OCaml function to evaluate addition and subtraction - recursion

I'm currently trying to write an OCaml function that will evaluate expressions and return a Boolean value. I've tried to do research online, and the closest answer that I could find is this one. However, I'm still having trouble which led me to ask my own question.
Here's the basic code:
type equation =
| True
| False
| Equal of exp * exp
and exp =
| Val of int
| Add of exp * exp
| Sub of exp * exp
let rec eval : equation -> bool
= fun f ->
match f with
| True -> true
| False -> false
| Equal (x, y) -> match (x, y) with
| (Val a, Val b) -> if (x = y) then true else false
| ((Add (Val a, Val b), c) -> eval (Equal (Val (a + b), c))
The program is incomplete, and the recursive call to eval in the last line is where I got stuck. A specific input example that I've thought of is:
eval (Equal (Add (Add (Val 1, Val 2), Val 3), Val 6))
This should evaluate to true, since the two Add's add up to 6, and Equal compares Val 6 with Val 6. The trouble that I'm experiencing is how to recursively call the function to evaluate the second Add inside the expression, so that Add (Val 2, Val 2) first evaluates to Val 3, then the first Add adds Val 3 with Val 3. The program that I've written right now only evaluates one of the two Add's.
Is there anything that I should be thinking of or keeping in mind? Any feedback is welcome. Thank you.

As #Amadan mentioned, it's easier to define a function that would first evaluate expression to an int eval_exp: exp -> int. Then you can just evaluate both expressions in the tuple Equal(e1, e2) and compare them (eval: equation -> bool).
You also do not need values True and False in type equation, because you can just return bool from function without pattern-matching. Note that you could need True and False if you passed those, for some reason, again to eval function.
type equation =
Equal of exp * exp
and exp =
| Val of int
| Add of exp * exp
| Sub of exp * exp
let rec eval (e: equation) : bool =
let rec eval_exp e =
match e with
| Val i -> i
| Add (e1, e2) -> (eval_exp e1) + (eval_exp e2)
| Sub (e1, e2) -> (eval_exp e1) - (eval_exp e2)
in
match e with
| Equal (e1, e2) ->
if (eval_exp e1) = (eval_exp e2)
then true
else false

