Relación 7

header {* R7: Recorridos de árboles *}

theory R7
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] 
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
fun preOrden :: "'a arbol ⇒ 'a list" where
  "preOrden (H x) = [x]"
 |"preOrden (N x yy zz) = x#(preOrden yy @ preOrden zz)"

value "preOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
 = [e,c,a,d,g,f,h]" -- "True"

value "preOrden (N e (N c (N a1 (H a2) (H a3)) (H d)) (N g (H f) (H h))) 
 = [e, c, a1, a2, a3, d, g, f, h]" -- "True"

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)))
     = [a,d,c,f,h,g,e]
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
fun postOrden :: "'a arbol ⇒ 'a list" where
  "postOrden (H x) = [x]"
 |"postOrden (N x yy zz)= postOrden yy @ postOrden zz @ [x]"

value "postOrden (N e (N c (H a) (H d)) (N g (H f) (H h)))
 = [a,d,c,f,h,g,e]" -- "True"

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]
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
fun inOrden :: "'a arbol ⇒ 'a list" where
  "inOrden (H x) = [x]"
 |"inOrden (N x yy zz) = inOrden yy @ [x] @ inOrden zz"

value "inOrden (N e (N c (H a) (H d)) (N g (H f) (H h))) 
 = [a,c,d,e,f,g,h]" -- "True"

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))
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
fun espejo :: "'a arbol ⇒ 'a arbol" where
  "espejo (H x) = (H x)"
 |"espejo (N x yy zz) = (N x (espejo zz) (espejo yy))"

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))" -- "True"

text {*  
  --------------------------------------------------------------------- 
  Ejercicio 6. Demostrar que
     preOrden (espejo a) = rev (postOrden a)
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
lemma  "preOrden (espejo a) = rev (postOrden a)"
by (induct a, simp_all)

--"jeshorcob,davoremar"
lemma  "preOrden (espejo a) = rev (postOrden a)" (is "?P a")
proof (induct a)
  fix x show "?P (H x)" by simp
next
  fix x yy zz
  assume h1: "?P yy"
  assume h2: "?P zz"
  have "preOrden (espejo (N x yy zz))
    = preOrden (N x (espejo zz) (espejo yy))" by simp
  also have "... = x#(preOrden(espejo zz)@preOrden(espejo yy))" by simp
  also have "... = x#rev(postOrden zz)@rev(postOrden yy)" 
    using h1 h2 by simp
  also have "... = rev((postOrden yy)@(postOrden zz)@[x])" by simp
  finally show "?P (N x yy zz)" by simp
qed

text {*  
  --------------------------------------------------------------------- 
  Ejercicio 7. Demostrar que
     postOrden (espejo a) = rev (preOrden a)
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
lemma "postOrden (espejo a) = rev (preOrden a)"
by (induct a, simp_all)

--"jeshorcob,davoremar"
lemma "postOrden (espejo a) = rev (preOrden a)" (is "?P a")
proof (induct a)
  fix x show "?P (H x)" by simp
next
  fix x y z
  assume h1: "?P y"
  assume h2: "?P z"
  have "postOrden (espejo (N x y z)) = 
    postOrden (N x (espejo z) (espejo y))" by simp
  also have "... = postOrden(espejo z)@postOrden(espejo y)@[x]" by simp
  also have "... = rev(preOrden z)@rev(preOrden y)@[x]" 
    using h1 h2 by simp
  also have "... = rev(x#(preOrden y)@(preOrden z))" by simp
  finally show "?P (N x y z)" by simp
qed

text {*  
  --------------------------------------------------------------------- 
  Ejercicio 8. Demostrar que
     inOrden (espejo a) = rev (inOrden a)
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
theorem "inOrden (espejo a) = rev (inOrden a)"
by (induct a, simp_all)

--"jeshorcob,davoremar"
theorem "inOrden (espejo a) = rev (inOrden a)" (is "?P a")
proof (induct a)
  fix x show "?P (H x)" by simp
next
  fix x y z
  assume h1: "?P y"
  assume h2: "?P z"
  have "inOrden(espejo (N x y z)) = 
    inOrden(N x (espejo z) (espejo y))" by simp
  also have "... = inOrden(espejo z)@[x]@inOrden(espejo y)" by simp
  also have "... = rev(inOrden z)@[x]@rev(inOrden y)" using h1 h2 by simp
  also have "... = rev(inOrden y @ [x] @ inOrden z)" by simp
  finally show "?P (N x y z)" by simp
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
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
fun raiz :: "'a arbol ⇒ 'a" where
  "raiz (H x) = x"
 |"raiz (N x _ _ ) = x"

value "raiz (N e (N c (H a) (H d)) (N g (H f) (H h))) = e" -- "True"

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
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
fun extremo_izquierda :: "'a arbol ⇒ 'a" where
  "extremo_izquierda (H x) = x"
 |"extremo_izquierda (N _ yy _) = extremo_izquierda yy"

