Solutions. CS 2800 Fall 2017 Final exam Friday, December 8. NetID: 1. Modular arithmetic [9 pts]

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1 S 28 Fall 27 Final exam Friday, December 8. Modular arithmetic [9 pts] Solutions (a) [5 pts] Let d j d j... d 2 d d be the base representation of n. Use equivalence classes to prove that if n is a multiple of 3, then the sum of the d i is also a multiple of 3. We have n = d i i. Reducing this equation mod 3 we have [n] = [ d i i ] = [di ][] i. Since [] = [], this is just [ d i ]. Since n is a multiple of 3, we have [n] =, so [ d i ] = [], and thus the sum of the digits is a multiple of 3. (b) [ pt] List the units of Z 9., 2, 4, 5, 7, 8 (c) [3 pts] Use Euler s theorem to compute [7 65 ] 9. For full credit, use 3 or fewer multiplications of single digit numbers (and some division). y above, φ(9) = 6. y Euler s theorem, we know [7 65 ] 9 = [7 5 ] 9 since [65] 6 = [5] 6. Now we find [7] 5 by repeated squaring. [7] 2 = [49] = [4] 9. [7] 4 = [4] 2 = [6] = [7]. Thus [7] 5 = [7][7] 4 = [7][7] = [4]. 2. Equivalence relations and cardinality [9 pts] Let be the relation on sets given by if and only if =. Prove that is an equivalence relation. You may not use any facts about cardinality without proof (other than the definitions), but you may use other facts about sets or functions without proving them, as long as you clearly state what facts you are using. We must show is reflexive, symmetric, and transitive. To see is reflexive, note that the identity function is a bijection from to ; thus there exists a bijection so =. To see that is symmetric, assume that =. Then there exists a bijection f :. Since f is a bijection, it has a two-sided inverse g : ; since g has a two-sided inverse (f) it is a bijection. Therefore, there exists a bijection from to so =. To see that is transitive, assume = and =. Then there are bijections f : and g :. The composition of bijections is a bijection, so g f : is a bijection, and thus =.

2 3. ézout s identity [9 pts] (a) [6 pts] Let P (b) be the statement for any a N, there exists s, t Z such that gcd(a, b) = sa + tb. Prove P (b) for all b. We will prove b, P (b) by strong induction. To see P (), note that gcd(a, ) = a = a + ; choosing s = and t = gives the result. To see P (b), assume P (k) for all k < b. Now, gcd(a, b) = gcd(b, r) where q and r are the quotient and remainder of a over b (so a = qb + r). Since r < b, we know P (r), so that gcd(b, r) = s b + t r. Since a = qb + r, we have r = a qb; plugging this in gives gcd(a, b) = gcd(b, r) by definition = s b + t r by P (r) = s b + t (a qb) since a = qb + r = t a + (s t q)b algebra = sa + tb if we defined s = t and t = s t q. (b) [3 pts] Use this to show that [a] is a unit mod m if gcd(a, m) =. If gcd(a, m) =, then = sa + tm. Reducing this equation mod m we have [] = [s][a] + [t][] = [s][a] so [a] has an inverse (namely [s]). Therefore [a] is a unit. 4. Proof trees [8 pts] (a) [3 pts] Use a truth table to argue that Q (P P ) Q. Here is the truth table for Q (P P ) and Q: P Q Q (P P ) Q T T F F T F T T F T F F F F T T We can see that every line I in which eval(q (P P ), I) = T, we also have eval( Q, I) = T, thus Q (P P ) Q. (b) [5 pts] Prove Q (P P ) Q (without relying on completeness). 2

3 Q Q below P P elim below P P elim elim elim P P absurd, Q Q Q (P P ) Q, Q Q assum elim Q (P P ) assum Q assum elim Q (P P ), Q P P 5. Logic [6 pts] Suppose we wished to extend propositional logic with an if-then-else formula: we interpret φ? φ 2 : φ 3 as if φ then φ 2, otherwise φ 3. Here is the truth table for φ? φ 2 : φ 3 : φ φ 2 φ 3 φ? φ 2 : φ 3 T T T T T T F T T F T F T F F F F T T T F T F F F F T T F F F F (a) [2 pts] Extend the inductive definition of the eval function for? / : expressions. eval(φ? φ 2 : φ 3, I) ::= { eval(φ2 ), if eval(φ ) = T eval(φ 3 ), otherwise (b) [4 pts] omplete the following partial rules to give reasonable introduction and elimination rules for? / : expressions:, φ φ 2, φ φ 3 φ? φ 2 : φ 3 (?/: intro) φ? φ 2 : φ 3 φ φ 2 (?/: elim) 3

