Graph function in Haskell - haskell

I have such a data type :
data Node a = Node
{ label :: a,
adjacent :: [(a,Int)] } deriving Show
Example : ( Node 'a' [ ( 'b' , 3 ) , ( 'c' ,2 ) ] )
I want to get the label from this structure, I wrote this function (and several other combinations which I thought might work) :
giveLabel Node a [(c,b)] = a;
But I keep getting errors. Can you tell me how should I change my function? Thanks

giveLabel (Node a [(c,b)]) = a
Is the syntax you want - defining functions uses the same rules of precedence as calling them, and according to those rules, you defined a function giveLabel with three arguments (Node, a, and [c,b]); and that was illegal because in that context Node was missing arguments.
Even that probably isn't what you want - the pattern [(c,b)] only matches lists with exactly one item in. Since you don't care about the list of neighbours you can write:
giveLabel (Node a xs) = a
...where xs will bind to the whole list of neighbours; but actually since you don't even care about that, you can write:
giveLabel (Node a _) = a
...where _ is a useful way of pattern matching against a parameter you aren't going to use.

Related

Need help storing the previous element of a list (Haskell)

I'm currently working on an assignment. I have a function called gamaTipo that converts the values of a tuple into a data type previously defined by my professor.
The problem is: in order for gamaTipo to work, it needs to receive some preceding element. gamaTipo is defined like this: gamaTipo :: Peca -> (Int,Int) -> Peca where Peca is the data type defined by my professor.
What I need to do is to create a funcion that takes a list of tuples and converts it into Peca data type. The part that im strugling with is taking the preceding element of the list. i.e : let's say we have a list [(1,2),(3,4)] where the first element of the list (1,2) always corresponds to Dirt Ramp (data type defined by professor). I have to create a function convert :: [(Int,Int)] -> [Peca] where in order to calculate the element (3,4) i need to first translate (1,2) into Peca, and use it as the previous element to translate (3,4)
Here's what I've tried so far:
updateTuple :: [(Int,Int)] -> [Peca]
updateTuple [] = []
updateTuple ((x,y):xs) = let previous = Dirt Ramp
in (gamaTipo previous (x,y)): updateTuple xs
Although I get no error messages with this code, the expected output isn't correct. I'm also sorry if it's not easy to understand what I'm asking, English isn't my native tongue and it's hard to express my self. Thank you in advance! :)
If I understand correctly, your program needs to have a basic structure something like this:
updateTuple :: [(Int, Int)] -> [Peca]
updateTuple = go initialValue
where
go prev (xy:xys) =
let next = getNextValue prev xy
in prev : (go next xys)
go prev [] = prev
Basically, what’s happening here is:
updateTuple is defined in terms of a helper function go. (Note that ‘helper function’ isn’t standard terminology, it’s just what I’ve decided to call it).
go has an extra argument, which is used to store the previous value.
The implementation of go can then make use of the previous value.
When go recurses, the recursive call can then pass the newly-calculated value as the new ‘previous value’.
This is a reasonably common pattern in Haskell: if a recursive function requires an extra argument, then a new function (often named go) can be defined which has that extra argument. Then the original function can be defined in terms of go.

Haskell # type annotation [duplicate]

I've come across a piece of Haskell code that looks like this:
ps#(p:pt)
What does the # symbol mean in this context? I can't seem to find any info on Google (it's unfortunately hard to search for symbols on Google), and I can't find the function in the Prelude documentation, so I imagine it must be some sort of syntactic sugar instead.
Yes, it's just syntactic sugar, with # read aloud as "as". ps#(p:pt) gives you names for
the list: ps
the list's head : p
the list's tail: pt
Without the #, you'd have to choose between (1) or (2):(3).
This syntax actually works for any constructor; if you have data Tree a = Tree a [Tree a], then t#(Tree _ kids) gives you access to both the tree and its children.
The # Symbol is used to both give a name to a parameter and match that parameter against a pattern that follows the #. It's not specific to lists and can also be used with other data structures.
This is useful if you want to "decompose" a parameter into it's parts while still needing the parameter as a whole somewhere in your function. One example where this is the case is the tails function from the standard library:
tails :: [a] -> [[a]]
tails [] = [[]]
tails xxs#(_:xs) = xxs : tails xs
I want to add that # works at all levels, meaning you can do this:
let a#(b#(Just c), Just d) = (Just 1, Just 2) in (a, b, c, d)
Which will then produce this: ((Just 1, Just 2), Just 1, 1, 2)
So basically it's a way for you to bind a pattern to a value. This also means that it works with any kind of pattern, not just lists, as demonstrated above. This is a very useful thing to know, as it means you can use it in many more cases.
In this case, a is the entire Maybe Tuple, b is just the first Just in the tuple, and c and d are the values contained in the first and second Just in the tuple respectively
To add to what the other people have said, they are called as-patterns (in ML the syntax uses the keyword "as"), and are described in the section of the Haskell Report on patterns.

