--- a/src/HOL/Library/Library.thy Sun May 01 17:26:27 2016 +0200
+++ b/src/HOL/Library/Library.thy Wed May 04 10:19:01 2016 +0200
@@ -73,6 +73,7 @@
Saturated
Set_Algebras
State_Monad
+ Stirling
Stream
Sublist
Sum_of_Squares
--- /dev/null Thu Jan 01 00:00:00 1970 +0000
+++ b/src/HOL/Library/Stirling.thy Wed May 04 10:19:01 2016 +0200
@@ -0,0 +1,281 @@
+(* Authors: Amine Chaieb & Florian Haftmann, TU Muenchen
+ with contributions by Lukas Bulwahn and Manuel Eberl*)
+
+section {* Stirling numbers of first and second kind *}
+
+theory Stirling
+imports Binomial
+begin
+
+subsection {* Stirling numbers of the second kind *}
+
+fun Stirling :: "nat \<Rightarrow> nat \<Rightarrow> nat"
+where
+ "Stirling 0 0 = 1"
+| "Stirling 0 (Suc k) = 0"
+| "Stirling (Suc n) 0 = 0"
+| "Stirling (Suc n) (Suc k) = Suc k * Stirling n (Suc k) + Stirling n k"
+
+lemma Stirling_1 [simp]:
+ "Stirling (Suc n) (Suc 0) = 1"
+ by (induct n) simp_all
+
+lemma Stirling_less [simp]:
+ "n < k \<Longrightarrow> Stirling n k = 0"
+ by (induct n k rule: Stirling.induct) simp_all
+
+lemma Stirling_same [simp]:
+ "Stirling n n = 1"
+ by (induct n) simp_all
+
+lemma Stirling_2_2:
+ "Stirling (Suc (Suc n)) (Suc (Suc 0)) = 2 ^ Suc n - 1"
+proof (induct n)
+ case 0 then show ?case by simp
+next
+ case (Suc n)
+ have "Stirling (Suc (Suc (Suc n))) (Suc (Suc 0)) =
+ 2 * Stirling (Suc (Suc n)) (Suc (Suc 0)) + Stirling (Suc (Suc n)) (Suc 0)" by simp
+ also have "\<dots> = 2 * (2 ^ Suc n - 1) + 1"
+ by (simp only: Suc Stirling_1)
+ also have "\<dots> = 2 ^ Suc (Suc n) - 1"
+ proof -
+ have "(2::nat) ^ Suc n - 1 > 0" by (induct n) simp_all
+ then have "2 * ((2::nat) ^ Suc n - 1) > 0" by simp
+ then have "2 \<le> 2 * ((2::nat) ^ Suc n)" by simp
+ with add_diff_assoc2 [of 2 "2 * 2 ^ Suc n" 1]
+ have "2 * 2 ^ Suc n - 2 + (1::nat) = 2 * 2 ^ Suc n + 1 - 2" .
