1 / 27

Counting

Counting. Often in CS we need to Enumerate/account for all possibilities of something, such as: generating passwords exhaustive testing of an algorithm Calculate a probability In general, there are two ways to count:

nakia
Download Presentation

Counting

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Counting • Often in CS we need to • Enumerate/account for all possibilities of something, such as: • generating passwords • exhaustive testing of an algorithm • Calculate a probability • In general, there are two ways to count: • Chicken method  We’ll get an answer eventually, but we won’t necessarily understand why • Analytically  but we need to learn some rules

  2. Counting outline • Some types of problems to address • Definition of probability • Based on some “experiment” or “event” that has a set of outcomes called our sample space. • P = (# favorable outcomes)/ (total in sample space) • Multiplication rule, i.e. fundamental counting principle • Number of possible phone numbers or license plate numbers • Permutations • Combinations • *** See handout for details • Common technique required: break a problem down into distinct cases.

  3. Multiplication rule • If you need to make consecutive (or independent) choices, just multiply. • If you have x ways of making one decision, and y ways of making a second decision, the total number of outcomes is xy. • We can generalize to a “lining up” situation. If you have x items that need to be arranged in some order, this can be done in x! ways. • Sometimes we have different cases to add. • E.g. At 3:00, you may choose from either the lunch or dinner menu. Do lunch and dinner separately, then add.

  4. Distinguishable • The number of ways to line up n objects, but a of them are identical, b of them are identical, etc. = n! / (a! b! …) • Think of 3 people sitting in a row of 3 chairs versus 5 chairs. • Examples: How many ways can we arrange the letters in the following words? • NEEDED 6! / (3! 2!) • DIVISIBILITY 12! / 5! • AAABBB 6! / (3! 3!)

  5. Handling restriction • (This is a permutation problem with a twist.) • We need to select a president, vice president and secretary from 10 people. However, A can’t be president and B cannot be secretary. How many ways can this be done? • Focus on cases that are legal: • B is president, A is VP, anyone else can by sec’y • B is president, A is secretary, anyone else can be VP • B is VP, A is sec’y, anyone else can be president • Any more?

  6. Combinations • Definition • Contrast with permutations • Evaluating • Applications • Practice

  7. When to use • Questions dealing with making some selection, membership, inclusion, presence vs. absence…. In which order does not matter • Combinations answer this type of question: A set has n elements. How many subsets contain r elements? • Try examples when n = 4. • Contrast with permutations: • How many ways can we select 3 letters from the alphabet? • How many 3 letter words can we make?

  8. Points to remember • Order doesn’t matter • Subsets, committees • What matters is whether something is chosen or not. • Evaluation: • The number of ways to choose r items out of a set of n is n! / (r!(n – r)!). • Shorthand notation: C(n, r) or • Shortcut to evaluate: C(10, 3) = 10 * 9 * 8 / (3 * 2 * 1) • Incidentally: choosing r is the same as (not) choosing n – r.

  9. Examples • How many 8-bit representations have exactly three 0’s? • Choose which positions get a 0: choose 3 out of 8: C(8, 3). • Six people need to be divided into two teams A and B. How many ways can this be done? • Choose 3 to go to team A: C(6, 3). • Choose 3 remaining people for B: C(3, 3) = 1. • What is the probability that exactly 4 out of 7 coin tosses will be heads? • Choose which 4 occurrences are heads: C(7, 4). • Divide by the total size of sample space: 27.

  10. Watch the wording • Sometimes we want “at least” or “at most” • Both • Neither • Be careful with “or”. There could be overlap. • King or club • Number or red • Are objects/categories distinguishable? • “two teams” may be indistinguishable • Selecting cards from a deck: almost always the order that we deal cards does not matter

  11. Indirect probability • Sometimes it’s easier to find the probability that an even will NOT happen, and then subtract from 1. • Ex. Suppose we roll 5 dice, as in Yahtzee. • P(all sixes) = 1/7776 • P(didn’t get all sixes) = 1 – 1/7776 • P(1000 tries, didn’t get all sixes) = (1 – 1/7776)1000 • P(1000 tries, got all sixes at least once) = 1 – (1 – 1/7776)1000 • Ex. The “birthday problem.” In a group of n people, what is the probability that at least two of them have the same birthday? What is n if p > ½?

  12. Poker hands • How many straight flushes are there? • The high card can be anything from 5 up to Ace. • All cards of the same suit, so select a suit. • How many 4-of-a-kind hands? • Choose which denomination you want 4 of. • Choose those 4 cards. • Choose the 5th card. • Full house? • Choose two denominations. • Choose three of one and two of the other.

  13. More counting problems • Indistinguishable subsets • Ball-in-urn: the objects being selected are identical • Pascal’s triangle, a combinatorical argument • Stirling numbers

  14. Indistinguishable subsets • How many ways can 4 people divide into 2 teams of 2, as in a doubles match? • The teams are the same size and don’t have inherent name, place, uniform, etc. • AB/CD, AC/BD, AD/BC • Is the answer 3 or 6? • How many ways to divide … ? • 20 people into 4 teams of 5 • 20 people into teams of 3, 4, 5, 8 • 20 people into teams of 4, 5, 5, 6

  15. Ball in urn • The number of ways to distribute • n distinct objects into r categories = rn • n identical objects into r categories = C(n + r – 1, n) • Give 4 cards to 2 people. • If face up, they are distinct. Each card has 2 possible destinations  24. • If face down, then the only thing that matters is how many cards you have. Your cards and my cards are separated by a divider CARD CARDCARD / CARD 0 0 0 1 0 2 categories  1 divider that can go anywhere: answer = C(5, 1) or C(5, 4).

