C LAUS B RABRAND S EMANTICS (Q1,’05) S EP 8, 2005 C LAUS B RABRAND © 2005, University of Aarhus [...

CLAUS BRABRAND SEMANTICS (Q1,’05) SEP 8, 2005

CLAUS BRABRAND

© 2005, University of Aarhus

[ [email protected] ] [ http://www.daimi.au.dk/~brabrand/ ]

SEMANTICS (Q1,’05)

WEEK 2: ”STRUCTURAL OPERATIONAL SEMANTICS”

CLAUS BRABRAND © SEMANTICS (Q1,’05)[ 2 ]

SEP 8, 2005

Week 2 - Outline

Repetition (week 1) Structural Operational Semantics

Expressions (Exp) Big-step vs. Small-step semantics Side-effects Behavior and Equivalence, …

Boolean Expressions (BExp) Lazy evaluation, …

Commands (Com) …

Induction and Structural Induction


REPETITION

Keywords:

inference systems (relations), transition systems, virtual machine semantics


SEP 8, 2005

Repetition (“3x3 things…”)

Class X: 1x. Transition Systems: determinism vs. non-determinism ? 2x. Inference Systems: premise(s) vs. side-condition(s) ? 3x. SMC Machine: physical memory vs. abstract memory ?

Class Y: 1y. Homepage: program relationship figure (example c) ? 2y. Inference Systems: name all applications ! 3y. Prolog: how much Prolog are we expected to know ?

Class Z: 1z. Functions: total functions vs. partial functions ? 2z. Prolog: did Prolog really answer “no” for nat(-3) ? 3z. Exercise classes: active learning vs. passive learning… !


SEP 8, 2005

Program Relationship (Example c)

Program worldModel world

ConcreteAbstract

~

P

P’

M

M’

1. P ~ P’ ?2. abstract

3. M ~ M’ ?

4. relate

5. M ~ M’ !6. concretize7. P ~ P’ !


SEP 8, 2005

Should be blatantly obvious, but here goes: Correlation:

Investment ~ Benefit Example:

Studying ~ Exam result

Main Entry: os·mo·sis Pronunciation: äz-'mO-s&s, äs-Function: nounEtymology: New Latin, short for endosmosis1 : movement of a solvent through a semipermeable membrane (as of a living cell) […]2 : a process of absorption […] usually effortless often unconscious assimilation <learned a number of languages by osmosis >

Active Learning vs. Passive Learning (aka. Learning-by-Osmosis)

“I am sorry for saying this, but…”


SEP 8, 2005

Learn by Osmosis

It’s Amazing…

Do Active Learning: Study = read + make exercises + reflect

(for your own sake)!

“Learn while you sleep!”

“The Semantics Pillow”

Only$19,95


ARITHMETIC EXPRESSIONS

Keywords:

structural operational semantics (sos) big-step & small-step semantics, side-effects


SEP 8, 2005

VM Semantics: Major Drawbacks!

Advantage: Easy to implement (and efficient)

Drawbacks:

Non-intuitive Too concrete (e.g., stack) Indirect semantics (not syntax directed) Computational step?

“High-level language understood in terms of low-level machine code”

“Many other machine along these lines […]. They all have a tendency to pull the syntax into pieces or at any rate to wander around the syntax creating various complex symbolic structures which do not seem particularly forced by the demands of the language itself”

- Gordon Plotkin, ‘81


SEP 8, 2005

Given program: and memory:

VM: Processing of Additions

< , M, ((1 + (2 + 3)) + 4) > < , M, (1 + (2 + 3)) 4 + >

< , M, (1 + (2 + 3)) 4 + > < , M, 1 (2 + 3) + 4 + >

< , M, 1 (2 + 3) + 4 + > < 1, M, (2 + 3) + 4 + >

< 1, M, (2 + 3) + 4 + > < 1, M, 2 3 + + 4 + >

< 1, M, 2 3 + + 4 + > < 2 1, M, 3 + + 4 + >

< 2 1, M, 3 + + 4 + > < 3 2 1, M, + + 4 + >

< 3 2 1, M, + + 4 + > < 5 1, M, + 4 + >

< 5 1, M, + 4 + > < 6, M, 4 + >

< 6, M, 4 + > < 4 6, M, + >

< 4 6, M, + > < 10, M, >

((1 + (2 + 3)) + 4) M

rearrange

rearrange

addition1!

addition2!

rearrange

addition3!

