CS 310 (sec 20) - Winter Final Exam (solutions) SOLUTIONS

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1 CS 310 (sec 20) - Winter Finl Exm (solutions) SOLUTIONS 1. (Logic) Use truth tles to prove the following logicl equivlences: () p q (p p) (q q) () p q (p q) (p q) () p q p q p p q q (q q) (p p) (q q) T T T F F T T T F T F T F T F T T T F T T F F F T T F F They hve the sme truth vlues () p q p q p q (p q) (p q) (p q) T T T F T T T F F T F F F T F T F F F F F T F F They hve the sme truth vlues 1

2 2. (Reltions) Let X e the set of ll 4-it strings (e.g., 0011, 0101, 1000). Define reltion R on X s s 1 R s 2 if some sustring of s 1 of length 2 is equl to some sustring of s 2 of length 2. 1 Exmples: 0111 R 1010 (ecuse oth 0111 nd 1010 contin 01), ut 1110 R 0001 (ecuse 1110 nd 0001 do not hve common sustring of length 2). Determine whether R is: reflexive, symmetric, ntisymmetric, trnsitive, prtil order nd/or n equivlence reltion. 1. Reflexive: YES. Given string α X, α indeed hs some common sustring of size 2 with itself, sy its sustring consisting of the first two its of α. 2. Symmetric: YES. If α, β X, nd α hs some sustring of size 2 in common with β, lso β hs tht sme sustring in common with α. 3. Antisymmetric: NO. Two strings in X my e relted to ech other without eing equl, for instnce 0111 R 1010 (ecuse oth 0111 nd 1010 contin 01), ut Trnsitive: NO. Counterexmple: 1110 R 1100 (oth contin 11) nd 1100 R 0001 (oth contin 00), however 1110 R Prtil order: NO, since the reltion is not ntisymmetric nor trnsitive. 6. Equivlence reltion: NO, since since the reltion is not trnsitive. 1 By definition s is sustring of t if there re strings u nd v such tht t = usv. 2

3 3. (Algorithms) Let T (n) e the numer of times the sttement x := x + 1 is executed in the pieces of pseudocode elow. Select thet nottion for T (n) from mong Θ(1), Θ(log 2 (n)), Θ(n), Θ(n log 2 (n)), Θ(n), Θ(n 2 ), Θ(n 3 ),..., Θ(2 n ), Θ(n!). 1: procedure proc1(n) 2: for i := 1 to n do 3: for j := 1 to n do 4: for k := 1 to n do 5: x := x + 1 6: end proc1 1: procedure proc2(n) 2: if n = 1 then 3: x := x + 1 4: else 5: egin 6: cll proc2(n-1) // (two recursive clls 7: cll proc2(n-1) // in row) 8: end 9: end proc2 1: procedure proc3(n) 2: for i := 1 to n-1 do 3: for j := 1 to n-i do 4: x := x + 1 5: end proc3 1. proc1(n): T (n) = n 3 = Θ(n 3 ). 2. proc2(n): T (1) = 1, T (n) = 2T (n 1) = 4T (n 2) = = 2 n T (1) = 2 n = Θ(2 n ). 3. proc3(n): T (n) = (n 1) + (n 2) = n(n 1) 2 = Θ(n 2 ). 3

4 4. (Proility) We hve two coins. One of them is fir, i.e., the proilities of hed nd tils re oth equl to 1/2. The other one is loded, so tht the proility of getting tils fter tossing it is 1/3 nd the proility of hed is 2/3. We choose one of the coins t rndom (with proility 1/2) nd toss it. 1. Wht is the proility of getting tils. 2. Assume we get tils. Wht is the proility tht the coin we just tossed is the loded one? Give the proilities s frctions (no decimls!). Let F represent fir, L loded, H hed, T tils. 1. Proility of tils : P (T ) = P (T F ) P (F ) + P (T L) P (L) = = Byes theorem: P (L T ) = P (T L) P (L) P (T ) = =

