Maybe the question is not suiting the topic i am going to approach
But i will try to explain the best i can:
I have a genealogical tree that has this structure:
data GT = Person Name Father Mother
| unknown
type Father = GT
type Mother = GT
type Name = String
I need to find out the grandfathers of a given name:
grandfathers :: Name -> GT -> [Name]
this is the best i could do:
grandfathers :: Name -> GT -> [Name]
grandfathers s (Person x f m) = if (searchgrandson s f 0) then [x] else (grandfathers s f)
where
searchgrandson s unknown k = False
searchgrandson s (Person x f m) k = if s==x && k<2 then True
else searchgrandson s f (k+1)
of course that for this tree it works, because my code goes all the way through the left side, ignoring the mother side, and giving only the grandfather, and not the grandmother right?
$grandfathers "grandson" (Person "grandpa" (Person "son" (Person "grandson" unknown unknown ) unknown ) unknown )
["grandpa"]
EDIT:
after following the dfeuer advice:
avos_ :: Nome -> AG -> Int
avos_ s Desconhecida = 0
avos_ s l#(Pessoa x p m) = encontraneto s l 0
where
encontraneto s Desconhecida k = 0
encontraneto s (Pessoa x p m) k = if s==x then k
else encontraneto s p (k+1) + encontraneto s m (k+1)
This gives me the depth of the tree where the grandson is, searching both sides. After this would be simple...
Thanks!
A good way to go about solving this problem is to decompose it into two functions. One of the functions will search the tree for zero or more matches against a name and return zero or more trees, starting from the place where the match occurred; the root of each tree will be the person that matches the name. The other function will simply take a tree and return zero, one, two, three or four names corresponding to the grandparents of the person at the root of the tree.
The types of the first function is easy:
searchPerson :: Name -> GT -> [GT]
The type of the second function is also straightforward
getGrandparents :: GT -> [Name]
Now, since the first function returns a list of GT, rather than a single GT, the second function should be mapped over the results of the first function (or, said other way, lifted to operate on lists of GT):
map getGrandparents (searchPerson x myGenTree)
Writing the second function (getGrandParents) is trivial and can be achieved with pattern matching, deconstructing the GT type up to the second level of nesting, but you would probably like to save yourself some typing and create a helper function to operate on each parent and return their parents.
The first function (searchPerson) is also trivial, and can be done in a couple of ways. One way is to simply use recursion looking for a match on a name and returning a list of all the subtrees that had a match. The other option is to simply return a list of every possible subtree in that tree, starting from any possible offset, and to use a filter to only keep those subtrees whose root is a match. The second option is more "wholesome" and equivalent to addressing a similar problem in lists (the function that returns all sublists is tails) so would probably appeal to some people more. You may think that the second approach is wasteful, but Haskell immutability and lazyness should make it pretty efficient too.
I quickly put together the code in my fpcomplete account and will gladly post the URL in a comment below if you ask, but I don't want to spoil the fun of coming up with the solution on your own.
Related
I am trying to learn haskell and saw a exercise which says
Write two different Haskell functions having the same type:
[a] -> [b] -> Int -> (a,b)
So from my understanding the expressions should take in two lists, an int and return a tuple of the type of the lists.
What i tried so far was
together :: [a] -> [b] -> Int -> (a,b)
together [] [] 0 = (0,0)
together [b] [a] x = if x == a | b then (b,a) else (0,0)
I know I am way off but any help is appreciated!
First you need to make your mind up what the function should return. That is partly determined by the signature. But still you can come up with a lot of functions that return different things, but have the same signature.
Here one of the most straightforward functions is probably to return the elements that are placed on the index determined by the third parameter.
It makes no sense to return (0,0), since a and b are not per se numerical types. Furthermore if x == a | b is not semantically valid. You can write this as x == a || x == b, but this will not work, since a and b are not per se Ints.
We can implement a function that returns the heads of the two lists in case the index is 0. In case the index is negative, or at least one of the two lists is exhausted, then we can raise an error. I leave it as an exercise what to do in case the index is greater than 0:
together :: [a] -> [b] -> Int -> (a,b)
together [] _ = error "List exhausted"
together _ [] = error "List exhausted"
together (a:_) (b:_) 0 = (a, b)
together (a:_) (b:_) n | n < 0 = error "Negative index!"
| …
you thus still need to fill in the ….
I generally dislike those "write any function with this signature"-type excercises precisely because of how arbitrary they are. You're supposed to figure out a definition that would make sense for that particular signature and implement it. In a lot of cases, you can wing it by ignoring as many arguments as possible:
fa :: [a] -> [b] -> Int -> (a,b)
fa (a:_) (b:_) _ = (a,b)
fa _ _ _ = error "Unfortunately, this function can't be made total because lists can be empty"
The error here is the important bit to note. You attempted to go around that problem by returning 0s, but this will only work when 0 is a valid value for types of a and b. Your next idea could be some sort of a "Default" value, but not every type has such a concept. The key observation is that without any knowledge about a type, in order to produce a value from a function, you need to get this value from somewhere else first*.
