(* Author: Tobias Nipkow *)
section \<open>Creating Balanced Trees\<close>
theory Balance
imports
Complex_Main
"~~/src/HOL/Library/Tree"
begin
(* The following two lemmas should go into theory \<open>Tree\<close>, except that that
theory would then depend on \<open>Complex_Main\<close>. *)
lemma min_height_balanced: assumes "balanced t"
shows "min_height t = nat(floor(log 2 (size1 t)))"
proof cases
assume *: "complete t"
hence "size1 t = 2 ^ min_height t"
by (simp add: complete_iff_height size1_if_complete)
hence "size1 t = 2 powr min_height t"
using * by (simp add: powr_realpow)
hence "min_height t = log 2 (size1 t)"
by simp
thus ?thesis
by linarith
next
assume *: "\<not> complete t"
hence "height t = min_height t + 1"
using assms min_height_le_height[of t]
by(auto simp add: balanced_def complete_iff_height)
hence "2 ^ min_height t \<le> size1 t \<and> size1 t < 2 ^ (min_height t + 1)"
by (metis * min_height_size1 size1_height_if_incomplete)
hence "2 powr min_height t \<le> size1 t \<and> size1 t < 2 powr (min_height t + 1)"
by(simp only: powr_realpow)
(metis of_nat_less_iff of_nat_le_iff of_nat_numeral of_nat_power)
hence "min_height t \<le> log 2 (size1 t) \<and> log 2 (size1 t) < min_height t + 1"
by(simp add: log_less_iff le_log_iff)
thus ?thesis by linarith
qed
lemma height_balanced: assumes "balanced t"
shows "height t = nat(ceiling(log 2 (size1 t)))"
proof cases
assume *: "complete t"
hence "size1 t = 2 ^ height t"
by (simp add: size1_if_complete)
hence "size1 t = 2 powr height t"
using * by (simp add: powr_realpow)
hence "height t = log 2 (size1 t)"
by simp
thus ?thesis
by linarith
next
assume *: "\<not> complete t"
hence **: "height t = min_height t + 1"
using assms min_height_le_height[of t]
by(auto simp add: balanced_def complete_iff_height)
hence 0: "2 ^ min_height t < size1 t \<and> size1 t \<le> 2 ^ (min_height t + 1)"
by (metis "*" min_height_size1_if_incomplete size1_height)
hence "2 powr min_height t < size1 t \<and> size1 t \<le> 2 powr (min_height t + 1)"
by(simp only: powr_realpow)
(metis of_nat_less_iff of_nat_le_iff of_nat_numeral of_nat_power)
hence "min_height t < log 2 (size1 t) \<and> log 2 (size1 t) \<le> min_height t + 1"
by(simp add: log_le_iff less_log_iff)
thus ?thesis using ** by linarith
qed
(* mv *)
text \<open>The lemmas about \<open>floor\<close> and \<open>ceiling\<close> of \<open>log 2\<close> should be generalized
from 2 to \<open>n\<close> and should be made executable. In the end they should be moved
to theory \<open>Log_Nat\<close> and \<open>floorlog\<close> should be replaced.\<close>
lemma floor_log_nat_ivl: fixes b n k :: nat
assumes "b \<ge> 2" "b^n \<le> k" "k < b^(n+1)"
shows "floor (log b (real k)) = int(n)"
proof -
have "k \<ge> 1"
using assms(1,2) one_le_power[of b n] by linarith
show ?thesis
proof(rule floor_eq2)
show "int n \<le> log b k"
using assms(1,2) \<open>k \<ge> 1\<close>
by(simp add: powr_realpow le_log_iff of_nat_power[symmetric] del: of_nat_power)
next
have "real k < b powr (real(n + 1))" using assms(1,3)
by (simp only: powr_realpow) (metis of_nat_less_iff of_nat_power)
thus "log b k < real_of_int (int n) + 1"
using assms(1) \<open>k \<ge> 1\<close> by(simp add: log_less_iff add_ac)
qed
qed
lemma ceil_log_nat_ivl: fixes b n k :: nat
assumes "b \<ge> 2" "b^n < k" "k \<le> b^(n+1)"
shows "ceiling (log b (real k)) = int(n)+1"
proof(rule ceiling_eq)
show "int n < log b k"
using assms(1,2)
by(simp add: powr_realpow less_log_iff of_nat_power[symmetric] del: of_nat_power)
next
have "real k \<le> b powr (real(n + 1))"
using assms(1,3)
by (simp only: powr_realpow) (metis of_nat_le_iff of_nat_power)
thus "log b k \<le> real_of_int (int n) + 1"
using assms(1,2) by(simp add: log_le_iff add_ac)
qed
lemma ceil_log2_div2: assumes "n \<ge> 2"
shows "ceiling(log 2 (real n)) = ceiling(log 2 ((n-1) div 2 + 1)) + 1"
proof cases
assume "n=2"
thus ?thesis by simp
next
let ?m = "(n-1) div 2 + 1"
assume "n\<noteq>2"
hence "2 \<le> ?m"
using assms by arith
then obtain i where i: "2 ^ i < ?m" "?m \<le> 2 ^ (i + 1)"
using ex_power_ivl2[of 2 ?m] by auto
have "n \<le> 2*?m"
by arith
also have "2*?m \<le> 2 ^ ((i+1)+1)"
using i(2) by simp
finally have *: "n \<le> \<dots>" .
