plumbing/format/packfile/diff.go

package packfile

/*

Copyright (c) 2013, Patrick Mezard
All rights reserved.

Redistribution and use in source and binary forms, with or without
modification, are permitted provided that the following conditions are
met:

    Redistributions of source code must retain the above copyright
notice, this list of conditions and the following disclaimer.
    Redistributions in binary form must reproduce the above copyright
notice, this list of conditions and the following disclaimer in the
documentation and/or other materials provided with the distribution.
    The names of its contributors may not be used to endorse or promote
products derived from this software without specific prior written
permission.

THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS
IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED
TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A
PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT
HOLDER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED
TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR
PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF
LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING
NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS
SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.

*/

// Code based on https://github.com/pmezard/go-difflib
// Removed unnecessary code for this use case and changed string inputs to byte

type match struct {
	A    int
	B    int
	Size int
}

type opCode struct {
	Tag byte
	I1  int
	I2  int
	J1  int
	J2  int
}

// SequenceMatcher compares sequence of bytes. The basic
// algorithm predates, and is a little fancier than, an algorithm
// published in the late 1980's by Ratcliff and Obershelp under the
// hyperbolic name "gestalt pattern matching".  The basic idea is to find
// the longest contiguous matching subsequence that contains no "junk"
// elements (R-O doesn't address junk).  The same idea is then applied
// recursively to the pieces of the sequences to the left and to the right
// of the matching subsequence.  This does not yield minimal edit
// sequences, but does tend to yield matches that "look right" to people.
//
// SequenceMatcher tries to compute a "human-friendly diff" between two
// sequences.  Unlike e.g. UNIX(tm) diff, the fundamental notion is the
// longest *contiguous* & junk-free matching subsequence.  That's what
// catches peoples' eyes.  The Windows(tm) windiff has another interesting
// notion, pairing up elements that appear uniquely in each sequence.
// That, and the method here, appear to yield more intuitive difference
// reports than does diff.  This method appears to be the least vulnerable
// to synching up on blocks of "junk lines", though (like blank lines in
// ordinary text files, or maybe "<P>" lines in HTML files).  That may be
// because this is the only method of the 3 that has a *concept* of
// "junk" <wink>.
//
// Timing:  Basic R-O is cubic time worst case and quadratic time expected
// case.  SequenceMatcher is quadratic time for the worst case and has
// expected-case behavior dependent in a complicated way on how many
// elements the sequences have in common; best case time is linear.
type sequenceMatcher struct {
	a              []byte
	b              []byte
	b2j            map[byte][]int
	IsJunk         func(byte) bool
	autoJunk       bool
	bJunk          map[byte]struct{}
	matchingBlocks []match
	fullBCount     map[byte]int
	bPopular       map[byte]struct{}
	opCodes        []opCode
}

func newMatcher(a, b []byte) *sequenceMatcher {
	m := sequenceMatcher{autoJunk: true}
	m.SetSeqs(a, b)
	return &m
}

// Set two sequences to be compared.
func (m *sequenceMatcher) SetSeqs(a, b []byte) {
	m.SetSeq1(a)
	m.SetSeq2(b)
}

// Set the first sequence to be compared. The second sequence to be compared is
// not changed.
//
// SequenceMatcher computes and caches detailed information about the second
// sequence, so if you want to compare one sequence S against many sequences,
// use .SetSeq2(s) once and call .SetSeq1(x) repeatedly for each of the other
// sequences.
//
// See also SetSeqs() and SetSeq2().
func (m *sequenceMatcher) SetSeq1(a []byte) {
	if &a == &m.a {
		return
	}
	m.a = a
	m.matchingBlocks = nil
	m.opCodes = nil
}

// Set the second sequence to be compared. The first sequence to be compared is
// not changed.
func (m *sequenceMatcher) SetSeq2(b []byte) {
	if &b == &m.b {
		return
	}
	m.b = b
	m.matchingBlocks = nil
	m.opCodes = nil
	m.fullBCount = nil
	m.chainB()
}

func (m *sequenceMatcher) chainB() {
	// Populate line -> index mapping
	b2j := map[byte][]int{}
	for i, s := range m.b {
		indices := b2j[s]
		indices = append(indices, i)
		b2j[s] = indices
	}

	// Purge junk elements
	m.bJunk = map[byte]struct{}{}
	if m.IsJunk != nil {
		junk := m.bJunk
		for s := range b2j {
			if m.IsJunk(s) {
				junk[s] = struct{}{}
			}
		}
		for s := range junk {
			delete(b2j, s)
		}
	}

