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VerifyThis 2016: Binary Tree Traversal

solution to VerifyThis 2016 challenge 2

Authors: Martin Clochard

Topics: Trees / Pointer Programs

Tools: Why3

References: VerifyThis @ ETAPS 2016

see also the index (by topic, by tool, by reference, by year)

VerifyThis @ ETAPS 2016 competition, Challenge 2: Binary Tree Traversal

The following is the original description of the verification task, reproduced verbatim from the competition web site 


Challenge 2: Binary Tree Traversal

Consider a binary tree:

  class Tree {
    Tree left, right, parent;
    bool mark;
  }

We are given a binary tree with the following properties:
It is well formed, in the sense that following a child pointer (left or right) and then following a parent pointer brings us to the original node. Moreover, the parent pointer of the root is null.
It has at least one node, and each node has 0 or 2 children.
We do not know the initial value of the mark fields.

Our goal is to set all mark fields to true. The algorithm below (Morris 1979) works in time linear in the number of nodes, as usual, but uses only a constant amount of extra space.

  void markTree(Tree root) {
    Tree x, y;
    x = root;
    do {
          x.mark = true;
          if (x.left == null && x.right == null) {
                  y = x.parent;
          } else {
                y = x.left;
                x.left = x.right;
                x.right = x.parent;
                x.parent = y;
          }
          x = y;
    } while (x != null);
  }

Tasks. Prove that:
upon termination of the algorithm, all mark fields are set
the tree shape does not change
the code does not crash, and
the code terminates.

As a bonus, prove that the nodes are visited in depth-first order

The following is a solution by Martin Clochard (Université Paris-Sud) who entered the competition.

module Memory

Component-as-array memory model with null pointers.

  use map.Map

loc is the type of memory locations (e.g pointers)

  type loc
  val predicate eq (x y:loc) ensures { result <-> x = y }

Kinds of pointer fields.

  type kind = Parent | Left | Right

Convenience alias for a pointer field table.

  type pmem = map kind (map loc loc)
  type memory = abstract {
    mutable accessor : pmem;
    mutable mark : map loc bool;
  }

memo represent memory.

  val memo : memory

the memory model has a null pointer.

  constant null : loc
  val null () : loc ensures { result = null }
  val is_null (l:loc) : bool ensures { result <-> l = null }

Memory getter & setters. non-null preconditions to check absence of null pointer access.

  val get_mark (l:loc) : bool
    requires { l <> null }
    reads { memo }
    ensures { result = memo.mark[l] }
  val set_mark (l:loc) (b:bool) : unit
    requires { l <> null }
    writes { memo.mark }
    ensures { memo.mark = (old memo).mark[l <- b] }
  val get_acc (p:kind) (l:loc) : loc
    requires { l <> null }
    reads { memo }
    ensures { result = memo.accessor[p][l] }
  val set_acc (p:kind) (l d:loc) : unit
    requires { l <> null }
    writes { memo.accessor }
    ensures { memo.accessor =
      (old memo).accessor[p <- (old memo).accessor[p][l <- d]] }

Ghost global accessors. Those are technical tools to create ghost witnesses of past states.

  val ghost get_all_accs () : pmem
    reads { memo }
    ensures { result = memo.accessor }
  val ghost get_all_marks () : map loc bool
    reads { memo }
    ensures { result = memo.mark }

  type non_det_magic
  type non_det = abstract { mutable non_det_field : non_det_magic }
  val non_det : non_det
  val anyloc () : loc writes { non_det }

Non-deterministic initialization. In the original alorithm, the variable y starts uninitialized.

end

module TreeShape

In module TreeShape we describe the correlation between the memory and the binary tree model. We also gives elements of the algorithm specification, and prove some frame/separation properties.

  use int.Int
  use set.Set
  use map.Map
  use bintree.Tree
  use bintree.Size
  use Memory

  type treel = tree loc

The described structure can be modeled as a tree of locations

  predicate is_tree (memo:pmem) (t:treel) (c p:loc) = match t with
    | Empty -> c = null
    | Node l m r -> c <> null /\ c = m /\ memo[Parent][c] = p /\
      let cl = memo[Left][c] in
      let cr = memo[Right][c] in
      (cl = null <-> cr = null) /\
      is_tree memo l cl c /\ is_tree memo r cr c
    end

is_tree memo t c p describe the fact that c points to a well-formed tree modeled by t, whose root parent node is p.

