Sol 12
De Lógica matemática y fundamentos (2018-19)
Revisión del 19:06 26 jun 2019 de Jalonso (discusión | contribuciones)
chapter {* R12: Recorridos de árboles *}
theory R12_sol
imports Main
begin
text {*
---------------------------------------------------------------------
Ejercicio 1. Definir el tipo de datos arbol para representar los
árboles binarios que tiene información en los nodos y en las hojas.
Por ejemplo, el árbol
e
/ \
/ \
c g
/ \ / \
a d f h
se representa por "N e (N c (H a) (H d)) (N g (H f) (H h))".
---------------------------------------------------------------------
*}
datatype 'a arbol = H "'a" | N "'a" "'a arbol" "'a arbol"
value "N e (N c (H a) (H d)) (N g (H f) (H h))"
text {*
---------------------------------------------------------------------
Ejercicio 2. Definir la función
preOrden :: "'a arbol ⇒ 'a list"
tal que (preOrden a) es el recorrido pre orden del árbol a. Por
ejemplo,
preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}
fun preOrden :: "'a arbol ⇒ 'a list" where
"preOrden (H x) = [x]"
| "preOrden (N x i d) = x#((preOrden i)@(preOrden d))"
value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]"
text {*
---------------------------------------------------------------------
Ejercicio 3. Definir la función
postOrden :: "'a arbol ⇒ 'a list"
tal que (postOrden a) es el recorrido post orden del árbol a. Por
ejemplo,
postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [e,c,a,d,g,f,h]
---------------------------------------------------------------------
*}
fun postOrden :: "'a arbol ⇒ 'a list" where
"postOrden (H x) = [x]"
| "postOrden (N x i d) = (postOrden i)@(postOrden d)@[x]"
value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,d,c,f,h,g,e]"
text {*
---------------------------------------------------------------------
Ejercicio 4. Definir la función
inOrden :: "'a arbol ⇒ 'a list"
tal que (inOrden a) es el recorrido in orden del árbol a. Por
ejemplo,
inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]
---------------------------------------------------------------------
*}
fun inOrden :: "'a arbol ⇒ 'a list" where
"inOrden (H x) = [x]"
| "inOrden (N x i d) = (inOrden i)@[x]@(inOrden d)"
value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
= [a,c,d,e,f,g,h]"
text {*
---------------------------------------------------------------------
Ejercicio 5. Definir la función
espejo :: "'a arbol ⇒ 'a arbol"
tal que (espejo a) es la imagen especular del árbol a. Por ejemplo,
espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))
---------------------------------------------------------------------
*}
fun espejo :: "'a arbol ⇒ 'a arbol" where
"espejo (H x) = (H x)"
| "espejo (N x i d) = (N x (espejo d) (espejo i))"
value "espejo (N e (N c (H a) (H d)) (N g (H f) (H h)))
= N e (N g (H h) (H f)) (N c (H d) (H a))"
text {*
---------------------------------------------------------------------
Ejercicio 6. Demostrar que
preOrden (espejo a) = rev (postOrden a)
---------------------------------------------------------------------
*}
lemma "preOrden (espejo a) = rev (postOrden a)"
by (induct a) simp_all
(* 1ª demostración: sin patrones *)
lemma "preOrden (espejo a) = rev (postOrden a)"
proof (induct a)
fix x :: 'a
show "preOrden (espejo (H x)) = rev (postOrden (H x))" by simp
next
fix x :: 'a and i d :: "'a arbol"
assume HI1: "preOrden (espejo i) = rev (postOrden i)"
assume HI2: "preOrden (espejo d) = rev (postOrden d)"
show "preOrden (espejo (N x i d)) = rev (postOrden (N x i d))"
proof -
have "preOrden (espejo (N x i d)) =
preOrden (N x (espejo d) (espejo i))" by simp
also have "… = [x] @ preOrden (espejo d) @ preOrden (espejo i)"
by simp
also have "… = rev [x] @ rev (postOrden d) @ rev (postOrden i)"
