Relación 1

chapter {* R1: Programación funcional en Isabelle *}

theory R1
imports Main 
begin

text {* ----------------------------------------------------------------
  Ejercicio 0. Definir, por recursión, la función
     factorial :: nat ⇒ nat
  tal que (factorial n) es el factorial de n. Por ejemplo,
     factorial 4 = 24
  ------------------------------------------------------------------- *}

(* danrodcha, anaprarod, ivamenjim, serrodcal, manmorjim1 *) 
fun factorial :: "nat ⇒ nat" where
  "factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"


(*wilmorort, pablucoto,marcarmor13, crigomgom, rubgonmar, jeamacpov *)
fun factorial1 :: "nat ⇒ nat" where
  "factorial1 0  = 1 "
| "factorial1 n  = n * factorial1(n-1)"

 
value "factorial 4" -- "24"

text {* ----------------------------------------------------------------
  Ejercicio 1. Definir, por recursión, la función
     longitud :: 'a list ⇒ nat
  tal que (longitud xs) es la longitud de la listas xs. Por ejemplo,
     longitud [4,2,5] = 3
  ------------------------------------------------------------------- *}

(*wilmorort*)
(* Para usar las lista en forma de [a,b,c] *)

translations
  "[x, xs]" == "x#[xs]"
  "[x]" == "x#[]"

(*wilmorort, serrodcal,crigomgom,rubgonmar*)
fun longitud :: "'a  list  ⇒ nat" where
  "longitud  []  = 0" |
  "longitud (x # xs) = 1 + longitud xs "

(*pablucoto*)
fun longitud0 :: "'a list ⇒ nat " where
" longitud0 [] = 0"
|"longitud0 xs = 1 + longitud0 ((butlast xs)) "

fun longitud0_1 :: "'a list ⇒ nat " where
"longitud0_1 xs = (if xs =[] then 0 else 1 + longitud0_1((butlast xs))) "  


(*marcarmor13, manmorjim1, jeamacpov *)
fun longitud1 :: "'a list ⇒ nat" where
  "longitud1 []  = 0 "
| "longitud1  xs = (1+ longitud2 (tl xs))"

(*danrodcha*)
fun longitud2 :: "'a list ⇒ nat" where
  "longitud2 [] = 0"
| "longitud2 (x#xs) = Suc (longitud2 xs)"

(*serrodcal*)
fun longitud3 :: "'a list ⇒ nat" where
  "longitud xs = length (xs)"

value "longitud [4,2,5] " -- "= 3"

text {* --------------------------------------------------------------- 
  Ejercicio 2. Definir la función
     fun intercambia :: 'a × 'b ⇒ 'b × 'a
  tal que (intercambia p) es el par obtenido intercambiando las
  componentes del par p. Por ejemplo,
     intercambia (u,v) = (v,u)
  ------------------------------------------------------------------ *}

(*wilmorort,marcarmor13,danrodcha,crigomgom,pablucoto, rubgonmar, manmorjim1*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
  "intercambia (x,y) = (y,x) "

(* jeamacpov *)
fun intercambia2 :: "'a × 'b ⇒ 'b × 'a" where
 "intercambia2 (x,y) = (snd (x,y), fst (x,y))"


value "intercambia (u,v)"-- "= (v,u)"
value "intercambia2 (u,v) = (v,u)"

text {* --------------------------------------------------------------- 
  Ejercicio 3. Definir, por recursión, la función
     inversa :: 'a list ⇒ 'a list
  tal que (inversa xs) es la lista obtenida invirtiendo el orden de los
  elementos de xs. Por ejemplo,
     inversa [a,d,c] = [c,d,a]
  ------------------------------------------------------------------ *}

(*wilmorort*)
(* @ :: "'a list => 'a list => 'a list", función agregación definida
 en Theory Main, concatena dos listas: [a,b] @ [c,d] = [a,b,c,d] *)
fun inversa :: "'a list ⇒ 'a list" where
  "inversa [] = []" |
  "inversa (x # xs) = (inversa xs)@(x#[]) "


(*marcarmor13, manmorjim1*)
fun inversa1 :: "'a list ⇒ 'a list" where
  "inversa1 [] = []"
| "inversa1 xs =  inversa1 (tl xs)@ ((hd xs)#[])"

(*danrodcha, pablucoto*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
   "inversa2 [] = []"
|  "inversa2 (x#xs) = (inversa2 xs)@[x] "

(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
   "inversa3 [] = []"
|  "inversa3 (x#xs) = concat [(inversa3 xs),[x]] "

(*crigomgom*)
fun inversa4 :: "'a list ⇒ 'a list" where
  "inversa4 [] = []" |
  "inversa4 xs = (last xs)#(inversa4(butlast xs)) "

