Program Termination, and Well Partial Orderings

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Program Termination,and Well Partial Orderings
Andreas Blass and Yuri Gurevich
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Full version of the paper

• Andreas Blass and Yuri GurevichProgram
  Termination, and Well Partial OrderingsTech
  report MSR-TR-2006-27, Microsoft Research, March
  2006
• 178 at YGs website

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Agenda

• Program termination and the covering question
• Well partial orderings and a game
• Stature ?P? of a wpo set P, and reduction to the
  question what ??1?...??n? is
• More about the stature
• ??1?...??n? ?1?...??n
• Extra Linearizations of an arbitrary well
  partial ordering
• Related work

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• A (nondeterministic) program ? terminates ?
  the successor relation ySx (y succeeds x)
  on (possibly some abstraction of) states is
  well founded ? the transitive closure S is well
  founded? there is a ranking function for ?

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Ranking functions

• A ranking function for ? is a function f from
  the states to ordinals that is monotone ySx ?
  fy
• The ranking height of f is the smallest ordinal
  ? such that there is a ranking function f for ?
  with values

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Covering observation

• A transitive relation covered by a finitely many
  well-founded relations is well-founded if R ? U1
  ? U2 ? ? Unand U1, U2,, Un are well-founded
  and R is transitive then R is well-founded.
• Ref Geser, Podelski RybalchenkoWe learned it
  from Byron Cook.
• S is often over-approximated by a disjunction of
  relations.

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Program ?1

• if a a,b a1, b 1/2 a,b a-1, b - 1
• Use the covering observation whereyUx is
  ax
• Ranking height of is ? as 3b-2a 1000 and
  decreases at every step.

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Program ?2

• if a a1 a,b any int, b-1
• By covering observation, ?2 terminates.
• Ranking height is ?2.

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Program ?3

• In each state, some points (a,b,c) of N3 may be
  "forbidden initially, none. Once forbidden a
  point remains so forever. As long as possible,
  chose a free point (a,b,c) and forbid points
  (a?,b?,c?) with a?? a b? ? or c? ? c.
• By covering observation, ?3 terminates. What is
  the ranking height of ?3?

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Ordinal height

• Consider a well founded poset (or digraph) P and
  let x, y range over the elements of P.
• x min ? ? y for all y
• P min ? ? x for all x

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Ranking height ordinal height

• Suppose that ? terminates and consider the well
  founded poset P (states, S).
• Ordinal height x is a ranking function, and x
  ? fx for any ranking function.
• Hence P is the ranking height of ?.

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Covering question

• Suppose that U1,..,Un are well founded and R ?
  U1 ? ... ? Un is transitive.By covering
  observation, R is well founded. How to bound R
  in terms of U1,.., Un?

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Well partially ordered sets

• Df. A sequence ?x1,x2,...? in a poset P is bad
  if there are no ipartially ordered (wpo) if every bad sequence is
  finite.
• E.g. ordinals, strings with the (not necessarily
  contiguous) substring relation
• wpo ? well-founded, but not the other way round
  e.g. ? ? ? ? ? ? ???
• If P is wpo then, for every filter F, the set X
  Min(F) is finite and F y y ? some x in X.

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A frequently rediscovered concept (Kruskal 1972)

• The bad terminology harks back to Cantors well
  ordered sets. Some alternatives
• Finitely based posets (Higman, apparently the
  original discover, 1952)
• Tight posets (think boots, YG 1969)

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Games ?(P,Q)

• P and Q are well-founded posets. Positions of P
  are elements of P plus the summit above P. Same
  for Q.
• Players 1, 2 play at P, Q resp. and move
  alternately 1 moves first.
• Each player has a pebble, initially at the summit
  position.
• Move put the pebble to a position lower than the
  old.
• Win/lose if you cant move, you lose.

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Game criterion for height comparison

• 1 wins ?(P,Q) ? P Q
• 2 wins ?(P,Q) ? P ? Q
• here wins means has a winning strategy in.

