63071
|
1 |
(* Authors: Amine Chaieb & Florian Haftmann, TU Muenchen
|
|
2 |
with contributions by Lukas Bulwahn and Manuel Eberl*)
|
|
3 |
|
63145
|
4 |
section \<open>Stirling numbers of first and second kind\<close>
|
63071
|
5 |
|
|
6 |
theory Stirling
|
|
7 |
imports Binomial
|
|
8 |
begin
|
|
9 |
|
63145
|
10 |
subsection \<open>Stirling numbers of the second kind\<close>
|
63071
|
11 |
|
|
12 |
fun Stirling :: "nat \<Rightarrow> nat \<Rightarrow> nat"
|
|
13 |
where
|
|
14 |
"Stirling 0 0 = 1"
|
|
15 |
| "Stirling 0 (Suc k) = 0"
|
|
16 |
| "Stirling (Suc n) 0 = 0"
|
|
17 |
| "Stirling (Suc n) (Suc k) = Suc k * Stirling n (Suc k) + Stirling n k"
|
|
18 |
|
|
19 |
lemma Stirling_1 [simp]:
|
|
20 |
"Stirling (Suc n) (Suc 0) = 1"
|
|
21 |
by (induct n) simp_all
|
|
22 |
|
|
23 |
lemma Stirling_less [simp]:
|
|
24 |
"n < k \<Longrightarrow> Stirling n k = 0"
|
|
25 |
by (induct n k rule: Stirling.induct) simp_all
|
|
26 |
|
|
27 |
lemma Stirling_same [simp]:
|
|
28 |
"Stirling n n = 1"
|
|
29 |
by (induct n) simp_all
|
|
30 |
|
|
31 |
lemma Stirling_2_2:
|
|
32 |
"Stirling (Suc (Suc n)) (Suc (Suc 0)) = 2 ^ Suc n - 1"
|
|
33 |
proof (induct n)
|
|
34 |
case 0 then show ?case by simp
|
|
35 |
next
|
|
36 |
case (Suc n)
|
|
37 |
have "Stirling (Suc (Suc (Suc n))) (Suc (Suc 0)) =
|
|
38 |
2 * Stirling (Suc (Suc n)) (Suc (Suc 0)) + Stirling (Suc (Suc n)) (Suc 0)" by simp
|
|
39 |
also have "\<dots> = 2 * (2 ^ Suc n - 1) + 1"
|
|
40 |
by (simp only: Suc Stirling_1)
|
|
41 |
also have "\<dots> = 2 ^ Suc (Suc n) - 1"
|
|
42 |
proof -
|
|
43 |
have "(2::nat) ^ Suc n - 1 > 0" by (induct n) simp_all
|
|
44 |
then have "2 * ((2::nat) ^ Suc n - 1) > 0" by simp
|
|
45 |
then have "2 \<le> 2 * ((2::nat) ^ Suc n)" by simp
|
|
46 |
with add_diff_assoc2 [of 2 "2 * 2 ^ Suc n" 1]
|
|
47 |
have "2 * 2 ^ Suc n - 2 + (1::nat) = 2 * 2 ^ Suc n + 1 - 2" .
