CSE Discrete Structures

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1 CSE Discrete Structures Homework 3- Solution - Fall 2010 Due Date: Oct , 3:30 pm Sets 1. Rewrite the following sets as a list of elements. (8 points) a) {x ( y)(y N x = y 3 x < 30)} {0, 1, 8, 27} ) {x x is a palindrome over the alphaet a, of length less than 5} {a,, aa,, aaa, aa,, a, aaaa, aa,, aa} c) {x x Z x < 6} {0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5} d) {x x < 0 x 3 Z} { 3, 6, 9, 12, 15,...} Rewrite the following sets y giving a characteristic property. e) {,,,,,...} {x ( y)(y N y > 0 x = n )} f) { 3, 5, 7, 11, 13, 17, 19, 23, 29 } {x x is a prime numer x > 2} g) { Red, Green, Blue } {x x is a primary color} h) {0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5,...} {x ( y)(y N x = 0.5 y)} 2. Given the sets A, B, C, and D elow, indicate which of the following statements are true. (8 points) A = {x y Z x = y 2 } B = {x x Z x 0} C = {2, 3, 5, 8, 12, 17, 23, 30} D = {x ( y)(y Z x = 2 y)} a) A B True 2010 Manfred Huer Page 1

2 ) B A False c) C B True d) (A C) = True e) (A D) B True f) 25 (A D) True g) D B False h) A (D B) False 3. Determine which of the following are operations on the given set S. You have to justify your decision (a short justification is sufficient). (16 points) a) x y = x 3, S = {x x Z x > 5} Is an operation. Third power is well defined on the integers larger than 5 and the third power of any integer larger than 5 is an integer larger than 5. ) x y = x 2 y 2, S = {x x N} Is not an operation. It is not closed on the natural numers since for any y larger than x the difference of the squares is negative and thus not a natural numer. c) x y = x y, S = {x x Q x > 0} x+y Is an operation. It is well defined for the positive rational numers (since the denominator can never ecome 0) and is closed since the result of multiplication and addition of positive rational numers are positive rational numers and teh division of two positive rational numers also results in a positive rational numer. d) x y = xyx, S = {x x is a palindrome over the alphaet a, } Is not an operation. xyx is not closed for palindromes since, e.g., for x = a and y = a it results in xyx = aaa which is not a palindrome. e) x y = x y, S = (Z) Is an operation. Union is well defined for the power set and the poser set of all integers is closed under union. f) x # = x S, S = N Is not an operation. returns a truth value and the set of natural numers is therefore not closed under. g) x # = x + x, S = R Is not an operation. Square root is not well defined on the real numers Manfred Huer Page 2

3 h) x # = xx, S = {a,, aa,, aaa,, aaaa,,...} Is an operation. String concatenation is well defined over the set of strings containing only one kind of symol and this set is closed under concatenation with itself since concatenation of a string containing only one symol type with itself again results in a string containing only one symol type. 4. Prove the following set identities. (18 points) a) A A = S A A S : x (A A) x (S A) x A x S x S x S S A A : x S x ((S A) A) x (A A) ) A B = A B A B A B : x A B x (A B) x (A B) x A x B x A x B x A x B x A x B x (A B) A B A B : x (A B x A x B x A x B x A x B x A x B x A B x (A B) x A B c) A B = B A x (A B) x A x B x B x A x (B A) Using the asic set identities listed in the tale prove the following set identities. You have to indicate at each step which identity was used. d) (A B) ((C A B) (B C A)) = B (A B) ((C A B) (B C A)) (A B) ((C A B) (C A B)) comm (A B) ((C C) (A B)) dist (A B) (S (A B)) comp (A B) (A B) id (A A) B dist S B comp B id e) (A (B C)) (A B C) C B = 2010 Manfred Huer Page 3

4 (A (B C)) (A B C) C B (A (B C)) (A B C) (C B) DeMorgan( from part ) (A (B C)) (A B C) (C B) dn A (B C) A B C (C B) comm A (B C) (C B) A B C comm A (B C) (B C) A B C comm A (B (C C)) A B C dist A (B ) A B C comp A B A B C id ((A B) A B) C C id id f) (((A (B C)) (A C))) (A (C (B S))) = C 2010 Manfred Huer Page 4

