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Axiomatic Semantics. Hoare’s Correctness Triplets Dijkstra’s Predicate Transformer s. gcd -lcm algorithm w/ invariant. {PRE: (x = n) and (y = m)} u := x; v := y; while {INV: 2*m*n = x*v + y*u} (x y) do if x > y then x := x - y; u := u + v else y := y - x; v := v + u fi - PowerPoint PPT Presentation
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Hoare’s Correctness TripletsDijkstra’s Predicate Transformers
Axiomatic Semantics
gcd-lcm algorithm w/ invariant
{PRE: (x = n) and (y = m)}u := x; v := y;while {INV: 2*m*n = x*v + y*u} (x <> y) do if x > y then
x := x - y; u := u + v else
y := y - x; v := v + u fi
od{POST:(x = gcd(m, n)) and (lcm(m, n) = (u+v) div 2)}
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Goal of a program = IO Relation• Problem Specification
– Properties satisfied by the input and expected of the output (usually described using “assertions”).
– E.g., Sorting problem• Input : Sequence of numbers• Output: Permutation of input that is ordered.
• View Point– All other properties are ignored.
• Timing behavior• Resource consumption• …
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ax·i·om
• n.1. A self-evident or universally recognized truth; a maxim
• 2. An established rule, principle, or law.• 3. A self-evident principle or one that is
accepted as true without proof as the basis for argument; a postulate.
• From a dictionary
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Axiomatic Semantics
• Capture the semantics of the elements of the PL as axioms
• Capture the semantics of composition as a rule of inference.
• Apply the standard rules/logic of inference.• Consider termination separately.
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States and Assertions• States: Variables mapped to Values
– Includes all variables– Files etc. are considered “global” variables.– No notion of value-undefined variables– At a given moment in execution
• An assertion is a logic formula involving program variables, arithmetic/boolean operations, etc.
• All assertions are attached to a control point.• Assertions: States mapped to Boolean
– Boolean connectives: and, or, not, implies, …– For-all, There-exists– Special predicates defined just for use in assertions (not for use in the
program).
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Hoare’s Logic• Hoare Triplets: {P} S {Q}
– P, pre-condition assertion; S, statements of a PL; Q, post-condition assertion
– If S begins executing in a state satisfying P, upon completion of S, the resulting state satisfies Q.
– {P} S {Q} has no relevance if S is begun otherwise.• A Hoare triplet is either true or false. Never undefined.• The entire {P}S{Q} is considered true if the resulting
state satisfies Q if and when S terminates.• If not, the entire {P}S{Q} is false.
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Hoare Triplet Examples• true triplets
– {x = 11 } x := 0 { x = 0 }• we can give a weaker precondition
– {x = 0 } x := x + 1 { x = 1 }– {y = 0} if x <> y then x:= y fi { x = 0 }– {false } x := 0 { x = 111 }
• correct because “we cannot begin”– no state satisfies false
• post condition can be any thing you dream– {true} while true do od {x = 0}
• true is the weakest of all predicates• correct because control never reaches post
– {false} while true do od {x = 0}• false is the strongest of all predicates
• false triplet– {true} if x < 0 then x:= -x fi { x > 0 }
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Weaker/Stronger
• An assertion R is said to be weaker than assertion P if – the truth of P implies the truth of R
• written: P→R– equivalently
• not P or R.• For arbitrary A, B we have: A and B → B
– This general idea is from Propositional Calculus• n > 0 is of course weaker than n = 1, but this follows
from Number Theory.
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Weaker/Stronger
P’
StatesP’
P
States
Q’
Q
P’ weakerP P’
Q’ strongerQ’ Q
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Partial vs Total Correctness
• Are P and S such that termination is guaranteed?
• S is partially correct for P and Q iff whenever the execution terminates, the resulting state satisfies Q.
• S is totally correct for P and Q iff the execution is guaranteed to terminate, and the resulting state satisfies Q.
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Hoare Triplet Examples
1. Totally correct (hence, partially correct)a. {x = 11} x := 0 {x = 0}b. {x = 0} x := x + 1 {x = 1}c. {y = 0}if x <> y then x:= y fi {x = 0}d. {false} while true do S od {x = 0}e. {false} x := 0 {x = 111}
2. Not totally correct, but partially correct– {true} while true do S od {x = 0}
3. Not partially correct– {true} if x < 0 then x:= -x fi {x > 0}
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Assignment axiom
• {Q(e)} x := e {Q(x)}• Q(x) has free occurrences of x.• Q(e): every free x in Q replaced with e• Assumption: e has no side effects.• Caveats
– If x is not a “whole” variable (e.g., a[2]), we need to work harder.
– PL is assumed to not facilitate aliasing.cs784(tk/pm) 13
Inference Rules
• Rules are written as Hypotheses: H1, H2, H3
------------------------------Conclusion: C1
• Can also be stated as:– H1 and H2 and H3 implies C1– Given H1, H2, and H3, we can conclude C1.
