Program Analysisrayna/pa1617-lectures/lecture1.pdfProgram analysis Software I has bugs I is often too slow Applications of program analysis I software veri cation I compiler optimization

Program AnalysisLecture 1

Rayna Dimitrova

WS 2016/2017

AdministrativeLectures

I Slides/lecture notes will be posted onhttp://www.mpi-sws.org/~rayna/pa1617.html

I We will follow closely the WS 2015/2016 edition of the course(lecture notes in German)

I Written exam

Exercises

I Time and place of tutorial to be determined

I Exercise sheet will be posted after each lecture onhttp://www.mpi-sws.org/~rayna/pa1617.html

I Solutions are due at the beginning of the next lecture

I You need 60% of the total number of points on the exercisesheets during the semester to qualify for the exam

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http://www.mpi-sws.org/~rayna/pa1617.html

http://www.mpi-sws.org/~rayna/pa1617.html

Program analysis

Software

I has bugs

I is often too slow

Applications of program analysis

I software verification

I compiler optimization

Static program analysisestablish properties of a program without executing it

Theorem (Rice)All non-trivial semantic properties of programs are undecidable.

=⇒ develop algorithms that give conservative answers

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A simple program

x and y are integer variables

r e a d ( x ) ;

i f ( x < 0){y : = 0 ;whi le x < 0 do{

x++;y++;

}}e l s e {

y := x ;}

p r i n t ( y ) ;

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A simple sign lattice

∅

(n, nn) (nn, n)(n, n) (nn, nn)

(n, ?) (?, n) (?, nn) (nn, ?)

(?, ?)

I n : negative value, nn: non-negative value; ?: any value in ZI (n, nn) represents the set {(x, y) ∈ Z× Z | x < 0, y ≥ 0}I an arrow a −→ b denotes the fact that the set represented by

a is a subset of the set represented by bI Note: we have omitted arrows such as

I (n, nn) −→ (n, nn) (self-loop),I (n, nn) −→ (?, ?), induced by the arrows

(n, nn) −→ (n, ?) and (n, ?) −→ (?, ?) 5/18

Total orders

(N,≤) is a total order: it satisfies the following conditions

reflexivity: a ≤ a for all a ∈ Nantisymmetry: a ≤ b and b ≤ a implies a = b

transitivity: a ≤ b and b ≤ c implies b ≤ c

comparability: For any a, b ∈ N, either a ≤ b or b ≤ a

What about (P({1, 2, 3}),⊆)?(where P({1, 2, 3}) is the set of all subsets of {1,2,3})

{1, 2} 6⊆ {1, 3} and {1, 3} 6⊆ {1, 2}{1, 2} and {1, 3} are incomparable

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Total orders

(N,≤) is a total order: it satisfies the following conditions



comparability: For any a, b ∈ N, either a ≤ b or b ≤ a

What about (P({1, 2, 3}),⊆)?(where P({1, 2, 3}) is the set of all subsets of {1,2,3})

{1, 2} 6⊆ {1, 3} and {1, 3} 6⊆ {1, 2}{1, 2} and {1, 3} are incomparable

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Partial orders

A partial order (D,≤) consists of a set D 6= ∅ and arelation ≤ ⊆ D ×D which has the following properties



(P({1, 2, 3}),⊆) is a partial order

partial order + comparability = total order

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A Hasse diagram for (P({1, 2, 3}),⊆)I leave out reflexivity self-loops

I leave out edges induced by transitivity

{1, 2, 3}

{1, 3}{1, 2} {2, 3}

{1} {2} {3}

∅

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Least and greatest elements of a setLet (D,≤) be a partial order and X ⊆ D.

a ∈ X is a minimal element of Xif there does not exist x ∈ X such that x ≤ a and x 6= a.

l ∈ X is the least element of Xif for all x ∈ X it holds that l ≤ x.

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ExampleConsider (P({1, 2, 3}),⊆).The set {{1, 2}, {2, 3}} has two minimal elements.The set {{1, 2}, {2, 3}} does not have a least element.

