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Program Termination, and Well Partial Orderings Andreas Blass and Yuri Gurevich

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2 Full version of the paper Andreas Blass and Yuri Gurevich Program Termination, and Well Partial Orderings Tech report MSR-TR , Microsoft Research, March 2006 #178 at YG s website

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3 Agenda 1. Program termination and the covering question 2. Well partial orderings and a game 3. Stature P of a wpo set P, and reduction to the question what 1... n i s 4. More about the stature n = 1... n 6. Extra: Linearizations of an arbitrary well partial ordering 7. Related work

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4 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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5 Ranking functions A ranking function for is a function f from the states to ordinals that is monotone: ySx fy

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6 Covering observation A transitive relation covered by a finitely many well-founded relations is well- founded: if R U 1 U 2 … U n and U 1, U 2, …, U n are well-founded and R is transitive then R is well-founded. Ref: Geser, Podelski & Rybalchenko We learned it from Byron Cook. S + is often over-approximated by a disjunction of relations.

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7 Program 1 if a < 1000 < b then choose between a,b:= a+1, b + 1/2 a,b:= a-1, b - 1 Use the covering observation where yUx is a x 1000 and decreases at every step.

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8 Program 2 if a < 1000 < b then choose between a := a+1 a,b:= any int, b-1 By covering observation, 2 terminates. Ranking height is 2.

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9 Program 3 In each state, some points (a,b,c) of N 3 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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10 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 < x} |P| = min { : > |x| for all x}

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11 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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12 Covering question Suppose that U 1,..,U n are well founded and R U 1... U n is transitive. By covering observation, R is well founded. How to bound |R| in terms of |U 1 |,.., |U n |?

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13 Well partially ordered sets Df. A sequence x 1,x 2,... in a poset P is bad if there are no i

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14 A frequently rediscovered concept (Kruskal 1972) The bad terminology harks back to Cantor s well ordered sets. Some alternatives: Finitely based posets (Higman, apparently the original discover, 1952) Tight posets (think boots, YG 1969)

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15 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 can t move, you lose.

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16 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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17 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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18 Example Order componentwise. A sequence (x 1,y 1 ),(x 2,y 2 ),... without repetitions is bad iff there are no i

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19 Let R and U i be as in the covering question, and let i =|U i |. Theorem. |R| = 1... n.

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20 Upper bound Let R and U i be as in the covering question, and let i =|U i |. 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,U i ).

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21 The bound is tight 1.. n R,U 1,..,U n as in the question with |U i | n and |R|= 1... n. R is the lower relation of BS( 1... n ), and..., (x 1,..,x n ) U i..., (y 1,..,y n ) if x i < y i

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22 New question What is 1... n ?

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23 More about the stature

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24 Ideals IDL(P) is the set of proper ideals of P ordered by inclusion. IDL is well founded. Pf. A descending sequence D 1,D 2,... of ideals gives rise to a bad sequence x 1,x 2,... where x i D i -D i+1.

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25 Antichains Let ACH(P) be the set of nonempty antichains of a wpo set P ordered thus: A 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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26 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) < (E,y) if D E – {y}. PI is well founded. Pf. A descending seq of pointed ideals gives a descending seq of ideals.

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27 Equivalent defs of stature Proposition. |ACH(P)| = |IDL(P)| = |PI(P)| = P. Pf. 1 st 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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28 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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29 Natural sums and products = 3 < The last one is in Cantor s normal form. Natural sum: add as polynomials in. Natural product: multiply as polynomials in using natural sum for the exponents.

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n = 1... n

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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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n is large enough 1... n 1... n Pf. Induction on 1... n.

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

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34 Extra: Linearizations of an arbitrary well partial ordering

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35 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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36 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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37 The most relevant related work De Jongh and Parikh showed that 1. among the lengths of linearizations of a wpo set P there is always a largest one, and 2. in case P =, the largest length is.

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