Discrete math with computers_3

•n1/m R → n1/m+1 R

•Nothing is in R unless it can be shown to be in R by a finite number of uses of the base and induction rules.

24.If n is a positive multiple of 2 then yes, otherwise no.

25.The odd positive integers.

26.By rule (1), 0 S

By rule (2) 0 S → 2 S and so by Modus Ponens 2 S. By rule (2) 2 S → 4 S so 4 S.

27.

1. "" Y Y S

2. "" Y Y S → "yy" Y Y S and so by Modus Ponens, "yy" Y Y S.

3. "yy" Y Y S → "yyyy" Y Y S and so by Modus Ponens, "yyyy" Y Y S.

28.

zs = "" : map (’z’:) zs

29. The set SS of strings of spaces of length less than or equal to n is defined as follows:

1. ”” SS

2. ss SS length ss < n → ’ ’: ss SS

3.Nothing is in SS unless it can be shown to be in SS by a finite number of uses of rules (1) and (2).

30.

ss :: Integer -> [String] ss 0 = []

ss n = take n (repeat ’ ’) : ss (n-1)

31. The set of sets of naturals SSN, each of which is missing a distinct natural number, is defined inductively as follows:

1.N − {0} SSN

2.(N − {n}) SSN → (N − {n + 1} SSN

3.Nothing is in SSN unless it can be shown to be in SSN by a finite number of uses of rules (1) and (2).

32.

1.I − {−1} SSI−

2.I − {−1} → I − {−2} SSI−

3.I − {−2} → I − {−3} SSI−

4.By Modus Ponens, I − {−3} SSI−.

33.

1.−1 ON I

2.−1 ON I → −3 ON I so −3 ON I.

3.−3 ON I → −5 ON I so −5 ON I.

4.−5 ON I → −7 ON I so −7 ON I.

34.

decrement :: Integer -> Integer decrement x = x - 1

ni = -1 : map decrement ni

35.No, because 0 will be paired first with every element in [0..], which is infinitely long.

36.The set is {0, 1}.

C.10 Relations

45.

•(a) Yes.

•(b) This depends upon whether the relationship is defined by DNA, in which case all of us are somewhat related to each other, or last name, in which case some people named Smith are not actually closely related physically. ’No’ seems a reasonable answer.

•(c) No, because equivalence relations are reflexive, and you cannot be bigger than yourself.

•(d) Yes.

46.Yes. Here is an example:

[(1,2),(2,3),(3,1)]

47. The empty relation.

48.

checkPowers :: (Eq a, Show a) => Digraph a -> Bool checkPowers (set,relation)

= any (setEq relation) [relationalPower (set,relation) n

|n <- [2..length (domain relation)]]

49.No. If it did, then the end of each loop would have to have a reflexive loop, because the relation is transitive. Thus it would not be irreflexive.

50.Yes. For example

[(1,2),(2,3),(3,1)]

has a transitive closure that is reflexive and symmetric.

51.

fewerArcs :: (Eq a, Show a) => Digraph a -> Bool fewerArcs (set,relation)

= all

(< (length relation))

[length (relationalPower (set,relation) n) | n <- [2..length (domain relation)]]

52.

isSmaller :: (Ord a, Show a) => Digraph a -> Bool isSmaller (set,relation)

= let (symset,symrelation) = symmetricClosure (set,relation)

in length relation < length symrelation

53.The last is not a topological sort.

54.Yes, for example this is reflexive, symmetric, and antisymmetric:

[(1,1),(2,2),(3,3)]

55. The number of arcs the transitive closure will add is 1 + 2 + ... + n − 1. The symmetric closure will double each of these, so the total is 2(1+2+...+n−1).

56.Power n.

57.No, because the function could not examine every arc in the relation.

58.Given a partial order, assume that it has a cycle of length n. Because it is

transitive, it also has cycles of lengths 1 to n − 1. But that means that it has a cycle of length 2, which cannot be because a partial order is antisymmetric. So it cannot both be a partial order and have cycles greater than 1.

59.No. Some have powers that repeat, such as the relation

[(1,2),(2,1)]

60.A linear order.

