Packet #6: Counting & Graph Theory. Applied Discrete Mathematics

Transcription

1 Packet #6: Counting & Graph Theory Applied Discrete Mathematics Table of Contents Counting Pages 1-8 Graph Theory Pages 9-16 Exam Study Sheet Page 17

2 Counting Information I. Product Rule: A B C = A * B * C II. Union Rule: A B = A + B - A B III. Inclusion/Exclusion Principle: The Union Rule applied inductively to: A 1 A 2 A n IV. Selection with Replacement A. Permutations: n B. Combinations: C(n+r-1,r) or C(n+r-1,n-1) V. Selection without Replacement A. Permutations: P(n,r) B. Combinations: C(n,r) VI. Pigeonhole Principle If (k+1) or more objects are placed in k boxes, then there is at least one box containing two or more objects. 1

3 Counting Product Rule A B C = A * B * C Examples Example 1: A = {1,2,3} B = {a,b} C = {a,b,g,d} How many 3-character words with 1st character from A, 2nd from B, 3rd from C? Answer: A * B * C = 3*2*4 = 24 Example 2: 3 Independents, 2 Democrats, 4 Republicans How many ways to form 3-person committee with one member from each party? Method: Count number of triples <Independent, Democrat, Republican> Answer: 3*2*4 = 24 Union Rule A B = A + B - A B Example S = {a,b,c} How many 3-letter words in S * start with a or end in b? Method: Let A = 3 letter words starting with "a" Let B = 3 letter words ending with "b" Find: A B Answer: A = (1) (3) (3) = 1*3*3 = 9 B = (3) (3) (1) = 3*3*1 = 9 A B = (1) (3) (1) = 1*3*1 = 3 A B = = 15 2

4 Inclusion/Exclusion Principle To calculate A 1 A 2 A n, 1. Calculate size of all possible intersections 2. Add results from of odd numbered sets. 3. Subtract results from of even numbered sets. Ie. A 1 A 2 A 3 = A 1 + A 2 + A 3 - A 1 A 2 - A 1 A 3 - A 2 A 3 + A 1 A 2 A 3 Example: How many integers in {1,,1000} are divisible by either 2, 3, or 5? A = divisible by 2 B = divisible by 3 C = divisible by 5 Find: A B C A = {2,4,,1000} = 500 B = {3,6,,999} = 333 C = {5,10,,1000} = 200 A B = {6,12,18,,996} = 166 A C = {10,20,,1000} = 100 B C = {15,30,,990} = 66 A B C = {30,60,,990} = 33 So, A B C = = 734 Selection with Replacement Permutations (Order Matters), with replacement: n r Choose r times from a set of n items with replacement and order matters. Drawing with replacement (sequence can repeat items): If there are n distinct objects and we draw r times. Each drawing has n distinct possibilities. If order matters, then by the product rule we have n x n x... x n possibilities = n r. 3

5 Examples Example 1: Create a sequence of 10 numbers by rolling a die 10 times. How many possible sequences? Method: Choose 10 times from a set of 6. Answer: 6 10 Example 2: S = {0,1,,7} How many strings of length 5 are there in S *? Method: Choose 5 times from a set of 8. Answer: 8 5 Combinations (Order Does Not Matter), with replacement: C(n+r-1,r) Choose r times from a set of n items with replacement and order does not matter. The number of possible outcomes is written as C(n+r-1,r). Combinations are discussed in more detail later. Example: You have 4 different kinds of cookies. How many different ways can 6 cookies be chosen? Method: Choose 6 times from a set of 4 with replacement and order does not matter. Answer: C(4+6-1,6) = C(9,6) = 9*8*7 3*2*1 = 84 Selection without Replacement Permutations (Order Matters): P(n,r) Choose r times from a set of n items without replacement and order matters. If we do not have replacement but order matters: Then first drawing has n possibilities; the second drawing has (n-1), the third has n! (n-2), etc. By the product rule we have: n x (n-1) x (n-2) x... x (n-r+1) = (n r)!. This is usually called permutations of n things taken r at a time. The number of n! possible outcomes is P(n,r) = (n r)! 4