ExampleConsider (Z∪{#},≤) where ≤ is the less-than-or-equal relation onZ, and # is incomparable with each of the elements of Z. The setZ∪{#} has exactly one minimal element, #, and no least element.

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b ∈ X is a maximal element of Xif there does not exist x ∈ X such that b ≤ x and x 6= b.

g ∈ X is the greatest element of Xif for all x ∈ X it holds that x ≤ g.

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Join

Let (D,≤) be a partial order and let X ⊆ D

u ∈ D is called an upper bound for X if x ≤ u for all x ∈ X

u ∈ D is called a least upper bound for X (also join of X) if

I u is an upper bound for X

I u ≤ u′ for all upper bounds u′ for X

By definition, if it exists, the least upper bound of X is unique.

The least upper bound of X is denoted by⊔X.

Instead of⊔{a, b} we write a t b.

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Join









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Join









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Join









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Meet


l ∈ D is called a lower bound for X if l ≤ x for all x ∈ X

l ∈ D is called a greatest lower bound for X (also meet of X) if

I l is a lower bound for X

I l′ ≤ l for all lower bounds l′ for X

By definition, if it exists, the greatest lower bound of X is unique.

The greatest lower bound of X is denoted bydX.

Instead ofd{a, b} we write a u b.

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Meet









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Meet









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Meet









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LatticesA lattice is a partial order (D,≤) in which for every pair a, b ∈ Dthere exist least upper bound a t b and greatest lower bound a u b.

A lattice is called complete if for every subset X ⊆ D there existthe least upper bound

⊔X and the greatest lower bound

dX.

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dX.

Examples

x

x1 x2

not a lattice:{x1, x2} does nothave a lower bound

a1

a2 a3

a4 a5

a6

not a lattice: {a2, a3} hasupper bounds a4, a5, a6but no least upper bound(a4, a5 not comparable)

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dX.

Example

(N,≤) is a lattice but not a complete lattice:

I a t b = max (a, b), a u b = min (a, b)

I the set N does not have an upper bound

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dX.

Lemma

Every complete lattice (D,≤) has a unique least element (bottom)

⊥ :=⊔∅ =

lD,

Every complete lattice (D,≤) has a unique greatest element (top)

> :=l∅ =

⊔D

Every lattice (D,≤) where D is finite is complete.

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Fixpoints

Let (D,≤) be a partial order and let f : D → D be a function.

A fixpoint of f is an element x ∈ D such that f(x) = x.

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Fixpoints



A function f : D → D is called monotonic if

a ≤ b implies f(a) ≤ f(b).

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Fixpoints



Theorem (Knaster, Tarski’55)Let (D,≤) be a complete lattice andlet f : D → D a monotonic function. Then,

(1) there exists a (unique) least fixpoint of f given by

lfp(f) :=l{x ∈ D | f(x) ≤ x}

(2) there exists a (unique) greatest fixpoint of f given by

gfp(f) :=⊔{x ∈ D | x ≤ f(x)}

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Proof

Let (D,≤) be a complete lattice and let f : D → D be monotonic.We show that l := lfp(f) is the least fixpoint of f , meaning that l is afixpoint of f , and for every fixpoint d of f , we have that l ≤ d.

I We show that f(l) ≤ l for l =dD′, D′ = {x ∈ D | f(x) ≤ x}.

Since l is the greatest lower bound of D′, if we show that f(l) is alower bound of D′, then we have f(l) ≤ l.

l ≤ x for all x ∈ D′(l is a lower bound for D′)

f(l) ≤ f(x) for all x ∈ D′(f is monotonic)

f(l) ≤ x for all x ∈ D′(definition of D′, transitivity)

Thus, f(l) is a lower bound of D′. So, we have f(l) ≤d

D′ = l.

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Proof continued

I We show that l ≤ f(l) for l =dD′, D′ = {x ∈ D | f(x) ≤ x}.

If we show f(l) ∈ D′, then l ≤ f(l), because l is a lower bound for D′.

Above we showed that f(l) ≤ l.

By monotonicity of f , f(f(l)) ≤ f(l).

By definition of D′, f(l) ∈ D′. Thus, l ≤ f(l).

