Download - Lecture 4.2: Relations Basics
Lecture 4.2: Relations Basics
CS 250, Discrete Structures, Fall 2011
Nitesh Saxena
*Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag
10/27/2011 Lecture 4.2 -- Relations
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10/27/2011 Lecture 4.2 -- Relations
Outline
Relation Examples and Definitions Matrix Representation Closures
10/27/2011
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 SR ?
Lecture 4.2 -- Relations
10/27/2011
Composing RelationsxRy iff y = x 2 xSy iff y = x 3
A: These are functions (squaring and cubing) so the composite SR is just the function composition (raising to the 6th power). xSRy iff y = x 6 (in this odd case SR = RS)
Q: Compose the following:1 1 1 12 2 2 23 3 3 34 4 4
5 5Lecture 4.2 -- Relations
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Composing Relations1 1 12 2 23 3 34 4
5A: Draw all possible shortcuts. In our case,
all shortcuts went through 1:1 12 23 34
Lecture 4.2 -- Relations
10/27/2011
Composing Relations: Picture1 1 12 2 23 3 34 4
5A: Draw all possible shortcuts. In our case,
all shortcuts went through 1:1 12 23 34
Lecture 4.2 -- Relations
10/27/2011
Composing Relations: Picture1 1 12 2 23 3 34 4
5A: Draw all possible shortcuts. In our case,
all shortcuts went through 1:1 12 23 34
Lecture 4.2 -- Relations
10/27/2011
Composing Relations: Picture1 1 12 2 23 3 34 4
5A: Draw all possible shortcuts. In our case,
all shortcuts went through 1:1 12 23 34
Lecture 4.2 -- Relations
10/27/2011
Composing Relations: Picture1 1 12 2 23 3 34 4
5A: Draw all possible shortcuts. In our case,
all shortcuts went through 1:1 12 23 34
Lecture 4.2 -- Relations
10/27/2011
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 ?
Lecture 4.2 -- Relations
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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
Lecture 4.2 -- Relations
Lecture 4.2 -- Relations
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.
u v w
1 1 1 0
2 0 0 1
3 0 0 1
4 1 0 0
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Lecture 4.2 -- Relations
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, ______________.
A. All entries in MR are 1.
B. The \ diagonal of MR contains only 1s.
C. The first column of MR contains no 0s.
D. None of the above.
1***
*1**
**1*
***1
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Lecture 4.2 -- Relations
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, ______________.
A. All entries above the \ are 1.
B. The first and last columns of MR contain an equal # of 0s.
C. MR is visually symmetric about the \ diagonal.
D. None of the above.
*101
1*01
00*0
110*
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Lecture 4.2 -- Relations
Relations – matrix representation
Suppose we have R1 and R2 defined on A:
R1 u v w
u 1 0 1
v 0 0 1
w 1 1 0
R2 u v w
u 1 1 0
v 0 1 1
w 0 0 1
Then R1 R2 is the bitwise “or” of the entries (Join By):
Then R1 R2 is the bitwise “and” of the entries (Meet):
1 1 1
0 1 1
1 1 1
1 0 0
0 0 1
0 0 0
MR1R2 = MR1 v MR2
MR1R2 = MR1 MR2
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Lecture 4.2 -- Relations
Relations – composition using matrices
Suppose we have R and S defined on A:
R u v w
u 1 0 1
v 0 0 1
w 1 1 0
S u v w
u 1 1 0
v 0 1 1
w 0 0 1
Then SR corresponds to the boolean product1 1 1
0 0 1
1 1 1
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Cs173 - Spring 2004
Relations - A TheoremTheorem: If R is a transitive relation, then Rn R, n.
How to prove? What strategy or technique should we use?
Cs173 - Spring 2004
Relations - A TheoremIf R is a transitive relation, then Rn R, n.
Typical way of provin
g subset.
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
Relations - Another TheoremIf R is a reflexive relation, then Rn is reflexive relation, n.
Whiteboard!
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
10/27/2011 Lecture 4.2 -- Relations
Lecture 4.2 -- Relations
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?
No
(1,1), (4,4)
R’ = R U {(1,1),(4,4)} is called the reflexive closure of
R.
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10/27/2011 Lecture 4.2 -- Relations
Today’s Reading Rosen 9.1 and 9.3