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Stature Definition

• Let x, y be bad sequences. y is lower than x
  if x is a proper initial segment of y.
• Clearly BS(P) is well founded.
• The stature ?P? of a wpo set P is the height
  BS(P) of the forest BS(P) of nonempty bad
  sequences of P.

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Example

• Order ??? componentwise.
• A sequence ?(x1,y1),(x2,y2),...? without
  repetitions is bad iff there are no i(xi,yi)
• BS(???) is the well-founded forest of nonempty
  bad sequences
• ????? BS(???)

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• Let R and Ui be as in the covering question, and
  let ?iUi.
• Theorem. R ??1?...??n?.

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Upper bound

• Let R and Ui be as in the covering question, and
  let ?iUi.
• Proposition. R ? ??1?...??n?.
• Suffices to prove 2 wins ?(R,BS)). Strategy
  when 1 moves to a point x of the domain D, append
  (?1,, ?n) to the existing bad sequence where ?i
  is the height of x in the poset (D,Ui).

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The bound is tight

• ??1..?n ?R,U1,..,Un as in the question with
  Ui??n and R ??1?...??n?.
• R is the lower relation of BS(?1?...??n), and
  ?...,(x1,..,xn)? Ui ?...,(y1,..,yn)? if xi

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New question

• What is ??1?...??n??

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More about the stature
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Ideals

• IDL(P) is the set of proper ideals of P ordered
  by inclusion.
• IDL is well founded.
• Pf. A descending sequence ?D1,D2,...? of ideals
  gives rise to a bad sequence ?x1,x2,...? where
  xi?Di-Di1.

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Antichains

• Let ACH(P) be the set of nonempty antichains of a
  wpo set P ordered thusA?B if ?b in B ?a ? b in
  A.
• ACH is well founded.
• Pf. The mapping X ? P - Filter(X) is a poset
  isomorphism.

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Pointed ideals

• A pointed ideal is a pair (D,x) where D is an
  ideal and x a max element of D.
• PI(P) is the set of pointed ideals of P ordered
  thus(D,x)
• PI is well founded.
• Pf. A descending seq of pointed ideals gives a
  descending seq of ideals.

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Equivalent defs of stature

• Proposition.ACH(P) IDL(P) PI(P) ?P?.
• Pf. 1st equality by isomorphism. The rest by
  games. E.g. PI?IDL. When 1 moves to (D,x),
  move to D - x.

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Linearizations of wpo set P

• Every linearization A is well ordered and A ?
  ?P?
• Pf. 2 wins ?(A,BS(P)) append the new position to
  the current bad sequence.

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Natural sums and products

• ?2 5 ?3 ?3
• The last one is in Cantors normal form.
• Natural sum add as polynomials in ?.
• Natural product multiply as polynomials in ?
  using natural sum for the exponents.

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??1?...??n? ?1?...??n
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?1?...??n is small enough

• ?1?...??n ? ??1?...??n?
• Pf. The length of any linearization of ?1?...??n
  is ? ??1?...??n?. By induction on ?1?...??n,
  construct a linearization of ?1?...??n of
  length ?1?...??n.

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?1?...??n is large enough

• ?1?...??n ? ??1?...??n?
• Pf. Induction on ?1?...??n.

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Program ?3, again

• As long as possible, chose a free (a,b,c) in N3
  and forbid all (a?,b?,c?) with a?? a b? ? or
  c? ? c.
• The ranking height is ?3.
• Replace N3 with the direct product of arbitrary
  wpo sets.

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Extra Linearizations of an arbitrary well
partial ordering
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Motivation

• Let P be an arbitrary wpo set, and let A range
  over linearizations of P.
• We know that A ? ?P?.
• In the case P ???, we have an A with A
  ?P?.
• What is the supremum of linearization lengths in
  general?
• Is the supremum attainable?

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Theorem

• For every wpo set P, there is a linearization A
  of P such that A ?P?.
• Cor. The supremum of linearization lengths is
  ?P?.
• Cor. The supremum is attainable.

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The most relevant related work

• De Jongh and Parikh showed that
• among the lengths of linearizations of a wpo set
  P there is always a largest one, and
• in case P ???, the largest length is ???.