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

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En presentation över ämnet: "1 CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 0 - Intro Mälardalen University 2005."— Presentationens avskrift:

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

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

3 3 Lecturer & Examiner Gordana Dodig-Crnkovic

4 4 Teaching Assistent Andreas Ermedahl

5 5 kurser/cd5560/05_04 visit home page regularly! Course Home Page

6 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 7 Mathematical Preliminaries

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

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

10 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 11 A = { 1, 2, 3, 4, 5 } Universal Set: All possible elements U = { 1, …, 10 } A U

12 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 } U A B A-B

13 13 Complement Universal set = {1, …, 7} A = { 1, 2, 3 } A = { 4, 5, 6, 7} A A A = A

14 14 { even integers } = { odd integers } even odd Integers

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

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

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

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

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

20 20 Powersets A powerset is a set of sets Powerset of S = the set of all the subsets of S S = { a, b, c } 2 S = {, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } Observation: | 2 S | = 2 |S| ( 8 = 2 3 )

21 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 22 PROOF TECHNIQUES Proof by construction Proof by induction Proof by contradiction

23 23 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 = 4 n = 6

24 24 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 Proof by Construction END OF PROOF

25 25 Induction We have statements P 1, P 2, P 3, … If we know for some k that P 1, P 2, …, P k are true for any n  k that P 1, P 2, …, P n imply P n+1 Then Every P i is true

26 26 Proof by Induction Inductive basis Find P 1, P 2, …, P k which are true Inductive hypothesis Let’s assume P 1, P 2, …, P n are true, for any n  k Inductive step Show that P n+1 is true

27 27 Example Theorem A binary tree of height n has at most 2 n leaves. Proof let L(i) be the number of leaves at level i L(0) = 1 L(3) = 8

28 28 We want to show: L(i)  2 i Inductive basis L(0) = 1 (the root node) Inductive hypothesis Let’s assume L(i)  2 i for all i = 0, 1, …, n Induction step we need to show that L(n + 1)  2 n+1

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

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

31 31 Inductionsbevis: Potensmängdens kardinalitet Påstående: En mängd med n element har 2 n delmängder Kontroll Tomma mängden {} (med noll element) har bara en delmängd: {}. Mängden {a} (med ett element) har två delmängder: {} och {a}

32 32 Påstående: En mängd med n element har 2 n 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 33 Bassteg Enklaste fallet är en mängd med noll element (det finns bara en sådan), som har 2 0 = 1 delmängder.

34 34 Induktionssteg Antag att påståendet gäller för alla mängder med k element, dvs antag att varje mängd med k element har 2 k 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 2 k+1 delmängder.

35 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) 2 k stycken.

36 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 2 k delmängder utan element nr k+1 kan man även skapa 2 k delmängder med detta element. Totalt har man 2 k + 2 k = 2. 2 k = 2 k+1 delmängder till den betraktade mängden. END OF PROOF (Exempel från boken: Diskret matematik och diskreta modeller, K Eriksson, H. Gavel)

37 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 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 39 Therefore, n 2 is even n is even n = 2 k 2 m 2 = 4k 2 m 2 = 2k 2 m is even m = 2 p Thus, m and n have common factor 2 Contradiction! = n/m 2 m 2 = n 2 END OF PROOF

40 40 Languages, Alphabets and Strings

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

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

43 43 Operations on Strings

44 44 String Operations m n bbbv aaaw     y  bbbaaa x  abba Concatenation (sammanfogning) xy  abbabbbaaa

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

46 46 String Length Length: Examples:

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

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

49 49 Prefix and Suffix Suffixes prefix suffix Prefixes

50 50 Repetition Example: Definition: n n  } (String repeated n times)

51 51 The * (Kleene star) Operation the set of all possible strings from alphabet [Kleene is pronounced "clay-knee“]

52 52 The + Operation : the set of all possible strings from alphabet except ,ba  ,,,,,,,,,*aabaaabbbaabaaba  


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