I'm writing a mergesort function in F#, but I am receiving this error code and I don't understand why.
"error FS0030: Value restriction. The value 'it' has been inferred to have generic type
val it : '_a list when '_a : comparison
Either define 'it' as a simple data term, make it a function with explicit arguments or, if you do not intend for it to be generic, add a type annotation."
I get the error code when I try calling, for example, mergesort [1; 2; 3; 3; 2; 6];;
Here is the code snippet
let rec merge l =
match l with
| ([], ys) -> ys
| (xs, []) -> xs
| (x::xs, y::ys) -> if x < y then x :: merge (xs, y::ys)
else y :: merge (x::xs, ys)
let rec split l =
match l with
| [] -> ([], [])
| [a] -> ([a], [])
| a::b::cs -> let (M,N) = split cs
(a::M, b::N)
let rec mergesort l =
match l with
| [] -> []
| L -> let (M, N) = split L
merge (mergesort M, mergesort N)
Tried the code in VS2015. The error reproduces in FSI but if I run it as a program the program crashes with StackOverflowException.
Turns out there's a slight mistake in mergesort
let rec mergesort l =
match l with
| []
| [_] -> l // Without this case mergesort crashes with `StackOverflowException`
| L -> let (M, N) = split L
merge (mergesort M, mergesort N)
Fixing this error fixed it for me in FSI. I have noticed that occassionally FSI shows the wrong error message. Seems to be one of those cases.
Here is quite a typical make a century problem.
We have a natural number list [1;2;3;4;5;6;7;8;9].
We have a list of possible operators [Some '+'; Some '*';None].
Now we create a list of operators from above possibilities and insert each operator into between each consecutive numbers in the number list and compute the value.
(Note a None b = a * 10 + b)
For example, if the operator list is [Some '+'; Some '*'; None; Some '+'; Some '+'; Some '+'; Some '+'; Some '+'], then the value is 1 + 2 * 34 + 5 + 6 + 7 + 8 + 9 = 104.
Please find all possible operator lists, so the value = 10.
The only way I can think of is brute-force.
I generate all possible operator lists.
Compute all possible values.
Then filter so I get all operator lists which produce 100.
exception Cannot_compute
let rec candidates n ops =
if n = 0 then [[]]
else
List.fold_left (fun acc op -> List.rev_append acc (List.map (fun x -> op::x) (candidates (n-1) ops))) [] ops
let glue l opl =
let rec aggr acc_l acc_opl = function
| hd::[], [] -> (List.rev (hd::acc_l), List.rev acc_opl)
| hd1::hd2::tl, None::optl -> aggr acc_l acc_opl (((hd1*10+hd2)::tl), optl)
| hd::tl, (Some c)::optl -> aggr (hd::acc_l) ((Some c)::acc_opl) (tl, optl)
| _ -> raise Cannot_glue
in
aggr [] [] (l, opl)
let compute l opl =
let new_l, new_opl = glue l opl in
let rec comp = function
| hd::[], [] -> hd
| hd::tl, (Some '+')::optl -> hd + (comp (tl, optl))
| hd1::hd2::tl, (Some '-')::optl -> hd1 + (comp ((-hd2)::tl, optl))
| hd1::hd2::tl, (Some '*')::optl -> comp (((hd1*hd2)::tl), optl)
| hd1::hd2::tl, (Some '/')::optl -> comp (((hd1/hd2)::tl), optl)
| _, _ -> raise Cannot_compute
in
comp (new_l, new_opl)
let make_century l ops =
List.filter (fun x -> fst x = 100) (
List.fold_left (fun acc x -> ((compute l x), x)::acc) [] (candidates ((List.length l)-1) ops))
let rec print_solution l opl =
match l, opl with
| hd::[], [] -> Printf.printf "%d\n" hd
| hd::tl, (Some op)::optl -> Printf.printf "%d %c " hd op; print_solution tl optl
| hd1::hd2::tl, None::optl -> print_solution ((hd1*10+hd2)::tl) optl
| _, _ -> ()
I believe my code is ugly. So I have the following questions
computer l opl is to compute using the number list and operator list. Basically it is a typical math evaluation. Is there any nicer implementation?
I have read Chapter 6 in Pearls of Functional Algorithm Design. It used some techniques to improve the performance. I found it really really obscurity and hard to understand. Anyone who read it can help?
Edit
I refined my code. Basically, I will scan the operator list first to glue all numbers where their operator is None.
Then in compute, if I meet a '-' I will simply negate the 2nd number.
