(* Author: Tobias Nipkow *) section ‹Creating Balanced Trees› theory Balance imports "HOL-Library.Tree_Real" begin fun bal :: "nat ⇒ 'a list ⇒ 'a tree * 'a list" where "bal n xs = (if n=0 then (Leaf,xs) else (let m = n div 2; (l, ys) = bal m xs; (r, zs) = bal (n-1-m) (tl ys) in (Node l (hd ys) r, zs)))" declare bal.simps[simp del] definition bal_list :: "nat ⇒ 'a list ⇒ 'a tree" where "bal_list n xs = fst (bal n xs)" definition balance_list :: "'a list ⇒ 'a tree" where "balance_list xs = bal_list (length xs) xs" definition bal_tree :: "nat ⇒ 'a tree ⇒ 'a tree" where "bal_tree n t = bal_list n (inorder t)" definition balance_tree :: "'a tree ⇒ 'a tree" where "balance_tree t = bal_tree (size t) t" lemma bal_simps: "bal 0 xs = (Leaf, xs)" "n > 0 ⟹ bal n xs = (let m = n div 2; (l, ys) = bal m xs; (r, zs) = bal (n-1-m) (tl ys) in (Node l (hd ys) r, zs))" by(simp_all add: bal.simps) text‹Some of the following lemmas take advantage of the fact that ‹bal xs n› yields a result even if ‹n > length xs›.› lemma size_bal: "bal n xs = (t,ys) ⟹ size t = n" proof(induction n xs arbitrary: t ys rule: bal.induct) case (1 n xs) thus ?case by(cases "n=0") (auto simp add: bal_simps Let_def split: prod.splits) qed lemma bal_inorder: "⟦ bal n xs = (t,ys); n ≤ length xs ⟧ ⟹ inorder t = take n xs ∧ ys = drop n xs" proof(induction n xs arbitrary: t ys rule: bal.induct) case (1 n xs) show ?case proof cases assume "n = 0" thus ?thesis using 1 by (simp add: bal_simps) next assume [arith]: "n ≠ 0" let ?n1 = "n div 2" let ?n2 = "n - 1 - ?n1" from "1.prems" obtain l r xs' where b1: "bal ?n1 xs = (l,xs')" and b2: "bal ?n2 (tl xs') = (r,ys)" and t: "t = ⟨l, hd xs', r⟩" by(auto simp: Let_def bal_simps split: prod.splits) have IH1: "inorder l = take ?n1 xs ∧ xs' = drop ?n1 xs" using b1 "1.prems" by(intro "1.IH"(1)) auto have IH2: "inorder r = take ?n2 (tl xs') ∧ ys = drop ?n2 (tl xs')" using b1 b2 IH1 "1.prems" by(intro "1.IH"(2)) auto have "drop (n div 2) xs ≠ []" using "1.prems"(2) by simp hence "hd (drop ?n1 xs) # take ?n2 (tl (drop ?n1 xs)) = take (?n2 + 1) (drop ?n1 xs)" by (metis Suc_eq_plus1 take_Suc) hence *: "inorder t = take n xs" using t IH1 IH2 using take_add[of ?n1 "?n2+1" xs] by(simp) have "n - n div 2 + n div 2 = n" by simp hence "ys = drop n xs" using IH1 IH2 by (simp add: drop_Suc[symmetric]) thus ?thesis using * by blast qed qed corollary inorder_bal_list[simp]: "n ≤ length xs ⟹ inorder(bal_list n xs) = take n xs" unfolding bal_list_def by (metis bal_inorder eq_fst_iff) corollary inorder_balance_list[simp]: "inorder(balance_list xs) = xs" by(simp add: balance_list_def) corollary inorder_bal_tree: "n ≤ size t ⟹ inorder(bal_tree n t) = take n (inorder t)" by(simp add: bal_tree_def) corollary inorder_balance_tree[simp]: "inorder(balance_tree t) = inorder t" by(simp add: balance_tree_def inorder_bal_tree) corollary size_bal_list[simp]: "size(bal_list n xs) = n" unfolding bal_list_def by (metis prod.collapse size_bal) corollary size_balance_list[simp]: "size(balance_list xs) = length xs" by (simp add: balance_list_def) corollary size_bal_tree[simp]: "size(bal_tree n t) = n" by(simp add: bal_tree_def) corollary size_balance_tree[simp]: "size(balance_tree t) = size t" by(simp add: balance_tree_def) lemma min_height_bal: "bal n xs = (t,ys) ⟹ min_height t = nat(⌊log 2 (n + 1)⌋)" proof(induction n xs arbitrary: t ys rule: bal.induct) case (1 n xs) show ?case proof cases assume "n = 0" thus ?thesis using "1.prems" by (simp add: bal_simps) next assume [arith]: "n ≠ 0" from "1.prems" obtain l r xs' where b1: "bal (n div 2) xs = (l,xs')" and b2: "bal (n - 1 - n div 2) (tl xs') = (r,ys)" and t: "t = ⟨l, hd xs', r⟩" by(auto simp: bal_simps Let_def split: prod.splits) let ?log1 = "nat (floor(log 2 (n div 2 + 1)))" let ?log2 = "nat (floor(log 2 (n - 1 - n div 2 + 1)))" have IH1: "min_height l = ?log1" using "1.IH"(1) b1 by simp have IH2: "min_height r = ?log2" using "1.IH"(2) b1 b2 by simp have "(n+1) div 2 ≥ 1" by arith hence 0: "log 2 ((n+1) div 2) ≥ 0" by simp have "n - 1 - n div 2 + 1 ≤ n div 2 + 1" by arith hence le: "?log2 ≤ ?log1" by(simp add: nat_mono floor_mono) have "min_height t = min ?log1 ?log2 + 1" by (simp add: t IH1 IH2) also have "… = ?log2 + 1" using le by (simp add: min_absorb2) also have "n - 1 - n div 2 + 1 = (n+1) div 2" by linarith also have "nat (floor(log 2 ((n+1) div 2))) + 1 = nat (floor(log 2 ((n+1) div 2) + 1))" using 0 by linarith also have "… = nat (floor(log 2 (n + 1)))" using floor_log2_div2[of "n+1"] by (simp add: log_mult) finally show ?thesis . qed qed lemma height_bal: "bal n xs = (t,ys) ⟹ height t = nat ⌈log 2 (n + 1)⌉" proof(induction n xs arbitrary: t ys rule: bal.induct) case (1 n xs) show ?case proof cases assume "n = 0" thus ?thesis using "1.prems" by (simp add: bal_simps) next assume [arith]: "n ≠ 0" from "1.prems" obtain l r xs' where b1: "bal (n div 2) xs = (l,xs')" and b2: "bal (n - 1 - n div 2) (tl xs') = (r,ys)" and t: "t = ⟨l, hd xs', r⟩" by(auto simp: bal_simps Let_def split: prod.splits) let ?log1 = "nat ⌈log 2 (n div 2 + 1)⌉" let ?log2 = "nat ⌈log 2 (n - 1 - n div 2 + 1)⌉" have IH1: "height l = ?log1" using "1.IH"(1) b1 by simp have IH2: "height r = ?log2" using "1.IH"(2) b1 b2 by simp have 0: "log 2 (n div 2 + 1) ≥ 0" by auto have "n - 1 - n div 2 + 1 ≤ n div 2 + 1" by arith hence le: "?log2 ≤ ?log1" by(simp add: nat_mono ceiling_mono del: nat_ceiling_le_eq) have "height t = max ?log1 ?log2 + 1" by (simp add: t IH1 IH2) also have "… = ?log1 + 1" using le by (simp add: max_absorb1) also have "… = nat ⌈log 2 (n div 2 + 1) + 1⌉" using 0 by linarith also have "… = nat ⌈log 2 (n + 1)⌉" using ceiling_log2_div2[of "n+1"] by (simp) finally show ?thesis . qed qed lemma balanced_bal: assumes "bal n xs = (t,ys)" shows "balanced t" unfolding balanced_def using height_bal[OF assms] min_height_bal[OF assms] by linarith lemma height_bal_list: "n ≤ length xs ⟹ height (bal_list n xs) = nat ⌈log 2 (n + 1)⌉" unfolding bal_list_def by (metis height_bal prod.collapse) lemma height_balance_list: "height (balance_list xs) = nat ⌈log 2 (length xs + 1)⌉" by (simp add: balance_list_def height_bal_list) corollary height_bal_tree: "n ≤ length xs ⟹ height (bal_tree n t) = nat⌈log 2 (n + 1)⌉" unfolding bal_list_def bal_tree_def using height_bal prod.exhaust_sel by blast corollary height_balance_tree: "height (balance_tree t) = nat⌈log 2 (size t + 1)⌉" by (simp add: bal_tree_def balance_tree_def height_bal_list) corollary balanced_bal_list[simp]: "balanced (bal_list n xs)" unfolding bal_list_def by (metis balanced_bal prod.collapse) corollary balanced_balance_list[simp]: "balanced (balance_list xs)" by (simp add: balance_list_def) corollary balanced_bal_tree[simp]: "balanced (bal_tree n t)" by (simp add: bal_tree_def) corollary balanced_balance_tree[simp]: "balanced (balance_tree t)" by (simp add: balance_tree_def) lemma wbalanced_bal: "bal n xs = (t,ys) ⟹ wbalanced t" proof(induction n xs arbitrary: t ys rule: bal.induct) case (1 n xs) show ?case proof cases assume "n = 0" thus ?thesis using "1.prems" by(simp add: bal_simps) next assume "n ≠ 0" with "1.prems" obtain l ys r zs where rec1: "bal (n div 2) xs = (l, ys)" and rec2: "bal (n - 1 - n div 2) (tl ys) = (r, zs)" and t: "t = ⟨l, hd ys, r⟩" by(auto simp add: bal_simps Let_def split: prod.splits) have l: "wbalanced l" using "1.IH"(1)[OF ‹n≠0› refl rec1] . have "wbalanced r" using "1.IH"(2)[OF ‹n≠0› refl rec1[symmetric] refl rec2] . with l t size_bal[OF rec1] size_bal[OF rec2] show ?thesis by auto qed qed text‹An alternative proof via @{thm balanced_if_wbalanced}:› lemma "bal n xs = (t,ys) ⟹ balanced t" by(rule balanced_if_wbalanced[OF wbalanced_bal]) lemma wbalanced_bal_list[simp]: "wbalanced (bal_list n xs)" by(simp add: bal_list_def) (metis prod.collapse wbalanced_bal) lemma wbalanced_balance_list[simp]: "wbalanced (balance_list xs)" by(simp add: balance_list_def) lemma wbalanced_bal_tree[simp]: "wbalanced (bal_tree n t)" by(simp add: bal_tree_def) lemma wbalanced_balance_tree: "wbalanced (balance_tree t)" by (simp add: balance_tree_def) hide_const (open) bal end