Lecture 4.2: Relations CS 250, Discrete Structures, Fall 2013 Nitesh Saxena Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag

1/20/2016 Course Admin Mid-Term 2 Exam Solution posted Should have the results this week HW3 Solution posted Should have the results this week HW4 should follow by the end of the week If you haven’t picked up your previous graded submissions, please do so from my office 2

Traveling this Week Presenting at conference (Nov 13-15): No class this Thursday (Nov 14) 1/20/2016

Outline Relation Examples and Definitions Matrix Representation 4

1/20/2016 Composing Relations Q: Suppose R defined on N by: xRy iff y = x 2 and S defined on N by: xSy iff y = x 3 What is the composition S  R ? 5

1/20/2016 Composing Relations xRy iff y = x 2 xSy iff y = x 3 A: These are functions (squaring and cubing) so the composite S  R is just the function composition (raising to the 6 th power). xSRy iff y = x 6 (in this odd case S  R = R  S) Q: Compose the following:

1/20/2016 Composing Relations A: Draw all possible shortcuts. In our case, all shortcuts went through 1:

1/20/2016 Composing Relations: Picture A: Draw all possible shortcuts. In our case, all shortcuts went through 1:

1/20/2016 Composing Relations: Picture A: Draw all possible shortcuts. In our case, all shortcuts went through 1:

1/20/2016 Composing Relations: Picture A: Draw all possible shortcuts. In our case, all shortcuts went through 1:

1/20/2016 Composing Relations: Picture A: Draw all possible shortcuts. In our case, all shortcuts went through 1:

1/20/2016 Inverting Relations Relational inversion amounts to just reversing all the tuples of a binary relation. DEF: If R is a relation from A to B, then the relation R -1 from B to A defined by setting bR -1 a if and only aRb. Q: Suppose R defined from Z to N by: xRy iff y = x 2. What is the inverse R -1 ? 12

1/20/2016 Inverting Relations A: xRy iff y = x 2. R is the square function so R -1 is square root: i.e. the union of the two square-root branches. I.e: yR -1 x iff y = x 2 or in terms of square root: xR -1 y iff y = ±  x where x is non-negative 13

Relations – matrix representation Suppose we have a relation R on AxB, where A={1,2,3,4}, and B={u,v,w}, R={(1,u),(1,v),(2,w),(3,w),(4,u)}. Then we can represent R as: The labels on the outside are for clarity. It’s really the matrix in the middle that’s important. This is a |A| x |B| matrix whose entries indicate membership in R. uvw /20/201614

Relations – matrix representation Some things to think about. Let R be a relation on a set A, and let M R be the matrix representation of R. Then R is reflexive if, ______________. A.All entries in M R are 1. B.The \ diagonal of M R contains only 1s. C.The first column of M R contains no 0s. D.None of the above. 1/20/201615

Relations – matrix representation Some things to think about. Let R be a relation on a set A, and let M R be the matrix representation of R. Then R is symmetric if, ______________. A.All entries above the \ are 1. B.The first and last columns of M R contain an equal # of 0s. C.M R is visually symmetric about the \ diagonal. D.None of the above. 1/20/201616

Relations – matrix representation Suppose we have R 1 and R 2 defined on A: R1R1 uvw u101 v001 w110 R2R2 uvw u110 v011 w001 Then R 1  R 2 is the bitwise “or” of the entries (Join By): Then R 1  R 2 is the bitwise “and” of the entries (Meet): M R1  R2 = M R1 v M R2 M R1  R2 = M R1  M R2 1/20/201617

Relations – composition using matrices Suppose we have R and S defined on A: Ruvw u101 v001 w110 Suvw u110 v011 w001 Then S  R corresponds to the boolean product /20/201618

Relations - A Theorem Theorem: If R is a transitive relation, then R n  R,  n. How to prove? What strategy or technique should we use? 1/20/201619

Relations - A Theorem If R is a transitive relation, then R n  R,  n. Typical way of proving subset. Proof by induction on n. Base case (n=1): R 1  R because by definition, R 1 = R. Induction case: if R is transitive, then R k  R. Prove: if R is transitive, then R k+1  R. We are trying to prove that R k+1  R. To do this, we select an element of R k+1 and show that it is also an element of R. Let (a,b) be an element of R k+1. Since R k+1 = R k  R, we know there is an x so that (a,x)  R and (x,b)  R k. By assumption at the induction step, since R k  R, (x,b)  R. But wait, if (a,x)  R, and (x,b)  R, and R is transitive, then (a,b)  R. 1/20/201620

Relations - Another Theorem If R is a reflexive relation, then R n is reflexive relation,  n. Whiteboard! 1/20/201621

N-ary Relations So far, we were talking about binary relations – defined on two sets. Can be generalized to N sets Ex: R = {(a, b, c): a < b < c}, defined on set of integers – a 3-ary relation Applications in databases 1/20/201622

1/20/2016 Today’s Reading Rosen 9.1 and