|
|
48 |
then show ?thesis by (simp add: nat_distrib)
|
|
49 |
qed
|
|
50 |
finally show ?case by simp
|
|
51 |
qed
|
|
52 |
|
|
53 |
lemma Stirling_2:
|
|
54 |
"Stirling (Suc n) (Suc (Suc 0)) = 2 ^ n - 1"
|
|
55 |
using Stirling_2_2 by (cases n) simp_all
|
|
56 |
|
63145
|
57 |
subsection \<open>Stirling numbers of the first kind\<close>
|
63071
|
58 |
|
|
59 |
fun stirling :: "nat \<Rightarrow> nat \<Rightarrow> nat"
|
|
60 |
where
|
|
61 |
"stirling 0 0 = 1"
|
|
62 |
| "stirling 0 (Suc k) = 0"
|
|
63 |
| "stirling (Suc n) 0 = 0"
|
|
64 |
| "stirling (Suc n) (Suc k) = n * stirling n (Suc k) + stirling n k"
|
|
65 |
|
|
66 |
lemma stirling_0 [simp]: "n > 0 \<Longrightarrow> stirling n 0 = 0"
|
|
67 |
by (cases n) simp_all
|
|
68 |
|
|
69 |
lemma stirling_less [simp]:
|
|
70 |
"n < k \<Longrightarrow> stirling n k = 0"
|
|
71 |
by (induct n k rule: stirling.induct) simp_all
|
|
72 |
|
|
73 |
lemma stirling_same [simp]:
|
|
74 |
"stirling n n = 1"
|
|
75 |
by (induct n) simp_all
|
|
76 |
|
|
77 |
lemma stirling_Suc_n_1:
|
|
78 |
"stirling (Suc n) (Suc 0) = fact n"
|
|
79 |
by (induct n) auto
|
|
80 |
|
|
81 |
lemma stirling_Suc_n_n:
|
|
82 |
shows "stirling (Suc n) n = Suc n choose 2"
|
|
83 |
by (induct n) (auto simp add: numerals(2))
|
|
84 |
|
|
85 |
lemma stirling_Suc_n_2:
|
|
86 |
assumes "n \<ge> Suc 0"
|
|
87 |
shows "stirling (Suc n) 2 = (\<Sum>k=1..n. fact n div k)"
|
|
88 |
using assms
|
|
89 |
proof (induct n)
|
|
90 |
case 0 from this show ?case by simp
|
|
91 |
next
|
|
92 |
case (Suc n)
|
|
93 |
show ?case
|
|
94 |
proof (cases n)
|
|
95 |
case 0 from this show ?thesis by (simp add: numerals(2))
|
|
96 |
next
|
|
97 |
case Suc
|
|
98 |
from this have geq1: "Suc 0 \<le> n" by simp
|
|
99 |
have "stirling (Suc (Suc n)) 2 = Suc n * stirling (Suc n) 2 + stirling (Suc n) (Suc 0)"
|
|
100 |
by (simp only: stirling.simps(4)[of "Suc n"] numerals(2))
|
|
101 |
also have "... = Suc n * (\<Sum>k=1..n. fact n div k) + fact n"
|
|
102 |
using Suc.hyps[OF geq1]
|
|
103 |
by (simp only: stirling_Suc_n_1 of_nat_fact of_nat_add of_nat_mult)
|
|
104 |
also have "... = Suc n * (\<Sum>k=1..n. fact n div k) + Suc n * fact n div Suc n"
|
|
105 |
by (metis nat.distinct(1) nonzero_mult_divide_cancel_left)
|
|
106 |
also have "... = (\<Sum>k=1..n. fact (Suc n) div k) + fact (Suc n) div Suc n"
|
|
107 |
by (simp add: setsum_right_distrib div_mult_swap dvd_fact)
|
|
108 |
also have "... = (\<Sum>k=1..Suc n. fact (Suc n) div k)" by simp
|
|
109 |
finally show ?thesis .
