| 1 | // Copyright 2022 The Go Authors. All rights reserved. |
|---|---|
| 2 | // Use of this source code is governed by a BSD-style |
| 3 | // license that can be found in the LICENSE file. |
| 4 | |
| 5 | package lcs |
| 6 | |
| 7 | // TODO(adonovan): remove unclear references to "old" in this package. |
| 8 | |
| 9 | import ( |
| 10 | "fmt" |
| 11 | ) |
| 12 | |
| 13 | // A Diff is a replacement of a portion of A by a portion of B. |
| 14 | type Diff struct { |
| 15 | Start, End int // offsets of portion to delete in A |
| 16 | ReplStart, ReplEnd int // offset of replacement text in B |
| 17 | } |
| 18 | |
| 19 | // DiffStrings returns the differences between two strings. |
| 20 | // It does not respect rune boundaries. |
| 21 | func DiffStrings(a, b string) []Diff { return diff(stringSeqs{a, b}) } |
| 22 | |
| 23 | // DiffBytes returns the differences between two byte sequences. |
| 24 | // It does not respect rune boundaries. |
| 25 | func DiffBytes(a, b []byte) []Diff { return diff(bytesSeqs{a, b}) } |
| 26 | |
| 27 | // DiffRunes returns the differences between two rune sequences. |
| 28 | func DiffRunes(a, b []rune) []Diff { return diff(runesSeqs{a, b}) } |
| 29 | |
| 30 | func diff(seqs sequences) []Diff { |
| 31 | // A limit on how deeply the LCS algorithm should search. The value is just a guess. |
| 32 | const maxDiffs = 30 |
| 33 | diff, _ := compute(seqs, twosided, maxDiffs/2) |
| 34 | return diff |
| 35 | } |
| 36 | |
| 37 | // compute computes the list of differences between two sequences, |
| 38 | // along with the LCS. It is exercised directly by tests. |
| 39 | // The algorithm is one of {forward, backward, twosided}. |
| 40 | func compute(seqs sequences, algo func(*editGraph) lcs, limit int) ([]Diff, lcs) { |
| 41 | if limit <= 0 { |
| 42 | limit = 1 << 25 // effectively infinity |
| 43 | } |
| 44 | alen, blen := seqs.lengths() |
| 45 | g := &editGraph{ |
| 46 | seqs: seqs, |
| 47 | vf: newtriang(limit), |
| 48 | vb: newtriang(limit), |
| 49 | limit: limit, |
| 50 | ux: alen, |
| 51 | uy: blen, |
| 52 | delta: alen - blen, |
| 53 | } |
| 54 | lcs := algo(g) |
| 55 | diffs := lcs.toDiffs(alen, blen) |
| 56 | return diffs, lcs |
| 57 | } |
| 58 | |
| 59 | // editGraph carries the information for computing the lcs of two sequences. |
| 60 | type editGraph struct { |
| 61 | seqs sequences |
| 62 | vf, vb label // forward and backward labels |
| 63 | |
| 64 | limit int // maximal value of D |
| 65 | // the bounding rectangle of the current edit graph |
| 66 | lx, ly, ux, uy int |
| 67 | delta int // common subexpression: (ux-lx)-(uy-ly) |
| 68 | } |
| 69 | |
| 70 | // toDiffs converts an LCS to a list of edits. |
| 71 | func (lcs lcs) toDiffs(alen, blen int) []Diff { |
| 72 | var diffs []Diff |
| 73 | var pa, pb int // offsets in a, b |
| 74 | for _, l := range lcs { |
| 75 | if pa < l.X || pb < l.Y { |
