827 lines
28 KiB
JavaScript
827 lines
28 KiB
JavaScript
'use strict';
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Object.defineProperty(exports, '__esModule', {
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value: true
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});
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exports.default = void 0;
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/**
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* Copyright (c) Facebook, Inc. and its affiliates. All Rights Reserved.
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*
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* This source code is licensed under the MIT license found in the
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* LICENSE file in the root directory of this source tree.
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*
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*/
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// This diff-sequences package implements the linear space variation in
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// An O(ND) Difference Algorithm and Its Variations by Eugene W. Myers
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// Relationship in notation between Myers paper and this package:
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// A is a
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// N is aLength, aEnd - aStart, and so on
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// x is aIndex, aFirst, aLast, and so on
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// B is b
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// M is bLength, bEnd - bStart, and so on
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// y is bIndex, bFirst, bLast, and so on
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// Δ = N - M is negative of baDeltaLength = bLength - aLength
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// D is d
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// k is kF
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// k + Δ is kF = kR - baDeltaLength
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// V is aIndexesF or aIndexesR (see comment below about Indexes type)
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// index intervals [1, N] and [1, M] are [0, aLength) and [0, bLength)
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// starting point in forward direction (0, 0) is (-1, -1)
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// starting point in reverse direction (N + 1, M + 1) is (aLength, bLength)
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// The “edit graph” for sequences a and b corresponds to items:
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// in a on the horizontal axis
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// in b on the vertical axis
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//
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// Given a-coordinate of a point in a diagonal, you can compute b-coordinate.
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//
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// Forward diagonals kF:
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// zero diagonal intersects top left corner
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// positive diagonals intersect top edge
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// negative diagonals insersect left edge
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//
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// Reverse diagonals kR:
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// zero diagonal intersects bottom right corner
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// positive diagonals intersect right edge
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// negative diagonals intersect bottom edge
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// The graph contains a directed acyclic graph of edges:
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// horizontal: delete an item from a
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// vertical: insert an item from b
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// diagonal: common item in a and b
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//
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// The algorithm solves dual problems in the graph analogy:
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// Find longest common subsequence: path with maximum number of diagonal edges
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// Find shortest edit script: path with minimum number of non-diagonal edges
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// Input callback function compares items at indexes in the sequences.
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// Output callback function receives the number of adjacent items
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// and starting indexes of each common subsequence.
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// Either original functions or wrapped to swap indexes if graph is transposed.
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// Indexes in sequence a of last point of forward or reverse paths in graph.
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// Myers algorithm indexes by diagonal k which for negative is bad deopt in V8.
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// This package indexes by iF and iR which are greater than or equal to zero.
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// and also updates the index arrays in place to cut memory in half.
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// kF = 2 * iF - d
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// kR = d - 2 * iR
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// Division of index intervals in sequences a and b at the middle change.
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// Invariant: intervals do not have common items at the start or end.
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const pkg = 'diff-sequences'; // for error messages
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const NOT_YET_SET = 0; // small int instead of undefined to avoid deopt in V8
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// Return the number of common items that follow in forward direction.
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// The length of what Myers paper calls a “snake” in a forward path.
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const countCommonItemsF = (aIndex, aEnd, bIndex, bEnd, isCommon) => {
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let nCommon = 0;
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while (aIndex < aEnd && bIndex < bEnd && isCommon(aIndex, bIndex)) {
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aIndex += 1;
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bIndex += 1;
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nCommon += 1;
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}
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return nCommon;
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}; // Return the number of common items that precede in reverse direction.
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// The length of what Myers paper calls a “snake” in a reverse path.
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const countCommonItemsR = (aStart, aIndex, bStart, bIndex, isCommon) => {
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let nCommon = 0;
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while (aStart <= aIndex && bStart <= bIndex && isCommon(aIndex, bIndex)) {
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aIndex -= 1;
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bIndex -= 1;
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nCommon += 1;
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}
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return nCommon;
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}; // A simple function to extend forward paths from (d - 1) to d changes
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// when forward and reverse paths cannot yet overlap.
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const extendPathsF = (d, aEnd, bEnd, bF, isCommon, aIndexesF, iMaxF) => {
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// Unroll the first iteration.
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let iF = 0;
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let kF = -d; // kF = 2 * iF - d
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let aFirst = aIndexesF[iF]; // in first iteration always insert
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let aIndexPrev1 = aFirst; // prev value of [iF - 1] in next iteration
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aIndexesF[iF] += countCommonItemsF(
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aFirst + 1,
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aEnd,
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bF + aFirst - kF + 1,
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bEnd,
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isCommon
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); // Optimization: skip diagonals in which paths cannot ever overlap.