Related

Lambda calculus implementation using CBV small step operational semantics

I'm trying to implement an interpreter for the lambda calculus that has constant intergers and supports the addition operation. The interpreter should use the call-by-value small-step operational semantics. So I've implemented a step that should be able to reduce a lambda term by one step. However, the stepper is losing the surrounding program of the reduced subterm when reduced.
This is my implementation in F#:
type Exp =
| Cst of int
| Var of string
| Abs of string * Exp
| App of Exp * Exp
| Arith of Oper * Exp * Exp
and Oper =
Plus
and the stepper looks like this:
let rec step (exp : Exp) (env : Map<string, Exp>) : Exp =
match exp with
| Cst _ | Abs(_) -> exp
| Var x ->
match Map.tryFind x env with
| Some v -> v
| None -> failwith "Unbound variable"
| App(e1, e2) ->
match step e1 env with
| Abs(x, e) ->
let newEnv = Map.add x (step e2 env) env
step e newEnv
| e1' -> failwithf "%A is not a lambda abstraction" e1'
| Arith(Plus, Cst a, Cst b) -> Cst (a + b)
| Arith(Plus, e1, Cst b) -> Arith(Plus, step e1 env, Cst b)
| Arith(Plus, Cst a, e2) -> Arith(Plus, Cst a, step e2 env)
| Arith(Plus, a, b) -> Arith(Plus, step a env, step b env)
So, given the following example of a program (\x.(\y.y x) 21 + 21) \x.x + 1
App
(Abs
("x", App (Abs ("y", App (Var "y", Var "x")), Arith (Plus, Cst 21, Cst 21))),
Abs ("x", Arith (Plus, Var "x", Cst 1)))
I expect the step function to only reduce the 21 + 21 while keeping the rest of the program i.e. I expect the following output after one step (\x.(\y.y x) 42) \x.x + 1. However, I'm not able to retain the surrounding code around the Cst 42. How should I modify the program such that it reduction only steps once while maintaining the rest of the program?
I think there are two things that you should do differently if you want to implement standard small-step CBV lambda calculus.
First, you want to always perform just one step. This means that you should always call step recursively only once. For example, you have Arith(Plus, step a env, step b env) - but this means that if you have an expression representing (1+2)+(2+3), you will reduce this in "one step" to 3+5 but this is really two steps in one.
Second, I don't think your way of handling variables will work. If you have (\x.x+2) 1, this should reduce to 1+2 using variable substitution. You could reduce this to x+2 and remember the assignment x=1 on the side, but then your function would need to work on expression alongside with variable assignment Exp * Map<string, Exp> -> Exp * Map<string, Exp>. It is easier to use normal substitution, at least for the start.
So, I would first define subst x repl exp which substitutes all free occurences of x in the expression exp with repl:
let rec subst (n : string) (repl : Exp) (exp : Exp) =
match exp with
| Var x when x = n -> repl
| Cst _ | Var _ -> exp
| Abs(x, _) when x = n -> exp
| Abs(x, b) -> Abs(x, subst n repl b)
| App(e1, e2) -> App(subst n repl e1, subst n repl e2)
| Arith(op, e1, e2) -> Arith(op, subst n repl e1, subst n repl e2)
Now you can implement your step function.
let rec step (exp : Exp) =
match exp with
// Values - do nothing & return
| Cst _ | Abs _ -> exp
// There should be no variables, because we substituted them
| Var x -> failwith "Unbound variable"
// App #1 - e1 is function, e2 is a value, apply
| App(Abs(x, e1), (Cst _ | Abs _)) -> subst x e2 e1
// App #2 - e1 is not a value, reduce that first
| App(e1, e2) -> App(step e1, e2)
// App #3 - e1 is value, but e2 not, reduce that
| App(Abs(x,e1), e2) -> App(Abs(x,e1), step e2)
// Similar to App - if e1 or e2 is not value, reduce e1 then e2
| Arith(Plus, Cst a, Cst b) -> Cst (a + b)
| Arith(Plus, Cst a, e2) -> Arith(Plus, Cst a, step e2)
| Arith(Plus, a, b) -> Arith(Plus, step a, b)
Using your example:
App
(Abs
("x", App (Abs ("y", App (Var "y", Var "x")), Arith (Plus, Cst 21, Cst 21))),
Abs ("x", Arith (Plus, Var "x", Cst 1)))
|> step
|> step
|> step
|> step
I get:
App (Cst 42, Abs ("x", Arith (Plus, Var "x", Cst 1)))
And if I'm correctly making sense of your example, this is correct - because now you are trying to treat a number as a function, which gets stuck.

Unbound value in recursive function (Beginner OCaml)

Code:
let isPrime x =
let checkZero d = match (x mod d, x mod d + 2, intRoot x < d) with
| (0,_,_) -> false
| (_,0,_) -> false
| (_,_,true) -> true
| _ -> checkZero (d + 6) in
match x with
| 0 -> false
| 1 -> true
| 2 -> true
| 3 -> true
| _ -> match (x mod 2, x mod 3) with
| (0,_) -> false
| (_,0) -> false
| _ -> checkZero 5
Error:
line 9, characters 24-33:
Error: Unbound value checkZero
Which refers to the recursive call checkZero (d+6)
I've tried placing the checkZero function as a let ... in in the final checkZero 5 call and added/removed the x parameter in the checkZero function in case there was an error with the definition.
(Running OCaml downloaded in the past week on OSX through homebrew)
If you want a function to be able to call itself you need to declare it as recursive:
let rec checkZero d ...

