mirror of
https://github.com/shivammathur/setup-php.git
synced 2024-11-23 12:21:07 +07:00
827 lines
28 KiB
JavaScript
827 lines
28 KiB
JavaScript
'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;
|