Refinement Planning: Status and Prospectus

Refinement Planning: Status and Prospectus

9 Machine Learning: Symbol-based 9.0 Introduction 9.5 Knowledge and Learning 9.1 A Framework for Symbol-based Learning 9.6 Unsupervised Learning 9.7 Reinforcement Learning

9.8 Epilogue and References 9.9 Exercises 9.2 Version Space Search 9.3 The ID3 Decision Tree Induction Algorithm 9.4 Inductive Bias and Learnability Additional source used in preparing the slides: Jean-Claude Latombes CS121 (Introduction to Artificial Intelligence) lecture

notes, http://robotics.stanford.edu/~latombe/cs121/2003/home.htm (version spaces, decision trees) Tom Mitchells machine learning notes (explanation based learning) 1 Chapter Objectives Learn about several paradigms of symbolbased learning Learn about the issues in implementing and using learning algorithms The agent model: can learn, i.e., can use prior experience to perform better in the future 2 A learning agent Critic Learning element KB sensors environment actuators

3 A general model of the learning process 4 A learning game with playing cards I would like to show what a full house is. I give you examples which are/are not full houses: 6 6 6 9 9 is a full house 6 6 6 6 9 is not a full house 3 3 3 6 6 is a full house 1 1 1 6 6 is a full house Q Q Q 6 6 is a full house 1 2 3 4 5 is not a full house 1 1 3 4 5 is not a full house 1 1 1 4 5 is not a full house 1 1 1 4 4 is a full house 5 A learning game with playing cards If you havent guessed already, a full house is three of a kind and a pair of another kind.

6 6 6 9 9 is a full house 6 6 6 6 9 is not a full house 3 3 3 6 6 is a full house 1 1 1 6 6 is a full house Q Q Q 6 6 is a full house 1 2 3 4 5 is not a full house 1 1 3 4 5 is not a full house 1 1 1 4 5 is not a full house 1 1 1 4 4 is a full house 6 Intuitively, Im asking you to describe a set. This set is the concept I want you to learn. This is called inductive learning, i.e., learning a generalization from a set of examples. Concept learning is a typical inductive learning problem: given examples of some concept, such as cat, soybean disease, or good stock investment, we attempt to infer a definition that will allow the learner to correctly recognize future instances of that concept. 7

Supervised learning This is called supervised learning because we assume that there is a teacher who classified the training data: the learner is told whether an instance is a positive or negative example of a target concept. 8 Supervised learning? This definition might seem counter intuitive. If the teacher knows the concept, why doesnt s/he tell us directly and save us all the work? The teacher only knows the classification, the learner has to find out what the classification is. Imagine an online store: there is a lot of data concerning whether a customer returns to the store. The information is there in terms of attributes and whether they come back or not. However, it is up to the learning system to characterize the concept, e.g, If a customer bought more than 4 books, s/he will return. If a customer spent more than $50, s/he will return.

9 Rewarded card example Deck of cards, with each card designated by [r,s], its rank and suit, and some cards rewarded Background knowledge in the KB: ((r=1) (r=10)) NUM (r) ((r=J) (r=Q) (r=K)) FACE (r) ((s=S) (s=C)) BLACK (s) ((s=D) (s=H)) RED (s) Training set: REWARD([4,C]) REWARD([7,C]) REWARD([2,S]) REWARD([5,H]) REWARD([J,S]) 10 Rewarded card example Training set: REWARD([4,C]) REWARD([7,C]) REWARD([2,S]) REWARD([5,H]) REWARD([J,S]) Card 4 7 2

5 J In the target set? yes yes yes no no Possible inductive hypothesis, h,: h = (NUM (r) BLACK (s) REWARD([r,s]) 11 Learning a predicate Set E of objects (e.g., cards, drinking cups, writing instruments) Goal predicate CONCEPT (X), where X is an object in E, that takes the value True or False (e.g., REWARD, MUG, PENCIL, BALL) Observable predicates A(X), B(X), (e.g., NUM, RED, HAS-HANDLE, HAS-ERASER) Training set: values of CONCEPT for some combinations of values of the observable predicates Find a representation of CONCEPT of the form

CONCEPT(X) A(X) ( B(X) C(X) ) 12 How can we do this? Go with the most general hypothesis possible: any card is a rewarded card This will cover all the positive examples, but will not be able to eliminate any negative examples. Go with the most specific hypothesis possible: the rewarded cards are 4, 7, 2 This will correctly sort all the examples in the training set, but it is overly specific, will not be sort any new examples. But the above two are good starting points. 13 Version space algorithm What we want to do is start with the most general and specific hypotheses, and when we see a positive example, we minimally generalize the most specific hypotheses when we see a negative example, we

minimally specialize the most general hypothesis When the most general hypothesis and the most specific hypothesis are the same, the algorithm has converged, this is the target concept 14 Pictorially + - - - + + - + - ++ +

- + + ? ? ? ? + + ? ? ? - -

? + boundary of G - - ? ? - boundary of S + + + ? - ?