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const nF = d < iMaxF ? d : iMaxF; // The diagonals kF are odd when d is odd and even when d is even.
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for (iF += 1, kF += 2; iF <= nF; iF += 1, kF += 2) {
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// To get first point of path segment, move one change in forward direction
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// from last point of previous path segment in an adjacent diagonal.
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// In last possible iteration when iF === d and kF === d always delete.
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if (iF !== d && aIndexPrev1 < aIndexesF[iF]) {
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aFirst = aIndexesF[iF]; // vertical to insert from b
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} else {
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aFirst = aIndexPrev1 + 1; // horizontal to delete from a
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if (aEnd <= aFirst) {
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// Optimization: delete moved past right of graph.
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return iF - 1;
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}
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} // To get last point of path segment, move along diagonal of common items.
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aIndexPrev1 = aIndexesF[iF];
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aIndexesF[iF] =
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aFirst +
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countCommonItemsF(aFirst + 1, aEnd, bF + aFirst - kF + 1, bEnd, isCommon);
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}
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return iMaxF;
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}; // A simple function to extend reverse paths from (d - 1) to d changes
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// when reverse and forward paths cannot yet overlap.
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const extendPathsR = (d, aStart, bStart, bR, isCommon, aIndexesR, iMaxR) => {
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// Unroll the first iteration.
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let iR = 0;
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let kR = d; // kR = d - 2 * iR
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let aFirst = aIndexesR[iR]; // in first iteration always insert
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let aIndexPrev1 = aFirst; // prev value of [iR - 1] in next iteration
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aIndexesR[iR] -= countCommonItemsR(
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aStart,
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aFirst - 1,
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bStart,
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bR + aFirst - kR - 1,
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isCommon
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); // Optimization: skip diagonals in which paths cannot ever overlap.
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const nR = d < iMaxR ? d : iMaxR; // The diagonals kR are odd when d is odd and even when d is even.
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for (iR += 1, kR -= 2; iR <= nR; iR += 1, kR -= 2) {
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// To get first point of path segment, move one change in reverse direction
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// from last point of previous path segment in an adjacent diagonal.
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// In last possible iteration when iR === d and kR === -d always delete.
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if (iR !== d && aIndexesR[iR] < aIndexPrev1) {
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aFirst = aIndexesR[iR]; // vertical to insert from b
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} else {
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aFirst = aIndexPrev1 - 1; // horizontal to delete from a
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if (aFirst < aStart) {
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// Optimization: delete moved past left of graph.
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return iR - 1;
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}
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} // To get last point of path segment, move along diagonal of common items.
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aIndexPrev1 = aIndexesR[iR];
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aIndexesR[iR] =
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aFirst -
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countCommonItemsR(
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aStart,
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aFirst - 1,
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bStart,
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bR + aFirst - kR - 1,
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isCommon
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);
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}
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return iMaxR;
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}; // A complete function to extend forward paths from (d - 1) to d changes.
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// Return true if a path overlaps reverse path of (d - 1) changes in its diagonal.
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const extendOverlappablePathsF = (
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d,
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aStart,
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aEnd,
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bStart,
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bEnd,
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isCommon,
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aIndexesF,
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iMaxF,
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aIndexesR,
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iMaxR,
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division
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) => {
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const bF = bStart - aStart; // bIndex = bF + aIndex - kF
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const aLength = aEnd - aStart;
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const bLength = bEnd - bStart;
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const baDeltaLength = bLength - aLength; // kF = kR - baDeltaLength
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// Range of diagonals in which forward and reverse paths might overlap.
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const kMinOverlapF = -baDeltaLength - (d - 1); // -(d - 1) <= kR
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const kMaxOverlapF = -baDeltaLength + (d - 1); // kR <= (d - 1)
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let aIndexPrev1 = NOT_YET_SET; // prev value of [iF - 1] in next iteration
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// Optimization: skip diagonals in which paths cannot ever overlap.
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const nF = d < iMaxF ? d : iMaxF; // The diagonals kF = 2 * iF - d are odd when d is odd and even when d is even.
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for (let iF = 0, kF = -d; iF <= nF; iF += 1, kF += 2) {
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// To get first point of path segment, move one change in forward direction
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// from last point of previous path segment in an adjacent diagonal.
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// In first iteration when iF === 0 and kF === -d always insert.
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// In last possible iteration when iF === d and kF === d always delete.
