1. CS 415 – A.I. Slide Set 12

2. Chapter 5 – Stochastic Learning Heuristic – apply to problems who either don’t have an exact solution, or whose state spaces are prohibitively large Stochastic Methodology – also good for these situations  Based on counting the elements of an application domain

3. Addition and Multiplication Rule Set A |A| - cardinality of A (number of elts)‏  A could be: empty, finite, countably infinite, or uncountably infinite U – Universe (a.k.a. Domain)‏  The set of ALL elements that could be in A A’ - Compliment Example  U – people in a room  A – males from U  A’ – females in the room

4. Other Notations Subset, Union, Intersection

5. Permutations and Combinations Permutation – an arranged sequence of elements of a set (each used only once)‏  Question: how many unique permutations are there of a set of size n?  n * n-1 * n-2 * n-3 * … * 1  Question: how many ways can we arrange a set of 10 books on a shelf where only 6 books can fit?  nPr

6. Combination – Any subset of the elements that can be formed  Question: How many combinations given a set of items?  1 combination for n elements  Order DOES NOT MATTER  Question: How many combinations taken r at a time (How many ways can I form a four person committee from 10 people?)

7. Elements of Probability Theory

8. Examples What is the probability of rolling a 7 or 11 from two fair die?  Sample Space Size?  36  Event Size?  8   For 6: 1,6; 2,5; 3,4; 4,3; 5,2; 6,1  For 11: 10,1; 1,10  Probability  8/36 = 2/9  Add them together because they are “independent”

9. How many four-of-a-kind hands can be dealt in all possible five card hands?  Sample Space?  52 cards taken 5 at a time  Event Space?  Multiply number of combinations of 13 cards 1 at a time * Combination of 4 taken 4 at a time * 48  (number of different kinds of cards) * (number of ways to pick all four cards of same kind) * (times the number of ways the fifth card can be chosen)‏  See top of pg 172  Approx. 0.00024

10. The probability of any event E from the sample space S is: The sum of the probabilities of all possible outcomes is 1 The probability of the compliment of an event is The probability of the contradictory or false outcome of an event Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 8

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12. Are Two Events Independent? Random bit strings of length four 2 Events 1. String has even number of ones 2. Bit string ends with a zero A total of 24=16 bit strings – 8 strings end with zero – 8 strings have even number of ones Are independent events

13. All of Probability Theory In a Nut Shell

14. Probabilistic Inference: Example 3 boolean random variables  All either true or false  S – traffic is slowing down  A – probability of an accident  C – probability of road construction  Given state traffic data in Table 5.1  Next slide  Note: all possibilities sum to 1  Can use these numbers to calculate  Probability traffic slowdown  Probability of construction without slowdown, etc.

15. Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 Table 5.1The joint probability distribution for the traffic slowdown, S, accident, A, and construction, C, variable of the example of Section 5.3.2 Fig 5.1 A Venn diagram representation of the probability distributions of Table 5.1; S is traffic slowdown, A is accident, C is construction. 11

16. Random Variables Individual Probability Computation 1. Combinatorical Methods (Analytical)‏  Ex: Probability of rolling a 5 on a 6-sided die 2. Sampling Events (Empirical)‏  For times when it isn’t that simple to analyze  Assumptions  Not all events are equally likely  (Easier if they are)‏  Probability of event lies between 0 and 1  Probabilities of union of sets still holds  Use “Random Variables”

17. Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 12

18. Random Variable Example Random variable – Season  Domain – {spring, summer, fall, winter} Discrete Random Variable  p(Season=spring) =.75 Boolean Random Variable  p(Season=spring) = true

19. Expectation Expectation – the notion of expected payoff or cost of an outcome

20. Example A fair roulette wheel  Integers 0 – 36 equally spaced  Each player places $1 on any number  Wheel stops on that number, wins $35  Else – loses the $1 Reward of win $35  Probability 1/37 Cost of loss $1  Probability 36/37 Ex(E) = 35(1/37) + (-1)(36/37) = -0.027

21. Conditional Probability 2 kinds of Probablities 1. Prior Probabilities  What’s the probability of getting a 2 or a 3 on a fair die? 2. Conditional (Posterior) Probabilities  If a patient has system X, Y and Z then what is the probability that he has the flu?

22. Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 14

23. Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 Fig 5.2A Venn diagram illustrating the calculations of P(d|s) as a function of p(s|d). 15

24. The chain rule for two sets: The generalization of the chain rule to multiple sets Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 We make an inductive argument to prove the chain rule, consider the nth case: We apply the intersection of two sets of rules to get: And then reduce again, considering that: Until is reached, the base case, which we have already demonstrated. 16

25. Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 17

26. Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 18

27. You say [to ow m ey tow] and I say [t ow m aa t ow]… - Ira Gershwin, Lets Call The Whole Thing Off Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 Fig 5.3A probabilistic finite state acceptor for the pronunciation of “tomato”, adapted from Jurafsky and Martin (2000). 19

28. Phoneme Recognition Problem Use the Tomato style stochastic finite state acceptor  Interpret ambiguous collections of phonemes  See how well the phonemes match the path through the state machine for this and other words phoneme is the smallest structural unit that distinguishes meaning. Phonemes are not the physical segments themselves, but, in theoretical terms, cognitive abstractions or categorizations of them.  the /t/ sound in the words tip, stand, water, and cat

29. Phoneme Recognition Problem Suppose an algorithm has identified the phoneme /ni/  Occurs just after other recognized speech, /l/ Need to associate phoneme with either a word or the first part of a word How?  Brown corpus  1 Million word collection of sentences from 500 texts  Switchboard corpus  1.4 Million word collection of phone conversations  Together: ~2.5 Million words that let us sample written and spoken words

30. How to proceed? Can determine which word with the phoneme is most likely used  See Table 5.2  Most likely, “the”

31. Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 Table 5.2 The ni words with their frequencies and probabilities from the Brown and Switchboard corpora of 2.5M words, adapted from Jurafsky and Martin (2000). 20

32. Use Bayes’ theorem  p(word | [ni]) = p([ni]|word) x p(word)‏  See Table 5.3  Most likely, “new”  “I new” doesn’t make sense  “I need”, does

33. Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 Table 5.3 The ni phone/word probabilities from the Brown and Switchboard corpora (Jurafsky and Martin, 2000). 21

34. Bayes’ Theorem Review: one disease and one symptom  Individual hypotheses, hi  Each is disjoint  Set of hypotheses, H  Set of evidence, E P(hi|E) = (p(E|hi) x p(hi))/p(E)‏ Can use this to determine which hypothesis is strongest given E  Drop the denominator  Arg max (maximum likelihood) hypothesis

35. Finding p(E)‏ Given: entire space is partitioned by the set of hypotheses hi  Partition of a set = split of set into disjoint subsets p(E) = Σip(E|hi)p(hi)‏

36. Bayes’ Theorem : General Form The general form of Bayes’ theorem where we assume the set of hypotheses H partition the evidence set E :

37. Example Suppose you want to buy a car  Prob go to dealer 1, d1  Prob purchasing a car at d1, a1 Necessary for using Bayes’  All probabilities with various hi must be known  All relationships between evidence and hypothesis {p(E|hi)} must be known

38. The application of Bayes’ rule to the car purchase problem: Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 23

39. Naïve Bayes, or the Bayes classifier, that uses the partition assumption, even when it is not justified: Luger: Artificial Intelligence, 5th edition. © Pearson Education Limited, 2005 24