have "2^(i+1) < n"
using i(1) by (auto simp add: less_Suc_eq_0_disj)
from ceil_log_nat_ivl[OF _ this *] ceil_log_nat_ivl[OF _ i]
show ?thesis by simp
qed
lemma floor_log2_div2: fixes n :: nat assumes "n \<ge> 2"
shows "floor(log 2 n) = floor(log 2 (n div 2)) + 1"
proof cases
assume "n=2"
thus ?thesis by simp
next
let ?m = "n div 2"
assume "n\<noteq>2"
hence "1 \<le> ?m"
using assms by arith
then obtain i where i: "2 ^ i \<le> ?m" "?m < 2 ^ (i + 1)"
using ex_power_ivl1[of 2 ?m] by auto
have "2^(i+1) \<le> 2*?m"
using i(1) by simp
also have "2*?m \<le> n"
by arith
finally have *: "2^(i+1) \<le> \<dots>" .
have "n < 2^(i+1+1)"
using i(2) by simp
from floor_log_nat_ivl[OF _ * this] floor_log_nat_ivl[OF _ i]
show ?thesis by simp
qed
lemma balanced_Node_if_wbal1:
assumes "balanced l" "balanced r" "size l = size r + 1"
shows "balanced \<langle>l, x, r\<rangle>"
proof -
from assms(3) have [simp]: "size1 l = size1 r + 1" by(simp add: size1_def)
have "nat \<lceil>log 2 (1 + size1 r)\<rceil> \<ge> nat \<lceil>log 2 (size1 r)\<rceil>"
by(rule nat_mono[OF ceiling_mono]) simp
hence 1: "height(Node l x r) = nat \<lceil>log 2 (1 + size1 r)\<rceil> + 1"
using height_balanced[OF assms(1)] height_balanced[OF assms(2)]
by (simp del: nat_ceiling_le_eq add: max_def)
have "nat \<lfloor>log 2 (1 + size1 r)\<rfloor> \<ge> nat \<lfloor>log 2 (size1 r)\<rfloor>"
by(rule nat_mono[OF floor_mono]) simp
hence 2: "min_height(Node l x r) = nat \<lfloor>log 2 (size1 r)\<rfloor> + 1"
using min_height_balanced[OF assms(1)] min_height_balanced[OF assms(2)]
by (simp)
have "size1 r \<ge> 1" by(simp add: size1_def)
then obtain i where i: "2 ^ i \<le> size1 r" "size1 r < 2 ^ (i + 1)"
using ex_power_ivl1[of 2 "size1 r"] by auto
hence i1: "2 ^ i < size1 r + 1" "size1 r + 1 \<le> 2 ^ (i + 1)" by auto
from 1 2 floor_log_nat_ivl[OF _ i] ceil_log_nat_ivl[OF _ i1]
show ?thesis by(simp add:balanced_def)
qed
lemma balanced_sym: "balanced \<langle>l, x, r\<rangle> \<Longrightarrow> balanced \<langle>r, y, l\<rangle>"
by(auto simp: balanced_def)
lemma balanced_Node_if_wbal2:
assumes "balanced l" "balanced r" "abs(int(size l) - int(size r)) \<le> 1"
shows "balanced \<langle>l, x, r\<rangle>"
proof -
have "size l = size r \<or> (size l = size r + 1 \<or> size r = size l + 1)" (is "?A \<or> ?B")
using assms(3) by linarith
thus ?thesis
proof
assume "?A"
thus ?thesis using assms(1,2)
apply(simp add: balanced_def min_def max_def)
by (metis assms(1,2) balanced_optimal le_antisym le_less)
next
assume "?B"
thus ?thesis
by (meson assms(1,2) balanced_sym balanced_Node_if_wbal1)
qed
qed
lemma balanced_if_wbalanced: "wbalanced t \<Longrightarrow> balanced t"
proof(induction t)
case Leaf show ?case by (simp add: balanced_def)
next
case (Node l x r)
thus ?case by(simp add: balanced_Node_if_wbal2)
qed
(* end of mv *)
fun bal :: "nat \<Rightarrow> 'a list \<Rightarrow> '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 \<Rightarrow> 'a list \<Rightarrow> 'a tree" where
"bal_list n xs = fst (bal n xs)"
definition balance_list :: "'a list \<Rightarrow> 'a tree" where
"balance_list xs = bal_list (length xs) xs"
definition bal_tree :: "nat \<Rightarrow> 'a tree \<Rightarrow> 'a tree" where
"bal_tree n t = bal_list n (inorder t)"
definition balance_tree :: "'a tree \<Rightarrow> 'a tree" where
"balance_tree t = bal_tree (size t) t"
lemma bal_simps:
"bal 0 xs = (Leaf, xs)"
"n > 0 \<Longrightarrow>
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\<open>Some of the following lemmas take advantage of the fact
that \<open>bal xs n\<close> yields a result even if \<open>n > length xs\<close>.\<close>
