mirror of https://github.com/docker/cli.git
424 lines
13 KiB
Go
424 lines
13 KiB
Go
/*Package difflib is a partial port of Python difflib module.
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Original source: https://github.com/pmezard/go-difflib
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This file is trimmed to only the parts used by this repository.
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*/
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package difflib // import "gotest.tools/v3/internal/difflib"
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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// Match stores line numbers of size of match
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type Match struct {
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A int
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B int
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Size int
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}
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// OpCode identifies the type of diff
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type OpCode struct {
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Tag byte
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I1 int
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I2 int
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J1 int
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J2 int
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}
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// SequenceMatcher compares sequence of strings. The basic
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// algorithm predates, and is a little fancier than, an algorithm
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// published in the late 1980's by Ratcliff and Obershelp under the
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// hyperbolic name "gestalt pattern matching". The basic idea is to find
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// the longest contiguous matching subsequence that contains no "junk"
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// elements (R-O doesn't address junk). The same idea is then applied
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// recursively to the pieces of the sequences to the left and to the right
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// of the matching subsequence. This does not yield minimal edit
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// sequences, but does tend to yield matches that "look right" to people.
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//
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// SequenceMatcher tries to compute a "human-friendly diff" between two
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// sequences. Unlike e.g. UNIX(tm) diff, the fundamental notion is the
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// longest *contiguous* & junk-free matching subsequence. That's what
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// catches peoples' eyes. The Windows(tm) windiff has another interesting
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// notion, pairing up elements that appear uniquely in each sequence.
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// That, and the method here, appear to yield more intuitive difference
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// reports than does diff. This method appears to be the least vulnerable
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// to synching up on blocks of "junk lines", though (like blank lines in
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// ordinary text files, or maybe "<P>" lines in HTML files). That may be
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// because this is the only method of the 3 that has a *concept* of
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// "junk" <wink>.
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//
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// Timing: Basic R-O is cubic time worst case and quadratic time expected
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// case. SequenceMatcher is quadratic time for the worst case and has
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// expected-case behavior dependent in a complicated way on how many
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// elements the sequences have in common; best case time is linear.
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type SequenceMatcher struct {
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a []string
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b []string
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b2j map[string][]int
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IsJunk func(string) bool
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autoJunk bool
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bJunk map[string]struct{}
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matchingBlocks []Match
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fullBCount map[string]int
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bPopular map[string]struct{}
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opCodes []OpCode
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}
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// NewMatcher returns a new SequenceMatcher
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func NewMatcher(a, b []string) *SequenceMatcher {
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m := SequenceMatcher{autoJunk: true}
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m.SetSeqs(a, b)
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return &m
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}
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// SetSeqs sets two sequences to be compared.
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func (m *SequenceMatcher) SetSeqs(a, b []string) {
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m.SetSeq1(a)
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m.SetSeq2(b)
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}
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// SetSeq1 sets the first sequence to be compared. The second sequence to be compared is
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// not changed.
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//
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// SequenceMatcher computes and caches detailed information about the second
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// sequence, so if you want to compare one sequence S against many sequences,
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// use .SetSeq2(s) once and call .SetSeq1(x) repeatedly for each of the other
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// sequences.
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//
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// See also SetSeqs() and SetSeq2().
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func (m *SequenceMatcher) SetSeq1(a []string) {
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if &a == &m.a {
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return
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}
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m.a = a
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m.matchingBlocks = nil
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m.opCodes = nil
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}
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// SetSeq2 sets the second sequence to be compared. The first sequence to be compared is
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// not changed.
