Theory HOL-Number_Theory.Fib
section ‹The fibonacci function›
theory Fib
imports Complex_Main
begin
subsection ‹Fibonacci numbers›
fun fib :: "nat ⇒ nat"
where
fib0: "fib 0 = 0"
| fib1: "fib (Suc 0) = 1"
| fib2: "fib (Suc (Suc n)) = fib (Suc n) + fib n"
subsection ‹Basic Properties›
lemma fib_1 [simp]: "fib 1 = 1"
by (metis One_nat_def fib1)
lemma fib_2 [simp]: "fib 2 = 1"
using fib.simps(3) [of 0] by (simp add: numeral_2_eq_2)
lemma fib_plus_2: "fib (n + 2) = fib (n + 1) + fib n"
by (metis Suc_eq_plus1 add_2_eq_Suc' fib.simps(3))
lemma fib_add: "fib (Suc (n + k)) = fib (Suc k) * fib (Suc n) + fib k * fib n"
by (induct n rule: fib.induct) (auto simp add: field_simps)
lemma fib_neq_0_nat: "n > 0 ⟹ fib n > 0"
by (induct n rule: fib.induct) auto
lemma fib_Suc_mono: "fib m ≤ fib (Suc m)"
by(induction m) auto
lemma fib_mono: "m ≤ n ⟹ fib m ≤ fib n"
by (simp add: fib_Suc_mono lift_Suc_mono_le)
subsection ‹More efficient code›
text ‹
The naive approach is very inefficient since the branching recursion leads to many
values of \<^term>‹fib› being computed multiple times. We can avoid this by ``remembering''
the last two values in the sequence, yielding a tail-recursive version.
This is far from optimal (it takes roughly $O(n\cdot M(n))$ time where $M(n)$ is the
time required to multiply two $n$-bit integers), but much better than the naive version,
which is exponential.
›
fun gen_fib :: "nat ⇒ nat ⇒ nat ⇒ nat"
where
"gen_fib a b 0 = a"
| "gen_fib a b (Suc 0) = b"
| "gen_fib a b (Suc (Suc n)) = gen_fib b (a + b) (Suc n)"
lemma gen_fib_recurrence: "gen_fib a b (Suc (Suc n)) = gen_fib a b n + gen_fib a b (Suc n)"
by (induct a b n rule: gen_fib.induct) simp_all
lemma gen_fib_fib: "gen_fib (fib n) (fib (Suc n)) m = fib (n + m)"
by (induct m rule: fib.induct) (simp_all del: gen_fib.simps(3) add: gen_fib_recurrence)
lemma fib_conv_gen_fib: "fib n = gen_fib 0 1 n"
using gen_fib_fib[of 0 n] by simp
declare fib_conv_gen_fib [code]
subsection ‹A Few Elementary Results›
text ‹
┉ Concrete Mathematics, page 278: Cassini's identity. The proof is
much easier using integers, not natural numbers!