- potential target concepts + - ? - - - + + + + + - - -

15 Hypothesis space When we shrink G, or enlarge S, we are essentially conducting a search in the hypothesis space A hypothesis is any sentence h of the form CONCEPT(X) A(X) ( B(X) C(X) ) where, the right hand side is built with observable predicates The set of all hypotheses is called the hypothesis space, or H A hypothesis h agrees with an example if it gives the correct value of CONCEPT 16 Size of the hypothesis space n observable predicates 2^n entries in the truth table A hypothesis is any subset of observable predicates with the associated truth tables: so there are 2^(2^n) hypotheses to choose from: BIG!

22n n=6 2 ^ 64 = 1.8 x 10 ^ 19 BIG! Generate-and-test wont work. 17 Simplified Representation for the card problem For simplicity, we represent a concept by rs, with: r = a, n, f, 1, , 10, j, q, k s = a, b, r, , , , For example: n represents: NUM(r) (s=) REWARD([r,s]) aa represents: ANY-RANK(r) ANY-SUIT(s) REWARD([r,s]) 18 Extension of an hypothesis The extension of an hypothesis h is the set of objects that verifies h.

For instance, the extension of f is: {j, q, k}, and the extension of aa is the set of all cards. 19 More general/specific relation Let h1 and h2 be two hypotheses in H h1 is more general than h2 iff the extension of h1 is a proper superset of the extension of h2 For instance, aa is more general than f, f is more general than q, fr and nr are not comparable 20 More general/specific relation (contd) The inverse of the more general relation is the more specific relation The more general relation defines a partial ordering on the hypotheses in H 21

A subset of the partial order for cards aa na 4a ab nb 4b a n 4 22 G-Boundary / S-Boundary of V An hypothesis in V is most general iff no hypothesis in V is more general G-boundary G of V: Set of most general hypotheses in V An hypothesis in V is most specific iff no hypothesis in V is more general S-boundary S of V: Set of most specific hypotheses in V

23 Example: The starting hypothesis space G aa na 4a ab nb 4b S 1 a n

4 k 24 4 is a positive example We replace every hypothesis in S whose extension does not contain 4 by its generalization set The generalization set of a hypothesis h is the set of the hypotheses that are immediately more general than h aa na 4a

ab nb 4b a n 4 Generalization set of 4 25 7 is the next positive example Minimally generalize the most specific hypothesis set We replace every hypothesis in S whose extension does not contain 7 by its generalization set

aa na 4a ab nb 4b a n 4 26 7 is positive(contd) Minimally generalize the most specific hypothesis set aa na 4a

ab nb 4b a n 4 27 7 is positive (contd) Minimally generalize the most specific hypothesis set aa na 4a ab nb

4b a n 4 28 5 is a negative example Minimally specialize the most general hypothesis set aa na 4a Specialization set of aa ab nb 4b

a n 4 29 5 is negative(contd) Minimally specialize the most general hypothesis set aa na 4a ab nb 4b a n

4 30 After 3 examples (2 positive,1 negative) G and S, and all hypotheses in between form exactly the version space ab nb 1. If an hypothesis between G and S disagreed with an example x, then an hypothesis G or S would also disagree with x, hence would have been removed a n 31 After 3 examples (2 positive,1 negative) G and S, and all hypotheses in between form exactly the version space ab nb 2. If there were an hypothesis

n not in this set which agreed with all examples, then it would have to be either no more specific than any member of G but then it would be in G or no more general than some member of S but then it would be in a 32 At this stage ab nb No Yes a n Maybe Do 8, 6, j

satisfy CONCEPT? 33 2 is the next positive example Minimally generalize the most specific hypothesis set ab nb a n 34 j is the next negative example Minimally specialize the most general hypothesis set ab nb

