Binary Relations Part II

Transcription

1 Binary Relations Part II

2 Outline for Today Finish from Last Time Pt. 3 of our proof that ~ is an equivalence relation Properties of Equivalence Relations What s so special about those three rules? Strict Orders A diferent type of mathematical structure Hasse Diagrams How to visualize rankings

3 Finish from Last Time

4 a A. ara a A. b A. (arb bra) a A. b A. c A. (arb brc arc)

5 a~b if a+b is even Lemma 1: The binary relation ~ is refexive. Proof: Consider an arbitrary a Z. We need to prove that a~a. From the defnition of the ~ relation, this means that we need to prove that a+a is even. To see this, notice that a+a = 2a, so the sum a+a can be written as 2k for some integer k (namely, a), so a+a is even. Therefore, a~a holds, as required.

6 a~b if a+b is even Lemma 2: The binary relation ~ is symmetric. Proof: Consider any integers a and b where a~b. We need to show that b~a. Since a~b, we know that a+b is even. Because a+b = b+a, this means that b+a is even. Since b+a is even, we know that b~a, as required.

7 New Stuf!

8 a~b if a+b is even Lemma 3: The binary relation ~ is transitive. Proof: Consider arbitrary integers a, b and c where a~b and b~c. We need to prove that a~c, meaning that we need to show that a+c is even. Which Since a~b Which of and b~c, of the the following we following works know that works best a~b best as and as the b~c the are even. introduction This means there introduction to are integers to this this proof? k and proof? m where a+b = 2k A. A. and Pick Pick b+c an an arbitrary arbitrary = 2m. Notice a and and b that from from A where where a~b. a~b. We ll We ll prove prove b~a. b~a. B. B. Consider Consider any any a, a, b, b, c (a+b) c A where + where a~b, (b+c) a~b, b~c, = 2k b~c, and + 2m. and a~c. a~c. C. C. Choose Choose an an a, a, b, b, c c A. A. We We will will prove prove a~b, a~b, b~c, b~c, and and a~c. a~c. D. D. Rearranging, Take Take any any a, a, b, b, c we c A see where where thata~b a~b and and b~c; b~c; we ll we ll prove prove a~c. a~c. so a+c + 2b = 2k + 2m, a+c = 2k + 2m 2b = 2(k+m b). Answer Answer at at PollEv.com/cs103 PollEv.com/cs103 or or So there text text is CS103 CS103 an integer to to r, once once namely to to join, join, k+m b, then then A, A, B, B, such C, C, or or D. that D. a+c = 2r. Thus a+c is even, so a~c, as required.

9 a~b if a+b is even Lemma 3: The binary relation ~ is transitive. Proof: Consider arbitrary integers a, b and c where a~b and b~c. We need to prove that a~c, meaning that we need to show that a+c is even. Since a~b and b~c, we know that a~b and b~c are even. This means there are integers k and m where a+b = 2k and b+c = 2m. Notice that (a+b) + (b+c) = 2k + 2m. Rearranging, a we see that a Z. Z. b b Z. Z. c c Z. Z. (a (a ~ b b ~ c a ~ c) c) a+c + 2b = 2k + 2m, so What What is is the the formal formal defnition defnition of of transitivity? transitivity? Therefore, Therefore, we'll we'll choose choose arbitrary arbitrary integers integers a, a, b, b, and and c where where a ~ a+c b and = and 2k b + ~ c, 2m c, then then prove 2b = prove that 2(k+m b). that a ~ c. c. So there is an integer r, namely k+m b, such that a+c = 2r. Thus a+c is even, so a~c, as required.

10 a~b if a+b is even Lemma 3: The binary relation ~ is transitive. Proof: Consider arbitrary integers a, b and c where a~b and b~c. We need to prove that a~c, meaning that we need to show that a+c is even. Since a~b and b~c, we know that a+b and b+c are even. This means there are integers k and m where a+b = 2k and b+c = 2m. Notice that Rearranging, we see that so (a+b) + (b+c) = 2k + 2m. a+c + 2b = 2k + 2m, a+c = 2k + 2m 2b = 2(k+m b). So there is an integer r, namely k+m b, such that a+c = 2r. Thus a+c is even, so a~c, as required.