  function footprint (t:treel) : set loc = match t with
    | Empty -> empty
    | Node l m r -> add m (union (footprint l) (footprint r))
    end

footprint t is the memory footprint of t, e.g the set of locations occuring in the tree.

  predicate ext (s:set loc) (memo1 memo2:pmem) =
    forall k x. mem x s -> memo1[k][x] = memo2[k][x]

ext s memo1 memo2 mean that the pointer fields associated to locations in set s are identical in memo1 and memo2.

unchanged memo1 memo2 mean that all pointer fields in memo1 and memo2 are identical.

  predicate unchanged (memo1 memo2:pmem) =
    forall k x. memo1[k][x] = memo2[k][x]

was_marked t memo1 memo2 mean that memo2 is the update of memo1 obtained by marking all elements in t.

  predicate was_marked (t:treel) (memo1 memo2:map loc bool) =
    (forall l. mem l (footprint t) -> memo2[l]) /\
    (forall l. not mem l (footprint t) -> memo2[l] = memo1[l])

  let rec ghost ext_set (s:set loc) (memo1 memo2:pmem)
                                    (t:treel) (c p:loc) : unit
    requires { ext s memo1 memo2 }
    requires { subset (footprint t) s }
    requires { is_tree memo1 t c p }
    ensures { is_tree memo2 t c p }
    variant { t }
  = let ghost rc = ext_set s memo1 memo2 in
    match t with
    | Empty -> ()
    | Node l _ r -> rc l (memo1[Left][c]) c; rc r (memo1[Right][c]) c
    end

General extensionality property.

  let ghost ext_cur (memo1:pmem) (t:treel) (c p:loc) : unit
    reads { memo }
    requires { ext (footprint t) memo1 memo.accessor }
    requires { is_tree memo1 t c p }
    ensures { is_tree memo.accessor t c p }
  = ext_set (footprint t) memo1 (get_all_accs ()) t c p

Specialized for our use case.

  let rec ghost unicity (memo:pmem) (t1 t2:treel) (c p1 p2:loc) : unit
    requires { is_tree memo t1 c p1 /\ is_tree memo t2 c p2 }
    ensures { t1 = t2 }
    variant { t1 }
  = match t1, t2 with
    | Empty, Empty -> ()
    | Empty, _ | _, Empty -> absurd
    | Node l1 _ r1, Node l2 _ r2 ->
        unicity memo l1 l2 (memo[Left][c]) c c;
        unicity memo r1 r2 (memo[Right][c]) c c
    end

The tree model corresponding to a given pointer is unique.

  let ghost not_below (memo:pmem) (lf rg:treel) (c p:loc) : unit
    requires { is_tree memo (Node lf c rg) c p }
    ensures { not mem c (footprint lf) /\ not mem c (footprint rg) }
  = let t0 = Node lf c rg in
    let rec aux (t:treel) (c2 p2:loc) : unit
      requires { size t < size t0 }
      requires { is_tree memo t c2 p2 }
      ensures { not mem c (footprint t) }
      variant { t }
    = match t with
      | Empty -> ()
      | Node l _ r ->
        if eq c2 c then (unicity memo t t0 c p2 p;absurd);
        aux l (memo[Left][c2]) c2; aux r (memo[Right][c2]) c2
      end in
    aux lf (memo[Left][c]) c; aux rg (memo[Right][c]) c

In a non-empty well formed tree, the top pointer cannot occur in the child subtree. Otherwise, their would be an infinite branch in the tree, which is impossible in our inductive tree setting.

  type phase =
    | GoLeft
    | GoRight
    | GoBack
    | Finish

Algorithm phases. GoLeft mean that the algorithm will try to explore the left subtree GoRight mean that the algorithm will explore the right subtree GoBack mean that the algorithm will go back to the parent node Finish mean that the alogrithm will exit

  function next_phase (ph:phase) : phase =
    match ph with
    | GoLeft -> GoRight
    | GoRight -> GoBack
    | GoBack -> GoLeft
    | Finish -> Finish
    end

rotated memo1 memo2 ph c describe how c is rotated in memo2 with respect to its initial position in memo1 for current phase ph. In final phase, it is not rotated but null instead.