4 φ? φ 2 : φ 3 φ φ 3 (?/: elim) 6. Graphs [5 pts] (a) [2 pts] One of the following two graphs contains an eulerian cycle; the other does not. Number the edges of an eulerian cycle in the graph that contains one, and prove that the other does not. D E F D 8 7 E F 5 G G 2 The first graph has no eulerian cycle, because vertex F has in-degree but outdegree 2; thus any eulerian cycle must leave F twice but enter only once, an impossibility. (b) [3 pts] Prove or disprove: all 3-colorable graphs with 3 vertices are isomorphic. False: the following two graphs are 3-colorable but not isomorphic: They are 3-colorable because we can color both of them with c() =, c() = 2, and c() = 3, but they are not isomorphic because (, ) the first graph has 3 edges and the second has utomata [ pts] 4

5 Recall that when proving that the intersection of regular languages is regular, we constructed a machine M from two machines M = (Q, Σ, δ, q, F ) and M 2 = (Q 2, Σ, δ 2, q 2, F 2 ) as follows: M = (Q Q 2, Σ, δ, (q, q 2 ), F F 2 ), where δ((q, q 2 ), a) := (δ (q, a), δ 2 (q 2, a)). (a) [6 pts] Prove that for all x Σ, δ((q, q 2 ), x) = ( δ (q, x), δ 2 (q 2, x)). We prove this by induction on the structure of x. δ((q, q 2 ), x) = ( δ (q, x), δ 2 (q 2, x)). To see P (ɛ), we use the definition of δ to get δ((q, q 2 ), ɛ) = (q, q 2 ) Let P (x) be the statement and ( δ (q, ɛ), δ 2 (q 2, ɛ)) = (q, q 2 ) These are the same. To see P (xa), assume P (x). We compute δ((q, q 2 ), xa) = δ(ˆδ((q, q 2 ), x), a) definition of ˆδ = δ((ˆδ (q, x), ˆδ 2 (q 2, x)), a) ( ) by P (x) = δ (ˆδ (q, x), a), δ 2 (ˆδ 2 (q 2, x), a) by definition of δ = (ˆδ (q, xa), ˆδ ) 2 (q 2, xa) by definition of ˆδ and ˆδ 2 (b) [4 pts] Use this to show that L(M) = L(M ) L(M 2 ). L(M) = {x ˆδ((q, q 2 ), x) F F 2 } by definition = {x (ˆδ (q, x), ˆδ 2 (q 2, x)) F F 2 } by part (a) = {x ˆδ (q, x) F and ˆδ 2 (q 2, x) F 2 } by definition of F F 2 = {x ˆδ (q, x) F } {x ˆδ 2 (q 2, x) F 2 } by definition of = L(M ) L(M 2 ) by definition of L(M ) and L(M 2 ) 8. Regular expressions [7 pts] Use the proof of Kleene s theorem to give a regular expression that matches strings with an even number of s and an odd number of s. 5

6 Note: make sure your reasoning is clear; small mistakes will not be penalized heavily if your process is correct. Step. Draw an automata. Step 2. dd a new final state q oe q oe q oo ɛ q oo Step 3. Remove states. + () ɛ + () ɛ q oo () + () + ( + () )(() ) ( + () )(() )ɛ Step 4. Read the RE: ( + () + ( + () )(() ) ) ( + () )(() )ɛ 9. Probability [2 pts] 6

7 onsider an experiment described by the following sample space and random variables: s Pr({s}) X(s) Y (s) a /8 b 2/8 2 2 c 3/8 3 d /8 3 4 e /8 5 Indicate whether each of the following is an event, an outcome, a number, or a random variable, and give its value. (a) [2 pts] The sample space, S an event, {a, b, c, d, e}. (b) [2 pts] Y = 4 an event, {d}. (d) [2 pts] Pr(X > Y > 2) a number, Pr({e})/P r({c, d, e}) = /5 (e) [2 pts] X + Y a random variable, s X + Y (s) a b 4 c 3 d 7 e 6 (c) [2 pts] Pr(Y 2) a number, 3/8 (f) [2 pts] a number. V ar(x) V ar(x) = E(X 2 ) E(X) 2 E(X) = 2 2/8 + 3 /8 + /8 = E(X 2 ) = 4 2/8 + 9 /8 + /8 = 8/8 V ar(x) = /8 (This page intentionally left blank) 7