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Algorithms

object Algorithms

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  6. def commonSubstringIndicesToDiff(a: String, b: String, initialCommonSubstringIndices: List[(Int, Int, Int)]): Seq[Diff[Char]]

    Convert a list of common substring indices into a diff.

    Convert a list of common substring indices into a diff.

    returns

    a diff of a and b based on initialCommonSubstringIndices. If initialCommonSubstringIndices contains no empty runs, the result will not contain an Insert(_) followed by a Delete(_), i.e. in difference regions it puts characters from a before those from b. If initialCommonSubstringIndices does not correspond to a common subsequence of a and b, hilarity may ensue.

  7. final def eq(arg0: AnyRef): Boolean
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  8. def equals(arg0: AnyRef): Boolean
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  9. def findMiddle(a: String, b: String, zero: Int, vForward: Array[Int], vReverse: Array[Int], i: Int, n: Int, j: Int, m: Int): (Int, (Int, Int, Int))

    Find a (dist, (xlo, xhi, ylo)) such that dist is the LCS edit distance between a.substring(i, n) and b.substring(j, m), and such that a(i) == b(i - xlo + ylo) for all i in (xlo until xhi).

    Find a (dist, (xlo, xhi, ylo)) such that dist is the LCS edit distance between a.substring(i, n) and b.substring(j, m), and such that a(i) == b(i - xlo + ylo) for all i in (xlo until xhi).

    If 0 <= i <= n <= a.size and 0 <= j <= m <= b.size don't both hold, hilarity may ensue.

    a

    the observed/old sequence

    b

    the expected/new sequence

    zero

    an index into vForward and vReverse such that ((zero - (a.size + b.size)) to (zero + (a.size + b.size))) is a range of valid (in-bounds) indices into vForward and vReverse

    vForward

    an array, some entries of which will be overwritten

    vReverse

    an array, some entries of which will be overwritten

    i

    the start (inclusive) of the substring of a to search in

    n

    the end (exclusive) of the substring of a to search in

    j

    the start (inclusive) of the substring of b to search in

    m

    the end (exclusive) of the substring of b to search in

    returns

    see description

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  12. final def isInstanceOf[T0]: Boolean
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  13. def lcsSubstringIndices(a: String, b: String): List[(Int, Int, Int)]

    returns

    a list of substring indices (xlo, xhi, ylo) which together make up a (the) longest common subsequence between a and b. The substring index triples are such that

    lcsRuns(a, b).forall {
    case (xlo, xhi, ylo) =>
        (xlo until xhi).forall(i => a(i) == b(i - xlo + ylo))
}

    furthermore lcsRuns(a, b).flatMap(t => Seq(t._1, t._2)) is sorted, and so is lcsRuns(a, b).map(_._3)

  14. def longestCommonSubsequenceEdits(a: String, b: String): Seq[Diff[Char]]

    See "An O(ND) Difference Algorithm and Its Variations" by Eugene W.

    See "An O(ND) Difference Algorithm and Its Variations" by Eugene W. Myers at http://xmailserver.org/diff2.pdf; this implements the linear space recursive bidirectional search.

    returns

    a char-by-char diff describing how to turn a into b. Specifically,

    a == longestCommonSubsequenceEdits(a, b).collect {
    case Common(c) => c
    case Delete(c) => c
}.mkString

    and

    b == longestCommonSubsequenceEdits(a, b).collect {
    case Common(c) => c
    case Insert(c) => c
}.mkString

    and

    // this is the longest common subsequence of a and b:
longestCommonSubsequenceEdits(a, b).collect { case Common(c) => c }
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  18. def recurseAroundTheMiddle(a: String, b: String, zero: Int, vForward: Array[Int], vReverse: Array[Int], i: Int, n: Int, j: Int, m: Int, commonSubstringIndices: List[(Int, Int, Int)] = Nil): List[(Int, Int, Int)]

    Use findMiddle to find a common substring with (approximately) half the differences before it and half the differences after it, such that the common substring is part of the longest common subsequence.

    Use findMiddle to find a common substring with (approximately) half the differences before it and half the differences after it, such that the common substring is part of the longest common subsequence. This substring may be empty. Then recursively solve the two subproblems on each side of the substring.

    The parameters are the same as findMiddle with one exception.

    commonSubstringIndices

    return value accumulator, initially Nil

    returns

    a list of common substring indices making up the longest common subsequence.

  19. final def synchronized[T0](arg0: => T0): T0
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