11 First-Order Logic and Proofs First-order logic is an excellent tool for giving formal defnitions to key terms. While frst-order logic guides the structure of proofs, it is exceedingly rare to see frst-order logic in written proofs. Follow the example of these proofs: Use the FOL defnitions to determine what to assume and what to prove. Write the proof in plain English using the conventions we set up in the frst week of the class. Please, please, please, please, please internalize the contents of this slide!

12 a A. ara a A. b A. (arb bra) a A. b A. c A. (arb brc arc)

13 Properties of Equivalence Relations

14 xry if x and y have the same shape

15 xty if x and y have the same color

16 Equivalence Classes Given an equivalence relation R over a set A, for any x A, the equivalence class of x is the set [x] R = { y A xry } Intuitively, the set [x]r contains all elements of A that are related to x by relation R.

17 [ ] R [ ] R [ ] R xry if x and y have the same shape

18 The Fundamental Theorem of Equivalence Relations: Let R be an equivalence relation over a set A. Then every element a A belongs to exactly one equivalence class of R.

19 How d We Get Here? We discovered equivalence relations by thinking about partitions of a set of elements. We saw that if we had a binary relation that tells us whether two elements are in the same group, it had to be refexive, symmetric, and transitive. The FToER says that, in some sense, these rules precisely capture what it means to be a partition. Question: What s so special about these three rules?

20 a A. b A. c A. (arb brc cra)

21 A binary binary relation relation with with this this property property is is called called cyclic. cyclic. a A. b A. c A. (arb brc cra)

22 Let Let R be be the the relation relation depicted depicted here. here. How How many many of of the the following following claims claims are are true? true? R is is refexive. refexive. R is is symmetric. symmetric. R is is transitive. transitive. R is is an an equivalence equivalence relation. relation. Answer Answer at at PollEv.com/cs103 PollEv.com/cs103 or or text text CS103 CS103 to to once once to to join, join, then then 0, 0, 1, 1, 2, 2, 3, 3, or or a A. b A. c A. (arb brc cra)

23 Theorem: A binary relation R over a set A is an equivalence relation if and only if it is refexive and cyclic.

24 Lemma 1: If R is an equivalence relation over a set A, then R is refexive and cyclic. Lemma 2: If R is a binary relation over a set A that is refexive and cyclic, then R is an equivalence relation.

25 Lemma 1: If R is an equivalence relation over a set A, then R is refexive and cyclic. What We re Assuming What We Need To Show R is an equivalence relation. R is refexive. R is cyclic. R is refexive. R is symmetric. R is transitive.

26 Lemma 1: If R is an equivalence relation over a set A, then R is refexive and cyclic. What We re Assuming What We Need To Show R is an equivalence relation. R is refexive. R is symmetric. R is refexive. R is cyclic. If arb and brc, then cra. R is transitive.

27 Lemma 1: If R is an equivalence relation over a set A, then R is refexive and cyclic. What We re Assuming What We Need To Show R is an equivalence relation. If arb and brc, then cra. R is refexive. R is symmetric. R is transitive. a b c

28 Lemma 1: If R is an equivalence relation over a set A, then R is refexive and cyclic. What We re Assuming What We Need To Show R is an equivalence relation. If arb and brc, then cra. R is refexive. R is symmetric. R is transitive. a b c

29 Lemma 1: If R is an equivalence relation over a set A, then R is refexive and cyclic. Proof: Let R be an arbitrary equivalence relation over some set A. We need to prove that R is refexive and cyclic. Since R is an equivalence relation, we know that R is refexive, symmetric, and transitive. Consequently, we already know that R is refexive, so we only need to show that R is cyclic. To prove that R is cyclic, consider any arbitrary a, b, c A where arb and brc. We need to prove that cra holds. Since R is transitive, from arb and brc we see that arc. Then, since R is symmetric, from arc we see that cra, which is what we needed to prove.