  predicate rotated (memo1 memo2:pmem) (ph:phase) (c:loc) =
    (forall k x. x <> c -> memo1[k][x] = memo2[k][x]) /\
    (ph <> Finish -> c <> null) /\
    match ph with
    | GoLeft -> unchanged memo1 memo2
    | GoRight -> memo2[Left][c] = memo1[Right][c] /\
        memo2[Right][c] = memo1[Parent][c] /\
        memo2[Parent][c] = memo1[Left][c]
    | GoBack -> memo1[Left][c] = memo2[Right][c] /\
        memo1[Right][c] = memo2[Parent][c] /\
        memo1[Parent][c] = memo2[Left][c]
    | Finish -> c = null
    end

end

module Recursive

In this module, we prove the algorithm by modifying the code to put it in recursive form. We justify that our limited usage of recursion makes the code equivalent to the one proposed in the challenge.

  use map.Map
  use bintree.Tree
  use ref.Ref
  use Memory
  use TreeShape

Recursion-based proof of the algorithm. The principal idea is the following: in its recursive fashion, the algorithm is a standard tree traversal (which is easy to prove correct). Hence we nest the algorithm inside a recursive procedure to link it with its recursive version. However, the obtained algorithm does not really need the call stack to run. Here is how we achieve this: - We create a non-recursive sub-routine bloc corresponding to one turn of the while loop in the challenge's code. It uses an exception to simulate exit. Also, this sub-routine does not have any non-ghost argument, so calling bloc really amounts to advancing in the loop execution. - We forbids use of any side-effect in the recursive procedure except those obtained by calling bloc. Hence calling the recursive procedure is equivalent to run a certain amounts of turns in the loop execution. Note that since we will prove that the recursive procedure terminates, it might not run a finite amount of turns and diverges afterward. - After the topmost call to the recursive procedure, we add an infinite loop who only calls bloc at every turns. This call and the loop are enclosed in an exception catching construction to detect loop termination. Hence it is justifiable that the algorithm with the recursive procedure is equivalent (in the sense that they have the same side-effects) as an infinite loop of bloc calls, encapsulated in an exception catching expression. And this algorithm is evidently equivalent to the original challenge code. 