| 76 | diffs = append(diffs, Diff{pa, l.X, pb, l.Y}) |
| 77 | } |
| 78 | pa = l.X + l.Len |
| 79 | pb = l.Y + l.Len |
| 80 | } |
| 81 | if pa < alen || pb < blen { |
| 82 | diffs = append(diffs, Diff{pa, alen, pb, blen}) |
| 83 | } |
| 84 | return diffs |
| 85 | } |
| 86 | |
| 87 | // --- FORWARD --- |
| 88 | |
| 89 | // fdone decides if the forwward path has reached the upper right |
| 90 | // corner of the rectangle. If so, it also returns the computed lcs. |
| 91 | func (e *editGraph) fdone(D, k int) (bool, lcs) { |
| 92 | // x, y, k are relative to the rectangle |
| 93 | x := e.vf.get(D, k) |
| 94 | y := x - k |
| 95 | if x == e.ux && y == e.uy { |
| 96 | return true, e.forwardlcs(D, k) |
| 97 | } |
| 98 | return false, nil |
| 99 | } |
| 100 | |
| 101 | // run the forward algorithm, until success or up to the limit on D. |
| 102 | func forward(e *editGraph) lcs { |
| 103 | e.setForward(0, 0, e.lx) |
| 104 | if ok, ans := e.fdone(0, 0); ok { |
| 105 | return ans |
| 106 | } |
| 107 | // from D to D+1 |
| 108 | for D := 0; D < e.limit; D++ { |
| 109 | e.setForward(D+1, -(D + 1), e.getForward(D, -D)) |
| 110 | if ok, ans := e.fdone(D+1, -(D + 1)); ok { |
| 111 | return ans |
| 112 | } |
| 113 | e.setForward(D+1, D+1, e.getForward(D, D)+1) |
| 114 | if ok, ans := e.fdone(D+1, D+1); ok { |
| 115 | return ans |
| 116 | } |
| 117 | for k := -D + 1; k <= D-1; k += 2 { |
| 118 | // these are tricky and easy to get backwards |
| 119 | lookv := e.lookForward(k, e.getForward(D, k-1)+1) |
| 120 | lookh := e.lookForward(k, e.getForward(D, k+1)) |
| 121 | if lookv > lookh { |
| 122 | e.setForward(D+1, k, lookv) |
| 123 | } else { |
| 124 | e.setForward(D+1, k, lookh) |
| 125 | } |
| 126 | if ok, ans := e.fdone(D+1, k); ok { |
| 127 | return ans |
| 128 | } |
| 129 | } |
| 130 | } |
| 131 | // D is too large |
| 132 | // find the D path with maximal x+y inside the rectangle and |
| 133 | // use that to compute the found part of the lcs |
| 134 | kmax := -e.limit - 1 |
| 135 | diagmax := -1 |
| 136 | for k := -e.limit; k <= e.limit; k += 2 { |
| 137 | x := e.getForward(e.limit, k) |
| 138 | y := x - k |
| 139 | if x+y > diagmax && x <= e.ux && y <= e.uy { |
| 140 | diagmax, kmax = x+y, k |
| 141 | } |
| 142 | } |
| 143 | return e.forwardlcs(e.limit, kmax) |
| 144 | } |
| 145 | |
| 146 | // recover the lcs by backtracking from the farthest point reached |
| 147 | func (e *editGraph) forwardlcs(D, k int) lcs { |
| 148 | var ans lcs |
| 149 | for x := e.getForward(D, k); x != 0 || x-k != 0; { |
| 150 | if ok(D-1, k-1) && x-1 == e.getForward(D-1, k-1) { |
| 151 | // if (x-1,y) is labelled D-1, x--,D--,k--,continue |
| 152 | D, k, x = D-1, k-1, x-1 |
| 153 | continue |
| 154 | } else if ok(D-1, k+1) && x == e.getForward(D-1, k+1) { |
| 155 | // if (x,y-1) is labelled D-1, x, D--,k++, continue |
| 156 | D, k = D-1, k+1 |
