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1 CSCI 2400 section 3 Models of Computation Instructor: Costas Busch
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2 Computation CPU input memory output memory Program memory temporary memory
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3 A Model Machine CPU input memory output memory Program memory temporary memory Automaton Internal state
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4 Different Kinds of Automata Are dinstinguished by the temporary memory Finite Automata No temporary memory Pushdown Automata Stack Turing Machines Unlimited memory
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5 input memory output memory temporary memory Finite Automaton Finite Automaton Vending Machines (small computing power)
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6 input memory output memory stack Pushdown Automaton Pushdown Automaton Programming Languages (medium computing power)
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7 input memory output memory memory Turing Machine Turing Machine Algorithms (highest computing power)
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8 Accepter input memory Yes or No memory Automaton Language = inputs for which answer is Yes
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9 Finite Automaton Languages Pushdown Automaton Languages Turing Machine Languages We will show
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10 We will show How to parse Programming Languages Some problems have no algorithms (cannot be solved)
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11 Mathematical Preliminaries Sets Functions Relations Graphs Proof Techniques
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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
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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
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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
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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
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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
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17 0 2 4 6 1 3 5 7 even { even integers } = { odd integers } odd Integers
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18 DeMorgan’s Laws A U B = A B U A B = A U B U
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19 Empty, Null Set: = { } S U = S S = S - = S - S = U = Universal Set
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20 Subset A = { 1, 2, 3} B = { 1, 2, 3, 4, 5 } A B U Proper Subset:A B U A B
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21 Disjoint Sets A = { 1, 2, 3 } B = { 5, 6} A B = U AB
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22 Set Cardinality For finite sets A = { 2, 5, 7 } |A| = 3
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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 )
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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
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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
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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
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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
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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}
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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
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30 Labeled Graph a b c d e 1 3 5 6 2 6
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31 Walk a b c d e Walk is a sequence of adjacent edges (e, d), (d, c), (c, a)
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32 Path a b c d e Path is a walk where no edge is repeated Simple path: no node is repeated
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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
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34 Euler Tour a b c d e 1 2 3 4 5 6 7 8 base A cycle that contains each edge once
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35 Hamiltonian Cycle a b c d e 1 2 3 4 5 base A simple cycle that contains all nodes
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36 Finding All Simple Paths a b c d e f
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37 a b c d e (c, a) (c, e) f Step 1
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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
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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
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40 Trees root leaf parent child Trees have no cycles
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41 root leaf Level 0 Level 1 Level 2 Level 3 Height 3
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42 Binary Trees
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43 PROOF TECHNIQUES Proof by induction Proof by contradiction
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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
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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
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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
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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
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48 Induction Step hypothesis: l(n) <= 2 n Level n n+1 l(n+1) <= 2 l(n) <= 2 2 n = 2 n+1
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49 Remark Recursion is another thing Example of recursive function: f(n) = f(n-1) + f(n-2) f(0) = 1, f(1) = 1
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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
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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
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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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