i was given a homework in Haskell in which i should program a module, which helps detect prime numbers from a list, say :
[2,3,4,5,6,7,8,9,10]
For the homework, I should iterate through every elements of this list, and eliminate all of it's multiples. Example, I go at number 2, I should eliminate 4,6,8,10. Then go to number 3 and delete 6 and 9, and so on until the end, return the list with prime numbers only.
I have an idea of using function map, but I'm stuck at this place (I'm pretty new to Haskell, though)
Yes, it is my homework, but no, i don't have to do it, it's just practicing. So I'm thankful for any help.
Instead of using a map (I don't think that's possible without doing some pre-processing), you can roll your own function:
sieveWith _ [] = []
sieveWith ss (x:xs) | any ((==) 0 . mod x) ss = sieveWith ss xs
| otherwise = x : (sieveWith (x:ss) xs)
and:
sieve = sieveWith []
Now if you call sieve:
*Main> sieve [2,3,4,5,6,7,8,9,10]
[2,3,5,7]
The function works with a variable (the first one) that is passed through the function calls and each time a value is picked, added to the list. A value is picked if no modulo operation on the variable list yields a zero (second guard). In case any of the modulo's yields zero, the value is simply omitted.
Related
I'm starting learning haskell and i'm stuck in a problem.
I read from the standard input a string like "1234" or "azer"
and I want to make a list like ["123", "234", "341", "412"] or ["aze", "zer", "era", "raz"].
I probably must use map but i don't know how to proceed.
Is someone can help me to do that ? Thanks
Let's start with a list, [1..4]. Let's repeat it for eternity:
>>> cycle [1..4]
[1,2,3,4,1,2,3,4,1,2,3,4,...
Now let's take a slice of it, at say, the 2nd index:
>>> take 4 $ drop (2-1) $ cycle [1..4]
[2,3,4,1]
We can generalize this by naming a function:
slice n = take 4 $ drop n $ cycle [1..4]
To obtain all possible cyclic permutations, we only need to sample n from 1 to 4:
>>> map slice [1..4]
[[2,3,4,1],[3,4,1,2],[4,1,2,3],[1,2,3,4]]
Now, how can we make this work with an arbitrary string? Let's redefine slice to accept a string:
slice s n = take (length s) $ drop n $ cycle s
And so our cyclic permutations function can be defined as follows:
cyclicPerms s = map (slice s) [1..(length s)]
Testing:
>>> cyclicPerms "abcde"
["bcdea","cdeab","deabc","eabcd","abcde"]
I had originally posted an answer totally misunderstanding the specification. I, like a Haskell enumeration, read only the first two numbers so I thought it continued as such. Oops. In any event I just adapted a chunks function I wrote to produce the repetitions. When I get home, I think I have another that cycles lists. I'll post it as well if it's not the same. Who knows.
This function allows you to specify the chunk size as well as the list.
cychnks n ls = [take n.drop x$ls2|(x,y) <-zip [0..] ls]
where ls2 = ls++ls
cychnks 5 "abcde"
["abcde","bcdea","cdeab","deabc","eabcd"]
cychnks 3 "abcde"
["abc","bcd","cde","dea","eab"]
So i am currently working with Haskell at my college, but kinda struggling with pattern-matching and to be more specific i'll give the program, that i am to solve:
My function is awaiting a list of lists ( with each list containing at least 3 elements ) and in each list the second element is replaced with the number of the third element.
Example:
[[1,2,3],[4,5,6,7],[8,9,10]] should become [[1,3,3],[4,6,6,7],[8,10,10]
So far I've made the following:
f [] = []
f [[x]] = []
f [[x,y]] = []
f [[x,y,z]] = [[x,z,z]]
f ([x,y,z]:xs) = [x,z,z]:f(xs)
My questions are:
How can i identify, that some lists may contain more than 3 elements and that the list must remain the same, only the 2nd element changes.
How can i make the recursion, so that it handles exceptions (for example the first list has 4 elements).
Thank you in advance!
It may help to first write the function that swaps the second value in a list with the third.
swap (x:y:z:rest) = x:z:z:rest
swap xs = xs
In the above, x:y:z:rest matches a list of at least length 3. The value of rest, since it is at the end of the pattern, will match any kind of list, both empty and full. The second pattern match of xs is a catch-all for any other type of list, which will just return the same list if it is less than 3 items long.