| 157 | continue |
| 158 | } |
| 159 | // if (x-1,y-1)--(x,y) is a diagonal, prepend,x--,y--, continue |
| 160 | y := x - k |
| 161 | ans = ans.prepend(x+e.lx-1, y+e.ly-1) |
| 162 | x-- |
| 163 | } |
| 164 | return ans |
| 165 | } |
| 166 | |
| 167 | // start at (x,y), go up the diagonal as far as possible, |
| 168 | // and label the result with d |
| 169 | func (e *editGraph) lookForward(k, relx int) int { |
| 170 | rely := relx - k |
| 171 | x, y := relx+e.lx, rely+e.ly |
| 172 | if x < e.ux && y < e.uy { |
| 173 | x += e.seqs.commonPrefixLen(x, e.ux, y, e.uy) |
| 174 | } |
| 175 | return x |
| 176 | } |
| 177 | |
| 178 | func (e *editGraph) setForward(d, k, relx int) { |
| 179 | x := e.lookForward(k, relx) |
| 180 | e.vf.set(d, k, x-e.lx) |
| 181 | } |
| 182 | |
| 183 | func (e *editGraph) getForward(d, k int) int { |
| 184 | x := e.vf.get(d, k) |
| 185 | return x |
| 186 | } |
| 187 | |
| 188 | // --- BACKWARD --- |
| 189 | |
| 190 | // bdone decides if the backward path has reached the lower left corner |
| 191 | func (e *editGraph) bdone(D, k int) (bool, lcs) { |
| 192 | // x, y, k are relative to the rectangle |
| 193 | x := e.vb.get(D, k) |
| 194 | y := x - (k + e.delta) |
| 195 | if x == 0 && y == 0 { |
| 196 | return true, e.backwardlcs(D, k) |
| 197 | } |
| 198 | return false, nil |
| 199 | } |
| 200 | |
| 201 | // run the backward algorithm, until success or up to the limit on D. |
| 202 | func backward(e *editGraph) lcs { |
| 203 | e.setBackward(0, 0, e.ux) |
| 204 | if ok, ans := e.bdone(0, 0); ok { |
| 205 | return ans |
| 206 | } |
| 207 | // from D to D+1 |
| 208 | for D := 0; D < e.limit; D++ { |
| 209 | e.setBackward(D+1, -(D + 1), e.getBackward(D, -D)-1) |
| 210 | if ok, ans := e.bdone(D+1, -(D + 1)); ok { |
| 211 | return ans |
| 212 | } |
| 213 | e.setBackward(D+1, D+1, e.getBackward(D, D)) |
| 214 | if ok, ans := e.bdone(D+1, D+1); ok { |
| 215 | return ans |
| 216 | } |
| 217 | for k := -D + 1; k <= D-1; k += 2 { |
| 218 | // these are tricky and easy to get wrong |
| 219 | lookv := e.lookBackward(k, e.getBackward(D, k-1)) |
| 220 | lookh := e.lookBackward(k, e.getBackward(D, k+1)-1) |
| 221 | if lookv < lookh { |
| 222 | e.setBackward(D+1, k, lookv) |
| 223 | } else { |
| 224 | e.setBackward(D+1, k, lookh) |
| 225 | } |
| 226 | if ok, ans := e.bdone(D+1, k); ok { |
| 227 | return ans |
| 228 | } |
| 229 | } |
| 230 | } |
| 231 | |
| 232 | // D is too large |
| 233 | // find the D path with minimal x+y inside the rectangle and |
| 234 | // use that to compute the part of the lcs found |
| 235 | kmax := -e.limit - 1 |
| 236 | diagmin := 1 << 25 |
| 237 | for k := -e.limit; k <= e.limit; k += 2 { |
| 238 | x := e.getBackward(e.limit, k) |
| 239 | y := x - (k + e.delta) |
| 240 | if x+y < diagmin && x >= 0 && y >= 0 { |
| 241 | diagmin, kmax = x+y, k |
| 242 | } |
| 243 | } |
| 244 | if kmax < -e.limit { |