30 Lemma 1: If R is an equivalence relation over a set A, then R is refexive and cyclic. Proof: Let R be an arbitrary equivalence relation over some set A. We need to prove that R is refexive and cyclic. Since R is an equivalence relation, we know that R is refexive, symmetric, Notice and transitive. Consequently, we Notice how how the the frst frst few few sentences sentences of of this this already know that R proof is refexive, so we only need to show proof mirror mirror the the structure structure of of what what needs needs that R is cyclic. to to be be proved. proved. We re We re just just following following the the To prove that R is cyclic, templates templates consider from from the any the frst frst arbitrary week week of of a, class! class! b, c A where arb and brc. We need to prove that cra holds. Since R is transitive, from arb and brc we see that arc. Then, since R is symmetric, from arc we see that cra, which is what we needed to prove.

31 Notice Notice how how this this setup setup mirrors mirrors the the frst-order frst-order Lemma 1: If R is an equivalence defnition relation over a set A, then R defnition of of cyclicity: cyclicity: is refexive and cyclic. Proof: Let a a R be A. A. an b b arbitrary A. A. c c equivalence A. A. (arb (arb relation brc brc over cra) cra) some set A. We need to prove that R is refexive and cyclic. When When writing writing proofs proofs about about terms terms with with frst-order frst-order Since R is an defnitions, equivalence relation, we know that R is defnitions, it s it s critical critical to to call call back back to to those those refexive, symmetric, and defnitions! transitive. Consequently, we defnitions! already know that R is refexive, so we only need to show that R is cyclic. To prove that R is cyclic, consider any arbitrary a, b, c A where arb and brc. We need to prove that cra holds. Since R is transitive, from arb and brc we see that arc. Then, since R is symmetric, from arc we see that cra, which is what we needed to prove.

32 Although Although this this proof proof is is deeply deeply informed informed by by the the frst-order frst-order defnitions, defnitions, notice notice that that there there is is no no frst-order frst-order logic logic notation notation anywhere anywhere in in the the proof. proof. That s That s normal normal it s it s actually actually Lemma 1: If R is an equivalence relation over a set A, then R quite quite rare rare to to see see frst-order frst-order logic logic in in written written proofs. proofs. is refexive and cyclic. Proof: Let R be an arbitrary equivalence relation over some set A. We need to prove that R is refexive and cyclic. Since R is an equivalence relation, we know that R is refexive, symmetric, and transitive. Consequently, we already know that R is refexive, so we only need to show that R is cyclic. To prove that R is cyclic, consider any arbitrary a, b, c A where arb and brc. We need to prove that cra holds. Since R is transitive, from arb and brc we see that arc. Then, since R is symmetric, from arc we see that cra, which is what we needed to prove.

33 Lemma 1: If R is an equivalence relation over a set A, then R is refexive and cyclic. Proof: Let R be an arbitrary equivalence relation over some set A. We need to prove that R is refexive and cyclic. Since R is an equivalence relation, we know that R is refexive, symmetric, and transitive. Consequently, we already know that R is refexive, so we only need to show that R is cyclic. To prove that R is cyclic, consider any arbitrary a, b, c A where arb and brc. We need to prove that cra holds. Since R is transitive, from arb and brc we see that arc. Then, since R is symmetric, from arc we see that cra, which is what we needed to prove.

34 Lemma 2: If R is a binary relation over a set A that is refexive and cyclic, then R is an equivalence relation. What We re Assuming What We Need To Show R is refexive. R is an equivalence relation. R is cyclic. R is refexive. R is symmetric. R is transitive.

35 Lemma 2: If R is a binary relation over a set A that is refexive and cyclic, then R is an equivalence relation. What We re Assuming What We Need To Show R is refexive. R is cyclic. R is an equivalence relation. R is refexive. R is symmetric. R is transitive.

