added lemmas image_mset_eq_{image_mset_plus,plus,plus_image_mset}D, and multp_image_mset_image_msetD
(*
File: Power_By_Squaring.thy
Author: Manuel Eberl, TU München
Fast computing of funpow (applying some functon n times) for weakly associative binary
functions using exponentiation by squaring. Yields efficient exponentiation algorithms on
monoid_mult and for modular exponentiation "b ^ e mod m" (and thus also for "cong")
*)
section \<open>Exponentiation by Squaring\<close>
theory Power_By_Squaring
imports Main
begin
context
fixes f :: "'a \<Rightarrow> 'a \<Rightarrow> 'a"
begin
function efficient_funpow :: "'a \<Rightarrow> 'a \<Rightarrow> nat \<Rightarrow> 'a" where
"efficient_funpow y x 0 = y"
| "efficient_funpow y x (Suc 0) = f x y"
| "n \<noteq> 0 \<Longrightarrow> even n \<Longrightarrow> efficient_funpow y x n = efficient_funpow y (f x x) (n div 2)"
| "n \<noteq> 1 \<Longrightarrow> odd n \<Longrightarrow> efficient_funpow y x n = efficient_funpow (f x y) (f x x) (n div 2)"
by force+
termination by (relation "measure (snd \<circ> snd)") (auto elim: oddE)
lemma efficient_funpow_code [code]:
"efficient_funpow y x n =
(if n = 0 then y
else if n = 1 then f x y
else if even n then efficient_funpow y (f x x) (n div 2)
else efficient_funpow (f x y) (f x x) (n div 2))"
by (induction y x n rule: efficient_funpow.induct) auto
end
lemma efficient_funpow_correct:
assumes f_assoc: "\<And>x z. f x (f x z) = f (f x x) z"
shows "efficient_funpow f y x n = (f x ^^ n) y"
proof -
have [simp]: "f ^^ 2 = (\<lambda>x. f (f x))" for f :: "'a \<Rightarrow> 'a"
by (simp add: eval_nat_numeral o_def)
show ?thesis
by (induction y x n rule: efficient_funpow.induct[of _ f])
(auto elim!: evenE oddE simp: funpow_mult [symmetric] funpow_Suc_right f_assoc
simp del: funpow.simps(2))
qed
(*
TODO: This could be used as a code_unfold rule or something like that but the
implications are not quite clear. Would this be a good default implementation
for powers?
*)
context monoid_mult
begin
lemma power_by_squaring: "efficient_funpow (*) (1 :: 'a) = (^)"
proof (intro ext)
fix x :: 'a and n
have "efficient_funpow (*) 1 x n = ((*) x ^^ n) 1"
by (subst efficient_funpow_correct) (simp_all add: mult.assoc)
also have "\<dots> = x ^ n"
by (induction n) simp_all
finally show "efficient_funpow (*) 1 x n = x ^ n" .
qed
end
end