61.Yes. For example, the composition of these two relations

[(Red,Blue),(Blue,Green),(Green,Yellow),(Yellow,Red)]

[(Blue,Red),(Green,Blue),(Yellow,Green),(Red,Yellow)]

yields

[(Red,Red),(Blue,Blue),(Green,Green),(Yellow,Yellow)]

62. The empty relation and the equality relation.

C.11 Functions

26. The functions surjExp, injExp, and bijExp each take two domains and images and test the corresponding hypothesis.

surjectiveExperiment :: (Eq a, Show a) => Set a -> Set a -> Set (a, FunVals a) ->

Set a -> Set a -> Set (a, FunVals a) -> Bool surjectiveExperiment dom_f co_f f dom_g co_g g

=isSurjective dom_f co_f f /\ isSurjective dom_g co_g g ==>

isSurjective dom_g co_f (functionalComposition f g)

surjExp :: Set Int -> Set Int ->

Set Int -> Set Int -> Bool surjExp domain_f image_f domain_g image_g

= let f = [(x,Value y) | x <- domain_f, y <- image_f] g = [(x,Value y) | x <- domain_g, y <- image_g]

in

surjectiveExperiment [1..10] [1..10] f [1..10] [1..10] g

injectiveExperiment :: (Eq a, Show a) => Set a -> Set a -> Set (a, FunVals a) ->

Set a -> Set a -> Set (a, FunVals a) -> Bool injectiveExperiment dom_f co_f f dom_g co_g g

=isInjective dom_f co_f f /\ isInjective dom_g co_g g ==>

isInjective dom_g co_f (functionalComposition f g)

injExp :: Set Int -> Set Int ->

Set Int -> Set Int -> Bool injExp domain_f image_f domain_g image_g

= let f = [(x,Value y) | x <- domain_f, y <- image_f] g = [(x,Value y) | x <- domain_g, y <- image_g]

in injectiveExperiment [1..10] [1..10] f [1..10] [1..10] g

bijectiveExperiment :: (Eq a, Show a) => Set a -> Set a -> Set (a, FunVals a) ->

Set a -> Set a -> Set (a, FunVals a) -> Bool bijectiveExperiment dom_f co_f f dom_g co_g g = isBijective dom_g co_f

(functionalComposition f g) ==>

isSurjective dom_f co_f f /\ isInjective dom_g co_g g

bijExp :: Set Int -> Set Int ->

Set Int -> Set Int -> Bool bijExp domain_f image_f domain_g image_g

= let f = [(x,Value y) | x <- domain_f, y <- image_f] g = [(x,Value y) | x <- domain_g, y <- image_g]

in bijectiveExperiment [1..10] [1..10] f [1..10] [1..10] g

27.1,1,4,4

28.The function f is a permutation, g is an identity function, and h is the inverse of f.

29.constant function, injection, constant function.

30.Only g is a function. f maps 3 to two values, and g does not map 5 to any value.

31.g is a total function, and the other two are partial functions.

32.This can be solved by making a simple table for each function showing the results it returns for all inputs, and then referring to the definition of surjection.

35. The first function is partial, because it does not terminate when given False, and so does not produce a result given that domain value. The second function is total.

36.

isFun :: (Eq a, Show a) => Set (a,b) -> Bool isFun ps = normalForm (map fst ps)

37.

isInjection :: (Eq a, Eq b, Show a, Show b) => Set (a,b) -> Bool

isInjection ps

=normalForm (map fst ps) /\ normalForm (map snd ps)

39.You only need to know its type. If the domain type is the same as that of the codomain, then it might be the identity function, otherwise it certainly isn’t. There are many properties of functions that can be deduced solely from the function type.

40.

compare f g h = ((g.f) a) == (h a)

41.Yes, for natural numbers. If it receives a negative integer, it will loop indefinitely.

42.Yes, for all integers.

43.

For problems 2 and 3, we will use the bullet/target metaphor of functions. In this metaphor, the domain of the function is regarded as the set of bullets. The codomain (also known as the range) is regarded as the set of targets. When x comes from the domain and y comes from the codomain and y = f (x ), the target y is said to be hit by the bullet x .

Following this metaphor, we can define some of the terminology as follows:

•Function. A function is a subset of the Cartesian product of the set of bullets with the set of targets that does not use the same bullet for more than one target.

•Surjective. A function is surjective if it hits all of the targets.

•Injective. A function is injective if it hits no targets more than once.

1.Yes, f is a function because it is a subset of A × B with the property that whenever both (a, b) and (a, c) are elements of f , it is the case that b = c. That’s what it means to be a function of type A → B . Or, using the metaphor, f is a function because no single bullet hits two or more targets.

2.No, f is not injective because f (1) = 7 and f (4) = 7. So, f −1({7}) contains two elements (1 and 4), and inverse images of singleton sets never have more than one element for injective functions. Or, using the metaphor, f is not injective because it hits the target 7 with bullet 1 and bullet 4.

3.No, f is not surjective because f (A) is {6, 7, 9, 10}, which is lacking an element of B (8). The image of the domain is the entire range when a function is surjective. Or, using the metaphor, f is not surjective because it fails to hit the target 8.

