1 / 60

Program verification: flowchart programs

Program verification: flowchart programs. (Book: chapter 7). History. Verification of flowchart programs: Floyd, 1967 Hoare’s logic: Hoare, 1969 Linear Temporal Logic: Pnueli, Krueger, 1977 Model Checking: Clarke & Emerson, 1981. Program Verification. Predicate (first order) logic.

lolita
Download Presentation

Program verification: flowchart programs

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Program verification: flowchart programs (Book: chapter 7)

  2. History • Verification of flowchart programs: Floyd, 1967 • Hoare’s logic: Hoare, 1969 • Linear Temporal Logic: Pnueli, Krueger, 1977 • Model Checking: Clarke & Emerson, 1981

  3. Program Verification • Predicate (first order) logic. • Partial correctness, Total correctness • Flowchart programs • Invariants, annotated programs • Well founded ordering (for termination) • Hoare’s logic

  4. Predicate (first order logic) • Variables, functions, predicates • Terms • Formulas (assertions)

  5. Signature • Variables: v1, x, y18 Each variable represents a value of some given domain (int, real, string, …). • Function symbols: f(_,_), g2(_), h(_,_,_). Each function has an arity (number of paramenters), a domain for each parameter, and a range. f:int*int->int (e.g., addition), g:real->real (e.g., squareroot) A constant is a predicate with arity 0. • Relation symbols: R(_,_), Q(_). Each relation has an arity, and a domain for each parameter. R : real*real (e.g., greater than). Q : int (e.g., is a prime).

  6. Terms • Terms are objects that have values. • Each variable is a term. • Applying a function with arity n to n terms results in a new term. Examples: v1, 5.0, f(v1,5.0), g2(f(v1,5.0)) More familiar notation: sqr(v1+5.0)

  7. Formulas • Applying predicates to terms results in a formula. R(v1,5.0), Q(x) More familiar notation: v1>5.0 • One can combine formulas with the boolean operators (and, or, not, implies). R(v1,5.0)->Q(x) x>1 -> x*x>x • One can apply existentail and universal quantification to formulas. x Q(X) x1 R(x1,5.0) x y R(x,y)

  8. A model, A proofs • A model gives a meaning (semantics) to a first order formula: • A relation for each relation symbol. • A function for each function symbol. • A value for each variable. • An important concept in first order logic is that of a proof. We assume the ability to prove that a formula holds for a given model. • Example proof rule (MP) : 

  9. Flowchart programs Input variables: X=x1,x2,…,xl Program variables: Y=y1,y2,…,ym Output variables: Z=z1,z2,…,zn start Z=h(X,Y) Y=f(X) halt

  10. Assignments and tests T F Y=g(X,Y) t(X,Y)

  11. Initial condition start Initial condition: the values for the input variables for which the program must work. x1>=0 /\ x2>0 (y1,y2)=(0,x1) y2>=x2 F T (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) halt

  12. The input-output claim start The relation between the values of the input and the output variables at termination. x1=z1*x2+z2 /\ 0<=z2<x2 (y1,y2)=(0,x1) y2>=x2 T F (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) halt

  13. Partial correctness, Termination, Total correctness • Partial correctness: if the initial condition holds and the program terminates then the input-output claim holds. • Termination: if the initial condition holds, the program terminates. • Total correctness: if the initial condition holds, the program terminates and the input-output claim holds.

  14. start (y1,y2)=(0,x1) y2>=x2 F T (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) halt Subtle point: The program is partially correct with respect to x1>=0/\x2>=0 and totally correct with respect to x1>=0/\x2>0

  15. Annotating a scheme start A Assign an assertion for each pair of nodes. The assertion expresses the relation between the variable when the program counter is located between these nodes. (y1,y2)=(0,x1) B T F y2>=x2 C D (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E halt

  16. Invariants • Invariants are assertions that hold at each state throughout the execution of the program. • One can attach an assertion to a particular location in the code:e.g., at(B) (B).This is also an invariant; in other locations, at(B) does not hold hence the implication holds. • If there is an assertion attached to each location, (A), (B),  (C), (D), (E), then their disjunction is also an invariant:(A)\/(B)\/ (C)\/(D)\/(E)(since location is always at one of these locations).

  17. Annotating a scheme with invariants start A A): x1>=0 /\ x2>=0 B): x1=y1*x2+y2 /\ y2>=0 C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2 D):x1=y1*x2+y2 /\ y2>=0 /\ y2<x2 E):x1=z1*x2+z2 /\ 0<=z2<x2 Notice: (A) is the initial condition,  Eis the input-output condition. (y1,y2)=(0,x1) B T F y2>=x2 C D (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E A)Is theprecondition of (y1,y2)=(0,x1) and B) is its postcondition halt

