Result of this function:
list = [x*x | x <- [0..100]]
res = id
. filter (\(x,y,sum) -> sum > 9)
. takeWhile (\(x,y,sum) -> sum < 100)
. nub
$ [(x, y, (sqrt (x) + sqrt (y))) | x <- list, y <- list]
is [(0, 100, 10), (0, 121, 11 ... (0, 9801 ,99)].
How can I make it work with other items in x's list?
A takeWhile :: (a -> Bool) -> [a] -> [a] is a function that - like the name says - takes elements while the condtions holds. From the moment one such element fails, then the rest of the list is not emitted anymore, regardless whether it yields valueable data or not.
Since your list is however finite, and you are interested in all configurations, you do not have to use takeWhile here, you can use filter, so:
res = id
. filter (\(x,y,sum) -> sum > 9)
. filter (\(x,y,sum) -> sum < 100)
. nub
$ [(x, y, (sqrt (x) + sqrt (y))) | x <- list, y <- list]
Now we have fixed this. But the code still has some redundant and inelegant aspects.
We better avoid calculating the square root, by modifying list to emit squares together with the original value. Furthermore we can simply make it an infinite list, which is usally better if we want to reuse it in other parts of the code:
squares = [ (x, x*x) | x <- [0..]]
Next we thus will take values for as long as the value square root is less than 100:
where sq100 = takeWhile (\(r,_) -> r < 100) squares
Now we can use list comprehension that takes square roots and squares of the two lists and calculate the sum of the square roots, such that the sum is greater than 9 and less than 100. We can thus write:
res = [ (x,y,sqq) | (sqx,x) <- sq100, (sqy,y) <- sq100, x <= y,
let sqq = sqx+sqy, sqq > 9, sqq < 100 ]
where sq100 = takeWhile (\(r,_) -> r < 100) squares
Here we thus take an x and square root sqx of x, we take an y and square root sqy of y, then we constraint x <= y to remove duplicates, calculate the sum of the square roots sqq = sqx+sqy and finally we check whether that sum if greater than 9 and less than 100.
This produces 2520 results.
The takeWhile is the culprit because the sum will be > 100 at some point, and then the list gets cut off.
Related
I would like to know the number of cases in which 1 dollar can be expressed in 1,5,10,20,50 cents.
For example, the count(100,[50,25]) is:
Because 50 * 1 + 25 * 2, it = 3:int is printed.
However, in my code, only the front part of the list is printed, so even if I count (100,[50,25]), it = 2:int is printed.
In other words, My code is not taking advantage of the whole list.
How do I solve this?
SML coin count function:
fun count(x,[]) = 0
| count (x,y::ys) =
let val cnt = 0
in if y*2 = x then cnt+2
else if y*4 = x then cnt + 4
else if y*10 = x then cnt + 10
else if y*10 = x then cnt + 10
else if y*20 = x then cnt + 20
else count(x-y,ys)
end;
Consider what happens as you evaluate your test expression of count (100, [50, 25]).
cnt is 0, y is 50, and ys is [25].
y times 2 does equal 100, so it returns cnt+2 which is 2. Nothing further happens.
When it comes to recursion, remember than the parameter list to a function is your means of communication. It seems like cnt is something that should be passed as a parameter so you can update it between recursive calls.
With count(x, []) = 0 you already have an exit point that will stop the recursion.
Edit: Based on comments, it looks like you're trying to figure out how many times each value in a list goes into a value x.
So the end result of your recursive function isn't a single integer. It's a list of integers. Or better yet, of tuples containing the value to look for, and the number of times it goes into x.
So if the list is empty, the result is obvious.
fun count(x, []) = []
It's an empty list. Otherwise, we need to append something onto a list.
fun count(x, []) = []
| count(x, y::ys) =
(y, x div y) :: count(x, ys)
Of course, we also have functions like map that basically do this for us.
fun count(x, lst) = List.map (fn y => (y, x div y)) lst
I need to generate infinite sorted list of all coprimes.
The first element in each pair must be less than the second.
The sorting must be done in ascending order -- by the sum of pair's elements; and if two sums are equal, then by the pair's first element.
So, the resulting list must be
[(2,3),(2,5),(3,4),(3,5),(2,7),(4,5),(3,7),(2,9),(3,8),(4,7)...`
Here's my solution.
coprimes :: [(Int, Int)]
coprimes = sortBy (\t1 t2 -> if uncurry (+) t1 <= uncurry (+) t2 then LT else GT) $ helper [2..]
where helper xs = [(x,y) | x <- xs, y <- xs, x < y, gcd x y == 1]
The problem is that I can't take n first pairs. I realize that sorting can't be done on infinite lists.
