CSE 21 Mathematics for Algorithm and System Analysis

Transcription

1 CSE 21 Mathematics for Algorithm and System Analysis Unit 1: Basic Count and List Section 3: Set CSE21: Lecture 3 1

2 Reminder Piazza forum address: e Notes on the course website will be updated after class. CSE21: Lecture 3 2

3 Review : Set Set: a collection of distinct objects where order does not matter. Cardinality of a set A: A. k-set: a set with k elements, also called a set with cardinality k. CSE21: Lecture 3 3

4 Review : Rule of Product and Sum When to use the Two Rules? Count the number of structures in a set and describe how to construct the structures in terms of subconstructions that are connected by and and or. If this leads to the construction of each structure in a unique way, then the Rules of Sum and Product apply. Which Rule to use? and means products if the two parts are independent; or means sums. CSE21: Lecture 3 4

5 Learning Outcomes By the end of this lesson, you should be able to Understand and apply the operations and algebraic rules of Sets. Understand the meanings of Binomial Coefficient and Multinomial Coefficients, and calculate their values. Understand how recursion works. CSE21: Lecture 3 5

6 Why do we need to learn them? Set operation and listing are useful in many real world problems. Recursion is one of the central ideas of computer science. CSE21: Lecture 3 6

7 Definition 3: Set Notation Two ways to describe a set Condition Form: S = {x condition(x)} Enumeration Form: S = {1, 2, 3} Membership x S: x is an element of S x S: x is not an element of S Subsets: A B: if every member of set A is also a member of set B. It is the same with B A. Empty set: CSE21: Lecture 3 7

8 Universal Set: U When dealing with sets, we usually have a universal set U in mind, and the sets we discuss are subsets of U. Universal Set Example U = Z = {all integers}, its subsets include: N = {0, 1, 2,...}: all natural numbers N+ = {1, 2, 3, } : all positive integers CSE21: Lecture 3 8

9 Set Equality Checking Two sets A and B are equal if and only if every element of A is an element of B and every element of B is an element of A. Examples: {a, b, c} = {a, b, c} {a, b, c} = {b, a, c} {a, b, c} = {a, b, b, c, b} CSE21: Lecture 3 9

10 Set Operation Intersection: A B = {x x Aand x B} Union: A B= {x x Aor x B} Set Difference: A \ B or A B = {x x Aand x B} Complement of A (relative to U): U \ A or A c or A = {x x A} = U A Symmetric Difference: A B= (A \ B) (B \ A). Cartesian Product (also called product): A B = {(x, y) x A, y B} CSE21: Lecture 3 10

11 Theorem 6: Algebraic rules for sets Name Intersection Union Associative (P Q) R = P (Q R) (P Q) R = P (Q R) Distributive P (Q R) = (P Q) (P R) P (Q R) = (P Q) (P R) Idempotent P P = P P P = P Double Negation P = P DeMorgan: (P Q) = P Q (P Q) = P Q Absorption P (P Q) = P P (P Q) = P Commutative P Q = Q P P Q= Q P CSE21: Lecture 3 11

12 Subset and Subset Number P(A) : the set of all subsets of A P(A) = 2 A : for each element of A we have two choices - include the element in the subset or not include it. P k (A) : the set of all subsets of A of size (or cardinality) k n P k (A) is denoted as C(n, k) or k, called binomial coefficients, are read n choose k. CSE21: Lecture 3 12

13 Theorem 7: Binomial coefficient formula n k n(n 1) (n k +1) k! =C(n, k) = = n! k!(n k)! k = n n n k Reason for the formula k-list number from n-set: (n) k =n!/(n-k)! k-list number from k-set: k! Based on Rule of Product (k-set number from n-set k-list number from k-set = k-list number from n-set), result is n! k!(n k)! CSE21: Lecture 3 13

14 Example of Binomial Coefficient Calculation 5 3 = Divide out common factor Generate 3-sets from a 5-set: {a, b, c, d, e} {a, b, c}, {a, b, d}, {a, b, e}, {a, c, d}, {a, c, e}, {a, d, e} {b, c, d}, {b, c, e}, {b, d, e} {c, d, e} 5! = 3!(5 3)! (1 2 3) (1 2) = 10 CSE21: Lecture 3 14

15 Example 15: Full House Card Hands A standard deck of cards contains 52 cards, each marked with two labels. The first label, called the suit, suits = {,,, } called club, heart, diamond and spade. The second label, called the value, belongs to set {2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A} A hand is a subset of a deck. Pair: two cards have the same value. Triple: three cards have the same value. How many 5 card hands consist of a pair and a triple (called full house in poker)? CSE21: Lecture 3 15

16 Solution of Example 15 To construct such a hand: Choose the value for the pair AND Choose the value for the triple different from the pair AND Choose the 2 suits for the pair AND Choose the 3 suits for the triple Based on Rule of Product, result is C(4, 2) C(4, 3) = 3,744 CSE21: Lecture 3 16

