Base case not getting picked up in my F# function - f#

This function is supposed to just return the index of a list. That part works. However when a element is not in a list it must return -1.
For some reason it does not return -1.
let rec search f list =
match list with
| head::tail ->
if f head then 0
else 1 + search f tail
| [] -> -1
printfn "%A" (search (fun x -> x = 5) [ 5; 4; 3; 2 ])
//>> return index 0 for #5
printfn "%A" (search (fun x -> x = 6) [ 5; 4; 3; 2 ])
//>> should return -1 but it returns 3 which is the len of the list not -1
EDIT: Can not use nested functions.

You could use e.g.
let search f list =
let rec where at list =
match list with
| [] -> -1
| head::tail ->
if f head then at
else where (at + 1) tail
where 0 list
which has the benefit of being tail-recursive. Regarding your comment:
let rec search f list =
match list with
| [] -> -1
| head::tail ->
if f head then 0 else
match search f tail with
| -1 -> -1
| i -> i + 1

Related

recursion with several functions F#

I need some help with my hometask: to express one function (sort) through others (smallest, delete, insert). If you know how, please, tell me, how I can do running my recursion cicle? it doing now only one step. maybe something like this: val4 -> head :: tail |> sort tail on line 25 (val4)?
let rec smallest = function
| x :: y :: tail when x <= y -> smallest (x :: tail)
| x :: y :: tail when x > y -> smallest (y :: tail)
| [x] -> Some x
| _ -> None
let rec delete (n, xs) =
match (n, xs) with
| (n, x :: xs) when n <> x -> x :: delete (n, xs)
| (n, x :: xs) when n = x -> xs
| (n, _) -> []
let rec insert (xs, n) =
match (xs, n) with
| ([x], n) when x < n -> [x]#[n]
| (x :: xs, n) when x < n -> x :: insert (xs, n)
| (x :: xs, n) when x >= n -> n :: x :: xs
| (_, _) -> []
let rec sort = function
| xs -> let val1 = smallest xs
let val2 = val1.[0]
let val3 = delete (val2, xs)
let val4 = insert (val3, val2)
val4
let res = sort [5; 4; 3; 2; 1; 1]
printfn "%A" res
This is sort of like insertion sort, but since you're always finding the smallest number in the whole list instead of the next highest number, it will recurse forever unless you skip whatever you've already found to be the smallest.
Furthermore, your insert and delete functions act not on the item index, but on equality to the value, so it won't be able to handle repeated numbers.
Keeping most of your original code the same, usually you have an inner recursive function to help you keep track of state. This is a common FP pattern.
let sort lst =
let size = lst |> List.length
let rec sort' xs = function
| index when index = size -> xs
| index ->
let val1 = smallest (xs |> List.skip index)
let val2 = val1.[0]
let val3 = delete (val2, xs)
let val4 = insert (val3, val2)
sort' val4 (index + 1)
sort' lst 0
let res = sort [5; 3; 2; 4; 1; ]
printfn "%A" res
Needless to say, this isn't correct or performant, and each iteration traverses the list multiple times. It probably runs in cubic time.
But keep learning!
I found it... I only had changed 4 & 5 lines above in the "smallest" on this: | [x] -> Some x
| _ -> None, when there was: | [x] -> [x]
| _ -> []
let rec sort = function
| xs -> match xs with
| head :: tail -> let val1 = smallest xs
match val1 with
| Some x -> let val2 = delete (x, xs)
let val3 = insert (val2, x)
let val4 = (fun list -> match list with head :: tail -> head :: sort tail | _ -> [])
val4 val3
| None -> []
| _ -> []
// let res = sort [5; 4; 3; 2; 1]
// printfn "%A" res

Multiplication in F#

I can't get my answer correctly from my multiplication function
My code is:
let List = [77; 14; 89; 93; 201]
let rec Mult =
match n with
| 24 -> 24
| _-> n * n
for i = 1 to 5 do
printfn "Multiplication: %A" (Mult i)
My question is how do I get I get it to call my List?
let List = [24; 103; 7; 13; 445]
let rec Mult = function
| head :: tail -> head * (Mult tail)
| [] -> 1
let result = Mult List
printfn "%A" result