By antisymmetry, from f(l) ≤ l and l ≤ f(l) we get f(l) = l.

Now, suppose that d ∈ D is such that f(d) = d. We show that l ≤ d.By the choice of d and reflexivity, we have f(d) ≤ d. Hence, d ∈ D′.Then, l ≤ d, because l is a lower bound for D′.

From this we conclude that l is the least fixpoint of f .

The proof for gfp(f) is analogous.

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Chains

Let (D,≤) be a partial order.

I A chain in D is a totally ordered non-empty subset C ⊆ D.(For any a, b ∈ C, either a ≤ b or b ≤ a.)

I A sequence (ci)i∈N is called

I ascending chain if ci ≤ ci+1 for all i ∈ N.I descending chain if ci+1 ≤ ci for all i ∈ N.

I A partial order (D,≤) has

I finite height if every chain C in D is finite.I bounded height if there exists h ∈ N such that every chain C

in D has at most h elements.

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t-continuous and u-continuous functions

Let (D,≤) be a complete lattice and let f : D → D be a function.

f is called t-continuous if for every chain C in D,

f(⊔

C) =⊔{f(c) | c ∈ C}.

f is called u-continuous if for every chain C in D,

f(l

C) =l{f(c) | c ∈ C}.

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Ascending and descending chain conditions

A partial order (D,≤)

I satisfies the ascending chain condition (ACC) iffor every ascending chain c0 ≤ c1 ≤ . . .there exists n ∈ N such that for all i ≥ n we have ci = cn.

I satisfies the descending chain condition (DCC) iffor every descending chain c0 ≥ c1 ≥ . . .there exists n ∈ N such that for all i ≥ n we have ci = cn.

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Example(N,≤) is a totally ordered set that satisfies the (DCC) condition.Such totally ordered sets are called well-ordered.

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TheoremLet (D,≤) be a complete lattice andlet f : D → D be a monotonic function. Then

(1) if (D,≤) satisfies (ACC), then f is t-continuous.

(2) if (D,≤) satisfies (DCC), then f is u-continuous.

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Proof

Let (D,≤) be a complete lattice satisfying (ACC) and f : D → D bemonotonic. We show that f is t-continuous. Let C ⊆ D be a chain.

I We show that f(⊔

C) ≤⊔{f(c) | c ∈ C}.

Since⊔{f(c) | c ∈ C} is an upper bound for {f(c) | c ∈ C} it

suffices to show that f(⊔C) ∈ {f(c) | c ∈ C}. We show

⊔C ∈ C.

First, we show that there exists c ∈ C such that c ≤ c for all c ∈ C.

Assume this is not true. Then, for every d ∈ C there exists c ∈ Csuch that d < c (d ≤ c and d 6= c), or c and d are incomparable.Since C is a chain, c and d cannot be incomparable. Thus, we canconstruct a sequence c0 < c1 < . . ., which contradicts (ACC).

⊔C is an upper bound for C =⇒ c ≤

⊔C (since c ∈ C)

c is an upper bound for C =⇒⊔C ≤ c

By antisymmetry we get c =⊔C, and thus,

⊔C ∈ C.

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Proof continued

I We show that⊔{f(c) | c ∈ C} ≤ f(

⊔C).

Since⊔{f(c) | c ∈ C} is the least upper bound for {f(c) | c ∈ C} it

suffices to show that f(⊔C) is an upper bound for {f(c) | c ∈ C}.

c ≤⊔

C for all c ∈ C (⊔

C is an upper bound for C)

f(c) ≤ f(⊔

C) for all c ∈ C (f is monotonic)

Thus, f(⊔C) is an upper bound for {f(c) | c ∈ C}, which

concludes the proof.

From f(⊔

C) ≤⊔{f(c) | c ∈ C} and

⊔{f(c) | c ∈ C} ≤ f(

⊔C)

by antisymmetry we conclude that f(⊔C) =

⊔{f(c) | c ∈ C}.

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Documents

Program Analysisrayna/pa1617-lectures/lecture1.pdfProgram analysis Software I has bugs I is often too slow Applications of program analysis I software veri cation I compiler optimization