5 5. (Comintorics) 1. How mny strings cn e formed y ordering the letters of NORTHWESTERN so tht ll E s pper etween the 2 N s? 2. Find the coefficient of x 3 yz 4 in the expnsion of (2x + y + z) The word NORTHWESTERN hs 12 letters: 2 E s, 1 H, 2 N s, 1 O, 2 R s, 1 S, 2 T s, nd 1 W. First we choose 4 plces for the 2 E s nd the 2 N s, which cn e done in ( ) 12 4 wys. Then we plce the 2 E s nd the 2 N s so tht the E s re etween the N s; this cn e done in just 1 wy:... N... E... E... N.... Then we plce the remining 8 letters in the remining plces, which cn e done in P (8; 2, 2, 1, 1, 1, 1, 1, 1) = 8! ( ) 2!2! 12 8! wys. So the finl nswer is 4 2!2! = ! 3!1!4! =

6 6. (Grphs) In the following grph with its vertices in lpheticl order find spnning tree using 1. Bredth-first serch. 2. Depth-first serch. f g c e d f f g g c e c e d d Bredth first Depth first 6

7 7. (Trees) Construct n optiml Huffmn code for the set of letters in the following tle (to reduce the numer of possile nswers lwys lel 1 the edge going from vertex to its child of lowest frequency nd 0 the edge tht goes to the child with highest frequency): letter frequency A 3 B 4 C 10 D 5 E 6 Find the verge length of it strings encoding 28-letter words with the Huffmn code. The following is the tree corresponding to the optiml Huffmn code: A B C D E The resulting code is s follows: letter code A 011 B 010 C 00 D 11 E 10 The verge length of it strings encoding 28-letter words is = 63 (= 2.25 per it). 7

8 8. (Comintoril Circuits nd Boolen Algers) Write the output f(x, y, z) of the following comintoril circuit s Boolen expression involving x, y nd z. Simplify tht Boolen expression. Design n equivlent simpler circuit sed on the simplified expression using the minimum possile numer of gtes. x y z f(x,y f(x, y, z) = (x y z) (x y z) = x y. x y x y 8

9 9. (Finite-Stte Mchines) In the following questions the input is lwys string over {, }. () Find the output string of the following finite-stte mchine for the input : strt /2 σ 1 /0 /1 /1 σ 0 σ 2 /1 /2 () Design finite-stte mchine tht outputs 1 if string with two or more s (n ny numer of s) is input; otherwise, outputs 0; e.g., input would produce output (it strts outputting 1 s right fter the second hs een input). () () strt σ 0 /0 /0 /1 /0 σ 1 /1 σ 2 /1 9

10 10. (Lnguges) Let G e the grmmr with non terminl symols {E, T, F }, terminl symol {, +,, (, )}, productions E T, E E + T, T F, T T F, F (E), F, nd strting symol E. 1. Find derivtion for the following string: + ( + ) 2. Prove tht the following string is not in the lnguge L(G) ssocited to the given grmmr: + ( + ( + ) 1. There re mny wys to derive the given string, the following is one: E E +T T +T T F +T F F +T F +T +T + F + (E) + (E + T ) + (T + T ) +(F +T ) +(+T ) +(+T F ) +(+F F ) + ( + F ) + ( + ) 2. Prentheses cn e produced only y using the rule F (E), so ech time we produce left prenthesis we lso produce right prenthesis, hence in the finl string the numer of left prentheses must e equl to the numer of right prentheses. However the given string hs two left prentheses nd only one right prenthesis. 10

11 11. (Extr-Credit) 1. Design finite-stte utomton (y giving its trnsition digrm) tht recognizes the strings represented y the regulr expression ( + ). 2. Is your finite-stte utomton deterministic or non-deterministic? If it is non-deterministic design n equivlent deterministic utomton (nd simplify it y removing unrechle sttes). 1. strt σ 0 σ 1 σ 2 2. {σ 0, σ 1 } strt {σ 0 } {σ 1 } {σ 1, σ 2 } {σ 0, σ 2 } {σ 0, σ 1, σ 2 } {σ 2 } After removing unrechle sttes we get the following: strt {σ 0 } {σ 1 } {σ 1, σ 2 } 11

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