36 Lemma 2: If R is a binary relation over a set A that is refexive and cyclic, then R is an equivalence relation. What We re Assuming What We Need To Show R is refexive. R is symmetric. x A. xrx If arb, then bra. R is cyclic. xry yrz zrx a b

37 Lemma 2: If R is a binary relation over a set A that is refexive and cyclic, then R is an equivalence relation. What We re Assuming What We Need To Show R is refexive. R is symmetric. x A. xrx If arb, then bra. R is cyclic. y x z xry yrz zrx a b

38 Lemma 2: If R is a binary relation over a set A that is refexive and cyclic, then R is an equivalence relation. What We re Assuming What We Need To Show R is refexive. R is transitive. x A. xrx If arb and brc, R is cyclic. xry yrz zrx then arc. a b c

39 Lemma 2: If R is a binary relation over a set A that is refexive and cyclic, then R is an equivalence relation. What We re Assuming What We Need To Show R is refexive. R is transitive. x A. xrx If arb and brc, R is cyclic. then arc. xry yrz zrx a b c

40 Lemma 2: If R is a binary relation over a set A that is refexive and cyclic, then R is an equivalence relation. What We re Assuming What We Need To Show R is refexive. R is transitive. x A. xrx If arb and brc, R is cyclic. xry yrz zrx R is symmetric xry yrx then arc. a b c

41 Lemma 2: If R is a binary relation over a set A that is cyclic and refexive, then R is an equivalence relation. Proof: Let R be an arbitrary binary relation over a set A that is cyclic and refexive. We need to prove that R is an equivalence relation. To do so, we need to show that R is refexive, symmetric, and transitive. Since we already know by assumption that R is refexive, we just need to show that R is symmetric and transitive. First, we'll prove that R is symmetric. To do so, pick any arbitrary a, b A where arb holds. We need to prove that bra is true. Since R is refexive, we know that ara holds. Therefore, by cyclicity, since ara and arb, we learn that bra, as required. Next, we'll prove that R is transitive. Let a, b, and c be any elements of A where arb and brc. We need to prove that arc. Since R is cyclic, from arb and brc we see that cra. Earlier, we showed that R is symmetric. Therefore, from cra we see that arc is true, as required.

42 Lemma Notice 2: If R is a binary relation over a set A that is cyclic Notice how how this this setup setup mirrors mirrors the the frst-order frst-order defnition defnition and refexive, then R is an equivalence relation. of of symmetry: symmetry: Proof: Let R be an arbitrary binary relation over a set A that is cyclic and refexive. a a A. A. b b We need A. A. (arb (arb to prove bra) bra) that R is an equivalence relation. To do so, we need to show that R is refexive, When When symmetric, writing writing proofs proofs and about about transitive. terms terms with with Since frst-order frst-order we already know defnitions, defnitions, by assumption it s it s critical critical that to to R call call is refexive, back back to to those those we defnitions! defnitions! just need to show that R is symmetric and transitive. First, we'll prove that R is symmetric. To do so, pick any arbitrary a, b A where arb holds. We need to prove that bra is true. Since R is refexive, we know that ara holds. Therefore, by cyclicity, since ara and arb, we learn that bra, as required. Next, we'll prove that R is transitive. Let a, b, and c be any elements of A where arb and brc. We need to prove that arc. Since R is cyclic, from arb and brc we see that cra. Earlier, we showed that R is symmetric. Therefore, from cra we see that arc is true, as required.

43 Lemma 2: If R is a binary relation over a set A that is cyclic and refexive, then R is an equivalence relation. Proof: Let R be an arbitrary binary relation over a set A that is cyclic and refexive. We need to prove that R is an equivalence relation. To do so, we need to show that R is refexive, Notice Notice symmetric, how how this this setup setup and mirrors transitive. mirrors the the frst-order frst-order Since we defnition defnition already know by assumption that of of R transitivity: transitivity: is refexive, we just need to show that R is symmetric and transitive. a First, we'll a A. prove A. b b that A. A. R c is symmetric. A. A. (arb (arb To brc brc do so, arc) pick arc) any arbitrary a, b A where arb holds. We need to prove that When bra is true. When writing Since writing proofs R is proofs about refexive, about terms we terms with know with frst-order frst-order that ara holds. defnitions, Therefore, defnitions, it s by cyclicity, it s critical critical to since to call ara call back back to and to those arb, those defnitions! we defnitions! learn that bra, as required. Next, we'll prove that R is transitive. Let a, b, and c be any elements of A where arb and brc. We need to prove that arc. Since R is cyclic, from arb and brc we see that cra. Earlier, we showed that R is symmetric. Therefore, from cra we see that arc is true, as required.

