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qwen/nodejs/node_modules/diff-sequences/build/index.js

@@ -0,0 +1,798 @@
+'use strict';
+
+Object.defineProperty(exports, '__esModule', {
+  value: true
+});
+exports.default = diffSequence;
+/**
+ * Copyright (c) Meta Platforms, Inc. and affiliates.
+ *
+ * This source code is licensed under the MIT license found in the
+ * LICENSE file in the root directory of this source tree.
+ *
+ */
+
+// This diff-sequences package implements the linear space variation in
+// An O(ND) Difference Algorithm and Its Variations by Eugene W. Myers
+
+// Relationship in notation between Myers paper and this package:
+// A is a
+// N is aLength, aEnd - aStart, and so on
+// x is aIndex, aFirst, aLast, and so on
+// B is b
+// M is bLength, bEnd - bStart, and so on
+// y is bIndex, bFirst, bLast, and so on
+// Δ = N - M is negative of baDeltaLength = bLength - aLength
+// D is d
+// k is kF
+// k + Δ is kF = kR - baDeltaLength
+// V is aIndexesF or aIndexesR (see comment below about Indexes type)
+// index intervals [1, N] and [1, M] are [0, aLength) and [0, bLength)
+// starting point in forward direction (0, 0) is (-1, -1)
+// starting point in reverse direction (N + 1, M + 1) is (aLength, bLength)
+
+// The “edit graph” for sequences a and b corresponds to items:
+// in a on the horizontal axis
+// in b on the vertical axis
+//
+// Given a-coordinate of a point in a diagonal, you can compute b-coordinate.
+//
+// Forward diagonals kF:
+// zero diagonal intersects top left corner
+// positive diagonals intersect top edge
+// negative diagonals insersect left edge
+//
+// Reverse diagonals kR:
+// zero diagonal intersects bottom right corner
+// positive diagonals intersect right edge
+// negative diagonals intersect bottom edge
+
+// The graph contains a directed acyclic graph of edges:
+// horizontal: delete an item from a
+// vertical: insert an item from b
+// diagonal: common item in a and b
+//
+// The algorithm solves dual problems in the graph analogy:
+// Find longest common subsequence: path with maximum number of diagonal edges
+// Find shortest edit script: path with minimum number of non-diagonal edges
+
+// Input callback function compares items at indexes in the sequences.
+
+// Output callback function receives the number of adjacent items
+// and starting indexes of each common subsequence.
+// Either original functions or wrapped to swap indexes if graph is transposed.
+// Indexes in sequence a of last point of forward or reverse paths in graph.
+// Myers algorithm indexes by diagonal k which for negative is bad deopt in V8.
+// This package indexes by iF and iR which are greater than or equal to zero.
+// and also updates the index arrays in place to cut memory in half.
+// kF = 2 * iF - d
+// kR = d - 2 * iR
+// Division of index intervals in sequences a and b at the middle change.
+// Invariant: intervals do not have common items at the start or end.
+const pkg = 'diff-sequences'; // for error messages
+const NOT_YET_SET = 0; // small int instead of undefined to avoid deopt in V8
+
+// Return the number of common items that follow in forward direction.
+// The length of what Myers paper calls a “snake” in a forward path.
+const countCommonItemsF = (aIndex, aEnd, bIndex, bEnd, isCommon) => {
+  let nCommon = 0;
+  while (aIndex < aEnd && bIndex < bEnd && isCommon(aIndex, bIndex)) {
+    aIndex += 1;
+    bIndex += 1;
+    nCommon += 1;
+  }
+  return nCommon;
+};
+
+// Return the number of common items that precede in reverse direction.
+// The length of what Myers paper calls a “snake” in a reverse path.
+const countCommonItemsR = (aStart, aIndex, bStart, bIndex, isCommon) => {
+  let nCommon = 0;
+  while (aStart <= aIndex && bStart <= bIndex && isCommon(aIndex, bIndex)) {
+    aIndex -= 1;
+    bIndex -= 1;
+    nCommon += 1;
+  }
+  return nCommon;
+};
+
+// A simple function to extend forward paths from (d - 1) to d changes
+// when forward and reverse paths cannot yet overlap.
+const extendPathsF = (
+  d,
+  aEnd,
+  bEnd,
+  bF,
+  isCommon,
+  aIndexesF,
+  iMaxF // return the value because optimization might decrease it
+) => {
+  // Unroll the first iteration.
