CS 250, Discrete Structures, Fall 2014 Lecture 4.2: Relations CS 250, Discrete Structures, Fall 2014 Nitesh Saxena Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag

Course Admin Mid-Term 2 Exam HW3 How was it? Solution will be posted Should have the results this week HW3 Should have the results soon If you haven’t picked up your previous graded submissions, please do so from my office 1/17/2019

Roadmap PLEASE DO ATTEND THE LECTURES Done with about 70% of the course We are left with one more homework 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 Still a lot of scope of improvement PLEASE DO ATTEND THE LECTURES 1/17/2019

Outline Relation Examples and Definitions Matrix Representation 1/17/2019

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 ? 1/17/2019

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 6th power). xSRy iff y = x 6 (in this odd case SR = RS) Q: Compose the following: 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 5 5 More detail on: xRSy iff y = x 6 1) xRx2 2) ySy3 so x2S(x2) 3 3) Composing (1) and (2) we get xRSx6 1/17/2019

Composing Relations 1 1 1 2 2 2 3 3 3 4 4 5 A: Draw all possible shortcuts. In our case, all shortcuts went through 1: 1 1 2 2 3 3 4 1/17/2019

Composing Relations: Picture 1 1 1 2 2 2 3 3 3 4 4 5 A: Draw all possible shortcuts. In our case, all shortcuts went through 1: 1 1 2 2 3 3 4 1/17/2019

Composing Relations: Picture 1 1 1 2 2 2 3 3 3 4 4 5 A: Draw all possible shortcuts. In our case, all shortcuts went through 1: 1 1 2 2 3 3 4 1/17/2019

Composing Relations: Picture 1 1 1 2 2 2 3 3 3 4 4 5 A: Draw all possible shortcuts. In our case, all shortcuts went through 1: 1 1 2 2 3 3 4 1/17/2019

Composing Relations: Picture 1 1 1 2 2 2 3 3 3 4 4 5 A: Draw all possible shortcuts. In our case, all shortcuts went through 1: 1 1 2 2 3 3 4 1/17/2019

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 -1a if and only aRb. Q: Suppose R defined from Z to N by: xRy iff y = x 2. What is the inverse R -1 ? 1/17/2019

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 -1x iff y = x 2 or in terms of square root: xR -1y iff y = ±x where x is non-negative 1/17/2019

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. u v w 1 2 3 4 This is a |A| x |B| matrix whose entries indicate membership in R. 1/17/2019

Relations – matrix representation Some things to think about. Let R be a relation on a set A, and let MR be the matrix representation of R. Then R is reflexive if, ______________. All entries in MR are 1. The \ diagonal of MR contains only 1s. The first column of MR contains no 0s. None of the above. 1/17/2019

Relations – matrix representation Some things to think about. Let R be a relation on a set A, and let MR be the matrix representation of R. Then R is symmetric if, ______________. All entries above the \ are 1. The first and last columns of MR contain an equal # of 0s. MR is visually symmetric about the \ diagonal. None of the above. 1/17/2019

Relations – matrix representation Suppose we have R1 and R2 defined on A: R1 u v w 1 R2 u v w 1 1 Then R1  R2 is the bitwise “or” of the entries (Join By): MR1R2 = MR1 v MR2 1 Then R1  R2 is the bitwise “and” of the entries (Meet): MR1R2 = MR1  MR2 1/17/2019

Relations – composition using matrices Suppose we have R and S defined on A: R u v w 1 S u v w 1 1 Then SR corresponds to the boolean product   1/17/2019

Relations - A Theorem Theorem: If R is a transitive relation, then Rn  R, n. How to prove? What strategy or technique should we use? 1/17/2019

Typical way of proving subset. Relations - A Theorem If R is a transitive relation, then Rn  R, n. Proof by induction on n. Base case (n=1): R1  R because by definition, R1 = R. Induction case: if R is transitive, then Rk  R. Prove: if R is transitive, then Rk+1  R. Typical way of proving subset. We are trying to prove that Rk+1  R. To do this, we select an element of Rk+1 and show that it is also an element of R. Let (a,b) be an element of Rk+1. Since Rk+1 = Rk  R, we know there is an x so that (a,x)  R and (x,b)  Rk. By assumption at the induction step, since Rk  R, (x,b)  R. But wait, if (a,x)  R, and (x,b)  R, and R is transitive, then (a,b)  R. 1/17/2019

Relations - Another Theorem If R is a reflexive relation, then Rn is reflexive relation, n. Whiteboard! 1/17/2019

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 1/17/2019

Today’s Reading Rosen 9.1 and 9.3 1/17/2019