|
|
110 |
qed
|
|
111 |
qed
|
|
112 |
|
|
113 |
lemma of_nat_stirling_Suc_n_2:
|
|
114 |
assumes "n \<ge> Suc 0"
|
|
115 |
shows "(of_nat (stirling (Suc n) 2)::'a::field_char_0) = fact n * (\<Sum>k=1..n. (1 / of_nat k))"
|
|
116 |
using assms
|
|
117 |
proof (induct n)
|
|
118 |
case 0 from this show ?case by simp
|
|
119 |
next
|
|
120 |
case (Suc n)
|
|
121 |
show ?case
|
|
122 |
proof (cases n)
|
|
123 |
case 0 from this show ?thesis by (auto simp add: numerals(2))
|
|
124 |
next
|
|
125 |
case Suc
|
|
126 |
from this have geq1: "Suc 0 \<le> n" by simp
|
|
127 |
have "(of_nat (stirling (Suc (Suc n)) 2)::'a) =
|
|
128 |
of_nat (Suc n * stirling (Suc n) 2 + stirling (Suc n) (Suc 0))"
|
|
129 |
by (simp only: stirling.simps(4)[of "Suc n"] numerals(2))
|
|
130 |
also have "... = of_nat (Suc n) * (fact n * (\<Sum>k = 1..n. 1 / of_nat k)) + fact n"
|
|
131 |
using Suc.hyps[OF geq1]
|
|
132 |
by (simp only: stirling_Suc_n_1 of_nat_fact of_nat_add of_nat_mult)
|
|
133 |
also have "... = fact (Suc n) * (\<Sum>k = 1..n. 1 / of_nat k) + fact (Suc n) * (1 / of_nat (Suc n))"
|
|
134 |
using of_nat_neq_0 by auto
|
|
135 |
also have "... = fact (Suc n) * (\<Sum>k = 1..Suc n. 1 / of_nat k)"
|
|
136 |
by (simp add: distrib_left)
|
|
137 |
finally show ?thesis .
|
|
138 |
qed
|
|
139 |
qed
|
|
140 |
|
|
141 |
lemma setsum_stirling:
|
|
142 |
"(\<Sum>k\<le>n. stirling n k) = fact n"
|
|
143 |
proof (induct n)
|
|
144 |
case 0
|
|
145 |
from this show ?case by simp
|
|
146 |
next
|
|
147 |
case (Suc n)
|
|
148 |
have "(\<Sum>k\<le>Suc n. stirling (Suc n) k) = stirling (Suc n) 0 + (\<Sum>k\<le>n. stirling (Suc n) (Suc k))"
|
|
149 |
by (simp only: setsum_atMost_Suc_shift)
|
|
150 |
also have "\<dots> = (\<Sum>k\<le>n. stirling (Suc n) (Suc k))" by simp
|
|
151 |
also have "\<dots> = (\<Sum>k\<le>n. n * stirling n (Suc k) + stirling n k)" by simp
|
|
152 |
also have "\<dots> = n * (\<Sum>k\<le>n. stirling n (Suc k)) + (\<Sum>k\<le>n. stirling n k)"
|
|
153 |
by (simp add: setsum.distrib setsum_right_distrib)
|
|
154 |
also have "\<dots> = n * fact n + fact n"
|
|
155 |
proof -
|
|
156 |
have "n * (\<Sum>k\<le>n. stirling n (Suc k)) = n * ((\<Sum>k\<le>Suc n. stirling n k) - stirling n 0)"
|
|
157 |
by (metis add_diff_cancel_left' setsum_atMost_Suc_shift)
|
|
158 |
also have "\<dots> = n * (\<Sum>k\<le>n. stirling n k)" by (cases n) simp+
|
|
159 |
also have "\<dots> = n * fact n" using Suc.hyps by simp
|
|
160 |
finally have "n * (\<Sum>k\<le>n. stirling n (Suc k)) = n * fact n" .
|
|
161 |
moreover have "(\<Sum>k\<le>n. stirling n k) = fact n" using Suc.hyps .
|
|
162 |
ultimately show ?thesis by simp
|
|
163 |
qed
|
|
164 |
also have "\<dots> = fact (Suc n)" by simp
|
|
165 |
finally show ?case .
