Answer: ``Choose" designates a non-deterministic choice point. All possible choices must be considered. Different choices lead to different solutions.
``Pick'' designates an arbitrary choice point. Any one of the possible options can be used; they all lead to the same solutions, though possibly in a different order. If the option picked does not lead to a solution, no other option will do any better.
Example: A non-deterministic algorithm for solving the map-coloring problem can be written as follows:
repeat pick an uncolored country X; choose a color C for X that is different from any of the colors for countries that border X; until all the countries have been colored.Here, the algorithm needs to consider all consistent partial colorings of the map, so ``choose'' is used for the color. If one color does not lead to a solution, you consider an alternative color. The ``pick'' among countries, on the other hand, determines the order in which countries are considered. This may effect the efficiency of search, but certainly cannot effect the existence or non-existence of a solution.
Answer: (i) It must be (i) zero-sum; (ii) perfect knowledge; (iii) deterministic; (iv) two-player; (v) discrete.
s(X,L) --- Person X speaks language L. c(X,Y) --- Persons X and Y can communicate. i(W,X,Y) --- Person W can serve as an interpreter between persons X and Y. j,p,e,f --- Constants: Joe, Pierre, English, and French respectively.A. Express the following statements in L:
B. Show that (vi) can be proven from (i)---(v) using backward-chaining resolution. You must show the Skolemized form of each statement, and every resolution that is used in the final proof. You need not show the intermediate stages of Skolemization, or show resolutions that are not used in the final proof.
Answer: The Skolemized forms of (i)-(v) are
(sk1(L,M) is a Skolem function mapping L and M to a person who speaks both L and M.)
The Skolemized form of the negation of (vi) is
The backward chaining proof is then as follows:
This is a deterministic decision tree for a data set with two Boolean predictive attributes, A and B, and a Boolean classification C. The tree first tests an example X on attribute A. If X.A is T, then the tree tests on the tree tests on attribute B. If X.B is T, then the tree predicts that X.C is T; otherwise it predicts that X.C is F. If, in the first test, X.A is F, then the tree predicts that X.C is F.
B. Describe the ID3 algorithm to construct decision
trees from training data.
Answer: The ID3 algorithm builds the decision tree recursively from the top down. At each stage of the recursion, an internal node N is passed a subtable, with those instances that would reach N in the decision procedure, and those attributes that have not already been tested. If the table has more than one classification value, and has some predictive attributes left to test, and is more than a specified minimum splitting size, then the algorithm chooses the best attribute A to split on, namely the one that gives the minimum expected entropy. If the expected entropy of the classification C after splitting on A would be less than the entropy of C at node N, then node N is labelled as a test for A, the table is partitioned according to the value of A, and the partitions are passed down to the next nodes recursively.
C. What is the entropy of a classification C in table T? What is
the expected entropy of classification C if the table is split on
predictive attribute A?
Let pc be the fraction of instances X in T where X.C=c. Then the entropy of C in T is the sum over c of -pc log(pc).
Let qa be the fraction of instance X in T where X.A=a. Let Ta be the subtable of T of instances X such that X.A=a. Let ra,c be the fraction of instances X in Ta where X.C=c. Then the expected entropy of C after splitting on A is
suma qa (sumc -ra,c log(ra,c))
D. What kinds of techniques can be used to counter the problem of over-fitting in decision trees?
Answer: Prepruning techniques prune the decision tree as it is being built; e.g. by enforcing a minimum splitting size on nodes. Postpruning techniques build the entire decision tree, and then eliminate tests that do not seem to be useful.
W X Y C ---------------- T T T T T F T F T F F F F T T F F F F TWe now encounter a new example: W=F, X=T, Y=F. If we apply the Naive Bayes method, what probability is assigned to the two values of C?
Answer: We wish to compute
argmaxc Prob(C=c | W=F, X=T, Y=F) = (by Bayes' Law) argmaxc Prob(W=F, X=T, Y=F | C=c) Prob(C=c) / Prob(W=F, X=T, Y=F) = (dropping the normalizing factor, which does not depend on c) argmaxc Prob(W=F, X=T, Y=F | C=c) Prob(C=c) = (assuming conditional independence) argmaxc Prob(W=F|C=c) Prob(X=T|C=c) Prob(Y=F|C=c) Prob(C=c)Using the frequencies in the tables to estimate the probabilities, for c=T, this product evaluates to (1/2) * (1/2) * (1/2) * (2/5) = 1/20. For c=F, it evaluates to (1/3) * (1/3) * (1/3) * (3/5) = 1/45. The Bayesian estimate, therefore, is that c=T is more likely than c=F by a factor of 2.25. To be exact the estimated probability that C=T is (1/20)/((1/20)+(1/45)) = 9/13; the estimated probability that C=F is 4/13.
Answer: Local minima of the error function over the corpus of labelled examples, as a function of the weights on the links. The learning algorithm is in effect doing a hill-climbing search for the value of the weights that gives the minimal error function. If the hill-climbing search finds a local mininum of this function, it will get stuck at that suboptimal state, and will not find the true solution.
Answer: Task execution is carried out through process (a). Learning is carried out through process (d).
Answer: When you wish to evaluate a learning program using a fixed corpus of examples, it is common to divide the corpus into a training set and a test set. The program is first trained by running it on the training set. Then the learning module is turned off, and the quality of the current state of the execution module is tested by running it on the test set. In this way you guard against the possibility of overlearning, in which the program simply learns the examples in the corpus, rather than the underlying concept.
A. The original data can be encoded by storing the predictive attributes for each example plus the hypothesis. If the space saved on recording the classification is less than the space required for the hypothesis, then the overall description length is reduced.
B. The original data can be encoded by storing (i) the predictive attributes; (ii) the hypothesis; (iii) the classification X.C' which states either that the hypothesis suceeds on example X or that it fails and gives the correct value of C. If the entropy of C' is less than that of C, and if the data set is large enough, then the space saved storing C' in an optimal coding as opposed to storing C in an optimal coding will be enough to make up for the cost of the space to store the hypothesis.
C. Let A,B,C be the predictive attributes; let D be the classification; let f(A,B,C) be the hypothesis. For each example X, record X.A, X.B, X.C and the error X.D - f(A,B,C). If f is a useful hypothesis, then the range of values of the error will be much less than the range of values of X.D; hence the error will require fewer bits, for a given level of precision, than X.D. If the space thus saved is more than the space required to store f, then the description length is reduced.