1. Week 15 - Monday

2.  What did we talk about last time?  Context free grammars

4.  Three men in a cafe order food totaling $15  They each contribute $5 to the bill  The waiter takes the money to the manager who recognizes the three as friends and asks the waiter to return $5 to the men  However, the waiter is dishonest  Instead of splitting the $5 evenly (and who would get the last penny, anyway?), he simply gives each man $1 and pockets the remaining $2  Now, each diner effectively paid $4  Thus, the amount paid is $12  Add the $2 in the waiter's pocket and the total comes to $14  Where has the other $1 gone from the original $15?

6.  The following is a grammar that produces the language of strings of 1s and 0s that end in a 1 S  A1S  A1  A  1 | 0 | AA | є  This language is regular and is equivalent to (0 | 1)* 1  The following is a grammar that produces the language of a k b k where k ≥ 1 (which is not regular) S  AS  A  A  ab | aAb

7.  Write a grammar that specifies legal nesting of parentheses and braces in Java  Don't worry about the stuff that goes inside  Write a grammar for legal mathematical expressions in Java using variables and integers, +, -, *, /, and parentheses  Write a grammar that corresponds to the same language defined by the regular expression ab* (a | bb) (ba)*

8.  As you know, regular languages can be expressed by regular expressions and finite state automata  Thus, regular expressions are equal to FSAs in power  Context free languages can be expressed by context free grammars  There is also a class of automata called pushdown automata that correspond to context free languages

10.  To make the PDA for the language 0 k 1 k, k ≥ 1, we have the following:  In this case, the notation X/Y means that if X is on the top of the stack, replace it with Y  The top of the stack is Z  Notation is not well standardized for PDAs  It's awkward keeping track of the stack s1s1 s1s1 s2s2 s2s2 s0s0 s0s0 0: Z/AZ 0: A/AA 1: A/ε 1: A/A 1: Z/Z

12.  Noam Chomsky is a brilliant linguist who has recently focused mostly on political activism  Remember that a grammar consists of terminals (alphabet symbols), non-terminals, production rules, and a start symbol  He noted that grammars can be divided into four levels in terms of expressiveness:  Type-0 (Unrestricted grammars)  Type-1 (Context sensitive grammars)  Type-2 (Context free grammars)  Type-3 (Regular grammars)

13.  Each grammar has rules for what is a legal production rule  Let α, β, and γ be any combinations of terminals and non- terminals, where γ is non-empty  Unrestricted grammars  α  β (anything to anything)  Context-sensitive grammars  αAβ  αγβ (non-terminal in a particular context to anything)  Context-free grammars  A  γ (a single non-terminal to anything)  Regular grammars  A  a and A  aB (a single non-terminal to a single terminal or a terminal and a single non-terminal on the right side)

14.  Every kind of language has a particular kind of machine associated with it GrammarLanguagesAutomaton Production Rules Examples Type-0 Recursively enumerable Turing machine α  βα  β All languages, any computable functions Type-1 Context- sensitive Linear-bounded non- deterministic Turing machine αAβ  αγβ Most natural human languages Type-2Context-free Non-deterministic pushdown automaton A  γ Most programming languages Type-3RegularFinite state automaton A  a and A  aB Regular expressions

15.  Each set of languages in the hierarchy strictly contains the sets beneath it  Regular languages  Can be accepted by nondeterministic or deterministic finite automata (or a read-0nly Turing machine)  Are closed under union, intersection, complement, concatenation, and Kleene star  Context-free languages  Are defined by those languages accepted by nondeterministic pushdown automata  Are closed under union, concatenation, and Kleene star (but not under intersection or complement)  Deciding whether a string is in a context-free language can be determined in polynomial time  Deciding whether a language is empty is decidable  Context-sensitive languages  Are very rarely used  Are closed under union, intersection, complement, and Kleene star  Deciding whether a string is in a context-sensitive language is a PSPACE-complete problem Recursively enumerable Context- sensitive Context-free Regular

