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Lecture 4.1: Relations Basics. CS 250, Discrete Structures, Fall 2011 Nitesh Saxena * Adopted from previous lectures by Cinda Heeren. Course Admin. Mid-Term 2 Exam How was it? Solution will be posted soon Should have the results by the coming weekend HW3 Was due yesterday
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Lecture 4.1: Relations Basics CS 250, Discrete Structures, Fall 2011 NiteshSaxena *Adopted from previous lectures by CindaHeeren
Course Admin • Mid-Term 2 Exam • How was it? • Solution will be posted soon • Should have the results by the coming weekend • HW3 • Was due yesterday • Bonus problem? • Solution will be posted soon • Results should be ready by the coming weekend Lecture 4.1: Relations Basics
Roadmap • Done with about 60% of the course • We are left with two more homeworks • We are left with one final exam • Cumulative – will cover everything • Focus more on material covered after the two mid-terms • Please try to do your best in the remainder of the time; I know you have been working hard • Please submit your homeworks on time • Do you need periodic reminders? • Still a lot of scope of improvement • Done with travel for this semester – I will be fully at your disposal for the rest of the semester Lecture 4.1: Relations Basics
Outline • Relation Definition and Examples • Types of Relations • Operations on Relations Lecture 4.1: Relations Basics
Relations • Recall the definition of the Cartesian (Cross) Product: The Cartesian Product of sets A and B, A x B, is the set A x B = {<x,y> : xA and yB}. • A relation is just any subset of the CP!! R AxB • Ex: A = students; B = courses. R = {(a,b) | student a is enrolled in class b} We often say: aRb if (a,b) belongs to R Lecture 4.1: Relations Basics
Yes, a function is a special kind of relation. Relations vs. Functions • Recall the definition of a function: f = {<a,b> : b = f(a) , aA and bB} • Is every function a relation? • Draw venn diagram of cross products, relations, functions Lecture 4.1: Relations Basics
Properties of Relations • Reflexivity: A relation R on AxA is reflexive if for all aA, (a,a) R. • Symmetry: A relation R on AxA is symmetric if (a,b) R implies (b,a) R . Lecture 4.1: Relations Basics
Properties of Relations • Transitivity: A relation on AxA is transitive if (a,b) R and (b,c) R imply (a,c) R. • Anti-symmetry: A relation on AxA is anti-symmetric if (a,b) R implies (b,a) R. Lecture 4.1: Relations Basics
1 2 3 Loops must exist on EVERY vertex. 4 Properties of Relations - Techniques How can we check for the reflexive property? Draw a picture of the relation (called a “graph”). • Vertex for every element of A • Edge for every element of R Now, what’s R? {(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} Lecture 4.1: Relations Basics
1 2 3 EVERY edge must have a return edge. 4 Properties of Relations - Techniques How can we check for the symmetric property? Draw a picture of the relation (called a “graph”). • Vertex for every element of A • Edge for every element of R Now, what’s R? {(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} Lecture 4.1: Relations Basics
1 2 3 A “short cut” must be present for EVERY path of length 2. 4 Properties of Relations - Techniques How can we check for transitivity? Draw a picture of the relation (called a “graph”). • Vertex for every element of A • Edge for every element of R Now, what’s R? {(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} Lecture 4.1: Relations Basics
1 2 3 No edge can have a return edge. 4 Properties of Relations - Techniques How can we check for the anti-symmetric property? Draw a picture of the relation (called a “graph”). • Vertex for every element of A • Edge for every element of R Now, what’s R? {(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} Lecture 4.1: Relations Basics
? 2 1 2 1 1 ? ? 3 Yes Yes No No Properties of Relations - Examples Let R be a relation on People, R={(x,y): x and y have lived in the same country} Is R transitive? Is it symmetric? Is it reflexive? Is it anti-symmetric? Lecture 4.1: Relations Basics
Yes Definition of “divides” Is (x,z) in R? Properties of Relations - Examples Let R be a relation on positive integers, R={(x,y): 3|(x-y)} Suppose (x,y) and (y,z) are in R. Then we can write 3j = (x-y) and 3k = (y-z) Can we say 3m = (x-z)? Add prev eqn to get: 3j + 3k = (x-y) + (y-z) 3(j + k) = (x-z) Is R transitive? Lecture 4.1: Relations Basics
Yes Yes Definition of “divides” Yes, for k=0. Properties of Relations - Techniques Let R be a relation on positive integers, R={(x,y): 3|(x-y)} Is (x,x) in R, for all x? Does 3k = (x-x) for some k? Is R transitive? Is it reflexive? Lecture 4.1: Relations Basics
Yes Yes Yes Definition of “divides” Yes, for k=-j. Properties of Relations - Techniques Let R be a relation on positive integers, R={(x,y): 3|(x-y)} Suppose (x,y) is in R. Then 3j = (x-y) for some j. Does 3k = (y-x) for some k? Is R transitive? Is it symmetric? Is it reflexive? Lecture 4.1: Relations Basics
Yes Yes Yes No Definition of “divides” Yes, for k=-j. Properties of Relations - Techniques Let R be a relation on positive integers, R={(x,y): 3|(x-y)} Suppose (x,y) is in R. Then 3j = (x-y) for some j. Does 3k = (y-x) for some k? Is R transitive? Is it symmetric? Is it reflexive? Is it anti-symmetric? Lecture 4.1: Relations Basics
More than one relation Suppose we have 2 relations, R1 and R2, and recall that relations are just sets! So we can take unions, intersections, complements, symmetric differences, etc. There are other things we can do as well… Lecture 4.1: Relations Basics
B C A R S 1 x s 2 y t 3 z u 4 v More than one relation Let R be a relation from A to B (R AxB), and let S be a relation from B to C (S BxC). The composition of R and S is the relation from A to C (SR AxC): SR = {(a,c): bB, (a,b) R, (b,c) S} SR = {(1,u),(1,v),(2,t),(3,t),(4,u)} Lecture 4.1: Relations Basics
More than one relation Let R be a relation on A. Inductively define R1 = R Rn+1 = Rn R A A A R R1 1 1 1 2 2 2 3 3 3 4 4 4 R2 = R1R = {(1,1),(1,2),(1,3),(2,3),(3,3),(4,1), (4,2)} Lecture 4.1: Relations Basics
… = R4 = R5 = R6… More than one relation Let R be a relation on A. Inductively define R1 = R Rn+1 = Rn R A A A R R2 1 1 1 2 2 2 3 3 3 4 4 4 R3 = R2R = {(1,1),(1,2),(1,3),(2,3),(3,3),(4,1),(4,2),(4,3)} Lecture 4.1: Relations Basics
Today’s Reading • Rosen 9.1 Lecture 4.1: Relations Basics