|
|
166 |
qed
|
|
167 |
|
|
168 |
lemma stirling_pochhammer:
|
|
169 |
"(\<Sum>k\<le>n. of_nat (stirling n k) * x ^ k) = (pochhammer x n :: 'a :: comm_semiring_1)"
|
|
170 |
proof (induction n)
|
|
171 |
case (Suc n)
|
|
172 |
have "of_nat (n * stirling n 0) = (0 :: 'a)" by (cases n) simp_all
|
|
173 |
hence "(\<Sum>k\<le>Suc n. of_nat (stirling (Suc n) k) * x ^ k) =
|
|
174 |
(of_nat (n * stirling n 0) * x ^ 0 +
|
|
175 |
(\<Sum>i\<le>n. of_nat (n * stirling n (Suc i)) * (x ^ Suc i))) +
|
|
176 |
(\<Sum>i\<le>n. of_nat (stirling n i) * (x ^ Suc i))"
|
|
177 |
by (subst setsum_atMost_Suc_shift) (simp add: setsum.distrib ring_distribs)
|
|
178 |
also have "\<dots> = pochhammer x (Suc n)"
|
|
179 |
by (subst setsum_atMost_Suc_shift [symmetric])
|
|
180 |
(simp add: algebra_simps setsum.distrib setsum_right_distrib pochhammer_Suc Suc [symmetric])
|
|
181 |
finally show ?case .
|
|
182 |
qed simp_all
|
|
183 |
|
|
184 |
|
|
185 |
text \<open>A row of the Stirling number triangle\<close>
|
|
186 |
|
|
187 |
definition stirling_row :: "nat \<Rightarrow> nat list" where
|
|
188 |
"stirling_row n = [stirling n k. k \<leftarrow> [0..<Suc n]]"
|
|
189 |
|
|
190 |
lemma nth_stirling_row: "k \<le> n \<Longrightarrow> stirling_row n ! k = stirling n k"
|
|
191 |
by (simp add: stirling_row_def del: upt_Suc)
|
|
192 |
|
|
193 |
lemma length_stirling_row [simp]: "length (stirling_row n) = Suc n"
|
|
194 |
by (simp add: stirling_row_def)
|
|
195 |
|
|
196 |
lemma stirling_row_nonempty [simp]: "stirling_row n \<noteq> []"
|
|
197 |
using length_stirling_row[of n] by (auto simp del: length_stirling_row)
|
|
198 |
|
|
199 |
(* TODO Move *)
|
|
200 |
lemma list_ext:
|
|
201 |
assumes "length xs = length ys"
|
|
202 |
assumes "\<And>i. i < length xs \<Longrightarrow> xs ! i = ys ! i"
|
|
203 |
shows "xs = ys"
|
|
204 |
using assms
|
|
205 |
proof (induction rule: list_induct2)
|
|
206 |
case (Cons x xs y ys)
|
|
207 |
from Cons.prems[of 0] have "x = y" by simp
|
|
208 |
moreover from Cons.prems[of "Suc i" for i] have "xs = ys" by (intro Cons.IH) simp
|
|
209 |
ultimately show ?case by simp
|
|
210 |
qed simp_all
|
|
211 |
|
|
212 |
|
|
213 |
subsubsection \<open>Efficient code\<close>
|
|
214 |
|
|
215 |
text \<open>
|
|
216 |
Naively using the defining equations of the Stirling numbers of the first kind to
|
|
217 |
compute them leads to exponential run time due to repeated computations.
|
|
218 |
We can use memoisation to compute them row by row without repeating computations, at
|
|
219 |
the cost of computing a few unneeded values.
|
|
220 |
|
|
221 |
As a bonus, this is very efficient for applications where an entire row of
|
|
222 |
Stirling numbers is needed..