If you actually wanted a more sensible definition, you'd need to think up a use for that Int parameter; maybe it's the nth element from each
list? With the help of take :: Int -> [a] -> [a] and head :: [a] -> a this should be doable as an excercise.
Again, your idea of comparing x with a won't work for all types; not every type is comparable with an Int. You might think that this would make generic functions awfully limited; that's the point where you typically learn about how to express certain expectations about the types you get, which will allow you to operate only on certain subsets of all possible types.
* That's also the reason why id :: a -> a has only one possible implementation.
Write two different Haskell functions having the same type:
[a] -> [b] -> Int -> (a,b)
As Willem and Bartek have pointed out, there's a lot of gibberish functions that have this type.
Bartek took the approach of picking two based on what the simplest functions with that type could look like. One was a function that did nothing but throw an error. And one was picking the first element of each list, hoping they were not empty and failing otherwise. This is a somewhat theoretical approach, since you probably don't ever want to use those functions in practice.
Willem took the approach of suggesting an actually useful function with that type and proceeded to explore how to exhaust the possible patterns of such a function: For lists, match the empty list [] and the non-empty list a:_, and for integers, match some stopping point, 0 and some categories n < 0 and ….
A question that arises to me is if there is any other equally useful function with this type signature, or if a second function would necessarily have to be hypothetically constructed. It would seem natural that the Int argument has some relation to the positions of elements in [a] and [b], since they are also integers, especially because a pair of single (a,b) is returned.
But the only remotely useful functions (in the sense of not being completely silly) that I can think of are small variations of this: For example, the Int could be the position from the end rather than from the beginning, or if there's not enough elements in one of the lists, it could default to the last element of a list rather than an error. Neither of these are very pleasing to make ("from the end" conflicts with the list being potentially infinite, and having a fall-back to the last element of a list conflicts with the fact that lists don't necessarily have a last element), so it is tempting to go with Bartek's approach of writing the simplest useless function as the second one.
So basically I've past learning this part way back a month ago, and I can do more complicated stuff but I still don't understand when I need "Ord" or "Eq" etc in my type definitions. When I look it up online its just so confusing to me for some reason.
E.g.
my_min :: Ord a => a -> a -> a
my_min n1 n2 = if n1<n2 then n1 else n2
Why does this need Ord? Can you give an example of when you need Eq as well (a very simple one)? I need a very clear, basic explanation of when you need to put these, what to look out for to do that, and how to avoid using these at all if possible.
Usually I just need something like "[Int] -> Int -> [Int]" so i know ok this function takes an integer list and an integer, and returns an integer list. But when it includes Eq or Ord I have no idea.
What about this Eq type in this example I found in my lecture notes about finding two lists is identical, how does it apply?
identical :: Eq a =>[a]->[a]->Bool
identical [] [] = True
identical [] _ = False
identical _ [] = False
identical (h1:t1) (h2:t2) =
if h1==h2
then (identical t1 t2)
else False
Thank you.
Ord implies that the thing can be ordered, which means that you can say a is smaller (or greater) than b. Using only Eq is like saying: I know that these two items are not the same, but I cannot say which one is greater or smaller. For example if you take a traffic light as a data type:
data TLight = Red | Yellow | Green deriving (Show, Eq)
instance Eq TLight where
Green == Green = True
Yellow == Yellow = True
Red == Red = True
_ == _ = False
Now we can say: Red is unequal to Yellow but we cannot say what is greater. This is the reason why you could not use TLight in your my_min. You cannot say which one is greater.
To your second question: "Is there any case where you have to use Eq and Ord?":
Ord implies Eq. This means that if a type can be ordered, you can also check it for equality.
You said you have mostly dealt with [Int] -> Int -> [Int] and you then knew it takes a list of integer and an integer and returns an integer. Now if you want to generalise your function you have to ask yourself: Do the possible types I want to use in my function need any special functionality? like if they have to be able to be ordered or equated.
Lets do a few examples: Say we want to write a function which takes a list of type a and an element of type a and returns the lisy with the element consed onto it. How would it's type signature look like? Lets start with simply this:
consfunc :: [a] -> a -> [a]
Do we need any more functionality? No! Our type a can be anything because we do not need it to be able to be ordered simple because that is mot what our function should do.