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
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
fun extremo_derecha :: "'a arbol ⇒ 'a" where
  "extremo_derecha (H x) = x"
 |"extremo_derecha (N _ _ zz) = extremo_derecha zz"

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
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
(*jeshorcob: al intentar hacer la prueba automática, el sistema se 
pregunta por el caso en el que la lista sea vacía. Debemos indicar que
esto no puede ocurrir y para ello añadimos un lema auxiliar.*)
lemma a1: "inOrden a ≠ []"
by (induct a, simp_all)

--"jeshorcob,davoremar"
theorem "last (inOrden a) = extremo_derecha a"
by (induct a, simp_all add: a1)

--"jeshorcob,davoremar"
lemma  "inOrden a ≠ []"
proof (induct a)
  fix x::"'a" show "inOrden (H x) ≠ []" by simp
next
  fix x::"'a" and y z::"'a arbol"
  assume h1: "inOrden y ≠ []"
  assume h2: "inOrden z ≠ []"
  show "inOrden (N x y z) ≠ []" using h1 h2 by simp  
qed

--"jeshorcob,davoremar"
theorem "last (inOrden a) = extremo_derecha a" (is "?P a")
proof (induct a)
  fix x show "?P (H x)" by simp
next
  fix x y z
  assume h2: "?P z"
  have "last(inOrden(N x y z))=last(inOrden y @[x]@ inOrden z)" by simp
  also have "... = last(inOrden z)" by (simp add: a1)
  also have "... = extremo_derecha z" using h2 by simp
  finally show "?P (N x y z)" by simp
qed

text {*  
  --------------------------------------------------------------------- 
  Ejercicio 13. Demostrar o refutar
     hd (inOrden a) = extremo_izquierda a
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
(*jeshorcob: aquí necesitamos el mismo lema por un motivo similar*)
theorem "hd (inOrden a) = extremo_izquierda a"
by (induct a, simp_all add: a1)

--"jeshorcob,davoremar"
theorem "hd (inOrden a) = extremo_izquierda a" (is "?P a")
proof (induct a)
  fix x show "?P (H x)" by simp
next
  fix x y z
  assume h1: "?P y"
  have "hd(inOrden(N x y z))=hd(inOrden y @[x]@ inOrden z)" by simp
  also have "... = hd(inOrden y)" by (simp add: a1)
  also have "... = extremo_izquierda y" using h1 by simp
  finally show "?P (N x y z)" by simp
qed

text {*  
  --------------------------------------------------------------------- 
  Ejercicio 14. Demostrar o refutar
     hd (preOrden a) = last (postOrden a)
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
theorem "hd (preOrden a) = last (postOrden a)"
by (induct a, simp_all)

--"jeshorcob,davoremar"
theorem "hd (preOrden a) = last (postOrden a)" (is "?P a")
proof (induct a)
  fix x show "?P (H x)" by simp
next
  fix x::"'a" and y z
  have "hd(preOrden(N x y z)) = hd(x#(preOrden y @ preOrden z))" by simp
  also have "... = x" by simp
  also have "... = last(postOrden y @ postOrden z @[x])" by simp
  finally show "?P (N x y z)" by simp
qed

text {*  
  --------------------------------------------------------------------- 
  Ejercicio 15. Demostrar o refutar
     hd (preOrden a) = raiz a
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
theorem "hd (preOrden a) = raiz a"
by (induct a, simp_all)

--"jeshorcob,davoremar"
theorem "hd (preOrden a) = raiz a" (is "?P a")
proof (induct a)
  fix x show "?P (H x)" by simp
next
  fix x::"'a" and y z
  have "hd(preOrden(N x y z)) = hd(x#(preOrden y @ preOrden z))" by simp
  also have "... = x" by simp
  finally show "?P (N x y z)" by simp
qed

text {*  
  --------------------------------------------------------------------- 
  Ejercicio 16. Demostrar o refutar
     hd (inOrden a) = raiz a
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
theorem "hd (inOrden a) = raiz a"
quickcheck
oops
(*Encuentra el contrajemplo:
a = N a⇩1 (H a⇩2) (H a⇩1)

ya que evaluando se ve que son distindos:
hd (inOrden a) = a⇩2
raiz a = a⇩1
*)

text {*  
  --------------------------------------------------------------------- 
  Ejercicio 17. Demostrar o refutar
     last (postOrden a) = raiz a
  --------------------------------------------------------------------- 
*}

--"jeshorcob,davoremar"
theorem "last (postOrden a) = raiz a"
by (induct a, simp_all)

--"jeshorcob,davoremar"
theorem "last (postOrden a) = raiz a" (is "?P a")
proof (induct a)
  fix x show "?P (H x)" by simp
next
  fix x::"'a" and y z
  have "last(postOrden(N x y z)) = 
    last(postOrden y @ postOrden z @ [x])" by simp
  also have "... = x" by simp
  finally show "?P (N x y z)" by simp
qed

end