To output a verb phrase that has an object as a question, then as it seems RGL only offers two functions:
VQ -> QS -> VP
V2Q -> NP -> QS -> VP
And in these two functions, the verb type was divided into two different categories. But the type V2Q has a parameter that requires adding a preposition to the sentence. In order to generate the sentence Tell me who I am I used the following code:
MySentence = {s = (mkPhr
(mkImp
(mkVP
(mkV2Q
(mkV "tell")
(mkPrep ""))
(i_NP)
(mkQS
(mkQCl
(mkIComp (who_IP))
(i_NP)))))).s };
The code above generates the output I desire without a problem. So my question is, is there any reason the preposition was added to the verb V2Q？ Or was this output generated in a wrong way?
First, yes you constructed the sentence correctly.
Why is there a slot for preposition in V2Q
In general, all V2* (and V3*) may take their NP object as a direct object, like eat ___, see ___, or with a preposition, like believe in ___.
This is more flexible than forcing all transitive verbs only take direct objects, and all prepositional phrases to be analysed as optional adverbials. Take a VP like "believe in yourself", it's not that you're believing (something) and also your location is yourself. It's nice to be able to encode that believe_V2 takes an obligatory argument, and that argument is introduced by the preposition in.
(Side note: for a VP like "sleep in a soft bed", "in a soft bed" is not an obligatory argument of sleep. So then we just make sleep into an intransitive verb, sleep_V, and make in a soft bed into an Adv.)
So, this generalises to all verbs that take some NP argument (V2V, V2S, V2Q, V2A). Take a VP like "lie [to children] [that moon is made of cheese]": the verb lie is a V2S that introduces its NP object with the preposition to.
In fact, many RGL languages offer a noPrep in their Paradigms module—you can Ctrl+F in the RGL synopsis page to see examples.
The constructors of V2Q
So why are you forced to make your V2Q with mkV2Q (mkV "tell") (mkPrep ""), even when there is no preposition?
More common verb types, like V2, have several overload instances of mkV2. The simplest is just mkV2 : Str -> V2. Since it's such a common thing for transitive verbs to have a direct object (i.e. not introduce their object with a preposition), and there are so many simple V2s, it would be pretty annoying to have to always specify a noPrep for most of them.
V2Q is rarer than V2, so nobody just hasn't bothered creating an instance that doesn't take a preposition. The constructor that takes preposition is more general than the constructor that doesn't, since you can always choose the preposition to be noPrep. Well, I just pushed a few new additions, see here, so if you get the latest RGL, you can now just do mkV2Q "tell".
This kind of thing is completely customisable: if you want more overload instances of some mkX oper, you can just make them.
Related
I am new to Functional programming.
The challenge I have is regarding the mental map of how a binary search tree works in Haskell.
In other programs (C,C++) we have something called root. We store it in a variable. We insert elements into it and do balancing etc..
The program takes a break does other things (may be process user inputs, create threads) and then figures out it needs to insert a new element in the already created tree. It knows the root (stored as a variable) and invokes the insert function with the root and the new value.
So far so good in other languages. But how do I mimic such a thing in Haskell, i.e.
I see functions implementing converting a list to a Binary Tree, inserting a value etc.. That's all good
I want this functionality to be part of a bigger program and so i need to know what the root is so that i can use it to insert it again. Is that possible? If so how?
Note: Is it not possible at all because data structures are immutable and so we cannot use the root at all to insert something. in such a case how is the above situation handled in Haskell?
It all happens in the same way, really, except that instead of mutating the existing tree variable we derive a new tree from it and remember that new tree instead of the old one.
For example, a sketch in C++ of the process you describe might look like:
int main(void) {
Tree<string> root;
while (true) {
string next;
cin >> next;
if (next == "quit") exit(0);
root.insert(next);
doSomethingWith(root);
}
}
A variable, a read action, and loop with a mutate step. In haskell, we do the same thing, but using recursion for looping and a recursion variable instead of mutating a local.
main = loop Empty
where loop t = do
next <- getLine
when (next /= "quit") $ do
let t' = insert next t
doSomethingWith t'
loop t'
If you need doSomethingWith to be able to "mutate" t as well as read it, you can lift your program into State:
main = loop Empty
where loop t = do
next <- getLine
when (next /= "quit") $ do
loop (execState doSomethingWith (insert next t))
Writing an example with a BST would take too much time but I give you an analogous example using lists.
