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1 CSCI 2400 section 3 Models of Computation Instructor: Costas Busch.

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Presentation on theme: "1 CSCI 2400 section 3 Models of Computation Instructor: Costas Busch."— Presentation transcript:

1 1 CSCI 2400 section 3 Models of Computation Instructor: Costas Busch

2 2 Computation CPU input memory output memory Program memory temporary memory

3 3 A Model Machine CPU input memory output memory Program memory temporary memory Automaton Internal state

4 4 Different Kinds of Automata Are dinstinguished by the temporary memory Finite Automata No temporary memory Pushdown Automata Stack Turing Machines Unlimited memory

5 5 input memory output memory temporary memory Finite Automaton Finite Automaton Vending Machines (small computing power)

6 6 input memory output memory stack Pushdown Automaton Pushdown Automaton Programming Languages (medium computing power)

7 7 input memory output memory memory Turing Machine Turing Machine Algorithms (highest computing power)

8 8 Accepter input memory Yes or No memory Automaton Language = inputs for which answer is Yes

9 9 Finite Automaton Languages Pushdown Automaton Languages Turing Machine Languages We will show

10 10 We will show How to parse Programming Languages Some problems have no algorithms (cannot be solved)

11 11 Mathematical Preliminaries Sets Functions Relations Graphs Proof Techniques

12 12 SETS A set is a collection of elements A = { 1, 2, 3 } B = { train, car, bicycle, airplane } We say: 1 is in A ship is not in B

13 13 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

14 14 A = { 1, 2, 3, 4, 5 } Universal Set: all possible elements U = { 1, …, 10 } 1 2 3 4 5 A U 6 7 8 9 10

15 15 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

16 16 Complement Universal set = {1, …, 7} A = { 1, 2, 3 } A = { 4, 5, 6, 7} 1 2 3 4 5 6 7 A A A = A

17 17 0 2 4 6 1 3 5 7 even { even integers } = { odd integers } odd Integers

18 18 DeMorgan’s Laws A U B = A B U A B = A U B U

19 19 Empty, Null Set: = { } S U = S S = S - = S - S = U = Universal Set

20 20 Subset A = { 1, 2, 3} B = { 1, 2, 3, 4, 5 } A B U Proper Subset:A B U A B

21 21 Disjoint Sets A = { 1, 2, 3 } B = { 5, 6} A B = U AB

22 22 Set Cardinality For finite sets A = { 2, 5, 7 } |A| = 3

23 23 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 )

24 24 Cartesian Product A = { 2, 4 } B = { 2, 3, 5 } A X B = { (2, 2), (2, 3), (2, 5), ( 4, 2), (4, 3), (4, 4) } |A X B| = |A| |B| Generalizes to more than two sets A X B X … X Z

25 25 FUNCTIONS domain 1 2 3 a b c range f : A -> B A B If A = domain then f is a total function otherwise f is a partial function f(1) = a

26 26 RELATIONS R = {(x 1, y 1 ), (x 2, y 2 ), (x 3, y 3 ), …} x i R y i e. g. if R = ‘>’: 2 > 1, 3 > 2, 3 > 1 In relations a x i can be repeated In functions a x i cannot be repeated

27 27 Equivalence Relations Reflexive: x R x Symmetric: x R y y R x Transitive: x R Y and y R z x R z Example: R = ‘=‘ x = x x = y y = x x = y and y = z x = z

28 28 Equivalence Classes For equivalence relation R equivalence class of x = {y : x R y} Example: R = { (1, 1), (2, 2), (1, 2), (2, 1), (3, 3), (4, 4), (3, 4), (4, 3) } Equivalence class of 1 = {1, 2} Equivalence class of 3 = {3, 4}

29 29 GRAPHS A directed graph Nodes (Vertices) V = { a, b, c, d, e } Edges E = { (a, b), (b, c), (c, a), (b, d), (d, c), (e, d) } e a b c d node edge

30 30 Labeled Graph a b c d e 1 3 5 6 2 6

31 31 Walk a b c d e Walk is a sequence of adjacent edges (e, d), (d, c), (c, a)

32 32 Path a b c d e Path is a walk where no edge is repeated Simple path: no node is repeated

33 33 Cycle a b c d e 1 2 3 Cycle: a walk from a node (base) to itself Simple cycle: only the base node is repeated base

34 34 Euler Tour a b c d e 1 2 3 4 5 6 7 8 base A cycle that contains each edge once

35 35 Hamiltonian Cycle a b c d e 1 2 3 4 5 base A simple cycle that contains all nodes

36 36 Finding All Simple Paths a b c d e f

37 37 a b c d e (c, a) (c, e) f Step 1

38 38 a b c d e (c, a) (c, a), (a, b) (c, e) (c, e), (e, b) (c, e), (e, d) Step 2 f

39 39 Step 3 a b c d e f (c, a) (c, a), (a, b) (c, e) (c, e), (e, b) (c, e), (e, d) (c, e), (e, d), (d, f) Repeat the same for each starting node

40 40 Trees root leaf parent child Trees have no cycles

41 41 root leaf Level 0 Level 1 Level 2 Level 3 Height 3

42 42 Binary Trees

43 43 PROOF TECHNIQUES Proof by induction Proof by contradiction

44 44 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

45 45 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

46 46 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

47 47 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

48 48 Induction Step hypothesis: l(n) <= 2 n Level n n+1 l(n+1) <= 2 l(n) <= 2 2 n = 2 n+1

49 49 Remark Recursion is another thing Example of recursive function: f(n) = f(n-1) + f(n-2) f(0) = 1, f(1) = 1

50 50 Proof by Contradiction We want to prove that a statement P is true we assume that P is false then we arrive at an incorrect conclusion therefore, statement P must be true

51 51 Example Theorem: sqrt(2) is not rational Proof: Assume by contradiction that it is rational sqrt(2) = n/m n and m have no common factors We will show that this is impossible

52 52 Sqrt(2) = n/m 2 m 2 = n 2 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 k’ m and n have common factor 2 Contradiction!


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