Ranking Functions. Linear-Constraint Loops

Transcription

1 for Linear-Constraint Loops Amir Ben-Amram 1

2 for Loops Example 1 (GCD program): while (x > 1, y > 1) if x<y then (x,y) := (x, y-x) else (x,y) := (y, x-y) Here f(x,y) = x+y is a ranking function non-negative in all (enabled) states strictly decreasing proves termination 4

3 for Loops Example 2: while (x > 1, y > 1) if then (x,y) := (x, y-1) else (x,y) := (x-1, f(x,y)) Here f(x,y) = x,y is a ranking function strictly decreasing lexicographically 5

4 Outline Ranking functions for linear-constraint loops: a wellunderstood tool in termination and resource analysis In this talk: linear ranking functions & lexicographic linear ranking functions For each type: review one major technique proceed to recent work (B. & Genaim) 6

5 Why? Termination of imperative programs Termination of functional and logic programs Complexity analysis (execution time, etc) The ranking function bounds the number of iterations / length of call chain Loop parallelization how to schedule computations that depend on previous results 7

6 Loops A loop: while (1< x+y+z) { x := x + 1; y := y 1; z := z 1; } Guard update Consider numerical variables (most often integer) The update is straight-line code 8

7 Loops A loop can have multiple paths while (1< x+y+z) { if (y>0) x := x + 1; y := y 1; else z := z 1; } 9

8 Loops A loop in linear constraint representation: while (y x, x+y 1) while (B ) do next state x x 0, y y+2x 1 A Update may be non-deterministic, and can model non-linear operations, e.g., integer division x := (2 x)/5 5x' <= 2x, 5x >= 2x 4 10

9 Loops A multiple path loop: loop (B 1 ) A (B 2 ) A 11

10 States & Transitions State: =,, (typically) perhaps of (as an over approximation easier to solve) Transition: = =,,,,, Each path of a loop is a subset of ( ) May be written this specifies a convex polyhedron, the transition polyhedron the set of its integer points is I( ) we can analyze the same loop over rationals or over integers 12

11 with values in f: 0 1 (Bounded) (Descending) Essentially a function into r.f. over integers: consider only I( ) 16

12 Linear (LRF) = + for rational, x + 0 and 1 (B) (D) r.f. over integers: I( ) The Linar Ranking Function problem [over ] Given a loop (as constraints), find a LRF (satisfying (B),(D)) for [for I( )] 19

13 LRF by Linear Programming (LP) linear constraints x + 0 and 1 (B) (D) linear constraints 20

14 LRF by Linear Programming (LP) Sohn and van Gelder (1991) Feautrier (1992) Colόn and Sipma (2001) Podelski and Rybalchenko (2004) Mesnard and Serebrenik (2008) Logic programming Farkas Lemma parallelization imperative programs see also survey by Bagnara et al. (2012) 21

15 The Farkas based solution How to draw an implication from inequalities + + ( + ) Farkas lemma says that all implications are formed this way 22

16 The Farkas based solution transition polyhedron - should imply (B)+(D)

17 The Farkas based solution transition polyhedron - should imply (B)+(D) We just have to find the Farkas multipliers - which is an LP problem - every solution of which yields a LRF - it specifies the set of all LRFs for 24

18 Multipath loops An LRF has to be valid for all the paths A conjunction of LP problems is an LP problem Solved! The Farkas based solution is: polynomial time complete, in some sense 25

19 Completeness We call a method complete if it is guaranteed to find a LRF, when one exists The LP based methods are complete over the rationals (B)+(D) have to hold over, not just I( ) 26

20 Understanding the integer LRF problem (B. & Genaim, POPL 13) The decision problem is conp-complete we show: conp hardness inclusion in conp synthesis algorithm in exponential time PTIME-solvable classes (DBMs? octagons? TVPI?) 27

21 Examples: Integer vs Rational while (x y, x+y 1) do y := y+1-2x Here f(x,y) = x+y is a ranking function for integers but not for rationals! (consider (½,½)) while (x y, x+y 1, 4x 1) do y := y+1-2x Here f(x,y) = 2(x+y) is a ranking function over the rationals, while for integers x+y is valid 28

22 The source of hardness Easy (PTIME) hard (NPC) 29

23 Hardness proof NP-hard problem: Given constraints: integer solution?, is there any Reduction to termination problem: while ( x z b, z 0) do x = x, z = 0 30 This loop has a LRF there is no integer solution to Bx

24 Clue to solution polyehdron I integer hull = convex hull of I( ) 31

25 while (x 2 x 1, x 1 +x 2 1) do x 2 = x x 1, x 1 = x 1 I obnoxious point I has a LRF 32

26 Solution If I( ) has a LRF, has one. I Compute a representation of Find LRF over rationals I Complete solution, exponential complexity 33

27 Inclusion in conp THM The LRF existence problem over integers is conp-complete. There are polynomially-checkable witnesses to non-existence of a LRF 34

28 Witnesses Consider a candidate function Point witnesses against f if f fails to satisfy (B) or (D) + 0 (B) 1 (D) Let no LRF space of coefficient vectors 35

