CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 1 - Intro

1. CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 1 - Intro School of Innovation, Design and Engineering Mälardalen University 2011

2. Content • Adminstrivia • Mathematical Preliminaries • Countable Sets (Uppräkneliga mängder) • Uncountable sets (Överuppräkneliga mängder)

3. Lecturer&Examiner • Gordana Dodig-Crnkovic • Lessons: Leo HatvaniAssistant: Svetlana Girs

4. Course Home Page http://www.idt.mdh.se/kurser/cd5560/11_11/ visit home page regularly!

5. How Much Work? 20 hours a week for this type of course (norm) 4 hours lectures 2 hours exercises 14 hours own work a week!

6. Why Theory of Computation? • A real computer can be modelled by a mathematical object: a theoretical computer. • A formal language is a set of strings, and can represent a computational problem. • A formal language can be described in many different ways that ultimately prove to be identical. • Simulation: the relative power of computing models can be based on the ease with which one model can simulate another.

7. Why Theory of Computation? • 5. Robustness of a general computational model. • 6. The Church-Turing thesis: "everything algorithmically computable is computable by a Turing machine." • 7. Study of non-determinism: languages can be described by the existence or no-nexistence of computational paths. • 8. Understanding unsolvability: for some computational problems there is no corresponding algorithm that will unerringly solve them.

8. Practical Applications • Efficient compilation of computer languages • String search • Investigation of the limits of computation,recognizing difficult/unsolvable problems • Applications to other areas: • circuit verification • economics and game theory (finite automata as strategy models in decision-making); • theoretical biology (L-systems as models of organism growth) • computer graphics (L-systems) • linguistics (modelling by grammars)

9. History • Euclid's attempt to axiomatize geometry • (Archimedes realized, during his own efforts to define the area of a planar figure, that Euclid's attempt had failed and that additional postulates were needed.) • Leibniz's (1646 - 1716) dream reasoning as calculus - "Characteristica Universalis" aiming at: • "a general method in which all truths of the reason would be reduced to a kind of calculation. At the same time this would be a sort of universal language or script, but infinitely different from all those projected hitherto; for the symbols and even the words in it would direct reason; and errors, except those of fact, would be mere mistakes in calculation.“ • de Morgan, Boole, Frege, Russell, Whitehead: • Mathematics as a branch of symbolic logic!

10. 1900 Hilberts program for axiomatization of mathematics, redefined "proof" to become a completely rigorous notion, totally different from the psycho/sociological "A proof is something that convinces other mathematicians.“ He confirms the prediction Leibniz made, that "the symbols would direct reason" 1880 -1936 first programming languages 1931 Gödels incompleteness theorems 1936 Turing maschine (showed to be equivalent with recursive functions). Commonly accepted: TM as ultimate computer 1950 automata 1956 language/automata hierarchy

11. Formalization of Mathematics • Every mathematical truth expressed in a formal language is consisting of • a fixed alphabet of admissible symbols, and • explicit rules of syntax for combining those symbols • into meaningful words and sentences • Gödel's two incompleteness theorems of mathematical logic show limitations of all but the most trivial axiomatic systems. The theorems are widely interpreted as showing that Hilbert's program to find a complete and consistent set of axioms for all of mathematics is impossible, thus giving a negative answer to Hilbert's second problem.

12. The main goal of Hilbert's program was to provide secure foundations for all mathematics. In particular this should include: • A formalization of all mathematics; in other words all mathematical statements should be written in a precise formal language, and manipulated according to well defined rules. • Completeness: a proof that all true mathematical statements can be proved in the formalism. • Consistency: a proof that no contradiction can be obtained in the formalism of mathematics. This consistency proof should preferably use only "finitistic" reasoning about finite mathematical objects. • Decidability: there should be an algorithm for deciding the truth or falsity of any mathematical statement. Statement of Hilbert’s program

