RA2016: Verificación de propiedades de recorridos en árboles binarios con Isabelle/HOL
En tercera parte de la clase de hoy del curso de Razonamiento automático se han comentado las soluciones de los ejercicios de la relación 6. En dicha relación se definen funciones para recorrer árboles binarios y se demuestran con Isabelle/HOL algunas de sus propiedades.
Los ejercicios y sus soluciones se muestran a continuación
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chapter {* R6: Recorridos de árboles *} theory R6_Recorridos_de_arboles_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))) = [a,d,c,f,h,g,e] --------------------------------------------------------------------- *} 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) --------------------------------------------------------------------- *} -- "La demostración automática es" lemma "preOrden (espejo a) = rev (postOrden a)" by (induct a) auto -- "La demostración estructurada es" 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 "… = x # (rev (postOrden d) @ rev (postOrden i))" using HI1 HI2 by simp also have "… = x # rev (postOrden i @ postOrden d)" by simp also have "… = rev ((postOrden i) @ (postOrden d) @ [x])" by simp also have "… = rev (postOrden (N x i d))" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 7. Demostrar que postOrden (espejo a) = rev (preOrden a) --------------------------------------------------------------------- *} -- "La demostración automática es" lemma "postOrden (espejo a) = rev (preOrden a)" by (induct a) auto -- "La demostración estructurada es" 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) @ [x]" using HI1 HI2 by simp also have "… = rev (preOrden i @ preOrden d) @ [x]" by simp also have "… = rev ([x] @ preOrden i @ preOrden d)" by simp also have "… = rev (preOrden (N x i d))" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 8. Demostrar que inOrden (espejo a) = rev (inOrden a) --------------------------------------------------------------------- *} -- "La demostración automática es" theorem "inOrden (espejo a) = rev (inOrden a)" by (induct a) auto -- "La demostración estructurada es" theorem "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) @ [x] @ rev (inOrden i)" using HI1 HI2 by simp also have "… = rev (inOrden i @ [x] @ inOrden d)" by simp also have "… = rev (inOrden (N x i d))" by simp finally show ?thesis . 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 x) = x" | "extremo_izquierda (N x i d) = extremo_izquierda i" 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 --------------------------------------------------------------------- *} -- "La demostración automática, basada en un lema, es" lemma inOrdenNoVacio: "inOrden a ≠ []" by (induct a) auto theorem "last (inOrden a) = extremo_derecha a" by (induct a) (auto simp add: inOrdenNoVacio) -- "La demostración estructurada es" 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 HI: "?P d" show "?P (N x i d)" proof - have "last (inOrden (N x i d)) = last (inOrden d @ [x] @ inOrden d)" by simp also have "… = last (inOrden d)" by (simp add: inOrdenNoVacio) also have "… = extremo_derecha d" using HI by simp also have "… = extremo_derecha (N x i d)" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 13. Demostrar o refutar hd (inOrden a) = extremo_izquierda a --------------------------------------------------------------------- *} -- "La demostración automática es" theorem "hd (inOrden a) = extremo_izquierda a" by (induct a) (auto simp add: inOrdenNoVacio) -- "La demostración estructurada es" 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 HI: "?P i" 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 HI by simp also have "… = extremo_izquierda (N x i d)" by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 14. Demostrar o refutar hd (preOrden a) = last (postOrden a) --------------------------------------------------------------------- *} -- "La demostración automática es" theorem "hd (preOrden a) = last (postOrden a)" by (cases a) auto -- "La demostración estructurada es" theorem "hd (preOrden a) = last (postOrden a)"(is "?P a") proof (cases a) fix x assume "a = H x" then show "?P a" by simp next fix x i d assume H: "a = N x i d" show "?P a" proof - have "hd (preOrden a) = hd (preOrden (N x i d))" using H by simp also have "… = 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 also have "… = last (postOrden a)" using H by simp finally show ?thesis . qed qed text {* --------------------------------------------------------------------- Ejercicio 15. Demostrar o refutar hd (preOrden a) = raiz a --------------------------------------------------------------------- *} -- "La demostración automática es" theorem "hd (preOrden a) = raiz a" by (cases a) auto -- "La demostración estructurada es" theorem "hd (preOrden a) = raiz a" (is "?P a") proof (cases a) fix x assume "a = H x" then show "?P a" by simp next fix x i d assume H: "a = N x i d" show "?P a" proof - have "hd (preOrden a) = hd (preOrden (N x i d))" using H by simp also have "… = hd (x#(preOrden i @ preOrden d))" by simp also have "… = x" by simp also have "… = raiz (N x i d)" by simp also have "… = raiz a" using H by simp finally show ?thesis . 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 --------------------------------------------------------------------- *} -- "La demostración automática es" theorem "last (postOrden a) = raiz a" by (cases a) auto -- "La demostración estructurada es" theorem "last (postOrden a) = raiz a" (is "?P a") proof (cases a) fix x assume "a = H x" then show "?P a" by simp next fix x i d assume H: "a = N x i d" show "?P a" proof - have "last (postOrden a) = last (postOrden (N x i d))" using H by simp also have "… = last (postOrden i @ preOrden d @ [x])" by simp also have "… = x" by simp also have "… = raiz a" using H by simp finally show ?thesis . qed qed end |