But how can I generate the same sequence in a lazy way?
While probably not the most optimal way it should works if you first generate all possible pairs and then filter them.
So using your criteria:
pairs :: [(Integer,Integer)]
pairs = [ (i,l-i) | l <- [1..], i <- [1..l-1] ]
coprimes :: [(Integer,Integer)]
coprimes = [ (i,j) | (i,j) <- pairs, 1 < i, i < j,gcd i j == 1]
produces
λ> take 10 coprimes
[(2,3),(2,5),(3,4),(3,5),(2,7),(4,5),(3,7),(2,9),(3,8),(4,7)]
now of course you can put some of the stuff 1 < i and i < j comes to mind into the pairs definition or even join them but I think here it's more obvious what's going on
Here's a possible solution following Chapter 9 of Richard Bird's Thinking Functionally in Haskell:
coprimes = mergeAll $ map coprimes' [2..]
coprimes' n = [(n, m) | m <- [n+1..], gcd m n == 1]
merge (x:xs) (y:ys)
| s x < s y = x:merge xs (y:ys)
| s x == s y = x:y:merge xs ys
| otherwise = y:merge (x:xs) ys
where s (x, y) = x+y
xmerge (x:xs) ys = x:merge xs ys
mergeAll = foldr1 xmerge
And the result is:
> take 10 $ coprimes
[(2,3),(2,5),(3,4),(3,5),(2,7),(4,5),(3,7),(2,9),(3,8),(4,7)]
Note that the natural definition of mergeAll would be foldr1 merge, but this doesn't work because it will try to find the minimum of the first elements of all the list before returning the first element, and hence you end up in an infinite loop. However, since we know that the lists are in ascending order and the minimum is the first element of the first list xmerge does the trick.
Note: this solution appears to be significantly (like 2 order of magnitudes) slower than Carsten "naive" answer. So I advise to avoid this if you are interested in performance. Yet it still is an interesting approach that might be effective in other situations.
As #Bakuriu suggests, merging an infinite list of infinite lists is a solution, but the devil is in the details.
The diagonal function from the universe-base package can do this, so you could write:
import Data.Universe.Helpers
coprimes = diagonal [ go n | n <- [2..] ]
where go n = [ (n,k) | k <- [n+1..], gcd n k == 1 ]
Note - this doesn't satisfy your sorted criteria, but I mention it because the functions in that package are useful to know about, and implementing a function like diagonal correctly is not easy.
If you want to write your own, consider decomposing the infinite grid N x N (where N is the natural numbers) into diagonals:
[ (1,1) ] ++ [ (1,2), (2,1) ] ++ [ (1,3), (2,2), (3,1) ] ++ ...
and filtering this list.
I need to generate infinite sorted list of all coprimes. The first element in each pair must be less than the second. The sorting must be done in ascending order -- by the sum of pair's elements; and if two sums are equal, then by the pair's first element.
So, we generate ascending pairs of sum and first element, and keep only the coprimes. Easy cheesy!
[ (first, second)
| sum <- [3..]
, first <- [2..sum `div` 2]
, let second = sum-first
, gcd first second == 1
]
listX n = xs
if sum[x | x <- [2, 4..n-1], y <- [1..n-1], y `rem` x == 0] == y
then insert y xs
else return ()
Alright, first time trying to work with Haskell, and only having novice Java knowledge has led to some problems.
What I was trying to do, was to define the result of the function listX n as a list called xs.
My idea was that the program would grab every number of from 1 to n, and check if it was equal to the sum of its positive divisors.
Clearly, I have failed horribly and need help, pointers to concepts I haven't understood is extremely appreciated.
Your main problem seems to be that you still think imperative (with the insert) - also () is the value unit - you probably wanted to write [] (the empty list) instead - but still the xs here is totally undefined so you would have to fix this too (and I don't see how to be honest).
perfect numbers
I think I can see a basic idea in there, and I think the best way to fix this is to go full list-comprehension (as you seem to understand them quite well) - here is a version that should work:
listX n = [ x | x <- [1..n], sum [ y | y <- [1..x-1], x `mod` y == 0] == x]
As you can see I changed this a bit - first I check all x from 1 to n if they could be perfect - and I do this by checking by summing up all proper divisors and checking if the sum is equal to x (that's the job of the sum [...] == x part) - in case you don't know this works because you can add guards to list comprehensions (the sum [..] == x filters out all values of x where this is true).
a nicer version
to make this a bit more readable (and separate the concerns) I would suggest writing it that way:
properDivisors :: Integer -> [Integer]
properDivisors n = [ d | d <- [1..n-1], n `mod` d == 0]
isPerfect :: Integer -> Bool
isPerfect n = sum (properDivisors n) == n
perfectNumbers :: [Integer]
perfectNumbers = filter isPerfect [1..]
perfectNumbersUpTo :: Integer -> [Integer]
perfectNumbersUpTo n = takeWhile (<= n) perfectNumbers
Here is my problem: I need a Haskell function that computes an approximation of the sine of some number, using the associated Taylor serie ...