	// Purge remaining popular elements
	popular := map[byte]struct{}{}
	n := len(m.b)
	if m.autoJunk && n >= 200 {
		ntest := n/100 + 1
		for s, indices := range b2j {
			if len(indices) > ntest {
				popular[s] = struct{}{}
			}
		}
		for s := range popular {
			delete(b2j, s)
		}
	}
	m.bPopular = popular
	m.b2j = b2j
}

func (m *sequenceMatcher) isBJunk(s byte) bool {
	_, ok := m.bJunk[s]
	return ok
}

// Find longest matching block in a[alo:ahi] and b[blo:bhi].
//
// If IsJunk is not defined:
//
// Return (i,j,k) such that a[i:i+k] is equal to b[j:j+k], where
//     alo <= i <= i+k <= ahi
//     blo <= j <= j+k <= bhi
// and for all (i',j',k') meeting those conditions,
//     k >= k'
//     i <= i'
//     and if i == i', j <= j'
//
// In other words, of all maximal matching blocks, return one that
// starts earliest in a, and of all those maximal matching blocks that
// start earliest in a, return the one that starts earliest in b.
//
// If IsJunk is defined, first the longest matching block is
// determined as above, but with the additional restriction that no
// junk element appears in the block.  Then that block is extended as
// far as possible by matching (only) junk elements on both sides.  So
// the resulting block never matches on junk except as identical junk
// happens to be adjacent to an "interesting" match.
//
// If no blocks match, return (alo, blo, 0).
func (m *sequenceMatcher) findLongestMatch(alo, ahi, blo, bhi int) match {
	// CAUTION:  stripping common prefix or suffix would be incorrect.
	// E.g.,
	//    ab
	//    acab
	// Longest matching block is "ab", but if common prefix is
	// stripped, it's "a" (tied with "b").  UNIX(tm) diff does so
	// strip, so ends up claiming that ab is changed to acab by
	// inserting "ca" in the middle.  That's minimal but unintuitive:
	// "it's obvious" that someone inserted "ac" at the front.
	// Windiff ends up at the same place as diff, but by pairing up
	// the unique 'b's and then matching the first two 'a's.
	besti, bestj, bestsize := alo, blo, 0

	// find longest junk-free match
	// during an iteration of the loop, j2len[j] = length of longest
	// junk-free match ending with a[i-1] and b[j]
	j2len := map[int]int{}
	for i := alo; i != ahi; i++ {
		// look at all instances of a[i] in b; note that because
		// b2j has no junk keys, the loop is skipped if a[i] is junk
		newj2len := map[int]int{}
		for _, j := range m.b2j[m.a[i]] {
			// a[i] matches b[j]
			if j < blo {
				continue
			}
			if j >= bhi {
				break
			}
			k := j2len[j-1] + 1
			newj2len[j] = k
			if k > bestsize {
				besti, bestj, bestsize = i-k+1, j-k+1, k
			}
		}
		j2len = newj2len
	}

	// Extend the best by non-junk elements on each end.  In particular,
	// "popular" non-junk elements aren't in b2j, which greatly speeds
	// the inner loop above, but also means "the best" match so far
	// doesn't contain any junk *or* popular non-junk elements.
	for besti > alo && bestj > blo && !m.isBJunk(m.b[bestj-1]) &&
		m.a[besti-1] == m.b[bestj-1] {
		besti, bestj, bestsize = besti-1, bestj-1, bestsize+1
	}
	for besti+bestsize < ahi && bestj+bestsize < bhi &&
		!m.isBJunk(m.b[bestj+bestsize]) &&
		m.a[besti+bestsize] == m.b[bestj+bestsize] {
		bestsize += 1
	}

	// Now that we have a wholly interesting match (albeit possibly
	// empty!), we may as well suck up the matching junk on each
	// side of it too.  Can't think of a good reason not to, and it
	// saves post-processing the (possibly considerable) expense of
	// figuring out what to do with it.  In the case of an empty
	// interesting match, this is clearly the right thing to do,
	// because no other kind of match is possible in the regions.
	for besti > alo && bestj > blo && m.isBJunk(m.b[bestj-1]) &&
		m.a[besti-1] == m.b[bestj-1] {
		besti, bestj, bestsize = besti-1, bestj-1, bestsize+1
	}
	for besti+bestsize < ahi && bestj+bestsize < bhi &&
		m.isBJunk(m.b[bestj+bestsize]) &&
		m.a[besti+bestsize] == m.b[bestj+bestsize] {
		bestsize += 1
	}

