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
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 thought these two function were the same, but it seems that I was wrong.
I define two function f and g in this way:
let rec f n k =
match k with
|_ when (k < 0) || (k > n) -> 0
|_ when k = n -> 100
|_ -> (f n (k+1)) + 1
let rec g n k =
match k with
|_ when (k < 0) || (k > n) -> 0
| n -> 100
|_ -> (g n (k+1)) + 1
let x = f 10 5
let y = g 10 5
The results are:
val x : int = 105
val y : int = 100
Could anyone tell me what's the difference between these two functions?
EDIT
Why does it work here?
let f x =
match x with
| 1 -> 100
| 2 -> 200
|_ -> -1
List.map f [-1..3]
and we get
val f : x:int -> int
val it : int list = [-1; -1; 100; 200; -1]
The difference is that
match k with
...
when k = n -> 100
is a case that matches when some particular condition is true (k = n). The n used in the condition refers to the n that is bound as the function parameter. On the other hand
match k with
...
n -> 100
is a case that only needs to match k against a pattern variable n, which can always succeed. The n in the pattern isn't the same n as the n passed into the function.
For comparison, try the code
let rec g n k =
match k with
|_ when (k < 0) || (k > n) -> 0
| n -> n
|_ -> (g n (k+1)) + 1
and you should see that when you get to the second case, the value returned is the value of the pattern variable n, which has been bound to the value of k.
This behavior is described in the Variable Patterns section of the MSDN F# Language Reference, Pattern Matching:
Variable Patterns
The variable pattern assigns the value being matched to a variable
name, which is then available for use in the execution expression to
the right of the -> symbol. A variable pattern alone matches any
input, but variable patterns often appear within other patterns,
therefore enabling more complex structures such as tuples and arrays
to be decomposed into variables. The following example demonstrates a
variable pattern within a tuple pattern.
let function1 x =
match x with
| (var1, var2) when var1 > var2 -> printfn "%d is greater than %d" var1 var2
| (var1, var2) when var1 < var2 -> printfn "%d is less than %d" var1 var2
| (var1, var2) -> printfn "%d equals %d" var1 var2
function1 (1,2)
function1 (2, 1)
function1 (0, 0)
The use of when is described in more depth in Match Expressions.
The first function is ok, it calls recursively itself n-k times and returns 100 when matches with the conditional where k = n. So, it returns all the calls adding 1 n-k times. with your example, with n=10 and k=5 it is ok the result had been 105.
The problem is the second function. I tested here. See I changed the pattern n->100 to z->100 and it still matches there and never calls itself recursively. So, it always returns 100 if it does not fail in the first conditional. I think F# don't allow that kind of match so it is better to put a conditional to get what you want.
let rec f n =
match n with
| 0 | 1 | 2 -> 1
| _ -> f (n - 2) + f (n - 3)
Without CPS or Memoization, how could it be made tail recursive?
let f n = Seq.unfold (fun (x, y, z) -> Some(x, (y, z, x + y))) (1I, 1I, 1I)
|> Seq.nth n
Or even nicer:
let lambda (x, y, z) = x, (y, z, x + y)
let combinator = Seq.unfold (lambda >> Some) (1I, 1I, 1I)
let f n = combinator |> Seq.nth n
To get what's going on here, refer this snippet. It defines Fibonacci algorithm, and yours is very similar.
UPD There are three components here:
The lambda which gets i-th element;
The combinator which runs recursion over i; and
The wrapper that initiates the whole run and then picks up the value (from a triple, like in #Tomas' code).
You have asked for a tail-recursive code, and there are actually two ways for that: make your own combinator, like #Tomas did, or utilize the existing one, Seq.unfold, which is certainly tail-recursive. I preferred the second approach as I can split the entire code into a group of let statements.
The solution by #bytebuster is nice, but he does not explain how he created it, so it will only help if you're solving this specific problem. By the way, your formula looks a bit like Fibonacci (but not quite) which can be calculated analytically without any looping (even without looping hidden in Seq.unfold).
You started with the following function:
let rec f0 n =
match n with
| 0 | 1 | 2 -> 1
| _ -> f0 (n - 2) + f0 (n - 3)
The function calls f0 for arguments n - 2 and n - 3, so we need to know these values. The trick is to use dynamic programming (which can be done using memoization), but since you did not want to use memoization, we can write that by hand.
We can write f1 n which returns a three-element tuple with the current and two past values values of f0. This means f1 n = (f0 (n - 2), f0 (n - 1), f0 n):
let rec f1 n =
match n with
| 0 -> (0, 0, 1)
| 1 -> (0, 1, 1)
| 2 -> (1, 1, 1)
| _ ->
// Here we call `f1 (n - 1)` so we get values
// f0 (n - 3), f0 (n - 2), f0 (n - 1)
let fm3, fm2, fm1 = (f1 (n - 1))
(fm2, fm1, fm2 + fm3)
This function is not tail recurisve, but it only calls itself recursively once, which means that we can use the accumulator parameter pattern:
let f2 n =
let rec loop (fm3, fm2, fm1) n =
match n with
| 2 -> (fm3, fm2, fm1)
| _ -> loop (fm2, fm1, fm2 + fm3) (n - 1)
match n with
| 0 -> (0, 0, 1)
| 1 -> (0, 1, 1)
| n -> loop (1, 1, 1) n
We need to handle arguments 0 and 1 specially in the body of fc. For any other input, we start with initial three values (that is (f0 0, f0 1, f0 2) = (1, 1, 1)) and then loop n-times performing the given recursive step until we reach 2. The recursive loop function is what #bytebuster's solution implements using Seq.unfold.
So, there is a tail-recursive version of your function, but only because we could simply keep the past three values in a tuple. In general, this might not be possible if the code that calculates which previous values you need does something more complicated.
Better even than a tail recursive approach, you can take advantage of matrix multiplication to reduce any recurrence like that to a solution that uses O(log n) operations. I leave the proof of correctness as an exercise for the reader.
module NumericLiteralG =
let inline FromZero() = LanguagePrimitives.GenericZero
let inline FromOne() = LanguagePrimitives.GenericOne
// these operators keep the inferred types from getting out of hand
let inline ( + ) (x:^a) (y:^a) : ^a = x + y
let inline ( * ) (x:^a) (y:^a) : ^a = x * y
let inline dot (a,b,c) (d,e,f) = a*d+b*e+c*f
let trans ((a,b,c),(d,e,f),(g,h,i)) = (a,d,g),(b,e,h),(c,f,i)
let map f (x,y,z) = f x, f y, f z
type 'a triple = 'a * 'a * 'a
// 3x3 matrix type
type 'a Mat3 = Mat3 of 'a triple triple with
static member inline ( * )(Mat3 M, Mat3 N) =
let N' = trans N
map (fun x -> map (dot x) N') M
|> Mat3
static member inline get_One() = Mat3((1G,0G,0G),(0G,1G,0G),(0G,0G,1G))
static member (/)(Mat3 M, Mat3 N) = failwith "Needed for pown, but not supported"
let inline f n =
// use pown to get O(log n) time
let (Mat3((a,b,c),(_,_,_),(_,_,_))) = pown (Mat3 ((0G,1G,0G),(0G,0G,1G),(1G,1G,0G))) n
a + b + c
// this will take a while...
let bigResult : bigint = f 1000000