›
lemma fib_Cassini_int: "int (fib (Suc (Suc n)) * fib n) - int((fib (Suc n))⇧2) = - ((-1)^n)"
by (induct n rule: fib.induct) (auto simp add: field_simps power2_eq_square power_add)
lemma fib_Cassini_nat:
"fib (Suc (Suc n)) * fib n =
(if even n then (fib (Suc n))⇧2 - 1 else (fib (Suc n))⇧2 + 1)"
using fib_Cassini_int [of n] by (auto simp del: of_nat_mult of_nat_power)
subsection ‹Law 6.111 of Concrete Mathematics›
lemma coprime_fib_Suc_nat: "coprime (fib n) (fib (Suc n))"
apply (induct n rule: fib.induct)
apply (simp_all add: coprime_iff_gcd_eq_1 algebra_simps)
apply (simp add: add.assoc [symmetric])
done
lemma gcd_fib_add:
"gcd (fib m) (fib (n + m)) = gcd (fib m) (fib n)"
proof (cases m)
case 0
then show ?thesis
by simp
next
case (Suc q)
from coprime_fib_Suc_nat [of q]
have "coprime (fib (Suc q)) (fib q)"
by (simp add: ac_simps)
have "gcd (fib q) (fib (Suc q)) = Suc 0"
using coprime_fib_Suc_nat [of q] by simp
then have *: "gcd (fib n * fib q) (fib n * fib (Suc q)) = fib n"
by (simp add: gcd_mult_distrib_nat [symmetric])
moreover have "gcd (fib (Suc q)) (fib n * fib q + fib (Suc n) * fib (Suc q)) =
gcd (fib (Suc q)) (fib n * fib q)"
using gcd_add_mult [of "fib (Suc q)"] by (simp add: ac_simps)
moreover have "gcd (fib (Suc q)) (fib n * fib (Suc q)) = fib (Suc q)"
by simp
ultimately show ?thesis
using Suc ‹coprime (fib (Suc q)) (fib q)›
by (auto simp add: fib_add algebra_simps gcd_mult_right_right_cancel)
qed
lemma gcd_fib_diff: "m ≤ n ⟹ gcd (fib m) (fib (n - m)) = gcd (fib m) (fib n)"
by (simp add: gcd_fib_add [symmetric, of _ "n-m"])
lemma gcd_fib_mod: "0 < m ⟹ gcd (fib m) (fib (n mod m)) = gcd (fib m) (fib n)"
proof (induct n rule: less_induct)
case (less n)
show "gcd (fib m) (fib (n mod m)) = gcd (fib m) (fib n)"
proof (cases "m < n")
case True
then have "m ≤ n" by auto
with ‹0 < m› have "0 < n" by auto
with ‹0 < m› ‹m < n› have *: "n - m < n" by auto
have "gcd (fib m) (fib (n mod m)) = gcd (fib m) (fib ((n - m) mod m))"
by (simp add: mod_if [of n]) (use ‹m < n› in auto)
also have "… = gcd (fib m) (fib (n - m))"
by (simp add: less.hyps * ‹0 < m›)
also have "… = gcd (fib m) (fib n)"
by (simp add: gcd_fib_diff ‹m ≤ n›)
finally show "gcd (fib m) (fib (n mod m)) = gcd (fib m) (fib n)" .
next
case False
then show "gcd (fib m) (fib (n mod m)) = gcd (fib m) (fib n)"
by (cases "m = n") auto
qed
qed
lemma fib_gcd: "fib (gcd m n) = gcd (fib m) (fib n)"
by (induct m n rule: gcd_nat_induct) (simp_all add: gcd_non_0_nat gcd.commute gcd_fib_mod)
theorem fib_mult_eq_sum_nat: "fib (Suc n) * fib n = (∑k ∈ {..n}. fib k * fib k)"
by (induct n rule: nat.induct) (auto simp add: field_simps)
subsection ‹Closed form›
lemma fib_closed_form:
fixes φ ψ :: real
defines "φ ≡ (1 + sqrt 5) / 2"
and "ψ ≡ (1 - sqrt 5) / 2"
shows "of_nat (fib n) = (φ ^ n - ψ ^ n) / sqrt 5"
proof (induct n rule: fib.induct)
fix n :: nat
assume IH1: "of_nat (fib n) = (φ ^ n - ψ ^ n) / sqrt 5"
assume IH2: "of_nat (fib (Suc n)) = (φ ^ Suc n - ψ ^ Suc n) / sqrt 5"
have "of_nat (fib (Suc (Suc n))) = of_nat (fib (Suc n)) + of_nat (fib n)" by simp
also have "… = (φ^n * (φ + 1) - ψ^n * (ψ + 1)) / sqrt 5"
by (simp add: IH1 IH2 field_simps)
also have "φ + 1 = φ⇧2" by (simp add: φ_def field_simps power2_eq_square)
also have "ψ + 1 = ψ⇧2" by (simp add: ψ_def field_simps power2_eq_square)
also have "φ^n * φ⇧2 - ψ^n * ψ⇧2 = φ ^ Suc (Suc n) - ψ ^ Suc (Suc n)"
by (simp add: power2_eq_square)
finally show "of_nat (fib (Suc (Suc n))) = (φ ^ Suc (Suc n) - ψ ^ Suc (Suc n)) / sqrt 5" .
