/// Computes the SES (shortest edit script), i.e. the shortest sequence of diffs (`Free>`) for two arrays of terms (`Fix`) which would suffice to transform `a` into `b`. /// /// 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. public func SES(a: [Fix], _ b: [Fix], equals: (A, A) -> Bool, recur: (Fix, Fix) -> Free>?) -> [Free>] { typealias Term = Fix typealias Diff = Free> if a.isEmpty { return b.map { Diff.Pure(Patch.Insert($0)) } } if b.isEmpty { return a.map { Diff.Pure(Patch.Delete($0)) } } func cost(diff: Diff) -> Int { return diff.map(const(1)).iterate { syntax in switch syntax { case .Leaf: return 0 case let .Indexed(costs): return costs.reduce(0, combine: +) case let .Keyed(costs): return costs.values.reduce(0, combine: +) } } } func cons(diff: Diff, rest: Memo>) -> Stream<(Diff, Int)> { return .Cons((diff, cost(diff) + costOfStream(rest)), rest) } func costOfStream(stream: Memo>) -> Int { return stream.value.first?.1 ?? 0 } func min(a: A, _ rest: A..., _ isLessThan: (A, A) -> Bool) -> A { return rest.reduce(a, combine: { isLessThan($0, $1) ? $0 : $1 }) } // A matrix whose values are streams representing paths through the edit graph, carrying both the diff & the cost of the remainder of the path. var matrix: Matrix>! matrix = Matrix(width: a.count + 1, height: b.count + 1) { i, j in // Some explanation is warranted: // // 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. // // 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. let right = matrix[i + 1, j] let down = matrix[i, j + 1] let diagonal = matrix[i + 1, j + 1] let recur = { Term.equals(equals)($0, $1) ? Diff($1) : recur($0, $1) } if let right = right, down = down, diagonal = diagonal { let right = (right, Diff.Pure(Patch.Delete(a[i])), costOfStream(right)) let down = (down, Diff.Pure(Patch.Insert(b[j])), costOfStream(down)) let diagonal = recur(a[i], b[j]).map { (diagonal, $0, costOfStream(diagonal)) } // nominate the best edge to continue along, not considering diagonal if `recur` returned `nil`. let (best, diff, _) = diagonal .map { min($0, right, down) { $0.2 < $1.2 } } ?? min(right, down) { $0.2 < $1.2 } return cons(diff, rest: best) } // right extent of the edit graph; can only move down if let down = down { return cons(Diff.Pure(Patch.Insert(b[j])), rest: down) } // bottom extent of the edit graph; can only move right if let right = right { return cons(Diff.Pure(Patch.Delete(a[i])), rest: right) } // bottom-right corner of the edit graph return Stream.Nil } return Array(matrix[0, 0]!.value.map { diff, _ in diff }) } import Memo import Prelude import Stream