  exception Stop

  let markTree (ghost t:treel) (root:loc) : unit
    (* Requires a well-formed non-empty tree *)
    requires { is_tree memo.accessor t root null /\ root <> null }
    writes { memo, non_det }
    (* Tree shape is unchanged *)
    ensures { is_tree memo.accessor t root null }
    ensures { unchanged (old memo).accessor memo.accessor }
    (* All nodes are marked. *)
    ensures { was_marked t (old memo).mark memo.mark }
  = let x = ref (anyloc ()) in (* ==> Tree x, y *)
    let y = ref (anyloc ()) in
    x := root; (* ==> x = root *)
    let entered = ref false in
    (* ^ Used to encode the do { .. } while() loop as a while() {..}.
       Iterations will set the flag. *)
    let ghost z = ref (null ()) in (* Store previously explored node. *)
    (* Loop body. `mem0` is memory state at start of recursive call,
       `ph` is phase. *)
    let bloc (ghost mem0:pmem) (ghost ph:phase) : unit
      (* current node `!x` is rotated correctly with
         respect to the phase. *)
      requires { rotated mem0 memo.accessor ph !x }
      requires { ph = Finish -> !entered }
      writes { memo, x, y, z, entered }
      (* Caracterise `x` and `z` reference updates. *)
      ensures { !z = !(old x) /\ !x = memo.accessor[Parent][!z] }
      (* `Finish` phase never ends normally *)
      ensures { ph <> Finish /\ !entered }
      (* Node is marked, and other nodes do not change. *)
      ensures { memo.mark[!z] }
      ensures { forall l. l <> !z -> memo.mark[l] = (old memo).mark[l] }
      (* Describe phase shift. *)
      ensures {
        if (old memo).accessor[Left][!z] = null
           = (old memo).accessor[Right][!z]
        then memo.accessor = (old memo).accessor
        else rotated mem0 memo.accessor (next_phase ph) !z }
      (* In `Finish` phase, the code throws `Stop` without changing
         anything. *)
      raises { Stop -> ph = Finish /\ memo = old memo }
    = if !entered && is_null !x then raise Stop; (* ==> do { BODY         *)
      entered := true;                           (*  } while (x != null); *)
      (ghost z := !x);
      set_mark !x true; (* ==> x.mark = true; *)
      if is_null (get_acc Left !x) && is_null (get_acc Right !x)
         (* ==> if (x.left = null && x.right == null) { *)
      then y := get_acc Parent !x (* ==> y = x.parent *)
      else begin (* ==> } else { *)
        y := get_acc Left !x; (* ==> y = x.left; *)
        set_acc Left !x (get_acc Right !x); (* ==> x.left = x.right; *)
        set_acc Right !x (get_acc Parent !x); (* ==> x.right = x.parent; *)
        set_acc Parent !x !y; (* ==> x.parent = y; *)
      end; (* ==> } *)
      x := !y (* ==> x = y; *)
    in
    (* 'Recursive proof', corresponding to the
        expected depth-first traversal. *)
    let rec aux (ghost t:treel) : unit
      requires { !x <> null }
      requires { is_tree memo.accessor t !x !z }
      writes { memo, x, y, z, entered }
      ensures { unchanged (old memo).accessor memo.accessor }
      ensures { !x = old !z /\ !z = old !x }
      ensures { was_marked t (old memo).mark memo.mark }
      ensures { !entered }
      raises { Stop -> false }
      variant { t }
    = let ghost mem0 = get_all_accs () in
      let ghost _c = !x in
      let _lf = get_acc Left !x in
      let _rg = get_acc Right !x in
      let lf, rg = ghost match t with
        | Empty -> absurd
        | Node l _ r -> l, r
        end in
      (ghost not_below mem0 lf rg _c !z);
      bloc mem0 GoLeft;
      let b = begin ensures { result <-> _lf = null /\ _rg = null }
        ensures { result -> lf = Empty /\ rg = Empty }
        is_null _lf && is_null _rg end in
      if not b then begin
        (ghost ext_cur mem0 lf _lf _c);
        aux lf;
        bloc mem0 GoRight;
        (ghost ext_cur mem0 rg _rg _c);
        aux rg;
        bloc mem0 GoBack
      end
    in
    let ghost mem0 = get_all_accs () in
    try aux t;
      label I in
      while true do
        invariant { memo = (memo at I) }
        invariant { rotated mem0 memo.accessor Finish !x /\ !entered }
        variant { 0 }
        bloc mem0 Finish
      done
    with Stop -> (ghost ext_cur mem0 t root (null ())); () end

end

module Iterative

In this module we provide a proof of the initial algorithm by derecursiving the previous one

  use int.Int
  use map.Map
  use map.Const
  use bintree.Tree
  use bintree.Size
  use option.Option
  use ref.Ref
  use Memory
  use TreeShape

  type snap = {
    pointers : pmem;
    cursor : loc;
    parent : loc;
    marks : map loc bool;
  }

Snapshot of all relevant memory values in the program

  type frame = {
    (* Memory & ghost argument at call site (pc>=0). *)
    memo0 : snap;
    tree : treel;
    (* right & left trees/pointers (pc>=1) *)
    tleft : treel;
    pleft : loc;
    tright : treel;
    pright : loc;
    (* Memory after first bloc call (pc>=2) *)
    memo1 : snap;
    (* Memory after first recursive call (pc >= 4) *)
    memo2 : snap;
    (* Memory after second bloc call (pc >= 5) *)
    memo3 : snap;
    (* Memory after second recursive call (pc >= 6) *)
    memo4 : snap;
  }

Stack frame in the recursive version. Fields are assigned as the code pointer increase.

  function frame_memo (f:frame) (pc:int) : snap =
    if pc <= 0 then f.memo0
    else if pc <= 1 then f.memo1
    else if pc <= 2 then f.memo2
    else if pc <= 3 then f.memo3
    else f.memo4

