DISCRETE STRUCTURES EXAM 2 FALL 2005, SECTION 0201 (PICKETT)

Transcription

1 DISCRETE STRUCTURES EXAM FALL 005, SECTION 001 (PICKETT) NAME: This exam is worth 15% of your final grade (375/500 points) There are 7 problems You have until 11:15 to finish You may use the exam cheat sheet I ve provided, and calculators should be unnecessary Be sure to show your work, as I ll be giving partial credit 1

2 DISCRETE STRUCTURES EXAM FALL 005, SECTION 001 (PICKETT) (1) (50 points total) Consider the sequence f where f 1, f 6, f 3 1, f 4 0, f 5 30, f 6 4, f 7 56, f 8 7, f 9 90, etc (a) (5 points) Write a closed form equation for the nth term of this sequence Answer: If we subtract f n 1 from f n we get n, which tells us that this is a quadratic series (of the form n +cn+d) Solving for the constants c 1, d 0, so we get f n n + n (b) (5 points) Write a closed form equation for n i1 f i Answer: Using our result from part a, and the summations from the cheat (1) () (3) sheet, we get n f i i1 n ( i + i ) i1 n i + i1 n i i1 n (n + 1) (n + 1) 6 + n + n

3 DISCRETE STRUCTURES EXAM FALL 005, SECTION 001 (PICKETT) 3 () (50 points) Give a formal definition of KBAG n, which is the set of all people whose Kevin Bacon number is n (The Kevin Bacon number (KB#) of an actor is the minimum number of acting links separating that actor from Kevin Bacon For example Kevin Bacon s KB# is 0 (so KBAG 0 {KB}) Tom Hanks KB# is 1 because Tom Hanks acted in a movie (Apollo 13) with Kevin Bacon (so Tom Hanks would be in KBAG 1 (and in KBAG, KBAG 3, etc), but not KBAG 0 )) Your definition should have no English in it, but it may use the following terms: KB means Kevin Bacon, A (x, y) means that x and y acted in a movie together, and P is the set of all people Your definition may use recurrence relations, but remember to ground them out in base cases If you use the term KB#, you ll need to formally define it in terms of A (x, y) and KB Answer: We can define KBAG n using a cross between a recurrence relation and a recursive set definition: KBAG n {x P x KBAG n 1 ya (x, y) y KBAG n 1 } where KBAG 0 is defined in the example above to be {KB} Alternatively, if we assume that A is a reflexive relation (ie, everyone has acted with themself), then we can shorten our definition to: KBAG n {x P ya (x, y) y KBAG n 1 }

4 4 DISCRETE STRUCTURES EXAM FALL 005, SECTION 001 (PICKETT) (3) (50 points) The global summit on Discrete Structures is scheduled to take place this year in Marc s fantasy world, Marcland Leaders from the world s 19 countries are expected to show up with their nation s flag At the conference center, there is a flagpole tall enough to fly 7 flags at once The flag of Marcland will always be flown at the top, and a flag from each of the 6 geographic regions (listed below) will be flown in one of the 6 positions below the Marcland flag If a leader s country s flag is flying on a particular day (at position n from the top), he or she will be granted nth pick of that day s 7 Starburst candies The flags are changed daily, and, for fairness, the summit will last as long as it takes to fly every allowable combination of flags in every allowable configuration The world s regions are the following: Africa: 53 countries Europe: 46 countries Asia: 44 countries North America: 3 countries Oceania: 14 countries South America: 1 countries How many days will the summit last? Answer: This will be the number of ways to choose a representative from each region multiplied by the number of permutations Since Marcland s flag will always be at the top, the number of permutations is 6! The number of ways to choose a representative from each region is 53 (choices for the African flag) 46 (choices for the European flag) ,499,008 So the summit will take ! 98, 439, 85, 760 days, which is about 8 million years

