Diferencia entre revisiones de «Relación 1»
De Razonamiento automático (2016-17)
Línea 59: | Línea 59: | ||
| "longitud1 xs = (1+ longitud2 (tl xs))" | | "longitud1 xs = (1+ longitud2 (tl xs))" | ||
− | + | (*danrodcha*) | |
+ | fun longitud2 :: "'a list ⇒ nat" where | ||
+ | "longitud2 [] = 0" | ||
+ | | "longitud2 (x#xs) = Suc (longitud xs)" | ||
Línea 72: | Línea 75: | ||
------------------------------------------------------------------ *} | ------------------------------------------------------------------ *} | ||
− | (*wilmorort,marcarmor13*) | + | (*wilmorort,marcarmor13,danrodcha*) |
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where | fun intercambia :: "'a × 'b ⇒ 'b × 'a" where | ||
"intercambia (x,y) = (y,x) " | "intercambia (x,y) = (y,x) " | ||
Línea 99: | Línea 102: | ||
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])" | | "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])" | ||
+ | (*danrodcha*) | ||
+ | (* es igual que inversa sustituyendo x#[] por [x] *) | ||
+ | fun inversa2 :: "'a list ⇒ 'a list" where | ||
+ | "inversa2 [] = []" | ||
+ | | "inversa2 (x#xs) = (inversa xs)@[x] " | ||
+ | |||
+ | (*danrodcha*) | ||
+ | fun inversa3 :: "'a list ⇒ 'a list" where | ||
+ | "inversa3 [] = []" | ||
+ | | "inversa3 (x#xs) = concat [(inversa xs),[x]] " | ||
value "inversa [a,d,c]" -- "= [c,d,a]" | value "inversa [a,d,c]" -- "= [c,d,a]" | ||
Línea 114: | Línea 127: | ||
"repite 0 x = [] " | | "repite 0 x = [] " | | ||
"repite n x = x # (repite(n-1) 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]" | value "repite 3 a" -- "= [a,a,a]" | ||
Línea 129: | Línea 147: | ||
"conc xs ys = xs@ys" | "conc xs ys = xs@ys" | ||
+ | (*danrodcha*) | ||
+ | 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)" | ||
value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]" | value "conc [a,d] [b,d,a,c]" -- "= [a,d,b,d,a,c]" | ||
Línea 144: | Línea 167: | ||
| "coge n xs = (hd xs)#(coge (n-1) (tl 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)" | ||
value "coge 2 [a,c,d,b,e]" -- "= [a,c]" | value "coge 2 [a,c,d,b,e]" -- "= [a,c]" | ||
Línea 158: | Línea 186: | ||
"elimina 0 xs = xs" | "elimina 0 xs = xs" | ||
| "elimina n xs = (elimina (n-1) (tl xs ))" | | "elimina n xs = (elimina (n-1) (tl xs ))" | ||
+ | |||
+ | (*danrodcha*) | ||
+ | fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where | ||
+ | "elimina1 0 xs = xs" | ||
+ | | "elimina1 _ [] = []" | ||
+ | | "elimina1 (Suc n) (x#xs) = elimina1 n xs" | ||
value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]" | value "elimina 2 [a,c,d,b,e]" -- "= [d,b,e]" | ||
Línea 169: | Línea 203: | ||
------------------------------------------------------------------ *} | ------------------------------------------------------------------ *} | ||
− | (*marcarmor13, rubgonmar*) | + | (*marcarmor13, rubgonmar, danrodcha*) |
fun esVacia :: "'a list ⇒ bool" where | fun esVacia :: "'a list ⇒ bool" where | ||
"esVacia [] = True " | "esVacia [] = True " | ||
| "esVacia xs = False" | | "esVacia xs = False" | ||
+ | |||
+ | (*danrodcha*) | ||
+ | fun esVacia1 :: "'a list ⇒ bool" where | ||
+ | "esVacia1 xs = (xs = [])" | ||
value "esVacia []" -- "= True" | value "esVacia []" -- "= True" | ||
Línea 190: | Línea 228: | ||
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) " | | "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) " | ||
+ | (*danrodcha*) | ||
+ | fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where | ||
+ | "inversaAcAux1 [] ys = ys" | ||
+ | | "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)" | ||
+ | |||
+ | (*rubgonmar, danrodcha*) | ||
fun inversaAc :: "'a list ⇒ 'a list" where | fun inversaAc :: "'a list ⇒ 'a list" where | ||