| 245 | panic(fmt.Sprintf("no paths when limit=%d?", e.limit)) |
| 246 | } |
| 247 | return e.backwardlcs(e.limit, kmax) |
| 248 | } |
| 249 | |
| 250 | // recover the lcs by backtracking |
| 251 | func (e *editGraph) backwardlcs(D, k int) lcs { |
| 252 | var ans lcs |
| 253 | for x := e.getBackward(D, k); x != e.ux || x-(k+e.delta) != e.uy; { |
| 254 | if ok(D-1, k-1) && x == e.getBackward(D-1, k-1) { |
| 255 | // D--, k--, x unchanged |
| 256 | D, k = D-1, k-1 |
| 257 | continue |
| 258 | } else if ok(D-1, k+1) && x+1 == e.getBackward(D-1, k+1) { |
| 259 | // D--, k++, x++ |
| 260 | D, k, x = D-1, k+1, x+1 |
| 261 | continue |
| 262 | } |
| 263 | y := x - (k + e.delta) |
| 264 | ans = ans.append(x+e.lx, y+e.ly) |
| 265 | x++ |
| 266 | } |
| 267 | return ans |
| 268 | } |
| 269 | |
| 270 | // start at (x,y), go down the diagonal as far as possible, |
| 271 | func (e *editGraph) lookBackward(k, relx int) int { |
| 272 | rely := relx - (k + e.delta) // forward k = k + e.delta |
| 273 | x, y := relx+e.lx, rely+e.ly |
| 274 | if x > 0 && y > 0 { |
| 275 | x -= e.seqs.commonSuffixLen(0, x, 0, y) |
| 276 | } |
| 277 | return x |
| 278 | } |
| 279 | |
| 280 | // convert to rectangle, and label the result with d |
| 281 | func (e *editGraph) setBackward(d, k, relx int) { |
| 282 | x := e.lookBackward(k, relx) |
| 283 | e.vb.set(d, k, x-e.lx) |
| 284 | } |
| 285 | |
| 286 | func (e *editGraph) getBackward(d, k int) int { |
| 287 | x := e.vb.get(d, k) |
| 288 | return x |
| 289 | } |
| 290 | |
| 291 | // -- TWOSIDED --- |
| 292 | |
| 293 | func twosided(e *editGraph) lcs { |
| 294 | // The termination condition could be improved, as either the forward |
| 295 | // or backward pass could succeed before Myers' Lemma applies. |
| 296 | // Aside from questions of efficiency (is the extra testing cost-effective) |
| 297 | // this is more likely to matter when e.limit is reached. |
| 298 | e.setForward(0, 0, e.lx) |
| 299 | e.setBackward(0, 0, e.ux) |
| 300 | |
| 301 | // from D to D+1 |
| 302 | for D := 0; D < e.limit; D++ { |
| 303 | // just finished a backwards pass, so check |
| 304 | if got, ok := e.twoDone(D, D); ok { |
| 305 | return e.twolcs(D, D, got) |
| 306 | } |
| 307 | // do a forwards pass (D to D+1) |
| 308 | e.setForward(D+1, -(D + 1), e.getForward(D, -D)) |
| 309 | e.setForward(D+1, D+1, e.getForward(D, D)+1) |
| 310 | for k := -D + 1; k <= D-1; k += 2 { |
| 311 | // these are tricky and easy to get backwards |
| 312 | lookv := e.lookForward(k, e.getForward(D, k-1)+1) |
| 313 | lookh := e.lookForward(k, e.getForward(D, k+1)) |
| 314 | if lookv > lookh { |
| 315 | e.setForward(D+1, k, lookv) |
| 316 | } else { |
| 317 | e.setForward(D+1, k, lookh) |
| 318 | } |
| 319 | } |
| 320 | // just did a forward pass, so check |
| 321 | if got, ok := e.twoDone(D+1, D); ok { |
| 322 | return e.twolcs(D+1, D, got) |
