Assume there is a NxN board, I want to create a seq<seq<int*int>> containing all of the lines on the board. I cannot use mutable variable.
For example, input N is 2, the output will be:
seq [seq[(0,0);(0,1)];seq[(1,0);(1,1)];seq[(0,0);
(1,0)];seq[(0,1);(1,1)];seq[(0,0);(1,1)];seq[(0,1);(1,0)]]
There are six lines including Horizontal, Vertical and Diagonal lines on the board.
let f (n:int) : seq<seq<int*int>> =
let seq1 = seq{ for x in 0 .. n-1 do yield 0,x}
...
I can create a single line. I cannot combine all lines into a seq[] by a loop.
let loop =
for j in 0 .. n-1 do
let seq1 (j:int) = seq{ for x in 0 .. n-1 do yield j,x}
I have an idea that is use for loop to create the lines. But I have no idea how to store the seq1 value and combine with a new seq1 value.
There is more than one way to do this, but I like the seq builder:
let positions =
seq {
for x in 0..10 do
for y in 0..10 do
yield (x, y)
}
Related
I have recently started learning f# and I have a problem with a task like the one in the subject line. I managed to solve this task but not using a recursive function. I have tried to convert my function to a recursive function but it does not work because in the function I create arrays which elements I then change. Please advise me how to convert my function to a recursive function or how else to perform this task.
let list = [8;4;3;3;5;9;-7]
let comp (a,b) = if a>b then a elif b = a then a else b
let maks (b: _ list) =
let x = b.Length
if x % 2 = 0 then
let tab = Array.create ((x/2)) 0
for i = 0 to (x/2)-1 do
tab.[i] <- (comp(b.Item(2*i),b.Item(2*i+1)))
let newlist = tab |> Array.toList
newlist
else
let tab = Array.create (((x-1)/2)+1) 0
tab.[(((x-1)/2))] <- b.Item(x-1)
for i = 0 to ((x-1)/2)-1 do
tab.[i] <- (comp(b.Item(2*i),b.Item(2*i+1)))
let newlist = tab |> Array.toList
newlist
It is worth noting that, if you were doing this not for learning purposes, there is a nice way of doing this using the chunkBySize function:
list
|> List.chunkBySize 2
|> List.map (fun l -> comp(l.[0], l.[l.Length-1]))
This splits the list into chunks of size at most 2. For each chunk, you can then compare the first element with the last element and that is the result you wanted.
If this is a homework question, I don't want to give away the answer, so consider this pseudocode solution instead:
If the list contains at least two elements:
Answer a new list consisting of:
The greater of the first two elements, followed by
Recursively applying the function to the rest of the list
Else the list contains less than two elements:
Answer the list unchanged
Hint: F#'s pattern matching ability makes this easy to implement.
Thanks to your guidance I managed to create the following function:
let rec maks2 (b: _ list,newlist: _ list,i:int) =
let x = b.Length
if x >= 2 then
if x % 2 = 0 then
if i < ((x/2)-1)+1 then
let d = (porownaj(b.Item(2*i),b.Item(2*i+1)))
let list2 = d::newlist
maks2(b,list2,i+1)
else
newlist
else
if i < ((x/2)-1)+1 then
let d = (porownaj(b.Item(2*i),b.Item(2*i+1)))
let list2 = d::newlist
maks2(b,list2,i+1)
else
let list3 = b.Item(x-1)::newlist
list3
else
b
The function works correctly, it takes as arguments list, empty list and index.
The only problem is that the returned list is reversed, i.e. values that should be at the end are at the beginning. How to add items to the end of the list?
You can use pattern matching to match and check/extract lists in one step.A typical recursive function, would look like:
let rec adjGreater xs =
match xs with
| [] -> []
| [x] -> [x]
| x::y::rest -> (if x >= y then x else y) :: adjGreater rest
It checks wether the list is empty, has one element, or has two elements and the remaining list in rest.
Then it builds a new list by either using x or y as the first element, and then compute the result of the remaing rest recursivly.
This is not tail-recursive. A tail-call optimized version would be, that instead of using the result of the recursive call. You would create a new list, and pass the computed valuke so far, to the recursive function. Usually this way, you want to create a inner recursive loop function.
