1 / 21

Exam 2 Review

Exam 2 Review. 7.5, 7.6, 8.1-8.6. 7.5. |A1  A2  A3| =∑|Ai| - ∑|Ai ∩ Aj | + |A1∩ A2 ∩ A3| |A1  A2  A3  A4| =∑|Ai| - ∑|Ai ∩ Aj | + ∑ |Ai∩ Aj ∩ Ak | - |A1∩ A2 ∩ A3∩ A4|. 7.6. Let A i =subset containing elements with property P i

doyle
Télécharger la présentation

Exam 2 Review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Exam 2 Review 7.5, 7.6, 8.1-8.6

  2. 7.5 |A1  A2  A3| =∑|Ai| - ∑|Ai ∩ Aj| + |A1∩ A2 ∩ A3| |A1  A2  A3  A4| =∑|Ai| - ∑|Ai ∩ Aj| + ∑ |Ai∩ Aj ∩ Ak| - |A1∩ A2 ∩ A3∩ A4|

  3. 7.6 Let Ai=subset containing elements with property Pi N(P1P2P3…Pn)=|A1∩A2∩…∩An| N(P1’ P 2 ‘ P 3 ‘…Pn ‘)= number of elements with none of the properties P1, P2, …Pn =N - |A1  A2  …  An| =N- (∑|Ai| - ∑|Ai ∩ Aj| + … +(-1)n+1|A1∩ A2 ∩…∩ An|) = N - ∑ N (Pi) + ∑(PiPj) -∑N(PiPjPk) +… +(-1)nN(P1P2…Pn)

  4. Sample applications • Ex 1: How many solutions does x1+x2+x3= 11 have where xi is a nonnegative integer with x1≤ 3, x2≤ 4, x3≤ 6 (note: harder than previous > problems) • Ex: 2: How many onto functions are there from a set A of 7 elements to a set B of 3 elements

  5. … • Ex. 3: Sieve- primes • Ex. 4: Hatcheck-- The number of derangements of a set with n elements is Dn= n![1 - ] • Derangement formula will be given.

  6. 8.1- Relations • Def. of Function: f:A→B assigns a unique element of B to each element of A • Def of Relation?

  7. RSAT A relation R on a set A is called: • reflexive if (a,a)  R for every a A • symmetric if (b,a) R whenever (a,b)  R for a,b A • antisymmetric : (a,b)  R and (b,a)  R only if a=b for a,b A • transitive if whenever (a,b)  R and (b,c) R, then (a,c) R for a,b,c A

  8. RSAT A relation R on a set A is called: • reflexive if aRa for every a  A • symmetric if bRa whenever aRb for every a,b A • antisymmetric : aRb and bRa only if a=b for a,b A • transitive if whenever aRb and bRc, then aRc for every a, b, c  A • Do Proofs of these****

  9. Combining relations R∩S RS R – S S – R S ο R = {(a,c)| a  A, c C, and there exists b  B such that (a,b)  R and (b,c)  S} Rn+1=Rn⃘ R

  10. Thm 1 on 8.1 • Theorem 1: Let R be a transitive relation on a set A. Then Rn is a subset of R for n=1,2,3,… • Proof • 8.2– not much on this – just joins and projections

  11. 8.3 • Representing relations R on A as both matrices and as digraphs (directed graphs) • Zero-one matrix operations: join, meet, Boolean product • MR R6 = MR5 v MR6 • MR5∩R6 = MR5 ^ MR6 • MR6 °R5 = MR5 MR6

  12. 8.4 • Def: Let R be a relation on a set A that may or may not have some property P. (Ex: Reflexive,…) If there is a relation S with property P containing R such that S is a subset of every relation with property P containing R, then S is called the closure of R with respect to P. • Find reflexive and symmetric closures

  13. Transitive closures 8.4: Theorem 1: Let R be a relation on a set A. There is a path of length n from a to b iff (a,b)Rn --In examples, find paths of length n that correspond to elements inRn

  14. R* • Find R*= • Sample mid-level proofs: • R* is transitive

  15. 8.5 and 8.6 • Equivalence Relations: R, S, T • Partial orders: R, A, T • (see definitions in other notes)

  16. Definitions to thoroughly know and use • a divides b • ab mod m • Relation • Reflexive, symmetric, antisymmetric, transitive (not ones like asymmetric, from hw) • Equivalence Relation- RST • Partial Order- RAT • Comparable • Total Order

  17. Definitions to be apply to apply • You won’t have to state word for word, but may need to apply: • Maximal, minimal, greatest, least element • Formulas in 7.5 and 7.6

  18. Thereoms to know well and use • 8.1: Theorem 1: Let R be a transitive relation on a set A. Then Rn is a subset of R for n=1,2,3,… • 8.4: Theorem 1: Let R be a relation on a set A. There is a path of length n from a to b iff (a,b)Rn • 8.4: Thm. 2: The transitive closure of a relation R is R* =

  19. Mid-level proofs to be able to do • Prove that a given relation R, S, A, or T using the definitions • Ex: Show (Z+,|) is antisymmemetric • Ex: Show R={(a,b)|ab mod m} on Z+ is transitive • Some basic proofs by induction • Let R be a transitive relation on a set A. Then Rn is a subset of R for n=1,2,3,… • R* is transitive • Provide a counterexample to disprove that a relations is R, S, A, or T • Ex: Show R={(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,),(4,4)} on {1,2,3,4} is not transitive

  20. Procedures to do • Represent relations as ordered pairs, matrices, or digraphs • Find Pxy and Jx and composite keys (sed 8.2) • Create relations with designated properties (ex: reflexive, but not symmetric • Determine whether a relation has a designated property • Find closures (ex: reflexive, transitive) • Find paths and circuits of a certain length and apply section 8.4 Thm. 1 • Calculate R∩S,RS,R – S,S – R,S ο R,Rn+1=Rn⃘ R • Given R, describe an ordered pair in R3 • Given an equivalence R on a set S, find the partition… and vice versa • Identify examples and non-examples of eq. relations and of posets • Create and work with Hasse diagrams: max, min, lub,…

More Related