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'use strict';
Object.defineProperty(exports, '__esModule', {
value: true
});
exports.default = void 0;
/**
* Copyright (c) Facebook, Inc. and its affiliates. All Rights Reserved.
*
* 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) => {
// 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) => {
// 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
) => {
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
) => {
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=${bEnd}`
);
}; // Given index intervals and input function to compare items at indexes,
// return by output function the number of adjacent items and starting indexes
// of each common subsequence. Divide and conquer with only linear space.
//
// The index intervals are half open [start, end) like array slice method.
// DO NOT CALL if start === end, because interval cannot contain common items
// and because divide function will throw the “no overlap” error.
const findSubsequences = (
nChange,
aStart,
aEnd,
bStart,
bEnd,
transposed,
callbacks,
aIndexesF,
aIndexesR,
division // temporary memory, not input nor output
) => {
if (bEnd - bStart < aEnd - aStart) {
// Transpose graph so it has portrait instead of landscape orientation.
// Always compare shorter to longer sequence for consistency and optimization.
transposed = !transposed;
if (transposed && callbacks.length === 1) {
// Lazily wrap callback functions to swap args if graph is transposed.
const _callbacks$ = callbacks[0],
foundSubsequence = _callbacks$.foundSubsequence,
isCommon = _callbacks$.isCommon;
callbacks[1] = {
foundSubsequence: (function(_foundSubsequence) {
function foundSubsequence(_x, _x2, _x3) {
return _foundSubsequence.apply(this, arguments);
}
foundSubsequence.toString = function() {
return _foundSubsequence.toString();
};
return foundSubsequence;
})((nCommon, bCommon, aCommon) => {
foundSubsequence(nCommon, aCommon, bCommon);
}),
isCommon: (function(_isCommon) {
function isCommon(_x4, _x5) {
return _isCommon.apply(this, arguments);
}
isCommon.toString = function() {
return _isCommon.toString();
};
return isCommon;
})((bIndex, aIndex) => isCommon(aIndex, bIndex))
};
}
const tStart = aStart;
const tEnd = aEnd;
aStart = bStart;
aEnd = bEnd;
bStart = tStart;
bEnd = tEnd;
}
const _callbacks = callbacks[transposed ? 1 : 0],
foundSubsequence = _callbacks.foundSubsequence,
isCommon = _callbacks.isCommon; // Divide the index intervals at the middle change.
divide(
nChange,
aStart,
aEnd,
bStart,
bEnd,
isCommon,
aIndexesF,
aIndexesR,
division
);
const nChangePreceding = division.nChangePreceding,
aEndPreceding = division.aEndPreceding,
bEndPreceding = division.bEndPreceding,
nCommonPreceding = division.nCommonPreceding,
aCommonPreceding = division.aCommonPreceding,
bCommonPreceding = division.bCommonPreceding,
nCommonFollowing = division.nCommonFollowing,
aCommonFollowing = division.aCommonFollowing,
bCommonFollowing = division.bCommonFollowing,
nChangeFollowing = division.nChangeFollowing,
aStartFollowing = division.aStartFollowing,
bStartFollowing = division.bStartFollowing; // 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) => {
const type = typeof arg;
if (type !== 'number') {
throw new TypeError(`${pkg}: ${name} typeof ${type} is not a number`);
}
if (!Number.isSafeInteger(arg)) {
throw new RangeError(`${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.
var _default = (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);
}
}
};
exports.default = _default;