Find out current memory state with respect to code pointer.

  predicate bloc_rel (mem0:pmem) (ph:phase) (s1 s2:snap) =
    s2.parent = s1.cursor /\ s2.cursor = s2.pointers[Parent][s2.parent] /\
    s2.marks[s2.parent] /\
    (forall l. l <> s2.parent -> s2.marks[l] = s1.marks[l]) /\
    if s1.pointers[Left][s1.cursor] = null = s1.pointers[Right][s1.cursor]
    then s2.pointers = s1.pointers
    else rotated mem0 s2.pointers (next_phase ph) s2.parent

Postcondition relation for bloc calls.

  predicate rec_rel (t:treel) (s1 s2:snap) =
    unchanged s1.pointers s2.pointers /\
    s2.cursor = s1.parent /\ s2.parent = s1.cursor /\
    was_marked t s1.marks s2.marks

postcondition relation for recursive (aux) calls.

  type stack =
    | Bottom
    | Running stack int frame
    | Done

Call stack

  predicate is_stack (t:treel) (stop scur:snap)
                     (s:stack) (calls:option treel) =
    match s with
    | Bottom -> stop = scur /\ calls = Some t
    | Running s pc f ->
        0 <= pc <= 4 /\
        (* Correctness of caller stack. *)
        let m0 = f.memo0 in
        is_stack t stop m0 s (Some f.tree) /\
        (* Precondition for recursive call. *)
        m0.cursor <> null /\ is_tree m0.pointers f.tree m0.cursor m0.parent /\
        (* Initially obtained pointers & subtrees. *)
        f.tree = Node f.tleft m0.cursor f.tright /\
        f.pleft = m0.pointers[Left][m0.cursor] /\
        f.pright = m0.pointers[Right][m0.cursor] /\
        (* First bloc run, if completed normally. *)
        (pc >= 1 -> bloc_rel m0.pointers GoLeft m0 f.memo1 /\
                    f.pleft <> null /\ f.pright <> null) /\
        (* First recursive call. *)
        (pc >= 2 -> rec_rel f.tleft f.memo1 f.memo2) /\
        (* Second bloc run. *)
        (pc >= 3 -> bloc_rel m0.pointers GoRight f.memo2 f.memo3) /\
        (* Second recursive call. *)
        (pc >= 4 -> rec_rel f.tright f.memo3 f.memo4) /\
        (* Current memory state. *)
        frame_memo f pc = scur /\
        match calls with
        | None -> pc <> 1 /\ pc <> 3
        | Some u ->
            if pc = 1 then u = f.tleft else pc = 3 /\ u = f.tright
        end
    | Done -> rec_rel t stop scur /\ calls = None
    end

Describe a valid call stack. Mostly precise which relation between states holds. Note that in the previous version, Why3 did that part for us via the weakest precondition calculus.

  constant large_enough : int = 100
  function stack_size (st:stack) : int = match st with
    | Bottom -> 1
    | Done -> 0
    | Running s pc f ->
        stack_size s + (large_enough - pc) + if pc = 0
        then large_enough * (size f.tleft + size f.tright)
        else if pc <= 2
        then large_enough * size f.tright
        else 0
    end

Termination argument. In fully general cases, we would need a lexicographic ordering but here an integer size suffice.

  let rec lemma stack_size_pos (st:stack) : unit
    requires { exists t stop scur calls. is_stack t stop scur st calls }
    ensures { stack_size st >= 0 }
    variant { st }
  = match st with Running s _ _ -> stack_size_pos s | _ -> () end

  let ghost opening (t ct:treel) (stop scur:snap) (ghost st:ref stack)
    requires { is_stack t stop scur !st (Some ct) }
    requires { is_tree scur.pointers ct scur.cursor scur.parent }
    requires { scur.cursor <> null }
    writes { st }
    ensures { is_stack t stop scur !st None }
    ensures { stack_size !st <= stack_size !(old st) +
                                large_enough * size ct }
  = match ct with
    | Empty -> absurd
    | Node lf _ rg ->
      let f = {
        memo0 = scur; tree = ct; tleft = lf; tright = rg;
        pleft = scur.pointers[Left][scur.cursor];
        pright = scur.pointers[Right][scur.cursor];
        memo1 = scur; memo2 = scur; memo3 = scur; memo4 = scur;
      } in
      st := Running !st 0 f
    end

