(())
875 }
876}
877
878impl fmt::Debug for State {
879 fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
880 let rs = self
881 .transitions
882 .iter()
883 .map(|t| format!("{:?}", t))
884 .collect::<Vec<String>>()
885 .join(", ");
886 write!(f, "{}", rs)
887 }
888}
889
890impl fmt::Debug for Transition {
891 fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
892 if self.range.start == self.range.end {
893 write!(f, "{:02X} => {:02X}", self.range.start, self.next_id)
894 } else {
895 write!(
896 f,
897 "{:02X}-{:02X} => {:02X}",
898 self.range.start, self.range.end, self.next_id
899 )
900 }
901 }
902}
903
904/// Returns true if and only if the given ranges intersect.
905fn intersects(r1: Utf8Range, r2: Utf8Range) -> bool {
906 !(r1.end < r2.start || r2.end < r1.start)
907}
908
909#[cfg(test)]
910mod tests {
911 use std::ops::RangeInclusive;
912
913 use regex_syntax::utf8::Utf8Range;
914
915 use super::*;
916
917 fn r(range: RangeInclusive<u8>) -> Utf8Range {
918 Utf8Range { start: *range.start(), end: *range.end() }
919 }
920
921 fn split_maybe(
922 old: RangeInclusive<u8>,
923 new: RangeInclusive<u8>,
924 ) -> Option<Split> {
925 Split::new(r(old), r(new))
926 }
927
928 fn split(
929 old: RangeInclusive<u8>,
930 new: RangeInclusive<u8>,
931 ) -> Vec<SplitRange> {
932 split_maybe(old, new).unwrap().as_slice().to_vec()
933 }
934
935 #[test]
936 fn no_splits() {
937 // case 1
938 assert_eq!(None, split_maybe(0..=1, 2..=3));
939 // case 2
940 assert_eq!(None, split_maybe(2..=3, 0..=1));
941 }
942
943 #[test]
944 fn splits() {
945 let range = |r: RangeInclusive<u8>| Utf8Range {
946 start: *r.start(),
947 end: *r.end(),
948 };
949 let old = |r| SplitRange::Old(range(r));
950 let new = |r| SplitRange::New(range(r));
951 let both = |r| SplitRange::Both(range(r));
952
953 // case 3
954 assert_eq!(split(0..=0, 0..=0), vec![both(0..=0)]);
955 assert_eq!(split(9..=9, 9..=9), vec![both(9..=9)]);
956
957 // case 4
958 assert_eq!(split(0..=5, 0..=6), vec![both(0..=5), new(6..=6)]);
959 assert_eq!(split(0..=5, 0..=8), vec![both(0..=5), new(6..=8)]);
960 assert_eq!(split(5..=5, 5..=8), vec![both(5..=5), new(6..=8)]);
961
962 // case 5
963 assert_eq!(split(1..=5, 0..=5), vec![new(0..=0), both(1..=5)]);
964 assert_eq!(split(3..=5, 0..=5), vec![new(0..=2), both(3..=5)]);
965 assert_eq!(split(5..=5, 0..=5), vec![new(0..=4), both(5..=5)]);
966
967 // case 6
968 assert_eq!(split(0..=6, 0..=5), vec![both(0..=5), old(6..=6)]);
969 assert_eq!(split(0..=8, 0..=5), vec![both(0..=5), old(6..=8)]);
970 assert_eq!(split(5..=8, 5..=5), vec![both(5..=5), old(6..=8)]);
971
972 // case 7
973 assert_eq!(split(0..=5, 1..=5), vec![old(0..=0), both(1..=5)]);
974 assert_eq!(split(0..=5, 3..=5), vec![old(0..=2), both(3..=5)]);
975 assert_eq!(split(0..=5, 5..=5), vec![old(0..=4), both(5..=5)]);
976
977 // case 8
978 assert_eq!(
979 split(3..=6, 2..=7),
980 vec![new(2..=2), both(3..=6), new(7..=7)],
981 );
982 assert_eq!(
983 split(3..=6, 1..=8),
984 vec![new(1..=2), both(3..=6), new(7..=8)],
985 );
986
987 // case 9
988 assert_eq!(
989 split(2..=7, 3..=6),
990 vec![old(2..=2), both(3..=6), old(7..=7)],
991 );
992 assert_eq!(
993 split(1..=8, 3..=6),
994 vec![old(1..=2), both(3..=6), old(7..=8)],
995 );
996
997 // case 10
998 assert_eq!(
999 split(3..=6, 6..=7),
1000 vec![old(3..=5), both(6..=6), new(7..=7)],
1001 );
1002 assert_eq!(
1003 split(3..=6, 6..=8),
1004 vec![old(3..=5), both(6..=6), new(7..=8)],
1005 );
1006 assert_eq!(
1007 split(5..=6, 6..=7),
1008 vec![old(5..=5), both(6..=6), new(7..=7)],
1009 );
1010
1011 // case 11
1012 assert_eq!(
1013 split(6..=7, 3..=6),
1014 vec![new(3..=5), both(6..=6), old(7..=7)],
1015 );
1016 assert_eq!(
1017 split(6..=8, 3..=6),
1018 vec![new(3..=5), both(6..=6), old(7..=8)],
1019 );
1020 assert_eq!(
1021 split(6..=7, 5..=6),
1022 vec![new(5..=5), both(6..=6), old(7..=7)],
1023 );
1024
1025 // case 12
1026 assert_eq!(
1027 split(3..=7, 5..=9),
1028 vec![old(3..=4), both(5..=7), new(8..=9)],
1029 );
1030 assert_eq!(
1031 split(3..=5, 4..=6),
1032 vec![old(3..=3), both(4..=5), new(6..=6)],
1033 );
1034
1035 // case 13
1036 assert_eq!(
1037 split(5..=9, 3..=7),
1038 vec![new(3..=4), both(5..=7), old(8..=9)],
1039 );
1040 assert_eq!(
1041 split(4..=6, 3..=5),
1042 vec![new(3..=3), both(4..=5), old(6..=6)],
1043 );
1044 }
1045
1046 // Arguably there should be more tests here, but in practice, this data
1047 // structure is well covered by the huge number of regex tests.
1048}
1049