all_nat x = [ls| sum ls == x]
I'd like to write a function that given an integer x it returns all the lists that the result of their elements when summed is the integer x but I always get the error "not in scope: 'ls' " for both times it apperas. I'm new to haskell. What's the syntax error here?
The problem is that you need to define all used variables somewhere, but ls is undefined. Moreover, it can't be defined automatically, because the compiler doesn't know about the task — how the list should be generated? Ho long can it be? Are terms positive or not, integral or not? Unfortunately your code definition of the problem is quite vague for modern non-AI languages.
Let's help the compiler. To solve such problems, it's often useful to involve some math and infer the algorithm inductively. For example, let's write an algorithm with ordered lists (where [2,1] and [1,2] are different solutions):
Start with a basis, where you know the output for some given input. For example, for 0 there is only an empty list of terms (if 0 could be a term, any number could be decomposed as a sum in infinitely many ways). So, let's define that:
allNats 0 = [[]] --One empty list
An inductive step. Assuming we can decompose a number n, we can decompose any number n+k for any positive k, by adding k as a term to all decompositions of n. In other words: for numbers greater than 0, we can take any number k from 1 to n, and make it the first term of all decompositions of (n-k):
allNats n = [ k:rest --Add k as a head to the rest, where
| k <- [1 .. n] --k is taken from 1 to n, and
, rest <- allNats (n - k)] --rest is taken from solutions for (n—k)
That's all! Let's test it:
ghci> allNat 4
[[1,1,1,1],[1,1,2],[1,2,1],[1,3],[2,1,1],[2,2],[3,1],[4]]
Let's break this up into two parts. If I've understood your question correctly, the first step is to generate all possible (sub)lists from a list. There's a function to do this, called subsequences.
The second step is to evaluate the sum of each subsequence, and keep the subsequences with the sum you want. So your list comprehension looks like this:
all_nat x = [ls| ls <- subsequences [1..x], sum ls == x]
What about
getAllSums x = [(l,r)| l <- partial_nat, r <- partial_nat, l + r == x ]
where partial_nat = [1..x]
Related
Entering a list comprehension into GHCi does not generate a list, the final square brackets are missing, and the console freezes. This is what I have come up with:
[13*x + 3 | x <- [1..], rem (13*x + 3) 12 == 5, mod (13*x + 3) 11 == 0, 13*x + 3 <= 1000]
I believe the problem lies either with x <- [1..], or 13*x + 3 <= 1000. By 13*x + 3 <= 1000 I meant to determine the upper limit of the values x in x <- [1..] can take.
I'm given back a result [341, but it does the second square bracket is missing, and the console freezes.
Your program enters an infinite loop.
The first number is 341, but in order to produce the next number, your program keeps looking through all the subsequent values of x, evaluates all the guards for those values, and checks if all the guards are true. The very last guard, 13*x + 3 <= 1000 never becomes true again, so the program just keeps enumerating values of x forever. It's looking for the next such x for which all guards are true, and as soon as it finds one, it's going to print it. But such x never comes.
If you want the list to end once x*13 + 3 > 1000, you have to use takeWhile:
... | x <- takeWhile (\y -> y*13 + 3 <= 1000) [1..], ...
That way the list will actually stop when it reaches 1000. No more values of x would be produced.
You're giving the compiler way too much credit. It isn't going to carefully analyse your list comprehension in order to deduce that past a certain point there will be no more results, and it should call the list complete. It only does what you tell it to do.
In this case what you told it to do is:
[ 13*x + 3 -- produce numbers of the form 13*x + 3
| x <- [1..] -- by searching all x from [1..]
, rem (13*x + 3) 12 == 5 -- allowing only x that meet this condition
, mod (13*x + 3) 11 == 0 -- and this condition
, 13*x + 3 <= 1000 -- and this condition
]
So it prints [341 and "freezes" because it's still trying to compute the rest of that list. You don't see anything happening, but internally it's drawing ever bigger x from [1..] and diligently checking those conditions to realise that the number shouldn't be included. But it never hits the end of [1..] in order to stop, so it never gets up to printing the ] and waiting for more input.
With your code you are explicitly telling the compiler that you want to search every number in the infinite1 list [1..]. You are then expecting it to notice that 13*x + 3 <= 1000 can only be true for x drawn from a finite prefix of [1..] and thus actually not search the entire list [1..] as you instructed2.
That is a perfectly reasonable thing to want, and I can imagine a system capable of pulling that off (at least with simple conditions like this). So testing it out like this to see if it works is a good idea! However unless someone actually told you that figuring out enumeration upper bounds from conditions in list comprehensions is a feature that GHC can provide, it shouldn't be surprising that it never completes when you tell it to search an infinite list.
For this style of list comprehension (getting all numbers in a range meeting certain conditions) you normally shouldn't use [1..] and then try to impose a stopping condition. Just figure out that the last number that will pass 13*x + 3 <= 1000 and use [1..76] as your generator instead. You can even have Haskell figure it out for you with [1 .. (1000 - 3) div 13].
