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Combinatorics

Combinatorics. Introduction Combinatorics is the study of the arrangement of discrete objects Who Cares? Counting is important whenever we have finite resources Examples: Words of memory needed to store … Instructions needed to solve … Number of hits per second supported by …

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Combinatorics

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  1. Combinatorics

  2. Introduction • Combinatorics is the study of the arrangement of discrete objects • Who Cares? • Counting is important whenever we have finite resources • Examples: • Words of memory needed to store … • Instructions needed to solve … • Number of hits per second supported by … • A password must be between six and eight characters long. The characters can be a digit or a letter (case sensitive). Each password must include at least one digit. How many passwords can we support?

  3. Useful Counting Cheat Sheet • Summation Notation • Summation Properties • Summation Formulas

  4. Basic Counting Principles • Multiplication Principle • If there are n1 possible outcomes for a first event and n2 possible outcomes for a second even, then there are n1 n2 possible outcomes for the sequence of the two events. • Extends (by induction) to apply to a sequence of n events • Examples • How many different bit strings are there of length seven? • If A & B are finites sets, then |A  B| = _______. • How many possible 7 digit phone numbers are there? • What if you can’t use a digit more than once?

  5. Basic Counting Principles • Addition Principle • If A and B are disjoint events with n1 and n2 possible outcomes, then the total number of possible outcomes for event “A or B” is n1 + n2. • Extends (by induction) to n disjoint events • Examples • How many ways are there to elect an ACM president if the president must be a CS faculty member (from 7) or a CS major (from 250)? • If you have two disjoint sets, A & B, |A  B| =_________. • How many 7 digit phone numbers begin with 441 or 781?

  6. Basic Counting Principles • Practice Example • US Social Security Numbers are 9 digits (0..9) • How many Social Security numbers are there? • How many are even? • How many have all even digits? • How many are palindromes? • How many have none of their digits equal to 8? • How many have at least one digit equal to 8? • How many have exactly one 8?

  7. Basic Counting Principles • Inclusion-Exclusion Principle • If you have two arbitrary sets, A & B, • |A  B| = |A| + |B| - |A  B| • Example: How many bit strings of length eight start with a 1 or end with the two bits 00? • Answer: 27 + 26 – 25 = 160 • Subsumes the Sum Rule • Extends (by induction) to n arbitrary sets (see p. 228 of text) A B A–B B–A AB

  8. Example: Internet Address (IPv4) • Sample: 137.28.109.33 • Translation: 32-bits • 10001001 00011100 01101101 00100001 • Interpretation • Restrictions • 1111111 not available as netid of Class A network • All 0s or all 1s are not allowed as any hostid • How many possible computer Internet addresses? • Netid for Class A is 7 bits long, • Netid for Class B is 14 bits long, • Netid for Class C is 21 bits long Plus some additional classes

  9. Example: Internet Address • How many possible computer Internet addresses? • Class A: ((27 – 1) * (224 – 2)) + • 1111111 prohibited as netid • Class B: (214 * (216 – 2)) + • all 0’s and all 1’s prohibited • Class C: (221 * (28 – 2)) • = 3,737,091,842.

  10. Permutations • An ordered arrangement of n elements from a set • An r-permutation is an ordered arrangement of r elements taken from a set of n elements • The number of r-permutations of n items P(n,r) • P(n,r) = n! / (n – r)! • Examples: • How many ways are there to select a 1st, 2nd, and 3rd place winner from 100 different people? • How many 7 digit phone numbers exist where the numbers can’t be repeated? • Notice: • P(n,n) = n! • P(n,0) = 1 • P(n,1) = n

  11. Combinations • An unordered arrangement of n elements from a set • unordered means order does not matter! • An r-combination is an unordered arrangement of r elements taken from a set of n elements • The number of r-combinations of n items C(n,r) • C(n,r) = n! / r!(n – r)! = P(n,r) / r! • Examples: • Quality Control wants to test 25 microprocessor chips from the 300 manufacturers. In how many ways can this be done? • How many ways can I select 5 different fruits from choice of 10? • Notice: • C(n,n) = 1 • C(n,0) = 1 • C(n,1) = n • C(n,r) = C(n,n-r)

  12. Repetition • Thus far we have assumed that items can not be used twice in one draw • What if they can? • Letters in a password • Selecting from bag of items with copies • Permutations with repetition • PR(n,r) = nr • How many 7 digit phone numbers are possible? • Combinations with repetition • CR(n,r) = C(r+n–1, r) = (r+n-1)! / [r!  (n-1)!] • How many ways can I select 7 pieces of fruit from a selection of apples, oranges, and pears?

  13. Repetition • How many different strings can be made by reordering the letters of the word … • “FAILURE”? • “SUCCESS”? • We have three Ss, two Cs, one U, and one E. • Notice that the three Ss can be placed into the seven positions in C(7,3) ways, leaving four blank positions • The Cs can be placed into those four positions C(4,2) ways, leaving two free positions. • The U can be placed into those two positions in C(2,1) ways, leaving one free position (allowing E to be placed in C(1,1) ways). • From the product rule: C(7,3) * C(4,2) * C(2,1) * C(1,1) which equals • (7! * 4! * 2! * 1!) / (3! * 4! * 2! * 2! * 1! * 1! * 1! * 0!) which equals • 7! / (3! * 2! * 1! * 1!) which equals 420

  14. Repetition • Permutations with indistinguishable objects • PI(n,n1,n2,…,nk) = n! / (n1!  n2!  …  nk!)

  15. Repetition • Example: How many ways are there to build four hands of 5 cards from the standard deck of 52 cards? • Solution I: Combinations of Objects • Use Combinations for each player’s hand • Use Product Rule for total • C(52,5) * C(47,5) * C(42,5)* C(37,5) = 52! / (5!5!5!5!32!) • Solution II: Permutations of Indistinguishable Objects • 5 indistinguishable objects of each of four types (one for each hand) • 32 indistinguishable objects of a fifth type (undealt) • 52!/(5!5!5!5!32!)