35 Result + 4 7 2 5 j nb NUM(r) BLACK(s) REWARD([r,s]) 36 The version space algorithm Begin Initialize S to the first positive training instance N is the set of all negative instances seen so far; For each example x If x is positive, then (G,S) POSITIVE-UPDATE(G,S,x) else (G,S) NEGATIVE-UPDATE(G,S,x) If G = S and both are singletons, then the algorithm has found a single

concept that is consistent with all the data and the algorithm halts If G and S become empty, then there is no concept that covers all the positive instances and none of the negative instances End 37 The version space algorithm (contd) POSITIVE-UPDATE(G,S,x) Begin Delete all members of G that fail to match x For every s S, if s does not match x, replace s with its most specific generalizations that match x; Delete from S any hypothesis that is more general than some other hypothesis in S; Delete from S any hypothesis that is neither more specific than nor equal to a hypothesis in G; (different than the textbook) End; 38 The version space algorithm (contd) NEGATIVE-UPDATE(G,S,x) Begin Delete all members of S that match x

For every g G, that matches x, replace g with its most general specializations that do not match x; Delete from G any hypothesis that is more specific than some other hypothesis in G; Delete from G any hypothesis that is neither more general nor equal to hypothesis in S; (different than the textbook) End; 39 Comments on Version Space Learning (VSL) It is a bi-directional search. One direction is specific to general and is driven by positive instances. The other direction is general to specific and is driven by negative instances. It is an incremental learning algorithm. The examples do not have to be given all at once (as opposed to learning decision trees.) The version space is meaningful even before it converges. The order of examples matters for the speed of convergence As is, cannot tolerate noise (misclassified examples), the version space might collapse 40

Examples and near misses for the concept arch 41 More on generalization operators Replacing constants with variables. For example, color (ball,red) generalizes to color (X,red) Dropping conditions from a conjunctive expression. For example, shape (X, round) size (X, small) color (X, red) generalizes to shape (X, round) color (X, red) 42 More on generalization operators (contd) Adding a disjunct to an expression. For example, shape (X, round) size (X, small) color (X, red) generalizes to shape (X, round) size (X, small) ( color (X, red) (color (X, blue) ) Replacing a property with its parent in a class

hierarchy. If we know that primary_color is a superclass of red, then color (X, red) generalizes to color (X, primary_color) 43 Another example sizes = {large, small} colors = {red, white, blue} shapes = {sphere, brick, cube} object (size, color, shape) If the target concept is a red ball, then size should not matter, color should be red, and shape should be sphere If the target concept is ball, then size or color should not matter, shape should be sphere. 44 A portion of the concept space 45 Learning the concept of a red ball

G : { obj (X, Y, Z)} S:{} positive: obj (small, red, sphere) G: { obj (X, Y, Z)} S : { obj (small, red, sphere) } negative: obj (small, blue, sphere) G: { obj (large, Y, Z), obj (X, red, Z), obj (X, white, sphere) obj (X,Y, brick), obj (X, Y, cube) } S: { obj (small, red, sphere) } delete from G every hypothesis that is neither more general than nor equal to a hypothesis in S G: {obj (X, red, Z) } S: { obj (small, red, sphere) } 46 Learning the concept of a red ball (contd) G: { obj (X, red, Z) } S: { obj (small, red, sphere) } positive: obj (large, red, sphere) G: { obj (X, red, Z)} S : { obj (X, red, sphere) } negative: obj (large, red, cube) G: { obj (small, red, Z), obj (X, red, sphere), obj (X, red, brick)}

S: { obj (X, red, sphere) } delete from G every hypothesis that is neither more general than nor equal to a hypothesis in S G: {obj (X, red, sphere) } S: { obj (X, red, sphere) } converged to a single concept 47 LEX: a program that learns heuristics Learns heuristics for symbolic integration problems Typical transformations used in performing integration include OP1: r f(x) dx r f(x) dx OP2: u dv uv - v du OP3: 1 * f(x) f(x) OP4: (f1(x) + f2(x)) dx f1(x) dx + f2(x) dx A heuristic tells when an operator is particularly useful: If a problem state matches x transcendental(x) dx then apply OP2 with bindings u=x dv = transcendental (x) dx 48 A portion of LEXs hierarchy of symbols