13. Gödel showed that most of the goals of Hilbert's program were impossible to achieve. His second incompleteness theorem stated that any consistent theory powerful enough to encode addition and multiplication of integers cannot prove its own consistency. This wipes out most of Hilbert's program as follows: • It is not possible to formalize all of mathematics, as any attempt at such a formalism will omit some true mathematical statements. • The most interesting mathematical theories are not complete. • A theory such as Peano arithmetic cannot even prove its own consistency. Statement of Hilbert’s program

14. Turing’s Contribution • Turing used a Universal Turing machine (UTM) to prove an incompleteness theorem even more powerful than Gödel’s because it destroyed not one but two of Hilbert's dreams: • Finding a finite list of axioms from which all mathematical truths can be deduced • Solving the entscheidungsproblem, ("decision problem“) by producing a "fully automatic procedure" for deciding whether a given proposition (sentence) is true or false.

15. Mathematical Preliminaries

16. Sets • Functions • Relations • Proof Techniques • Languages, Alphabets and Strings • Strings & String Operations • Languages & Language Operations

17. SETS A set is a collection of elements We write

18. Set Representations • C = { a, b, c, d, e, f, g, h, i, j, k } • C = { a, b, …, k } • S = { 2, 4, 6, … } • S = { j : j > 0, and j = 2k for some k>0 } • S = { j : j is nonnegative and even } finite set infinite set

19. A = { 1, 2, 3, 4, 5 } U A 6 8 2 3 1 7 4 5 9 10 • Universal Set: All possible elements • U = { 1 , … , 10 }

20. Set Operations • A = { 1, 2, 3 } B = { 2, 3, 4, 5} • Union • A U B = { 1, 2, 3, 4, 5 } • Intersection • A B = { 2, 3 } • Difference • A - B = { 1 } • B - A = { 4, 5 } B A U A-B

21. Complement • Universal set = {1, …, 7} • A = { 1, 2, 3 } A = { 4, 5, 6, 7} 4 A A 6 3 1 2 5 7 A = A

22. { even integers } = { odd integers } Integers 1 odd 0 5 even 6 2 4 3 7

23. DeMorgan’s Laws A U B = A B U A B = A U B U

24. Empty, Null Set: = { } S U = S S = S - = S - S = U = Universal Set

25. Subset A = { 1, 2, 3} B = { 1, 2, 3, 4, 5 } U A B Proper Subset: B U A B A

26. Disjoint Sets A = { 1, 2, 3 } B = { 5, 6} A B = U A B

27. Set Cardinality For finite sets A = { 2, 5, 7 } |A| = 3

28. Powersets A powerset is a set of sets S = { a, b, c } Powerset of S = the set of all the subsets of S 2S = { , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } Observation: | 2S | = 2|S| ( 8 = 23 )

29. Cartesian Product A = { 2, 4 } B = { 2, 3, 5 } A X B = { (2, 2), (2, 3), (2, 5), ( 4, 2), (4, 3), (4, 5) } |A X B| = |A| |B| Generalizes to more than two sets A X B X … X Z

30. PROOF TECHNIQUES • Proof by construction • Proof by induction • Proof by contradiction

31. 2 2 3 1 1 4 0 0 5 3 Proof by Construction We define a graph to be k-regular if every node in the graph has degree k. Theorem. For each even number n > 2 there exists 3-regular graph with n nodes. n = 6 n = 4

32. Proof by Construction Construct a graph G = (V, E) with n > 2 nodes. V= { 0, 1, …, n-1 } E = { {i, i+1}  for 0  i  n-2}  {{n-1,0}} (*)  {{i, i+n/2 for 0  i  n/2 –1} (**) The nodes of this graph can be written consecutively around the circle. (*) edges between adjacent pairs of nodes (**) edges between nodes on opposite sides END OF PROOF

33. Inductive Proof We have statements P1, P2, P3, … • If we know • for some k that P1, P2, …, Pk are true • for any n k that • P1, P2, …, Pn imply Pn+1 • Then • Every Pi is true

34. Proof by Induction • Inductive basis • Find P1, P2, …, Pk which are true • Inductive hypothesis • Let’s assume P1, P2, …, Pn are true, • for any n  k • Inductive step • Show that Pn+1 is true