In C++ I wrote this:
double msin(double number, int counter = 0, double sum = 0)
{
// sin(x) = x - (x'3 / 3!) + (x'5 / 5!) - (x'7 / 7!) + (x'9 / 9!)
if (counter <= 20)
{
if (counter % 2 == 0)
sum += mpow(number, counter * 2 + 1) / mfak(counter * 2 + 1) ;
else
sum -= mpow(number, counter * 2 + 1) / mfak(counter * 2 + 1) ;
counter++;
sum = msin(number, counter, sum);
return sum;
}
return (sum* 180.0 / _PI);
}
Now I am trying to do it in Haskell and I have no idea how... For now I was trying something like this (it doesn't really work, but it is work in progress ;) ):
This works:
mfak number = if number < 2
then 1
else number *( mfak (number -1 ))
mpow number potenca = if potenca == 0
then 0
else if potenca == 1
then 1
else (number * (mpow number (potenca-1)))
This doesn't work:
msin :: Double -> Int -> Double -> Double
msin number counter sum = if counter <= 20
then if counter `mod` 2==0
then let sum = sum + (msin 1 (let counter = counter+1 in counter) sum) in sum
else let sum = sum + (msin 1 (let counter = counter+1 in counter) sum) in sum
else sum* 180.0 / 3.14
Updated....doesn't compile :/ "Couldn't match expected type Double' with actual type Int'"
msin :: Double -> Int -> Double -> Double
msin number counter sum = if counter <= 20
then if counter `mod` 2==0
then let sum' = sum + ((mpow number (counter*2+1))/(mfak counter*2+1)) in msin number (counter+1) sum'
else let sum' = sum - ((mpow number (counter*2+1))/(mfak counter*2+1)) in msin number (counter+1) sum'
else sum* 180.0 / 3.14
As you can see, the biggest problem is how to add something to "sum", increase "counter" and go in recursion again with these new values...
P.S. I am new to Haskell so try to explain your solution as much as you can please. I was reading some tutorials and that, but I can't find how to save the result of some expression into a value and then continue with other code after it... It just returns my value each time I try to do that, and I don't want that....
So thanks in advance for any help!
I would rework the algorithm a bit. First we can define the list of factorial inverses:
factorialInv :: [Double]
factorialInv = scanl (/) 1 [1..] -- 1/0! , 1/1! , 1/2! , 1/3! , ...
Then, we follow with the sine coefficients:
sineCoefficients :: [Double]
sineCoefficients = 0 : 1 : 0 : -1 : sineCoefficients
Then, given x, we multiply both the above lists with the powers of x, pointwise:
powerSeries :: [Double] -- ^ Coefficients
-> Double -- ^ Point x on which to compute the series
-> [Double] -- ^ Series terms
powerSeries cs x = zipWith3 (\a b c -> a * b * c) cs powers factorialInv
where powers = iterate (*x) 1 -- 1 , x , x^2 , x^3 , ...
Finally, we take the first 20 terms and sum them up.
sine :: Double -> Double
sine = sum . take 20 . powerSeries sineCoefficients
-- i.e., sine x = sum (take 20 (powerSeries sineCoefficients x))
The problem is expressions like let stevec = stevec+1 in stevec. Haskell is not an imperative language. This does not add one to stevec. Instead it defines stevec to be a number that is one more than itself. No such number exists, thus you will get an infinite loop or, if you are lucky, a crash.
Instead of
stevec++;
vsota = msin(stevilo, stevec, vsota);
You should use something like
let stevec' = stevec + 1
in msin stevilo stevec' vsota
or just
msin stevilo (stevec + 1) vsota
(There's also something here that I don't understand. You are going to need mpow and mfak. Where are they?)
As you can see the biggest problem is how to add something to "vsota",
In a functional language you would use recursion here - the variable vstota is implemented as a function parameter which is passed from call to call as a list is processed.
For example, to sum a list of numbers, we would write something like:
sum xs = go 0 xs
where go total [] = total
go total (x:xs) = go (total+x) xs
In an imperative language total would be a variable which gets updated. Here is is a function parameter which gets passed to the next recursive call to go.