44 Lemma 2: If R is a binary relation over a set A that is cyclic and refexive, then R is an equivalence relation. Proof: Let R be an arbitrary binary relation over a set A that is cyclic and refexive. We need to prove that R is an equivalence relation. To do so, we need to show that R is refexive, symmetric, and transitive. Since we already know by assumption that R is refexive, we just need to show that R is symmetric and transitive. First, we'll prove that R is symmetric. To do so, pick any arbitrary a, b A where arb holds. We need to prove that bra is true. Since R is refexive, we know that ara holds. Therefore, by cyclicity, since ara and arb, we learn that bra, as required. Next, we'll prove that R is transitive. Let a, b, and c be any elements of A where arb and brc. We need to prove that arc. Since R is cyclic, from arb and brc we see that cra. Earlier, we showed that R is symmetric. Therefore, from cra we see that arc is true, as required.

45 Refning Your Proofwriting When writing proofs about terms with formal defnitions, you must call back to those defnitions. Use the frst-order defnition to see what you ll assume and what you ll need to prove. When writing proofs about terms with formal defnitions, you must not include any frst-order logic in your proofs. Although you won t use any FOL notation in your proofs, your proof implicitly calls back to the FOL defnitions. You ll get a lot of practice with this on Problem Set Three. If you have any questions about how to do this properly, please feel free to ask on Piazza or stop by ofice hours!

46 Prerequisite Structures

47 Theory The CS Core CS106B CS103 Programming Abstractions Mathematical Foundations of Computing Systems CS107 Computer Organization and Systems CS109 Intro to Probability for Computer Scientists CS110 CS161 Principles of Computer Systems Design and Analysis of Algorithms

48

49 Pancakes Everyone's got a pancake recipe. This one comes from Food Wishes ( Ingredients 1 1/2 cups all-purpose four 3 1/2 tsp baking powder 1 tsp salt 1 tbsp sugar 1 1/4 cup milk 1 egg 3 tbsp butter, melted Directions 1. Sift the dry ingredients together. 2. Stir in the butter, egg, and milk. Whisk together to form the batter. 3. Heat a large pan or griddle on medium-high heat. Add some oil. 4. Make pancakes one at a time using 1/4 cup batter each. They're ready to fip when the centers of the pancakes start to bubble.

50 Measure Flour Measure Sugar Measure Baking Pwdr Measure Salt Heat Griddle Beat Egg Combine Dry Ingredients Melt Butter Measure Milk Oil Griddle Add Wet Ingredients Make Pancakes Serve Pancakes

51

52 Relations and Prerequisites Let's imagine that we have a prerequisite structure with no circular dependencies. We can think about a binary relation R where arb means a must happen before b What properties of R could we deduce just from this?

53 a a a b b c a c a b b a

54 a A. ara a A. b A. c A. (arb brc arc) a A. b A. (arb bra)

55 Irrefexivity Some relations never hold from any element to itself. As an example, x x for any x. Relations of this sort are called irrefexive. Formally speaking, a binary relation R over a set A is irrefexive if the following frst-order logic statement is true about R: a A. ar a ( No element is related to itself. )

56 Irrefexivity Visualized a A. ar a ( No element is related to itself. )

57 Let Let R be be the the relation relation depicted depicted here. here. How How many many of of the the following following claims claims are are true? true? R is is refexive. refexive. R is is not not refexive. refexive. R is is irrefexive. irrefexive. R is is not not irrefexive. irrefexive. Answer Answer at at PollEv.com/cs103 PollEv.com/cs103 or or text text CS103 CS103 to to once once to to join, join, then then 0, 0, 1, 1, 2, 2, 3, 3, or or 4. 4.