4.No, g is not a function because it contains the pairs (6,b) and (6,d ), and functions are not allowed to contain two pairs with the same first element and di erent second elements. Or, using the metaphor, g is not a function because the bullet 6 hits two targets (b and d ).

44.

1.The codomain of f is P (A), which contains 2n elements (where 2n stands for 2 to the power n). That is, the function f has 2n targets. The domain

of f contains n elements. That is, the function f has n bullets. Since a function must use at least one bullet per target, and since 2n always exceeds n, f doesn’t have enough bullets to hit all the targets, so it can’t be surjective.

2.The composition g.f is defined when g is defined on the image f (A). So, X is a subset of P (A). Furthermore, since (g.f )(A) is a subset of A, and since (g.f )(A) = g(f (A)), Y , the codomain of g, must contain at least those elements of A that are in (g.f )(A). That’s one way to look at it. Here’s another way: The composition (g.f ) is defined when the codomain of f is the domain of g. Taking this position, X = P (A) and Y , the codomain of g contains at least g(f (A)), which, by hypothesis, is a subset of A. Either position is ok. It depends on how the term “composition” is defined. Because we were not precise about when to consider the composition defined, either answer is ok.

3.Again, it depends on the details of the definition of composition. Taking the first view, the domain of g must contain, at least, the set f (A). It may contain only this set, in which case it can be injective because the number of targets f can hit cannot exceed the number of bullets in A, so g will not have more bullets than targets and can be defined to hit each potential target at most once. On the other hand, if P (A) is considered to be the domain of g, then g will have too many bullets for A. So, if g is to be injective, its codomain must contain elements other than those of A. The intersection of A and the codomain of g will contain, at least, (g.f )(A).

4.Define g : f (A) → A as follows. For each s in f (A), let g(s) be one of the elements in f −1(s). Any element will do. Just pick one, and let that be g(s). There must be at least one such element because s is in f (A). Then (f .g) : f (A) → f (A) will be bijective.

C.13 Discrete Mathematics in Circuit Design

1.

f a b c =

(¬a ¬b ¬c)

(¬a ¬b c)

(a ¬b c)

(a b ¬c)

(a b c)

4.The inputs are a and b. The outputs are the carry output, c, which comes from the truth table for and2, and the sum output, s, which comes from the truth table for xor2. The calculations show that for all inputs, the left-hand and right-hand sides of the equation in the theorem have the same value.

9.

Example: addition of 13+41=54 using 6-bit words 13 = 001101 41 = 101001

sum = 110110 = 54, with carry out = 0

rippleAdd False [(False,True),(False,False),(True,True), (True,False),(False,False),(True,True)]

The expected result is

(False, [True,True,False,True,True,False])

10.

Test cases for halfCmp

halfCmp (False,False) -- (False,True,False) since x=y halfCmp (False,True) -- (True,False,False) since x<y halfCmp (True,False) -- (False,False,True) since x>y halfCmp (True,True) -- (False,True,False) since x=y

11.

Test cases for rippleCmp

rippleCmp [(False,False),(False,True),(True,False)] -- (True,False,False) since x<y

rippleCmp [(False,False),(True,True),(False,False)] -- (False,True,False) since x=y

rippleCmp [(False,False),(True,False),(False,True)]

--(False,False,True) since x>y

12.Define

andfour a b c d = and2 (and2 a b) (and2 c d)

The output should be 1 if all of the inputs are 1; note that in this case, and2 a b gives 1, and so does and2 c d, so the circuit outputs 1 as it should. In all other cases, either and2 a b will be 0, or and2 c d will be 0, or both, so the andfour circuit produces 0, as it should.

13.

15.Yes. The rippleAdd circuit is defined using the higher order function mscanr, which is defined even when the word input is empty. In this case, the sum word is just [], and the carry output is the same as the carry input. Notice that you can connect an m-bit adder with an n-bit adder and obtain an (m + n)-bit adder, and this works for all natural numbers m and n.

16.No, because mscanr (and therefore rippleAdd ) requires a list input, and lists cannot have a negative length.

17.You can’t test an adder thoroughly because there is an exponential growth in the size of the truth table as the word size grows. For an n-bit word, there are 2 × 22n lines in the truth table. The exponent is 2n because there are two bits for each position, and the initial factor of 2 accounts for the possibility that the carry input is either 0 or 1. For a current generation processor, where the word size is 64, the truth table would have 2129 lines, which is larger than the size of the known universe (it hardly matters what unit of measurement you choose!)