49 The overall architecture A generalizer that uses candidate elimination to find heuristics A problem solver that produces positive and negative heuristics from a problem trace A critic that produces positive and negative instances from a problem traces (the credit assignment problem) A problem generator that produces new candidate problems 50 A version space for OP2 (Mitchell et al.,1983) 51 Comments on LEX The evolving heuristics are not guaranteed to be admissible. The solution path found by the problem solver may not actually be a shortest path solution. The problem generator is the least developed

part of the program. Empirical studies: before: 5 problems solved in an average of 200 steps train with 12 problems after: 5 problems solved in an average of 20 steps 52 More comments on VSL Still lots of research going on Uses breadth-first search which might be inefficient: might need to use beam-search to prune hypotheses from G and S if they grow excessively another alternative is to use inductive-bias and restrict the concept language How to address the noise problem? Maintain several G and S sets. 53 Decision Trees A decision tree allows a classification of an

object by testing its values for certain properties check out the example at: www.aiinc.ca/demos/whale.html The learning problem is similar to concept learning using version spaces in the sense that we are trying to identify a class using the observable properties. It is different in the sense that we are trying to learn a structure that determines class membership after a sequence of questions. This structure is a decision tree. 54 Reverse engineered decision tree of the whale watcher expert system see flukes? yes see dorsal fin? yes no size? vlg

med blue blow whale forward? yes no sperm whale humpback whale no (see next page) yes size med? blows? 1 gray whale no

Size? 2 lg right bowhead whale whale vsm narwhal whale 55 Reverse engineered decision tree of the whale watcher expert system (contd) see flukes? yes no (see previous page)

yes see dorsal fin?no blow? no yes lg size? dorsal fin and blow visible at the same time? yes no sei whale fin whale sm

dorsal fin tall and pointed? yes no killer whale northern bottlenose whale 56 What does the original data look like? Place Time Group Fluke Kaikora 17:00 Yes Yes

Kaikora 7:00 No Yes Kaikora 8:00 Yes Yes Kaikora 9:00 Yes Yes Cape Cod Cape Cod Newb. Port Cape

Cod 18:00 Yes Yes 20:00 No Yes 18:00 No No Dorsal Dorsal fin shape Yes

small triang. Yes small triang. Yes small triang. Yes squat triang. Yes Irregular Yes Irregular No Curved 6:00 Yes Yes No

None Size Blow Very large Very large Very large Medium Yes Blow Type fwd No Blue whale Yes No

Blue whale Yes No Blue whale Yes Yes Sperm whale Hump-back whale Hump-back whale Fin whale Right whale Medium Yes

No Medium Yes No Large Yes No Medium Yes No 57 The search problem Given a table of observable properties, search for a decision tree that correctly represents the data (assuming that the data is noise-free), and is as small as possible.

What does the search tree look like? 58 Comparing VSL and learning DTs A hypothesis learned in VSL can be represented as a decision tree. Consider the predicate that we used as a VSL example: NUM(r) BLACK(s) REWARD([r,s]) NUM? True False BLACK? True True False

False False The decision tree on the right represents it: 59 Predicate as a Decision Tree The predicate CONCEPT(x) A(x) (B(x) v C(x)) can be represented by the following decision tree: Example: A mushroom is poisonous iff it is yellow and small, or yellow, big and spotted x is a mushroom CONCEPT = POISONOUS A = YELLOW B = BIG C = SPOTTED True D = FUNNEL-CAP E = BULKY True

A? True B? True False False False True C? False False 60 Training Set Ex. # A B

C D E CONCEPT 1 False False True False True False 2

False True False False False False 3 False True True True True False

4 False False True False False False 5 False False False True

True False 6 True False True False False True 7 True False False

True False True 8 True False True False True True 9 True

True True False True True 10 True True True True True True 11

True True False False False False 12 True True False False True

False 13 True False True True True True 61 Possible Decision Tree D T E Ex. #

A B C D E CONCEPT 1 False False True False True False

2 False True False False False False 3 False True True True

True False 4 False False True False False False 5 False False False

True True False 6 True False True False False True 7 True

False False True False True 8 True False True False True True 9

True True True False True True 10 True True True True True

True 11 True True False False False False 12 True True False False

True False 13 True False True True True True T C T A

F F B F T E A F A T T F 62 Possible Decision Tree

CONCEPT (D (E v A)) v (C (B v ((E A) v A))) CONCEPT A (B v C) True T E A A? D F C T B F

False T F T tree KIS bias Build smallest decision E False B? True False True C? A A Computationally intractable problem False True True False

greedy algorithm F T T F 63 Getting Started The distribution of the training set is: True: 6, 7, 8, 9, 10,13 False: 1, 2, 3, 4, 5, 11, 12 64 Getting Started The distribution of training set is: True: 6, 7, 8, 9, 10,13 False: 1, 2, 3, 4, 5, 11,