A classic dynamic programming solution (which finds the = 104
solution instantly) that does not risk any problem with operators
associativity or precedence. It only returns a boolean saying whether
it's possible to come with the number; modifying it to return the
sequences of operations to get the solution is an easy but interesting
exercise, I was not motivated to go that far.
let operators = [ (+); ( * ); ]
module ISet = Set.Make(struct type t = int let compare = compare end)
let iter2 res1 res2 f =
res1 |> ISet.iter ## fun n1 ->
res2 |> ISet.iter ## fun n2 ->
f n1 n2
let can_make input target =
let has_zero = Array.fold_left (fun acc n -> acc || (n=0)) false input in
let results = Array.make_matrix (Array.length input) (Array.length input) ISet.empty in
for imax = 0 to Array.length input - 1 do
for imin = imax downto 0 do
let add n =
(* OPTIMIZATION: if the operators are known to be monotonous, we need not store
numbers above the target;
(Handling multiplication by 0 requires to be a bit more
careful, and I'm not in the mood to think hard about this
(I think one need to store the existence of a solution,
even if it is above the target), so I'll just disable the
optimization in that case)
*)
if n <= target && not has_zero then
results.(imin).(imax) <- ISet.add n results.(imin).(imax) in
let concat_numbers =
(* concatenates all number from i to j:
i=0, j=2 -> (input.(0)*10 + input.(1))*10 + input.(2)
*)
let rec concat acc k =
let acc = acc + input.(k) in
if k = imax then acc
else concat (10 * acc) (k + 1)
in concat 0 imin
in add concat_numbers;
for k = imin to imax - 1 do
let res1 = results.(imin).(k) in
let res2 = results.(k+1).(imax) in
operators |> List.iter (fun op ->
iter2 res1 res2 (fun n1 n2 -> add (op n1 n2););
);
done;
done;
done;
let result = results.(0).(Array.length input - 1) in
ISet.mem target result
Here is my solution, which evaluates according to the usual rules of precedence. It finds 303 solutions to find [1;2;3;4;5;6;7;8;9] 100 in under 1/10 second on my MacBook Pro.
Here are two interesting ones:
# 123 - 45 - 67 + 89;;
- : int = 100
# 1 * 2 * 3 - 4 * 5 + 6 * 7 + 8 * 9;;
- : int = 100
This is a brute force solution. The only slightly clever thing is that I treat concatenation of digits as simply another (high precedence) operation.
The eval function is the standard stack-based infix expression evaluation that you will find described many places. Here is an SO article about it: How to evaluate an infix expression in just one scan using stacks? The essence is to postpone evaulating by pushing operators and operands onto stacks. When you find that the next operator has lower precedence you can go back and evaluate what you pushed.
type op = Plus | Minus | Times | Divide | Concat
let prec = function
| Plus | Minus -> 0
| Times | Divide -> 1
| Concat -> 2
let succ = function
| Plus -> Minus
| Minus -> Times
| Times -> Divide
| Divide -> Concat
| Concat -> Plus
let apply op stack =
match op, stack with
| _, [] | _, [_] -> [] (* Invalid input *)
| Plus, a :: b :: tl -> (b + a) :: tl
| Minus, a :: b :: tl -> (b - a) :: tl
| Times, a :: b :: tl -> (b * a) :: tl
| Divide, a :: b :: tl -> (b / a) :: tl
| Concat, a :: b :: tl -> (b * 10 + a) :: tl
let rec eval opstack numstack ops nums =
match opstack, numstack, ops, nums with
| [], sn :: _, [], _ -> sn
| sop :: soptl, _, [], _ ->
eval soptl (apply sop numstack) ops nums
| [], _, op :: optl, n :: ntl ->
eval [op] (n :: numstack) optl ntl
| sop :: soptl, _, op :: _, _ when prec sop >= prec op ->
eval soptl (apply sop numstack) ops nums
| _, _, op :: optl, n :: ntl ->
eval (op :: opstack) (n :: numstack) optl ntl
| _ -> 0 (* Invalid input *)
let rec incr = function
| [] -> []
| Concat :: rest -> Plus :: incr rest
| x :: rest -> succ x :: rest
let find nums tot =
match nums with
| [] -> []
| numhd :: numtl ->
let rec try1 ops accum =
let accum' =
if eval [] [numhd] ops numtl = tot then
ops :: accum
else
accum
in
if List.for_all ((=) Concat) ops then
accum'
else try1 (incr ops) accum'
in
try1 (List.map (fun _ -> Plus) numtl) []
I came up with a slightly obscure implementation (for a variant of this problem) that is a bit better than brute force. It works in place, rather than generating intermediate data structures, keeping track of the combined values of the operators that have already been evaluated.