rearrange

rearrange

rearrange

rearrange

12 3

Note: Only three transitions are of real interest as “system events”


SEP 8, 2005

Ideally we would like:

…i.e., the following transition sequence:

Ideal Processing of Additions

((1 + (2 + 3)) + 4) ((1 + 5) + 4)

explanation

((1 + 5) + 4)

explanation

(6 + 4)

(6 + 4)

explanation

10

((1 + (2 + 3)) + 4)

explanation

((1 + 5) + 4)

explanation

(6 + 4)

explanation

10

(aka. derivation sequence)(aka. reduction sequence)


SEP 8, 2005

Informal vs. Formal Specification

Informally: “One evaluates from left to right”

Description (pseudo-formally): CONSTANTS:

Any constant, n, is already evaluated (with itself as value)

SUMS: Step 1. Evaluate to obtain result ; Step 2. Evaluate to obtain result ; Step 3. Add and to obtain final result .

n

e0 + e1

e0 n0

e1 n1

n0 n1 m


SEP 8, 2005

Inference System Semantics

Inference System Semantics: Abbreviate as

Meaning: “e terminates and evaluates to v”

e v

‘L’ Exp N

(e,v) ‘L’

Q: Did I just solve the halting problem here?!?

A: No; nobody can decide “c v” automatically!

A: Actually, for Exp termination is decidable…


SEP 8, 2005

Inference System Sem. (cont’d)

Inference System Semantics:

e0 + e1 mm = n0 + n1

e0 n0 e1 n1[SUM][CON] n n

‘L’ Exp N

Syntactic ‘+’Semantic ‘+’


SEP 8, 2005

Example Revisited (big-step)

Inference Tree:

((1 + (2 + 3)) + 4) 10

(1 + (2 + 3)) 6 4 4

1 1 (2 + 3) 5

2 2 3 3

e0 + e1 mm = n0 + n1

e0 n0 e1 n1[SUM][CON] n n

[SUM]

[CON]

[CON]

[SUM]

[SUM]

[CON] [CON]“big-step semantics”

“small-step semantics”vs.


SEP 8, 2005

Small-step Description

Description:

CONSTANTS: Any constant, n, is already evaluated (with itself as value)

SUMS: Step 1. If is not a constant,

evaluate it (one step) to obtain result ; Step 2. If is a constant, but is not,

evaluate it (one step) to obtain result ; Step 3. If and are both constants,

add them to obtain result, say .

n

e0 + e1

e0

e0’

e0 e1

e1’

e0 e1

m


SEP 8, 2005

Small-step Formalization

Transition System Semantics: Configurations: Final Configurations: Transition Relation:

Abbreviate as Meaning: “e evaluates to e’ in one step”

L := Exp

TL := N Exp

L Exp Exp

e e’(e,e’) ‘L’


SEP 8, 2005

Small-step Semantics

Transition System Semantics:

L := Exp

TL := N Exp

L Exp Exp

[SUM1]

e0 + e1 e0’ + e1

e0 e0’

[SUM2]

n0 + e1 n0 + e1’

e1 e1’

[SUM3]

n0 + n1 mm = n0 + n1

We call this a STRUCTURAL OPERATIONAL SEMANTICS


SEP 8, 2005

Example Revisited (small-step)

Transition sequence (with explanation):

((1 + (2 + 3)) + 4) ((1 + 5) + 4)

(1 + (2 + 3)) (1 + 5)

(2 + 3) 5

[SUM1]

[SUM2]

[SUM3]

“order of discovery”

find predicates make conclusions


SEP 8, 2005

Transition Sequence Information

A transition sequence

…specifies two things:

1. A “sequence of steps” themselves (here additions)

2. …and reasons why they should be performed

e e’ e’’ … m

((1 + (2 + 3)) + 4)

explanation

((1 + 5) + 4)


SEP 8, 2005

Adding Variables

Language L’: Arithmetic Expressions (e Exp)

Structural Operational Semantics: Configurations:

Final Configurations:

Transition Relation…

e ::= n | v | e0 + e1

L’ := Exp Store

TL’ := L’

Note the change in terminology: Store <= Memory <= MThe term Store is more generally accepted (as an abstraction of a computer’s physical memory)

Store = Var N (= Memory)where

A configuration now looks like: <e,><e,> L’ <e’,>


SEP 8, 2005

Small-step Semantics w/ Stores

Structural Operational Semantics (w/ Stores):

[SUM1]< e0 + e1 , > < e0’ + e1 , ’ >

< e0 , > < e0’ , ’ >

[SUM2]

[SUM3]< n0 + n1 , > < m , >

m = n0 + n1

< n0 + e1 , > < n0 + e1’ , ’ >< e1 , > < e1’ , ’ >

< v , > < m , >[VAR]

Store = Var Nm = (v)


SEP 8, 2005

Note: No Side-Effects!