12 Without testing any observable predicate, we could report that CONCEPT is False (majority rule) with an estimated probability of error P(E) = 6/13 65 Getting Started The distribution of training set is: True: 6, 7, 8, 9, 10,13 False: 1, 2, 3, 4, 5, 11, 12 Without testing any observable predicate, we could report that CONCEPT is False (majority rule) with an estimated probability of error P(E) = 6/13 Assuming that we will only include one observable predicate in the decision tree, which predicate should we test to minimize the probability of error? 66 Assume Its A A T True: 6, 7, 8, 9, 10, 13

False: 11, 12 F 1, 2, 3, 4, 5 If we test only A, we will report that CONCEPT is True if A is True (majority rule) and False otherwise The estimated probability of error is: Pr(E) = (8/13)x(2/8) + (5/8)x0 = 2/13 67 Assume Its B B T True: 9, 10 False: 2, 3, 11, 12 F 6, 7, 8, 13 1, 4, 5 If we test only B, we will report that CONCEPT is False if B is True and True otherwise The estimated probability of error is:

Pr(E) = (6/13)x(2/6) + (7/13)x(3/7) = 5/13 68 Assume Its C C T True: 6, 8, 9, 10, 13 False: 1, 3, 4 F 7 1, 5, 11, 12 If we test only C, we will report that CONCEPT is True if C is True and False otherwise The estimated probability of error is: Pr(E) = (8/13)x(3/8) + (5/13)x(1/5) = 4/13 69 Assume Its D D T True: 7, 10, 13 False: 3, 5

F 6, 8, 9 1, 2, 4, 11, 12 If we test only D, we will report that CONCEPT is True if D is True and False otherwise The estimated probability of error is: Pr(E) = (5/13)x(2/5) + (8/13)x(3/8) = 5/13 70 Assume Its E E T True: 8, 9, 10, 13 False: 1, 3, 5, 12 F 6, 7 2, 4, 11 to test is A If weSo, test the only Ebest

we willpredicate report that CONCEPT is False, independent of the outcome The estimated probability of error is unchanged: Pr(E) = (8/13)x(4/8) + (5/13)x(2/5) = 6/13 71 Choice of Second Predicate A F T C T True: 6, 8, 9, 10, 13 False: False F 7 11, 12 The majority rule gives the probability of error Pr(E|A) = 1/ and Pr(E) = 1/13

72 Choice of Third Predicate A F T False C F T True T True: False: 11,12 B F 7

73 Final Tree True True C? True True A True C True B? A? False

False False False False True False False False B True True False False True L CONCEPT A (C v B) 74

Learning a decision tree Function induce_tree (example_set, properties) begin if all entries in example_set are in the same class then return a leaf node labeled with that class else if properties is empty then return leaf node labeled with disjunction of all classes in example_set else begin select a property, P, and make it the root of the current tree; delete P from properties; for each value, V, of P begin create a branch of the tree labeled with V; let partitionv be elements of example_set with values V for property P; call induce_tree (partitionv, properties), attach result to branch V end end end If property V is Boolean: the partition will contain two sets, one with property V true and one with false

75 What happens if there is noise in the training set? The part of the algorithm shown below handles this: if properties is empty then return leaf node labeled with disjunction of all classes in example_set Consider a very small (but inconsistent) training set: A T F F classification T F T A? True True False

False True 76 Using Information Theory Rather than minimizing the probability of error, most existing learning procedures try to minimize the expected number of questions needed to decide if an object x satisfies CONCEPT. This minimization is based on a measure of the quantity of information that is contained in the truth value of an observable predicate and is explained in Section 9.3.2. We will skip the technique given there and use the probability of error approach. 77 % correct on test set Assessing performance 100

size of training set Typical learning curve 78 The evaluation of ID3 in chess endgame 79 Other issues in learning decision trees If data for some attribute is missing and is hard to obtain, it might be possible to extrapolate or use unknown. If some attributes have continuous values, groupings might be used. If the data set is too large, one might use bagging to select a sample from the training set. Or, one can use boosting to assign a weight showing importance to each instance. Or, one can divide the sample set into subsets and train on one, and test on others. 80