haskell: factors of a natural number

I'm trying to write a function in Haskell that calculates all factors of a given number except itself.
The result should look something like this:
factorlist 15 => [1,3,5]
I'm new to Haskell and the whole recursion subject, which I'm pretty sure I'm suppoused to apply in this example but I don't know where or how.
My idea was to compare the given number with the first element of a list from 1 to n div2
with the mod function but somehow recursively and if the result is 0 then I add the number on a new list. (I hope this make sense)
I would appreciate any help on this matter
Here is my code until now: (it doesn't work.. but somehow to illustrate my idea)
factorList :: Int -> [Int]
factorList n |n `mod` head [1..n`div`2] == 0 = x:[]
There are several ways to handle this. But first of all, lets write a small little helper:
isFactorOf :: Integral a => a -> a -> Bool
isFactorOf x n = n `mod` x == 0
That way we can write 12 `isFactorOf` 24 and get either True or False. For the recursive part, lets assume that we use a function with two arguments: one being the number we want to factorize, the second the factor, which we're currently testing. We're only testing factors lesser or equal to n `div` 2, and this leads to:
createList n f | f <= n `div` 2 = if f `isFactorOf` n
then f : next
else next
| otherwise = []
where next = createList n (f + 1)
So if the second parameter is a factor of n, we add it onto the list and proceed, otherwise we just proceed. We do this only as long as f <= n `div` 2. Now in order to create factorList, we can simply use createList with a sufficient second parameter:
factorList n = createList n 1
The recursion is hidden in createList. As such, createList is a worker, and you could hide it in a where inside of factorList.
Note that one could easily define factorList with filter or list comprehensions:
factorList' n = filter (`isFactorOf` n) [1 .. n `div` 2]
factorList'' n = [ x | x <- [1 .. n`div` 2], x `isFactorOf` n]
But in this case you wouldn't have written the recursion yourself.
Further exercises:
Try to implement the filter function yourself.
Create another function, which returns only prime factors. You can either use your previous result and write a prime filter, or write a recursive function which generates them directly (latter is faster).
#Zeta's answer is interesting. But if you're new to Haskell like I am, you may want a "simple" answer to start with. (Just to get the basic recursion pattern...and to understand the indenting, and things like that.)
I'm not going to divide anything by 2 and I will include the number itself. So factorlist 15 => [1,3,5,15] in my example:
factorList :: Int -> [Int]
factorList value = factorsGreaterOrEqual 1
where
factorsGreaterOrEqual test
| (test == value) = [value]
| (value `mod` test == 0) = test : restOfFactors
| otherwise = restOfFactors
where restOfFactors = factorsGreaterOrEqual (test + 1)
The first line is the type signature, which you already knew about. The type signature doesn't have to live right next to the list of pattern definitions for a function, (though the patterns themselves need to be all together on sequential lines).
Then factorList is defined in terms of a helper function. This helper function is defined in a where clause...that means it is local and has access to the value parameter. Were we to define factorsGreaterOrEqual globally, then it would need two parameters as value would not be in scope, e.g.
factorsGreaterOrEqual 4 15 => [5,15]
You might argue that factorsGreaterOrEqual is a useful function in its own right. Maybe it is, maybe it isn't. But in this case we're going to say it isn't of general use besides to help us define factorList...so using the where clause and picking up value implicitly is cleaner.
The indentation rules of Haskell are (to my tastes) weird, but here they are summarized. I'm indenting with two spaces here because it grows too far right if you use 4.
Having a list of boolean tests with that pipe character in front are called "guards" in Haskell. I simply establish the terminal condition as being when the test hits the value; so factorsGreaterOrEqual N = [N] if we were doing a call to factorList N. Then we decide whether to concatenate the test number into the list by whether dividing the value by it has no remainder. (otherwise is a Haskell keyword, kind of like default in C-like switch statements for the fall-through case)
Showing another level of nesting and another implicit parameter demonstration, I added a where clause to locally define a function called restOfFactors. There is no need to pass test as a parameter to restOfFactors because it lives "in the scope" of factorsGreaterOrEqual...and as that lives in the scope of factorList then value is available as well.