16.  A Turing machine is a mathematical model for computation  It consists of a head, an infinitely long tape, a set of possible states, and an alphabet of characters that can be written on the tape  A list of rules saying what it should write and should it move left or right given the current symbol and state 1011110000 A A

17.  You can specify a Turing machine with a table giving its behavior for a specific configuration  Turing's first example machine printed an infinite sequence of alternating 1s and 0s, separated by spaces: ConfigurationBehavior StateSymbolOperationResult State BBlankWrite 0, Move RightC CBlankMove RightE EBlankWrite 1, Move RightF FBlankMove RightB

18.  If an algorithm exists, a Turing machine can perform that algorithm  In essence, a Turing machine is the most powerful model we have of computation  Power, in this sense, means the ability to compute some function, not the speed associated with its computation  Do you own a Turing machine?

19.  Given a Turing machine and input x, does it reach the halt state?  As you know, there is no algorithm to determine this  If you had a Turing machine that could solve the halting problem, it would cause a logical contradiction

20.  Are two context-free languages the same?  Is the intersection of two context-free languages empty?  Is a context-free language equal to Σ *  Is a context-free language a subset of another context-free language?  Post correspondence problem:  You have lists a 1, a 2, … a n and b 1, b 2, … b n, where a i and b j are strings of some alphabet Σ with at least 2 symbols  Is there a value k ≥ 1 such that some sequence of k strings from the a list concatenated is equal to some sequence of k strings from the b list concatenated?  Is a given statement of first-order logic provable from a starting set of axioms?  Given a set of matrices, is there some sequence that they can be multiplied in (perhaps with repetitions) that will yield the zero matrix?

22.  Statements  AND, OR, NOT, IMPLIES  Truth tables  Logical equivalence  De Morgan's laws  Tautologies and contradictions

23. NameLawDual Commutative p  q  q  pp  q  q  pp  q  q  pp  q  q  p Associative (p  q)  r  p  (q  r)(p  q)  r  p  (q  r) Distributive p  (q  r)  (p  q)  (p  r)p  (q  r)  (p  q)  (p  r) Identity p  t  pp  c  p Negation p  ~p  tp  ~p  c Double Negative ~(~p)  p Idempotent p  p  pp  p  pp  p  pp  p  p Universal Bound p  t  tp  t  tp  c  c De Morgan’s ~(p  q)  ~p  ~q~(p  q)  ~p  ~q Absorption p  (p  q)  pp  (p  q)  p Negations of t and c ~t  c~t  c~c  t~c  t

24.  Can be used to write an if-then statement  Contrapositive is logically equivalent  Inverse and converse are not (though they are logically equivalent to each other)  Biconditional: p  q  q  pp  q  q  p

25.  A series of premises and a conclusion  Using the premises and rules of inference, an argument is valid if and only if you can show the conclusion  Rules of inference:  Modus Ponens  Modus Tollens  Generalization  Specialization  Conjunction  Elimination  Transitivity  Division into cases  Contradiction rule

26.  Show that the following is a valid argument  ~p ˅ q → r  s ˅ ~q ~t~t p → tp → t  ~p ˄ r → ~s ∴ ~q

27.  The following gates have the same function as the logical operators with the same names:  NOT gate:  AND gate:  OR gate:

28.  Make a truth table for:  ~(q → r) ˅ (s ˄ p) ↔ (p ˅ r)  Make a digital logic circuit for the same expression

30.  A predicate is a sentence with a fixed number of variables that becomes a statement when specific values are substituted for to the variables  The domain gives all the possible values that can be substituted  The truth set of a predicate P(x) are those elements of the domain that make P(x) true when they are substituted

