(* Author: Florian Haftmann, TU Muenchen *)
section ‹Common discrete functions
›
theory Discrete_Functions
imports Complex_Main
begin
subsection ‹Discrete logarithm
›
fun floor_log ::
"nat \ nat"
where [simp del]:
"floor_log n = (if n < 2 then 0 else Suc (floor_log (n div 2)))"
lemma floor_log_induct [consumes 1, case_names one double]:
fixes n :: nat
assumes "n > 0"
assumes one:
"P 1"
assumes double:
"\n. n \ 2 \ P (n div 2) \ P n"
shows "P n"
using ‹n > 0
› proof (induct n rule: floor_log.induct)
fix n
assume "\ n < 2 \
0 < n div 2
==> P (n div 2)
"
then have *:
"n \ 2 \ P (n div 2)" by simp
assume "n > 0"
show "P n"
proof (cases
"n = 1")
case True
with one
show ?thesis
by simp
next
case False
with ‹n > 0
› have "n \ 2" by auto
with *
have "P (n div 2)" .
with ‹n
≥ 2
› show ?thesis
by (rule double)
qed
qed
lemma floor_log_zero [simp]:
"floor_log 0 = 0"
by (simp add: floor_log.simps)
lemma floor_log_one [simp]:
"floor_log 1 = 0"
by (simp add: floor_log.simps)
lemma floor_log_Suc_zero [simp]:
"floor_log (Suc 0) = 0"
using floor_log_one
by simp
lemma floor_log_rec:
"n \ 2 \ floor_log n = Suc (floor_log (n div 2))"
by (simp add: floor_log.simps)
lemma floor_log_twice [simp]:
"n \ 0 \ floor_log (2 * n) = Suc (floor_log n)"
by (simp add: floor_log_rec)
lemma floor_log_half [simp]:
"floor_log (n div 2) = floor_log n - 1"
proof (cases
"n < 2")
case True
then have "n = 0 \ n = 1" by arith
then show ?thesis
by (auto simp del: One_nat_def)
next
case False
then show ?thesis
by (simp add: floor_log_rec)
qed
lemma floor_log_power [simp]:
"floor_log (2 ^ n) = n"
by (induct n) simp_all
lemma floor_log_mono:
"mono floor_log"
proof
fix m n :: nat
assume "m \ n"
then show "floor_log m \ floor_log n"
proof (induct m arbitrary: n rule: floor_log.induct)
case (1 m)
then have mn2:
"m div 2 \ n div 2" by arith
show "floor_log m \ floor_log n"
proof (cases
"m \ 2")
case False
then have "m = 0 \ m = 1" by arith
then show ?thesis
by (auto simp del: One_nat_def)
next
case True
then have "\ m < 2" by simp
with mn2
have "n \ 2" by arith
from True
have m2_0:
"m div 2 \ 0" by arith
with mn2
have n2_0:
"n div 2 \ 0" by arith
from ‹¬ m < 2
› "1.hyps" mn2
have "floor_log (m div 2) \ floor_log (n div 2)" by blast
with m2_0 n2_0
have "floor_log (2 * (m div 2)) \ floor_log (2 * (n div 2))" by simp
with m2_0 n2_0
‹m
≥ 2
› ‹n
≥ 2
› show ?thesis
by (simp only: floor_log_rec [of m] floor_log_rec [of n
]) simp
qed
qed
qed
lemma floor_log_exp2_le:
assumes "n > 0"
shows "2 ^ floor_log n \ n"
using assms
proof (induct n rule: floor_log_induct)
case one
then show ?case by simp
next
case (double n)
with floor_log_mono have "floor_log n \ Suc 0"
by (simp add: floor_log.simps)
assume "2 ^ floor_log (n div 2) \ n div 2"
with ‹n ≥ 2› have "2 ^ (floor_log n - Suc 0) \ n div 2" by simp
then have "2 ^ (floor_log n - Suc 0) * 2 ^ 1 \ n div 2 * 2" by simp
with ‹floor_log n ≥ Suc 0› have "2 ^ floor_log n \ n div 2 * 2"
unfolding power_add [symmetric] by simp
also have "n div 2 * 2 \ n" by (cases "even n") simp_all
finally show ?case .
