I am trying to write a tetranacci function using F# as efficiently as possible the first solution I came up with was really inefficient. can you help me come up with a better one? How would i be able to implement this in linear time?
let rec tetra n =
match n with
| 0 -> 0
| 1 -> 1
| 2 -> 1
| 3 -> 2
| _ -> tetra (n - 1) + tetra (n - 2) + tetra (n - 3) + tetra (n - 4)
You could economise by devising a function that computes the state for the next iteration on a 4-tuple. Then the sequence generator function Seq.unfold can be used to build a sequence that contains the first element of each state quadruple, an operation that is 'lazy` -- the elements of the sequence are only computed on demand as they are consumed.
let tetranacci (a3, a2, a1, a0) = a2, a1, a0, a3 + a2 + a1 + a0
(0, 1, 1, 2)
|> Seq.unfold (fun (a3, _, _, _ as a30) -> Some(a3, tetranacci a30))
|> Seq.take 10
|> Seq.toList
// val it : int list = [0; 1; 1; 2; 4; 8; 15; 29; 56; 108]
Note that the standard Tetranacci sequence (OEIS A000078) would usually be generated with the start state of (0, 0, 0, 1):
// val it : int list = [0; 0; 0; 1; 1; 2; 4; 8; 15; 29]
kaefer's answer is good, but why stop at linear time? It turns out that you can actually achieve logarithmic time instead, by noting that the recurrence can be expressed as a matrix multiplication:
[T_n+1] [0; 1; 0; 0][T_n]
[T_n+2] = [0; 0; 1; 0][T_n+1]
[T_n+3] [0; 0; 0; 1][T_n+2]
[T_n+4] [1; 1; 1; 1][T_n+3]
But then T_n can be achieved by applying the recurrence n times, which we can see as the first entry of M^n*[T_0; T_1; T_2; T_3] (which is just the upper right entry of M^n), and we can perform the matrix multiplication in O(log n) time by repeated squaring:
type Mat =
| Mat of bigint[][]
static member (*)(Mat arr1, Mat arr2) =
Array.init arr1.Length (fun i -> Array.init arr2.[0].Length (fun j -> Array.sum [| for k in 0 .. arr2.Length - 1 -> arr1.[i].[k]*arr2.[k].[j] |]))
|> Mat
static member Pow(m, n) =
match n with
| 0 ->
let (Mat arr) = m
Array.init arr.Length (fun i -> Array.init arr.Length (fun j -> if i = j then 1I else 0I))
|> Mat
| 1 -> m
| _ ->
let m2 = m ** (n/2)
if n % 2 = 0 then m2 * m2
else m2 * m2 * m
let tetr =
let m = Mat [| [|0I; 1I; 0I; 0I|]
[|0I; 0I; 1I; 0I|]
[|0I; 0I; 0I; 1I|]
[|1I; 1I; 1I; 1I|]|]
fun n ->
let (Mat m') = m ** n
m'.[0].[3]
for i in 0 .. 50 do
printfn "%A" (tetr i)
Here is a tail recursive version, which compiles to mostly loops (and its complexity should be O(n)):
let tetr n =
let rec t acc4 acc3 acc2 acc1 = function
| n when n = 0 -> acc4
| n when n = 1 -> acc3
| n when n = 2 -> acc2
| n when n = 3 -> acc1
| n -> t acc3 acc2 acc1 (acc1 + acc2 + acc3 + acc4) (n - 1)
t 0 1 1 2 n
acc1 corresponds to tetra (n - 1),
acc2 corresponds to tetra (n - 2),
acc3 corresponds to tetra (n - 3),
acc4 corresponds to tetra (n - 4)
Based on the Fibonacci example
Related
I'm trying to write Taylor series in F#.
Have a look at my code
let rec iter a b f i =
if a > b then i;
else f a (iter (a+1) b f i)
let sum a b = iter a b (+) 0 // from 0
// e^x = 1 + x + (x^2)/2 + ... (x^n)/n! + ...
let fact n = iter 1 n (*) 1 // factorial
let pow x n = iter 1 n (fun n acc -> acc * x) 1
let exp x =
iter 0 x
(fun n acc ->
acc + (pow x n) / float (fact n)) 0
In the last row I am trying cast int fact n to float, but seems like I'm wrong because this code isn't compileable :(
Am I doing the right algorithm?
Can I call my code functional-first?
The code doesn't compile, because:
You're trying to divide an integer pow x n by a float. Division has to have operands of the same type.
You're specifying the terminal case of the wrong type. Literal 0 is integer. If you want float zero, use 0.0 or abbreviated 0.
Try this:
let exp x =
iter 0 x
(fun n acc ->
acc + float (pow x n) / float (fact n)) 0.
P.S. In the future, please provide the exact error messages and/or unexpected results that you're getting. Simply saying "doesn't work" is not a good description of a problem.
I'm new to F# and I'm curious if this can still be optimized further. I am not particularly sure if I've done this correctly as well. I'm curious particularly on the last line as it looks really long and hideous.