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func (m *SequenceMatcher) SetSeq2(b []string) {
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if &b == &m.b {
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return
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}
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m.b = b
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m.matchingBlocks = nil
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m.opCodes = nil
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m.fullBCount = nil
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m.chainB()
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}
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func (m *SequenceMatcher) chainB() {
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// Populate line -> index mapping
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b2j := map[string][]int{}
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for i, s := range m.b {
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indices := b2j[s]
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indices = append(indices, i)
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b2j[s] = indices
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}
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// Purge junk elements
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m.bJunk = map[string]struct{}{}
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if m.IsJunk != nil {
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junk := m.bJunk
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for s := range b2j {
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if m.IsJunk(s) {
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junk[s] = struct{}{}
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}
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}
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for s := range junk {
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delete(b2j, s)
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}
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}
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// Purge remaining popular elements
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popular := map[string]struct{}{}
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n := len(m.b)
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if m.autoJunk && n >= 200 {
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ntest := n/100 + 1
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for s, indices := range b2j {
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if len(indices) > ntest {
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popular[s] = struct{}{}
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}
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}
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for s := range popular {
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delete(b2j, s)
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}
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}
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m.bPopular = popular
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m.b2j = b2j
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}
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func (m *SequenceMatcher) isBJunk(s string) bool {
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_, ok := m.bJunk[s]
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return ok
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}
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// Find longest matching block in a[alo:ahi] and b[blo:bhi].
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//
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// If IsJunk is not defined:
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//
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// Return (i,j,k) such that a[i:i+k] is equal to b[j:j+k], where
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// alo <= i <= i+k <= ahi
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// blo <= j <= j+k <= bhi
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// and for all (i',j',k') meeting those conditions,
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// k >= k'
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// i <= i'
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// and if i == i', j <= j'
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//
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// In other words, of all maximal matching blocks, return one that
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// starts earliest in a, and of all those maximal matching blocks that
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// start earliest in a, return the one that starts earliest in b.
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//
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// If IsJunk is defined, first the longest matching block is
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// determined as above, but with the additional restriction that no
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// junk element appears in the block. Then that block is extended as
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// far as possible by matching (only) junk elements on both sides. So
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// the resulting block never matches on junk except as identical junk
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// happens to be adjacent to an "interesting" match.
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//
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// If no blocks match, return (alo, blo, 0).
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func (m *SequenceMatcher) findLongestMatch(alo, ahi, blo, bhi int) Match {
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// CAUTION: stripping common prefix or suffix would be incorrect.
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// E.g.,
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// ab
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// acab
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// Longest matching block is "ab", but if common prefix is
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// stripped, it's "a" (tied with "b"). UNIX(tm) diff does so
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// strip, so ends up claiming that ab is changed to acab by
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// inserting "ca" in the middle. That's minimal but unintuitive:
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// "it's obvious" that someone inserted "ac" at the front.
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// Windiff ends up at the same place as diff, but by pairing up
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// the unique 'b's and then matching the first two 'a's.
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besti, bestj, bestsize := alo, blo, 0
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// find longest junk-free match
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// during an iteration of the loop, j2len[j] = length of longest
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// junk-free match ending with a[i-1] and b[j]
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j2len := map[int]int{}
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for i := alo; i != ahi; i++ {
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// look at all instances of a[i] in b; note that because
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// b2j has no junk keys, the loop is skipped if a[i] is junk
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newj2len := map[int]int{}
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for _, j := range m.b2j[m.a[i]] {
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// a[i] matches b[j]
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if j < blo {
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continue
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}
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if j >= bhi {
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break
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}
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k := j2len[j-1] + 1
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newj2len[j] = k
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if k > bestsize {
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besti, bestj, bestsize = i-k+1, j-k+1, k
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}
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}
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j2len = newj2len
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}
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// Extend the best by non-junk elements on each end. In particular,
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// "popular" non-junk elements aren't in b2j, which greatly speeds
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// the inner loop above, but also means "the best" match so far
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// doesn't contain any junk *or* popular non-junk elements.
