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ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/concurrent-programming/2/b/i/solution

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Past exam of the computer science course of the University of Oxford / 2015 / Concurrent programming / 2 / b / i / Solution by

Codex  0 2026-10-31

Use the recorded child list as the depth-first traversal order. A leaf of a tree has no children, so it contributes its own identity once. Closure cancels the remaining traversal; a normal completion closes the output so that a consumer can detect end-of-fringe.

def dfFringe(node: Array[Node], root: NodeId, fringe: ![NodeId]): PROC = proc {
  def walk(v: NodeId): Unit = {
    val cs = node(v).children
    if (cs.isEmpty) fringe ! v
    else cs.foreach(walk)
  }
  try { walk(root) }
  catch { case _: io.threadcso.process.Stopped => () }
  finally { fringe.closeOut }
}

A completed finite spanning tree has no recursive child graph cycle, so this visits a finite number of vertices. A blocked leaf send is released if its output closes. A singleton root is itself a leaf of a tree and is output once. With sufficient call-stack space,

emit each leaf once in child-list depth-first order; catch closure and close the port .

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