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CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 0 - Intro

CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 0 - Intro Mälardalen University 2005. Content Adminstrivia Mathematical Preliminaries Countable Sets (Uppräkneliga mängder) Uncountable sets (Överuppräkneliga mängder). Lecturer & E xami ner

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CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 0 - Intro

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  1. CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 0 - Intro Mälardalen University 2005

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

  3. Lecturer&Examiner • Gordana Dodig-Crnkovic

  4. Teaching Assistent • Andreas Ermedahl

  5. http://www.idt.mdh.se/kurser/cd5560/05_04 visit home page regularly! Course Home Page

  6. 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!

  7. Mathematical Preliminaries

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

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

  10. 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

  11. 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 }

  12. 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

  13. 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

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

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

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

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

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

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

  20. 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 )

  21. 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

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

  23. 2 2 3 1 1 4 0 0 5 3 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

  24. 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

  25. Induction 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

  26. 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

  27. 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

  28. 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

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

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

  31. 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}

  32. 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.

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

  34. 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.

  35. 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.

  36. 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)

  37. 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

  38. 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

  39. = 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

  40. Languages, Alphabets and Strings

  41. Languages • defined over an alphabet: A language is a set of strings A String is a sequence of letters • An alphabet is a set of symbols

  42. Alphabets and Strings • We will use small alphabets: Strings

  43. Operations on Strings

  44. = w a a a L x abba 1 2 n = v b b b y  bbbaaa L 1 2 m String Operations Concatenation (sammanfogning) xy abbabbbaaa

  45. Reverse (reversering) Example: Longest odd length palindrome in a natural language: saippuakauppias (Finnish: soap sailsman)

  46. Length: String Length Examples:

  47. Empty String • A string with no letters: • (Also denoted as ) • Observations:

  48. Substring (delsträng) • Substring of string: • a subsequence of consecutive characters • String Substring

  49. prefix suffix Prefix and Suffix • Suffixes Prefixes

  50. (String repeated n times) Repetition n = • Example: • Definition: w ww... w } n

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