5 DISCRETE STRUCTURES EXAM FALL 005, SECTION 001 (PICKETT) 5 (4) (50 points) A census of Kompleck City (commissioned by Vlad Urr) found that none of Kompleck City s Igs are Ogs The census also found that among Kompleck City s 100 Igs and Ogs: 5 are Male Igs that are Gogs 10 are Female Igs who are Gogs 5 are Male Ogs who are Gogs 0 are Female Ogs who are Gogs 15 are Male Igs that aren t Gogs 10 are Female Igs who aren t Gogs 5 are Male Ogs who aren t Gogs 10 are Female Ogs who aren t Gogs If all Gogs are either Igs or Ogs, what is the probability that a Gog (of Kompleck City) is an Og? Answer: There are 40 ( ) Gogs, of which 5 ( 5 + 0) are Ogs (and the rest are Igs) This gives a probability of 5 40 or 065

6 6 DISCRETE STRUCTURES EXAM FALL 005, SECTION 001 (PICKETT) (5) (75 points) If there are 6 students in the class, how many possible graphs can I make where the nodes are students, the edges (connecting students to students) are undirected, and there are neither multiple edges nor graph loops (ie, there s at most 1 edge between any pair of students, and no student is connected to him/her self)? (Hint, there s possibly an edge (or not) between each pair of students) Answer: If there are n students in the class, then there are n n (or ( n ) ) possible pairs of students in the class, each of which is a potential edge for the graph Since each potential edge will be activated or not, this gives n n possible graphs If n 6, then there are possible pairs of students, and possible graphs

7 DISCRETE STRUCTURES EXAM FALL 005, SECTION 001 (PICKETT) 7 (6) (50 points) Prove or disprove that a symmetric relation (over the natural numbers) composed with itself is reflexive Answer: Disproof by counter example The relation R {(1, 3),(3, 1)} is symmetric, and when composed with itself yields R R {(1, 1),(3, 3)}, but this isn t reflexive because (, ) (for example) isn t in R R

8 8 DISCRETE STRUCTURES EXAM FALL 005, SECTION 001 (PICKETT) (7) (50 points total) In Marc s Casino, my Mediocre Poker machine accepts only half dollars and quarters However, it distinguishes between 3 different types of quarters ( state quarters, the plain quarter, and the 1976 bicentennial quarter), and it distinguishes between different types of half dollars (bicentennial and plain ) (a) (5 points) Set up a recurrence relation for the number of different ways one can pay 5n cents to the machine, where the order in which the coins are inserted matters (Two different state quarters would be indistinguishable to the machine, but the machine can tell a state quarter from a bicentennial quarter So, there are 3 ways to insert $05, and 11 ways to insert $050 (so f 0 1, f 1 3, and f 11)) Answer: Following very similar reasoning to the solution of problem a on problem set 7, we get the base cases shown above (f 0 1, f 1 3, and f 11), and our recurrence relation is f n 3f n 1 + f n (We get the 3 constant because there are now 3 ways to pay the last $5) (b) (5 points) What characteristic equation would we have to solve to obtain a closed form solution for the above recurrence relation? Answer: This is a direct application of the characteristic equation definition that s on the cheat sheet: r 3r 0 (c) (5 BONUS points): Write a closed form expression for the number of different ways one can pay 5n cents to the Mediocre Poker machine Answer: We can use the quadratic formula to solve for r:

9 DISCRETE STRUCTURES EXAM FALL 005, SECTION 001 (PICKETT) 9 (4) (5) r r 3 So, our closed form equation is of the form f n α 1 r n 1 + α r n Solving the system of equations for α (6) (7) α 1 r α r 0 1 α 1 r α r 0 3 or (8) (9) (10) (11) (1) (13) (14) (15) (16) () (18) So our closed form is α 1 α 1 α 3 α 1r 1 r r r α 1 3 α 1 r 1 α 1 (r 1 r ) 3 r f n 3 r r 1 r r n 1 + r 1 3 r 1 r r n α 1 3 r r 1 r α r 1 3 r 1 r (3 r ) r1 n + (r 1 3)r n r 1 r ( )( ( ( 3+ )( ) n+1 ( ) n ( + 3+ )( ) n ( 3 3 ) n+1 )( 3 ) n 3 ) n