31.  We will frequently be referring to various sets of numbers in this class  Some typical notation used for these sets:  Some authors use Z + to refer to non-negative integers and only N for the natural numbers SymbolSetExamples RReal numbersVirtually everything ZIntegers{…, -2, -1, 0, 1, 2,…} Z-Z- Negative integers{-1, -2, -3, …} Z+Z+ Positive integers{1, 2, 3, …} NNatural numbers{1, 2, 3, …} QRational numbers a/b where a,b  Z and b  0

32.  The universal quantifier  means “for all”  The statement “All DJ’s are mad ill” can be written more formally as:   x  D, M(x)  Where D is the set of DJ’s and M(x) denotes that x is mad ill  The universal quantifier  means “there exists”  The statement “Some emcee can bust a rhyme” can be written more formally as:   y  E, B(y)  Where E is the set of emcees and B(y) denotes that y can bust a rhyme

33.  When doing a negation, negate the predicate and change the universal quantifier to existential or vice versa  Formally:  ~(  x, P(x))   x, ~P(x)  ~(  x, P(x))   x, ~P(x)  Thus, the negation of "Every dragon breathes fire" is "There is one dragon that does not breathe fire"

34.  Any statement with a universal quantifier whose domain is the empty set is vacuously true  When we talk about "all things" and there's nothing there, we can say anything we want  "All mythological creatures are real."  Every single one of the (of which there are none) is real

35.  Recall:  Statement: p  q  Contrapositive:~q  ~p  Converse:q  p  Inverse:~p  ~q  These can be extended to universal statements:  Statement:  x, P(x)  Q(x)  Contrapositive:  x, ~Q(x)  ~P(x)  Converse:  x, Q(x)  P(x)  Inverse:  x, ~P(x)  ~Q(x)  Similar properties relating a statement equating a statement to its contrapositive (but not to its converse and inverse) apply

36.  p is a sufficient condition for q means p  q  p is a necessary condition for q means q  p  These come over into universal conditional statements as well:   x, P(x) is a sufficient condition for Q(x) means  x, P(x)  Q(x)   x, P(x) is a necessary condition for Q(x) means  x, Q(x)  P(x)

37.  With multiple quantifiers, we imagine that corresponding “actions” happen in the same order as the quantifiers  The action for  x  A is something like, “pick any x from A you want”  Since a “for all” must work on everything, it doesn’t matter which you pick  The action for  y  B is something like, “find some y from B”  Since a “there exists” only needs one to work, you should try to find the one that matches

38.  For negation,  Simply switch every  to  and every  to   Then negate the predicate  Changing the order of quantifiers can change the truth of the whole statement but does not always  Furthermore, quantifiers of the same type are commutative:  You can reorder a sequence of  quantifiers however you want  The same goes for   Once they start overlapping, however, you can’t be sure anymore

39.  Use a statement of predicate logic to write the following statement:  "There is a person who can take every sad song and make it better"  Now negate the statement and read back the equivalent informal version  Rewrite the following without negations outside the predicates:  ∼ ( ∀ x ∈ D, ∀ y ∈ E, P(x, y)))

40.  Universal instantiation: If a property is true for everything in a domain (universal quantifier), it is true for any specific thing in the domain  Universal modus ponens:   x, P(x)  Q(x)  P(a) for some particular a Q(a)Q(a)  Universal modus tollens:   x, P(x)  Q(x)  ~Q(a) for some particular a   ~P(a)

42.  To prove  x  D  P(x) we need to find at least one element of D that makes P(x) true  To disprove  x  D, P(x)  Q(x), we need to find an x that makes P(x) true and Q(x) false

43.  If the domain is finite, we can use the method of exhaustion, by simply trying every element  Otherwise, we can use a direct proof 1. Express the statement to be proved in the form  x  D, if P(x) then Q(x) 2. Suppose that x is some specific (but arbitrarily chosen) element of D for which P(x) is true 3. Show that the conclusion Q(x) is true by using definitions, other theorems, and the rules for logical inference  Direct proofs should start with the word Proof, end with the word QED, and have a justification next to every step in the argument  For proofs with cases, number each case clearly and show that you have proved the conclusion for all possible cases