using HI1 HI2 by simp
also have "… = rev ((postOrden i) @ (postOrden d) @ [x])" by simp
finally show ?thesis by simp
qed
qed
(* 2ª demostración: con patrones *)
lemma "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
show "?P (N x i d)"
proof -
have "preOrden (espejo (N x i d)) =
preOrden (N x (espejo d) (espejo i))" by simp
also have "… = [x] @ preOrden (espejo d) @ preOrden (espejo i)"
by simp
also have "… = rev [x] @ rev (postOrden d) @ rev (postOrden i)"
using HI1 HI2 by simp
also have "… = rev ((postOrden i) @ (postOrden d) @ [x])" by simp
finally show ?thesis by simp
qed
qed
text {*
---------------------------------------------------------------------
Ejercicio 7. Demostrar que
postOrden (espejo a) = rev (preOrden a)
---------------------------------------------------------------------
*}
lemma "postOrden (espejo a) = rev (preOrden a)"
by (induct a) simp_all
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
show "?P (N x i d)"
proof -
have "postOrden (espejo (N x i d)) =
postOrden (N x (espejo d) (espejo i))" by simp
also have "... = postOrden (espejo d) @ postOrden (espejo i) @ [x]"
by simp
also have "... = rev (preOrden d) @ rev (preOrden i) @ rev [x]"
using HI1 HI2 by simp
also have "... = rev ([x] @ preOrden i @ preOrden d)" by simp
finally show ?thesis by simp
qed
qed
text {*
---------------------------------------------------------------------
Ejercicio 8. Demostrar que
inOrden (espejo a) = rev (inOrden a)
---------------------------------------------------------------------
*}
lemma "inOrden (espejo a) = rev (inOrden a)"
by (induct a) simp_all
lemma "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
show "?P (N x i d)"
proof -
have "inOrden (espejo (N x i d)) =
inOrden (N x (espejo d) (espejo i))" by simp
also have "... = inOrden (espejo d) @ [x] @ inOrden (espejo i)"
by simp
also have "... = rev (inOrden d) @ rev [x] @ rev (inOrden i)"
using HI1 HI2 by simp
also have "... = rev (inOrden i @ [x] @ inOrden d)" by simp
finally show ?thesis by simp
qed
qed
text {*
---------------------------------------------------------------------
Ejercicio 9. Definir la función
raiz :: "'a arbol ⇒ 'a"
tal que (raiz a) es la raiz del árbol a. Por ejemplo,
raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e
---------------------------------------------------------------------
*}
fun raiz :: "'a arbol ⇒ 'a" where
"raiz (H x) = x"
| "raiz (N x i d) = x"
value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e"
text {*
---------------------------------------------------------------------
Ejercicio 10. Definir la función
extremo_izquierda :: "'a arbol ⇒ 'a"
tal que (extremo_izquierda a) es el nodo más a la izquierda del árbol
a. Por ejemplo,
extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a
---------------------------------------------------------------------
*}
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
"extremo_izquierda (H a) = a"
| "extremo_izquierda (N f x y) = (extremo_izquierda x)"
value "extremo_izquierda (N e (N c (H a) (H d)) (N g (H f) (H h))) = a"
text {*
---------------------------------------------------------------------
Ejercicio 11. Definir la función
extremo_derecha :: "'a arbol ⇒ 'a"
tal que (extremo_derecha a) es el nodo más a la derecha del árbol
a. Por ejemplo,
extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h
---------------------------------------------------------------------
*}
fun extremo_derecha :: "'a arbol ⇒ 'a" where
"extremo_derecha (H x) = x"
| "extremo_derecha (N x i d) = (extremo_derecha d)"
value "extremo_derecha (N e (N c (H a) (H d)) (N g (H f) (H h))) = h"
text {*
---------------------------------------------------------------------