You use a generator like [1..] when you want to get all numbers of the right form. Then you can use functions like take or takeWhile to get a finite section at the point where you want to use it for something. e.g.
Prelude> let xs = [13*x + 3 | x <- [1..], rem (13*x + 3) 12 == 5, mod (13*x + 3) 11 == 0]
Prelude> takeWhile (<= 1000) xs
[341]
Prelude> take 5 xs
[341,2057,3773,5489,7205]
In fact the simplest and most direct way to express what you want in a single expression is this:
takeWhile (<= 1000) [13*x + 3 | x <- [1..], rem (13*x + 3) 12 == 5, mod (13*x + 3) 11 == 0]
Everything in a list comprehension (except the generator expression) is only talking about a single element at a time. There's just no way to express concepts that are talking about the returned list as a whole, like "stop searching once the returned numbers go out of this range". But that concept is trivial to express outside of list comprehension as a normal function (takeWhile (<= 1000)). Don't feel like you have to shoehorn your entire computation into a single list comprehension.
1 Strictly speaking it's infinite if you're using a type like Integer (which is the type Haskell will pick without any other code using the result to impose other constraints on the type). If you're using Int then it's technically finite, and your list comprehension will eventually end when it "runs out of numbers". [1..] as a list of Int is still impractically vast for an exhaustive search, however.
But if you use a smaller type, like Word16 (needs to be imported from Data.Word) then you can in fact finish your original list comprehension in a practical amount of time. (Though I had to tweak it a little to make sure the 13*x stuff was computed in a larger type so it doesn't overflow)
Prelude> import Data.Word
Prelude Data.Word> [13*x + 3 | x <- [1 :: Word16 ..], let x' = fromIntegral x, rem (13*x' + 3) 12 == 5, mod (13*x' + 3) 11 == 0, 13*x' + 3 <= 1000]
[341]
2 While I'm being pedantic in the footnotes, if your original list comprehension is being evaluated as a list of Int it wouldn't even be valid to just stop after x grows high enough that 13*x + 3 <= 1000 fails for the first time. Try this:
Prelude Data.Word> let x = 768614336404564650 :: Int
Prelude Data.Word> 13*x + 3 <= 1000
True
This happens because Int does in fact have an upper bound, so a large enough Int will overflow back to negative when you multiply it by 13. So when searching [1..] as [Int] the compiler is in fact right to keep looking past x = 77; there are almost certainly more numbers in your original list comprehension if it's [Int], they just take a long time to reach.
Again a good way to demonstrate is to use a smaller finite type, like Word16. If I use your original list comprehension as [Word16] without modifying it to avoid overflow in the conditions, you get this:
Prelude Data.Word> [13*x + 3 | x <- [1..], rem (13*x + 3) 12 == 5, mod (13*x + 3) 11 == 0, 13*x + 3 <= 1000] :: [Word16]
[341,605,209,869,473,77,737]
Even if the compiler was smart enough to know the regions of [1..] that could possibly pass 13*x + 3 <= 1000 condition, it's never going to be able to read your mind and know whether the overflow-produced numbers are solutions you intended or are the result of a bug in your code. It just does what you tell it to do.
I want to take five consecutive primes generated as an infinite list by primes and check them if they summed make another prime. I want to have something like this:
consecutivePrimes = [ a+b+c+d+e | a:b:c:d:e <- primes, prime a+b+c+d+e]
This a:b:c:d:e <- primes however doesn't work and I can't find any way as to get multiple elements at once in a list comprehension.
Since a list comprehension can be thought of as a map combined with a filter (at least for lists), you can only get one element at a time inside of it.
But you can still do this by making primes into a list of lists using tails, then taking 5 elements from each of the lists. The single element you map over (ps) in this case is a list.
import Data.List (tails)
consecutivePrimes = [ a+b+c+d+e | ps <- tails primes, let [a,b,c,d,e] = take 5 ps, prime a+b+c+d+e]
Pattern matching on the list of 5 elements will always succeed if your input list is infinite.
This is my solution which works but I find it quite ugly:
consecutivePrimes = [x | x <- consecutivePrimes' primes, prime x]
consecutivePrimes' (a:b:c:d:e:xs) | prime (a+b+c+d+e) = (a+b+c+d+e) : consecutivePrimes' (b:c:d:e:xs)
| otherwise = consecutivePrimes' (b:c:d:e:xs))
Can someone tell me the formal reason why list/arrays and such are considered more secure when it comes to incremental steps i.e (List.fold > loops).