Let's invent a updateListN which updates the n-th element in a list.
updateListN :: Int -> a -> [a] -> [a]
updateListN i n l = take (i - 1) l ++ n : drop i l
Now for our program:
list = [1,2,3,4,5,6,7,8,9,10] -- The big data structure we might want to use multiple times
main = do
-- only for shows
print $ updateListN 3 30 list -- [1,2,30,4,5,6,7,8,9,10]
print $ updateListN 8 80 list -- [1,2,3,4,5,6,7,80,9,10]
-- now some illustrative complicated processing
let list' = foldr (\i l -> updateListN i (i*10) l) list list
-- list' = [10,20,30,40,50,60,70,80,90,100]
-- Our crazily complicated illustrative algorithm still needs `list`
print $ zipWith (-) list' list
-- [9,18,27,36,45,54,63,72,81,90]
See how we "updated" list but it was still available? Most data structures in Haskell are persistent, so updates are non-destructive. As long as we have a reference of the old data around we can use it.
As for your comment:
My program is trying the following a) Convert a list to a Binary Search Tree b) do some I/O operation c) Ask for a user input to insert a new value in the created Binary Search Tree d) Insert it into the already created list. This is what the program intends to do. Not sure how to get this done in Haskell (or) is am i stuck in the old mindset. Any ideas/hints welcome.
We can sketch a program:
data BST
readInt :: IO Int; readInt = undefined
toBST :: [Int] -> BST; toBST = undefined
printBST :: BST -> IO (); printBST = undefined
loop :: [Int] -> IO ()
loop list = do
int <- readInt
let newList = int : list
let bst = toBST newList
printBST bst
loop newList
main = loop []
"do balancing" ... "It knows the root" nope. After re-balancing the root is new. The function balance_bst must return the new root.
Same in Haskell, but also with insert_bst. It too will return the new root, and you will use that new root from that point forward.
Even if the new root's value is the same, in Haskell it's a new root, since one of its children has changed.
See ''How to "think functional"'' here.
Even in C++ (or other imperative languages), it would usually be considered a poor idea to have a single global variable holding the root of the binary search tree.
Instead code that needs access to a tree should normally be parameterised on the particular tree it operates on. That's a fancy way of saying: it should be a function/method/procedure that takes the tree as an argument.
So if you're doing that, then it doesn't take much imagination to figure out how several different sections of code (or one section, on several occasions) could get access to different versions of an immutable tree. Instead of passing the same tree to each of these functions (with modifications in between), you just pass a different tree each time.
It's only a little more work to imagine what your code needs to do to "modify" an immutable tree. Obviously you won't produce a new version of the tree by directly mutating it, you'll instead produce a new value (probably by calling methods on the class implementing the tree for you, but if necessary by manually assembling new nodes yourself), and then you'll return it so your caller can pass it on - by returning it to its own caller, by giving it to another function, or even calling you again.
Putting that all together, you can have your whole program manipulate (successive versions of) this binary tree without ever having it stored in a global variable that is "the" tree. An early function (possibly even main) creates the first version of the tree, passes it to the first thing that uses it, gets back a new version of the tree and passes it to the next user, and so on. And each user of the tree can call other subfunctions as needed, with possibly many of new versions of the tree produced internally before it gets returned to the top level.
Note that I haven't actually described any special features of Haskell here. You can do all of this in just about any programming language, including C++. This is what people mean when they say that learning other types of programming makes them better programmers even in imperative languages they already knew. You can see that your habits of thought are drastically more limited than they need to be; you could not imagine how you could deal with a structure "changing" over the course of your program without having a single variable holding a structure that is mutated, when in fact that is just a small part of the tools that even C++ gives you for approaching the problem. If you can only imagine this one way of dealing with it then you'll never notice when other ways would be more helpful.
Haskell also has a variety of tools it can bring to this problem that are less common in imperative languages, such as (but not limited to):
Using the State monad to automate and hide much of the boilerplate of passing around successive versions of the tree.