+  let iF = 0;
+  let kF = -d; // kF = 2 * iF - d
+  let aFirst = aIndexesF[iF]; // in first iteration always insert
+  let aIndexPrev1 = aFirst; // prev value of [iF - 1] in next iteration
+  aIndexesF[iF] += countCommonItemsF(
+    aFirst + 1,
+    aEnd,
+    bF + aFirst - kF + 1,
+    bEnd,
+    isCommon
+  );
+
+  // Optimization: skip diagonals in which paths cannot ever overlap.
+  const nF = d < iMaxF ? d : iMaxF;
+
+  // The diagonals kF are odd when d is odd and even when d is even.
+  for (iF += 1, kF += 2; iF <= nF; iF += 1, kF += 2) {
+    // To get first point of path segment, move one change in forward direction
+    // from last point of previous path segment in an adjacent diagonal.
+    // In last possible iteration when iF === d and kF === d always delete.
+    if (iF !== d && aIndexPrev1 < aIndexesF[iF]) {
+      aFirst = aIndexesF[iF]; // vertical to insert from b
+    } else {
+      aFirst = aIndexPrev1 + 1; // horizontal to delete from a
+
+      if (aEnd <= aFirst) {
+        // Optimization: delete moved past right of graph.
+        return iF - 1;
+      }
+    }
+
+    // To get last point of path segment, move along diagonal of common items.
+    aIndexPrev1 = aIndexesF[iF];
+    aIndexesF[iF] =
+      aFirst +
+      countCommonItemsF(aFirst + 1, aEnd, bF + aFirst - kF + 1, bEnd, isCommon);
+  }
+  return iMaxF;
+};
+
+// A simple function to extend reverse paths from (d - 1) to d changes
+// when reverse and forward paths cannot yet overlap.
+const extendPathsR = (
+  d,
+  aStart,
+  bStart,
+  bR,
+  isCommon,
+  aIndexesR,
+  iMaxR // return the value because optimization might decrease it
+) => {
+  // Unroll the first iteration.
+  let iR = 0;
+  let kR = d; // kR = d - 2 * iR
+  let aFirst = aIndexesR[iR]; // in first iteration always insert
+  let aIndexPrev1 = aFirst; // prev value of [iR - 1] in next iteration
+  aIndexesR[iR] -= countCommonItemsR(
+    aStart,
+    aFirst - 1,
+    bStart,
+    bR + aFirst - kR - 1,
+    isCommon
+  );
+
+  // Optimization: skip diagonals in which paths cannot ever overlap.
+  const nR = d < iMaxR ? d : iMaxR;
+
+  // The diagonals kR are odd when d is odd and even when d is even.
+  for (iR += 1, kR -= 2; iR <= nR; iR += 1, kR -= 2) {
+    // To get first point of path segment, move one change in reverse direction
+    // from last point of previous path segment in an adjacent diagonal.
+    // In last possible iteration when iR === d and kR === -d always delete.
+    if (iR !== d && aIndexesR[iR] < aIndexPrev1) {
+      aFirst = aIndexesR[iR]; // vertical to insert from b
+    } else {
+      aFirst = aIndexPrev1 - 1; // horizontal to delete from a
+
+      if (aFirst < aStart) {
+        // Optimization: delete moved past left of graph.
+        return iR - 1;
+      }
+    }
+
+    // To get last point of path segment, move along diagonal of common items.
+    aIndexPrev1 = aIndexesR[iR];
+    aIndexesR[iR] =
+      aFirst -
+      countCommonItemsR(
+        aStart,
+        aFirst - 1,
+        bStart,
+        bR + aFirst - kR - 1,
+        isCommon
+      );
+  }
+  return iMaxR;
+};
+
+// A complete function to extend forward paths from (d - 1) to d changes.
+// Return true if a path overlaps reverse path of (d - 1) changes in its diagonal.
+const extendOverlappablePathsF = (
+  d,
+  aStart,
+  aEnd,
+  bStart,
+  bEnd,
+  isCommon,
+  aIndexesF,
+  iMaxF,
+  aIndexesR,
+  iMaxR,
+  division // update prop values if return true
+) => {
+  const bF = bStart - aStart; // bIndex = bF + aIndex - kF
+  const aLength = aEnd - aStart;
+  const bLength = bEnd - bStart;
+  const baDeltaLength = bLength - aLength; // kF = kR - baDeltaLength
+
+  // Range of diagonals in which forward and reverse paths might overlap.
