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86 lines
3.4 KiB
Swift
86 lines
3.4 KiB
Swift
/// Computes the SES (shortest edit script), i.e. the shortest sequence of diffs (`Free<A, Patch<A>>`) for two arrays of terms (`Fix<A>`) which would suffice to transform `a` into `b`.
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///
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/// This is computed w.r.t. an `equals` function, which computes the equality of leaf nodes within terms, and a `recur` function, which produces diffs representing matched-up terms.
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public func SES<A>(a: [Fix<A>], _ b: [Fix<A>], equals: (A, A) -> Bool, recur: (Fix<A>, Fix<A>) -> Free<A, Patch<A>>?) -> [Free<A, Patch<A>>] {
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typealias Term = Fix<A>
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typealias Diff = Free<A, Patch<A>>
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if a.isEmpty { return b.map { Diff.Pure(Patch.Insert($0)) } }
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if b.isEmpty { return a.map { Diff.Pure(Patch.Delete($0)) } }
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func cost(diff: Diff) -> Int {
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return diff.map { $0.cost }.iterate { syntax in
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switch syntax {
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case .Leaf:
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return 0
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case let .Indexed(costs):
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return costs.reduce(0, combine: +)
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case let .Keyed(costs):
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return costs.values.reduce(0, combine: +)
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}
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}
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}
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func cons(diff: Diff, rest: Memo<Stream<(Diff, Int)>>) -> Stream<(Diff, Int)> {
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return .Cons((diff, cost(diff) + costOfStream(rest)), rest)
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}
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func costOfStream(stream: Memo<Stream<(Diff, Int)>>) -> Int {
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return stream.value.first?.1 ?? 0
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}
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func min<A>(a: A, _ rest: A..., _ isLessThan: (A, A) -> Bool) -> A {
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return rest.reduce(a, combine: { isLessThan($0, $1) ? $0 : $1 })
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}
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// A matrix whose values are streams representing paths through the edit graph, carrying both the diff & the cost of the remainder of the path.
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var matrix: Matrix<Stream<(Diff, Int)>>!
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matrix = Matrix(width: a.count + 1, height: b.count + 1) { i, j in
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// Some explanation is warranted:
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//
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// 1. `matrix` captures itself during construction, because each vertex in the edit graph depends on other vertices. This is safe, because a) `Matrix` populates its fields lazily, and b) vertices only depend on those vertices downwards and rightwards of them.
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//
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// 2. `matrix` is sized bigger than `a.count` x `b.count`. This is safe, because a) we only get a[i]/b[j] when right/down are non-nil (respectively), and b) right/down are found by looking up elements (i + 1, j) & (i, j + 1) in the matrix, which returns `nil` when out of bounds. So we only access a[i] and b[j] when i and j are in bounds.
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let right = matrix[i + 1, j]
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let down = matrix[i, j + 1]
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let diagonal = matrix[i + 1, j + 1]
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let recur = {
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Term.equals(equals)($0, $1)
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? Diff($1)
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: recur($0, $1)
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}
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if let right = right, down = down, diagonal = diagonal {
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let right = (right, Diff.Pure(Patch.Delete(a[i])), costOfStream(right))
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let down = (down, Diff.Pure(Patch.Insert(b[j])), costOfStream(down))
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let diagonal = recur(a[i], b[j]).map { (diagonal, $0, costOfStream(diagonal)) }
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// nominate the best edge to continue along, not considering diagonal if `recur` returned `nil`.
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let (best, diff, _) = diagonal
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.map { min($0, right, down) { $0.2 < $1.2 } }
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?? min(right, down) { $0.2 < $1.2 }
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return cons(diff, rest: best)
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}
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// right extent of the edit graph; can only move down
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if let down = down {
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return cons(Diff.Pure(Patch.Insert(b[j])), rest: down)
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}
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// bottom extent of the edit graph; can only move right
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if let right = right {
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return cons(Diff.Pure(Patch.Delete(a[i])), rest: right)
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}
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// bottom-right corner of the edit graph
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return Stream.Nil
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}
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return Array(matrix[0, 0]!.value.map { diff, _ in diff })
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}
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import Memo
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import Stream
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