6 Examples Example 1: Six CD's with 57 songs. CD player can be programmed to play any 20 songs in any order. How many ways can 20 different songs be played? Method: Choose 20 from 57 without replacement, order matters Answer: P(57,20) = 57! 37! Example 2: Given n integers x 1, x 2,, x n, all distinct. How many sorted arrangements are possible (i.e. how many permutations of x 1 x n?) Method: Choose all n items from n, without replacement, order matters n! Answer: P(n,n) = (n n)! = n! 0! = n! Example 3: S = {a,b,c,d,e,f} How many 3-letter words are in S * which have no character repeated? Method: Choose 3 from 6 without replacement and order matters. Answer: P(6,3) = 6! = 6*5*4 = 120 3! Combinations (Order Does Not Matter) without replacement: C(n,r) Choose r times from a set of n items without replacement and order doesn't matter. If order does not count and there is no replacement, then we must determine how many of the permutations are the same objects but only differ in order. That is: In how many orders can n distinct objects be arranged? P(n,n) = n! = n!. Therefore, if order does not matter and we do not have 0! replacement, the number of possibilities for n objects taken r at a time is P(n,r) P(r,r) = n!. This is usually referred to as combinations of n things taken r at a time (n r)!r! = C(n,r). There is another important type of combination, combinations with replacement. The sequence can repeat, but order does not matter. (Discussed Earlier) Number of ways to choose r from n is C(n,r) = n! (n r)!r! 5

7 Example How many ways can you choose a committee of 5 from 7 people? 7! Answer: C(n,r) = 2!5! = 7*6 2 = 21 Summary of Permutations and Combinations Type (n things, r times) Order Counts Replacement Symbol Expression Permutations w/ repl. Yes Yes None n r Permutations - no repl. Yes No P(n,r) n! (n-r)! Combinations - no repl. No No C(n,r) n! (n-r)! r! Combinations - w/ repl. No Yes C(n+r-1,r) (n+r-1)! r! (n-1)! More Examples Example 1: Place n indistinguishable objects into r distinguishable boxes. Number of possible outcomes is C(n+r-1,r-1) (n+r-1) bits, (r-1) ones: C(n+r-1,r-1) Example 2: 12 identical flyers are to be placed into 4 mailboxes. How many ways can this be done? Answer: C(12+4-1,4-1) = C(15,3) = 15! 15 * 14 * 13 = = !3! 3*2*1 What if each mailbox must receive at least 2 letters? Method: Put 2 letters in every box. Then, how many ways are there to distribute remaining 4 among 4 mailboxes? 7! Answer: C(4+4-1,4-1) = C(7,3) = 4!3! = 7*6*5 3*2*1 = 35 6

8 Example 3a: 15 basketball players are to be drafted by 3 professional teams (A,B,C). Each team will draft 5 players. In how many ways can this be done? Method: Choose 5 from 15 for A: C(15,5), 5 from remaining 10 for B: C(10,5), C has remaining 5: C(5,5) = 1 (multiply all three together). This determines all the ways we can partition the players where A chooses first, B second, and C last. Answer: C(15,5)*C(10,5) Example 3b: Now, if we cannot tell which team is which, we say the teams are indistinguishable. If this is the case, how many ways can 15 players be partitioned into 3 teams of 5 each? (i.e. - we want to know how many different sets {A,B,C} there are. In 3a, we could have one division with (A,B,C) = ({1,2,3,4,5},{6,7,8,9,10},{11,12,13,14,15}) and another with (A,B,C) = ({6,7,8,9,10},{1,2,3,4,5},{11,12,13,14,15}). But these two examples divide the fifteen players up into the same groups.) C(15,5) *C(10,5) Answer: (since the teams are indistinguishable) 3! Example 4: In how many ways can 2n people be divided into n pairs without replacement and order does not matter? C(2 n, 2 ) * C(2 n 2,2) *... * C(2,2) Answer: n! Example 5: In how many ways can 10 boys and 5 girls stand in a line if no 2 girls can stand next to each other? Boy1 Boy2... Boy10 Method: Choose one slot for each girl (Select 5 slots from 11, without replacement and order matters.). Answer: There are 10! ways to place the boys and P(11,5) ways to place the girls, so by the product rule the answer is 10! * P(11,5). Example 6: In how many ways can 2 distinct numbers be selected from 1,2,...,100 so that their sum is even? Answer: (2 Odds) C(50,2) + (2 Evens) C(50,2) = 50*49 Example 7: In how many ways can 2 distinct numbers be selected from 1,2,...,100 so that their sum is odd? Answer: Choose 1 even and 1 odd = C(50,1) *C(50,1)= 50*50 7