  18. Preliminary:Relativizing assertions A (B) : x1= y1 * x2 + y2 /\ y2 >= 0 Relativize B) w.r.t. the assignment, obtaining B) [Y\g(X,Y)] (I.e., (B) expressed w.r.t. variables at A.)  (B)A =x1=0 * x2 + x1 /\ x1>=0 Think about two sets of variables,before={x, y, z, …} after={x’,y’,z’…}. Rewrite (B) using after, and the assignment as a relation between the set of variables. Then eliminate after by substitution. Here: x1’=y1’ * x2’ + y2’ /\ y2’>=0 /\x1’=x1 /\ x2’=x2 /\ y1’=0 /\ y2’=x1now eliminate x1’, x2’, y1’, y2’. Y=g(X,Y) (y1,y2)=(0,x1) B A (y1,y2)=(0,x1) B

  19. Preliminary:Relativizing assertions A (B)A Y=g(X,Y) (y1,y2)=(0,x1) B A): Y=g(X,Y) A (B) (y1,y2)=(0,x1) B

  20. Verification conditions: assignment A A)  B)A whereB)A =B)[Y\g(X,Y)] A): x1>=0 /\ x2>=0 B): x1=y1*x2+y2 /\ y2>=0 B)A= x1=0*x2+x1 /\ x1>=0 Y=g(X,Y) (y1,y2)=(0,x1) B A (y1,y2)=(0,x1) B

  21. Second assignment C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2 B): x1=y1*x2+y2 /\ y2>=0 B)C: x1=(y1+1)*x2+y2-x2 /\ y2-x2>=0 C (y1,y2)=(y1+1,y2-x2) B

  22. Third assignment D):x1=y1*x2+y2 /\ y2>=0 /\ y2<x2 E):x1=z1*x2+z2 /\ 0<=z2<x2 E)D: x1=y1*x2+y2 /\ 0<=y2<x2 D (z1,z2)=(y1,y2) E

  23. Verification conditions: tests B T F B) /\ t(X,Y) C) B) /\¬t(X,Y) D) B): x1=y1*x2+y2 /\y2>=0 C): x1=y1*x2+y2 /\ y2>=0 /\ y2>=x2 D):x1=y1*x2+y2 /\ y2>=0 /\ y2<x2 t(X,Y) C D B T F y2>=x2 C D

  24. C) B) t(X,Y) ¬t(X,Y) Verification conditions: tests B T F t(X,Y) C D B T F y2>=x2 C D

  25. Partial correctness proof:An induction on length of execution Initially, states satisfy the initial conditions. Then, passing from one set of states to another, we preserve the invariants at the appropriate location. We prove: starting with a state satisfying the initial conditions, if are at a point in the execution, the invariant there holds. Not a proof of termination! A) no B) yes start A C) (y1,y2)=(0,x1) B T F no B) y2>=x2 C D yes (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E D) halt

  26. Initial condition: x>=0 Input-output claim: z=x! Exercise: prove partial correctness start (y1,y2)=(0,1) F T y1=x (y1,y2)=(y1+1,(y1+1)*y2) z=y2 halt

  27. What have we achieved? • For each statement S that appears between points X and Y we showed that if the control is in X when (X) holds (the precondition of S) and S is executed, then (Y) (the postcondition of S) holds. • Initially, we know that (A) holds. • The above two conditions can be combined into an induction on the number of statements that were executed: • If after n steps we are at point X, then (X) holds.

  28. (A) : x>=0 (F) : z^2<=x<(z+1)^2 z is the biggest number that is not greater than sqrt x. start A (y1,y2,y3)=(0,0,1) B y2=y2+y3 C false true y2>x D E (y1,y3)=(y1+1,y3+2) z=y1 F halt Another example

  29. 1+3+5+…+(2n+1)=(n+1)^2 y2 accumulates the above sum, until it is bigger than x. y3 ranges over odd numbers 1,3,5,… y1 is n-1. start A (y1,y2,y3)=(0,0,1) B y2=y2+y3 C false true y2>x D E (y1,y3)=(y1+1,y3+2) z=y1 F halt Some insight

  30. It is usually sufficient to have one invariant for every loop (cycle in the program’s graph). We will have (C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 start A (y1,y2,y3)=(0,0,1) B y2=y2+y3 C false true y2>x D E (y1,y3)=(y1+1,y3+2) z=y1 F halt Invariants

  31. By backwards substitution in (C). (C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 (B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\ y3=2*y1+1 start A (y1,y2,y3)=(0,0,1) B y2=y2+y3 C false true y2>x D E (y1,y3)=(y1+1,y3+2) z=y1 F halt Obtaining (B)

  32. (A)=x>=0 (B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\ y3=2*y1+1 (B) relativized is 0^2<=x /\ 0+1=(0+1)^2 /\ 1=2*0+1 Simplified: x>=0 start A (y1,y2,y3)=(0,0,1) B y2=y2+y3 C false true y2>x D E (y1,y3)=(y1+1,y3+2) z=y1 F halt Check assignment condition

  33. By backwards substitution in (B). (B)=y1^2<=x /\ y2+y3=(y1+1)^2 /\ y3=2*y1+1 (D)=(y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1 start A (y1,y2,y3)=(0,0,1) B y2=y2+y3 C false true y2>x D E (y1,y3)=(y1+1,y3+2) z=y1 F halt Obtaining (D)

