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Semantics Q1 2007 S EMANTICS (Q1,’07) Week 1 Jacob Andersen PhD student

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1 Semantics Q1 2007 S EMANTICS (Q1,’07) Week 1 Jacob Andersen PhD student andersen@daimi.au.dk

2 Semantics Q1 2007 Week 1 - Outline Introduction Welcome and introduction Short-film about teaching (20')... Course Presentation [ homepage ]homepage Prerequisitional (discrete) Mathematics Relations Inference Systems Transition Systems The Language “L”

3 Semantics Q1 2007 Introduction Semantics –Merriam-Webster: –Example: “I would like a cup of coffee.” “Colourless green ideas sleep furiously.” Question: Would you fly the space shuttle if you had written the guidance software??? In this course we shall explore formal ways of reasoning about the meaning / behaviour of computer programs. Main Entry: se·man·tics Pronunciation: si-'man-tiks Function: noun plural but singular or plural in construction 1 : the study of meanings

4 Semantics Q1 2007 Short-film about teaching...

5 Semantics Q1 2007 Course Introduction I SOLO 2, 3, 4 Verbs identifies skills / competences Actor

6 Semantics Q1 2007 Course Introduction II No-one can force new skills into your head –You do not learn how to ride a bike by observing others (!!) Expected weekly time consumption: Lecture3 Reading1 Exercise class3 Hand-in2 Sum13 Fitting the pieces together Lecture3 Reading1 Reflecting / Solving exercises4 Exercise class3 Hand-in2 Sum13

7 Semantics Q1 2007 Course Introduction III Course Retrospection dSem ’05:

8 Semantics Q1 2007 Learning vs. Breaks #Breaks, |Breaks|,...? Conclusion: more (shorter) breaks

9 Semantics Q1 2007 R ELATIONS AND I NFERENCE S YSTEMS

10 Semantics Q1 2007 Relations I 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’ 

11 Semantics Q1 2007 Relations II In general, a k-ary relation, L, over the sets S 1, S 2, …, S k is a subset L  S 1  S 2  …  S k k is called the “arity” of L, and “L  S 1  S 2  …  S k ” is the “signature”.

12 Semantics Q1 2007 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

13 Semantics Q1 2007 Terminology Meaning: Inductive: “If n is even, then m is even (provided m = n+2)”; or Deductive: “m is even, if n is even (provided m = n+2)” | _ even n | _ even m m = n+2 premise(s) conclusion side-condition(s) Question: Why have both premises and side-conditions?? Note: The “ “ is actually just another way of writing a “  ” !

14 Semantics Q1 2007 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 Even so, this is what we mean Warning: Be careful !

15 Semantics Q1 2007 Relation Membership (Even)? 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

16 Semantics Q1 2007 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 ]

17 Semantics Q1 2007 Activation Exercise 1 Activation Exercise: 1. Specify the signature of the relation: ' << ' »x << y " y is-double-that-of x " 2. Specify the relation via an inference system »i.e. axioms and rules 3. Prove that indeed: »3 << 6 "6 is-double-that-of 3"

18 Semantics Q1 2007 Activation Exercise 2 Activation Exercise: 1. Specify the signature of the relation: ' // ' »x // y " x is-half-that-of y " 2. Specify the relation via an inference system »i.e. axioms and rules 3. Prove that indeed: »3 // 6 "3 is-half-that-of 6" Syntactically different Semantically the same relation

19 Semantics Q1 2007 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! |_R 0|_R 0 | _ R n | _ R n+2 [rule 1 ] [axiom 1 ] | _ R  Z (0  ‘| _ R ’)  (  n  ‘| _ R ’  n+2  ‘| _ R ’)

20 Semantics Q1 2007 Inductive Interpretation A inference system: …induces a function: Definition: ‘lfp’ (least fixed point) ~ least solution: | _ R 0 | _ R n | _ R n+2 [rule 1 ] [axiom 1 ] F R : P(Z)  P(Z) | _ R  Z F R (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 R ) =  F R n (Ø) n | _ R  P(Z) From rel. to rel. 2N   = F n (Ø) ~ “Anything that can be proved in ‘n’ steps” 

21 Semantics Q1 2007 T RANSITION S YSTEMS

22 Semantics Q1 2007 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 might use the following notations:  ,     ’  ’( ,  ’)     ’  ’  |_ ’ |_ ’   ’  ’,, …

23 Semantics Q1 2007 The transition may be illustrated as: In this course, we will be (mostly) using: For instance: A Transition   ’  ’ ’’  = system configuration = ...and have "  " describe "one step of computation"

24 Semantics Q1 2007 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  T :     :  t  

25 Semantics Q1 2007 E.g.: Finite Automaton A Non-det Finite Automaton (NFA) is a 5tuple: 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} }

26 Semantics Q1 2007 E.g.: NFA Transition System (1) Define Term. Trans. Syst. 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:      M =  Q, , q 0, F,   Given

27 Semantics Q1 2007 E.g.: NFA Trans. Sys. (Cont’d) Define behavior of M:  M  T M L (M) := { w   * |  T :  *  }

28 Semantics Q1 2007 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,  ’)  ‘  ’

29 Semantics Q1 2007 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

30 Semantics Q1 2007 E.g.: NFA 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  M := {(q,a,q’)|q’  (q,a)} A M :=  M =  Q, , q 0, F,   X ??? “Relaxed” v

31 Semantics Q1 2007 E.g.: NFA Trans. Sys. (Cont’d) Behavior: Define ”  *” as the reflexive transitive closure of ”  ” on sequences of labels: 0  M 1  M 0  M 2  M 0 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’’

32 Semantics Q1 2007 NFA: 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

33 Semantics Q1 2007 More Examples… More Examples in [Plotkin, p. 22 – 26]: –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…

34 Semantics Q1 2007 T HE L ANGUAGE “L”

35 Semantics Q1 2007 The Language ”L” Basic Syntactic Sets: Truthvalues: –Setranged over by: t, t’, t 0, … Numbers: –Setranged over by: m, n, … Variables: –Setranged over by: v, v’, … T = { tt, ff } N = { 0, 1, 2, …} VAR = { a, b, c, …, z } Meta-variables

36 Semantics Q1 2007 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

37 Semantics Q1 2007 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”

38 Semantics Q1 2007 Next week Use everything…: Inference Systems, Transition Systems, Syntax, … …to: –describe –explain –analyse –compare …semantics of Expressions:

39 Semantics Q1 2007 "Three minutes paper" Please spend three minutes writing down the most important things that you have learned today (now). After 1 day After 1 week After 3 weeks After 2 weeks Right away

40 Semantics Q1 2007 Exercises... [ homepage ]

41 Semantics Q1 2007 See you next week… Questions?

42 Semantics Q1 2007


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