lemma size_bal: "bal n xs = (t,ys) \<Longrightarrow> 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:
"\<lbrakk> bal n xs = (t,ys); n \<le> length xs \<rbrakk>
\<Longrightarrow> inorder t = take n xs \<and> 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 \<noteq> 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 = \<langle>l, hd xs', r\<rangle>"
by(auto simp: Let_def bal_simps split: prod.splits)
have IH1: "inorder l = take ?n1 xs \<and> xs' = drop ?n1 xs"
using b1 "1.prems" by(intro "1.IH"(1)) auto
have IH2: "inorder r = take ?n2 (tl xs') \<and> ys = drop ?n2 (tl xs')"
using b1 b2 IH1 "1.prems" by(intro "1.IH"(2)) auto
have "drop (n div 2) xs \<noteq> []" 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 \<le> length xs \<Longrightarrow> 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 \<le> size t \<Longrightarrow> 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) \<Longrightarrow> min_height t = nat(floor(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 \<noteq> 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 = \<langle>l, hd xs', r\<rangle>"
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 \<ge> 1" by arith
hence 0: "log 2 ((n+1) div 2) \<ge> 0" by simp
have "n - 1 - n div 2 + 1 \<le> n div 2 + 1" by arith
hence le: "?log2 \<le> ?log1"
by(simp add: nat_mono floor_mono)
have "min_height t = min ?log1 ?log2 + 1" by (simp add: t IH1 IH2)
also have "\<dots> = ?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 "\<dots> = 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) \<Longrightarrow> height t = nat \<lceil>log 2 (n + 1)\<rceil>"
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 \<noteq> 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 = \<langle>l, hd xs', r\<rangle>"
by(auto simp: bal_simps Let_def split: prod.splits)
let ?log1 = "nat \<lceil>log 2 (n div 2 + 1)\<rceil>"
let ?log2 = "nat \<lceil>log 2 (n - 1 - n div 2 + 1)\<rceil>"
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) \<ge> 0" by auto
have "n - 1 - n div 2 + 1 \<le> n div 2 + 1" by arith
hence le: "?log2 \<le> ?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 "\<dots> = ?log1 + 1" using le by (simp add: max_absorb1)
also have "\<dots> = nat \<lceil>log 2 (n div 2 + 1) + 1\<rceil>" using 0 by linarith
also have "\<dots> = nat \<lceil>log 2 (n + 1)\<rceil>"
using ceil_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 \<le> length xs \<Longrightarrow> height (bal_list n xs) = nat \<lceil>log 2 (n + 1)\<rceil>"
unfolding bal_list_def by (metis height_bal prod.collapse)
lemma height_balance_list:
"height (balance_list xs) = nat \<lceil>log 2 (length xs + 1)\<rceil>"
by (simp add: balance_list_def height_bal_list)
corollary height_bal_tree:
"n \<le> length xs \<Longrightarrow> height (bal_tree n t) = nat(ceiling(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(ceiling(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) \<Longrightarrow> 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 \<noteq> 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 = \<langle>l, hd ys, r\<rangle>"
by(auto simp add: bal_simps Let_def split: prod.splits)
have l: "wbalanced l" using "1.IH"(1)[OF \<open>n\<noteq>0\<close> refl rec1] .
have "wbalanced r" using "1.IH"(2)[OF \<open>n\<noteq>0\<close> refl rec1[symmetric] refl rec2] .
with l t size_bal[OF rec1] size_bal[OF rec2]
show ?thesis by auto
qed
qed
text\<open>An alternative proof via @{thm balanced_if_wbalanced}:\<close>
lemma "bal n xs = (t,ys) \<Longrightarrow> 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