Consider Slide 55 in "8 Reinforcement Learning.ppt". We have a part of another trajectory: (c1,E, -1) -> (c2, N, -1). 。 Under SARSA, the new Q(c1,E) = • Under Q-Learning, the new Q(c1,E) = answer) Picture below is of the slide N(s,a) (keep one decimal value in your answer) (keep one decimal value in your Q(s, a) 6 3 1 -2.4 -1.8 41.0 10 10 55 12 0 25 25 +100 C-2.4 16.4-1.8 60.8 0.0 88.8 +100 11 3 1 -2.4 -1.8 3.4 18 5 13 9 -2.9 45.8 9 55 55 2 -100 b-2.9 -2.9 12.2 -101.0 -100 4 -2.9 -51.5 6 4 28 4 -2.6 -1.9 11.0 -101.0 00 8 17 3 20 1 88 8a-2.5 -2.2-2.0 0.5 -2.0 -2.1 -15.0 -14.9 6 4 4 77 -2.4 -2.1 -1.9 -3.7 1 2 3 4 1 2 3 4

Consider Slide 55 in "8 Reinforcement Learning.ppt". We have a part of another trajectory: (c1,E, -1) -> (c2, N, -1). 。 Under SARSA, the new Q(c1,E) = • Under Q-Learning, the new Q(c1,E) = answer) Picture below is of the slide N(s,a) (keep one decimal value in your answer) (keep one decimal value in your Q(s, a) 6 3 1 -2.4 -1.8 41.0 10 10 55 12 0 25 25 +100 C-2.4 16.4-1.8 60.8 0.0 88.8 +100 11 3 1 -2.4 -1.8 3.4 18 5 13 9 -2.9 45.8 9 55 55 2 -100 b-2.9 -2.9 12.2 -101.0 -100 4 -2.9 -51.5 6 4 28 4 -2.6 -1.9 11.0 -101.0 00 8 17 3 20 1 88 8a-2.5 -2.2-2.0 0.5 -2.0 -2.1 -15.0 -14.9 6 4 4 77 -2.4 -2.1 -1.9 -3.7 1 2 3 4 1 2 3 4

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Suppose we are trying to model a classification problem with two labels: 'sick' and 'healthy.' For the purpose of this test, we consider a positive result to be testing sick and a negative to test as healthy. After fitting the model with LDA in R, we compare predicted values from the actual values, as shown below: > lda. fit = 1da (test.result x1 + x2, data-mydata) > lda. pred > lda.class=lda.pred$class > table (lda.class, mydata$test.result) = predict (1da.fit) mydata lda.pred sick healthy sick 40 healthy 25 32 121
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A valid argument is given below. Fill in the missing values of the proof. vx (R(x) v Q(x)) ax (P(x) v -Q(x)) 1. ax (P(x) v -Q(x)) Hypothesis 2. (b is a particular element) ^ (P(b) v ¬Q(b)) (A) : 3x (P(x) v R(x)) 3. P(b) v -Q(b) (B) 4. vx (R(x) v Q(x)) Hypothesis 5. b is a particular element (C) 6. (D) Universal instantiation, 4, 5 7. P(b) v R(b) Resolution, 3, 6 8. 3x (P(x) v R(X)) Existential generalization, 5, 7 (A) Pick (C) Pick (B) Pick (D) Pick Check Next
Consider Slide 55 in "8 Reinforcement Learning.ppt". We have a part of another trajectory: (c1,E, -1) -> (c2, N, -1). 。 Under SARSA, the new Q(c1,E) = • Under Q-Learning, the new Q(c1,E) = answer) Picture below is of the slide N(s,a) (keep one decimal value in your answer) (keep one decimal value in your Q(s, a) 6 3 1 -2.4 -1.8 41.0 10 10 55 12 0 25 25 +100 C-2.4 16.4-1.8 60.8 0.0 88.8 +100 11 3 1 -2.4 -1.8 3.4 18 5 13 9 -2.9 45.8 9 55 55 2 -100 b-2.9 -2.9 12.2 -101.0 -100 4 -2.9 -51.5 6 4 28 4 -2.6 -1.9 11.0 -101.0 00 8 17 3 20 1 88 8a-2.5 -2.2-2.0 0.5 -2.0 -2.1 -15.0 -14.9 6 4 4 77 -2.4 -2.1 -1.9 -3.7 1 2 3 4 1 2 3 4
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