opose that we have three variables X1, X2 and X3, which are defined on the same domain of 2,3}. Two binary constraints for these three variables are defined as follows: SR12 {(X1, X2),[(2, 1), (2,3), (3, 2), (3,3)]} R13 {(X1, X3), [(1,2), (2, 1), (3, 1), (3,3)|} %3D (12 points) Is X1 arc-consistent with respect to X2? And is X1 arc-consistent with respect to X3? and why? In addition, is X3 arc-consistent with respect X1 if the constraints between R and R3 are undirected (i.e., R31 is defined as {(X3, X1), [(2, 1), (1, 2), (1,3), (3,3)]} that switches the clement order of every two-tuple of R13)? and why? (12 points) Suppose that, after some inference, the domain of X1 is reduced as (2,3} and the constrains in R12 and R13 for X1 is removed from R13 due to reducing the domain of X1. Now is X1 still arc-consistent with = 1 are removed accordingly. To be more specific, (1,2)

opose that we have three variables X1, X2 and X3, which are defined on the same domain of 2,3}. Two binary constraints for these three variables are defined as follows: SR12 {(X1, X2),[(2, 1), (2,3), (3, 2), (3,3)]} R13 {(X1, X3), [(1,2), (2, 1), (3, 1), (3,3)|} %3D (12 points) Is X1 arc-consistent with respect to X2? And is X1 arc-consistent with respect to X3? and why? In addition, is X3 arc-consistent with respect X1 if the constraints between R and R3 are undirected (i.e., R31 is defined as {(X3, X1), [(2, 1), (1, 2), (1,3), (3,3)]} that switches the clement order of every two-tuple of R13)? and why? (12 points) Suppose that, after some inference, the domain of X1 is reduced as (2,3} and the constrains in R12 and R13 for X1 is removed from R13 due to reducing the domain of X1. Now is X1 still arc-consistent with = 1 are removed accordingly. To be more specific, (1,2)

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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Given the definition in Haskell: fmap :: (a -> b) -> [a] -> [b] fmap g xs = reverse (map g xs) With Functor law: (fmap g . fmap h) xs = fmap (g . h) xs Using list xs :: [Int], prove the definition does not satisfy the law:
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True or false? 6 ⊆ {2, 4, 6}
i need help for 6, 7, and 9.
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