+ then show ?thesis by (simp add: nat_distrib)
+ qed
+ finally show ?case by simp
+qed
+
+lemma Stirling_2:
+ "Stirling (Suc n) (Suc (Suc 0)) = 2 ^ n - 1"
+ using Stirling_2_2 by (cases n) simp_all
+
+subsection {* Stirling numbers of the first kind *}
+
+fun stirling :: "nat \<Rightarrow> nat \<Rightarrow> nat"
+where
+ "stirling 0 0 = 1"
+| "stirling 0 (Suc k) = 0"
+| "stirling (Suc n) 0 = 0"
+| "stirling (Suc n) (Suc k) = n * stirling n (Suc k) + stirling n k"
+
+lemma stirling_0 [simp]: "n > 0 \<Longrightarrow> stirling n 0 = 0"
+ by (cases n) simp_all
+
+lemma stirling_less [simp]:
+ "n < k \<Longrightarrow> stirling n k = 0"
+ by (induct n k rule: stirling.induct) simp_all
+
+lemma stirling_same [simp]:
+ "stirling n n = 1"
+ by (induct n) simp_all
+
+lemma stirling_Suc_n_1:
+ "stirling (Suc n) (Suc 0) = fact n"
+ by (induct n) auto
+
+lemma stirling_Suc_n_n:
+ shows "stirling (Suc n) n = Suc n choose 2"
+by (induct n) (auto simp add: numerals(2))
+
+lemma stirling_Suc_n_2:
+ assumes "n \<ge> Suc 0"
+ shows "stirling (Suc n) 2 = (\<Sum>k=1..n. fact n div k)"
+using assms
+proof (induct n)
+ case 0 from this show ?case by simp
+next
+ case (Suc n)
+ show ?case
+ proof (cases n)
+ case 0 from this show ?thesis by (simp add: numerals(2))
+ next
+ case Suc
+ from this have geq1: "Suc 0 \<le> n" by simp
+ have "stirling (Suc (Suc n)) 2 = Suc n * stirling (Suc n) 2 + stirling (Suc n) (Suc 0)"
+ by (simp only: stirling.simps(4)[of "Suc n"] numerals(2))
+ also have "... = Suc n * (\<Sum>k=1..n. fact n div k) + fact n"
+ using Suc.hyps[OF geq1]
+ by (simp only: stirling_Suc_n_1 of_nat_fact of_nat_add of_nat_mult)
+ also have "... = Suc n * (\<Sum>k=1..n. fact n div k) + Suc n * fact n div Suc n"
+ by (metis nat.distinct(1) nonzero_mult_divide_cancel_left)
+ also have "... = (\<Sum>k=1..n. fact (Suc n) div k) + fact (Suc n) div Suc n"
+ by (simp add: setsum_right_distrib div_mult_swap dvd_fact)
+ also have "... = (\<Sum>k=1..Suc n. fact (Suc n) div k)" by simp
+ finally show ?thesis .
+ qed
+qed
+
+lemma of_nat_stirling_Suc_n_2:
+ assumes "n \<ge> Suc 0"
+ shows "(of_nat (stirling (Suc n) 2)::'a::field_char_0) = fact n * (\<Sum>k=1..n. (1 / of_nat k))"
+using assms
+proof (induct n)
+ case 0 from this show ?case by simp
+next
+ case (Suc n)
+ show ?case
+ proof (cases n)
+ case 0 from this show ?thesis by (auto simp add: numerals(2))
+ next
+ case Suc
+ from this have geq1: "Suc 0 \<le> n" by simp
+ have "(of_nat (stirling (Suc (Suc n)) 2)::'a) =
+ of_nat (Suc n * stirling (Suc n) 2 + stirling (Suc n) (Suc 0))"
+ by (simp only: stirling.simps(4)[of "Suc n"] numerals(2))
+ also have "... = of_nat (Suc n) * (fact n * (\<Sum>k = 1..n. 1 / of_nat k)) + fact n"
+ using Suc.hyps[OF geq1]
+ by (simp only: stirling_Suc_n_1 of_nat_fact of_nat_add of_nat_mult)
+ also have "... = fact (Suc n) * (\<Sum>k = 1..n. 1 / of_nat k) + fact (Suc n) * (1 / of_nat (Suc n))"
+ using of_nat_neq_0 by auto
+ also have "... = fact (Suc n) * (\<Sum>k = 1..Suc n. 1 / of_nat k)"
+ by (simp add: distrib_left)
+ finally show ?thesis .