35. L(0) = 1 L(3) = 8 Example TheoremA binary tree of height n has at most 2n leaves. Proof let L(i) be the number of leaves at level i

36. We want to show: L(i)  2i • Inductive basis • L(0) = 1 (the root node) • Inductive hypothesis • Let’s assume L(i)  2i for all i = 0, 1, …, n • Induction step • we need to show that L(n + 1)  2n+1

37. Induction Step Level hypothesis: L(n)  2n n n+1

38. Induction Step Level hypothesis: L(n)  2n n n+1 L(n+1)  2 * L(n)  2 * 2n = 2n+1 END OF PROOF

39. Proof by induction: Cardinality of a power set Let S be a finite set with N elements. Then the powerset of S (that is the set of all subsets of S ) contains 2^N elements. In other words, S has 2^N subsets. This statement can be proved by induction. It's true for N=0,1,2,3 as can be shown by examination. The notation 2^N means 2 to the power N, i.e., the product of N factors all of which equal 2. 2^0 is defined to be 1

40. The size of a finite power set For the induction step suppose that the statement is true for a set with N-1 elements, and let S be a set with N elements. Remove on element x from S to obtain a set T with N-1 elements. There are two types of subsets of S: those that contain x and those that do not contain x. The latter are subsets of T, of which there are 2^ N-1. Every subset P of S that does contain x can be obtained from a subset Q of T by adding x. The set Q is simply the set P with x removed. Clearly there is a unique and distinct set Q for each set P and every subset Q of T gives rise to a unique and distinct subset P of S. There are thus also 2^ (N-1) subsets of S that contain x, for a total of 2^ (N-1) + 2^ (N-1) = 2^ N subsets of S.

41. Inductionsbevis: Potensmängdens kardinalitet Påstående:En mängd med n element har 2n delmängder • Kontroll • Tomma mängden {} (med noll element) har bara en delmängd: {}. • Mängden {a} (med ett element) har två delmängder: {} och {a}

42. Påstående:En mängd med n element har 2n delmängder • Kontroll (forts.) • Mängden {a, b} (med två element) har fyra delmängder: {}, {a}, {b} och {a,b} • Mängden {a, b, c} (med tre element) har åtta delmängder: {}, {a}, {b}, {c} och {a,b}, {a,c}, {b,c}, {a,b,c} • Påstående stämmer så här långt.

43. Bassteg • Enklaste fallet är en mängd med noll element (det finns bara en sådan), som har 20 = 1 delmängder.

44. Induktionssteg • Antag att påståendet gäller för alla mängder med k element, dvs antag att varje mängd med k element har 2k delmängder. • Visa att påståendet i så fall också gäller för alla mängder med k+1 element, dvs visa att varje mängd med k+1 element har 2k+1 delmängder.

45. Vi betraktar en godtycklig mängd med k+1 element. Delmängderna till mängden kan delas upp i två sorter: • Delmängder som inte innehåller element nr k+1:En sådan delmängd är en delmängd till mängden med de k första elementen, och delmängder till en mängd med k element finns det (enligt antagandet) 2k stycken.

46. Delmängder som innehåller element nr k+1:En sådan delmängd kan man skapa genom att ta en delmängd som inte innehåller element nr k+1 och lägga till detta element. Eftersom det finns 2k delmängder utan element nr k+1 kan man även skapa 2k delmängder med detta element. • Totalt har man 2k +2k = 2. 2k= 2k+1 delmängder till den betraktade mängden. • END OF PROOF • (Exempel från boken: Diskret matematik och diskreta modeller, K Eriksson, H. Gavel)

47. Proof by Contradiction • We want to prove that a statement P is true • we assume that P is false • then we arrive at a conclusion that contradicts our assumptions • therefore, statement P must be true

48. Example • Theorem is not rational • Proof • Assume by contradiction that it is rational • = n/m • n and m have no common factors • We will show that this is impossible

49. = n/m 2 m2 = n2 n is even n = 2 k Therefore, n2 is even m is even m = 2 p 2 m2 = 4k2 m2 = 2k2 Thus, m and n have common factor 2 Contradiction! END OF PROOF

50. Countable Sets