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for besti > alo && bestj > blo && !m.isBJunk(m.b[bestj-1]) &&
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m.a[besti-1] == m.b[bestj-1] {
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besti, bestj, bestsize = besti-1, bestj-1, bestsize+1
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}
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for besti+bestsize < ahi && bestj+bestsize < bhi &&
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!m.isBJunk(m.b[bestj+bestsize]) &&
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m.a[besti+bestsize] == m.b[bestj+bestsize] {
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bestsize += 1
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}
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// Now that we have a wholly interesting match (albeit possibly
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// empty!), we may as well suck up the matching junk on each
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// side of it too. Can't think of a good reason not to, and it
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// saves post-processing the (possibly considerable) expense of
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// figuring out what to do with it. In the case of an empty
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// interesting match, this is clearly the right thing to do,
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// because no other kind of match is possible in the regions.
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for besti > alo && bestj > blo && m.isBJunk(m.b[bestj-1]) &&
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m.a[besti-1] == m.b[bestj-1] {
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besti, bestj, bestsize = besti-1, bestj-1, bestsize+1
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}
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for besti+bestsize < ahi && bestj+bestsize < bhi &&
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m.isBJunk(m.b[bestj+bestsize]) &&
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m.a[besti+bestsize] == m.b[bestj+bestsize] {
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bestsize += 1
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}
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return Match{A: besti, B: bestj, Size: bestsize}
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}
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// GetMatchingBlocks returns a list of triples describing matching subsequences.
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//
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// Each triple is of the form (i, j, n), and means that
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// a[i:i+n] == b[j:j+n]. The triples are monotonically increasing in
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// i and in j. It's also guaranteed that if (i, j, n) and (i', j', n') are
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// adjacent triples in the list, and the second is not the last triple in the
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// list, then i+n != i' or j+n != j'. IOW, adjacent triples never describe
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// adjacent equal blocks.
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//
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// The last triple is a dummy, (len(a), len(b), 0), and is the only
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// triple with n==0.
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func (m *SequenceMatcher) GetMatchingBlocks() []Match {
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if m.matchingBlocks != nil {
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return m.matchingBlocks
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}
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var matchBlocks func(alo, ahi, blo, bhi int, matched []Match) []Match
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matchBlocks = func(alo, ahi, blo, bhi int, matched []Match) []Match {
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match := m.findLongestMatch(alo, ahi, blo, bhi)
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i, j, k := match.A, match.B, match.Size
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if match.Size > 0 {
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if alo < i && blo < j {
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matched = matchBlocks(alo, i, blo, j, matched)
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}
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matched = append(matched, match)
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if i+k < ahi && j+k < bhi {
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matched = matchBlocks(i+k, ahi, j+k, bhi, matched)
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}
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}
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return matched
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}
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matched := matchBlocks(0, len(m.a), 0, len(m.b), nil)
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// It's possible that we have adjacent equal blocks in the
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// matching_blocks list now.
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nonAdjacent := []Match{}
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i1, j1, k1 := 0, 0, 0
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for _, b := range matched {
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// Is this block adjacent to i1, j1, k1?
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i2, j2, k2 := b.A, b.B, b.Size
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if i1+k1 == i2 && j1+k1 == j2 {
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// Yes, so collapse them -- this just increases the length of
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// the first block by the length of the second, and the first
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// block so lengthened remains the block to compare against.
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k1 += k2
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} else {
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// Not adjacent. Remember the first block (k1==0 means it's
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// the dummy we started with), and make the second block the
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// new block to compare against.
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if k1 > 0 {
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nonAdjacent = append(nonAdjacent, Match{i1, j1, k1})
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}
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i1, j1, k1 = i2, j2, k2
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}
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}
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if k1 > 0 {
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nonAdjacent = append(nonAdjacent, Match{i1, j1, k1})
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}
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nonAdjacent = append(nonAdjacent, Match{len(m.a), len(m.b), 0})
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m.matchingBlocks = nonAdjacent
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return m.matchingBlocks
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}
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// GetOpCodes returns a list of 5-tuples describing how to turn a into b.