Ejercicio 12. Demostrar o refutar
last (inOrden a) = extremo_derecha a
---------------------------------------------------------------------
*}
lemma inOrdenNoVacio: "inOrden a ≠ []"
apply (induct a)
apply simp_all
done
theorem ultimoInOrden:
"last (inOrden a) = extremo_derecha a"
apply (induct a)
apply simp
apply (simp add: inOrdenNoVacio)
done
theorem "last (inOrden a) = extremo_derecha a"
by (induct a) (simp_all add: inOrdenNoVacio)
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
show "?P (N x i d)"
proof -
have "last (inOrden (N x i d)) = last ((inOrden i)@[x]@(inOrden d))"
by simp
also have "... = last (inOrden d)" by (simp add: inOrdenNoVacio)
also have "... = extremo_derecha d" using HI2 by simp
also have "... = extremo_derecha (N x i d)" by simp
finally show ?thesis by simp
qed
qed
text {*
---------------------------------------------------------------------
Ejercicio 13. Demostrar o refutar
hd (inOrden a) = extremo_izquierda a
---------------------------------------------------------------------
*}
theorem "hd (inOrden a) = extremo_izquierda a"
apply (induct a)
apply simp
apply (simp add: inOrdenNoVacio)
done
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
show "?P (N x i d)"
proof -
have "hd (inOrden (N x i d)) = hd((inOrden i)@[x]@(inOrden d))"
by simp
also have "... = hd (inOrden i)" by (simp add: inOrdenNoVacio)
also have "... = extremo_izquierda i" using HI1 by simp
also have "... = extremo_izquierda (N x i d)" by simp
finally show ?thesis by simp
qed
qed
text {*
---------------------------------------------------------------------
Ejercicio 14. Demostrar o refutar
hd (preOrden a) = last (postOrden a)
---------------------------------------------------------------------
*}
theorem "hd (preOrden a) = last (postOrden a)"
by (induct a) simp_all
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
show "?P (N x i d)"
proof -
have "hd (preOrden (N x i d)) = hd (x#((preOrden i)@(preOrden d)))"
by simp
also have "... = x" by simp
also have "... = last ((postOrden i)@(postOrden d)@[x])" by simp
also have "... = last (postOrden (N x i d))" by simp
finally show ?thesis by simp
qed
qed
text {*
---------------------------------------------------------------------
Ejercicio 15. Demostrar o refutar
hd (preOrden a) = raiz a
---------------------------------------------------------------------
*}
theorem "hd (preOrden a) = raiz a"
by (induct a) simp_all
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
show "?P (N x i d)"
proof -
have "hd (preOrden (N x i d)) = hd (x#((preOrden i)@(preOrden d)))"
by simp
also have "... = x" by simp
also have "... = raiz (N x i d)" by simp
finally show ?thesis by simp
qed
qed
text {*
---------------------------------------------------------------------
Ejercicio 16. Demostrar o refutar
hd (inOrden a) = raiz a
---------------------------------------------------------------------
*}
theorem "hd (inOrden a) = raiz a"
quickcheck
oops
text {*
Quickcheck found a counterexample:
a = N a1 (H a2) (H a1)
Evaluated terms:
hd (inOrden a) = a2
raiz a = a1
*}
text {*
---------------------------------------------------------------------
Ejercicio 17. Demostrar o refutar
last (postOrden a) = raiz a
---------------------------------------------------------------------
*}
lemma "last (postOrden a) = raiz a"
by (induct a) simp_all
lemma "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
fix x
show "?P (H x)" by simp
next
fix x i d
assume HI1: "?P i"
assume HI2: "?P d"
show "?P (N x i d)"
proof -
have "last (postOrden (N x i d)) =
last ((postOrden i)@(postOrden d)@[x])" by simp
also have "... = last [x]" by simp
also have "... = x" by simp
also have "... = raiz (N x i d)" by simp
finally show ?thesis by simp
qed
qed
end