哈工大深圳机器学习复习4丁宇新

哈工大深圳机器学习复习4丁宇新Machine Learning Question one: Question two: Initilize: G=?,?,?,?,?,? S=, Step 1: G=?,?,?,?,?,? S=sunny,warm,normal,strong,warm,same Step2: coming one positive instance 2 G=?,?,?,?,?,? S=sunny,warm,?,strong,warm,same Step3: coming one negative instance 3 G= S=sunny,warm,?,strong,warm,same Step4: coming one positive instance 4 S= sunny,warm,?,strong,?,? G= Question three: (a) Entropy(S)=og(3/5) og(2/5)= 0.971 (b) Gain(S,sky) = Entropy(S) (4/5) Entropy(Ssunny) + (1/5) Entropy(Srainny) = 0.322 Gain(S,AirTemp) = Gain(S,wind) = Gain(S,sky) =0.322 Gain(S,Humidity) = Gain(S,Forcast) = 0.02 Gain(S,water) = 0.171 Choose any feature of AirTemp, wind and sky as the top node. The decision tree as follow: (If choose sky as the top node) Question Four: Answer: Inductive bias: give some proor assumption for a target concept made by the learner to have a basis for classifying unseen instances. Suppose L is a machine learning algorithm and x is a set of training examples. L(xi, Dc) denotes the classification assigned to xi by L after training examples on Dc. Then the inductive bias is a minimal set of assertion B, given an arbitrary target concept C and set of training examples Dc: (Xi ) (BDcXi) -| L(xi, Dc) C_E: the target concept is contained in the given gypothesis space H, and the training examples are all positive examples. ID3: a, small trees are preferred over larger trees. B, the trees that place high information gain attribute close to root are preferred over those that do not. BP:Smooth interpolation beteen data points. Question Five: Answer: In nave bayes classification, we assump that all attributes are independent given the tatget value, while in bayes belif net, it specifes a set of conditional independence along with a set of probability distribution. Question Six： 随即梯度下降算法 Question Seven：朴素贝叶斯例子 Question Eight：The definition of three types of fitness functions in genetic algorithm Answer: In order to select one hypothese according to fitness function, there are always three methods: roulette wheel selection, tournament selection and rank selection. Question nine: Single-point crossover: Crossover mask: 11111100000 or 11111000000 or 1111 0000000 or 00001111111 Two-point crossover: Offspring: (11001011000, 00101000101) Uniform crossover: Crossover mask: 10011110011 or 10001110011 or 01111101100 or 10000010011 or 10011110001 01100001100 Point mutation: Any mutation is ok!