I've searched over google, but only Roman Numeral to Number solutions only show up, so I'm having a hard time comparing.
type RomanDigit = I | IV | V | IX
let rec romanNumeral number =
let values = [ 9; 5; 4; 1 ]
let capture number values =
values
|> Seq.find ( fun x -> number >= x )
let toRomanDigit x =
match x with
| 9 -> IX
| 5 -> V
| 4 -> IV
| 1 -> I
match number with
| 0 -> []
| int -> Seq.toList ( Seq.concat [ [ toRomanDigit ( capture number values ) ]; romanNumeral ( number - ( capture number values ) ) ] )
Thanks for anyone who can help with this problem.
A slightly shorter way of recursively finding the largest digit representation that can be subtracted from the value (using List.find):
let units =
[1000, "M"
900, "CM"
500, "D"
400, "CD"
100, "C"
90, "XC"
50, "L"
40, "XL"
10, "X"
9, "IX"
5, "V"
4, "IV"
1, "I"]
let rec toRomanNumeral = function
| 0 -> ""
| n ->
let x, s = units |> List.find (fun (x,s) -> x <= n)
s + toRomanNumeral (n-x)
If I had to use a Discriminated Union to represent the roman letters I would not include IV and IX.
type RomanDigit = I|V|X
let numberToRoman n =
let (r, diff) =
if n > 8 then [X], n - 10
elif n > 3 then [V], n - 5
else [], n
if diff < 0 then I::r
else r # (List.replicate diff I)
Then, based in this solution you can go further and extend it to all numbers.
Here's my first attempt, using fold and partial application:
type RomanDigit = I|V|X|L|C|D|M
let numberToRoman n i v x =
let (r, diff) =
if n > 8 then [x], n - 10
elif n > 3 then [v], n - 5
else [], n
if diff < 0 then i::r
else r # (List.replicate diff i)
let allDigits (n:int) =
let (_, f) =
[(I,V); (X,L); (C,D)]
|> List.fold (fun (n, f) (i, v) ->
(n / 10, fun x -> (numberToRoman (n % 10) i v x) # f i)) (n, (fun _ -> []))
f M
Here's a tail-recursive version of #Philip Trelford's answer:
let toRomanNumeral n =
let rec iter acc n =
match n with
| 0 -> acc
| n ->
let x, s = units |> List.find (fun (x, _) -> x <= n)
iter (acc + s) (n-x)
iter "" n
I need to implement a simple dynamic programming algorithm in 2D in F#. For simple 1D cases Seq.unfold seems to be the way to go, see e.g. https://stackoverflow.com/a/7986083/5363
Is there a nice (and efficient) way to achieve a similar result in 2D, e.g. rewrite the following pseudo-code in functional style:
let alpha =
let result = Array2D.zeroCreate N T
for i in 0 .. N-1 do
result.[0, i] <- (initialPi i) * (b i observations.[0])
for t in 1 .. T-1 do
for i in 0 .. N-1 do
let s = row t-1 result |> Seq.mapi (fun j alpha_t_j -> alpha_t_j * initialA.[i, j]) () |> Seq.sum
result.[t, i] <- s * (b i observations.[t])
result
assume that all the missing functions and arrays are defined above.
EDIT: Actually read code, this is at least functional, does have a slightly different return type, although you could avoid that with a conversion
let alpha =
let rec build prev idx max =
match idx with
|0 ->
let r = (Array.init N (fun i -> (initialPi y) * (b i observations.[0]))
r:: (build r 1 max)
|t when t=max -> []
|_ ->
let s = prev |> Seq.mapi (fun j alpha_t_j -> alpha_t_j * initialA.[i, j]) () |> Seq.sum
let r = Array.init N (fun i -> s * (b i observations.[t]))
r:: build r (idx+1 max)
build [] 0 T |> List.toArray
My code (below) falls over with a stack overflow exception. Im assuming F# isnt like haskell and dosent play well with recursive lists. Whats the correct way of dealing with recursive lists like this in F# ? Should i pass it an int so it has a determined size?
let rec collatz num =
match num with
|x when x % 2 = 0 ->num :: collatz (x/2)
|x -> num :: collatz ((x * 3) + 1)
let smallList = collatz(4) |> Seq.take(4)
For an infinite list like this, you want to return a sequence. Sequences are lazy; lists are not.
let rec collatz num =
seq {
yield num
match num with
| x when x % 2 = 0 -> yield! collatz (x/2)
| x -> yield! collatz ((x * 3) + 1)
}
let smallList =
collatz 4
|> Seq.take 4
|> Seq.toList //[4; 2; 1; 4]
let collatz num =
let next x = if x % 2 = 0 then x / 2 else x * 3 + 1
(num, next num)
|>Seq.unfold (fun (n, x) -> Some (n, (x, next x)))
In F#, imagine we have an array of bytes representing pixel data with three bytes per pixel in RGB order:
[| 255; 0; 0; //Solid red
0; 255; 0; //Solid green
0; 0; 255; //Solid blue
1; 72; 9;
34; 15; 155
... |]
I'm having a hard time knowing how to functionally operate on this data as-is, since a single item is really a consecutive block of three elements in the array.