| 323 | } |
| 324 | // do a backward pass, D to D+1 |
| 325 | e.setBackward(D+1, -(D + 1), e.getBackward(D, -D)-1) |
| 326 | e.setBackward(D+1, D+1, e.getBackward(D, D)) |
| 327 | for k := -D + 1; k <= D-1; k += 2 { |
| 328 | // these are tricky and easy to get wrong |
| 329 | lookv := e.lookBackward(k, e.getBackward(D, k-1)) |
| 330 | lookh := e.lookBackward(k, e.getBackward(D, k+1)-1) |
| 331 | if lookv < lookh { |
| 332 | e.setBackward(D+1, k, lookv) |
| 333 | } else { |
| 334 | e.setBackward(D+1, k, lookh) |
| 335 | } |
| 336 | } |
| 337 | } |
| 338 | |
| 339 | // D too large. combine a forward and backward partial lcs |
| 340 | // first, a forward one |
| 341 | kmax := -e.limit - 1 |
| 342 | diagmax := -1 |
| 343 | for k := -e.limit; k <= e.limit; k += 2 { |
| 344 | x := e.getForward(e.limit, k) |
| 345 | y := x - k |
| 346 | if x+y > diagmax && x <= e.ux && y <= e.uy { |
| 347 | diagmax, kmax = x+y, k |
| 348 | } |
| 349 | } |
| 350 | if kmax < -e.limit { |
| 351 | panic(fmt.Sprintf("no forward paths when limit=%d?", e.limit)) |
| 352 | } |
| 353 | lcs := e.forwardlcs(e.limit, kmax) |
| 354 | // now a backward one |
| 355 | // find the D path with minimal x+y inside the rectangle and |
| 356 | // use that to compute the lcs |
| 357 | diagmin := 1 << 25 // infinity |
| 358 | for k := -e.limit; k <= e.limit; k += 2 { |
| 359 | x := e.getBackward(e.limit, k) |
| 360 | y := x - (k + e.delta) |
| 361 | if x+y < diagmin && x >= 0 && y >= 0 { |
| 362 | diagmin, kmax = x+y, k |
| 363 | } |
| 364 | } |
| 365 | if kmax < -e.limit { |
| 366 | panic(fmt.Sprintf("no backward paths when limit=%d?", e.limit)) |
| 367 | } |
| 368 | lcs = append(lcs, e.backwardlcs(e.limit, kmax)...) |
| 369 | // These may overlap (e.forwardlcs and e.backwardlcs return sorted lcs) |
| 370 | ans := lcs.fix() |
| 371 | return ans |
| 372 | } |
| 373 | |
| 374 | // Does Myers' Lemma apply? |
| 375 | func (e *editGraph) twoDone(df, db int) (int, bool) { |
| 376 | if (df+db+e.delta)%2 != 0 { |
| 377 | return 0, false // diagonals cannot overlap |
| 378 | } |
| 379 | kmin := -db + e.delta |
| 380 | if -df > kmin { |
| 381 | kmin = -df |
| 382 | } |
| 383 | kmax := db + e.delta |
| 384 | if df < kmax { |
| 385 | kmax = df |
| 386 | } |
| 387 | for k := kmin; k <= kmax; k += 2 { |
| 388 | x := e.vf.get(df, k) |
| 389 | u := e.vb.get(db, k-e.delta) |
| 390 | if u <= x { |
| 391 | // is it worth looking at all the other k? |
| 392 | for l := k; l <= kmax; l += 2 { |
| 393 | x := e.vf.get(df, l) |
| 394 | y := x - l |
| 395 | u := e.vb.get(db, l-e.delta) |
| 396 | v := u - l |
| 397 | if x == u || u == 0 || v == 0 || y == e.uy || x == e.ux { |
| 398 | return l, true |
| 399 | } |
| 400 | } |
| 401 | return k, true |
| 402 | } |
| 403 | } |
| 404 | return 0, false |
| 405 | } |
| 406 | |
| 407 | func (e *editGraph) twolcs(df, db, kf int) lcs { |