F# pattern-matching gone wrong

I am just starting out with F# so this might be a trivial question but I am not able to understand why the pattern matching in my code acts as it does.
Quick explanation of the code:
The func calcNextMatch should recurse a list and if 2 elements are equal they should be added together.
In the end the func should return a number that is the addition of all digits that has a match with the next digit in the list.
f.ex. [1;3;2;2;5] should return 4
Code:
let rec printList l =
match l with
| head :: tail -> printf "%d " head; printList tail
| [] -> printfn ""
let rec calcNextMatch list =
printList list
match list with
| [] -> 0
| _ :: tail ->
printList tail
let h = Seq.head list
let t = Seq.tryHead tail
printfn "h: %i" h
printfn "t: %O" t
match t with
| Some h ->
printfn "TAIL t: %i is equal to HEAD h: %i" t.Value h
printfn "Calculation is: %i" (t.Value + h)
(t.Value + h) + calcNextMatch tail
| _ -> calcNextMatch tail
let sequence = [ 1;3;2;2;5 ]
let run = calcNextMatch sequence
When I run this code the problem is that the pattern-matching
does not work as I expect it.
f.ex this print output from running the script.
h: 1
t: Some(3)
TAIL t: 3 is equal to HEAD h: 3
this means that F# has matched
match t with
| Some h ->
in a case where t = Some(3) and h = 1
which translates to
match 3 with
| Some 1 ->
and that I do not understand.
The print before the matching states the value of t and h to 3 and 1 but in the pattern-matching the value of h has change to 3
How is this possible?
You can only pattern match against constant literals, otherwise the value get bounded as if was a new let-binding.
In these cases what you do normally is to add a when condition:
match t with
| Some x when x = h ->
Also notice that you can use pattern match further to simplify your code, for instance here:
| _ :: tail ->
printList tail
let h = Seq.head list
You can write:
| h :: tail ->
printList tail
Also all this portion:
| _ :: tail ->
printList tail
let h = Seq.head list
let t = Seq.tryHead tail
printfn "h: %i" h
printfn "t: %O" t
match t with
| Some h ->
printfn "TAIL t: %i is equal to HEAD h: %i" t.Value h
printfn "Calculation is: %i" (t.Value + h)
(t.Value + h) + calcNextMatch tail
becomes:
| h :: tail ->
printList tail
//printfn "h: %i" h
//printfn "t: %O" t
match tail with
| t::_ when t = h ->
printfn "TAIL t: %i is equal to HEAD h: %i" t h
printfn "Calculation is: %i" (t + h)
(t + h) + calcNextMatch tail
And you can unify all matches in one, so your whole function becomes:
let rec calcNextMatch list =
printList list
match list with
| [] -> 0
| h::x::tail when x = h -> x + h + calcNextMatch (x::tail)
| _::tail -> calcNextMatch tail
Finally, when you're done with debugging, you can remove the prints and since the last parameter of your function is the one you match against, you can use the keyword function, also use an as pattern to avoid reconstructing the list:
let rec calcNextMatch = function
| [] -> 0
| h::((x::_) as tail) when x = h -> x + h + calcNextMatch tail
| _::tail -> calcNextMatch tail

Finding the Maximum element in a list with pattern matching and recursion F#

I'm trying to find the maximum element in a list without using List.Max for a school assignment using the below given template.
let findMax l =
let rec helper(l,m) = failwith "Not implemented"
match l with
| [] -> failwith "Error -- empty list"
| (x::xs) -> helper(xs,x)
The only solution to the problem I can think of, atm is
let rec max_value1 l =
match l with
|[] -> failwith "Empty List"
|[x] -> x
|(x::y::xs) -> if x<y then max_value1 (y::xs)
else max_value1 (x::xs)
max_value1 [1; 17; 3; 6; 1; 8; 3; 11; 6; 5; 9];;
Is there any way I can go from the function I built to one that uses the template? Thanks!
Your helper function should do the work, the outer function just validates that the list is not empty and if it's not, calls the helper, which should be something like this:
let rec helper (l,m) =
match (l, m) with
| [] , m -> m
| x::xs, m -> helper (xs, max m x)
Note, that you since you're matching against the last argument of the function you can remove it and use function instead of match with:
let rec helper = function
| [] , m -> m
| x::xs, m -> helper (xs, max m x)
let findMax l =
let rec helper(l,m) =
match l with
| [] -> m
| (x::xs) -> helper(xs, if (Some x > m) then Some x else m)
helper (l,None)
Example:
[-2;-6;-1;-9;-56;-3] |> findMax
val it : int option = Some -1
An empty list will return None.
You could go for a tuple to pass both, or simply apply the helper function in your main match (instead of the empty list guard clause). I'm including the answer for someone who might find this question in the future and not have a clear answer.
let findMax l =
let rec walk maxValue = function
| [] -> maxValue
| (x::xs) -> walk (if x > maxValue then x else maxValue) xs
match l with
| [] -> failwith "Empty list"
| (head::tail) -> walk head tail
findMax [1; 12; 3; ] //12
Using fold:
let findMax l = l |> List.fold (fun maxValue x -> if x > maxValue then x else maxValue) (List.head l)
I am not sure of what the exact rules of your assigment are but the max of a list is really just List.reduce max. So
let listMax : int list -> int = List.reduce max
You need the type annotation to please the typechecker.
let inline listMax xs = List.reduce max xs
also works and is generic so it works with e.g. floats and strings as well.

Further optimizing Number to Roman Numeral function in F#

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

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