qed
lemma floor_log_exp2_gt: "2 * 2 ^ floor_log n > n"
proof (cases "n > 0")
case True
thus ?thesis
proof (induct n rule: floor_log_induct)
case (double n)
thus ?case
by (cases "even n") (auto elim!: evenE oddE simp: field_simps floor_log.simps)
qed simp_all
qed simp_all
lemma floor_log_exp2_ge: "2 * 2 ^ floor_log n \ n"
using floor_log_exp2_gt[of n] by simp
lemma floor_log_le_iff: "m \ n \ floor_log m \ floor_log n"
by (rule monoD [OF floor_log_mono])
lemma floor_log_eqI:
assumes "n > 0" "2^k \ n" "n < 2 * 2^k"
shows "floor_log n = k"
proof (rule antisym)
from ‹n > 0› have "2 ^ floor_log n \ n" by (rule floor_log_exp2_le)
also have "\ < 2 ^ Suc k" using assms by simp
finally have "floor_log n < Suc k" by (subst (asm) power_strict_increasing_iff) simp_all
thus "floor_log n \ k" by simp
next
have "2^k \ n" by fact
also have "\ < 2^(Suc (floor_log n))" by (simp add: floor_log_exp2_gt)
finally have "k < Suc (floor_log n)" by (subst (asm) power_strict_increasing_iff) simp_all
thus "k \ floor_log n" by simp
qed
lemma floor_log_altdef: "floor_log n = (if n = 0 then 0 else nat \log 2 (real_of_nat n)\)"
proof (cases "n = 0")
case False
have "\log 2 (real_of_nat n)\ = int (floor_log n)"
proof (rule floor_unique)
from False have "2 powr (real (floor_log n)) \ real n"
by (simp add: powr_realpow floor_log_exp2_le)
hence "log 2 (2 powr (real (floor_log n))) \ log 2 (real n)"
using False by (subst log_le_cancel_iff) simp_all
also have "log 2 (2 powr (real (floor_log n))) = real (floor_log n)" by simp
finally show "real_of_int (int (floor_log n)) \ log 2 (real n)" by simp
next
have "real n < real (2 * 2 ^ floor_log n)"
by (subst of_nat_less_iff) (rule floor_log_exp2_gt)
also have "\ = 2 powr (real (floor_log n) + 1)"
by (simp add: powr_add powr_realpow)
finally have "log 2 (real n) < log 2 \"
using False by (subst log_less_cancel_iff) simp_all
also have "\ = real (floor_log n) + 1" by simp
finally show "log 2 (real n) < real_of_int (int (floor_log n)) + 1" by simp
qed
thus ?thesis by simp
qed simp_all
subsection ‹Discrete square root›
definition floor_sqrt :: "nat \ nat"
where "floor_sqrt n = Max {m. m\<^sup>2 \ n}"
lemma floor_sqrt_aux:
fixes n :: nat
shows "finite {m. m\<^sup>2 \ n}" and "{m. m\<^sup>2 \ n} \ {}"
proof -
have **: "m \ n" if "m\<^sup>2 \ n" for m
using that by (cases m) (simp_all add: power2_eq_square)
then have "{m. m\<^sup>2 \ n} \ {m. m \ n}" by auto
then show "finite {m. m\<^sup>2 \ n}" by (rule finite_subset) rule
have "0\<^sup>2 \ n" by simp
then show *: "{m. m\<^sup>2 \ n} \ {}" by blast
qed
lemma floor_sqrt_unique:
assumes "m^2 \ n" "n < (Suc m)^2"
shows "floor_sqrt n = m"
proof -
have "m' \ m" if "m'^2 \ n" for m'
proof -
note that
also note assms(2)
finally have "m' < Suc m" by (rule power_less_imp_less_base) simp_all
thus "m' \ m" by simp
qed
with ‹m^2 ≤ n› floor_sqrt_aux[of n] show ?thesis unfolding floor_sqrt_def
by (intro antisym Max.boundedI Max.coboundedI) simp_all
qed
lemma floor_sqrt_inverse_power2 [simp]: "floor_sqrt (n\<^sup>2) = n"
proof -
have "{m. m \ n} \ {}" by auto
then have "Max {m. m \ n} \ n" by auto
then show ?thesis
by (auto simp add: floor_sqrt_def power2_nat_le_eq_le intro: antisym)
qed
lemma floor_sqrt_zero [simp]: "floor_sqrt 0 = 0"
using floor_sqrt_inverse_power2 [of 0] by simp
lemma floor_sqrt_one [simp]: "floor_sqrt 1 = 1"
using floor_sqrt_inverse_power2 [of 1] by simp
lemma floor_sqrt_Suc_0 [simp]:
‹floor_sqrt (Suc 0) = 1›
using floor_sqrt_inverse_power2 [of 1] by simp
lemma mono_floor_sqrt: "mono floor_sqrt"
proof
fix m n :: nat
have *: "0 * 0 \ m" by simp
assume "m \ n"
then show "floor_sqrt m \ floor_sqrt n"
by (auto intro!: Max_mono ‹0 * 0 ≤ m› finite_less_ub simp add: power2_eq_square floor_sqrt_def)
qed
lemma mono_floor_sqrt': "m \ n \ floor_sqrt m \ floor_sqrt n"
using mono_floor_sqrt unfolding mono_def by auto
lemma floor_sqrt_greater_zero_iff [simp]: "floor_sqrt n > 0 \ n > 0"
proof -
have *: "0 < Max {m. m\<^sup>2 \ n} \ (\a\{m. m\<^sup>2 \ n}. 0 < a)"
by (rule Max_gr_iff) (fact floor_sqrt_aux)+
show ?thesis
proof
assume "0 < floor_sqrt n"
then have "0 < Max {m. m\<^sup>2 \ n}" by (simp add: floor_sqrt_def)
with * show "0 < n" by (auto dest: power2_nat_le_imp_le)
next
assume "0 < n"
then have "1\<^sup>2 \ n \ 0 < (1::nat)" by simp
then have "\q. q\<^sup>2 \ n \ 0 < q" ..