Function arguments allow a function to be given an unknown "tree-consumer" function, to which it can give a tree, without any one place both having the tree and knowing which function it's passing it to.
Lazy evaluation sometimes negates the need to even have successive versions of the tree; if the modifications are expanding branches of the tree as you discover they are needed (like a move-tree for a game, say), then you could alternatively generate "the whole tree" up front even if it's infinite, and rely on lazy evaluation to limit how much work is done generating the tree to exactly the amount you need to look at.
Haskell does in fact have mutable variables, it just doesn't have functions that can access mutable variables without exposing in their type that they might have side effects. So if you really want to structure your program exactly as you would in C++ you can; it just won't really "feel like" you're writing Haskell, won't help you learn Haskell properly, and won't allow you to benefit from many of the useful features of Haskell's type system.
The use of context is briefly mentioned in the K tutorial as a way to customize the order evaluation. But I'm also seeing other context statements that contain rewrite arrows in them, like this one in the untyped simple language.
context ++(HOLE => lvalue(HOLE))
rule <k> ++loc(L) => I +Int 1 ...</k>
<store>... L |-> (I => I +Int 1) ...</store> [increment]
Could someone explain how exactly context work in K? In particular, I'm interested in:
Is there a more general usage of context in K than just stating the order of evaluation?
How does the order in which context statements are declared affect the semantics?
Thank you!
More detailed information about context declarations in K can be found in K's documentation here. In particular, contexts with rewrite arrows mean that heating and cooling will wrap the term to be heated or cooled in a particular symbol. In your example, that symbol is lvalue.
To answer your questions specifically:
Context declarations, like strictness attributes, are primarily used in order to specify the evaluation strategy. While in theory they can be used for other things, in practice this rarely happens. That said, evaluation strategies can be complex, which is part of why K has so many different features relating to evaluation strategy. In the example you mentioned, we use rewrites in a context declaration in order to provide a separate set of rules for evaluating lvalues (ie, to avoid actually evaluating all the way to a value, and only evaluate to a location).
K's sentences are unordered. Within a single module, you can reorder any of its sentences (except import statements, which must appear first) and there will not be an effect on the intended semantics (although backends may result in slightly different behavior for concrete execution if your semantics is nondeterministic). This includes context declarations.
In HOTT and also in COQ one cannot prove UIP, i.e.
\Prod_{p:a=a} p = refl a
But one can prove:
\Prod_{p:a=a} (a,p) = (a, refl a)
Why is this defined as it is?
Is it, because one wants to have a nice homotopy interpretation?
Or is there some natural, deeper reason for this definition?
Today we know of a good reason for rejecting UIP: it is incompatible with the principle of univalence from homotopy type theory, which roughly says that isomorphic types can be identified. However, as far as I am aware, the reason that Coq's equality does not validate UIP is mostly a historical accident inherited from one of its ancestors: Martin-Löf's intensional type theory, which predates HoTT by many years.
The behavior of equality in ITT was originally motivated by the desire to keep type checking decidable. This is possible in ITT because it requires us to explicitly mark every rewriting step in a proof. (Formally, these rewriting steps correspond to the use of the equality eliminator eq_rect in Coq.) By contrast, Martin-Löf designed another system called extensional type theory where rewriting is implicit: whenever two terms a and b are equal, in the sense that we can prove that a = b, they can be used interchangeably. This relies on an equality reflection rule which says that propositionally equal elements are also definitionally equal. Unfortunately, there is a price to pay for this convenience: type checking becomes undecidable. Roughly speaking, the type-checking algorithm relies crucially on the explicit rewriting steps of ITT to guide its computation, whereas these hints are absent in ETT.
We can prove UIP easily in ETT because of the equality reflection rule; however, it was unknown for a long time whether UIP was provable in ITT. We had to wait until the 90's for the work of Hofmann and Streicher, which showed that UIP cannot be proved in ITT by constructing a model where UIP is not valid. (Check also these slides by Hofmann, which explain the issue from a historic perspective.)
Edit
This doesn' t mean that UIP is incompatible with decidable type checking: it was shown later that it can be derived in other decidable variants of Martin-Löf type theory (such as Agda), and it can be safely added as an axiom in a system like Coq.