(* jeamacpov *)
fun inversa5 :: "'a list ⇒ 'a list" where
  "inversa5 [] = []"
| "inversa5 xs =  (drop (length(xs)-1) xs)@inversa5(take (length(xs)-1) xs)"


value "inversa [a,d,c]" -- "= [c,d,a]"

text {* --------------------------------------------------------------- 
  Ejercicio 4. Definir la función
     repite :: nat ⇒ 'a ⇒ 'a list
  tal que (repite n x) es la lista formada por n copias del elemento
  x. Por ejemplo, 
     repite 3 a = [a,a,a]
  ------------------------------------------------------------------ *}

(*wilmorort,marcarmor13, crigomgom, pablucoto, rubgonmar, manmorjim1, jeamacpov *)
fun repite :: "nat ⇒ 'a ⇒ 'a list" where
  "repite 0 x = [] " |
  "repite n x = x # (repite(n-1) x) "

(*danrodcha*)
fun repite1 :: "nat ⇒ 'a ⇒ 'a list" where
  "repite1 0 x = []"
| "repite1 (Suc n) x = x#(repite1 n x)"

value "repite 3 a" -- "= [a,a,a]"

text {* --------------------------------------------------------------- 
  Ejercicio 5. Definir la función
     conc :: 'a list ⇒ 'a list ⇒ 'a list
  tal que (conc xs ys) es la concatención de las listas xs e ys. Por
  ejemplo, 
     conc [a,d] [b,d,a,c] = [a,d,b,d,a,c]
  ------------------------------------------------------------------ *}

(*marcarmor13*)
fun conc :: "'a list ⇒ 'a list ⇒ 'a list" where
  "conc xs ys = xs@ys"

(*danrodcha, crigomgom, rubgonmar*)
fun conc1 :: "'a list ⇒ 'a list ⇒ 'a list" where
  "conc1 [] ys = ys"
| "conc1 xs [] = xs"       (*esta no hace falta*)
| "conc1 (x#xs) ys = x#(conc1 xs ys)"

(*pablucoto, jeamacpov *)
fun conc2 :: "'a list ⇒ 'a list ⇒ 'a list" where
  "conc2 [] ys = ys" |
  "conc2 (x#xs) ys = x # (conc2 xs ys)"

(* manmorjim1 *)
fun conc3 :: "'a list ⇒ 'a list ⇒ 'a list" where
 "conc3 [] ys = ys"
| "conc3 xs ys = (hd xs)#[] @ (conc3 (tl xs) ys)"

value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]"

text {* --------------------------------------------------------------- 
  Ejercicio 6. Definir la función
     coge :: nat ⇒ 'a list ⇒ 'a list
  tal que (coge n xs) es la lista de los n primeros elementos de xs. Por 
  ejemplo, 
     coge 2 [a,c,d,b,e] = [a,c]
  ------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun coge :: "nat ⇒ 'a list ⇒ 'a list" where
  "coge 0 xs = []"
| "coge n xs = (hd xs)#(coge (n-1) (tl xs)) "

(*danrodcha*)
fun coge1 :: "nat ⇒ 'a list ⇒ 'a list" where
  "coge1 0 _ = []"
| "coge1 _ [] = []"
| "coge1 (Suc n) (x#xs) = x#(coge1 n xs)"

(*crimgomgom, jeamacpov*)
fun coge2 :: "nat ⇒ 'a list ⇒ 'a list" where
  "coge2 0 _ = []"|
  "coge2 _ [] = []" |
  "coge2 n (x#xs) = x#(coge2 (n-1) xs) "

(*pablucoto*)
fun coge3 :: "nat ⇒ 'a list ⇒ 'a list" where
  "coge3 0 xs = []" |
  "coge3 n (x#xs) = (if n>length(x#xs) then (x#xs) else x # (coge3 (n-1) xs))" 

value "coge 2 [a,c,d,b,e]" -- "= [a,c]"

text {* --------------------------------------------------------------- 
  Ejercicio 7. Definir la función
     elimina :: nat ⇒ 'a list ⇒ 'a list
  tal que (elimina n xs) es la lista obtenida eliminando los n primeros
  elementos de xs. Por ejemplo, 
     elimina 2 [a,c,d,b,e] = [d,b,e]
  ------------------------------------------------------------------ *}
(*marcarmor13, manmorjim1*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
  "elimina 0  xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"

(*danrodcha, crigomgom*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
  "elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n xs"