5 (((A (B C)) (A C))) (A (C (B S))) (((A (B C)) (A C))) (A (C S)) id (((A (B C)) (A C))) (A C) id (A (B C) A C) (A C) ass (A A C (B C)) (A C) comm (A C (B C)) (A C) id ((A C) (A C)) ((B C) (A C)) dist ((A A) C) ((B C) (A C)) dist ( C) ((B C) (A C)) comp C ((B C) (A C)) id (C (B C)) (C (A C)) dist (C (B C)) (C C A) comm (C (B C)) (C A) id ((S C) (B C)) (C A) id ((S B) C) (C A) dist (S C) (C A) id C (C A) id (C ) (C A) id C ( A) dist C id C id 2010 Manfred Huer Page 5

6 5. Show that the following sets are denumerale. (10 points) a) {x y Z x = 2 y + 1} This set contains all odd integers. To enumerate it we can use the following enumeration function: S(i) = ( 1) i i which enumerates them as: {1, 1, 3, 3, 5, 5,...} 2 ) {x y N z N x = y } z All elements of this set can e enumerated using dovetailing as pairs (y, z): {(0, 0), (0, 1), (1, 0), (2, 0), (1, 1), (0, 2), (0, 3), (1, 2),(2, 1),(3, 0), (4,0),...} c) {x x is a finite length alternating string of even length of the symols 1, and 0 starting with 0} All elements of this set can e enumerated as follows: {0, 01, 010, 0101, 01010, ,...} d) {(x, y) x N y Z} All elements of this set can e enumerated using dovetailing and the enumeration order for integers: {(0, 0), (0, 1), (1, 0), (2, 0), (1, 1), (0, 1), (0, 2), (1,1), (2, 1), (3, 0), (4,0), (3, 1), (2, 1), (1, 2), (0, 2), (0, 3), (1, 2),...} Use Cantor s diagonalization method to prove that the following set is not countale. e) {x x is an infinite length string over the english alphaet} Assuming that the set is countale we can write a list of all elements (strings) s i = s i,1 s i,2 s i,3... in this set in the following way: s 1,1 s 1,2 s 1,3 s 1,4 s 1,5... s 2,1 s 2,2 s 2,3 s 2,4 s 2,5... s 3,1 s 3,2 s 3,3 s 3,4 s 3,5... s 4,1 s 4,2 s 4,3 s 4,4 s 4,5.... where s i,j {a,, c, d,...}. Given this enumeration we can construct an element t = t 1 t 2 t 3... of the set as follows: { a if si,i = t i = otherwise Given this definition t is different from every s i since for all i the i th symol of t is different from the i th symol in s i. Therefore y is an element of the set ut not an element of the list and thus the assumption that the set is countale has to e false. Counting Principles 6. Use decision trees to determine the numer of possile outcomes for the following prolems. (8 points) 2010 Manfred Huer Page 6

7 a) Mary goes grocery shopping for Milk and Cookies. The Supermarket has 3 rands of milk and 4 types of cookies. How many choices does Mary have? M1 M2 M3 C1 C2 C3 C4 C1 C2 C3 C4 C1 C2 C3 C4 Mary has 12 different choices. ) How many different strings of length 3 can e formed over the alphaet {a,, c}? a c a c a c a c a c a c a c a c a c a c a c a c a c There are 27 different strings. c) If there are 3 types of pens, 4 types of pencils, and 2 types of paper, how many different ways are there to uy paper and something to write? P1 P l1 l2 l3 l l1 l2 l3 l4 There are 14 different choices. d) There are three colored chips on the tale, one red, one lue, and one green. How many ways are there to put the three chips in a row? r g g r r g g r g r There are 6 different ways Manfred Huer Page 7

