Lecture 4.2: Relations Basics

1. Lecture 4.2: Relations Basics CS 250, Discrete Structures, Fall 2011 NiteshSaxena *Adopted from previous lectures by CindaHeeren, ZephGrunschlag

2. Course Admin • Mid-Term 2 Exam • Solution will be posted soon • Should have the results by the coming weekend • HW3 • Solution will be posted soon • Results should be ready by the coming weekend Lecture 4.2 -- Relations

3. Outline • Relation Examples and Definitions • Matrix Representation • Closures Lecture 4.2 -- Relations

4. 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? Lecture 4.2 -- Relations

5. Composing Relations xRy iff y = x 2 xSy iff y = x 3 A: These are functions (squaring and cubing) so the composite SRis 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 Lecture 4.2 -- Relations

6. 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 Lecture 4.2 -- Relations

7. 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 Lecture 4.2 -- Relations

8. 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 Lecture 4.2 -- Relations

9. 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 Lecture 4.2 -- Relations

10. 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 Lecture 4.2 -- Relations

11. 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 -1from 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? Lecture 4.2 -- Relations

12. Inverting Relations A: xRy iff y = x 2. R is the square function so R -1is 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 Lecture 4.2 -- Relations

13. The labels on the outside are for clarity. It’s really the matrix in the middle that’s important. 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: This is a |A| x |B| matrix whose entries indicate membership in R. Lecture 4.2 -- Relations

14. 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. 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, ______________. Lecture 4.2 -- Relations

15. 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. 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, ______________. Lecture 4.2 -- Relations

16. MR1R2 = MR1 v MR2 MR1R2 = MR1  MR2 Relations – matrix representation Suppose we have R1 and R2 defined on A: Then R1  R2 is the bitwise “or” of the entries (Join By): Then R1  R2 is the bitwise “and” of the entries (Meet): Lecture 4.2 -- Relations

17. Relations – composition using matrices Suppose we have R and S defined on A: Then SR corresponds to the boolean product Lecture 4.2 -- Relations

18. Relations - A Theorem Theorem: If R is a transitive relation, then Rn R, n. How to prove? What strategy or technique should we use? Cs173 - Spring 2004

19. 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. 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. Cs173 - Spring 2004

20. Relations - Another Theorem If R is a reflexive relation, then Rn is reflexive relation, n. Whiteboard! Cs173 - Spring 2004

21. 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 Lecture 4.2 -- Relations

22. No (1,1), (4,4) R’ = R U {(1,1),(4,4)} is called the reflexive closure of R. Closure • Consider relation R={(1,2),(2,2),(3,3)} on the set A = {1,2,3,4}. • Is R reflexive? • What can we add to R to make it reflexive? Lecture 4.2 -- Relations

23. Today’s Reading • Rosen 9.1 and 9.3 Lecture 4.2 -- Relations