In your case, I would first write a function which generates the terms of the power series:
sinusTerms n x = ... -- the first n terms of x - (x'3 / 3!) + (x'5 / 5!) - (x'7 / 7!) ...
and then use the sum function above:
sinus n x = sum (sinusTerms n x)
You may also use recursive lists definitions to get [x, x^3, x^5 ...] and [1, 1/3!, 1/5! ...] infinite sequences. When they are done, the rest is to multiply their items each by other and take the sum.
sinus count x = sum (take count $ zipWith (*) ifactorials xpowers)
where xpowers = x : map ((x*x)*) xpowers
ifactorials = 1 : zipWith (/) ifactorials [i*(i+1) | i <- [2, 4 .. ]]
Also, it would be better to define xpowers = iterate ((x*x)*) x, as it seems to be much more readable.
I’ve tried to follow your conventions as much as I could. For mfak and mpow, you should avoid using if as it is clearer to write them using pattern matching :
mfak :: Int -> Int
mfak 0 = 1
mfak 1 = 1
mfak n = n * mfak (n - 1)
mpow :: Double -> Int -> Double
mpow _ 0 = 1
mpow x 1 = x
mpow x p = x * mpow x (p - 1)
Before calculating the sinus, we create a list of coefficients [(sign, power, factorial)] :
x - (x^3 / 3!) + (x^5 / 5!) - (x^7 / 7!) + (x^9 / 9!)
→ [(1,1,1), (-1,3,6), (1,5,120), (-1,7,5040), (1,9,362880)]
The list is created infinite by a list comprehension. First we zip the lists [1,-1,1,-1,1,-1...] and [1,3,5,7,9,11...]. This gives us the list [(1,1), (-1,3), (1,5), (-1,7)...]. From this list, we create the final list [(1,1,1), (-1,3,6), (1,5,120), (-1,7,5040)...]:
sinCoeff :: [(Double, Int, Double)]
sinCoeff = [ (fromIntegral s, i, fromIntegral $ mfak i)
| (s, i) <- zip (cycle [1, -1]) [1,3..]]
(cycle repeats a list indefinitely, [1,3..] creates an infinite list which starts at 1 with a step of 2)
Finally, the msin function is near the definition. It also uses a list comprehension to achieve its goeal (note that I kept the * 180 / pi though I’m not sure it should be there. Haskell knows pi).
msin :: Int -> Double -> Double
msin n x = 180 * sum [ s * mpow x p / f | (s, p, f) <- take n sinCoeff] / pi
(take n sinCoeff returns the first n elements of a list)
You may try the previous code with the following :
main = do
print $ take 10 sinCoeff
print $ msin 5 0.5
print $ msin 10 0.5
The expression is of the form x*P(x2).
For maximal efficiency, the polynomial in x2 must be evaluated using the Horner rule rather than computing the powers of x2 separately.
The coefficient serie with the factorial values can be expressed recursively in Haskell, just like is commonly done for the Fibonacci series. Using the ghci interpreter as our testbed, we have:
$ ghci
GHCi, version 8.8.4: https://www.haskell.org/ghc/ :? for help
λ>
λ>
λ> nextCoeffs d c = c : (nextCoeffs (d+1) ((-c)/(fromIntegral $ (2*d+2)*(2*d+3))))
λ>
λ> allCoeffs = nextCoeffs 0 1.0
λ>
where d is the depth inside the serie and c the current coefficient.
Sanity check: the coefficient at depth 3 must be the inverse of 7!
λ>
λ> 1.0 /(allCoeffs !! 3)
-5040.0
λ>
The Horner rule can be rendered in Haskell thru the foldr1 :: (a -> a -> a) -> [a] -> a library function.
As is customary in Haskell, I take the liberty to put the term count as the leftmost argument because it is the one most likely to be held constant. This is for currying (partial evaluation) purposes.
So we have:
λ> :{
|λ> msin count x = let { s = x*x ; cs = take count allCoeffs ;
|λ> stepFn c acc = acc*s + c ; }
|λ> in x * (foldr1 stepFn cs)
|λ> :}
Sanity checks, taking 20 terms:
λ>
λ> pi
3.141592653589793
λ>
λ> msin 20 (pi/6)
0.49999999999999994
λ>
λ> msin 20 (pi/2)
1.0
λ>
Side note 1: final multiplication by 180 / π is only of interest for inverse trigonometric functions.
Side note 2: in practice, to get a reasonably fast convergence, one should reduce the input variable x into the [-π,+π] interval using the periodicity of the sine function.
