Fall 2018 September 18, Written Homework 02

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1 S1800 Discrete Structures Profs. Gold, Pavlu, Rachlin & Sundaram Fall 2018 September 18, 2018 ssigned: Tue 18 Sep 2018 Due: Tue 25 Sep 2018 Instructions: Written Homework 02 The assignment has to be uploaded to lackboard by the due date. NO assignment will be accepted after 11:59pm on that day. We expect that you will study with friends and often work out problem solutions together, but you must write up your own solutions, in your own words. heating will not be tolerated. Professors, Ts, and peer tutors will be available to answer questions but will not do your homework for you. One of our course goals is to teach you how to think on your own. guitar You may turn in work to lackboard that is either handwritten and scanned, written in a word processor such as Word, or typeset in LaTeX. In the case of handwritten work, we may deduct points if the scan is upside down or thework isillegible. To get full credit, show INTERMEDITE steps leading to your answers, throughout. Problem 1 [26 pts (2,2,12,4,6)]: Sets asics i. Let S = {1, 3, 4, 8, 9, 11, 15, 17}. Describe the following sets by listing their elements between braces (listing method). 1. {2n + 1 prime : n S} The set of all prime values 2n + 1 such that n is in S: {3, 7, 17, 19, 23, 31} 2. {n S : 2n + 1 S} The set of all n in S such that 2n + 1 is also in S: {1, 4, 8} ( ) ( ) ii. Shade the indicated regions of the following Venn diagrams. ( ) ( ) ( ) 1

2 ) ( ) ( ) ( ) ( ) ( ) Two notes: (1) ( ) = = = (2) ( ) = ( ) = ( ) ( ) = iii. Let D = {(x, y) R 2 : x 2 + y 2 = 1}. re there sets, R such that D =? Your answer should either describe sets and with the required property or prove that they can t exist. Note that D is the disk of radius 1 centered at the origin. It includes the points (1,0) and (0,1). Were D to be a cartesian product D =, then we would have 1 as (1, 0) D and 1 as (0, 1) D. This would imply (1, 1) which is impossible as (1, 1) / D. Thus D is not a artesian product. iv. onsider the sets = {(x, y) R 2 : 4x y = 3}, = {(t + 1, 4t + 1) : t R}, = {(2a, 8a 3) : a R}. Prove that = =. (Hint: proving that two sets and are equal often involves showing the two inclusions and.) We prove and. : Suppose that (x, y). Note that then y = 4x 3. To show (x, y), we look for a value t such that (t + 1, 4t + 1) = (x, y). Setting t = x 1 we see that indeed x = t + 1 and y = 4x 3 = 4(t + 1) 3 = 4t + 1. Thus (x, y). : Suppose that (x, y). To show that (x, y), we need to verify that 4x y = 1. y definition of, the fact that (x, y) implies that there is some value t such that x = t + 1 and y = 4t + 1. Then 4x y = 4(t + 1) (4t + 1) = 3 and therefore (x, y). = : Since (x, y) in set contains all real values x and exactly one pair-value y = 4x 3 for each one of them, we can do a change in the iterator (variable) x = 2a which still iterates through all real values, and have corresponding y = 4x 3 = 4 2a 3 = 8a 3 2

3 pair value, thus iterating over the same pairs. This is a dangerous method to use without certain checks on variable domain (see change of variable for calculus). Problem 2 [22 pts (5, 4, 1, 3, 3, 6)]: ounting i. Students enrolled in S1800 are 300 women and 423 men. There are four sections, each of size at least 171, and each with at most 80 women enrolled. Prove that at least one section has at least 100 men, and that no section has more than 150 men enrolled. Proof. Pigeonhole principle gives us that at least one section has at least 423/4 = men. For the upper bound: every section has at least =91 men, so every three sections have together at least 91*3 = 273 men, leaving at most =150 men for the fourth section. ii. math professor wears for each class a combination of : one of his three hats (red, blue, green); one of his pants (black, blue, white, green); one of his shirts (white, red, black); and one of his pair of shoes (black, red, blue, white). He never wears combinations with three items of the same color. How many different combinations are wearable? There are 3*4*3*4 total combinations (product rule), including the following 13 invalid ones with three items of the same color: - 4 combinations with common red - 3 combinations with common blue - 3 combinations with common black - 3 combinations with common white Thus =131 valid combinations. iii. The professor is teaching 70 lectures for the term. an he wear a different combination for each lecture during one term? Yes, because 131 > 70 iv. If he does wear a different combination each lecture, prove that he must use either the red or the blue shoes that term. If the professor restricts combinations to only white and black shoes, he has only 3*4*3*2-3-3 =66 valid combinations, not enough for 70 lectures. v. There are 4 terms for the year. Prove that at least one combination will be used at least 3 times during the year. 3