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Soundness and Completeness• Soundness is about “validity”• Completeness is about “deducibililty”• Ideally in a formal system, we should have both.• Godel’s Incompleteness Theorem:
– Cannot have both• Inference Rules ought to be sound
– What we proved/ inferred/ deduced is valid• Examples of Unsound Rules
– A and B and C not B– x > y implies x > y+1 (in the context of numbers)
• All the rules we present from now on are sound
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Rule of Consequence
• Suppose: {P’} S {Q’}, P=>P’, Q’=>Q• Conclude: {P} S {Q}• Replace
– precondition by a stronger one– postcondition by a weaker one
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Statement Composition Rule
{P} S1 {R}, {R} S2 {Q}------------------------------{P} S1;S2 {Q}
Using Rule of Consequence {P} S1 {R1}, R1 R2, {R2} S2 {Q}
-----------------------------{P} S1;S2 {Q}
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if-then-else-fi Hoare’s Triplets
{P and B} S1 {Q}{P and not B} S2 {Q}-------------------------------------{P} if B then S1 else S2 fi {Q}• We assumed that B is side-effect free
– Execution of B does not alter state
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Invariants
• An invariant is an assertion whose truth-value does not change
• Recall: All assertions are attached to a control point.
• An Example: x > y– The values of x and y may or may not change each
time control reaches that point.– But suppose the > relationship remains true.– Then x > y is an invariant
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Loop Invariants
• Semantics of while-loop{I and B} S {I}-------------------------------------------{I} while B do S od {I and not B}
• Termination of while-loop is not included in the above.
• We assumed that B is side-effect free.cs784(tk/pm) 20
Data Invariants
• Well-defined OOP classes• Public methods ought to have a pre- and
post-conditions defined• There is a common portion across all public
methods• That common portion is known as the
data invariant of the class.
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while-loop: Hoare’s Approach • Wish to prove: {P} while B do S od {Q}
– Given: P, B, S and Q– Not given: a loop invariant I
• The I is expected to be true just before testing B• To prove {P} while B do S od {Q}, discover a strong enough
loop invariant I so that1. P => I2. {I and B} S {I}3. I and not B => Q
• We used the Rule of Consequence twicecs784(tk/pm)
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A while-loop example
{ n > 0 and x = 1 and y = 1}while (n > y) do
y := y + 1; x := x*y
od{x = n!}
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while-loop: Choose the Invariant
• Invariant I should be such that– I and not B Q– I and not (n > y) (x = n!)
• Choose (n ≥ y and x = y!) as our I• Precondition Invariant
– n > 0 and x=1 and y=1 n ≥ 1 and 1=1!
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while-loop: Verify Invariant
• I === n ≥ y and x = y!• Verify: {I and n > y} y:= y + 1; x:=x*y {I}
1. {I and n > y} y:= y + 1 {n ≥ y and x*y = y!}2. {I and n > y} y:= y + 1 {n ≥ y and x= (y-1)!}3. (I and n > y) (n ≥ y+1 and x= (y+1-1)!)4. (I and n > y) (n > y and x= y!)5. (n ≥ y and x = y! and n > y) (n > y and x= y!)6. QED
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while-loop: I and not B Q
• I === n ≥ y and x = y!• n ≥ y and x = y! and not (n > y) x = n! • n = y and x = y! x = n!• QED
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while-loop: Termination
• Termination is not part of Hoare’s Triplets• General technique:
– Find a quantity that decreases in every iteration.– And, has a lower bound– The quantity may or may not be computed by the
algorithm• For our example: Consider n – y
– values of y: 1, 2, …, n-1, n– values of n - y: n-1, n-2, …, 1, 0
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Weakest Preconditions• We want to determine minimally what must be true immediately
before executing S so that– assertion Q is true after S terminates.– S is guaranteed to terminate
• The Weakest-Precondition of S is a mathematical function mapping any post condition Q to the "weakest" precondition Pw. – Pw is a condition on the initial state ensuring that execution of S
terminates in a final state satisfying R.– Among all such conditions Pw is the weakest– wp(S, Q) = Pw
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Dijkstra’s wp(S, Q)• Let Pw = wp(S, Q)• Def of wp(S, Q): Weakest precondition such that if S is
started in a state satisfying Pw, S is guaranteed to terminate and Q holds in the resulting state.– Consider all predicates Pi so that {Pi}S{Q}.– Discard any Pi that does not guarantee termination of S.– Among the Pi remaining, choose the weakest. This is Pw.
• {P} S {Q} versus P => wp(S, Q)– {Pw} S {Q} is true.– But, the semantics of {Pw} S {Q} does not include termination.– If P => wp(S, Q) then {P}S{Q} also, and furthermore S terminates.