Explanation based learning Idea: can learn better when the background theory is known Use the domain theory to explain the instances taught Generalize the explanation to come up with a learned rule 81 Example We would like the system to learn what a cup is, i.e., we would like it to learn a rule of the form: premise(X) cup(X) Assume that we have a domain theory: liftable(X) holds_liquid(X) cup(X) part (Z,W) concave(W) points_up holds_liquid (Z) light(Y) part(Y,handle) liftable (Y) small(A) light(A) made_of(A,feathers) light(A) The training example is the following:

cup (obj1) small(obj1) small(obj1) part(obj1,handle) owns(bob,obj1) part(obj1,bottom) part(obj1, bowl) points_up(bowl) concave(bowl) color(obj1,red) 82 First, form a specific proof that obj1 is a cup cup (obj1) liftable (obj1) light (obj1) part (obj1, handle) holds_liquid (obj1) part (obj1, bowl)

points_up(bowl) concave(bowl) small (obj1) 83 Second, analyze the explanation structure to generalize it 84 Third, adopt the generalized the proof cup (X) liftable (X) light (X) part (X, handle) holds_liquid (X) part (X, W)

points_up(W) concave(W) small (X) 85 The EBL algorithm Initialize hypothesis = { } For each positive training example not covered by hypothesis: 1. Explain how training example satisfies target concept, in terms of domain theory 2. Analyze the explanation to determine the most general conditions under which this explanation (proof) holds 3. Refine the hypothesis by adding a new rule, whose premises are the above conditions, and whose consequent asserts the target concept 86 Wait a minute! Isnt this just a restatement of what the learner already knows?

Not really a theory-guided generalization from examples an example-guided operationalization of theories Even if you know all the rules of chess you get better if you play more Even if you know the basic axioms of probability, you get better as you solve more probability problems 87 Comments on EBL Note that the irrelevant properties of obj1 were disregarded (e.g., color is red, it has a bottom) Also note that irrelevant generalizations were sorted out due to its goal-directed nature Allows justified generalization from a single example Generality of result depends on domain theory Still requires multiple examples Assumes that the domain theory is correct (error-free)---as opposed to approximate domain theories which we will not cover. This assumption holds in chess and other search problems. It allows us to assume explanation = proof.

88 Two formulations for learning Inductive Analytical Given: Given: Instances Instances Hypotheses Hypotheses Target concept Target concept Training examples of the

target concept Training examples of the target concept Domain theory for explaining examples Determine: Determine: Hypotheses consistent with the training examples Hypotheses consistent with the training examples and the domain theory 89 Two formulations for learning (contd) Inductive Analytical Hypothesis fits data

Hypothesis fits domain theory Statistical inference Deductive inference Requires little prior knowledge Learns from scarce data Syntactic inductive bias Bias is domain theory DT and VS learners are similarity-based Prior knowledge is important. It might be one of the reasons for humans ability to generalize from as few as a single training instance. Prior knowledge can guide in a space of an unlimited number of generalizations that can be produced by training examples. 90

An example: META-DENDRAL Learns rules for DENDRAL Remember that DENDRAL infers structure of organic molecules from their chemical formula and mass spectrographic data. Meta-DENDRAL constructs an explanation of the site of a cleavage using structure of a known compound mass and relative abundance of the fragments produced by spectrography a half-order theory (e.g., double and triple bonds do not break; only fragments larger than two carbon atoms show up in the data) These explanations are used as examples for constructing general rules 91 Analogical reasoning Idea: if two situations are similar in some respects, then they will probably be in others Define the source of an analogy to be a problem solution. It is a theory that is relatively well understood. The target of an analogy is a theory that is not

completely understood. Analogy constructs a mapping between corresponding elements of the target and the source. 92 Example: atom/solar system analogy The source domain contains: yellow(sun) blue(earth) hotter-than(sun,earth) causes(more-massive(sun,earth), attract(sun,earth)) causes(attract(sun,earth), revolves-around(earth,sun)) The target domain that the analogy is intended to explain includes: more-massive(nucleus, electron) revolves-around(electron, nucleus) The mapping is: sun nucleus and earth electron The extension of the mapping leads to the inference: causes(more-massive(nucleus,electron), attract(nucleus,electron)) causes(attract(nucleus,electron), revolves-around(electron,nucleus)) 94

A typical framework Retrieval: Given a target problem, select a potential source analog. Elaboration: Derive additional features and relations of the source. Mapping and inference: Mapping of source attributes into the target domain. Justification: Show that the mapping is valid. 95

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