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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A unigram is a sequence of words of length one (i.e. a single word).• A bigram is a sequence of words of length two.• The conditional probability of an event E2 given another event E1, written p(E2|E1), is the probability that E2 will occur given that event E1 has already occurred.We write p(w(k)|w(k-1)) for the conditional probability of a word w in position k, w(k), given the immediately preceding word, w(k-1). You determine the conditional probabilities by determining unigram counts (the number of times each word appears, written c(w(k)), bigram counts (the number of times each pair of words appears, written c(w(k-1) w(k)), and then dividing each bigram count by the unigram count of the first word in the bigram:p(WORD(k)|WORD(k-1)) = c(WORD(k-1) WORD(k)) / c(WORD(k-1)) Apply and incorporate instrutions to code below. #include <stdio.h>//including headers#include <string.h>#include <stdlib.h> struct node{//structure intialization int data; struct node *next; };…
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Consider the doctor-patient problem. There is a Doctor's clinic which has one Doctor, one patient chair, and n chairs for waiting for patients if there are any to sit on the chair. If there is no patient, then the Doctor sleeps in his own chair. When a patient arrives, he has to wake up the Doctor. If there are many patients and the Doctor is treating a patient, then the remaining patients either wait if there are empty chairs in the waiting room or they leave if no chairs are empty. Write a solution (Algorithm / Pseudocode) using semaphores.
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Sometimes, like with 203, the sequence reaches (and stays at) 1. Numbers like this are called happy. Not all numbers are happy. If we started with 11, the sequence would be: 11 -> 2 -> 4 -> 16 -> ... This sequence will never reach 1, and so the number 11 is called unhappy. Given a positive whole number, you have to determine whether that number is happy or unhappy. Examples happy (203) → true happy (11) → false happy (107) → false Notes
Describe a Python program of ensemble methods by doing the following: If each base-learner is valid and correct with probability p > 1 / 2, what is the probability that a majority vote over L classifiers gives the correct answer?
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Given that we want to evaluate the performance of 'n' different machine learning models on the same data, why would the following splitting mechanism be incorrect: def get_splits(): df pd.DataFrame(...) rnd = np.random.rand(len(df)) train =df[ rnd = 0.8 & rnd = 0.9 ] return train, valid, test #Model 1 from sklearn.tree import Decision Tree Classifier train, valid, test = get_splits() #Model 2 from sklearn.linear_model import Logistic Regression train, valid, test get_splits() =
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