Intuitively, I tend to think of a = a as pi_1(A,a), i.e. the class of paths from a to itself modulo homotopy equivalence; whereas I think of { x:A | a = x } as the universal covering space of A, i.e. paths from a to some other point of A modulo homotopy equivalence. So, while pi_1(A,a) is often non-trivial, we do have that the universal covering space of A is contractible.
tl;dr: how do you write instances of Arbitrary that don't explode if your data type allows for way too much nesting? And how would you guarantee these instances produce truly random specimens of your data structure?
I want to generate random tree structures, then test certain properties of these structures after I've mangled them with my library code. (NB: I'm writing an implementation of a subtyping algorithm, i.e. given a hierarchy of types, is type A a subtype of type B. This can be made arbitrarily complex, by including multiple-inheritance and post-initialization updates to the hierarchy. The classical method that supports neither of these is Schubert Numbering, and the latest result known to me is Alavi et al. 2008.)
Let's take the example of rose-trees, following Data.Tree:
data Tree a = Node a (Forest a)
type Forest a = [Tree a]
A very simple (and don't-try-this-at-home) instance of Arbitray would be:
instance (Arbitrary a) => Arbitrary (Tree a) where
arbitrary = Node <$> arbitrary <$> arbitrary
Since a already has an Arbitrary instance as per the type constraint, and the Forest will have one, because [] is an instance, too, this seems straight-forward. It won't (typically) terminate for very obvious reasons: since the lists it generates are arbitrarily long, the structures become too large, and there's a good chance they won't fit into memory. Even a more conservative approach:
arbitrary = Node <$> arbitrary <*> oneof [arbitrary,return []]
won't work, again, for the same reason. One could tweak the size parameter, to keep the length of the lists down, but even that won't guarantee termination, since it's still multiple consecutive dice-rolls, and it can turn out quite badly (and I want the odd node with 100 children.)
Which means I need to limit the size of the entire tree. That is not so straight-forward. unordered-containers has it easy: just use fromList. This is not so easy here: How do you turn a list into a tree, randomly, and without incurring bias one way or the other (i.e. not favoring left-branches, or trees that are very left-leaning.)
Some sort of breadth-first construction (the functions provided by Data.Tree are all pre-order) from lists would be awesome, and I think I could write one, but it would turn out to be non-trivial. Since I'm using trees now, but will use even more complex stuff later on, I thought I might try to find a more general and less complex solution. Is there one, or will I have to resort to writing my own non-trivial Arbitrary generator? In the latter case, I might actually just resort to unit-tests, since this seems too much work.
Use sized:
instance Arbitrary a => Arbitrary (Tree a) where
arbitrary = sized arbTree
arbTree :: Arbitrary a => Int -> Gen (Tree a)
arbTree 0 = do
a <- arbitrary
return $ Node a []
arbTree n = do
(Positive m) <- arbitrary
let n' = n `div` (m + 1)
f <- replicateM m (arbTree n')
a <- arbitrary
return $ Node a f
(Adapted from the QuickCheck presentation).
P.S. Perhaps this will generate overly balanced trees...
You might want to use the library presented in the paper "Feat: Functional Enumeration of Algebraic Types" at the Haskell Symposium 2012. It is on Hackage as testing-feat, and a video of the talk introducing it is available here: http://www.youtube.com/watch?v=HbX7pxYXsHg
As Janis mentioned, you can use the package testing-feat, which creates enumerations of arbitrary algebraic data types. This is the easiest way to create unbiased uniformly distributed generators
for all trees of up to a given size.