No side-effects!

[SUM1]< e0 + e1 , > < e0’ + e1 , ’ >

< e0 , > < e0’ , ’ >

[SUM2]

[SUM3]< n0 + n1 , > < m , >

m = n0 + n1

< n0 + e1 , > < n0 + e1’ , ’ >< e1 , > < e1’ , ’ >

< v , > < m , >[VAR]

Store = Var Nm = (v)

e,: <e,> <e’,’> => = ’Easily proved by structural induction [later…]


SEP 8, 2005

Exp: Behavior and Equivalence

Definitions: Behavior:

Note: eval is only a partial functionif we have division (div-by-zero)

Equivalence:

Examples:

eval(e,) = m <=> <e,> E* <m,>

e e’ <=> : eval(e,) = eval(e’,)

1+1 2 x+y y+x

Can be proved by structural induction [soon…]

z+z 2*z


BOOLEAN EXPRESSIONS

Keywords:

eager evaluation, lazy evaluation,

parallel evaluation


SEP 8, 2005

Boolean Expressions BExp

Language B: Boolean Expressions (b BExp):


where

Final Configurations:

Transition Relation… as short-hand for

b ::= t | e = e’ | b or b’ | ~ b

B := BExp Env

TB := B

Env = Var N

|_ b B b’

We could also have used a Store (we actually have to if BExp had had side-effects)

(b,,b’,) B


SEP 8, 2005

Boolean Expressions BExp (cont’d)

Structural Operational Semantics:

Note that: The (this) boolean expression transition system...:

…uses the arithmetic expression transition system…:

...and does it transitively (E*) in “one step”!

|_ e0 = e1 B t

[EQ]B |_ e0 E

* n0 |_ e1 E* n1

B , TB , B

E , TE , E

t = tt, n0 = n1

ff, n0 n1

See [Plotkin, p. 31] for alternative semantics


SEP 8, 2005

Disjunction: Lazy Evaluation

Lazy Semantics for Disjunction (“or”):

|_ b0 or b1 L b0’ or b1

|_ b0 L b0’[OR1]L

[OR2]L

|_ ff or b1 L b1

[OR3]L

b: tt or b tt |_ tt or b1 L ttExploit:


SEP 8, 2005

Eager vs. Lazy Semantics for “or”

Eager Semantics: ‘B’ Lazy Semantics: ‘L’

Relationship: ? Stuck configurations? …e.g. div-by-zero? Assignment? Non-termination?

b := tt or (3 = 1-2)

b,: |_ b B* t <=> |_ b L

* t

b := tt or (1 = 2/0)

b := tt or (x := tt)“side-effects”

b := tt or loop()


SEP 8, 2005

For the language, BExp, we have that:

Whereas:

Eager vs. Lazy Semantics (cont’d)

b,: <b,> B* <t,’> ==> <b,> L

* <t,”>

b,: <b,> B* <t,’> <== <b,> L

* <t,”>


Easily disproved by a counterexample (e.g. )

Relationship: ? Stuck configurations? …e.g. div-by-zero? Assignment? Non-termination?

b := tt or (3 = 1-2)

b,: |_ b B* t <=> |_ b L

* t

b := tt or (1 = 2/0)

b := tt or (x := tt)“side-effects”

b := tt or loop()

b := tt or (3 = 1-2)


SEP 8, 2005

Parallel Evaluation of Disjunction

Idea: Evaluate non-deterministically (via. interleaving or in

parallel) left / right operands to or:

Live exercise… :) [Think 3 mins; then interactively on the blackboard]

‘P’


SEP 8, 2005

Negation: “~b”

Boolean Expressions (b BExp):


b ::= t | e = e’ | b or b’ | ~ b


SEP 8, 2005

BExp: Behavior and Equivalence


Note: eval is only a partial functionif we have division (div-by-zero)

Equivalence:

Examples:

eval(b,) = t <=> <b,> B* t

b b’ <=> : eval(b,) = eval(b’,)

b ~~b 1 = 2 ff


x = x tt


COMMANDS

Keywords:

Assignments, Conditionals, Loops,

Control flow


SEP 8, 2005

SOS of Commands

Commands (c Com):


where

Final Configurations: (?)