9 Pigeonhole Principle If (k+1) or more objects are placed in k boxes, then there is at least one box containing two or more objects. Examples Example 1: If you have a group of 367 people, there must be at least two people who have the same birthday (there are only 366 possible birthdays). Example 2: If you have a group of 27 English words, there must be at least two words that begin with the same letter (there are only 26 possible letters in the English language). 8

10 Graph Information I. Basic Definitions II. Euler Circuit and Fleury's Algorithm III. Hamilton Cycle IV. Kruskal's Algorithm V. Matrix Representation of Graphs VI. Hasse Diagrams 9

11 Graphs Basic Definitions path: (way of getting from one vertex to another via edges) - sequence of vertices in which successive vertices are joined by an edge length of path: number of edges in path Example paths: uvwxy (length 4) uxy (length 2) uxwustxzy (length 8) closed path: first vertex = last vertex cycle: closed path in which all vertices are distinct, except first = last acyclic graph: graph with no cycle Example closed path: xyztx closed path: swxtzyxs closed path: uvwvu acyclic graph: also cycle not cycle not cycle 10

12 A graph is connected if every pair of vertices is joined by a path. Example Connected Not connected The degree of a vertex is the number of edges touching the vertex. (loops count twice) Example Vertex Degree u 2 v 2 x 4 y 4 z 4 11

13 Euler Circuit and Fleury's Algorithm Euler circuit: closed path which uses each edge exactly once Can show there is an Euler circuit if and only if every vertex has even degree. ("easy" to find) Fleury's algorithm will always find Euler circuit in connected graph if every vertex has even degree. Fleury's Algorithm Start at any vertex v While (still more edges incident with v) do - if possible choose edge vw which does not disconnect - else choose only vw and delete v - delete edge vw - v := w end do Example of Fleury's Algorithm 12

14 etc. Euler circuit: Hamilton Cycle Hamilton cycle: cycle which uses each vertex once "hard" problem (version of Traveling Salesman Problem) Example Hamilton cycle: uvtwxysu 13

15 Kruskal's Algorithm Minimum cost network to connect sites. Given sites and link costs Kruskal's Algorithm Start with no edges. Examine edges in non-decreasing order of weight. If an edge does not create a cycle, add it. Example of Kruskal's Algorithm Matrix Representation of Graphs Label vertices v 1, v 2,, v n Adjacency matrix M: nxn matrix with M[x,y] = number of edges joining v x and v y. 14

16 Examples of Adjacency Matrices Graph isomorphism: Graphs G and H are isomorphic if there is some way to label the vertices of H so that H is G (ie, has same adjacency matrix). Examples 15

17 Suppose M is the adjacency matrix of a graph G. Can prove by induction: M k [i,j] = number of length-k paths joining i and j Reachability Matrix R 16

18 Fourth Hour Exam Study Sheet I. Counting A. Basic Rules (Product, Union, Inclusion/Exclusion) B. Selection with Replacement (Permutations, Combinations) C. Selection without Replacement (Permutations, Combinations) D. Pigeonhole Principle II. Graphs A. Basic Definitions B. Euler Circuit and Fleury's Algorithm C. Hamilton Cycle D. Kruskal's Algorithm E. Matrix Representation of Graphs F. Hasse Diagrams 17