58 Is Is this this relation relation refexive? refexive? Nope! a A. ara ( Every element is related to itself. )

59 Is Is this this relation relation irrefexive? irrefexive? Nope! a A. ar a ( No element is related to itself. )

60 Refexivity and Irrefexivity Refexivity and irrefexivity are not opposites! Here's the defnition of refexivity: a A. ara What is the negation of the above statement? a A. ar a What is the defnition of irrefexivity? a A. ar a

61 a A. ara Transitivity a A. b A. (arb bra)

62 Asymmetry In some relations, the relative order of the objects can never be reversed. As an example, if x y, then y x. These relations are called asymmetric. Formally: a binary relation R over a set A is called asymmetric if the following frst-order logic statement is true about R: a A. b A. (arb br a) ( If a relates to b, then b does not relate to a. )

63 Asymmetry Visualized a A. b A. (arb br a) ( If a relates to b, then b does not relate to a. )

64 Question to Ponder: Are symmetry and asymmetry opposites of one another?

65 Strict Orders A strict order is a relation that is irrefexive, asymmetric and transitive. Some examples: x y. a can run faster than b. A B (that is, A B and A B). Strict orders are useful for representing prerequisite structures and have applications in complexity theory (measuring notions of relative hardness) and algorithms (searching and sorting).

66 Drawing Strict Orders

67 Gold Silver Bronze

68 (46, 37, 38) (26, 18, 26) (19, 18, 19) (27, 23, 17) (12, 8, 21) (10, 18, 14) (17, 10, 15) (g₁, s₁, b₁) R (g₂, s₂, b₂) if g₁ g₂ s₁ s₂ b₁ b₂

69 (46, 37, 38) (26, 18, 26) (19, 18, 19) (27, 23, 17) (12, 8, 21) (10, 18, 14) (17, 10, 15) (g₁, s₁, b₁) R (g₂, s₂, b₂) if g₁ g₂ s₁ s₂ b₁ b₂

70 (46, 37, 38) (26, 18, 26) (19, 18, 19) (27, 23, 17) (12, 8, 21) (10, 18, 14) (17, 10, 15) (g₁, s₁, b₁) R (g₂, s₂, b₂) if g₁ g₂ s₁ s₂ b₁ b₂

71 More Medals (46, 37, 38) (26, 18, 26) (19, 18, 19) (27, 23, 17) (12, 8, 21) (17, 10, 15) (10, 18, 14) Fewer Medals (g₁, s₁, b₁) R (g₂, s₂, b₂) if g₁ g₂ s₁ s₂ b₁ b₂

72 Hasse Diagrams A Hasse diagram is a graphical representation of a strict order. Elements are drawn from bottom-to-top. No self loops are drawn, and none are needed! By irrefexivity we know they shouldn t be there. Higher elements are bigger than lower elements: by asymmetry, the edges can only go in one direction. No redundant edges: by transitivity, we can infer the missing edges.

73 (46, 37, 38) 379 (27, 23, 17) 221 (26, 18, 26) 210 (19, 18, 19) 168 (17, 10, 15) 130 (10, 18, 14) 118 (12, 8, 21) 105 (g₁, s₁, b₁) T (g₂, s₂, b₂) if 5g₁ + 3s₁ + b₁ 5g₂ + 3s₂ + b₂

74 (46, 37, 38) 121 (26, 18, 26) 72 (27, 23, 17) 67 (19, 18, 19) 56 (17, 10, 15) 42 (10, 18, 14) 42 (g₁, s₁, b₁) U (g₂, s₂, b₂) (12, 8, 21) 41 if g₁ + s₁ + b₁ g₂ + s₂ + b₂

75 xry if x must be eaten before y Hasse Artichokes

76 The Meta Strict Order Strict Order Equivalence Relation Asymmetry Transitivity Refexivity Symmetry Irrefexivity Question Question to to ponder: ponder: why why is is this this line line here? here? arb if a is less specifc than b

77 Next Time Functions How do we model transformations in a mathematical sense? Domains and Codomains Type theory meets mathematics! Injections, Surjections, and Bijections Three special classes of functions.