qed (simp_all add: φ_def ψ_def field_simps)
lemma fib_closed_form':
fixes φ ψ :: real
defines "φ ≡ (1 + sqrt 5) / 2"
and "ψ ≡ (1 - sqrt 5) / 2"
assumes "n > 0"
shows "fib n = round (φ ^ n / sqrt 5)"
proof (rule sym, rule round_unique')
have "¦φ ^ n / sqrt 5 - of_int (int (fib n))¦ = ¦ψ¦ ^ n / sqrt 5"
by (simp add: fib_closed_form[folded φ_def ψ_def] field_simps power_abs)
also {
from assms have "¦ψ¦^n ≤ ¦ψ¦^1"
by (intro power_decreasing) (simp_all add: algebra_simps real_le_lsqrt)
also have "… < sqrt 5 / 2" by (simp add: ψ_def field_simps)
finally have "¦ψ¦^n / sqrt 5 < 1/2" by (simp add: field_simps)
}
finally show "¦φ ^ n / sqrt 5 - of_int (int (fib n))¦ < 1/2" .
qed
lemma fib_asymptotics:
fixes φ :: real
defines "φ ≡ (1 + sqrt 5) / 2"
shows "(λn. real (fib n) / (φ ^ n / sqrt 5)) ⇢ 1"
proof -
define ψ :: real where "ψ ≡ (1 - sqrt 5) / 2"
have "φ > 1" by (simp add: φ_def)
then have *: "φ ≠ 0" by auto
have "(λn. (ψ / φ) ^ n) ⇢ 0"
by (rule LIMSEQ_power_zero) (simp_all add: φ_def ψ_def field_simps add_pos_pos)
then have "(λn. 1 - (ψ / φ) ^ n) ⇢ 1 - 0"
by (intro tendsto_diff tendsto_const)
with * have "(λn. (φ ^ n - ψ ^ n) / φ ^ n) ⇢ 1"
by (simp add: field_simps)
then show ?thesis
by (simp add: fib_closed_form φ_def ψ_def)
qed
subsection ‹Divide-and-Conquer recurrence›
text ‹
The following divide-and-conquer recurrence allows for a more efficient computation
of Fibonacci numbers; however, it requires memoisation of values to be reasonably
efficient, cutting the number of values to be computed to logarithmically many instead of
linearly many. The vast majority of the computation time is then actually spent on the
multiplication, since the output number is exponential in the input number.
›
lemma fib_rec_odd:
fixes φ ψ :: real
defines "φ ≡ (1 + sqrt 5) / 2"
and "ψ ≡ (1 - sqrt 5) / 2"
shows "fib (Suc (2 * n)) = fib n^2 + fib (Suc n)^2"
proof -
have "of_nat (fib n^2 + fib (Suc n)^2) = ((φ ^ n - ψ ^ n)⇧2 + (φ * φ ^ n - ψ * ψ ^ n)⇧2)/5"
by (simp add: fib_closed_form[folded φ_def ψ_def] field_simps power2_eq_square)
also
let ?A = "φ^(2 * n) + ψ^(2 * n) - 2*(φ * ψ)^n + φ^(2 * n + 2) + ψ^(2 * n + 2) - 2*(φ * ψ)^(n + 1)"
have "(φ ^ n - ψ ^ n)⇧2 + (φ * φ ^ n - ψ * ψ ^ n)⇧2 = ?A"
by (simp add: power2_eq_square algebra_simps power_mult power_mult_distrib)
also have "φ * ψ = -1"
by (simp add: φ_def ψ_def field_simps)
then have "?A = φ^(2 * n + 1) * (φ + inverse φ) + ψ^(2 * n + 1) * (ψ + inverse ψ)"
by (auto simp: field_simps power2_eq_square)
also have "1 + sqrt 5 > 0"
by (auto intro: add_pos_pos)
then have "φ + inverse φ = sqrt 5"
by (simp add: φ_def field_simps)