Now what if we want to take a list and an element and check if the element is in the list already? The beginning type signature is:
elemfunc :: [a] -> a -> Bool
Now does our element have to be able to do something special? Yes it does, we have to be able to check if it is equal to any element in the list, which says that our type a has to be equatable, so our type signature looks like this:
elemfunc :: (Eq a) => [a] -> a -> Bool
Now what if we want to take a list and a element and insert it if it is smaller than the first element? Can you guess how the type signature would look like?
Lets begin with the standard again and ask ourselves: Do we need more than just knowing that the element and the list have to be of the same type: Yes, becuase our condition needs to perform a test that requires our type to be ordered, we have to include Ord in our type signature:
conditionalconsfunc :: (Ord a) => [a] -> a -> [a]
Edit:
Well you want to see if two lists are identical, so there are two things you have to look out for:
Your lists have to contain the same type and the things inside the list have to be equatable, hence the Eq.
If you are working with fixed types like Int, you never need class constraints. These only arise when working with polymorphic code.
You need Eq if you ever use the == or /= functions, or if you call any other functions that do. (I.e., if you call a function that has Eq in its type, then your type needs to have Eq as well.)
You need Ord if you ever use <, >, compare or similar functions. (Again, or if you call something that does.)
Note that you do not need Eq if you only do pattern matching. Thus, the following are different:
factorial 1 = 1
factorial n = n * factorial (n-1)
-- Only needs Num.
factorial n = if n == 1 then 1 else n * factorial (n-1)
-- Needs Num and Eq.
factorial n = if n < 2 then 1 else n * factorial (n-1)
-- Needs Num, Eq and Ord. (But Ord implies Eq automatically.)
How can I search a tuple based on the first element or second or third?
I know how to search tuple of two but how do i do it for more than two?
type Stuff = (String, String, Int)
testStuff :: [Stuff]
testStuff = [
("beans","black",5),
("cod","fish",4),
("coke","diet",3)
]
How can I write a function that will search "Stuff" and return the "int" value?
e.g searchStuff "beans" should return 5.
Since you haven't provided your function for searching in a list of pairs, I'm going to assume that you used lookup. Lets focus on find for your new function. find can be found in Data.List (and a more general version in Data.Foldable) and has the following type:
find :: (a -> Bool) -> [a] -> Maybe a
Now, if you need to find something in a list of triples based on the first element, you could use
find (\(a,_,_) -> a == "beans") testStuff
However, this will leave you with a Maybe Stuff. But since Maybe is an instance of Functor, it is easy to change the result to Maybe Int (and is left as an exercise).
The Prelude defines lookup to handle searching a list of pairs. Here's the definition:
-- | 'lookup' #key assocs# looks up a key in an association list.
lookup :: (Eq a) => a -> [(a,b)] -> Maybe b
lookup _key [] = Nothing
lookup key ((x,y):xys)
| key == x = Just y
| otherwise = lookup key xys
Can you see how one would define a similar function to search a list of triples?
I am taking a class in Haskell, and we need to define the fold operation for a tree defined by:
data Tree a = Lf a | Br (Tree a) (Tree a)
I can not seem to find any information on the "tfold" operation or really what it supposed to do. Any help would be greatly appreciated.
I always think of folds as a way of systematically replacing constructors by other functions. So, for instance, if you have a do-it-yourself List type (defined as data List a = Nil | Cons a (List a)), the corresponding fold can be written as:
listfold nil cons Nil = nil
listfold nil cons (Cons a b) = cons a (listfold nil cons b)
or, maybe more concisely, as:
listfold nil cons = go where
go Nil = nil
go (Cons a b) = cons a (go b)
The type of listfold is b -> (a -> b -> b) -> List a -> b. That is to say, it takes two 'replacement constructors'; one telling how a Nil value should be transformed into a b, another replacement constructor for the Cons constructor, telling how the first value of the Cons constructor (of type a) should be combined with a value of type b (why b? because the fold has already been applied recursively!) to yield a new b, and finally a List a to apply the whole she-bang to - with a result of b.
In your case, the type of tfold should be (a -> b) -> (b -> b -> b) -> Tree a -> b by analogous reasoning; hopefully you'll be able to take it from there!
Imagine you define that a tree should be shown in the following manner,
<1 # <<2#3> # <4#5>>>
Folding such a tree means replacing each branch node with an actual supplied operation to be performed on the results of fold recursively performed on the data type's constituents (here, the node's two child nodes, which are themselves, each, a tree), for example with +, producing
(1 + ((2+3) + (4+5)))
So, for leaves you should just take the values inside them, and for branches, recursively apply the fold for each of the two child nodes, and combine the two results with the supplied function, the one with which the tree is folded. (edit:) When "taking" values from leaves, you could additionally transform them, applying a unary function. So in general, your folding will need two user-provided functions, one for leaves, Lf, and another one for combining the results of recursively folding the tree-like constituents (i.e. branches) of the branching nodes, Br.