+  const kMinOverlapF = -baDeltaLength - (d - 1); // -(d - 1) <= kR
+  const kMaxOverlapF = -baDeltaLength + (d - 1); // kR <= (d - 1)
+
+  let aIndexPrev1 = NOT_YET_SET; // prev value of [iF - 1] in next iteration
+
+  // Optimization: skip diagonals in which paths cannot ever overlap.
+  const nF = d < iMaxF ? d : iMaxF;
+
+  // The diagonals kF = 2 * iF - d are odd when d is odd and even when d is even.
+  for (let iF = 0, kF = -d; iF <= nF; iF += 1, kF += 2) {
+    // To get first point of path segment, move one change in forward direction
+    // from last point of previous path segment in an adjacent diagonal.
+    // In first iteration when iF === 0 and kF === -d always insert.
+    // In last possible iteration when iF === d and kF === d always delete.
+    const insert = iF === 0 || (iF !== d && aIndexPrev1 < aIndexesF[iF]);
+    const aLastPrev = insert ? aIndexesF[iF] : aIndexPrev1;
+    const aFirst = insert
+      ? aLastPrev // vertical to insert from b
+      : aLastPrev + 1; // horizontal to delete from a
+
+    // To get last point of path segment, move along diagonal of common items.
+    const bFirst = bF + aFirst - kF;
+    const nCommonF = countCommonItemsF(
+      aFirst + 1,
+      aEnd,
+      bFirst + 1,
+      bEnd,
+      isCommon
+    );
+    const aLast = aFirst + nCommonF;
+    aIndexPrev1 = aIndexesF[iF];
+    aIndexesF[iF] = aLast;
+    if (kMinOverlapF <= kF && kF <= kMaxOverlapF) {
+      // Solve for iR of reverse path with (d - 1) changes in diagonal kF:
+      // kR = kF + baDeltaLength
+      // kR = (d - 1) - 2 * iR
+      const iR = (d - 1 - (kF + baDeltaLength)) / 2;
+
+      // If this forward path overlaps the reverse path in this diagonal,
+      // then this is the middle change of the index intervals.
+      if (iR <= iMaxR && aIndexesR[iR] - 1 <= aLast) {
+        // Unlike the Myers algorithm which finds only the middle “snake”
+        // this package can find two common subsequences per division.
+        // Last point of previous path segment is on an adjacent diagonal.
+        const bLastPrev = bF + aLastPrev - (insert ? kF + 1 : kF - 1);
+
+        // Because of invariant that intervals preceding the middle change
+        // cannot have common items at the end,
+        // move in reverse direction along a diagonal of common items.
+        const nCommonR = countCommonItemsR(
+          aStart,
+          aLastPrev,
+          bStart,
+          bLastPrev,
+          isCommon
+        );
+        const aIndexPrevFirst = aLastPrev - nCommonR;
+        const bIndexPrevFirst = bLastPrev - nCommonR;
+        const aEndPreceding = aIndexPrevFirst + 1;
+        const bEndPreceding = bIndexPrevFirst + 1;
+        division.nChangePreceding = d - 1;
+        if (d - 1 === aEndPreceding + bEndPreceding - aStart - bStart) {
+          // Optimization: number of preceding changes in forward direction
+          // is equal to number of items in preceding interval,
+          // therefore it cannot contain any common items.
+          division.aEndPreceding = aStart;
+          division.bEndPreceding = bStart;
+        } else {
+          division.aEndPreceding = aEndPreceding;
+          division.bEndPreceding = bEndPreceding;
+        }
+        division.nCommonPreceding = nCommonR;
+        if (nCommonR !== 0) {
+          division.aCommonPreceding = aEndPreceding;
+          division.bCommonPreceding = bEndPreceding;
+        }
+        division.nCommonFollowing = nCommonF;
+        if (nCommonF !== 0) {
+          division.aCommonFollowing = aFirst + 1;
+          division.bCommonFollowing = bFirst + 1;
+        }
+        const aStartFollowing = aLast + 1;
+        const bStartFollowing = bFirst + nCommonF + 1;
+        division.nChangeFollowing = d - 1;
+        if (d - 1 === aEnd + bEnd - aStartFollowing - bStartFollowing) {
+          // Optimization: number of changes in reverse direction
+          // is equal to number of items in following interval,
+          // therefore it cannot contain any common items.