Unexpected output type

I am doing practice with F#. I am trying to create a simple program capable to find me out a couple of prime numbers that, summed together, equal a natural number input. It is the Goldbach conjecture. A single couple of primes will be enough. We will assume the input to be a even number.
I first created a function to check if a number is prime:
let rec isPrime (x: int) (i: int) :bool =
match x % i with
| _ when float i > sqrt (float x) -> true
| 0 -> false
| _ -> isPrime x (i + 1)
Then, I am trying to develop a function that (a) looks for prime numbers, (b) compare their sum with the input 'z' and (c) returns a tuple when it finds the two numbers. The function should not be correct yet, but I would get the reason behind this problem:
let rec sumPrime (z: int) (j: int) (k: int) :int * int =
match isPrime j, isPrime k with
| 0, 0 when j + k > z -> (0, 0)
| 0, 0 -> sumPrime (j + 1) (k + 1)
| _, 0 -> sumPrime j (k + 1)
| 0, _ -> sumPrime (j + 1) k
| _, _ -> if j + k < z then
sumPrime (j + 1) k
elif j + k = z then
(j, k)
The problem: even if I specified that the output should be a tuple :int * int the compiler protests, claiming that the expected output should be of type bool. When in trouble, I usually refer to F# for fun and profit, that i love, but this time I cannot find out the problem. Any suggestion is greatly appreciated.
Your code has three problems that I've spotted:
Your isPrime returns a bool (as you've specified), but your match expression in sumPrime is matching against integers (in F#, the Boolean value false is not the same as the integer value 0). Your match expression should look like:
match isPrime j, isPrime k with
| false, false when j + k > z -> (0, 0)
| false, false -> ...
| true, false -> ...
| false, true -> ...
| true, true -> ...
You have an if...elif expression in your true, true case, but there's no final else. By default, the final else of an if expression returns (), the unit type. So once you fix your first problem, you'll find that F# is complaining about a type mismatch between int * int and unit. You'll need to add an else condition to your final match case to say what to do if j + k > z.
You are repeatedly calling your sumPrime function, which takes three parameters, with just two parameters. That is perfectly legal in F#, since it's a curried language: calling sumPrime with two parameters produces the type int -> int * int: a function that takes a single int and returns a tuple of ints. But that's not what you're actually trying to do. Make sure you specify a value for z in all your recursive calls.
With those three changes, you should probably see your compiler errors go away.

OCaml error with types

I have the following error from OCaml and I don't understand why. I'm trying to define an interpreter in OCaml. I have some types and functions to evaluate these types. I paste the relevant code.
I have these types:
type ide = string
type exp = Eint of int
| Ebool of bool
| Var of ide
| Prod of exp * exp
| Sum of exp * exp
| Diff of exp * exp
| Eq of exp * exp
| Minus of exp
| Iszero of exp
| Or of exp * exp
| And of exp * exp
| Not of exp
| Ifthenelse of exp * exp * exp
| Let of ide * exp * exp
| Fun of ide list * exp
| Funval of exp * exp env
| Appl of exp * exp list
| Dot of ide * field_name
|Field of ide * exp
| Record of ide * exp list;;
type 'a env = Env of (ide * 'a) list;;
I have a function eval used to eval exp. It works correctly.
let rec eval ((e: exp), (r: exp env)) =
match e with
| Eint(n) -> Eint(n)
| Ebool(b) -> Ebool(b)
| Var(i) -> lookup r i
| Iszero(a) -> iszero(eval(a, r))
| Eq(a, b) -> equ(eval(a, r),eval(b, r))
| Prod(a, b) -> mult(eval(a, r), eval(b, r))
| Sum(a, b) -> plus(eval(a, r), eval(b, r))
| Diff(a, b) -> diff(eval(a, r), eval(b, r))
| Minus(a) -> minus(eval(a, r))
| And(a, b) -> et(eval(a, r), eval(b, r))
| Or(a, b) -> vel(eval(a, r), eval(b, r))
| Not(a) -> non(eval(a, r))
| Ifthenelse(a, b, c) -> let g = eval(a, r) in
if typecheck("bool", g) then
(if g = Ebool(true) then eval(b, r) else eval(c, r))
else failwith ("nonboolean guard")
| Let(i, e1, e2) ->
eval(e2, bind (r, i, eval(e1, r)))
| Fun(x, a) -> Funval(e, r)
| Appl(e1, e2) -> match eval(e1, r) with
| Funval(Fun(x, a), r1) ->
eval(a, bind_list r1 x e2)
| _ -> failwith("no funct in apply")
let eval_field (field:exp) (r: exp env)= match field with
| Field (id, e) -> Field (id, (eval e r))
| _ -> failwith ("Not a Field");;
And finally I have a function to evaluate fields of record:
let eval_field (field:exp) (r: exp env)= match field with
| Field (id, e) -> Field (id, (eval e r))
| _ -> failwith ("Not a Field");;
The problem is with eval_field: OCaml signals me ths error:
Characters 22-24:
let f1 = Field ("f1", e1);;
^^
Error: This expression has type exp/1542
but an expression was expected of type exp/2350
What could be wrong?
Thank you very much for your help.
The compiler is trying to tell you that you have two different types named exp and that you have one of them where the other is expected.
I only see one definition for the type exp in this code. I suspect your environment isn't clean. You might try loading the code up in a new OCaml interpreter. Or perhaps the problem is with some code you're not showing.
Here's a session showing how the error is produced:
$ ocaml
OCaml version 4.02.1
# type abc = A | B | C;;
type abc = A | B | C
# let f (x: abc) = x = A;;
val f : abc -> bool = <fun>
# type abc = A | B | C;;
type abc = A | B | C
# f (C: abc);;
Error: This expression has type abc/1024
but an expression was expected of type abc/1018
My guess is that you have a function (like f here) that was defined using an old definition of your exp type. But you're calling it with a value from the new definition of the type (as here).