Transition Relation… for

c ::= nil | v := e | c ; c’ | if b then c else c’ | while b do c

C := (Com Store)

TC := C

Store = Var N

<c,> C <c’,’>

Here we need a Store (since we have to model side-effects)

((c,),(c’,’)) C

Problem: how do we end computation?!?


SEP 8, 2005

SOS of Commands (cont’d)

Commands (c Com):


where

Final Configurations: (!)

Transition Relation… for

c ::= nil | v := e | c ; c’ | if b then c else c’ | while b do c

C := (Com Store) Store

TC := Store

Store = Var N

<c,> C <c’,’> ((c,),(c’,’)) C

<c,> C ’ ((c,),’) C(!)forand

Here we need a Store (since we have to model side-effects)


SEP 8, 2005

Nil and Assignment (SOS)

SOS for nil:

SOS for assignment:

v := e

< v := e , > C ’

< e , > A* < m , >

[ASS]C

’ = [m/v]

< nil , > C [NIL]C


SEP 8, 2005

Sequential Composition (SOS)

SOS for Sequential Composition:

Diagramatically: “(Control) flow-charts”

c0 ; c1

< c0 ; c1 , > C < c0’ ; c1 , ’ >

< c0 , > C < c0’ , ’ >[SEQ1]C

< c0 ; c1 , > C < c1 , ’ >

< c0 , > C ’[SEQ2]C

c c’


SEP 8, 2005

If-then-else (SOS)

SOS for “if-then-else”:


if b then c else c’

< if b then c else c’ , > C < c , >

< b , > B* < tt , >[IF1]C

< if b then c else c’ , > C < c’ , >

< b , > B* < ff , >[IF2]C

c

c’

b

tt

ff


SEP 8, 2005

While-do (SOS)

SOS for “while-do”:


while b do c

< while b do c, > C < c ; while b do c , >

< b , > B* < tt , >[WH1]C

< while b do c , > C

< b , > B* < ff , >[WH2]C

tt

ff

cb


SEP 8, 2005

Com: Behavior and Equivalence


Note: exec is only a partial function(e.g. non-termination, div-by-zero, …)

Equivalence:

Examples:

exec(c,) = ’ <=> <c,> C* ’

c c’ <=> : exec(c,) = exec(c’,)

if b then c else c’ if ~b then c’ else c

while ff do c nil

Can be proved by structural induction [v. soon…]


STRUCTURAL INDUCTION

Keywords:

Induction, base case, induction hypothesis, induction step structural induction


SEP 8, 2005

Principle of Mathematical Induction

Let P be a predicate (i.e. a boolean function): then we have that:

Intuitive: ?

P: N { true, false }

nN : P(n)

P(0)

induction stepbase case

Principle of mathematical induction:

P(n) P(n+1)

P(3)

P(0) P(0) => P(1) P(1) => P(2) P(2) => P(3)


SEP 8, 2005

Example Induction Proof

Example: Base case (i.e. prove P(0)):

Induction step (i.e. prove P(n) => P(n+1)): Assume the induction hypothesis (I.H.)

(i.e. assume P(n)):

Now prove P(n+1):

P(n) [ 20 + 21 + … + 2n = 2n+1 – 1 ]

P(0) [ 20 = 20+1 – 1 ]

[ 20 + 21 + … + 2n = 2n+1 – 1 ]

[ 20 + 21 + … + 2n+1 = 2(n+1)+1 – 1 ]

20 + 21 + … + 2n + 2n+1 (20 + 21 + … + 2n) + 2n+1=

(2n+1 – 1) + 2n+1 == 2*2n+1 – 1 = 2(n+1)+1 – 1 I.H.


SEP 8, 2005

Structural Induction (for Exp)

Given: Arithmetic Expressions (e Exp)

e ::= n | v | e0+e1

e Exp : P(e)

P(n)

composite (inductive) casebase cases

Principle of structural induction:

P(e0) P(e1) P(e0+e1) P(v) and


SEP 8, 2005

Structural Induction (for BExp)

Boolean Expressions (b BExp):


b ::= t | e = e’ | b or b’ | ~ b


SEP 8, 2005

More on this next week…

More on Structural Induction next week…


Next week: type checking, defn’s and decl’s

Any Questions?

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