17 Example 16: Two Pair Card Hands How many 5 card hands consist of two pairs? A description of a hand always means that there is nothing better in the hand, so two pairs means we do not have a full house or four of a kind. Tricky part: make sure there is no duplications; the calculation similar to the last example is C(4, 2) C(4, 2) C(4, 1), this calculation counts each hand exactly twice, e.g.: {5, 4} for the values of the two pair is counted by <5, 4> and <4, 5>. Solution: divide the above number by 2 or use solution from the textbook. CSE21: Lecture 3 17

18 Example 18: Multinomial Coefficient Given k boxes (labeled 1 to k) and an n-set S How many ways to distribute the elements of S among the boxes so that the ith box contains exactly m i elements? (m m k =n) Fill the first box (C(n, m 1 ) ways) AND Fill the second box (C(n m 1, m 2 ) ways) AND Fill the kth box (C(n (m m k-1 ), m k ) = C(m k, m k ) = 1 ways) CSE21: Lecture 3 18

19 Solution of Example 18 By Rule of Product and formula C(n, k)= n!/k!(n-k)!, the result is: C(n, m 1 ) C(n m 1, m 2 ) C(n (m m k-1 ), m k )= n! m!( n m )! ( n m1)! m!( n m1 m... )! ( n ( m )!( n ( m Divide out common factor, final result is called multinomial coefficient ( m n n! = m1, m2,, mk m1! m2! mk! m m k 1 ))! + m k 1 2 k 1 k 1 + m + m 2 ))! CSE21: Lecture 3 19

20 Example 17: Rearranging MISSISSIPPI Question: How many ways to rearrange the letters in the word MISSISSIPPI? How to describe/get one way? Describe a placement of letters by assigning positions to each letter, e.g., I {2, 5, 8, 11}, M {1}, P {9, 10}, S {3, 4, 6, 7} means the word MISSISSIPPI. We can get a eligible result by distributing 4 elements of set {1, 2,, 11} into a box labeled as I, 1 element to box M, 2 to box P, 4 to box S. How to get all the ways? It becomes a multinomial coefficient problem. Result: 11 4,1, 2, 4 = 11! 4!1!2!4! = ( ) 1 (2 1) ( ) CSE21: Lecture 3 20

21 Example 20: Words from a collection of Question: letters How many words of length k can be formed from the letters in TOOL when no letter may be used more often than it appears in TOOL? The same with Example 10. Let m 1, m 2, m 3 be the numbers of T, O, L, respectively. It becomes a multinomial coefficient problem. List and add possible values for m i for each possible k. k=1: k=2: k=3: k=4: 1 0, 0,1 2 0, 2,0 3 0, 2,1 4 1, 2, ,1, ,1, , 2,0 = , 0, , 0, ,1,1 = ,1, 0 =12 = 7 CSE21: Lecture 3 21

22 Example 22: Card hands and multinomial coefficients Full house: multiply the following numbers Face values: one for the triple, one for the pair, and leave eleven face values unused: 13 1,1,11 Choose the suits for the triple: Choose the suits for the pair: Two pair Face values: two for the pairs, one for the single card, and leave ten face values unused: 13 2,1,10 The suits for the two pairs and the single card: CSE21: Lecture 3 22

23 Recursion of Binomial Coefficient How to compute C(n, k) from values of the function with smaller arguments? Let S = {x 1,..., x n }, C(n, k) is k-subset number of S. x n may be in k-subset, may be not. If x n is not in k-subset, the number is the same with taking x n out of S : C(n-1, k) If x n is in k-subset, the number is the same with getting (k-1)-subset and add x n to every subset: C(n-1, k-1) 1 By the Rule of Sum: C(n, k) = C(n-1, k) + C(n-1, k-1) CSE21: Lecture 3 23

24 Two Ways of Deriving Recursion Breaking down (Top-down): How can I break down the things I want to count into smaller things of the same type? Constructing (Bottom-up): How can I construct the things I want to count of a given size by using the same type of things of a smaller size? Note: no matter which way to choose, the recursion only works when we have some (smallest) values to start with. CSE21: Lecture 3 24

25 Example 23 : Deriving Recursions How to construct the things of a given size by using the same type of things of a smaller size? A complete recursion formula of binomial coefficient. C(0,0) = 1, C(0,k)=0 for k 0 and C(n,k)=C(n 1,k 1)+C(n 1,k) for n > 0; or C(1,0) = C(1,1) = 1, C(1,k) = 0 for k 0,1 and C(n,k)=C(n 1,k 1)+C(n 1,k) for n > 1; CSE21: Lecture 3 25

26 Example 23 : Deriving Recursions Tabular representation of C(n,k), called Pascal s Triangle CSE21: Lecture 3 26

27 Homework and Pre-Reading Assignment Homework: Exercise 3.2, 3.3, 3.4, 3.5, 3.6, in page CL-26 to CL- 27 For next class, please read remaining pages of Section 3 (CL-25 to CL-26) and first six pages of Section 4 (CL-28 to CL-34). Try to understand set partition definition and examples. Try to understand probability and Venn diagram through examples. CSE21: Lecture 3 27