| 408 | // db==df || db+1==df |
| 409 | x := e.vf.get(df, kf) |
| 410 | y := x - kf |
| 411 | kb := kf - e.delta |
| 412 | u := e.vb.get(db, kb) |
| 413 | v := u - kf |
| 414 | |
| 415 | // Myers proved there is a df-path from (0,0) to (u,v) |
| 416 | // and a db-path from (x,y) to (N,M). |
| 417 | // In the first case the overall path is the forward path |
| 418 | // to (u,v) followed by the backward path to (N,M). |
| 419 | // In the second case the path is the backward path to (x,y) |
| 420 | // followed by the forward path to (x,y) from (0,0). |
| 421 | |
| 422 | // Look for some special cases to avoid computing either of these paths. |
| 423 | if x == u { |
| 424 | // "babaab" "cccaba" |
| 425 | // already patched together |
| 426 | lcs := e.forwardlcs(df, kf) |
| 427 | lcs = append(lcs, e.backwardlcs(db, kb)...) |
| 428 | return lcs.sort() |
| 429 | } |
| 430 | |
| 431 | // is (u-1,v) or (u,v-1) labelled df-1? |
| 432 | // if so, that forward df-1-path plus a horizontal or vertical edge |
| 433 | // is the df-path to (u,v), then plus the db-path to (N,M) |
| 434 | if u > 0 && ok(df-1, u-1-v) && e.vf.get(df-1, u-1-v) == u-1 { |
| 435 | // "aabbab" "cbcabc" |
| 436 | lcs := e.forwardlcs(df-1, u-1-v) |
| 437 | lcs = append(lcs, e.backwardlcs(db, kb)...) |
| 438 | return lcs.sort() |
| 439 | } |
| 440 | if v > 0 && ok(df-1, (u-(v-1))) && e.vf.get(df-1, u-(v-1)) == u { |
| 441 | // "abaabb" "bcacab" |
| 442 | lcs := e.forwardlcs(df-1, u-(v-1)) |
| 443 | lcs = append(lcs, e.backwardlcs(db, kb)...) |
| 444 | return lcs.sort() |
| 445 | } |
| 446 | |
| 447 | // The path can't possibly contribute to the lcs because it |
| 448 | // is all horizontal or vertical edges |
| 449 | if u == 0 || v == 0 || x == e.ux || y == e.uy { |
| 450 | // "abaabb" "abaaaa" |
| 451 | if u == 0 || v == 0 { |
| 452 | return e.backwardlcs(db, kb) |
| 453 | } |
| 454 | return e.forwardlcs(df, kf) |
| 455 | } |
| 456 | |
| 457 | // is (x+1,y) or (x,y+1) labelled db-1? |
| 458 | if x+1 <= e.ux && ok(db-1, x+1-y-e.delta) && e.vb.get(db-1, x+1-y-e.delta) == x+1 { |
| 459 | // "bababb" "baaabb" |
| 460 | lcs := e.backwardlcs(db-1, kb+1) |
| 461 | lcs = append(lcs, e.forwardlcs(df, kf)...) |
| 462 | return lcs.sort() |
| 463 | } |
| 464 | if y+1 <= e.uy && ok(db-1, x-(y+1)-e.delta) && e.vb.get(db-1, x-(y+1)-e.delta) == x { |
| 465 | // "abbbaa" "cabacc" |
| 466 | lcs := e.backwardlcs(db-1, kb-1) |
| 467 | lcs = append(lcs, e.forwardlcs(df, kf)...) |
| 468 | return lcs.sort() |
| 469 | } |
| 470 | |
| 471 | // need to compute another path |
| 472 | // "aabbaa" "aacaba" |
| 473 | lcs := e.backwardlcs(db, kb) |
| 474 | oldx, oldy := e.ux, e.uy |
| 475 | e.ux = u |
| 476 | e.uy = v |
| 477 | lcs = append(lcs, forward(e)...) |
| 478 | e.ux, e.uy = oldx, oldy |
| 479 | return lcs.sort() |
| 480 | } |
| 481 |
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