I am trying to convert a given decimal value its corresponding binary form. I am using Ocaml about which I don't know much and am quite confused. So far I have the following code
let dec_to_bin_helper function 1->'T' | 0->'F'
let dec_to_bin x =
List.fold_left(fun a z -> z mod 2 dec_to_bin_helper a) [] a ;;
I must include here that I want my output to be in the form of a list of T's and F's where T's represent the binary 1's and F's represent binary 0's
If I try to run the above code it gives me an error saying "Error: This expression is not a function; it cannot be applied"
I understand that the part where I am calling the helper function is wrong... Any help in the matter would be appreciated!
I don't really understand your second function at all. You are folding an empty list, and your function takes an argument x which it never uses. Am I correct in assuming that you want to take a number and return a list of 'T's and 'F's which represent the binary? If that is the case, this code should work:
let dec_to_bin x =
let rec d2b y lst = match y with 0 -> lst
| _ -> d2b (y/2) ((dec_to_bin_helper (y mod 2))::lst)
in
d2b x [];;
This function inserts (x mod 2) converted into a T/F into a list, then recursively calls the function on x/2 and the list. When x = 0 the list is returned. If call it on 0 an empty list will be returned (I'm not sure if that's what you want or not).
I think the problem that you had is that you are treating lists as if they are mutable and thinking that fold mutates the list. That is not the case, fold just goes through each element in a list and applies a function to it. Since your list is empty it didn't do anything.
Related
This is my program:
modify :: Integer -> Integer
modify a = a + 100
x = x where modify(x) = 101
In ghci, this compiles successfully but when I try to print x the terminal gets stuck. Is it not possible to find input from function output in Haskell?
x = x where modify(x) = 101
is valid syntax but is equivalent to
x = x where f y = 101
where x = x is a recursive definition, which will get stuck in an infinite loop (or generate a <<loop>> exception), and f y = 101 is a definition of a local function, completely unrelated to the modify function defined elsewhere.
If you turn on warnings you should get a message saying "warning: the local definition of modify shadows the outer binding", pointing at the issue.
Further, there is no way to invert a function like you'd like to do. First, the function might not be injective. Second, even if it were such, there is no easy way to invert an arbitrary function. We could try all the possible inputs but that would be extremely inefficient.
I'm studying project euler solutions and this is the solution of problem 4, which asks to
Find the largest palindrome made from the product of two 3-digit
numbers
problem_4 =
maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s]
I understand that this code creates a list such that x is a product of all possible z and y.
However I'm having a problem understanding what does s do here. Looks like everything after | is going to be executed everytime a new element from this list is needed, right?
I don't think I understand what's happening here. Shouldn't everything to the right of | be constraints?
A list comprehension is a rather thin wrapper around a do expression:
problem_4 = maximum $ do
y <- [100..999]
z <- [y..999]
let x = y*z
let s = show x
guard $ s == reverse s
return x
Most pieces translate directly; pieces that aren't iterators (<-) or let expressions are treated as arguments to the guard function found in Control.Monad. The effect of guard is to short-circuit the evaluation; for the list monad, this means not executing return x for the particular value of x that led to the false argument.
I don't think I understand what's happening here. Shouldn't everything to the right of | be constraints?
No, at the right part you see an expression that is a comma-separated (,) list of "parts", and every part is one of the following tree:
an "generator" of the form somevar <- somelist;
a let statement which is an expression that can be used to for instance introduce a variable that stores a subresult; and
expressions of the type boolean that act like a filter.
So it is not some sort of "constraint programming" where one simply can list some constraints and hope that Haskell figures it out (in fact personally that is the difference between a "programming language" and a "specification language": in a programming language you have "control" how the data flows, in a specification language, that is handled by a system that reads your specifications)
Basically an iterator can be compared to a "foreach" loop in many imperative programming languages. A "let" statement can be seen as introducing a temprary variable (but note that in Haskell you do not assign variable, you declare them, so you can not reassign values). The filter can be seen as an if statement.
So the list comprehension would be equivalent to something in Python like:
for y in range(100, 1000):
for z in range(y, 1000):
x = y * z
s = str(x)
if x == x[::-1]:
yield x
We thus first iterate over two ranges in a nested way, then we declare x to be the multiplication of y and z, with let s = show x, we basically convert a number (for example 15129) to its string counterpart (for example "15129"). Finally we use s == reverse s to reverse the string and check if it is equal to the original string.
Note that there are more efficient ways to test Palindromes, especially for multiplications of two numbers.
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.
I'd greatly appreciate if you could tell me how to make a single string from a range between two ints. Like [5..10] i would need to get a "5678910". And then I'd have to calculate how many (zeroes, ones ... nines) there are in a string.
For example: if i have a range from [1..10] i'd need to print out
1 2 1 1 1 1 1 1 1 1
For now i only have a function to search for a element in string.
`countOfElem elem list = length $ filter (\x -> x == elem) list`
But the part how to construct such a string is bugging me out, or maybe there is an easier way? Thank you.
I tried something like this, but it wouldn't work.
let intList = map (read::Int->String) [15..22]
I tried something like this, but it wouldn't work. let intList = map (read::Int->String) [15..22]
Well... the purpose of read is to parse strings to read-able values. Hence it has a type signature String -> a, which obviously doesn't unify with Int -> String. What you want here is the inverse1 of read, it's called show.
Indeed map show [15..22] gives almost the result you asked for – the numbers as decimal-encoded strings – but still each number as a seperate list element, i.e. type [String] while you want only String. Well, how about asking Hoogle? It gives the function you need as the fifth hit: concat.
If you want to get fancy you can then combine the map and concat stages: both the concatMap function and the >>= operator do that. The most compact way to achieve the result: [15..22]>>=show.
1show is only the right inverse of read, to be precise.
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.