  34. (C)=y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 (C)/\y2<=x) (D) (D)=(y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1 start A (y1,y2,y3)=(0,0,1) B y2=y2+y3 C false true y2>x D E (y1,y3)=(y1+1,y3+2) z=y1 F halt Checking

  35. y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 /\y2<=x  (y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1 y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 /\y2<=x  (y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1 y1^2<=x /\ y2=(y1+1)^2 /\ y3=2*y1+1 /\y2<=x (y1+1)^2<=x /\ y2+y3+2=(y1+2)^2 /\ y3+2=2*(y1+1)+1

  36. Still needs to: Calculate (E) by substituting backwards from (F). Check that (C)/\y2>x(E) Not finished! start A (y1,y2,y3)=(0,0,1) B y2=y2+y3 C false true y2>x D E (y1,y3)=(y1+1,y3+2) z=y1 F halt

  37. Exercise: prove partial correctness. Initially: x1>0/\x2>0. At termination: z1=gcd(x1,x2). start (y1,y2)=(x1,x2) y1=y2 F T y1>y2 F T z1=y1 y2=y2-y1 y1=y1-y2 halt

  38. Annotation of program with invariants gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1>y2 start gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1<y2 A x1>0 /\ x2>0 (y1,y2)=(x1,x2) gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0 B gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1y2 y1=y2 T F D G y1=gcd(x1,x2) F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H z1=gcd(x1,x2) halt

  39. Part 1 (A)=x1>0 /\ x2>0 start (B)=gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0 A (B)’rel= gcd(x1,x2)=gcd(x1,x2)/\x1>0/\x2>0 (y1,y2)=(x1,x2) (A) (B)’rel B y1=y2 T F D G F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H halt

  40. Part 2a (B)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0 start (D)=gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1y2 A (y1,y2)=(x1,x2) (B)/\¬(y1=y2) (D) B y1=y2 T F D G F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H halt

  41. Part 2b (B)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0 start (G)=y1=gcd(x1,x2) A (B)/\(y1=y2) (G) (y1,y2)=(x1,x2) B y1=y2 T F D G F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H halt

  42. Part 3 (F)=(gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1>y2 start (D)/\(y1>y2) (F) (E)=gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1<y2 A (D)/\¬(y1>y2) (E) (y1,y2)=(x1,x2) B (D)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1y2 F T y1=y2 G D F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H halt

  43. (B)’relE=gcd(y1,y2-y1)=gcd(x1,x2)/\y1>0/\y2-y1>0 Part 4 (F)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1>y2 (B)’relF=gcd(y1-y2,y2)=gcd(x1,x2)/\y1-y2>0/\y2>0 start (E)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0/\y1<y2 A x1>0 /\ x2>0 (y1,y2)=(x1,x2) (B)= gcd(y1,y2)=gcd(x1,x2)/\y1>0/\y2>0 B y1=y2 T F D G F T y1>y2 z1=y1 E F y2=y2-y1 y1=y1-y2 H halt (E) (B)’rel1 (F) (B)’rel2

  44. Annotation of program with invariants start (H)’rel= y1=gcd(x1,x2) A (y1,y2)=(x1,x2) B (G)= y1=gcd(x1,x2) y1=y2 T F D G F T y1>y2 z1=y1 E F (H)= z1=gcd(x1,x2) y2=y2-y1 y1=y1-y2 H halt (G) (H)’rel2

  45. Proving termination

  46. Well-founded sets • Partially ordered set (W,<): • If a<b and b<c then a<c (transitivity). • If a<b then not b<a (asymmetry). • Nota<a (irreflexivity). • Well-founded set (W,<): • Partially ordered. • No infinite decreasing chain a1>a2>a3>…

  47. Examples for well founded sets • Natural numbers with the bigger than relation. • Finite sets with the set inclusion relation. • Strings with the substring relation. • Tuples with alphabetic order: • (a1,b1)>(a2,b2) iff a1>a2 or [a1=a2 and b1>b2]. • (a1,b1,c1)>(a2,b2,c2) iff a1>a2 or [a1=a2 and b1>b2] or [a1=a2 and b1=b2 and c1>c2].

  48. y2 starts as x1. Each time the loop is executed, y2 is decremented. y2 is natural number The loop cannot be entered again when y2<x2. true y2>=x2 C Why does the program terminate start A (y1,y2)=(0,x1) B false D (y1,y2)=(y1+1,y2-x2) (z1,z2)=(y1,y2) E halt

  49. Proving termination • Choose a well-founded set (W,<). • Attach a function u(N) to each point N. • Annotate the flowchart with invariants, and prove their consistency conditions. • Prove that j(N)  (u(N) in W).

  50. Show that u(M)>=u(N)’rel. At least once in each loop, show that u(M)>u(N). How not to stay in a loop? M S N M T N

More Related