Here is how you would use it for rose trees:
import Test.Feat (Enumerable(..), uniform, consts, funcurry)
import Test.Feat.Class (Constructor)
import Data.Tree (Tree(..))
import qualified Test.QuickCheck as QC
-- We make an enumerable instance by listing all constructors
-- for the type. In this case, we have one binary constructor:
-- Node :: a -> [Tree a] -> Tree a
instance Enumerable a => Enumerable (Tree a) where
enumerate = consts [binary Node]
where
binary :: (a -> b -> c) -> Constructor c
binary = unary . funcurry
-- Now we use the Enumerable instance to create an Arbitrary
-- instance with the help of the function:
-- uniform :: Enumerable a => Int -> QC.Gen a
instance Enumerable a => QC.Arbitrary (Tree a) where
QC.arbitrary = QC.sized uniform
-- QC.shrink = <some implementation>
The Enumerable instance can also be generated automatically with TemplateHaskell:
deriveEnumerable ''Tree
So I'm trying to teach myself Haskell. I am currently on the 11th chapter of Learn You a Haskell for Great Good and am doing the 99 Haskell Problems as well as the Project Euler Problems.
Things are going alright, but I find myself constantly doing something whenever I need to keep track of "variables". I just create another function that accepts those "variables" as parameters and recursively feed it different values depending on the situation. To illustrate with an example, here's my solution to Problem 7 of Project Euler, Find the 10001st prime:
answer :: Integer
answer = nthPrime 10001
nthPrime :: Integer -> Integer
nthPrime n
| n < 1 = -1
| otherwise = nthPrime' n 1 2 []
nthPrime' :: Integer -> Integer -> Integer -> [Integer] -> Integer
nthPrime' n currentIndex possiblePrime previousPrimes
| isFactorOfAnyInThisList possiblePrime previousPrimes = nthPrime' n currentIndex theNextPossiblePrime previousPrimes
| otherwise =
if currentIndex == n
then possiblePrime
else nthPrime' n currentIndexPlusOne theNextPossiblePrime previousPrimesPlusCurrentPrime
where currentIndexPlusOne = currentIndex + 1
theNextPossiblePrime = nextPossiblePrime possiblePrime
previousPrimesPlusCurrentPrime = possiblePrime : previousPrimes
I think you get the idea. Let's also just ignore the fact that this solution can be made to be more efficient, I'm aware of this.
So my question is kind of a two-part question. First, am I going about Haskell all wrong? Am I stuck in the imperative programming mindset and not embracing Haskell as I should? And if so, as I feel I am, how do avoid this? Is there a book or source you can point me to that might help me think more Haskell-like?
Your help is much appreciated,
-Asaf
Am I stuck in the imperative programming mindset and not embracing
Haskell as I should?
You are not stuck, at least I don't hope so. What you experience is absolutely normal. While you were working with imperative languages you learned (maybe without knowing) to see programming problems from a very specific perspective - namely in terms of the van Neumann machine.
If you have the problem of, say, making a list that contains some sequence of numbers (lets say we want the first 1000 even numbers), you immediately think of: a linked list implementation (perhaps from the standard library of your programming language), a loop and a variable that you'd set to a starting value and then you would loop for a while, updating the variable by adding 2 and putting it to the end of the list.
See how you mostly think to serve the machine? Memory locations, loops, etc.!
In imperative programming, one thinks about how to manipulate certain memory cells in a certain order to arrive at the solution all the time. (This is, btw, one reason why beginners find learning (imperative) programming hard. Non programmers are simply not used to solve problems by reducing it to a sequence of memory operations. Why should they? But once you've learned that, you have the power - in the imperative world. For functional programming you need to unlearn that.)
In functional programming, and especially in Haskell, you merely state the construction law of the list. Because a list is a recursive data structure, this law is of course also recursive. In our case, we could, for example say the following:
constructStartingWith n = n : constructStartingWith (n+2)
And almost done! To arrive at our final list we only have to say where to start and how many we want:
result = take 1000 (constructStartingWith 0)
Note that a more general version of constructStartingWith is available in the library, it is called iterate and it takes not only the starting value but also the function that makes the next list element from the current one:
iterate f n = n : iterate f (f n)
constructStartingWith = iterate (2+) -- defined in terms of iterate
Another approach is to assume that we had another list our list could be made from easily. For example, if we had the list of the first n integers we could make it easily into the list of even integers by multiplying each element with 2. Now, the list of the first 1000 (non-negative) integers in Haskell is simply
[0..999]
And there is a function map that transforms lists by applying a given function to each argument. The function we want is to double the elements:
double n = 2*n
Hence:
result = map double [0..999]
Later you'll learn more shortcuts. For example, we don't need to define double, but can use a section: (2*) or we could write our list directly as a sequence [0,2..1998]
But not knowing these tricks yet should not make you feel bad! The main challenge you are facing now is to develop a mentality where you see that the problem of constructing the list of the first 1000 even numbers is a two staged one: a) define how the list of all even numbers looks like and b) take a certain portion of that list. Once you start thinking that way you're done even if you still use hand written versions of iterate and take.