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//
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// Each tuple is of the form (tag, i1, i2, j1, j2). The first tuple
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// has i1 == j1 == 0, and remaining tuples have i1 == the i2 from the
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// tuple preceding it, and likewise for j1 == the previous j2.
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//
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// The tags are characters, with these meanings:
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//
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// 'r' (replace): a[i1:i2] should be replaced by b[j1:j2]
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//
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// 'd' (delete): a[i1:i2] should be deleted, j1==j2 in this case.
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//
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// 'i' (insert): b[j1:j2] should be inserted at a[i1:i1], i1==i2 in this case.
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//
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// 'e' (equal): a[i1:i2] == b[j1:j2]
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func (m *SequenceMatcher) GetOpCodes() []OpCode {
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if m.opCodes != nil {
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return m.opCodes
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}
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i, j := 0, 0
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matching := m.GetMatchingBlocks()
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opCodes := make([]OpCode, 0, len(matching))
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for _, m := range matching {
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// invariant: we've pumped out correct diffs to change
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// a[:i] into b[:j], and the next matching block is
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// a[ai:ai+size] == b[bj:bj+size]. So we need to pump
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// out a diff to change a[i:ai] into b[j:bj], pump out
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// the matching block, and move (i,j) beyond the match
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ai, bj, size := m.A, m.B, m.Size
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tag := byte(0)
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if i < ai && j < bj {
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tag = 'r'
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} else if i < ai {
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tag = 'd'
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} else if j < bj {
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tag = 'i'
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}
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if tag > 0 {
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opCodes = append(opCodes, OpCode{tag, i, ai, j, bj})
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}
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i, j = ai+size, bj+size
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// the list of matching blocks is terminated by a
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// sentinel with size 0
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if size > 0 {
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opCodes = append(opCodes, OpCode{'e', ai, i, bj, j})
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}
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}
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m.opCodes = opCodes
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return m.opCodes
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}
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// GetGroupedOpCodes isolates change clusters by eliminating ranges with no changes.
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//
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// Return a generator of groups with up to n lines of context.
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// Each group is in the same format as returned by GetOpCodes().
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func (m *SequenceMatcher) GetGroupedOpCodes(n int) [][]OpCode {
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if n < 0 {
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n = 3
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}
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codes := m.GetOpCodes()
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if len(codes) == 0 {
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codes = []OpCode{{'e', 0, 1, 0, 1}}
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}
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// Fixup leading and trailing groups if they show no changes.
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if codes[0].Tag == 'e' {
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c := codes[0]
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i1, i2, j1, j2 := c.I1, c.I2, c.J1, c.J2
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codes[0] = OpCode{c.Tag, max(i1, i2-n), i2, max(j1, j2-n), j2}
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}
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if codes[len(codes)-1].Tag == 'e' {
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c := codes[len(codes)-1]
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i1, i2, j1, j2 := c.I1, c.I2, c.J1, c.J2
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codes[len(codes)-1] = OpCode{c.Tag, i1, min(i2, i1+n), j1, min(j2, j1+n)}
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}
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nn := n + n
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groups := [][]OpCode{}
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group := []OpCode{}
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for _, c := range codes {
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i1, i2, j1, j2 := c.I1, c.I2, c.J1, c.J2
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// End the current group and start a new one whenever
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// there is a large range with no changes.
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if c.Tag == 'e' && i2-i1 > nn {
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group = append(group, OpCode{c.Tag, i1, min(i2, i1+n),
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j1, min(j2, j1+n)})
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groups = append(groups, group)
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group = []OpCode{}
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i1, j1 = max(i1, i2-n), max(j1, j2-n)
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}
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group = append(group, OpCode{c.Tag, i1, i2, j1, j2})
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}
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if len(group) > 0 && !(len(group) == 1 && group[0].Tag == 'e') {
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groups = append(groups, group)
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}
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return groups
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}
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