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Properties of wp
• Law of the Excluded Miraclewp(S, false) = false
• Distributivity of Conjunctionwp(S, P and Q) = wp(S,P) and wp(S,Q)
• Law of Monotonicity(Q→R) → (wp(S,Q)→wp(S,R))
• Distributivity of Disjunctionwp(S,P) or wp(S, Q) → wp(S,P or Q)
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Predicate Transformers
• Assignmentwp(x := e, Q(x)) = Q(e)
• Compositionwp(S1;S2, Q) = wp(S1, wp(S2,Q))
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A Correctness Proof
• {x=0 and y=0} x:=x+1;y:=y+1 {x = y}• wp(x:=x+1;y:=y+1, x = y)• wp(x:=x+1, wp(y:=y+1, x = y))
=== wp(x:=x+1, x = y+1)=== x+1 = y+1=== x = y
• x = 0 and y = 0 x = y
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if-then-else-fi in Dijkstra’s wp
wp(if B then S1 else S2 fi, Q) ===(B wp(S1,Q)) and(not B wp(S2,Q)) ===(B and wp(S1,Q))or(not B and wp(S2,Q))
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wp-semantics of while-loops
• DO == while B do S od• IF == if B then S fi• Let k stand for the number of iterations of S• Clearly, k >= 0• If k > 0, while B do S od is the same as:
– if B then S fi; while B do S od
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while-loop: wp Approach
• wp(DO, Q)= P0 or there-exists k > 0: Pk
• States satisfying Pi cause i-iterations of while-loop before halting in a state in Q.
• Pi defined inductively– P0 = not B and Q– …
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wp(DO, Q)
• There exists a k, k ≥ 0, such that H(k, Q)• H is defined as follows• H(0, Q) = not B and Q• H(k, Q) = H(0, Q) or wp(IF, H(k-1, Q))
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Example (same as before)
{ n>0 and x=1 and y=1}while (n > y) do
y := y + 1; x := x*y
od{x = n!}
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Example: while-loop correctness
Pre === n>0 and x=1 and y=1P0 === y >= n and x = n!Pk === B and wp(S, Pk-1)P1 === y > n and y+1>=n and x*(y+1) = n!Pk === y=n-k and x=(n-k)!
Weakest Precondition:W === there exists k >= 0 such that P0 or Pk
Verification :For k = n-1: Pre => W
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Induction Proof• Hypothesis : Pk = (y=n-k and x=(n-k)!)• Pk+1 = B and wp(S,Pk)• = y<n and (y+1 = n-k) and (x*(y+1)=(n-k)!) • = y<n and (y = n-k-1) and (x = (n-k-1)!) • = y<n and (y = n- k+1) and (x = (n- k+1)!) • = (y = n - k+1) and (x = (n - k+1)!)
• Examples of Valid preconditions:– { n = 4 and y = 2 and x = 2 } (k = 2)– { n = 5 and x = 5! and y = 6} (no iteration)
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Detailed Work
wp(y:=y+1;x:=x*y, x=y!and n>=y)= wp(y:=y+1, x*y=y! and n>=y)= wp(y:=y+1, x=(y-1)! and n>=y)= x=(y+1-1)! and n>=y+1)= x=y! and n>y
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gcd-lcm algorithm w/ invariant
{PRE: (x = n) and (y = m)}u := x; v := y;while {INV: 2*m*n = x*v + y*u} (x <> y) do if x > y then
x := x - y; u := u + v else
y := y - x; v := v + u fi
od{POST:(x = gcd(m, n)) and (lcm(m, n) = (u+v) div 2)}
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gcd-lcm algorithm proof
• PRE implies Loop Invariant– (x = n) and (y = m) implies 2*m*n = x*v + y*u
• {Invariant and B} Loop-Body {Invariant} {2*n*m = x*v + y*u and x <> y}
loop-body{2*n*m = x*v + y*u}
• Invariant and not B implies POST 2*n*m = x*v + y*u and x == y
implies(x = gcd(n,m)) and (lcm(n,m) = (u+v) div 2)
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gcd-lcm algorithm proof
• Invariant and not B implies POST 2*m*n = x*v + y*u and x == y
implies(x = gcd(m, n)) and (lcm(m, n) = (u+v) div 2)
• Simplifying 2*m*n = x*(u + v) and x == y
implies(x = gcd(m, n)) and (lcm(m, n) = (u+v) div 2)
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gcd-lcm algorithm proof
• Simplifying 2*m*n = x*(u + v) and x == y
implies(x = gcd(m, n)) and (x*lcm(m, n) = m*n)
• Simplifying 2*m*n = x*(u + v) and x == y
implies(x = gcd(m, n)) and (x*lcm(m, n) = m*n)
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Some Properties of lcm-gcd
• gcd() and lcm() are symmetric– gcd(m, n) = gcd(n, m)– lcm(m, n) = lcm(n, m)
• gcd(m, n) = gcd(m + k*n, n)– where k is a natural number.
• gcd(m, n) * lcm(m, n) = m * n
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