+ qed
+qed
+
+lemma setsum_stirling:
+ "(\<Sum>k\<le>n. stirling n k) = fact n"
+proof (induct n)
+ case 0
+ from this show ?case by simp
+next
+ case (Suc n)
+ have "(\<Sum>k\<le>Suc n. stirling (Suc n) k) = stirling (Suc n) 0 + (\<Sum>k\<le>n. stirling (Suc n) (Suc k))"
+ by (simp only: setsum_atMost_Suc_shift)
+ also have "\<dots> = (\<Sum>k\<le>n. stirling (Suc n) (Suc k))" by simp
+ also have "\<dots> = (\<Sum>k\<le>n. n * stirling n (Suc k) + stirling n k)" by simp
+ also have "\<dots> = n * (\<Sum>k\<le>n. stirling n (Suc k)) + (\<Sum>k\<le>n. stirling n k)"
+ by (simp add: setsum.distrib setsum_right_distrib)
+ also have "\<dots> = n * fact n + fact n"
+ proof -
+ have "n * (\<Sum>k\<le>n. stirling n (Suc k)) = n * ((\<Sum>k\<le>Suc n. stirling n k) - stirling n 0)"
+ by (metis add_diff_cancel_left' setsum_atMost_Suc_shift)
+ also have "\<dots> = n * (\<Sum>k\<le>n. stirling n k)" by (cases n) simp+
+ also have "\<dots> = n * fact n" using Suc.hyps by simp
+ finally have "n * (\<Sum>k\<le>n. stirling n (Suc k)) = n * fact n" .
+ moreover have "(\<Sum>k\<le>n. stirling n k) = fact n" using Suc.hyps .
+ ultimately show ?thesis by simp
+ qed
+ also have "\<dots> = fact (Suc n)" by simp
+ finally show ?case .
+qed
+
+lemma stirling_pochhammer:
+ "(\<Sum>k\<le>n. of_nat (stirling n k) * x ^ k) = (pochhammer x n :: 'a :: comm_semiring_1)"
+proof (induction n)
+ case (Suc n)
+ have "of_nat (n * stirling n 0) = (0 :: 'a)" by (cases n) simp_all
+ hence "(\<Sum>k\<le>Suc n. of_nat (stirling (Suc n) k) * x ^ k) =
+ (of_nat (n * stirling n 0) * x ^ 0 +
+ (\<Sum>i\<le>n. of_nat (n * stirling n (Suc i)) * (x ^ Suc i))) +
+ (\<Sum>i\<le>n. of_nat (stirling n i) * (x ^ Suc i))"
+ by (subst setsum_atMost_Suc_shift) (simp add: setsum.distrib ring_distribs)
+ also have "\<dots> = pochhammer x (Suc n)"
+ by (subst setsum_atMost_Suc_shift [symmetric])
+ (simp add: algebra_simps setsum.distrib setsum_right_distrib pochhammer_Suc Suc [symmetric])
+ finally show ?case .
+qed simp_all
+
+
+text \<open>A row of the Stirling number triangle\<close>
+
+definition stirling_row :: "nat \<Rightarrow> nat list" where
+ "stirling_row n = [stirling n k. k \<leftarrow> [0..<Suc n]]"
+
+lemma nth_stirling_row: "k \<le> n \<Longrightarrow> stirling_row n ! k = stirling n k"
+ by (simp add: stirling_row_def del: upt_Suc)
+
+lemma length_stirling_row [simp]: "length (stirling_row n) = Suc n"
+ by (simp add: stirling_row_def)
+
+lemma stirling_row_nonempty [simp]: "stirling_row n \<noteq> []"
+ using length_stirling_row[of n] by (auto simp del: length_stirling_row)
+
+(* TODO Move *)
+lemma list_ext:
+ assumes "length xs = length ys"
+ assumes "\<And>i. i < length xs \<Longrightarrow> xs ! i = ys ! i"
+ shows "xs = ys"
+using assms
+proof (induction rule: list_induct2)
+ case (Cons x xs y ys)
+ from Cons.prems[of 0] have "x = y" by simp
+ moreover from Cons.prems[of "Suc i" for i] have "xs = ys" by (intro Cons.IH) simp
+ ultimately show ?case by simp
+qed simp_all
+
+
+subsubsection \<open>Efficient code\<close>
+
+text \<open>
+ Naively using the defining equations of the Stirling numbers of the first kind to
+ compute them leads to exponential run time due to repeated computations.