Create a stack frame for a recursive call

  let ghost closing (t ct:treel) (stop sprev scur:snap)
                    (ghost st:ref stack) : unit
    requires { is_stack t stop sprev !st (Some ct) }
    requires { rec_rel ct sprev scur }
    writes { st }
    ensures { is_stack t stop scur !st None }
    ensures { stack_size !st < stack_size !(old st) }
  = match !st with
    | Bottom -> st := Done
    | Done -> absurd
    | Running s pc f ->
      let f = if pc = 1 then { f with memo2 = scur }
      else { f with memo4 = scur } in
      st := Running s (pc+1) f
    end

Remove a stack frame at end of recursive call simulation

  let ghost continuing (t:treel) (stop sprev scur:snap)
                       (ghost st:ref stack)
    requires { is_stack t stop sprev !st None }
    requires { match !st with
      | Bottom | Done -> false
      | Running _ pc f ->
          let ph =
            if pc = 0 then GoLeft else if pc = 2 then GoRight else GoBack in
          bloc_rel f.memo0.pointers ph sprev scur
      end }
    writes { st }
    ensures { is_stack t stop scur !st None }
    ensures { stack_size !st < stack_size !(old st) }
  = match !st with
    | Bottom | Done -> absurd
    | Running s pc f ->
      not_below f.memo0.pointers f.tleft f.tright
                f.memo0.cursor f.memo0.parent;
      if pc = 0 then
        if is_null f.pleft && is_null f.pright
        then begin
          st := s;
          assert { match f.tleft, f.tright with
            Empty, Empty  -> true | _ -> false end };
          closing t f.tree stop f.memo0 scur st
        end else begin
          let f = { f with memo1 = scur } in
          st := Running s (pc+1) f;
          ext_set (footprint f.tleft) f.memo0.pointers scur.pointers
                  f.tleft f.pleft f.memo0.cursor;
          opening t f.tleft stop scur st;
        end
      else if pc = 2 then begin
        let f = { f with memo3 = scur } in
        st := Running s (pc+1) f;
        ext_set (footprint f.tright) f.memo0.pointers scur.pointers
                f.tright f.pright f.memo0.cursor;
        opening t f.tright stop scur st;
      end else if pc = 4 then begin
        st := s;
        assert { unchanged scur.pointers f.memo0.pointers };
        closing t f.tree stop f.memo0 scur st
      end
    end

Advance code pointer when a bloc is run. Takes care to open/close new frame as needed.

  let markTree (ghost t:treel) (root:loc) : unit
    (* Requires a well-formed non-empty tree *)
    requires { is_tree memo.accessor t root null /\ root <> null }
    writes { memo, non_det }
    (* Tree shape is unchanged *)
    ensures { is_tree memo.accessor t root null }
    ensures { unchanged (old memo).accessor memo.accessor }
    (* All nodes are marked. *)
    ensures { was_marked t (old memo).mark memo.mark }
  = let x = ref (anyloc ()) in (* ==> Tree x, y *)
    let y = ref (anyloc ()) in
    let ghost z = ref (null ()) in
    let ghost snapshot () : snap
      ensures { result.pointers = memo.accessor }
      ensures { result.cursor = !x }
      ensures { result.parent = !z }
      ensures { result.marks = memo.mark }
    = { pointers = get_all_accs ();
        cursor = !x; parent = !z;
        marks = get_all_marks () } in
    x := root; (* ==> x = root; *)
    let ghost stop = snapshot () in
    let ghost scur = ref stop in
    let ghost st = ref Bottom in
    ghost opening t t stop stop st;
    let entered = ref false in (* encode do-while loop. *)
    while not (begin ensures { result <-> !x = null }
        !entered && is_null !x end)
    do (* ==> do { BODY } while(x != null); *)
      invariant { !x = null -> !entered }
      invariant { !scur.pointers = memo.accessor }
      invariant { !scur.cursor = !x }
      invariant { !scur.parent = !z }
      invariant { !scur.marks = memo.mark }
      invariant { is_stack t stop !scur !st None }
      variant { stack_size !st }
      entered := true;
      (ghost z := !x);
      set_mark !x true; (* ==> x.mark = true; *)
      if is_null (get_acc Left !x) && is_null (get_acc Right !x)
        (* ==> if (x.left == null && x.right == null) { *)
      then begin
        y := get_acc Parent !x; (* ==> y = x.parent; *)
      end
      else begin (* ==> } else { *)
        y := get_acc Left !x; (* ==> y = x.left; *)
        set_acc Left !x (get_acc Right !x); (* ==> x.left = x.right; *)
        set_acc Right !x (get_acc Parent !x); (* ==> x.right = x.parent; *)
        set_acc Parent !x !y; (* ==> x.parent = y; *)
      end; (* ==> } *)
      x := !y; (* ==> x = y; *)
      let ghost sprev = !scur in
      ghost scur := snapshot ();
      ghost continuing t stop sprev !scur st;
    done; (* ==> end 'do while' *)
    ghost (match !st with Done -> () | _ -> absurd end);
    ghost ext_cur stop.pointers t root (null ())