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const insert = iF === 0 || (iF !== d && aIndexPrev1 < aIndexesF[iF]);
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const aLastPrev = insert ? aIndexesF[iF] : aIndexPrev1;
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const aFirst = insert
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? aLastPrev // vertical to insert from b
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: aLastPrev + 1; // horizontal to delete from a
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// To get last point of path segment, move along diagonal of common items.
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const bFirst = bF + aFirst - kF;
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const nCommonF = countCommonItemsF(
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aFirst + 1,
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aEnd,
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bFirst + 1,
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bEnd,
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isCommon
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);
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const aLast = aFirst + nCommonF;
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aIndexPrev1 = aIndexesF[iF];
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aIndexesF[iF] = aLast;
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if (kMinOverlapF <= kF && kF <= kMaxOverlapF) {
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// Solve for iR of reverse path with (d - 1) changes in diagonal kF:
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// kR = kF + baDeltaLength
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// kR = (d - 1) - 2 * iR
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const iR = (d - 1 - (kF + baDeltaLength)) / 2; // If this forward path overlaps the reverse path in this diagonal,
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// then this is the middle change of the index intervals.
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if (iR <= iMaxR && aIndexesR[iR] - 1 <= aLast) {
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// Unlike the Myers algorithm which finds only the middle “snake”
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// this package can find two common subsequences per division.
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// Last point of previous path segment is on an adjacent diagonal.
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const bLastPrev = bF + aLastPrev - (insert ? kF + 1 : kF - 1); // Because of invariant that intervals preceding the middle change
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// cannot have common items at the end,
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// move in reverse direction along a diagonal of common items.
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const nCommonR = countCommonItemsR(
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aStart,
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aLastPrev,
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bStart,
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bLastPrev,
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isCommon
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);
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const aIndexPrevFirst = aLastPrev - nCommonR;
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const bIndexPrevFirst = bLastPrev - nCommonR;
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const aEndPreceding = aIndexPrevFirst + 1;
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const bEndPreceding = bIndexPrevFirst + 1;
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division.nChangePreceding = d - 1;
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if (d - 1 === aEndPreceding + bEndPreceding - aStart - bStart) {
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// Optimization: number of preceding changes in forward direction
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// is equal to number of items in preceding interval,
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// therefore it cannot contain any common items.
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division.aEndPreceding = aStart;
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division.bEndPreceding = bStart;
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} else {
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division.aEndPreceding = aEndPreceding;
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division.bEndPreceding = bEndPreceding;
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}
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division.nCommonPreceding = nCommonR;
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if (nCommonR !== 0) {
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division.aCommonPreceding = aEndPreceding;
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division.bCommonPreceding = bEndPreceding;
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}
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division.nCommonFollowing = nCommonF;
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if (nCommonF !== 0) {
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division.aCommonFollowing = aFirst + 1;
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division.bCommonFollowing = bFirst + 1;
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}
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const aStartFollowing = aLast + 1;
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const bStartFollowing = bFirst + nCommonF + 1;
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division.nChangeFollowing = d - 1;
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if (d - 1 === aEnd + bEnd - aStartFollowing - bStartFollowing) {
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// Optimization: number of changes in reverse direction
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// is equal to number of items in following interval,
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// therefore it cannot contain any common items.
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division.aStartFollowing = aEnd;
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division.bStartFollowing = bEnd;
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} else {
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division.aStartFollowing = aStartFollowing;
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division.bStartFollowing = bStartFollowing;
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}
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return true;
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}
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}
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}
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return false;
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}; // A complete function to extend reverse paths from (d - 1) to d changes.
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// Return true if a path overlaps forward path of d changes in its diagonal.
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const extendOverlappablePathsR = (
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d,
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aStart,
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aEnd,
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bStart,
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bEnd,
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isCommon,
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aIndexesF,
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iMaxF,
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aIndexesR,
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iMaxR,
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division
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) => {
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const bR = bEnd - aEnd; // bIndex = bR + aIndex - kR
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const aLength = aEnd - aStart;
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const bLength = bEnd - bStart;
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const baDeltaLength = bLength - aLength; // kR = kF + baDeltaLength
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// Range of diagonals in which forward and reverse paths might overlap.
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const kMinOverlapR = baDeltaLength - d; // -d <= kF
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const kMaxOverlapR = baDeltaLength + d; // kF <= d
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let aIndexPrev1 = NOT_YET_SET; // prev value of [iR - 1] in next iteration
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// Optimization: skip diagonals in which paths cannot ever overlap.
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const nR = d < iMaxR ? d : iMaxR; // The diagonals kR = d - 2 * iR are odd when d is odd and even when d is even.