Back to the Euler problem: Here we can use the top down method (and a few basic list manipulation functions one should indeed know about: head, drop, filter, any). First, if we had the list of primes already, we can just drop the first 1000 and take the head of the rest to get the 1001th one:
result = head (drop 1000 primes)
We know that after dropping any number of elements form an infinite list, there will still remain a nonempty list to pick the head from, hence, the use of head is justified here. When you're unsure if there are more than 1000 primes, you should write something like:
result = case drop 1000 primes of
[] -> error "The ancient greeks were wrong! There are less than 1001 primes!"
(r:_) -> r
Now for the hard part. Not knowing how to proceed, we could write some pseudo code:
primes = 2 : {-an infinite list of numbers that are prime-}
We know for sure that 2 is the first prime, the base case, so to speak, thus we can write it down. The unfilled part gives us something to think about. For example, the list should start at some value that is greater 2 for obvious reason. Hence, refined:
primes = 2 : {- something like [3..] but only the ones that are prime -}
Now, this is the point where there emerges a pattern that one needs to learn to recognize. This is surely a list filtered by a predicate, namely prime-ness (it does not matter that we don't know yet how to check prime-ness, the logical structure is the important point. (And, we can be sure that a test for prime-ness is possible!)). This allows us to write more code:
primes = 2 : filter isPrime [3..]
See? We are almost done. In 3 steps, we have reduced a fairly complex problem in such a way that all that is left to write is a quite simple predicate.
Again, we can write in pseudocode:
isPrime n = {- false if any number in 2..n-1 divides n, otherwise true -}
and can refine that. Since this is almost haskell already, it is too easy:
isPrime n = not (any (divides n) [2..n-1])
divides n p = n `rem` p == 0
Note that we did not do optimization yet. For example we can construct the list to be filtered right away to contain only odd numbers, since we know that even ones are not prime. More important, we want to reduce the number of candidates we have to try in isPrime. And here, some mathematical knowledge is needed (the same would be true if you programmed this in C++ or Java, of course), that tells us that it suffices to check if the n we are testing is divisible by any prime number, and that we do not need to check divisibility by prime numbers whose square is greater than n. Fortunately, we have already defined the list of prime numbers and can pick the set of candidates from there! I leave this as exercise.
You'll learn later how to use the standard library and the syntactic sugar like sections, list comprehensions, etc. and you will gradually give up to write your own basic functions.
Even later, when you have to do something in an imperative programming language again, you'll find it very hard to live without infinte lists, higher order functions, immutable data etc.
This will be as hard as going back from C to Assembler.
Have fun!
It's ok to have an imperative mindset at first. With time you will get more used to things and start seeing the places where you can have more functional programs. Practice makes perfect.
As for working with mutable variables you can kind of keep them for now if you follow the rule of thumb of converting variables into function parameters and iteration into tail recursion.
Off the top of my head:
Typeclassopedia. The official v1 of the document is a pdf, but the author has moved his v2 efforts to the Haskell wiki.
What is a monad? This SO Q&A is the best reference I can find.
What is a Monad Transformer? Monad Transformers Step by Step.
Learn from masters: Good Haskell source to read and learn from.
More advanced topics such as GADTs. There's a video, which does a great job explaining it.
And last but not least, #haskell IRC channel. Nothing can even come close to talk to real people.
I think the big change from your code to more haskell like code is using higher order functions, pattern matching and laziness better. For example, you could write the nthPrime function like this (using a similar algorithm to what you did, again ignoring efficiency):
nthPrime n = primes !! (n - 1) where
primes = filter isPrime [2..]
isPrime p = isPrime' p [2..p - 1]
isPrime' p [] = True
isPrime' p (x:xs)
| (p `mod` x == 0) = False
| otherwise = isPrime' p xs
Eg nthPrime 4 returns 7. A few things to note:
The isPrime' function uses pattern matching to implement the function, rather than relying on if statements.
the primes value is an infinite list of all primes. Since haskell is lazy, this is perfectly acceptable.
filter is used rather than reimplemented that behaviour using recursion.