  16. Ball in urn (2) • Give 5 cards to 3 people. • If distinct, ask each card where it’s going: 35. • If identical: CARD CARD / CARD CARD / CARD 0 0 1 0 0 1 0 3 categories  2 dividers that can go anywhere Total = C(7, 2) • 3 divers find 5 gold coins. How many ways to share? • A case of 12 bottles of wine to distribute among 4 tables.

  17. Ball in urn (3) • What if we must give each table at least 1 bottle of wine? • We have less freedom to choose. • Only 8 bottles can be freely distributed. • Let’s buy 7 cans of soup. 3 varieties are available.

  18. Ride tickets • Buy 10 ride tickets at the fair. There are 3 possible rides. • If the issue is how many times you want to go on each, this is a ball in urn problem  C(10 + 3 – 1, 2) • If order matters  310 • If the tickets say we must go in haunted house 4 times, roller coaster 3 times and ferris wheel 3 times, and order matters. It’s a combination question. Choose when you do each type of ride  C(10, 4) * C(6, 3) * C(3, 3). • If the tickets are bought 4x, 3x and 3x, and order doesn’t matter  1. • *** More examples from book and handout.

  19. Understanding C(n, r) • Let’s justify: C(5, 2) = C(4, 1) + C(4, 2) • We want to select two elements from { A, B, C, D, E }. • Is E part of the selection? • If so, then we pick one of the others: C(4, 1) • If not, then we pick both from the others: C(4, 2). • This line of reasoning works in general for C(n, r). • Now that we understand Pascal’s triangle, let’s look at another one. 

  20. What good is Pascal’s Δ • Combinatorical arguments • C(n, r)  numbers in Pascal’s triangle  • Stirling numbers • Recap important ideas • When to use formulas: n!, P(n, r), C(n, r), C(n + r – 1, n) • Go over practice handout

  21. Stirling numbers • S(n, r) = the number of ways to break up a set of n elements into r groups. • If you study these numbers in detail, you can solve more complex counting problems, and determine coefficients of more Bernoulli-type summations!

  22. Starting simple • Let’s consider S(3, 2). How many ways can we split up a 3-element set { A, B, C } into two groups? Note: their sizes must be 1 and 2. • A / BC • B / AC • C / AB • Consider S(3, 3). Splitting { A, B, C } into 3 groups. • A / B / C • Next, let’s work out various cases of S(5, r) for 1  r  5.

  23. S(5, r) • S(5, 1): We need to “split up” a set of 5 elements into a single group. Degenerate case: we want 1 group of size 5. Answer = 1. • S(5, 2): Split into 2 groups • Their sizes could be 2 and 3. In this case, we have groups like A B / C D E. Select which 2 go in one group: C(5, 2). • Their sizes could be 1 and 4. In this case, we have groups like A / B C D E. Select 1 to go into the first group: C(5, 1). • Answer is the sum: 10 + 5 = 15.

  24. S(5, r) continued • S(5, 3): We need to split into 3 groups • The sizes could be 1, 1, 3. For example A / B / CDE. Think of it in terms of distinguishable permutations. The 5 elements can be ordered in 5! ways. But the 3 elements in the size-3 group have no inherent order, so we have to divide by 3!. Also, the two groups of size 1 are indistinguishable, so we divide by another 2!. 5!/(3!2!) = 10 • The sizes could be 1, 2, 2. For example A / BC / DE. The total number of permutations is 5!. But the 2 groups of 2 are indistinguishable. Also, the pairs of values inside each group of 2 are indistinguishable. 5!/(2!2!2!) = 15. • Total = 25.

  25. S(5, r) continued • S(5, 4): We need to split into 4 groups • The sizes could only be 1, 1, 1, 2. The two positions inside the size-2 group are not distinguishable, and the 3 groups of size 1 are not distinguishable. 5!/(2!3!) = 10. • S(5, 5): pulverize the set of 5 into singleton sets. Just one way to do it. (You can think of all 5 size-1 groups as being indistinguishable, and 5!/5! = 1) • So, row 5 of Stirling’s triangle is: 1, 15, 25, 10, 1

  26. General approach • It’s a bit tedious to keep evaluating each S(n, r) from scratch in terms of every possible case of group size. • Let’s make use of earlier/lower values of S. • Consider S(6, 3). Split { A, B, C, D, E, F } into 3 parts. • When we partition into groups, we may have F by itself. • F / (2 more groups of A-E). Answer = S(5, 2). • After partitioning A-E, just put F into any current group. There are 3 to choose from. Answer = 3 * S(5, 3) • Add it up: S(5, 2) + 3 * S(5, 3) • In general: S(n, r) = S(n – 1, r – 1) + r * S(n – 1, r)

  27. Using the formula • How can we use the Stirling formula to quickly determine S(n, r) for higher values of n? • Cell value = NW + N * (column number) • Observe 90 = 15 + 25*3 and 65 = 25 + 10*4

More Related