Manipulate unmutable variables inside of loop in OCaml

I have the following code in OCaml.I have defined all necesar functions and tested them step by step the evalution should work good but I didn't succed to manipulate the variables inside of while.How can I make x,vn,v to change their value?I think I should rewrite the while like a rec loop but can't figure out exactly:
Here is the rest of code: http://pastebin.com/Ash3xw6y
Pseudocode:
input : f formula
output: yes if f valid
else not
begin:
V =set of prop variables
eliminate from f => and <=>
while (V is not empty)
choose x from V
V =V -{x}
replace f with f[x->true]&&f[x->false]
simplify as much as possible f
if f is evaluated with true then return true
else if (not f) is evaluated true then return false
end if
end while
return false
end
type bexp = V of
| string
| B of bool
| Neg of bexp
| And of bexp * bexp
| Or of bexp * bexp
| Impl of bexp * bexp
| Eqv of bexp * bexp
module StringSet=Set.make(String)
let is_valide f=
let v= stringset_of_list (ens f []) in (*set of all variables of f *)
let g= elim f in (*eliminate => and <=> *)
let quit_loop=ref false in
while not !quit_loop
do
let x=StringSet.choose v in
let vn=StringSet.remove x v in
if StringSet.is_empty vn=true then quit_loop:=true;
let h= And( replace x (B true) g ,replace x (B false) g)in
let j=simplify h in
if (only_bools j) then
if (eval j) then print_string "yes"
else print_string "not"
done
(New form)
let tautology f =
let rec tautology1 x v g =
let h= And( remplace x (B true) g ,remplace x (B false) g)in
let j= simplify h in
if not (only_bools j) then tautology (StringSet.choose (StringSet.remove x v) (StringSet.remove x v) j
else
if (eval1 j) then print_string "yes \n " else
if (eval1 (Neg (j))) then print_string "not \n";
in tautology1 (StringSet.choose (stringset_of_list (ens f [])) (stringset_of_list (ens f [])) (elim f);;
while loop belongs to imperative programming part in OCaml.
Basically, you can't modify immutable variables in while or for loops or anywhere.
To let a variable to be mutable, you need to define it like let var = ref .... ref is the keyword for mutables.
Read these two chapters:
https://realworldocaml.org/v1/en/html/a-guided-tour.html#imperative-programming
https://realworldocaml.org/v1/en/html/imperative-programming-1.html
You can define x,vn,v as refs, but I guess it will be ugly.
I suggest you think your code in a functional way.
Since you haven't placed functions ens etc here, I can't produce an example refine for u.

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