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for (let iR = 0, kR = d; iR <= nR; iR += 1, kR -= 2) {
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// To get first point of path segment, move one change in reverse direction
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// from last point of previous path segment in an adjacent diagonal.
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// In first iteration when iR === 0 and kR === d always insert.
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// In last possible iteration when iR === d and kR === -d always delete.
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const insert = iR === 0 || (iR !== d && aIndexesR[iR] < aIndexPrev1);
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const aLastPrev = insert ? aIndexesR[iR] : aIndexPrev1;
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const aFirst = insert
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? aLastPrev // vertical to insert from b
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: aLastPrev - 1; // horizontal to delete from a
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// To get last point of path segment, move along diagonal of common items.
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const bFirst = bR + aFirst - kR;
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const nCommonR = countCommonItemsR(
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aStart,
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aFirst - 1,
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bStart,
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bFirst - 1,
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isCommon
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);
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const aLast = aFirst - nCommonR;
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aIndexPrev1 = aIndexesR[iR];
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aIndexesR[iR] = aLast;
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if (kMinOverlapR <= kR && kR <= kMaxOverlapR) {
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// Solve for iF of forward path with d changes in diagonal kR:
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// kF = kR - baDeltaLength
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// kF = 2 * iF - d
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const iF = (d + (kR - baDeltaLength)) / 2; // If this reverse path overlaps the forward path in this diagonal,
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// then this is a middle change of the index intervals.
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if (iF <= iMaxF && aLast - 1 <= aIndexesF[iF]) {
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const bLast = bFirst - nCommonR;
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division.nChangePreceding = d;
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if (d === aLast + bLast - aStart - bStart) {
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// Optimization: number of changes in reverse direction
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// is equal to number of items in preceding interval,
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// therefore it cannot contain any common items.
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division.aEndPreceding = aStart;
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division.bEndPreceding = bStart;
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} else {
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division.aEndPreceding = aLast;
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division.bEndPreceding = bLast;
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}
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division.nCommonPreceding = nCommonR;
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if (nCommonR !== 0) {
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// The last point of reverse path segment is start of common subsequence.
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division.aCommonPreceding = aLast;
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division.bCommonPreceding = bLast;
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}
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division.nChangeFollowing = d - 1;
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if (d === 1) {
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// There is no previous path segment.
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division.nCommonFollowing = 0;
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division.aStartFollowing = aEnd;
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division.bStartFollowing = bEnd;
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} else {
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// Unlike the Myers algorithm which finds only the middle “snake”
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// this package can find two common subsequences per division.
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// Last point of previous path segment is on an adjacent diagonal.
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const bLastPrev = bR + aLastPrev - (insert ? kR - 1 : kR + 1); // Because of invariant that intervals following the middle change
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// cannot have common items at the start,
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// move in forward direction along a diagonal of common items.
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const nCommonF = countCommonItemsF(
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aLastPrev,
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aEnd,
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bLastPrev,
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bEnd,
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isCommon
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);
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division.nCommonFollowing = nCommonF;
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if (nCommonF !== 0) {
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// The last point of reverse path segment is start of common subsequence.
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division.aCommonFollowing = aLastPrev;
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division.bCommonFollowing = bLastPrev;
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}
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const aStartFollowing = aLastPrev + nCommonF; // aFirstPrev
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const bStartFollowing = bLastPrev + nCommonF; // bFirstPrev
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if (d - 1 === aEnd + bEnd - aStartFollowing - bStartFollowing) {
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// Optimization: number of changes in forward direction
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// is equal to number of items in following interval,
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// therefore it cannot contain any common items.
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division.aStartFollowing = aEnd;
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division.bStartFollowing = bEnd;
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} else {
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division.aStartFollowing = aStartFollowing;
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division.bStartFollowing = bStartFollowing;
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}
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}
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return true;
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}
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}
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}
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return false;
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}; // Given index intervals and input function to compare items at indexes,
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// divide at the middle change.
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//
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// DO NOT CALL if start === end, because interval cannot contain common items
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// and because this function will throw the “no overlap” error.
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const divide = (
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nChange,
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aStart,
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aEnd,
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bStart,
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bEnd,
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isCommon,
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aIndexesF,
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aIndexesR,
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division // output
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) => {
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const bF = bStart - aStart; // bIndex = bF + aIndex - kF
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|
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const bR = bEnd - aEnd; // bIndex = bR + aIndex - kR
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const aLength = aEnd - aStart;
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const bLength = bEnd - bStart; // Because graph has square or portrait orientation,
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// 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;
|