(* Title: HOL/Library/Stirling.thy Author: Amine Chaieb Author: Florian Haftmann Author: Lukas Bulwahn Author: Manuel Eberl *) section ‹Stirling numbers of first and second kind› theory Stirling imports Main begin subsection ‹Stirling numbers of the second kind› fun Stirling :: "nat ⇒ nat ⇒ 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 ⟹ 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 "… = 2 * (2 ^ Suc n - 1) + 1" by (simp only: Suc Stirling_1) also have "… = 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 ≤ 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 ⇒ nat ⇒ 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 ⟹ stirling n 0 = 0" by (cases n) simp_all lemma stirling_less [simp]: "n < k ⟹ 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: "stirling (Suc n) n = Suc n choose 2" by (induct n) (auto simp add: numerals(2)) lemma stirling_Suc_n_2: assumes "n ≥ Suc 0" shows "stirling (Suc n) 2 = (∑k=1..n. fact n div k)" using assms proof (induct n) case 0 then show ?case by simp next case (Suc n) show ?case proof (cases n) case 0 then show ?thesis by (simp add: numerals(2)) next case Suc then have geq1: "Suc 0 ≤ 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 * (∑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 * (∑k=1..n. fact n div k) + Suc n * fact n div Suc n" by (metis nat.distinct(1) nonzero_mult_div_cancel_left) also have "… = (∑k=1..n. fact (Suc n) div k) + fact (Suc n) div Suc n" by (simp add: sum_distrib_left div_mult_swap dvd_fact) also have "… = (∑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 ≥ Suc 0" shows "(of_nat (stirling (Suc n) 2)::'a::field_char_0) = fact n * (∑k=1..n. (1 / of_nat k))" using assms proof (induct n) case 0 then show ?case by simp next case (Suc n) show ?case proof (cases n) case 0 then show ?thesis by (auto simp add: numerals(2)) next case Suc then have geq1: "Suc 0 ≤ 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 * (∑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) * (∑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) * (∑k = 1..Suc n. 1 / of_nat k)" by (simp add: distrib_left) finally show ?thesis . qed qed lemma sum_stirling: "(∑k≤n. stirling n k) = fact n" proof (induct n) case 0 then show ?case by simp next case (Suc n) have "(∑k≤Suc n. stirling (Suc n) k) = stirling (Suc n) 0 + (∑k≤n. stirling (Suc n) (Suc k))" by (simp only: sum_atMost_Suc_shift) also have "… = (∑k≤n. stirling (Suc n) (Suc k))" by simp also have "… = (∑k≤n. n * stirling n (Suc k) + stirling n k)" by simp also have "… = n * (∑k≤n. stirling n (Suc k)) + (∑k≤n. stirling n k)" by (simp add: sum.distrib sum_distrib_left) also have "… = n * fact n + fact n" proof - have "n * (∑k≤n. stirling n (Suc k)) = n * ((∑k≤Suc n. stirling n k) - stirling n 0)" by (metis add_diff_cancel_left' sum_atMost_Suc_shift) also have "… = n * (∑k≤n. stirling n k)" by (cases n) simp_all also have "… = n * fact n" using Suc.hyps by simp finally have "n * (∑k≤n. stirling n (Suc k)) = n * fact n" . moreover have "(∑k≤n. stirling n k) = fact n" using Suc.hyps . ultimately show ?thesis by simp qed also have "… = fact (Suc n)" by simp finally show ?case . qed lemma stirling_pochhammer: "(∑k≤n. of_nat (stirling n k) * x ^ k) = (pochhammer x n :: 'a::comm_semiring_1)" proof (induct n) case 0 then show ?case by simp next case (Suc n) have "of_nat (n * stirling n 0) = (0 :: 'a)" by (cases n) simp_all then have "(∑k≤Suc n. of_nat (stirling (Suc n) k) * x ^ k) = (of_nat (n * stirling n 0) * x ^ 0 + (∑i≤n. of_nat (n * stirling n (Suc i)) * (x ^ Suc i))) + (∑i≤n. of_nat (stirling n i) * (x ^ Suc i))" by (subst sum_atMost_Suc_shift) (simp add: sum.distrib ring_distribs) also have "… = pochhammer x (Suc n)" by (subst sum_atMost_Suc_shift [symmetric]) (simp add: algebra_simps sum.distrib sum_distrib_left pochhammer_Suc flip: Suc) finally show ?case . qed text ‹A row of the Stirling number triangle› definition stirling_row :: "nat ⇒ nat list" where "stirling_row n = [stirling n k. k ← [0..<Suc n]]" lemma nth_stirling_row: "k ≤ n ⟹ 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 ≠ []" using length_stirling_row[of n] by (auto simp del: length_stirling_row) (* TODO Move *) lemma list_ext: assumes "length xs = length ys" assumes "⋀i. i < length xs ⟹ xs ! i = ys ! i" shows "xs = ys" using assms proof (induction rule: list_induct2) case Nil then show ?case by simp next 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 subsubsection ‹Efficient code› text ‹ 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. › definition zip_with_prev :: "('a ⇒ 'a ⇒ 'b) ⇒ 'a ⇒ 'a list ⇒ 'b list" where "zip_with_prev f x xs = map2 f (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 ← [0..<length xs - 1]])" proof (cases xs) case Nil then show ?thesis by (simp add: zip_with_prev_def) next case (Cons y ys) then have "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 (λi. f (xs ! i) (xs ! (i + 1))) [0..<length xs - 1]" unfolding Cons by (induct ys arbitrary: y) (simp_all add: zip_with_prev_def upt_conv_Cons flip: map_Suc_upt del: upt_Suc) finally show ?thesis using Cons by simp qed 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 (λa b. a + n * b) y xs @ [1]" by (induct 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 goal_cases case 1 show ?case by (simp add: stirling_row_def) next case 2 have "stirling_row (Suc n) = 0 # [stirling_row n ! i + stirling_row n ! (i+1) * n. i ← [0..<n]] @ [1]" proof (rule list_ext, goal_cases length nth) case (nth i) from nth have "i ≤ Suc n" by simp then consider "i = 0 ∨ i = Suc n" | "i > 0" "i ≤ n" by linarith then show ?case proof cases case 1 then show ?thesis by (auto simp: nth_stirling_row nth_append) next case 2 then show ?thesis by (cases i) (simp_all add: nth_append nth_stirling_row) qed next case length then show ?case by simp qed also have "0 # [stirling_row n ! i + stirling_row n ! (i+1) * n. i ← [0..<n]] @ [1] = zip_with_prev (λa 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 "… = stirling_row_aux n 0 (stirling_row n)" by (simp add: stirling_row_aux_correct) finally show ?case . qed 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