86 lines
2.0 KiB
OCaml
86 lines
2.0 KiB
OCaml
type expr =
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| Val of int
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| Var of string
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| Add of expr * expr
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| Mul of expr * expr
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| Pow of expr * expr
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| Ln of expr;;
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let rec pown a n =
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if n == 0 then 1
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else if n == 1 then a
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else let b = pown a (n / 2) in
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b * b * (if n mod 2 == 0 then 1 else a);;
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let rec add n m =
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match (n, m) with
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| (Val n, Val m) -> Val (n+m)
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| (Val 0, f) -> f
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| (f, Val 0) -> f
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| (f, Val n) -> add (Val n) f
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| (Val n, Add(Val m, f)) -> add (Val (n+m)) f
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| (f, Add(Val n, g)) -> add (Val n) (add f g)
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| (Add(f, g), h) -> add f (add g h)
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| (f, g) -> Add (f, g);;
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let rec mul n m =
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match (n, m) with
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| (Val n, Val m) -> Val (n*m)
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| (Val 0, _) -> Val 0
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| (_, Val 0) -> Val 0
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| (Val 1, f) -> f
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| (f, Val 1) -> f
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| (f, Val n) -> mul (Val n) f
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| (Val n, Mul (Val m, f)) -> mul (Val (n*m)) f
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| (f, Mul (Val n, g)) -> mul (Val n) (mul f g)
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| (Mul (f, g), h) -> mul f (mul g h)
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| (f, g) -> Mul (f, g);;
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let rec pow m n =
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match (m, n) with
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| (Val m, Val n) -> Val (pown m n)
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| (_, Val 0) -> Val 1
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| (f, Val 1) -> f
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| (Val 0, _) -> Val 0
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| (f, g) -> Pow (f, g);;
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let rec ln n =
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match n with
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| (Val 1) -> Val 0
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| f -> Ln f;;
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let rec d x f =
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match f with
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| Val _ -> Val 0
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| Var y -> if x = y then Val 1 else Val 0
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| Add (f, g) -> add (d x f) (d x g)
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| Mul (f, g) -> add (mul f (d x g)) (mul g (d x f))
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| Pow (f, g) -> mul (pow f g) (add (mul (mul g (d x f)) (pow f (Val (-1)))) (mul (ln f) (d x g)))
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| Ln f -> mul (d x f) (pow f (Val (-1)));;
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let rec count f =
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match f with
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| Val _ -> 1
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| Var _ -> 1
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| Add (f, g) -> count f + count g
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| Mul (f, g) -> count f + count g
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| Pow (f, g) -> count f + count g
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| Ln f -> count f;;
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let rec nest_aux s f n x =
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if n == 0 then x
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else let x = f (s - n) x in
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nest_aux s f (n - 1) x;;
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let nest f n e =
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nest_aux n f n e;;
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let deriv i f =
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let d = d "x" f in
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Printf.printf "%8d count: %8d\n" (i+1) (count d);
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d;;
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let x = Var "x" in
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let f = pow x x in
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nest deriv (int_of_string Sys.argv.(1)) f;;
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