As you only can add values to the top of a list, you then need to reverse the result of the recursive function like this:
let adjGreater xs =
let rec loop xs result =
match xs with
| [] -> result
| [x] -> x :: result
| x::y::rest -> loop rest ((if x >= y then x else y) :: result)
List.rev (loop xs [])
I am attempting to generate a series of guesses for the second Taxicab number. What I want to do is is call the Attempt function on a series of integers in a finite sequence. I have my two questions about implementation in the comments.
A taxi cab number, in case your wondering, is the least number that satisfied the sum of 2 unique cubes in for n unique sets of 2 unique cubes. Ta(2) is 1729.
[<EntryPoint>]
let main argv =
let Attempt (start : int) =
let stop = start+20
let integerList = [start..stop]
let list = List.init 3 (fun x -> integerList.[x])
//Is there a simple way to make initialize the list with random indices of integerList?
let Cube x = x*x*x
let newlist = list |> List.map (fun x -> Cube x)
let partitionList (x : List<int>) (y : int) = List.sum [x.[y];x.[y+1]]
let intLIST = [0..2]
let partitionList' = [for i in intLIST do yield partitionList newlist i]
let x = Set.ofList partitionList'
let y = Set.ofList partitionList'
//I was going to try to use some kind of equality operator to determine whether the two sets were equal, which could tell me whether we had actually found a Taxicab number by the weakened definition.
System.Console.Write(list)
System.Console.Write(newlist)
let rnd = System.Random()
//My primary question is how can I convert a random to an integer to use in start for the function Attempt?
System.Console.ReadKey() |> ignore
printfn("%A") argv
0
Dirty way to initialize list with random indexes of another list:
let randomIndexes count myList =
let rand = System.Random()
seq {
for n = 1 to count do
yield rand.Next(List.length myList) }
|> Seq.distinct
//|> Seq.sort // if you need them sorted
|> List.ofSeq
let result = randomIndexes 5 [3;2;4;5]
printfn "%A" result
Consider the following:
It's a routine that takes a list, a threshold and a width and produces an array with the elements from the list where value are below the threshold.
If a value in the list at index i is higher than the threshold, the elements in the resulting array in width w around i is blanked out with -1.
let a = [1;4;1;4;3;3;2]
let w = 1
let thresh = 3
let res = Array.copy (a |> List.toArray)
let mutable i = 0
let N = a.Length
while i < N do
if a.[i] > thresh then
let lower = if i-w < 0 then 0 else i-w
let upper = if i+w > N-1 then N-1 else i+w
for j in lower..upper do
res.[j] <- -1
i <- i + 1
The output for this example should be
[|-1; -1; -1; -1; -1; 3; 2|]
While this works, I was wondering if this sort of width index manipulation with lists/seqs/arrays can be done in a more functional way with F#?
The key to translating this to a more functional method is to think in terms of data transformations. You want to return a value based on the values in a particular range, so the first thing to do is to transform your data into a set of those ranges and then perform the operation off of that.
This solution looks a bit funny because Windowed does not operate on partial windows, so you need to pre- and post- append the threshold value (or lower).
let replaceValues lst threshold width =
seq {
for n in 1 .. width -> threshold
yield! lst
for n in 1 .. width -> threshold
}
|> Seq.windowed (width * 2 + 1)
|> Seq.map (fun x->
if x |> Seq.exists (fun x->x > threshold) then -1
else x |> Seq.skip width |> Seq.head )
I'm writing some thing really easy, a program that finds all factors of an int. Here is what I have
let factor n=
let ls=[]
for i=1 to n do
if i % n =0 then ls = i::ls
l
If I do this then it pops an error This expression was expected to have the type unit. But I tried to put an expression that prints something after if..then.., which is suppose to return type unit, but it still gives the same error. I am lost about this. Can someone help please? Thanks
You are trying to make ls into a mutable variable nd assign it with =. While this is possible, by using mutable (1) or ref (2) along with <- or := assignment operators, this is generally discouraged in the functional world.