So, I need to first group the triples in the array into something like this:
[|
[| 255; 0; 0 |];
[| 0; 255; 0 |];
[| 0; 0; 255 |];
[| 1; 72; 9 |];
[| 34; 15; 155 |]
... |]
Now, gathering up the triples into sub-arrays is easy enough to do with a for loop, but I'm curious--is there a functional way to gather up groups of array elements in F#? My ultimate goal is not simply to convert the data as illustrated above, but to solve the problem in a more declarative and functional manner. But I have yet to find an example of how to do this without an imperative loop.
kvb's answer may not give you what you want. Seq.windowed returns a sliding window of values, e.g., [1; 2; 3; 4] becomes [[1; 2; 3]; [2; 3; 4]]. It seems like you want it split into contiguous chunks. The following function takes a list and returns a list of triples ('T list -> ('T * 'T * 'T) list).
let toTriples list =
let rec aux f = function
| a :: b :: c :: rest -> aux (fun acc -> f ((a, b, c) :: acc)) rest
| _ -> f []
aux id list
Here's the inverse:
let ofTriples triples =
let rec aux f = function
| (a, b, c) :: rest -> aux (fun acc -> f (a :: b :: c :: acc)) rest
| [] -> f []
aux id triples
EDIT
If you're dealing with huge amounts of data, here's a sequence-based approach with constant memory use (all the options and tuples it creates have a negative impact on GC--see below for a better version):
let (|Next|_|) (e:IEnumerator<_>) =
if e.MoveNext() then Some e.Current
else None
let (|Triple|_|) = function
| Next a & Next b & Next c -> Some (a, b, c) //change to [|a;b;c|] if you like
| _ -> None
let toSeqTriples (items:seq<_>) =
use e = items.GetEnumerator()
let rec loop() =
seq {
match e with
| Triple (a, b, c) ->
yield a, b, c
yield! loop()
| _ -> ()
}
loop()
EDIT 2
ebb's question about memory use prompted me to test and I found toSeqTriples to be slow and cause surprisingly frequent GCs. The following version fixes those issues and is almost 4x faster than the list-based version.
let toSeqTriplesFast (items:seq<_>) =
use e = items.GetEnumerator()
let rec loop() =
seq {
if e.MoveNext() then
let a = e.Current
if e.MoveNext() then
let b = e.Current
if e.MoveNext() then
let c = e.Current
yield (a, b, c)
yield! loop()
}
loop()
This has relatively constant memory usage vs a list or array-based approach because a) if you have a seq to start with the entire sequence doesn't have to be slurped into a list/array; and b) it also returns a sequence, making it lazy, and avoiding allocating yet another list/array.
I need to first group the triples in the array into something like this:
If you know they will always be triples then representing then as a tuple int * int * int is more "typeful" than using an array because it conveys the fact that there are only ever exactly three elements.
Other people have described various ways to massage the data but I would actually recommend not bothering (unless there is more to this than you have described). I would opt for a function to destructure your array as-is instead:
let get i = a.[3*i], a.[3*i+1], a.[3*i+2]
If you really want to change the representation then you can now do:
let b = Array.init (a.Length/3) get
The answer really depends upon what you want to do next though...
(Hat tip: Scott Wlaschin) As of F# 4.0, you can use Array.chunkBySize(). It does exactly what you want:
let bs = [| 255; 0; 0; //Solid red
0; 255; 0; //Solid green
0; 0; 255; //Solid blue
1; 72; 9;
34; 15; 155 |]
let grouped = bs |> Array.chunkBySize 3
// [| [|255; 0; 0|]
// [| 0; 255; 0|]
// [| 0; 0; 255|]
// [| 1; 72; 9|]
// [| 34; 15; 155|] |]
The List and Seq modules also have chunkBySize() in F# 4.0. As of this writing, the docs at MSDN don't show chunkBySize() anywhere, but it's there if you're using F# 4.0.
UPDATE: As pointed out by Daniel, this answer is wrong because it creates a sliding window.
You can use the Seq.windowed function from the library. E.g.
let rgbPix = rawValues |> Seq.windowed 3
This returns a sequence rather than an array, so if you need random access, you could follow that with a call to Seq.toArray.
Another approach, that takes and yields arrays directly:
let splitArrays n arr =
match Array.length arr with
| 0 ->
invalidArg "arr" "array is empty"
| x when x % n <> 0 ->
invalidArg "arr" "array length is not evenly divisible by n"
| arrLen ->
let ret = arrLen / n |> Array.zeroCreate
let rec loop idx =
ret.[idx] <- Array.sub arr (idx * n) n
match idx + 1 with
| idx' when idx' <> ret.Length -> loop idx'
| _ -> ret
loop 0
Or, yet another:
let splitArray n arr =
match Array.length arr with
| 0 ->
invalidArg "arr" "array is empty"
| x when x % n <> 0 ->
invalidArg "arr" "array length is not evenly divisible by n"
| arrLen ->
let rec loop idx = seq {
yield Array.sub arr idx n
let idx' = idx + n
if idx' <> arrLen then
yield! loop idx' }
loop 0 |> Seq.toArray