also have "ψ + inverse ψ = -sqrt 5"
by (simp add: ψ_def field_simps)
also have "(φ ^ (2 * n + 1) * sqrt 5 + ψ ^ (2 * n + 1) * - sqrt 5) / 5 =
(φ ^ (2 * n + 1) - ψ ^ (2 * n + 1)) * (sqrt 5 / 5)"
by (simp add: field_simps)
also have "sqrt 5 / 5 = inverse (sqrt 5)"
by (simp add: field_simps)
also have "(φ ^ (2 * n + 1) - ψ ^ (2 * n + 1)) * … = of_nat (fib (Suc (2 * n)))"
by (simp add: fib_closed_form[folded φ_def ψ_def] divide_inverse)
finally show ?thesis
by (simp only: of_nat_eq_iff)
qed
lemma fib_rec_even: "fib (2 * n) = (fib (n - 1) + fib (n + 1)) * fib n"
proof (induct n)
case 0
then show ?case by simp
next
case (Suc n)
let ?rfib = "λx. real (fib x)"
have "2 * (Suc n) = Suc (Suc (2 * n))" by simp
also have "real (fib …) = ?rfib n^2 + ?rfib (Suc n)^2 + (?rfib (n - 1) + ?rfib (n + 1)) * ?rfib n"
by (simp add: fib_rec_odd Suc)
also have "(?rfib (n - 1) + ?rfib (n + 1)) * ?rfib n = (2 * ?rfib (n + 1) - ?rfib n) * ?rfib n"
by (cases n) simp_all
also have "?rfib n^2 + ?rfib (Suc n)^2 + … = (?rfib (Suc n) + 2 * ?rfib n) * ?rfib (Suc n)"
by (simp add: algebra_simps power2_eq_square)
also have "… = real ((fib (Suc n - 1) + fib (Suc n + 1)) * fib (Suc n))" by simp
finally show ?case by (simp only: of_nat_eq_iff)
qed
lemma fib_rec_even': "fib (2 * n) = (2 * fib (n - 1) + fib n) * fib n"
by (subst fib_rec_even, cases n) simp_all
lemma fib_rec:
"fib n =
(if n = 0 then 0 else if n = 1 then 1
else if even n then let n' = n div 2; fn = fib n' in (2 * fib (n' - 1) + fn) * fn
else let n' = n div 2 in fib n' ^ 2 + fib (Suc n') ^ 2)"
by (auto elim: evenE oddE simp: fib_rec_odd fib_rec_even' Let_def)
subsection ‹Fibonacci and Binomial Coefficients›
lemma sum_drop_zero: "(∑k = 0..Suc n. if 0<k then (f (k - 1)) else 0) = (∑j = 0..n. f j)"
by (induct n) auto
lemma sum_choose_drop_zero:
"(∑k = 0..Suc n. if k = 0 then 0 else (Suc n - k) choose (k - 1)) =
(∑j = 0..n. (n-j) choose j)"
by (rule trans [OF sum.cong sum_drop_zero]) auto
lemma ne_diagonal_fib: "(∑k = 0..n. (n-k) choose k) = fib (Suc n)"
proof (induct n rule: fib.induct)
case 1
show ?case by simp
next
case 2
show ?case by simp
next
case (3 n)
have "(∑k = 0..Suc n. Suc (Suc n) - k choose k) =
(∑k = 0..Suc n. (Suc n - k choose k) + (if k = 0 then 0 else (Suc n - k choose (k - 1))))"
by (rule sum.cong) (simp_all add: choose_reduce_nat)
also have "… =
(∑k = 0..Suc n. Suc n - k choose k) +
(∑k = 0..Suc n. if k=0 then 0 else (Suc n - k choose (k - 1)))"
by (simp add: sum.distrib)
also have "… = (∑k = 0..Suc n. Suc n - k choose k) + (∑j = 0..n. n - j choose j)"
by (metis sum_choose_drop_zero)
finally show ?case using 3
by simp
qed
end