Your tree data type could have been defined differently, e.g. with possibly empty leaves, and with internal nodes also carrying the values. Then you'd have to provide a default value to be used instead of the empty leaf nodes, and a three-way combination operation. Still you'd have the fold defined by two functions corresponding to the two cases of the data type definition.
Another distinction to realize here is, what you fold, and how you fold it. I.e. you could fold your tree in a linear fashion, (1+(2+(3+(4+5)))) == ((1+) . (2+) . (3+) . (4+) . (5+)) 0, or you could fold a linear list in a tree-like fashion, ((1+2)+((3+4)+5)) == (((1+2)+(3+4))+5). It is all about how you parenthesize the resulting "expression". Of course in the classic take on folding the expression's structure follows that of the data structure being folded; but variations do exist. Note also, that the combining operation might not be strict, and the "result" type it consumes/produces might express compound (lists and such), as well as atomic (numbers and such), values.
(update 2019-01-26) This re-parenthesization is possible if the combining operation is associative, like +: (a1+a2)+a3 == a1+(a2+a3). A data type together with such associative operation and a "zero" element (a+0 == 0+a == a) is known as "Monoid", and the notion of folding "into" a Monoid is captured by the Foldable type class.
A fold on a list is a reduction from a list into a single element. It takes a function and then applies that function to elements, two at a time, until it has only one element. For example:
Prelude> foldl1 (+) [3,5,6,7]
21
...is found by doing operations one-by-one:
3 + 5 == 8
8 + 6 == 14
14 + 7 == 21
A fold can be written
ourFold :: (a -> a -> a) -> [a] -> a
ourFold _ [a] = a -- pattern-match for a single-element list. Our work is done.
ourFold aFunction (x0:x1:xs) = ourFold aFunction ((aFunction x0 x1):xs)
A tree fold would do this, but move up or down the branches of the tree. To do this, it first need to pattern-match to see whether you're operating on a Leaf or a Branch.
treeFold _ (Lf a) = Lf a -- You can't do much to a one-leaf tree
treeFold f (Br a b) = -- ...
The rest is left up to you, since it's homework. If you're stuck, try first thinking of what the type should be.
A fold is an operation which "compacts" a data structure into a single value using an operation. There are variations depending if you have a start value and execution order (e.g. for lists you have foldl, foldr, foldl1 and foldr1), so the correct implementation depends on your assignment.
I guess your tfold should simply replace all leafs with its values, and all branches with applications of the given operation. Draw an example tree with some numbers, an "collapse" him given an operation like (+). After this, it should be easy to write a function doing the same.
Im trying to make a funciton which allows me to add a new value to a tree IF the value at the given path is equal to ND (no data), this was my first attempt.
It checks the value etc, but the problem, is i want to be able to print the modified tree with the new data. can any one give me any pointers? I have also tried making a second function that checks the path to see if its ok to add data, but im just lost to how to print out the modified tree?
As iuliux points out, your problem is that you are treating your BTree as though it were a mutable structure. Remember functions in haskell take arguments and return a value. That is all. So when you "map over" a list, or traverse a tree your function needs to return a new tree.
The code you have is traversing the recursive tree and only returning the last leaf. Imagine for now that the leaf at the end of the path will always be ND. This is what you want:
add :: a -> Path -> Btree a -> Btree a
add da xs ND = Data da
add _ [] _ = error "You should make sure this doesn't happen or handle it"
add da (x:xs) (Branch st st2) =
case x of
L -> Branch (add da xs st) st2
R -> Branch st (add da xs st2)
Notice how in your original code you discard the Branch you pattern match against, when what you need to do is return it "behind you" as it were.
Now, on to the issue of handling situations where the leaf you arrive it is not a ND constructor:
This type of problem is common in functional programming. How can you return your recursive data structure "as you go" when the final result depends on a leaf far down the tree?
One solution for the trickiest of cases is the Zipper, which is a data structure that lets you go up down and sideways as you please. For your case that would be overkill.
I would suggest you change your function to the following:
add :: a -> Path -> Btree a -> Maybe (Btree a)
which means at each level you must return a Maybe (Btree a). Then use the Functor instance of Maybe in your recursive calls. Notice:
fmap (+1) (Just 2) == Just 3
fmap (+1) (Nothing) == Nothing
You should try to puzzle out the implementation for yourself!
I'm no expert in Haskell, but functional programming only works with functions. So kind of anything is a function.
Now, your function takes some input and returns something, not modifing the input. You have to retain the returned tree somewhere and that will be your new tree, the one with inserted element in it
We really need to see the Path and Error data types to answer your question, but you can print out your trees using the IO Monad:
main :: IO()
main = do let b = Branch ND (Branch (Data 1) (Data 2))
let b1 = add 10 [L] b --actual call depends on definition of Path
(putStrLn . show) b1