"inversaAc xs = inversaAcAux xs []" | "inversaAc xs = inversaAcAux xs []" | ||
+ | |||
value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]" | value "inversaAc [a,c,b,e]" -- "= [e,b,c,a]" | ||
Línea 207: | Línea 252: | ||
|"sum xs = hd xs + sum (tl xs)" | |"sum xs = hd xs + sum (tl xs)" | ||
+ | (*danrodcha*) | ||
+ | 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" | ||
value "sum [3,2,5]" -- "= 10" | value "sum [3,2,5]" -- "= 10" | ||
Línea 224: | Línea 277: | ||
− | (*wilmorort*) | + | (*wilmorort, danrodcha*) |
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where | fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where | ||
− | "map f [] = []" | | + | "map f [] = []" |
− | "map f (x # xs) = f x # map f xs" | + | |"map f (x # xs) = f x # map f xs" (*yo pondría paréntesis, pero sin ellos lo entiende*) |
value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]" | value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]" |
Revisión del 21:00 28 oct 2016
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*)
fun factorial :: "nat ⇒ nat" where
"factorial 0 = 1"
| "factorial (Suc n) = (Suc n) * factorial n"
(*wilmorort, pablucoto,marcarmor13*)
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#[]"
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*)
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 (longitud 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*)
fun intercambia :: "'a × 'b ⇒ 'b × 'a" where
"intercambia (x,y) = (y,x) "
value "intercambia (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*)
fun inversa1 :: "'a list ⇒ 'a list" where
"inversa1 [] = []"
| "inversa1 xs = inversa1 (tl xs)@ ((hd xs)#[])"
(*danrodcha*)
(* es igual que inversa sustituyendo x#[] por [x] *)
fun inversa2 :: "'a list ⇒ 'a list" where
"inversa2 [] = []"
| "inversa2 (x#xs) = (inversa xs)@[x] "
(*danrodcha*)
fun inversa3 :: "'a list ⇒ 'a list" where
"inversa3 [] = []"
| "inversa3 (x#xs) = concat [(inversa xs),[x]] "
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*)
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*)
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)"
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*)
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)"
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*)
fun elimina :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina 0 xs = xs"
| "elimina n xs = (elimina (n-1) (tl xs ))"
(*danrodcha*)
fun elimina1 :: "nat ⇒ 'a list ⇒ 'a list" where
"elimina1 0 xs = xs"
| "elimina1 _ [] = []"
| "elimina1 (Suc n) (x#xs) = elimina1 n 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*)
fun esVacia :: "'a list ⇒ bool" where
"esVacia [] = True "
| "esVacia xs = False"
(*danrodcha*)
fun esVacia1 :: "'a list ⇒ bool" where
"esVacia1 xs = (xs = [])"
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*)
fun inversaAcAux :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux [] ys = ys"
| "inversaAcAux xs ys = inversaAcAux (tl xs) (hd xs#ys) "
(*danrodcha*)
fun inversaAcAux1 :: "'a list ⇒ 'a list ⇒ 'a list" where
"inversaAcAux1 [] ys = ys"
| "inversaAcAux1 (x#xs) ys = inversaAcAux1 xs (x#ys)"
(*rubgonmar, danrodcha*)
fun inversaAc :: "'a list ⇒ 'a list" where
"inversaAc xs = inversaAcAux 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*)
fun sum :: "nat list ⇒ nat" where
"sum [] = 0"
|"sum xs = hd xs + sum (tl xs)"
(*danrodcha*)
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"
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*)
fun map :: "('a ⇒ 'b) ⇒ 'a list ⇒ 'b list" where
"map f [] = []"
| "map f xs = f(hd xs)#map f (tl xs)"
(*wilmorort, danrodcha*)
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*)
value "map (λx. 2*x) [3::nat,2,5]" -- "= [6,4,10]"
end