44.  If n is an integer, then:  n is even   k  Z  n = 2k  n is odd   k  Z  n = 2k + 1  If n is an integer where n > 1, then:  n is prime   r  Z +,  s  Z +, if n = r  s, then r = 1 or s = 1  n is composite   r  Z +,  s  Z +  n = r  s and r  1 and s  1  r is rational   a, b  Z  r = a/b and b  0  For n, d  Z,  n is divisible by d   k  Z  n = dk  For any real number x, the floor of x, written  x , is defined as follows:   x  = the unique integer n such that n ≤ x < n + 1  For any real number x, the ceiling of x, written  x , is defined as follows:   x  = the unique integer n such that n – 1 < x ≤ n

45.  Unique factorization theorem: For any integer n > 1, there exist a positive integer k, distinct prime numbers p 1, p 2, …, p k, and positive integers e 1, e 2, …, e k such that  Quotient remainder theorem: For any integer n and any positive integer d, there exist unique integers q and r such that  n = dq + r and 0 ≤ r < d

46.  Prove or disprove the following statements:  For all integers a, b, and c, if a | b and a | c then a | (2b − 3c)  For all integers a, b, and c, if a | (b + c) then a | b or a | c  Use the quotient-remainder theorem with d = 3 to prove that the square of any integer has the form 3k or 3k + 1 for some integer k

48.  In a proof by contradiction, you begin by assuming the negation of the conclusion  Then, you show that doing so leads to a logical impossibility  Thus, the assumption must be false and the conclusion true

49.  A proof by contradiction is different from a direct proof because you are trying to get to a point where things don't make sense  You should always mark such proofs clearly  Start your proof with the words Proof by contradiction  Write Negation of conclusion as the justification for the negated conclusion  Clearly mark the line when you have both p and ~p as a contradiction  Finally, state the conclusion with its justification as the contradiction found before

50.  Use a proof by contradiction to prove the following:  For all integers n, if n 2 is odd then n is odd  For all prime numbers a, b, and c, a 2 + b 2 ≠ c 2

52.  Mathematical sequences can be represented in expanded form or with explicit formulas  Examples:  2, 5, 10, 17, 26, …  a i = i 2 + 1, i ≥ 1  Summation notation is used to describe a summation of some part of a series  Product notation is used to describe a product of some part of a series

53.  To prove a statement of the following form:   n  Z, where n  a, property P(n) is true  Use the following steps: 1. Basis Step: Show that the property is true for P(a) 2. Induction Step: ▪ Suppose that the property is true for some n = k, where k  Z, k  a ▪ Now, show that, with that assumption, the property is also true for k + 1

55.  Using recursive definitions to generate sequences  Writing a recursive definition based on a sequence  Using mathematical induction to show that a recursive definition and an explicit definition are equivalent

56.  Expand the recursion repeatedly without combining like terms  Find a pattern in the expansions  When appropriate, employ formulas to simplify the pattern  Geometric series: 1 + r + r 2 + … + r n = (r n+1 – 1)/(r – 1)  Arithmetic series: 1 + 2 + 3 + … + n = n(n+ 1)/2

57.  Use the method of iteration to find an explicit formula for the following recursively defined sequence:  d k = 2d k−1 + 3, for all integers k ≥ 2  d 1 = 2  Use a proof by induction to show that your explicit formula is correct

58.  To solve sequence a k = Aa k-1 + Ba k-2  Find its characteristic equation t 2 – At – B = 0  If the equation has two distinct roots r and s  Substitute a 0 and a 1 into a n = Cr n + Ds n to find C and D  If the equation has a single root r  Substitute a 0 and a 1 into a n = Cr n + Dnr n to find C and D

59.  Find an explicit formula for the following:  r k = 2r k-1 − r k-2, for all integers k ≥ 2  r 0 = 1  r 1 = 4

62.  Review second third of the course

63.  Review chapters 6 - 9