+          division.aStartFollowing = aEnd;
+          division.bStartFollowing = bEnd;
+        } else {
+          division.aStartFollowing = aStartFollowing;
+          division.bStartFollowing = bStartFollowing;
+        }
+        return true;
+      }
+    }
+  }
+  return false;
+};
+
+// A complete function to extend reverse paths from (d - 1) to d changes.
+// Return true if a path overlaps forward path of d changes in its diagonal.
+const extendOverlappablePathsR = (
+  d,
+  aStart,
+  aEnd,
+  bStart,
+  bEnd,
+  isCommon,
+  aIndexesF,
+  iMaxF,
+  aIndexesR,
+  iMaxR,
+  division // update prop values if return true
+) => {
+  const bR = bEnd - aEnd; // bIndex = bR + aIndex - kR
+  const aLength = aEnd - aStart;
+  const bLength = bEnd - bStart;
+  const baDeltaLength = bLength - aLength; // kR = kF + baDeltaLength
+
+  // Range of diagonals in which forward and reverse paths might overlap.
+  const kMinOverlapR = baDeltaLength - d; // -d <= kF
+  const kMaxOverlapR = baDeltaLength + d; // kF <= d
+
+  let aIndexPrev1 = NOT_YET_SET; // prev value of [iR - 1] in next iteration
+
+  // Optimization: skip diagonals in which paths cannot ever overlap.
+  const nR = d < iMaxR ? d : iMaxR;
+
+  // The diagonals kR = d - 2 * iR are odd when d is odd and even when d is even.
+  for (let iR = 0, kR = d; iR <= nR; iR += 1, kR -= 2) {
+    // To get first point of path segment, move one change in reverse direction
+    // from last point of previous path segment in an adjacent diagonal.
+    // In first iteration when iR === 0 and kR === d always insert.
+    // In last possible iteration when iR === d and kR === -d always delete.
+    const insert = iR === 0 || (iR !== d && aIndexesR[iR] < aIndexPrev1);
+    const aLastPrev = insert ? aIndexesR[iR] : aIndexPrev1;
+    const aFirst = insert
+      ? aLastPrev // vertical to insert from b
+      : aLastPrev - 1; // horizontal to delete from a
+
+    // To get last point of path segment, move along diagonal of common items.
+    const bFirst = bR + aFirst - kR;
+    const nCommonR = countCommonItemsR(
+      aStart,
+      aFirst - 1,
+      bStart,
+      bFirst - 1,
+      isCommon
+    );
+    const aLast = aFirst - nCommonR;
+    aIndexPrev1 = aIndexesR[iR];
+    aIndexesR[iR] = aLast;
+    if (kMinOverlapR <= kR && kR <= kMaxOverlapR) {
+      // Solve for iF of forward path with d changes in diagonal kR:
+      // kF = kR - baDeltaLength
+      // kF = 2 * iF - d
+      const iF = (d + (kR - baDeltaLength)) / 2;
+
+      // If this reverse path overlaps the forward path in this diagonal,
+      // then this is a middle change of the index intervals.
+      if (iF <= iMaxF && aLast - 1 <= aIndexesF[iF]) {
+        const bLast = bFirst - nCommonR;
+        division.nChangePreceding = d;
+        if (d === aLast + bLast - aStart - bStart) {
+          // Optimization: number of changes in reverse direction
+          // is equal to number of items in preceding interval,
+          // therefore it cannot contain any common items.
+          division.aEndPreceding = aStart;
+          division.bEndPreceding = bStart;
+        } else {
+          division.aEndPreceding = aLast;
+          division.bEndPreceding = bLast;
+        }
+        division.nCommonPreceding = nCommonR;
+        if (nCommonR !== 0) {
+          // The last point of reverse path segment is start of common subsequence.
+          division.aCommonPreceding = aLast;
+          division.bCommonPreceding = bLast;
+        }
+        division.nChangeFollowing = d - 1;
+        if (d === 1) {
+          // There is no previous path segment.
+          division.nCommonFollowing = 0;
+          division.aStartFollowing = aEnd;
+          division.bStartFollowing = bEnd;
+        } else {
+          // Unlike the Myers algorithm which finds only the middle “snake”
+          // this package can find two common subsequences per division.
+          // Last point of previous path segment is on an adjacent diagonal.