with * have "0 < Max {m. m\<^sup>2 \ n}" by blast
then show "0 < floor_sqrt n" by (simp add: floor_sqrt_def)
qed
qed
lemma floor_sqrt_power2_le [simp]: "(floor_sqrt n)\<^sup>2 \ n" (* FIXME tune proof *)
proof (cases "n > 0")
case False then show ?thesis by simp
next
case True then have "floor_sqrt n > 0" by simp
then have "mono (times (Max {m. m\<^sup>2 \ n}))" by (auto intro: mono_times_nat simp add: floor_sqrt_def)
then have *: "Max {m. m\<^sup>2 \ n} * Max {m. m\<^sup>2 \ n} = Max (times (Max {m. m\<^sup>2 \ n}) ` {m. m\<^sup>2 \ n})"
using floor_sqrt_aux [of n] by (rule mono_Max_commute)
have "\a. a * a \ n \ Max {m. m * m \ n} * a \ n"
proof -
fix q
assume "q * q \ n"
show "Max {m. m * m \ n} * q \ n"
proof (cases "q > 0")
case False then show ?thesis by simp
next
case True then have "mono (times q)" by (rule mono_times_nat)
then have "q * Max {m. m * m \ n} = Max (times q ` {m. m * m \ n})"
using floor_sqrt_aux [of n] by (auto simp add: power2_eq_square intro: mono_Max_commute)
then have "Max {m. m * m \ n} * q = Max (times q ` {m. m * m \ n})" by (simp add: ac_simps)
moreover have "finite ((*) q ` {m. m * m \ n})"
by (metis (mono_tags) finite_imageI finite_less_ub le_square)
moreover have "\x. x * x \ n"
by (metis ‹q * q ≤ n›)
ultimately show ?thesis
by simp (metis ‹q * q ≤ n› le_cases mult_le_mono1 mult_le_mono2 order_trans)
qed
qed
then have "Max ((*) (Max {m. m * m \ n}) ` {m. m * m \ n}) \ n"
apply (subst Max_le_iff)
apply (metis (mono_tags) finite_imageI finite_less_ub le_square)
apply auto
apply (metis le0 mult_0_right)
done
with * show ?thesis by (simp add: floor_sqrt_def power2_eq_square)
qed
lemma floor_sqrt_le: "floor_sqrt n \ n"
using floor_sqrt_aux [of n] by (auto simp add: floor_sqrt_def intro: power2_nat_le_imp_le)
text ‹Additional facts about the discrete square root, thanks to Julian Biendarra, Manuel Eberl›
lemma Suc_floor_sqrt_power2_gt: "n < (Suc (floor_sqrt n))^2"
using Max_ge[OF floor_sqrt_aux(1), of "floor_sqrt n + 1" n]
by (cases "n < (Suc (floor_sqrt n))^2") (simp_all add: floor_sqrt_def)
lemma le_floor_sqrt_iff: "x \ floor_sqrt y \ x^2 \ y"
proof -
have "x \ floor_sqrt y \ (\z. z\<^sup>2 \ y \ x \ z)"
using Max_ge_iff[OF floor_sqrt_aux, of x y] by (simp add: floor_sqrt_def)
also have "\ \ x^2 \ y"
proof safe
fix z assume "x \ z" "z ^ 2 \ y"
thus "x^2 \ y" by (intro le_trans[of "x^2" "z^2" y]) (simp_all add: power2_nat_le_eq_le)
qed auto
finally show ?thesis .