(*pablucoto*)
fun elimina2 :: "nat ⇒ 'a list ⇒ 'a list" where
   "elimina2 0 xs = xs" |
  "elimina2 n (x#xs) = (if n>length(x#xs) then [] else (elimina2 (n-1) xs))" 

(* jeamacpov *)
fun elimina3 :: "nat ⇒ 'a list ⇒ 'a list" where
  "elimina3 0 xs = xs"
| "elimina3 n xs =  (elimina3 (n-1) ((drop 1 xs))) "


value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]"

text {* --------------------------------------------------------------- 
  Ejercicio 8. Definir la función
     esVacia :: 'a list ⇒ bool
  tal que (esVacia xs) se verifica si xs es la lista vacía. Por ejemplo,
     esVacia []  = True
     esVacia [1] = False
  ------------------------------------------------------------------ *}

(*marcarmor13, rubgonmar, danrodcha, crigomgom*)
fun esVacia :: "'a list ⇒ bool" where
  "esVacia [] = True "
| "esVacia xs = False"

(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
  "esVacia1 xs = (xs = [])"

(*pablucoto, jeamacpov *)
fun esVacia2 :: "'a list ⇒ bool" where
"esVacia2 xs = (if xs=[] then True else False)"

(*manmorjim1*)
fun esVacia3 :: "'a list ⇒ bool" where
  "esVacia3 xs = (length xs = 0)"

value "esVacia []"  -- "= True"
value "esVacia [1]" -- "= False"

text {* --------------------------------------------------------------- 
  Ejercicio 9. Definir la función
     inversaAc :: 'a list ⇒ 'a list
  tal que (inversaAc xs) es a inversa de xs calculada usando
  acumuladores. Por ejemplo, 
     inversaAc [a,c,b,e] = [e,b,c,a]
  ------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13, mamnorjim1*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
  "inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "

(*danrodcha, crigomgom*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
  "inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"

(*rubgonmar, danrodcha, crigomgom, manmorjim1*)
fun inversaAc :: "'a list ⇒ 'a list" where
  "inversaAc xs = inversaAcAux xs []"

(*pablucoto*)
fun inversaAcAux2 :: "'a list ⇒ 'a list ⇒ 'a list" where
  "inversaAcAux2 [] [] = []"| 
  "inversaAcAux2 xs (y#ys) = (inversaAcAux2 [] ys) @ [y]" 
fun inversaAc2 :: "'a list ⇒ 'a list" where
  "inversaAc2 xs = inversaAcAux2 [] xs "


(* jeamacpov *)
fun inversaAcAux3 :: "'a list ⇒ 'a list ⇒ 'a list" where
  "inversaAcAux3 ys [] = ys"
| "inversaAcAux3 ys (x#xs) = inversaAcAux3 (x#ys) xs"
fun inversaAc3 :: "'a list ⇒ 'a list" where
  "inversaAc3 xs = inversaAcAux3 [] xs"

value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]"

text {* --------------------------------------------------------------- 
  Ejercicio 10. Definir la función
     sum :: nat list ⇒ nat
  tal que (sum xs) es la suma de los elementos de xs. Por ejemplo,
     sum [3,2,5] = 10
  ------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13, manmorjim1*)
fun sum :: "nat list ⇒ nat" where
  "sum [] = 0"
 |"sum xs = hd xs + sum (tl xs)"

(*danrodcha, crigomgom, pablucoto*)
fun sum1 :: "nat list ⇒ nat" where
  "sum1 [] = 0"
| "sum1 (x#xs) = x + sum1 xs"

(*danrodcha*)
fun sum2 :: "nat list ⇒ nat" where
  "sum2 xs = fold (op +) xs 0"

(* jeamacpov *)
fun sum3 :: "nat list ⇒ nat" where
  "sum3 [] = 0"
| "sum3 xs = (hd xs)+(sum3(drop 1 xs))"

value "sum [3,2,5]" -- "= 10"

text {* --------------------------------------------------------------- 
  Ejercicio 11. Definir la función
     map :: ('a ⇒ 'b) ⇒ 'a list ⇒ 'b list
  tal que (map f xs) es la lista obtenida aplicando la función f a los
  elementos de xs. Por ejemplo,
     map (λx. 2*x) [3,2,5] = [6,4,10]
  ------------------------------------------------------------------ *}

(*rubgonmar,marcarmor13,manmorjim1*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
  "map f [] = []" 
| "map f xs = f(hd xs)#map f (tl xs)"


(*wilmorort, danrodcha, crigomgom, pablucoto*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
 "map f [] = []" 
|"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin
                                                         ellos lo entiende*)

(* jeamacpov *)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
  "map f [] = []"
| "map f xs =f(hd xs)#(map f (drop 1 xs)) "

value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"

end