The main algorithm.

end
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Why3 Proof Results for Project "verifythis_2016_tree_traversal"

Theory "verifythis_2016_tree_traversal.TreeShape": fully verified

ObligationsAlt-Ergo 2.0.0
VC for ext_set0.19
VC for ext_cur0.00
VC for unicity0.03
VC for not_below1.06

Theory "verifythis_2016_tree_traversal.Recursive": fully verified

ObligationsAlt-Ergo 2.0.0Alt-Ergo 2.4.1CVC4 1.5
VC for markTree---------
split_goal_right
exceptional postcondition0.00------
precondition0.01------
precondition0.00------
precondition0.01------
precondition0.00------
postcondition0.00------
postcondition0.01------
postcondition0.01------
postcondition0.01------
postcondition0.00------
precondition0.00------
precondition0.01------
precondition0.01------
precondition0.01------
precondition0.01------
precondition0.01------
postcondition0.01------
postcondition0.01------
postcondition0.01------
postcondition0.02------
postcondition0.10------
precondition0.01------
precondition0.01------
unreachable point0.00------
precondition0.01------
precondition0.01------
precondition0.00------
postcondition0.00------
postcondition------0.07
precondition0.02------
precondition0.01------
variant decrease0.01------
precondition0.02------
precondition0.02------
precondition0.05------
precondition0.01------
precondition0.03------
precondition0.01------
variant decrease0.02------
precondition0.03------
precondition0.04------
precondition0.13------
precondition0.01------
postcondition0.09------
postcondition0.12------
postcondition0.28------
postcondition0.01------
exceptional postcondition0.01------
exceptional postcondition0.01------
postcondition0.01------
postcondition0.01------
postcondition0.06------
postcondition0.01------
exceptional postcondition0.01------
precondition0.00------
precondition0.00------
loop invariant init0.01------
loop invariant init0.01------
precondition---0.01---
precondition---0.01---
loop variant decrease---0.01---
loop invariant preservation---0.00---
loop invariant preservation---0.00---
precondition---0.01---
precondition---0.01---
postcondition---0.01---
postcondition---0.01---
postcondition---0.01---

Theory "verifythis_2016_tree_traversal.Iterative": fully verified

ObligationsAlt-Ergo 2.0.0
VC for stack_size_pos0.07
VC for opening0.03
VC for closing0.27
VC for continuing---
split_goal_right
unreachable point0.02
unreachable point0.01
precondition0.03
assertion0.10
precondition0.03
precondition0.29
postcondition0.02
postcondition0.02
precondition0.25
precondition0.02
precondition0.07
precondition0.30
precondition0.12
precondition0.14
postcondition0.02
postcondition0.04
precondition0.76
precondition0.02
precondition0.06
precondition0.39
precondition0.30
precondition0.29
postcondition0.02
postcondition0.03
assertion0.27
precondition0.02
precondition0.62
postcondition0.02
postcondition0.02
postcondition0.03
postcondition0.02
VC for markTree0.86