+          const bLastPrev = bR + aLastPrev - (insert ? kR - 1 : kR + 1);
+
+          // Because of invariant that intervals following the middle change
+          // cannot have common items at the start,
+          // move in forward direction along a diagonal of common items.
+          const nCommonF = countCommonItemsF(
+            aLastPrev,
+            aEnd,
+            bLastPrev,
+            bEnd,
+            isCommon
+          );
+          division.nCommonFollowing = nCommonF;
+          if (nCommonF !== 0) {
+            // The last point of reverse path segment is start of common subsequence.
+            division.aCommonFollowing = aLastPrev;
+            division.bCommonFollowing = bLastPrev;
+          }
+          const aStartFollowing = aLastPrev + nCommonF; // aFirstPrev
+          const bStartFollowing = bLastPrev + nCommonF; // bFirstPrev
+
+          if (d - 1 === aEnd + bEnd - aStartFollowing - bStartFollowing) {
+            // Optimization: number of changes in forward direction
+            // is equal to number of items in following interval,
+            // therefore it cannot contain any common items.
+            division.aStartFollowing = aEnd;
+            division.bStartFollowing = bEnd;
+          } else {
+            division.aStartFollowing = aStartFollowing;
+            division.bStartFollowing = bStartFollowing;
+          }
+        }
+        return true;
+      }
+    }
+  }
+  return false;
+};
+
+// Given index intervals and input function to compare items at indexes,
+// divide at the middle change.
+//
+// DO NOT CALL if start === end, because interval cannot contain common items
+// and because this function will throw the “no overlap” error.
+const divide = (
+  nChange,
+  aStart,
+  aEnd,
+  bStart,
+  bEnd,
+  isCommon,
+  aIndexesF,
+  aIndexesR,
+  division // output
+) => {
+  const bF = bStart - aStart; // bIndex = bF + aIndex - kF
+  const bR = bEnd - aEnd; // bIndex = bR + aIndex - kR
+  const aLength = aEnd - aStart;
+  const bLength = bEnd - bStart;
+
+  // Because graph has square or portrait orientation,
+  // length difference is minimum number of items to insert from b.
+  // Corresponding forward and reverse diagonals in graph
+  // depend on length difference of the sequences:
+  // kF = kR - baDeltaLength
+  // kR = kF + baDeltaLength
+  const baDeltaLength = bLength - aLength;
+
+  // Optimization: max diagonal in graph intersects corner of shorter side.
+  let iMaxF = aLength;
+  let iMaxR = aLength;
+
+  // Initialize no changes yet in forward or reverse direction:
+  aIndexesF[0] = aStart - 1; // at open start of interval, outside closed start
+  aIndexesR[0] = aEnd; // at open end of interval
+
+  if (baDeltaLength % 2 === 0) {
+    // The number of changes in paths is 2 * d if length difference is even.
+    const dMin = (nChange || baDeltaLength) / 2;
+    const dMax = (aLength + bLength) / 2;
+    for (let d = 1; d <= dMax; d += 1) {
+      iMaxF = extendPathsF(d, aEnd, bEnd, bF, isCommon, aIndexesF, iMaxF);
+      if (d < dMin) {
+        iMaxR = extendPathsR(d, aStart, bStart, bR, isCommon, aIndexesR, iMaxR);
+      } else if (
+        // If a reverse path overlaps a forward path in the same diagonal,
+        // return a division of the index intervals at the middle change.
+        extendOverlappablePathsR(
+          d,
+          aStart,
+          aEnd,
+          bStart,
+          bEnd,
+          isCommon,
+          aIndexesF,
+          iMaxF,
+          aIndexesR,
+          iMaxR,
+          division
+        )
+      ) {
+        return;
+      }
+    }
+  } else {
+    // The number of changes in paths is 2 * d - 1 if length difference is odd.
+    const dMin = ((nChange || baDeltaLength) + 1) / 2;
+    const dMax = (aLength + bLength + 1) / 2;
+
+    // Unroll first half iteration so loop extends the relevant pairs of paths.
+    // Because of invariant that intervals have no common items at start or end,
+    // and limitation not to call divide with empty intervals,
+    // therefore it cannot be called if a forward path with one change
+    // would overlap a reverse path with no changes, even if dMin === 1.