  16. Probability • An area of mathematics that has its origin in counting • Developed to study gambling games • circa 1650 by Pascal • generalized circa 1812 by Laplace • Finite Probability • experiment – a procedure that yields one of a given finite set of possible outcomes • sample space – the finite set of possible outcomes • event – a subset of the sample space • The probability of an event E, which is a subset of a finite sample space S of equally likely outcomes, is p(E) = |E| / |S|

  17. Probability Examples • What is the probability of picking a winning combination of six numbers out of 40 (no repeats)? • S : set of all six-number sequences taken from 40 • E : the target number • |S| = C(40,6) = 3,838,380 • |E| = 1 • p(E) = 1/3,838,380 (approx. 0.00000026) • What is the probability that a hand of five cards in poker contains four of a kind? • S : set of all possible poker hands • E : set of all possible four-of-a-kind hands • |S| = C(52,5) = 2,598,960 • |E| = # of ways to pick rank * # of ways to pick other card • = C(13,1) * C(48,1) = 624 • p(E) = 624 / 2,598,960 (approx. 0.00024) product rule

  18. Probability • Properties • 0  p(E)  1 • p(S) = 1 • p(E) = 1 – p(E) • p(E1  E2) = p(E1) + p(E2) – p(E1  E2) • disjoint events • p(E1  E2) = 0 • p(E1  E2) = p(E1) + p(E2) • p(E2 | E1) = p(E1  E2) / p(E1) • independent events • p(E2 |E1) = p(E2) • p(E1  E2) = p(E1)  p(E2) Each outcome “equally likely” implies p(a) = 1/n where a S and |S| = n. Called uniform distribution Conditional Probability

  19. Probability Practice • What is the probability of drawing either an ace or a spade from a standard deck of cards? • What is the probability of drawing 2 aces in a row? • What is the probability of drawing 2 aces in a row (with replacement)?

  20. Expected Value • First an example • Find the average height of four students with heights 68, 72, 67, and 74 inches. • Avg height = (68+72+67+74)/4 = 70.25 inches • Find the average height of 1000 students with the following heights • Avg height = • 66(.2)+68(.25)+71(.3)+72(.1) • +74(.15) = 69.8 inches

  21. Expected Value • Definitions • Suppose we have a probability space with the sample space • {e1, e2, …, en} • A random variablef (ei) on the sample space is a real number associated with each outcome ei • The expected value (or average) of f (ei) is given by

  22. Expected Value • Example • Suppose we have an asymmetrical 6-side die: • p(1) = 0.2 p(2) = 0.15 p(3) = 0.1 • p(4) = 0.25 p(5) = 0.12 p(6) = 0.18 • A random variable on this sample space is the number written on the side • The expected value of this random variable is • 1(.2) + 2(.15) + 3(.1) + 4(.25) + 5(.12) + 6(.18) = 3.48 • A random variable on this sample space is the mod 2 function of the number on the side. • The expected value of this random variable is • 1(.2) + 0(.15) + 1(.1) + 0(.25) + 1(.12) + 0(.18) = .42

  23. Application: Monte Carlo Techniques • Deterministic algorithms don’t always exist or aren’t always practical • Use random sampling to compute results • Example: Value of  • Let P = set of randomly selected points • (x,y) | 0  x  1  0  y  1 • Let S = { (x,y) | (x,y)  P  } • Let a = | S | / | P | •   4a

  24. Application: Monte Carlo Techniques • Example: Efficiency of backtracking algs • Generate a “typical” path in the tree using random choices, then estimate the number of nodes in the tree based on this path. • Let m0 be the number of feasible children of the root • Randomly generate a feasible node at level 1. Let m1 be the number of feasible children of this node. • Randomly generate a feasible node at level 2. Let m2 be the number of feasible children of this node. • . . . • Randomly generate a feasible node at level i. Let mi be the number of feasible children of this node. • . . . • Estimate of total number nodes checked: • 1 + t0 + m0t1 + m0m1t2 + … + m0m1 … mi-1ti+ …

  25. Application: Monte Carlo Estimate • intMonteCarloEstimate( ) { • Node v = root of state space tree; • while(v has children and is not a solution) { • count number of children of v • add number of children into total count using • 1 + t0 + m0t1 + m0m1t2 + … + m0m1 … mi-1ti+ … • store number of feasible children of v • v = randomly selected feasible child of v • } • return total count • }

  26. Practice Problems • Section 3.2: 3, 5, 6,12, 52 – 56, 69 - 70 • Section 3.3: 7, 10, 12 • Section 3.4: 4, 5, 11, 12, 20 – 23, 29 – 37, 65 • Section 3.5: 17 – 24, 29 – 38, 44

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