4 Proof. 4*70 = 280 lectures. Pigeonhole principle gives that one dressing combination will be used at least 280/131 = 2.14 lectures, thus 3 lectures. vi. How many valid combinations have blue hat or shoes? 1) We compute the complement, i.e. number of combinations without blue hat or blue shoes, that is with hats red or green and shoes black red or white: there are 2*4*3*3 = 72 possibilites out of which invalid are - 4 for common red - 2 for common black - 2 for common white So 72-8 = 64 combinations without required blue. The final answer is = 67 combinations have blue hat or shoes (includes combinations with both blue) 2) by Inclusion-Exclusion H = set of combinations with blue hat S = set of combinations with blue shoes H S = set of combinations with blue hat ND shoes ounting H : 4*3*4 excluding 3 (blue) + 1 (black) + 1 (white) = 48-5=43 ounting S : 3*4*3 excluding 3 (blue) = 36-3=33 ounting H S : 4*3 excluding 3 (blue) = 9 Finally H S = H + S H S = = 67 Problem 3 [17 pts (4, 8, 5)]: ridges again Let edge (u, v) be part of undirected graph G, and we define the sets S u = {x vertex path(x, u) exists which does not pass through v} (includes u) S v = {x vertex path(x, v) exists which does not pass through u} (includes v) i. Prove that union S u S v is exactly one of the connected components of the graph. Proof. Let be the component that contains vertices u and v. ny other vertex x in this component is connected to u, thus a simple path(x, u) exists. If this path contains v, then x S v, otherwise x S u. Further, any vertex y in either of the sets is connected to both u and v, thus part of. ii. Prove that edge (u, v) is a bridge if and only if the two sets are disjoint (S u S v = ) Proof. (u, v) not bridge u, v is part of a simple cycle (u, v, x 1, x 2,..., x k, u) with k 1 there exists x i in both S u (connected to u without passing through v) and S v (connected to v without passing through u) S u S v 4

5 iii. Prove or disprove: the two sets are not disjoint if and only if they are they are equal. Not true, see counterexample. oth sets contain vertex x, but vertices a and b are part of only one set each. Problem 4 [24 pts (6, 5, 3, 10)]: ottleneck Two fixed vertices s and t in the undirected connected graph G=(V,E) have the shortest path p(s, t) of length L. Our goal in this problem is to show that if all paths p(s, t) are long, then there is a bottleneck common to all of them. i. Using set builder notation, list the sets of vertice organized by minimum distance from s: For example S 1 should be the set of vertices that has an edge to s since their shortest path to s is 1; S 2 should be the set of vertices reachable directly from S 1, but not directly from s, etc, up to S L Define the following sets of vertices S 1 = {x V x reachable from s with 1 edge} S 2 = {x V x reachable from s with 2 edges (but not fewer)} S 3 = {x V x reachable from s with 3 edges (but not fewer)} S k = {x V x reachable from s with k edges (but not fewer)} S L = {x V x reachable from s with L edges (but not fewer)} ii. Prove that the sets up to S L are disjoint and non-empty. These sets are non-empty because there is a path L long. Disjoint because they are defined that way: each set contains vertices not already listed in previous sets (therefore not listed in future sets). 5

6 iii. Show that one of the sets has at most ( V 1)/L vertices. If all sets had more than ( V 1)/L vertices (hypothetically), being disjoint sets, the union via sum rule would have more than V 1 vertices, not including s. Thats too many, contradiction. iv. If shortest path p(s, t) has length L > V /2 edges (all paths s t are long), show that all paths from s to t must have a common vertex v (not equal to either s or t) Sets S 1, S 2,..., S L 1 total at most V 2 vertices excluding s and t, and they are L 1 sets, so one must have at most V 2 /(L 1) elements. Now L > V /2 V 2 /(L 1) < 2 which means one must have exactly one element, say vertex v in set S k. ut every path must pass through set S k (because path length is more than k) so all paths must contain v. k > 0 means v is not the start s, and k < L means v is not the end t. n alternative derivation is below. There are others, and they worth full credit if complete. Starting from part (iii), the number of vertices in some set is at most ( V 1)/(L 1) ( V 1)/( V / )... 2( V 1)/ V... < 2 So, a set must have a size 1. If this is not the set S L containing t, that solo vertex is the bottleneck. 6

CS1800 Discrete Structures Fall 2016 Profs. Aslam, Gold, Ossowski, Pavlu, & Sprague December 16, CS1800 Discrete Structures Final

CS1800 Discrete Structures Fall 2016 Profs. Aslam, Gold, Ossowski, Pavlu, & Sprague December 16, 2016 Instructions: CS1800 Discrete Structures Final 1. The exam is closed book and closed notes. You may