+ We can use memoisation to compute them row by row without repeating computations, at
+ the cost of computing a few unneeded values.
+
+ As a bonus, this is very efficient for applications where an entire row of
+ Stirling numbers is needed..
+\<close>
+
+definition zip_with_prev :: "('a \<Rightarrow> 'a \<Rightarrow> 'b) \<Rightarrow> 'a \<Rightarrow> 'a list \<Rightarrow> 'b list" where
+ "zip_with_prev f x xs = map (\<lambda>(x,y). f x y) (zip (x # xs) xs)"
+
+lemma zip_with_prev_altdef:
+ "zip_with_prev f x xs =
+ (if xs = [] then [] else f x (hd xs) # [f (xs!i) (xs!(i+1)). i \<leftarrow> [0..<length xs - 1]])"
+proof (cases xs)
+ case (Cons y ys)
+ hence "zip_with_prev f x xs = f x (hd xs) # zip_with_prev f y ys"
+ by (simp add: zip_with_prev_def)
+ also have "zip_with_prev f y ys = map (\<lambda>i. f (xs ! i) (xs ! (i + 1))) [0..<length xs - 1]"
+ unfolding Cons
+ by (induction ys arbitrary: y)
+ (simp_all add: zip_with_prev_def upt_conv_Cons map_Suc_upt [symmetric] del: upt_Suc)
+ finally show ?thesis using Cons by simp
+qed (simp add: zip_with_prev_def)
+
+
+primrec stirling_row_aux where
+ "stirling_row_aux n y [] = [1]"
+| "stirling_row_aux n y (x#xs) = (y + n * x) # stirling_row_aux n x xs"
+
+lemma stirling_row_aux_correct:
+ "stirling_row_aux n y xs = zip_with_prev (\<lambda>a b. a + n * b) y xs @ [1]"
+ by (induction xs arbitrary: y) (simp_all add: zip_with_prev_def)
+
+lemma stirling_row_code [code]:
+ "stirling_row 0 = [1]"
+ "stirling_row (Suc n) = stirling_row_aux n 0 (stirling_row n)"
+proof -
+ have "stirling_row (Suc n) =
+ 0 # [stirling_row n ! i + stirling_row n ! (i+1) * n. i \<leftarrow> [0..<n]] @ [1]"
+ proof (rule list_ext, goal_cases length nth)
+ case (nth i)
+ from nth have "i \<le> Suc n" by simp
+ then consider "i = 0" | j where "i > 0" "i \<le> n" | "i = Suc n" by linarith
+ thus ?case
+ proof cases
+ assume i: "i > 0" "i \<le> n"
+ from this show ?thesis by (cases i) (simp_all add: nth_append nth_stirling_row)
+ qed (simp_all add: nth_stirling_row nth_append)
+ qed simp
+ also have "0 # [stirling_row n ! i + stirling_row n ! (i+1) * n. i \<leftarrow> [0..<n]] @ [1] =
+ zip_with_prev (\<lambda>a b. a + n * b) 0 (stirling_row n) @ [1]"
+ by (cases n) (auto simp add: zip_with_prev_altdef stirling_row_def hd_map simp del: upt_Suc)
+ also have "\<dots> = stirling_row_aux n 0 (stirling_row n)"
+ by (simp add: stirling_row_aux_correct)
+ finally show "stirling_row (Suc n) = stirling_row_aux n 0 (stirling_row n)" .
+qed (simp add: stirling_row_def)
+
+lemma stirling_code [code]:
+ "stirling n k = (if k = 0 then if n = 0 then 1 else 0
+ else if k > n then 0 else if k = n then 1
+ else stirling_row n ! k)"
+ by (simp add: nth_stirling_row)
+
+end