+    let d = 1;
+    iMaxF = extendPathsF(d, aEnd, bEnd, bF, isCommon, aIndexesF, iMaxF);
+    for (d += 1; d <= dMax; d += 1) {
+      iMaxR = extendPathsR(
+        d - 1,
+        aStart,
+        bStart,
+        bR,
+        isCommon,
+        aIndexesR,
+        iMaxR
+      );
+      if (d < dMin) {
+        iMaxF = extendPathsF(d, aEnd, bEnd, bF, isCommon, aIndexesF, iMaxF);
+      } else if (
+        // If a forward path overlaps a reverse path in the same diagonal,
+        // return a division of the index intervals at the middle change.
+        extendOverlappablePathsF(
+          d,
+          aStart,
+          aEnd,
+          bStart,
+          bEnd,
+          isCommon,
+          aIndexesF,
+          iMaxF,
+          aIndexesR,
+          iMaxR,
+          division
+        )
+      ) {
+        return;
+      }
+    }
+  }
+
+  /* istanbul ignore next */
+  throw new Error(
+    `${pkg}: no overlap aStart=${aStart} aEnd=${aEnd} bStart=${bStart} bEnd=${bEnd}`
+  );
+};
+
+// Given index intervals and input function to compare items at indexes,
+// return by output function the number of adjacent items and starting indexes
+// of each common subsequence. Divide and conquer with only linear space.
+//
+// The index intervals are half open [start, end) like array slice method.
+// DO NOT CALL if start === end, because interval cannot contain common items
+// and because divide function will throw the “no overlap” error.
+const findSubsequences = (
+  nChange,
+  aStart,
+  aEnd,
+  bStart,
+  bEnd,
+  transposed,
+  callbacks,
+  aIndexesF,
+  aIndexesR,
+  division // temporary memory, not input nor output
+) => {
+  if (bEnd - bStart < aEnd - aStart) {
+    // Transpose graph so it has portrait instead of landscape orientation.
+    // Always compare shorter to longer sequence for consistency and optimization.
+    transposed = !transposed;
+    if (transposed && callbacks.length === 1) {
+      // Lazily wrap callback functions to swap args if graph is transposed.
+      const {foundSubsequence, isCommon} = callbacks[0];
+      callbacks[1] = {
+        foundSubsequence: (nCommon, bCommon, aCommon) => {
+          foundSubsequence(nCommon, aCommon, bCommon);
+        },
+        isCommon: (bIndex, aIndex) => isCommon(aIndex, bIndex)
+      };
+    }
+    const tStart = aStart;
+    const tEnd = aEnd;
+    aStart = bStart;
+    aEnd = bEnd;
+    bStart = tStart;
+    bEnd = tEnd;
+  }
+  const {foundSubsequence, isCommon} = callbacks[transposed ? 1 : 0];
+
+  // Divide the index intervals at the middle change.
+  divide(
+    nChange,
+    aStart,
+    aEnd,
+    bStart,
+    bEnd,
+    isCommon,
+    aIndexesF,
+    aIndexesR,
+    division
+  );
+  const {
+    nChangePreceding,
+    aEndPreceding,
+    bEndPreceding,
+    nCommonPreceding,
+    aCommonPreceding,
+    bCommonPreceding,
+    nCommonFollowing,
+    aCommonFollowing,
+    bCommonFollowing,
+    nChangeFollowing,
+    aStartFollowing,
+    bStartFollowing
+  } = division;
+
+  // Unless either index interval is empty, they might contain common items.
+  if (aStart < aEndPreceding && bStart < bEndPreceding) {
+    // Recursely find and return common subsequences preceding the division.
+    findSubsequences(
+      nChangePreceding,
+      aStart,
+      aEndPreceding,
+      bStart,
+      bEndPreceding,
+      transposed,
+      callbacks,
+      aIndexesF,
+      aIndexesR,
+      division
+    );
+  }
+
+  // Return common subsequences that are adjacent to the middle change.
+  if (nCommonPreceding !== 0) {
+    foundSubsequence(nCommonPreceding, aCommonPreceding, bCommonPreceding);
+  }
+  if (nCommonFollowing !== 0) {
+    foundSubsequence(nCommonFollowing, aCommonFollowing, bCommonFollowing);
+  }
+
+  // Unless either index interval is empty, they might contain common items.
+  if (aStartFollowing < aEnd && bStartFollowing < bEnd) {
+    // Recursely find and return common subsequences following the division.