The trick is to keep track of a pending operator and value so that you can evaluate the "none" operator easily. That is, if the algorithm had just progressed though 1 + 23, the pending operator would be +, and the pending value would be 23, allowing you to easily generate either 1 + 23 + 4 or 1 + 234 as necessary.
type op = Add | Sub | Nothing
let print_ops ops =
let len = Array.length ops in
print_char '1';
for i = 1 to len - 1 do
Printf.printf "%s%d" (match ops.(i) with
| Add -> " + "
| Sub -> " - "
| Nothing -> "") (i + 1)
done;
print_newline ()
let solve k target =
let ops = Array.create k Nothing in
let rec recur i sum pending_op pending_value =
let sum' = match pending_op with
| Add -> sum + pending_value
| Sub -> if sum = 0 then pending_value else sum - pending_value
| Nothing -> pending_value in
if i = k then
if sum' = target then print_ops ops else ()
else
let digit = i + 1 in
ops.(i) <- Add;
recur (i + 1) sum' Add digit;
ops.(i) <- Sub;
recur (i + 1) sum' Sub digit;
ops.(i) <- Nothing;
recur (i + 1) sum pending_op (pending_value * 10 + digit) in
recur 0 0 Nothing 0
Note that this will generate duplicates - I didn't bother to fix that. Also, if you are doing this exercise to gain strength in functional programming, it might be beneficial to reject the imperative approach taken here and search for a similar solution that doesn't make use of assignments.
I need to make a function that takes a list and an element and returns a list in which the first occurrence of the element is removed: something like
removeFst [1,5,2,3,5,3,4,5,6] 5
[1,2,3,5,3,4,5,6]
What I tried is:
main :: IO()
main = do
putStr ( show $ removeFst [1,5,2,3,5,3,4,5,6] 5)
removeFst :: [Int] -> Int -> [Int]
removeFst [] m = []
removeFst [x] m
| x == m = []
| otherwise = [x]
removeFst (x:xs) m
| x == m = xs
| otherwise = removeFst xs m
But this doesn't work... it returns the list without the first elements. I think I should make the recursive call to make the list something like:
removeFst (x:xs) m
| x == m = xs
| otherwise = removeFst (-- return the whole list till element x) m
You are very close, what you miss is prepending the elements before the first found m to the result list,
removeFst :: [Int] -> Int -> [Int]
removeFst [] m = []
removeFst (x:xs) m
| x == m = xs
| otherwise = x : removeFst xs m
-- ^^^ keep x /= m
Note that the special case for one-element lists is superfluous.
Also note that removeFst = flip delete with delete from Data.List.
It should be mentioned that your function is equivalent to Data.List.delete.
Here another version:
import Data.List
removeFst xs x = front ++ drop 1 back where
(front, back) = break (==x) xs
I have written a sudoku solver in Haskell. It goes through a list and when it finds '0' (an empty cell) it will get the numbers that could fit and try them:
import Data.List (group, (\\), sort)
import Data.Maybe (fromMaybe)
row :: Int -> [Int] -> [Int]
row y grid = foldl (\acc x -> (grid !! x):acc) [] [y*9 .. y*9+8]
where y' = y*9
column :: Int -> [Int] -> [Int]
column x grid = foldl (\acc n -> (grid !! n):acc) [] [x,x+9..80]
box :: Int -> Int -> [Int] -> [Int]
box x y grid = foldl (\acc n -> (grid !! n):acc) [] [x+y*9*3+y' | y' <- [0,9,18], x <- [x'..x'+2]]
where x' = x*3
isValid :: [Int] -> Bool
isValid grid = and [isValidRow, isValidCol, isValidBox]
where isValidRow = isValidDiv row
isValidCol = isValidDiv column
isValidBox = and $ foldl (\acc (x,y) -> isValidList (box x y grid):acc) [] [(x,y) | x <- [0..2], y <- [0..2]]
isValidDiv f = and $ foldl (\acc x -> isValidList (f x grid):acc) [] [0..8]
isValidList = all (\x -> length x <= 1) . tail . group . sort -- tail removes entries that are '0'
isComplete :: [Int] -> Bool
isComplete grid = length (filter (== 0) grid) == 0
solve :: Maybe [Int] -> Maybe [Int]
solve grid' = foldl f Nothing [0..80]
where grid = fromMaybe [] grid'
f acc x
| isValid grid = if isComplete grid then grid' else f' acc x
| otherwise = acc
f' acc x
| (grid !! x) == 0 = case guess x grid of
Nothing -> acc
Just x -> Just x
| otherwise = acc
guess :: Int -> [Int] -> Maybe [Int]
guess x grid
| length valid /= 0 = foldl f Nothing valid
| otherwise = Nothing
where valid = [1..9] \\ (row rowN grid ++ column colN grid ++ box (fst boxN) (snd boxN) grid) -- remove numbers already used in row/collumn/box
rowN = x `div` 9 -- e.g. 0/9=0 75/9=8
colN = x - (rowN * 9) -- e.g. 0-0=0 75-72=3
boxN = (colN `div` 3, rowN `div` 3)
before x = take x grid
after x = drop (x+1) grid
f acc y = case solve $ Just $ before x ++ [y] ++ after x of
Nothing -> acc
Just x -> Just x
For some puzzles this works, for example this one:
sudoku :: [Int]
sudoku = [5,3,0,6,7,8,0,1,2,
6,7,0,0,0,0,3,4,8,
0,0,8,0,0,0,5,0,7,
8,0,0,0,0,1,0,0,3,
4,2,6,0,0,3,7,9,0,
7,0,0,9,0,0,0,5,0,
9,0,0,5,0,7,0,0,0,
2,8,7,4,1,9,6,0,5,
3,0,0,2,8,0,1,0,0]
Took under a second, however this one:
sudoku :: [Int]
sudoku = [5,3,0,0,7,0,0,1,2,
6,7,0,0,0,0,3,4,8,
0,0,0,0,0,0,5,0,7,
8,0,0,0,0,1,0,0,3,
4,2,6,0,0,3,7,9,0,
7,0,0,9,0,0,0,5,0,
9,0,0,5,0,7,0,0,0,
2,8,7,4,1,9,6,0,5,
3,0,0,2,8,0,1,0,0]
I have not seen finish. I don't think this is a problem with the method, as it does return correct results.