	return match{A: besti, B: bestj, Size: bestsize}
}

// Return list of triples describing matching subsequences.
//
// Each triple is of the form (i, j, n), and means that
// a[i:i+n] == b[j:j+n].  The triples are monotonically increasing in
// i and in j. It's also guaranteed that if (i, j, n) and (i', j', n') are
// adjacent triples in the list, and the second is not the last triple in the
// list, then i+n != i' or j+n != j'. IOW, adjacent triples never describe
// adjacent equal blocks.
//
// The last triple is a dummy, (len(a), len(b), 0), and is the only
// triple with n==0.
func (m *sequenceMatcher) GetMatchingBlocks() []match {
	if m.matchingBlocks != nil {
		return m.matchingBlocks
	}

	var matchBlocks func(alo, ahi, blo, bhi int, matched []match) []match
	matchBlocks = func(alo, ahi, blo, bhi int, matched []match) []match {
		match := m.findLongestMatch(alo, ahi, blo, bhi)
		i, j, k := match.A, match.B, match.Size
		if match.Size > 0 {
			if alo < i && blo < j {
				matched = matchBlocks(alo, i, blo, j, matched)
			}
			matched = append(matched, match)
			if i+k < ahi && j+k < bhi {
				matched = matchBlocks(i+k, ahi, j+k, bhi, matched)
			}
		}
		return matched
	}
	matched := matchBlocks(0, len(m.a), 0, len(m.b), nil)

	// It's possible that we have adjacent equal blocks in the
	// matching_blocks list now.
	nonAdjacent := []match{}
	i1, j1, k1 := 0, 0, 0
	for _, b := range matched {
		// Is this block adjacent to i1, j1, k1?
		i2, j2, k2 := b.A, b.B, b.Size
		if i1+k1 == i2 && j1+k1 == j2 {
			// Yes, so collapse them -- this just increases the length of
			// the first block by the length of the second, and the first
			// block so lengthened remains the block to compare against.
			k1 += k2
		} else {
			// Not adjacent.  Remember the first block (k1==0 means it's
			// the dummy we started with), and make the second block the
			// new block to compare against.
			if k1 > 0 {
				nonAdjacent = append(nonAdjacent, match{i1, j1, k1})
			}
			i1, j1, k1 = i2, j2, k2
		}
	}
	if k1 > 0 {
		nonAdjacent = append(nonAdjacent, match{i1, j1, k1})
	}

	nonAdjacent = append(nonAdjacent, match{len(m.a), len(m.b), 0})
	m.matchingBlocks = nonAdjacent
	return m.matchingBlocks
}

const (
	tagReplace = 'r'
	tagDelete  = 'd'
	tagInsert  = 'i'
	tagEqual   = 'e'
)

// Return list of 5-tuples describing how to turn a into b.
//
// Each tuple is of the form (tag, i1, i2, j1, j2).  The first tuple
// has i1 == j1 == 0, and remaining tuples have i1 == the i2 from the
// tuple preceding it, and likewise for j1 == the previous j2.
//
// The tags are characters, with these meanings:
//
// 'r' (replace):  a[i1:i2] should be replaced by b[j1:j2]
//
// 'd' (delete):   a[i1:i2] should be deleted, j1==j2 in this case.
//
// 'i' (insert):   b[j1:j2] should be inserted at a[i1:i1], i1==i2 in this case.
//
// 'e' (equal):    a[i1:i2] == b[j1:j2]
func (m *sequenceMatcher) GetOpCodes() []opCode {
	if m.opCodes != nil {
		return m.opCodes
	}
	i, j := 0, 0
	matching := m.GetMatchingBlocks()
	opCodes := make([]opCode, 0, len(matching))
	for _, m := range matching {
		//  invariant:  we've pumped out correct diffs to change
		//  a[:i] into b[:j], and the next matching block is
		//  a[ai:ai+size] == b[bj:bj+size]. So we need to pump
		//  out a diff to change a[i:ai] into b[j:bj], pump out
		//  the matching block, and move (i,j) beyond the match
		ai, bj, size := m.A, m.B, m.Size
		tag := byte(0)
		if i < ai && j < bj {
			tag = tagReplace
		} else if i < ai {
			tag = tagDelete
		} else if j < bj {
			tag = tagInsert
		}
		if tag > 0 {
			opCodes = append(opCodes, opCode{tag, i, ai, j, bj})
		}
		i, j = ai+size, bj+size
		// the list of matching blocks is terminated by a
		// sentinel with size 0
		if size > 0 {
			opCodes = append(opCodes, opCode{tagEqual, ai, i, bj, j})
		}
	}
	m.opCodes = opCodes
	return m.opCodes
}