Exampel code in F#
Functional way (list)
let rec sum lst =
match lst with
| [] -> 0
| x::xs -> x + sum xs
Imperative way (incremental)
let sum n m =
let mutable s = 0
for i=n to m do
s <- s + i
s
If by security you mean "safer" -- then I think this will explain it some. To begin with if you're summing a list, a fold should be somewhat safer as it removes the need for the programmer to correctly index the list:
let sum lst =
let mutable s = 0
for i=0 to (List.length lst - 1) do
s <- s + lst.[i]
s
You avoid a lot of pitfalls completely by using the library function:
let sum lst =
let folder acc element =
acc + element
List.fold folder 0 lst
The fold handles all the edge cases for you, in terms of indices, and list length. (note: this could also be done with a List.reduce (+) lst however that does not handle an empty list, where as a fold does).
The short of it all is that it keeps the programmer from making mistakes on silly index math, and keeps the focus on the actual logic of what is being done.
EDIT: I ironically messed up the index logic in my initial post
I have a quadratic Matrix of size n, say A, with non-negative real entries a_ij.
Furthermore I have a permutation tree. For n = 3 it looks like this: .
Now I would like to do a Depth-search (I don't know really, whether "Depth-search" is the correct description for this, but let's use it for now) along the branches of the tree in the following way:
On the first partial tree on the very left do the following starting with an "empty" Permutation (x,x,x):
If a_12 > a_21 set (1,2,x) and then check whether a_23 > a_32. If this is true as well, save (1,2,3) in a list, say P. Then go back to the first Level and check whether a_13 > a_31 and so on.
If a_21 > a_12 or a_32 > a_23 do not save the Permutation in P and go back to the first Level and check whether a_13 > a_31. If this is true set (1,3,x) and then check whether a_23 > a_32. If this is true save (1,3,2) in P and continue with the next partial tree. If a_31 > a_13 or a_32 > a_23 do not save the Permutation in P and continue with the same procedure for the next partial tree.
This procedure/algorithm I would like to implement for an arbitrary natural n > 0 with Input just the Matrix A and n and as an Output all permutations of size n that fullfill these conditions. By now I am not able to implement this in a general way.
Preferably in Python, but Pseudo Code would be nice as well. I also want to avoid functions like "itertools Permutation", because in the use case I Need to apply this for large n, for example n = 100, and then itertools Permutation is very slow.
If I understand correctly, this should get you what you want:
import numpy as np
from itertools import permutations
def fluboxing_permutations(a, n):
return [p for p in permutations(range(n))
if all(a[i, j] > a[j, i] for i, j in zip(p, p[1:]))]
n = 3
a = np.random.random([n, n])
fluboxing_permutations(a, n)
itertools.permutations will yield permutations in lexicographical order, which corresponds to your tree; then we check that for each consecutive pair of indices in the permutation, the element in the matrix is greater than the element at swapped indices. If so, we retain the permutation.
(No idea how to describe what the function does, so I made a new name. Hope you like it. If anyone knows a better way to describe it, please edit! :P )
EDIT: Here's a recursive function that should do the same, but with pruning:
def fluboxing_permutations_r(a, n):
nset = set(range(n))
def inner(p):
l = len(p)
if l > 1 and a[p[-2]][p[-1]] <= a[p[-1]][p[-2]]:
return []
if l == n:
return [p]
return [r for i in nset - set(p)
for r in inner(p + (i,))]
return inner(())
p starts as empty tuple, but it grows in recursion. Once there's at least two elements in the partial permutation, we can test the last two elements and see if it fails the test, and reject it if it does (pruning its subtree out of the search space). If it is a full permutation that wasn't rejected, we return it. If it's not full yet, we append to it all possible indices that are not already in there, and recurse.
tinyEDIT: BTW, parameter n is kind of redundant, because n = len(a) at the top of the function should take care of it.
I just started to learn Haskell today and is completely overwhelmed by its syntax.
I am trying to apply math calculation to a list of items.
For example, lets say I want to square every item in the list using list comprehension.
My attempt
myfunc (n:lis) = [ k | k <-lis, k == k^k]
result_list = myfunc[1..]
take 10 result_list
My understand of my myfunc code: take a list and loop through elements that is stored in variable k and set k equals to its square.
after i execute the take command, and hit enter, apparently the process is running but does not do anything.
Note that i want to use list comprehension as a way to do it. I can use map do achieve my goal already.
You misunderstand the list comprehension.
[ k | k <- lis, k == k^k ]
The k == k^k clause is a filter –– it only keeps elements of the list that satisfy this equation. (== is a comparison operator that returns a bool, which is one hint). The reason you see no output is that there are no numbers in [1..] that satisfy this equation. But we get an infinite loop because we keep checking ever higher numbers to see if they satisfy it.
Something to experiment with
[ k | k <- lis, k < 100 ]
As for how to get a list of squares, use a comprehension like this
[ k^2 | k <- lis ]
If you want something more like your original phrasing, you can make let bindings within a list comprehension:
[ r | k <- lis, let r = k^2 ]
There are other issues with your code, but one baby step at a time! Good luck!