A possibly more idiomatic implementation of the naive algorithm could be:
let factor n =
let rec factorLoop curr divs =
if curr > n then divs
else
if n % curr = 0
then factorLoop (curr+1) (curr::divs)
else factorLoop (curr+1) divs
factorLoop 1 [] |> List.rev
> factor 12;;
val it : int list = [1; 2; 3; 4; 6; 12]
Here the main function defines an inner factorLoop function that is recursive. Recursion is the way we can avoid many uses of mutable variables in functional languages. The recursive inner function threads along a curr variable that is the current divisor to be tested and a list divs of currently found divisors. The result includes 1 and n. This can be altered respectively by changing the initial value of curr and the terminating condition in the first line of factorLoop.
It is worth noting that it can all be shrunk down to one line by making use of the F# library:
let factor n =
[1..n] |> List.filter (fun x -> n % x = 0)
Here we build a list of values 1..n and feed them to List.filter which applies the given predicate (at the end of the line) to select only divisors on n. If n is large, however, the temp list will grow very large. We can use a lazily evaluated sequence instead, which won't blow the memory usage:
let factor n =
{1..n} |> Seq.filter (fun x -> n % x = 0) |> Seq.toList
Here we filter on a 'lazy' sequence and only convert the (much smaller) sequence of results to a list at the end:
> factor 10000000;;
val it : int list =
[1; 2; 4; 5; 8; 10; 16; 20; 25; 32; ... etc
= is comparison, not assignment. You want either
let factor n =
let mutable ls = []
for i = 1 to n do
if n % i = 0 then ls <- i::ls
ls
or
let factor n =
let ls = ref []
for i = 1 to n do
if n % i = 0 then ls := i::(!ls)
!ls
Note, however, that both of these solutions are highly unidiomatic, as there are equally easy immutable solutions to this problem.
I am trying to think of an elegant way of getting a random subset from a set in F#
Any thoughts on this?
Perhaps this would work: say we have a set of 2x elements and we need to pick a subset of y elements. Then if we could generate an x sized bit random number that contains exactly y 2n powers we effectively have a random mask with y holes in it. We could keep generating new random numbers until we get the first one satisfying this constraint but is there a better way?
If you don't want to convert to an array you could do something like this. This is O(n*m) where m is the size of the set.
open System
let rnd = Random(0);
let set = Array.init 10 (fun i -> i) |> Set.of_array
let randomSubSet n set =
seq {
let i = set |> Set.to_seq |> Seq.nth (rnd.Next(set.Count))
yield i
yield! set |> Set.remove i
}
|> Seq.take n
|> Set.of_seq
let result = set |> randomSubSet 3
for x in result do
printfn "%A" x
Agree with #JohannesRossel. There's an F# shuffle-an-array algorithm here you can modify suitably. Convert the Set into an array, and then loop until you've selected enough random elements for the new subset.
Not having a really good grasp of F# and what might be considered elegant there, you could just do a shuffle on the list of elements and select the first y. A Fisher-Yates shuffle even helps you in this respect as you also only need to shuffle y elements.
rnd must be out of subset function.
let rnd = new Random()
let rec subset xs =
let removeAt n xs = ( Seq.nth (n-1) xs, Seq.append (Seq.take (n-1) xs) (Seq.skip n xs) )
match xs with
| [] -> []
| _ -> let (rem, left) = removeAt (rnd.Next( List.length xs ) + 1) xs
let next = subset (List.of_seq left)
if rnd.Next(2) = 0 then rem :: next else next
Do you mean a random subset of any size?
For the case of a random subset of a specific size, there's a very elegant answer here:
Select N random elements from a List<T> in C#
Here it is in pseudocode:
RandomKSubset(list, k):
n = len(list)
needed = k
result = {}
for i = 0 to n:
if rand() < needed / (n-i)
push(list[i], result)
needed--
return result
Using Seq.fold to construct using lazy evaluation random sub-set:
let rnd = new Random()
let subset2 xs = let insertAt n xs x = Seq.concat [Seq.take n xs; seq [x]; Seq.skip n xs]
let randomInsert xs = insertAt (rnd.Next( (Seq.length xs) + 1 )) xs
xs |> Seq.fold randomInsert Seq.empty |> Seq.take (rnd.Next( Seq.length xs ) + 1)