|
|
223 |
\<close>
|
|
224 |
|
|
225 |
definition zip_with_prev :: "('a \<Rightarrow> 'a \<Rightarrow> 'b) \<Rightarrow> 'a \<Rightarrow> 'a list \<Rightarrow> 'b list" where
|
|
226 |
"zip_with_prev f x xs = map (\<lambda>(x,y). f x y) (zip (x # xs) xs)"
|
|
227 |
|
|
228 |
lemma zip_with_prev_altdef:
|
|
229 |
"zip_with_prev f x xs =
|
|
230 |
(if xs = [] then [] else f x (hd xs) # [f (xs!i) (xs!(i+1)). i \<leftarrow> [0..<length xs - 1]])"
|
|
231 |
proof (cases xs)
|
|
232 |
case (Cons y ys)
|
|
233 |
hence "zip_with_prev f x xs = f x (hd xs) # zip_with_prev f y ys"
|
|
234 |
by (simp add: zip_with_prev_def)
|
|
235 |
also have "zip_with_prev f y ys = map (\<lambda>i. f (xs ! i) (xs ! (i + 1))) [0..<length xs - 1]"
|
|
236 |
unfolding Cons
|
|
237 |
by (induction ys arbitrary: y)
|
|
238 |
(simp_all add: zip_with_prev_def upt_conv_Cons map_Suc_upt [symmetric] del: upt_Suc)
|
|
239 |
finally show ?thesis using Cons by simp
|
|
240 |
qed (simp add: zip_with_prev_def)
|
|
241 |
|
|
242 |
|
|
243 |
primrec stirling_row_aux where
|
|
244 |
"stirling_row_aux n y [] = [1]"
|
|
245 |
| "stirling_row_aux n y (x#xs) = (y + n * x) # stirling_row_aux n x xs"
|
|
246 |
|
|
247 |
lemma stirling_row_aux_correct:
|
|
248 |
"stirling_row_aux n y xs = zip_with_prev (\<lambda>a b. a + n * b) y xs @ [1]"
|
|
249 |
by (induction xs arbitrary: y) (simp_all add: zip_with_prev_def)
|
|
250 |
|
|
251 |
lemma stirling_row_code [code]:
|
|
252 |
"stirling_row 0 = [1]"
|
|
253 |
"stirling_row (Suc n) = stirling_row_aux n 0 (stirling_row n)"
|
|
254 |
proof -
|
|
255 |
have "stirling_row (Suc n) =
|
|
256 |
0 # [stirling_row n ! i + stirling_row n ! (i+1) * n. i \<leftarrow> [0..<n]] @ [1]"
|
|
257 |
proof (rule list_ext, goal_cases length nth)
|
|
258 |
case (nth i)
|
|
259 |
from nth have "i \<le> Suc n" by simp
|
|
260 |
then consider "i = 0" | j where "i > 0" "i \<le> n" | "i = Suc n" by linarith
|
|
261 |
thus ?case
|
|
262 |
proof cases
|
|
263 |
assume i: "i > 0" "i \<le> n"
|
|
264 |
from this show ?thesis by (cases i) (simp_all add: nth_append nth_stirling_row)
|
|
265 |
qed (simp_all add: nth_stirling_row nth_append)
|
|
266 |
qed simp
|
|
267 |
also have "0 # [stirling_row n ! i + stirling_row n ! (i+1) * n. i \<leftarrow> [0..<n]] @ [1] =
|
|
268 |
zip_with_prev (\<lambda>a b. a + n * b) 0 (stirling_row n) @ [1]"
|
|
269 |
by (cases n) (auto simp add: zip_with_prev_altdef stirling_row_def hd_map simp del: upt_Suc)
|
|
270 |
also have "\<dots> = stirling_row_aux n 0 (stirling_row n)"
|
|
271 |
by (simp add: stirling_row_aux_correct)
|
|
272 |
finally show "stirling_row (Suc n) = stirling_row_aux n 0 (stirling_row n)" .
|
|
273 |
qed (simp add: stirling_row_def)
|
|
274 |
|
|
275 |
lemma stirling_code [code]:
|
|
276 |
"stirling n k = (if k = 0 then if n = 0 then 1 else 0
|
|
277 |
else if k > n then 0 else if k = n then 1
|
|
278 |
else stirling_row n ! k)"
|
|
279 |
by (simp add: nth_stirling_row)
|
|
280 |
|
|
281 |
end
|