1. C LAUS B RABRAND © S EMANTICS (Q1,’05) S EP 1, 2005 C LAUS B RABRAND © 2005, University of Aarhus [ brabrand@daimi.au.dk ] [ http://www.daimi.au.dk/~brabrand/ ] S EMANTICS (Q1,’05) W EEK 1

2. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 2 ] S EP 1, 2005 Week 1 - Outline Introduction Welcome Course Presentation [ homepage ]homepage Prerequisitional (discrete) Mathematics Relations Inference Systems Transition Systems The Language “L” Virtual Machine Semantics

3. C LAUS B RABRAND © S EMANTICS (Q1,’05) S EP 1, 2005 I NFERENCE S YSTEMS Keywords: relations, axioms, rules, (co-)inductive interpretation, fixed-points

4. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 4 ] S EP 1, 2005 Relations Example 1 : “even” relation: Written as: as a short-hand for: … and as: as a short-hand for: Example 2 : “equals” relation: Written as: as a short-hand for: … and as: as a short-hand for: Example 3 : “DFA transition” relation: Written as: as a short-hand for: … and as: as a short-hand for: | _ even  Z | _ even 4 | _ even 5 4  | _ even 5  | _ even 2  3 (2,3)  ‘=’ ‘=’  Z  Z (2,2)  ‘=’ 2 = 2 ‘  ’  Q    Q q  q’  (q, , q’)  ‘  ’ (p, , p’)  ‘  ’p  p’ 

5. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 5 ] S EP 1, 2005 Inference System Inference System: Inductive (recursive) specification of relations Consists of axioms and rules Example: Axiom: “0 (zero) is even”! Rule: “If n is even, then m is even (where m = n+2)” | _ even 0 | _ even n | _ even m m = n+2 | _ even  Z

6. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 6 ] S EP 1, 2005 Terminology Meaning: “If n is even, then m is even (provided m = n+2)”; or “m is even, if n is even (provided m = n+2)” | _ even n | _ even m m = n+2 premise(s) conclusion side-condition(s)

7. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 7 ] S EP 1, 2005 Abbreviation Often, rules are abbreviated: Rule: “If n is even, then m is even (provided m = n+2)”; or “m is even, if n is even (provided m = n+2)” Abbreviated rule: “If n is even, then n+2 is even”; or “n+2 is even, if n is even” | _ even n | _ even n+2 | _ even n | _ even m m = n+2

8. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 8 ] S EP 1, 2005 Even Example (cont’d) Axiom: “0 (zero) is even”! Rule: “If n is even, then n+2 is even” Is 6 even?!? The inference tree proves that: | _ even 0 | _ even 2 | _ even 4 | _ even 6 | _ even 0 | _ even n | _ even n+2 [rule 1 ] [axiom 1 ] inference tree | _ even 6

9. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 9 ] S EP 1, 2005 Relation Definition (Interpretation) Actually, an inference system: …is a demand specification for a relation: The three relations: R = {0, 2, 4, 6, …}(aka., 2N) R’ = {0, 2, 4, 5, 6, 7, 8, …} R’’ = {…, -2, -1, 0, 1, 2, …}(aka., Z) …all satisfy the (above) specification! | _ even 0 | _ even n | _ even n+2 [rule 1 ] [axiom 1 ] | _ even  Z (0  ‘| _ even ’)  (  n  ‘| _ even ’  n+2  ‘| _ even ’)

10. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 10 ] S EP 1, 2005 Inductive Interpretation A relation: …induces a function: Definition: ‘lfp’ (least fixed point) ~ least solution: | _ even 0 | _ even n | _ even n+2 [rule 1 ] [axiom 1 ] F: P(Z)  P(Z) | _ even  Z F(R) = {0}  { n+2 | n  R } F(Ø) = {0}F 2 (Ø) = F({0}) = {0,2}F 3 (Ø) = F 2 ({0}) = F({0,2}) = {0,2,4} … | _ even := lfp(F) =  F n (Ø) n | _ even  P(Z) From rel. to rel. 2N   = F n (Ø) ~ “Anything that can be proved in ‘n’ steps” 

11. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 11 ] S EP 1, 2005 Co-inductive Interpretation (  ) A relation:  …induces a function: Definition: ‘gfp’ (greatest fixed point) ~ greatest solution: | _ even 0 | _ even n | _ even n+2 [rule 1 ] [axiom 1 ] F: P(Z)  P(Z) | _ even  Z F(R) = {0}  { n+2 | n  R } F(Z) = Z | _ even := gfp(F) =  F n (Z) n | _ even  P(Z) From rel. to rel. F 2 (Z) = F(Z) = Z  F 3 (Z) = F 2 (Z) = F(Z) = Z  … Z  = F n (Z) ~ “Anything that cannot be disproved in ‘n’ steps”

12. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 12 ] S EP 1, 2005 Example: “less-than-or-equal-to” Relation: Is ”1  2” ?!? Yes, because there exists an inference tree: In fact, it has two inference trees: 0  0 n  m n  m+1 [rule 1 ] [axiom 1 ] ‘  ’  N  N n  m n+1  m+1 [rule 2 ] 0  0 0  1 1  2 [rule 2 ] [rule 1 ] [axiom 1 ] 0  0 1  1 1  2 [rule 1 ] [rule 2 ] [axiom 1 ]

13. C LAUS B RABRAND © S EMANTICS (Q1,’05) S EP 1, 2005 T RANSITION S YSTEMS Keywords: Configuration, Final Configuration, Transition, Action / label

14. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 14 ] S EP 1, 2005 Definition: Transition System A Transition System is (just) a structure:  is a set of configurations      is a binary relation (called the transition relation) We will write instead of Other times we use the following notation:  ,     ’  ’( ,  ’)     ’  ’  |_ ’ |_ ’   ’  ’ …

15. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 15 ] S EP 1, 2005 The transition may be illustrated as: We will (often) be using: For instance: A Transition   ’  ’ ’’  = system configuration = 

16. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 16 ] S EP 1, 2005 Def: Terminal Transition System A Terminal Transition System is a structure:  is the set of configurations      is the transition relation T   is a set of final configurations …satisfying: i.e. “all configurations in ‘T’ really are terminal”.  , , T     T :   ’   :    ’

17. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 17 ] S EP 1, 2005 Example: Finite Automaton A Finite Automaton (FA) is a quintuplet: Q finite set of states  finite set of input symbols q 0  Qinitial state F  Qset of acceptance states  : Q    P(Q)state transition relation M =  Q, , q 0, F,   Q = { 0,1,2 }  = {  = { toss,heads,tails } (0,toss)  {1,2}, q 0 = 0 (1,heads)  {0}, F = { 0 } (2,tails)  {0} }

18. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 18 ] S EP 1, 2005 Example: FA Transition System (1) Define (FA) Trans. Sys. by: Configurations: Transition relation: i.e., we have whenever Final Configurations:  M := Q   *   M,  M, T M  T M := { | q  F }  M := { (, ) | q,q’  Q, a , w  *, q’  (q,a) } “State component” “Data component” q’   (q,a)  M Recall:     

19. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 19 ] S EP 1, 2005 Example: FA Trans. Sys. (Cont’d) Behavior:  M  T M L(M) := { w   * |  T :  *  }

20. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 20 ] S EP 1, 2005 Def: Labelled Transition System A Labelled Transition System is a structure:  is the set of configurations A is the set of actions (= labels)     A   is the transition relation Note: we will write instead of  , A,     ’  ’ a ( , a,  ’)  ‘  ’

21. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 21 ] S EP 1, 2005 The labelled transition may be illustrated as: The labels(/actions) add the opportunity of describing transitions: Internally (e.g., information about what went on internally) Externally (e.g., information about communication /w env) …or both. A Labelled Transition   ’  ’ ’’ a a

22. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 22 ] S EP 1, 2005 Example: FA Transition System (2) Given Finite Automaton: Define Labelled Terminal Transition System: Configurations: Labels: Transition relation: Final configurations:  M := Q   M, A M,  M, T M  T M := F q  M q’ := q’  (q,a) A M :=  a M =  Q, , q 0, F,  

23. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 23 ] S EP 1, 2005 Example: FA Trans. Sys. (Cont’d) Behavior: Define ”  *” as the transitive closure of ”  ” on sequences of labels: 0  M 1  M 0  M 2  M 0  T M L(M) := { w  A* |  q  T : q 0  * q w toss heads toss tails L(M) := { a a’ … a’’  A* |  q  T : q 0  q’  …  q a a’ a’’

24. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 24 ] S EP 1, 2005 FA: Machine 1 vs. Machine 2 The two transition systems are very different: Machine 1: ”I transitioned from to ” Implicit: ”…by consuming part of (internal) data component” Machine 2: ”I transitioned from q to q’ …by inputing an ’a’ symbol from the (external) environment!”  M q  M q’ a

25. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 25 ] S EP 1, 2005 More Examples… More Examples in [Plotkin, p. 6 – 13]: Three Counter Machine (***) Context-Free Grammars (**) Petri Nets (*) They illustrate expressive power of transition systems no new points...except formalizing input/output behavior (also later here…) Read these yourself…

26. C LAUS B RABRAND © S EMANTICS (Q1,’05) S EP 1, 2005 T HE L ANGUAGE “L” Keywords: Expressions, Boolean Expressions, Commands, Abstract Syntax, Concrete Syntax

27. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 27 ] S EP 1, 2005 The Language ”L” Basic Syntactic Sets: Truthvalues: Set ranged over by: t, t’, t 0, … Numbers: Set ranged over by: m, n, … Variables: Set ranged over by: v, v’, … T = { tt, ff } N = { 0, 1, 2, …} VAR = { a, b, c, …, z } Meta-variables

28. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 28 ] S EP 1, 2005 The Language ”L” Derived Syntactic Sets: Arithmetic Expressions ( e  Exp): Boolean Expressions ( b  BExp): Commands ( c  Com): e ::= n | v | e + e’ | e – e’ | e  e’ b ::= t | e = e’ | b or b’ | ~ b c ::= nil | v := e | c ; c’ | if b then c else c’ | while b do c

29. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 29 ] S EP 1, 2005 Consider program: It can be (ambiguously) understood: Note: Abstract- vs. Concrete Syntax while b do c ; c’  while b do c  ; c’while b do  c ; c’  …either as: …or as: “Concrete syntax” while b c ; c’ while b c ; c’ “Abstract syntax” Parsing: “Concrete syntax”  “Abstract syntax”

30. C LAUS B RABRAND © S EMANTICS (Q1,’05) S EP 1, 2005 V IRTUAL M ACHINE S EMANTICS Keywords: The “SMC” Machine, Virtual Machine Semantics, Machine Code Semantics

31. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 31 ] S EP 1, 2005 Value Stack: Set ranged over by: S Memories: Functions ranged over by: M Control Stack: Set ranged over by: C SMC Machine (Trans. Sys. Semantics): Configurations: Virtual Machines ( SMC Expressions )  =  Value Stack  Memories  Control Stack  ( T  N )* VAR  N ( Exp  { +, –,  } )* Initially Commands, and later bits of commands For accumulating partial results

32. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 32 ] S EP 1, 2005 Configurations: Notation: i.e., “Memory Update” More common syntax for this: Memory Update  =     ( T  N  VAR )* VAR  N ( Exp  { +, –,  } )* m, if v = v’ M’(v’) = M(v’), otherwise M[m/v] M[m/v] = M’ where M[m  v]

33. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 33 ] S EP 1, 2005 Virtual Machine Transitions Configurations: Transitions: Defined by case according to top of control stack:  =     ( T  N  VAR )* VAR  N ( Exp  { +, –,  } )* [case n ]:  [case v ]:  [case e  e’ ]:  [case + / - /  ]:  …where n = m  m’ Syntactic ‘+’ Semantic ‘+’

34. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 34 ] S EP 1, 2005 Virtual Machine: Example Given program: and memory: (((x + 1) – y)  7)  M = [x=9,y=4]         

35. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 35 ] S EP 1, 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

36. C LAUS B RABRAND © S EMANTICS (Q1,’05) S EP 1, 2005 N EXT W EEK… Semantics of…: “ Simple Commands and Expressions”

37. C LAUS B RABRAND S EMANTICS (Q1,’05) [ 37 ] S EP 1, 2005 Next week Use everything…: Inference Systems, Transition Systems, Syntax, … …to: describe explain analyse compare …semantics of Expressions:

38. C LAUS B RABRAND © S EMANTICS (Q1,’05) S EP 1, 2005 See you next week… Questions?