From there, you can write f in terms of mapping over the outer list and applying swap:
f = map swap
I'm a new student and I'm studying in Computer Sciences. We're tackling Haskell, and while I understand the idea of Haskell, I just can't seem to figure out how exactly the piece of code we're supposed to look at works:
module U1 where
double x = x + x
doubles (d:ds) = (double d):(doubles ds)
ds = doubles [1..]
I admit, it seems rather simple for someone that knows whats happening, but I can't wrap my head around it. If I write "take 5 ds", it obviously gives back [2,4,6,8,10]. What I dont get, is why.
Here's my train of thought : I call ds, which then looks for doubles. because I also submit the value [1..], doubles (d:ds) should mean that d = 1 and ds = [2..], correct? I then double the d, which returns 2 and puts it at the start of a list (array?). Then it calls upon itself, transferring ds = [2..] to d = 2 and ds = [3..], which then doubles d again and again calls upon itself and so on and so forth until it can return 5 values, [2,4,6,8,10].
So first of all, is my understanding right? Do I have any grave mistakes in my string of thought?
Second of all, since it seems to save all doubled d into a list to call for later, whats the name of that list? Where did I exactly define it?
Thanks in advance, hope you can help out a student to understand this x)
I think you are right about the recursion/loop part about how doubles goes through each element of the infinite list.
Now regarding
it seems to save all doubled d into a list to call for later, whats
the name of that list? Where did I exactly define it?
This relates to a feature that's called Lazy Evaluation in Haskell. The list isn't precomputed and stored any where. Instead, you can imagine that a list is a function object in C++ that can generate elements when needed. (The normal language you may see is that expressions are evaluated on demand). So when you do
take 5 [1..]
[1..] can be viewed as a function object that generates numbers when used with head, take etc. So,
take 5 [1..] == (1 : take 4 [2..])
Here [2..] is also a "function object" that gives you numbers. Similarly, you can have
take 5 [1..] == (1 : 2 : take 3 [3..]) == ... (1 : 2 : 3 : 4 : 5 : take 0 [6..])
Now, we don't need to care about [6..], because take 0 xs for any xs is []. Therefore, we can have
take 5 [1..] == (1 : 2 : 3 : 4 : 5 : [])
without needing to store any of the "infinite" lists like [2..]. They may be viewed as function objects/generators if you want to get an idea of how Lazy computation can actually happen.
Your train of thought looks correct. The only minor inaccuracy in it lies in describing the computation using expressions such has "it doubles 2 and then calls itself ...". In pure functional programming languages, such as Haskell, there actually is no fixed evaluation order. Specifically, in
double 1 : double [2..]
it is left unspecified whether doubling 1 happens before of after doubling the rest of the list. Theoretical results guarantee that order is indeed immaterial, in that -- roughly -- even if you evaluate your expression in a different order you will get the same result. I would recommend that you see this property at work using the Lambda Bubble Pop website: there you can pop bubbles in a different order to simulate any evaluation order. No matter what you do, you will get the same result.
Note that, because evaluation order does not matter, the Haskell compiler is free to choose any evaluation order it deems to be the most appropriate for your code. For instance, let ds be defined as in the final line in your code, and consider
take 5 (drop 5 ds)
this results in [12,14,16,18,20]. Note that the compiler has no need to double the first 5 numbers, since you are dropping them, so they can be dropped before they are completely computed (!!).
If you want to experiment, define yourself a function which is very expensive to compute (say, write fibonacci following the recursive definifion).
fibonacci 0 = 0
fibonacci 1 = 1
fibonacci n = fibonacci (n-1) + fibonacci (n-2)
Then, define
const5 n = 5
and compute
fibonacci 100
and observe how long that actually takes. Then, evaluate
const5 (fibonacci 100)
and see that the result is immediately reached -- the argument was not even computed (!) since there was no need for it.
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.
I am very bad at wording things, so please bear with me.
I am doing a problem that requires me to generate all possible numbers in the form of a lists of lists, in Haskell.
For example if I have x = 3 and y = 2, I have to generate a list of lists like this:
[[1,1,1], [1,2,1], [2,1,1], [2,2,1], [1,1,2], [1,2,2], [2,1,2], [2,2,2]]
x and y are passed into the function and it has to work with any nonzero positive integers x and y.