qed
lemma le_floor_sqrtI: "x^2 \ y \ x \ floor_sqrt y"
by (simp add: le_floor_sqrt_iff)
lemma floor_sqrt_le_iff:
‹floor_sqrt y ≤ x ⟷ (∀z. z🚫2 ≤ y ⟶ z ≤ x)›
using Max.bounded_iff [OF floor_sqrt_aux]
by (simp add: floor_sqrt_def)
lemma floor_sqrt_leI:
"(\z. z^2 \ y \ z \ x) \ floor_sqrt y \ x"
by (simp add: floor_sqrt_le_iff)
lemma floor_sqrt_less_eq_half:
‹floor_sqrt n ≤ Suc n div 2›
proof (rule floor_sqrt_leI)
fix m
assume ‹m🚫2 ≤ n›
have ‹m < Suc (Suc n div 2)›
proof (rule ccontr, unfold not_less)
assume ‹Suc (Suc n div 2) ≤ m›
then have ‹(Suc (Suc n div 2))🚫2 ≤ m🚫2›
by simp
then have ‹(Suc (Suc n div 2))🚫2 ≤ n›
using ‹m🚫2 ≤ n› by (rule order_trans)
then show False
by (simp only: Suc_eq_plus1 power2_sum algebra_simps) auto
qed
then show ‹m ≤ Suc n div 2›
by simp
qed
lemma floor_sqrt_Suc:
"floor_sqrt (Suc n) = (if \m. Suc n = m^2 then Suc (floor_sqrt n) else floor_sqrt n)"
proof cases
assume "\ m. Suc n = m^2"
then obtain m where m_def: "Suc n = m^2" by blast
then have lhs: "floor_sqrt (Suc n) = m" by simp
from m_def floor_sqrt_power2_le[of n]
have "(floor_sqrt n)^2 < m^2" by linarith
with power2_less_imp_less have lt_m: "floor_sqrt n < m" by blast
from m_def Suc_floor_sqrt_power2_gt[of "n"]
have "m^2 \ (Suc(floor_sqrt n))^2"
by linarith
with power2_nat_le_eq_le have "m \ Suc (floor_sqrt n)" by blast
with lt_m have "m = Suc (floor_sqrt n)" by simp
with lhs m_def show ?thesis by fastforce
next
assume asm: "\ (\ m. Suc n = m^2)"
hence "Suc n \ (floor_sqrt (Suc n))^2" by simp
with floor_sqrt_power2_le[of "Suc n"]
have "floor_sqrt (Suc n) \ floor_sqrt n" by (intro le_floor_sqrtI) linarith
moreover have "floor_sqrt (Suc n) \ floor_sqrt n"
by (intro monoD[OF mono_floor_sqrt]) simp_all
ultimately show ?thesis using asm by simp
qed
text ‹Computation by divide and conquer›
definition floor_sqrt_between :: ‹nat ==> nat ==> nat ==> nat›
where floor_sqrt_between_eq:
‹floor_sqrt_between m q n =
(if floor_sqrt n ∈ {m..<m + q} then floor_sqrt n else 0)›
🍋‹
The 🍋‹0::nat› is not for relevant regular computation
and can be chosen arbitrarily.
›
lemma floor_sqrt_between_out_of_bounds:
‹floor_sqrt_between m 0 n = 0›
by (simp add: floor_sqrt_between_eq)
lemma floor_sqrt_between_singleton:
‹floor_sqrt_between m (Suc 0) n =
(if m🚫2 ≤ n ∧ n < (Suc m)🚫2 then m else 0)›
by (auto simp add: floor_sqrt_between_eq Suc_floor_sqrt_power2_gt floor_sqrt_unique)
lemma floor_sqrt_between_rec:
‹floor_sqrt_between m q n = (
let
r = q div 2;
p = m + r;
s = p🚫2
in
if s = n
then p
else if s < n
then floor_sqrt_between (m + r) (q - r) n
else floor_sqrt_between m r n
)› if ‹q > 0›
using that le_floor_sqrt_iff [of ‹m + q div 2› n]
by (auto simp add: floor_sqrt_between_eq Let_def not_less)
lemma floor_sqrt_between_code [code]:
‹floor_sqrt_between m q n = (
if q = 0 then 0
else if q = 1
then if m🚫2 ≤ n ∧ n < (Suc m)🚫2
then m
else 0
else
let
r = q div 2;
p = m + r;
s = p🚫2
in
if s = n
then p
else if s < n
then floor_sqrt_between (m + r) (q - r) n
else floor_sqrt_between m r n
)›
proof -
consider ‹q = 0› | ‹q = 1› | ‹q ≥ 2›
by (cases ‹q ≥ 2›; cases q) simp_all
then show ?thesis
by cases
(simp_all add: floor_sqrt_between_out_of_bounds floor_sqrt_between_singleton floor_sqrt_between_rec)
qed
lemma [code]:
‹floor_sqrt n = floor_sqrt_between 0 (Suc (Suc n div 2)) n›
using floor_sqrt_less_eq_half [of n] by (simp add: floor_sqrt_between_eq)
end