Profiling showed that most of the time was spent in the "isValid" function. Is there something obviously inefficient/slow about that function?
The implementation is of course improvable, but that's not the problem. The problem is that for the second grid, the simple guess-and-check algorithm needs a lot of backtracking. Even if you speed up each of your functions 1000-fold, there will be grids where it still needs several times the age of the universe to find the (first, if the grid is not unique) solution.
You need a better algorithm to avoid that. A fairly efficient method to avoid such cases is to guess the square with the least number of possibilities first. That doesn't avoid all bad cases, but reduces them much.
One thing that you should also do is replace the length thing == 0 check with null thing. With the relatively short lists occurring here, the effect is limited, but in general it can be dramatic (and in general you should also not use length list <= 1, use null $ drop 1 list instead).
isValidList = all (\x -> length x <= 1) . tail . group . sort -- tail removes entries that are '0'
If the original list does not contain any zeros, tail will remove something else, perhaps a list of two ones. I'd replace tail . group. sort with group . sort . filter (/= 0).
I don't understand why isValidBox and isValidDiv use foldl as map appears to be adequate. Have I missed something / are they doing something terribly clever?
I'm trying to implement the following recursive definition for addition in F#
m + 0 := m
m + (n + 1) := (m + n) + 1
I can't seem to get the syntax correct, The closest I've come is
let rec plus x y =
match y with
| 0 -> x;
| succ(y) -> succ( plus(x y) );
Where succ n = n + 1. It throws an error on pattern matching for succ.
I'm not sure what succ means in your example, but it is not a pattern defined in the standard F# library. Using just the basic functionality, you'll need to use a pattern that matches any number and then subtract one (and add one in the body):
let rec plus x y =
match y with
| 0 -> x
| y -> 1 + (plus x (y - 1))
In F# (unlike e.g. in Prolog), you can't use your own functions inside patterns. However, you can define active patterns that specify how to decompose input into various cases. The following takes an integer and returns either Zero (for zero) or Succ y for value y + 1:
let (|Zero|Succ|) n =
if n < 0 then failwith "Unexpected!"
if n = 0 then Zero else Succ(n - 1)
Then you can write code that is closer to your original version:
let rec plus x y =
match y with
| Zero -> x
| Succ y -> 1 + (plus x y)
As Tomas said, you can't use succ like this without declaring it. What you can do is to create a discriminated union that represents a number:
type Number =
| Zero
| Succ of Number
And then use that in the plus function:
let rec plus x y =
match y with
| Zero -> x
| Succ(y1) -> Succ (plus x y1)
Or you could declare it as the + operator:
let rec (+) x y =
match y with
| Zero -> x
| Succ(y1) -> Succ (x + y1)
If you kept y where I have y1, the code would work, because the second y would hide the first one. But I think doing so makes the code confusing.
type N = Zero | Succ of N
let rec NtoInt n =
match n with
| Zero -> 0
| Succ x -> 1 + NtoInt x
let rec plus x y =
match x with
| Zero -> y
| Succ n -> Succ (plus n y)
DEMO:
> plus (Succ (Succ Zero)) Zero |> NtoInt ;;
val it : int = 2
> plus (Succ (Succ Zero)) (Succ Zero) |> NtoInt ;;
val it : int = 3
let rec plus x y =
match y with
| 0 -> x
| _ -> plus (x+1) (y-1)