I am completely lost and have no idea how to even begin.
For anyone kind enough to help me, please try to keep any math-heavy explanations as easy to understand as possible. I am really not good at math.
Assuming that this is homework, I'll give you the part of the answer, and show you how I think through this sort of problem. It's helpful to experiment in GHCi, and build up the pieces we need. One thing we need is to be able to generate a list of numbers from 1 through y. Suppose y is 7. Then:
λ> [1..7]
[1,2,3,4,5,6,7]
But as you'll see in a moment, what we really need is not a simple list, but a list of lists that we can build on. Like this:
λ> map (:[]) [1..7]
[[1],[2],[3],[4],[5],[6],[7]]
This basically says to take each element in the array, and prepend it to the empty list []. So now we can write a function to do this for us.
makeListOfLists y = map (:[]) [1..y]
Next, we need a way to prepend a new element to every element in a list of lists. Something like this:
λ> map (99:) [[1],[2],[3],[4],[5],[6],[7]]
[[99,1],[99,2],[99,3],[99,4],[99,5],[99,6],[99,7]]
(I used 99 here instead of, say, 1, so that you can easily see where the numbers come from.) So we could write a function to do that:
prepend x yss = map (x:) yss
Ultimately, we want to be able to take a list and a list of lists, and invoke prepend on every element in the list to every element in the list of lists. We can do that using the map function again. But as it turns out, it will be a little easier to do that if we switch the order of the arguments to prepend, like this:
prepend2 yss x = map (x:) yss
Then we can do something like this:
λ> map (prepend2 [[1],[2],[3],[4],[5],[6],[7]]) [97,98,99]
[[[97,1],[97,2],[97,3],[97,4],[97,5],[97,6],[97,7]],[[98,1],[98,2],[98,3],[98,4],[98,5],[98,6],[98,7]],[[99,1],[99,2],[99,3],[99,4],[99,5],[99,6],[99,7]]]
So now we can write that function:
supermap xs yss = map (prepend2 yss) xs
Using your example, if x=2 and y=3, then the answer we need is:
λ> let yss = makeListOfLists 3
λ> supermap [1..3] yss
[[[1,1],[1,2],[1,3]],[[2,1],[2,2],[2,3]],[[3,1],[3,2],[3,3]]]
(If that was all we needed, we could have done this more easily using a list comprehension. But since we need to be able to do this for an arbitrary x, a list comprehension won't work.)
Hopefully you can take it from here, and extend it to arbitrary x.
For the specific x, as already mentioned, the list comprehension would do the trick, assuming that x equals 3, one would write the following:
generate y = [[a,b,c] | a<-[1..y], b<-[1..y], c <-[1..y]]
But life gets much more complicated when x is not predetermined. I don't have much experience of programming in Haskell, I'm not acquainted with library functions and my approach is far from being the most efficient solution, so don't judge it too harshly.
My solution consists of two functions:
strip [] = []
strip (h:t) = h ++ strip t
populate y 2 = strip( map (\a-> map (:a:[]) [1..y]) [1..y])
populate y x = strip( map (\a-> map (:a) [1..y]) ( populate y ( x - 1) ))
strip is defined for the nested lists. By merging the list-items it reduces the hierarchy so to speak. For example calling
strip [[1],[2],[3]]
generates the output:
[1,2,3]
populate is the tricky one.
On the last step of the recursion, when the second argument equals to 2, the function maps each item of [1..y] with every element of the same list into a new list. For example
map (\a-> map (:a:[]) [1..2]) [1..2])
generates the output:
[[[1,1],[2,1]],[[1,2],[2,2]]]
and the strip function turns it into:
[[1,1],[2,1],[1,2],[2,2]]
As for the initial step of the recursion, when x is more than 2, populate does almost the same thing except this time it maps the items of the list with the list generated by the recursive call. And Finally:
populate 2 3
gives us the desired result:
[[1,1,1],[2,1,1],[1,2,1],[2,2,1],[1,1,2],[2,1,2],[1,2,2],[2,2,2]]
As I mentioned above, this approach is neither the most efficient nor the most readable one, but I think it solves the problem. In fact, theoritically the only way of solving this without the heavy usage of recursion would be building the string with list comprehension statement in it and than compiling that string dynamically, which, according to my short experience, as a programmer, is never a good solution.