Create an infinite list pairs :: [(Integer, Integer)] containing pairs of the form (m,n),
where each of m and n is a member of [0 ..]. An additional requirement is that if (m,n)
is a legit member of the list, then (elem (m,n) pairs) should return True in finite time.
An implementation of pairs that violates this requirement is considered a non- solution.
****Fresh edit Thank you for the comments, Lets see if I can make some progress****
pairs :: [(Integer, Integer)]
pairs = [(m,n) | t <- [0..], m <- [0..], n <-[0..], m+n == t]
Something like this? I just don't know where it's going to return True in finite time.
I feel the way the question is worded elem doesn't have to be part of my answer. Just if you call (elem (m,n) pairs) it should return true. Sound right?
Ignoring the helper method, the list comprehension you have will list out all pairs but the order of elements is a problem. You'll have a infinitely many pairs like (0, m) which are followed by infinitely many pairs like (1, m). Of course elem will forever iterate all the (0, m) pairs never reaching (1, m) or (2, m) etc.
I'm not sure why you have the helper method -- with it, you are only building a list of pairs like [(0,0), (1,1), (2,2), ...] because you've filtered on m = n. Was that part of the requirements?
Like #hammar suggested, start with 0 = m + n and list out the pairs (m, n). Then list pairs (m, n) where 1 = m + n. Then your list will look like [(0,0), (0,1), (1,0), (0,2), (1,1), (2,0), ...].
The helper function ensures that pairs is a list of the form [ (0,0) , (1,1) , (2,2) ... ].
So elem ( m , n ) pairs can be implemented as:
elem (m , n) _ | m == n = True
| otherwise = False
This is a constant time implementation.
I first posted
Prelude> let pairs = [(m, n) | t <- [0..]
, let m = head $ take 1 $ drop t [0..]
, let n = head $ take 1 $ drop (t + 1) [0..]]
Which, I believed answered the three conditions set by the professor. But hammar pointed out that if I chose this list as an answer, that is, the list of pairs of the form (t, t+1), then I might as well choose the list
repeat [(0,0)]
Well, both of these do seem to answer the professor's question, considering there seems to be no mention of the list having to contain all combinations of [0..] and [0..].
That aside, hammer helped me see how you can list all combinations, facilitating the evaluation of elem in finite time by building the infinite list from finite lists. Here are two other finite lists - less succinct than Hammar's suggestion of the diagonals - that seem to build all combinations of [0..] and [0..]:
edges = concat [concat [[(m,n),(n,m)] | let m = t, n <- take m [0..]] ++ [(t,t)]
| t <- [0..]]
*Main> take 9 edges
[(0,0),(1,0),(0,1),(1,1),(2,0),(0,2),(2,1),(1,2),(2,2)]
which construct the edges (t, 0..t) (0..t, t), and
oddSpirals size = concat [spiral m size' | m <- n] where
size' = if size < 3 then 3 else if even size then size - 1 else size
n = map (\y -> (fst y * size' + div size' 2, snd y * size' + div size' 2))
[(x, t-x) | let size' = 5, t <- [0..], x <- [0..t]]
spiral seed size = spiral' (size - 1) "-" 1 [seed]
spiral' limit op count result
| count == limit =
let op' = if op == "-" then (-) else (+)
m = foldl (\a b -> a ++ [(op' (fst $ last a) b, snd $ last a)]) result (replicate count 1)
nextOp = if op == "-" then "+" else "-"
nextOp' = if op == "-" then (+) else (-)
n = foldl (\a b -> a ++ [(fst $ last a, nextOp' (snd $ last a) b)]) m (replicate count 1)
n' = foldl (\a b -> a ++ [(nextOp' (fst $ last a) b, snd $ last a)]) n (replicate count 1)
in n'
| otherwise =
let op' = if op == "-" then (-) else (+)
m = foldl (\a b -> a ++ [(op' (fst $ last a) b, snd $ last a)]) result (replicate count 1)
nextOp = if op == "-" then "+" else "-"
nextOp' = if op == "-" then (+) else (-)
n = foldl (\a b -> a ++ [(fst $ last a, nextOp' (snd $ last a) b)]) m (replicate count 1)
in spiral' limit nextOp (count + 1) n
*Main> take 9 $ oddSpirals 3
[(1,1),(0,1),(0,2),(1,2),(2,2),(2,1),(2,0),(1,0),(0,0)]
which build clockwise spirals of length 'size' squared, superimposed on hammar's diagonals algorithm.
I believe the solution to your task is:
pairs = [(x,y) | u <- [0..], x <- [0..u], y <-[0..u] , u == x+y]