Bayesian network Graph theory exercise Exercise 6.3, 6.4, 6.5Q6.3) Does {X7, Xg} {X2} |{X₁, X3} hold?P(X7, X8X4, X₁, X3) = P(X7, X8|X₁, X3)Q6.4) Does {X4, X5} {X}| {X₂} hold?P(X4, X5 X3, X6) = P(X4, X51X3)Q6.5) Does {X4, X5} {X}| {X3, Xg} hold?P(X4, X5 X3, X6, X8) = P(X4, X5 X3, X8)X₁X3ХаX5X₂X6X₂X838

The condition for independence is "Xi is independent of its non-descendants given its parents" We…

Q6.3) Does {X₁, Xg} {X4}| {X₁, X3} hold? P(X7, X8X4, X1, X3) = P(X7, X8|X₁, X3) Q6.4) Does {X4, X5} {X}| {X₂} hold? P(X4, X5 X3, X6) = P(X4, X5|X3) Q6.5) Does {X4, X5} {X6}| {X3, Xg} hold? P(X4, X5 X3, X6, X8) = P(X4, X5 X3, X8)

Q6.3) Does {X₁, Xg} {X4}| {X₁, X3} hold? P(X7, X8X4, X1, X3) = P(X7, X8|X₁, X3) Q6.4) Does {X4, X5} {X}| {X₂} hold? P(X4, X5 X3, X6) = P(X4, X5|X3) Q6.5) Does {X4, X5} {X6}| {X3, Xg} hold? P(X4, X5 X3, X6, X8) = P(X4, X5 X3, X8)

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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Concept explainers

Parallel processing

The process of running two or more processors together is known as parallel processing. In parallel processing, a single process is divided into multiple parts, and each individual part is executed at the same time. It is used to handle various parts of a task at the same time. The main advantage of parallel processing is the reduction of the processing time of the program, where the processor in a system is the CPU. Hence, parallel processing can be performed by using multiple CPUs, and having more than one CPU creates a multi-processor. Another name of parallel processors is block processors. Here, the size of a block is the number of inputs that can be processed simultaneously in a single clock cycle.

Question

Bayesian network Graph theory exercise

Exercise 6.3, 6.4, 6.5
Q6.3) Does {X7, Xg} {X2} |{X₁, X3} hold?
P(X7, X8X4, X₁, X3) = P(X7, X8|X₁, X3)
Q6.4) Does {X4, X5} {X}| {X₂} hold?
P(X4, X5 X3, X6) = P(X4, X51X3)
Q6.5) Does {X4, X5} {X}| {X3, Xg} hold?
P(X4, X5 X3, X6, X8) = P(X4, X5 X3, X8)
X₁
X3
Ха
X5
X₂
X6
X₂
X8
38

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Step 1: Condition for independence

Step 2: Q6.3)

Step 3: Q6.4)

Step 4: Q6.5)

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