+    findSubsequences(
+      nChangeFollowing,
+      aStartFollowing,
+      aEnd,
+      bStartFollowing,
+      bEnd,
+      transposed,
+      callbacks,
+      aIndexesF,
+      aIndexesR,
+      division
+    );
+  }
+};
+const validateLength = (name, arg) => {
+  if (typeof arg !== 'number') {
+    throw new TypeError(`${pkg}: ${name} typeof ${typeof arg} is not a number`);
+  }
+  if (!Number.isSafeInteger(arg)) {
+    throw new RangeError(`${pkg}: ${name} value ${arg} is not a safe integer`);
+  }
+  if (arg < 0) {
+    throw new RangeError(`${pkg}: ${name} value ${arg} is a negative integer`);
+  }
+};
+const validateCallback = (name, arg) => {
+  const type = typeof arg;
+  if (type !== 'function') {
+    throw new TypeError(`${pkg}: ${name} typeof ${type} is not a function`);
+  }
+};
+
+// Compare items in two sequences to find a longest common subsequence.
+// Given lengths of sequences and input function to compare items at indexes,
+// return by output function the number of adjacent items and starting indexes
+// of each common subsequence.
+function diffSequence(aLength, bLength, isCommon, foundSubsequence) {
+  validateLength('aLength', aLength);
+  validateLength('bLength', bLength);
+  validateCallback('isCommon', isCommon);
+  validateCallback('foundSubsequence', foundSubsequence);
+
+  // Count common items from the start in the forward direction.
+  const nCommonF = countCommonItemsF(0, aLength, 0, bLength, isCommon);
+  if (nCommonF !== 0) {
+    foundSubsequence(nCommonF, 0, 0);
+  }
+
+  // Unless both sequences consist of common items only,
+  // find common items in the half-trimmed index intervals.
+  if (aLength !== nCommonF || bLength !== nCommonF) {
+    // Invariant: intervals do not have common items at the start.
+    // The start of an index interval is closed like array slice method.
+    const aStart = nCommonF;
+    const bStart = nCommonF;
+
+    // Count common items from the end in the reverse direction.
+    const nCommonR = countCommonItemsR(
+      aStart,
+      aLength - 1,
+      bStart,
+      bLength - 1,
+      isCommon
+    );
+
+    // Invariant: intervals do not have common items at the end.
+    // The end of an index interval is open like array slice method.
+    const aEnd = aLength - nCommonR;
+    const bEnd = bLength - nCommonR;
+
+    // Unless one sequence consists of common items only,
+    // therefore the other trimmed index interval consists of changes only,
+    // find common items in the trimmed index intervals.
+    const nCommonFR = nCommonF + nCommonR;
+    if (aLength !== nCommonFR && bLength !== nCommonFR) {
+      const nChange = 0; // number of change items is not yet known
+      const transposed = false; // call the original unwrapped functions
+      const callbacks = [
+        {
+          foundSubsequence,
+          isCommon
+        }
+      ];
+
+      // Indexes in sequence a of last points in furthest reaching paths
+      // from outside the start at top left in the forward direction:
+      const aIndexesF = [NOT_YET_SET];
+      // from the end at bottom right in the reverse direction:
+      const aIndexesR = [NOT_YET_SET];
+
+      // Initialize one object as output of all calls to divide function.
+      const division = {
+        aCommonFollowing: NOT_YET_SET,
+        aCommonPreceding: NOT_YET_SET,
+        aEndPreceding: NOT_YET_SET,
+        aStartFollowing: NOT_YET_SET,
+        bCommonFollowing: NOT_YET_SET,
+        bCommonPreceding: NOT_YET_SET,
+        bEndPreceding: NOT_YET_SET,
+        bStartFollowing: NOT_YET_SET,
+        nChangeFollowing: NOT_YET_SET,
+        nChangePreceding: NOT_YET_SET,
+        nCommonFollowing: NOT_YET_SET,
+        nCommonPreceding: NOT_YET_SET
+      };
+
+      // Find and return common subsequences in the trimmed index intervals.
+      findSubsequences(
+        nChange,
+        aStart,
+        aEnd,
+        bStart,
+        bEnd,
+        transposed,
+        callbacks,
+        aIndexesF,
+        aIndexesR,
+        division
+      );
+    }
+    if (nCommonR !== 0) {
+      foundSubsequence(nCommonR, aEnd, bEnd);
+    }
+  }
+}