Haskell graph data type representation

I want to represent a graph in Haskell in the following manner:
For each node I want to store it's value and a list of adjacent nodes. The problem which I'm having difficulties with is that I want the adjacent nodes to be stored as references to other nodes.
For example, I want node ny to be stored as („NY“ (l p)) where l and p are adjacent nodes,and not as („NY“ („London“ „Paris“)).
I tried something like this :
data Node a = Node { value :: a
, neighbors :: [Node a]
}deriving (Show)
let n1 = Node {value=1, neighbors=[n2]}
let n2 = Node {value=1, neighbors=[n1 n3]}
let n3 = Node {value=1, neighbors=[n2]}
But I get en error in let. What am I doing wrong ?
Two problems:
let is an expression form and at top level the compiler is expecting a declaration form.
You need a single nest of bindings; by using three lets, you've split the definitions into three separate scopes.
The following code compiles, and when I ask for n1, I get an infinite string printout as expected:
module Letnest
where
data Node a = Node { value :: a
, neighbors :: [Node a]
} deriving (Show)
n1 = Node {value=1, neighbors=[n2]}
n2 = Node {value=1, neighbors=[n1, n3]}
n3 = Node {value=1, neighbors=[n2]}
I wouldn't represent a graph this way. Store the nodes in a Map or an Array and refer to them by their keys instead of directly pointing to them. This would be much easier to save, load, maintain, and work with.
For some problems with your current representation:
Reid Barton commented:
Note that n1 and n3 are completely indistinguishable (since they have the same definition) which, depending on your application, may be a problem with this representation.
(there is no is comparison a la Python in Haskell)
Norman Ramsey noticed:
I get an infinite string printout

What does the "#" symbol mean in reference to lists in Haskell?

I've come across a piece of Haskell code that looks like this:
ps#(p:pt)
What does the # symbol mean in this context? I can't seem to find any info on Google (it's unfortunately hard to search for symbols on Google), and I can't find the function in the Prelude documentation, so I imagine it must be some sort of syntactic sugar instead.
Yes, it's just syntactic sugar, with # read aloud as "as". ps#(p:pt) gives you names for
the list: ps
the list's head : p
the list's tail: pt
Without the #, you'd have to choose between (1) or (2):(3).
This syntax actually works for any constructor; if you have data Tree a = Tree a [Tree a], then t#(Tree _ kids) gives you access to both the tree and its children.
The # Symbol is used to both give a name to a parameter and match that parameter against a pattern that follows the #. It's not specific to lists and can also be used with other data structures.
This is useful if you want to "decompose" a parameter into it's parts while still needing the parameter as a whole somewhere in your function. One example where this is the case is the tails function from the standard library:
tails :: [a] -> [[a]]
tails [] = [[]]
tails xxs#(_:xs) = xxs : tails xs
I want to add that # works at all levels, meaning you can do this:
let a#(b#(Just c), Just d) = (Just 1, Just 2) in (a, b, c, d)
Which will then produce this: ((Just 1, Just 2), Just 1, 1, 2)
So basically it's a way for you to bind a pattern to a value. This also means that it works with any kind of pattern, not just lists, as demonstrated above. This is a very useful thing to know, as it means you can use it in many more cases.
In this case, a is the entire Maybe Tuple, b is just the first Just in the tuple, and c and d are the values contained in the first and second Just in the tuple respectively
To add to what the other people have said, they are called as-patterns (in ML the syntax uses the keyword "as"), and are described in the section of the Haskell Report on patterns.

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