29 Where to look for witnesses? 1. If is bounded: x 1 x 2 x 3 x 4 If is not a RF, it must fail on one of the vertices. (If all vertices satisfy (B),(D), then so does any convex combination.) no LRF,, = 36

30 Where to look for witnesses? x 1 x 2 no LRF,, = x 3 (B),(D) fail on,, x 4 A corollary of Farkas Lemma (found in Schrijver): if a set of linear constraints on has no solution, there is a subset of +2of them that doesn t We have a small witness set The bit-size of the witnesses is polynomial (theorem on relation between bit-size of constraints and of vertices.) 37

31 Unbounded polyhedra Every polyhedron can be represented as =conv.hull,, +cone(,, ) A ray y is added to a point x conv.hull,, to form points x + y, 0 y witnesses against f if f fails to satisfy (B ) or (D ) 0 (B ) 0 (D ) 38

32 Inclusion in conp Witnesses can be found among the vertices and rays There are witness sets of polynomial cardinality and bit-size, and they are polynomially checkable 39

33 Cases we can solve in polynomial time Basic observation: Our problem becomes tractable if either is an integral polyhedron ( I = ) [CKRW 10] or we have a specialized procedure to compute I 41

34 Totally Unimodular matrices Matrix A is totally unimodular if each subdeterminant of A is in {0, 1} A is TUM the polyhedron is integral Example: difference-bound constraints ± yield TUM constraint matrices 42

35 Difference-bound constraints are also special cases of: ± ± Octagons ± Two-variable per inequality (TVPI) constraints + Octagons can be non-integral and their integer hull can be hard to compute 43

36 Using two-dimensional polyhedra Harvey (1999) shows how to compute in PTIME the integer hull of any two-dimensional polyhedron More than 2 variables? Try to decompose the constraint set into independent constraint-sets while (4x >= 1 & y>=1) do 5x' <= 2x + 1, 5x > 2x 4, y = y y' 44

37 Outline So far: linear ranking functions complete solutions, algorithms and complexity Next: lexicographic linear ranking functions For each type: review one major technique proceed to recent work (B. & Genaim) 47

38 L LRFs (Lexicographic Linear ) Very natural, at least for multipath-loops (and more general control-flow graphs) Stricly extend the class of loops (compared to LRFs) Known since the dawn of times [Turing 1948] Widely used in the termination literature Deduction from linear constraints begins with Colón and Sipma (2002); Bradley, Manna and Sipma ( 05) 50

39 The Farkas based solution Alias et al. [ADFG 2010] : Program is a control-flow graph with linear constraints (for simplicity: multipath loop), dimension variables,, Algorithm generates a LLRF,,,, Has polynomial time complexity based on LP Applies the Farkas technique times, where #of paths (for a multipath loop) Used to bound the no. of transitions Guaranteed to find the smallest dimension 51

40 Alias et al. prove completeness over the rationals Our plan: Solve the integer-restricted problem Our results: decision problem is conp-complete synthesis algorithm guarantees smallest dimension we find LLRFs for some single-path loops having no LRF a bit of a surprise more so because [ADFG] proves it cannot happen the reason is a difference in definitions 52

41 Think about integers Case 1: descends Case 2: does not descend In [ADFG] this is not allowed, since they would require to be non-negative at all times In our definition, if the 1 st component descends, 53 we do not care about the 2 nd

42 The algorithm at a glance Definition: a quasi-ranking function for is f: such that 0 Note that =0 satisfies it A function is non-trivial if. > 54

43 To build a LLRF for : Find a non-trivial quasi-ranking function (based on the Farkas method for LRF) Compute This gives the transitions where does not descend. If, proceed recursively. as in last example 55

44 Properties of the algorithm: Maximum recursion depth is because components are linearly independent It takes polynomial time (over the rationals) To get completeness over the integers, compute the integer hull first PTIME special cases apply Optimality of the dimension follows from a judicious selection of the quasi-lrf 56

45 LLRFs and the number of steps [ADFG 2010] use the LLRF to bound the number of steps. The idea: Since always decreases, the number of steps is bounded by the number of distinct values of ( ) Let be the state space (polyhedral invariant computation is used to find it) ( ) is a d-dimensional polyhedron The number of integer points in ( ) is estimated. 57

46 Example Bubble Sort loop (0 i<n 0 j<n) i = i + 1, j = 0, n = n i = i, j = j + 1, n = n (i,j) = n-i, n-j is a ranking function ={0 i < n 0 j < n n = } ( ) = 1, init [1, init ] - quadratic 58

47 Single-path example: Examples (i,j) =, is a ranking function = 0, 0, + ( ) = 0, [, + ] - quadratic We proved that for certain loops (including all SLC loops with LLRF) a linear bound can be found In the above loop, (max(, )) 59

48 Summary (Lexicographic) Linear ranking functions are a useful tool and there are solid theoretical results. The decision problems are PTIME over rationals, conp-complete over integers. The PTIME solution is sound for the integers, and for certain classes of constraints, it is complete. Synthesis algorithms have corresponding efficiency 63

49 Bibliography Please see our POPL paper and TR! 64

50 Demo program - irankfinder 65