8 7. Given the sets A, B, and C, use the addition principle, multiplication principle, and the principle of inclusion and exclusion to determine the numer of elements in the following sets. (12 points) A = {2, 3, 5, 7} B = {Red, Green, Blue} C = { 1, 1, 2, 2} D = {2, 3, 4} a) A C (B C) = A + C + B C A C A (B C) C (B C) + A C (B C) = B C = = 19. ) A C D = A + C + D A C A D C D + A C D = = 8. c) (B C) D = B C D = = 36. Use the addition and multiplication principles to determine the numer of possile outcomes for the following prolems. d) How many octal (i.e. ase 8) numers of length 4 are there that contain only odd digits? This is a sequence of 4 digit choices (multiplication principle) where in each choice there are 4 choices (odd digits - 1,3,5,7). Therefore there are = 256 numers. e) Consider an experiment where you either flip a coin or use a random numer generator to draw a random integer etween 1 and 10. If you repeat this experiment 4 times in a row, how many different outcomes can you otain? This is a sequence of experiments (multiplication principle) where in each experiment a choice etween flipping a coin and using a random numer generator is possile (addition principle). Therefore there are (2 + 10) (2 + 10) (2 + 10) (2 + 10) = different outcomes. f) If we randomly chain letters into strings of length 7, how many outcomes are there that contain only vowels? This is a sequence of letters (multiplication principle where in each position we have a choice of all the vowels (a, e, i, o, u). Therefore there are = possile outcomes that contain only vowels. Permutations and Cominations 8. Indicate if the following are permutation or comination prolems and determine the numer of possiilities. (20 points) a) How many possile outcomes are there if you flip a coin 10 times in a row? This is a permutation prolems since the sequence matters. There are repetitions here (you can get the same result multiple times) and all outcomes in each round are unique = Manfred Huer Page 8

9 ) How many different strings of length 5 can e formed over the english alphaet (26 distinct letters)? This is a permutation prolem since the order matters. There are repetitions since each letter can occur multiple times ut all elements are unique = 11, 881, 376 c) How many different sets of 5 distinct letters can you form from the letters of the english alphaet (26 distinct letters)? This is a comination prolem since the order does not matter when forming sets. There are no repetitions since the letters have to e distinct and no duplicates. C(26, 5) = 26! = 65, !5! d) How many distinct inary numers (potenitally with leading 0 s) can e formed from the digits in the numer ? This is a permutation prolem since the order matters. There is no repetition ut there are duplicate elements. 9! = 126 4! 5! e) How many different study groups of size 5 that contain at least one man and one woman can e formed in a class with 25 men and 12 women? This is a comination prolem since order does not matter. There are no repetitions or duplicates. This prolem can e solved using the addition principle and cominations y oserving that the numer of groups of size 5 containing at least one man and one woman is equal to the sum of the numer of groups containing exactly 1 woman and 4 men, exactly 2 women and 3 men, exactly 3 women and 2 men and exactly 4 women and 1 man. C(12, 1) C(25, 4) + C(12, 2) C(25, 3) + C(12, 3) C(25, 2) + C(12, 4) C(25, 1) = 12! 25! + 12! 25! + 12! 25! + 12! 25! = 381, !1! 21!4! 10!2! 22!3! 9!3! 23!2! 8!4! 24!1! f) In a lottery, 6 numers are drawn from a set of 49 numers (1-49). How many outcomes are there that contain an equal numer of odd and even numers? This is a comination prolem since the order does not matter. To solve it we oserve that this is the same as drawing 3 numers from the set of odd numers (25) and 3 numers from the set of even numers (24). C(25, 3) C(24, 3) = 4, 655, 200 g) At a party, 8 guests are distriuted throughout 3 rooms. How many different possiilities are there for the occupancy of the rooms (only the numer of guests in each room matters)? This is a comination prolem with repetition. It is easiest solved y interpreting it as a permutation prolem with duplicates (where all elements are duplicates and 2 additional - identical - elements are added to mark the rooms). (8+(3 1))! = 10! = 45 8!(3 1)! 8!2! h) How many decimal numers of length 6 (no leading 0 s) are there in which no digit occurs more than once? This is a permutation prolem since the order matters. There are no repetitions and the first digit is special since it is not allowed to e 0. This can either e solved y sutracting 2010 Manfred Huer Page 9

10 all length 6 permutations that start with 0 from the total. P(10, 6) P(10, 5) = 10! 9! = 151, , 120 = 136, 080 4! 5! 2010 Manfred Huer Page 10