With more experience you will find you will write more idiomatic haskell code - it sortof happens automatically with experience. So don't worry about it, just keep practicing, and reading other people's code.
Another approach, just for variety! Strong use of laziness...
module Main where
nonmults :: Int -> Int -> [Int] -> [Int]
nonmults n next [] = []
nonmults n next l#(x:xs)
| x < next = x : nonmults n next xs
| x == next = nonmults n (next + n) xs
| otherwise = nonmults n (next + n) l
select_primes :: [Int] -> [Int]
select_primes [] = []
select_primes (x:xs) =
x : (select_primes $ nonmults x (x + x) xs)
main :: IO ()
main = do
let primes = select_primes [2 ..]
putStrLn $ show $ primes !! 10000 -- the first prime is index 0 ...
I want to try to answer your question without using ANY functional programming or math, not because I don't think you will understand it, but because your question is very common and maybe others will benefit from the mindset I will try to describe. I'll preface this by saying I an not a Haskell expert by any means, but I have gotten past the mental block you have described by realizing the following:
1. Haskell is simple
Haskell, and other functional languages that I'm not so familiar with, are certainly very different from your 'normal' languages, like C, Java, Python, etc. Unfortunately, the way our psyche works, humans prematurely conclude that if something is different, then A) they don't understand it, and B) it's more complicated than what they already know. If we look at Haskell very objectively, we will see that these two conjectures are totally false:
"But I don't understand it :("
Actually you do. Everything in Haskell and other functional languages is defined in terms of logic and patterns. If you can answer a question as simple as "If all Meeps are Moops, and all Moops are Moors, are all Meeps Moors?", then you could probably write the Haskell Prelude yourself. To further support this point, consider that Haskell lists are defined in Haskell terms, and are not special voodoo magic.
"But it's complicated"
It's actually the opposite. It's simplicity is so naked and bare that our brains have trouble figuring out what to do with it at first. Compared to other languages, Haskell actually has considerably fewer "features" and much less syntax. When you read through Haskell code, you'll notice that almost all the function definitions look the same stylistically. This is very different than say Java for example, which has constructs like Classes, Interfaces, for loops, try/catch blocks, anonymous functions, etc... each with their own syntax and idioms.
You mentioned $ and ., again, just remember they are defined just like any other Haskell function and don't necessarily ever need to be used. However, if you didn't have these available to you, over time, you would likely implement these functions yourself when you notice how convenient they can be.
2. There is no Haskell version of anything
This is actually a great thing, because in Haskell, we have the freedom to define things exactly how we want them. Most other languages provide building blocks that people string together into a program. Haskell leaves it up to you to first define what a building block is, before building with it.
Many beginners ask questions like "How do I do a For loop in Haskell?" and innocent people who are just trying to help will give an unfortunate answer, probably involving a helper function, and extra Int parameter, and tail recursing until you get to 0. Sure, this construct can compute something like a for loop, but in no way is it a for loop, it's not a replacement for a for loop, and in no way is it really even similar to a for loop if you consider the flow of execution. Similar is the State monad for simulating state. It can be used to accomplish similar things as static variables do in other languages, but in no way is it the same thing. Most people leave off the last tidbit about it not being the same when they answer these kinds of questions and I think that only confuses people more until they realize it on their own.
3. Haskell is a logic engine, not a programming language
This is probably least true point I'm trying to make, but hear me out. In imperative programming languages, we are concerned with making our machines do stuff, perform actions, change state, and so on. In Haskell, we try to define what things are, and how are they supposed to behave. We are usually not concerned with what something is doing at any particular time. This certainly has benefits and drawbacks, but that's just how it is. This is very different than what most people think of when you say "programming language".
So that's my take how how to leave an imperative mindset and move to a more functional mindset. Realizing how sensible Haskell is will help you not look at your own code funny anymore. Hopefully thinking about Haskell in these ways will help you become a more productive Haskeller.