Data update
This commit is contained in:
Ingy döt Net
parent
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7735 changed files with 38060 additions and 199180 deletions
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@ -1,6 +1,8 @@
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from math import *
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def analytic_fibonacci(n):
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assert isinstance(n,int), "n must be an integer."
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assert n<=71 , "n must be <=71 due to floating point precision limitations."
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sqrt_5 = sqrt(5);
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p = (1 + sqrt_5) / 2;
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q = 1/p;
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@ -1,7 +1,31 @@
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def fib(n, c={0:1, 1:1}):
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if n not in c:
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x = n // 2
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c[n] = fib(x-1) * fib(n-x-1) + fib(x) * fib(n - x)
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return c[n]
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def prev_pow_two(n):
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"""Gets the power of two that is less than or equal to the given input
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"""
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if ((n & -n) == n):
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return n
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n -= 1
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n |= n >> 1
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n |= n >> 2
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n |= n >> 4
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n |= n >> 8
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n |= n >> 16
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n += 1
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return n//2
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fib(10000000) # calculating it takes a few seconds, printing it takes eons
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def crazy_fib(n):
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"""Crazy fast fibonacci number calculation
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"""
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pow_two = prev_pow_two(n)
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q = r = i = 1
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s = 0
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while i < pow_two:
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i *= 2
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q, r, s = q*q + r*r, r * (q + s), (r*r + s*s)
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while i < n:
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i += 1
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q, r, s = q+r, q, r
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return q
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@ -1,8 +1,7 @@
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F = {0: 0, 1: 1, 2: 1}
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def fib(n):
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if n in F:
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return F[n]
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f1 = fib(n // 2 + 1)
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f2 = fib((n - 1) // 2)
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F[n] = (f1 * f1 + f2 * f2 if n & 1 else f1 * f1 - f2 * f2)
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return F[n]
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def fib(n, c={0:1, 1:1}):
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if n not in c:
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x = n // 2
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c[n] = fib(x-1) * fib(n-x-1) + fib(x) * fib(n - x)
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return c[n]
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fib(10000000) # calculating it takes a few seconds, printing it takes eons
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@ -1,10 +1,8 @@
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def fib():
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"""Yield fib[n+1] + fib[n]"""
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yield 1 # have to start somewhere
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lhs, rhs = fib(), fib()
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yield next(lhs) # move lhs one iteration ahead
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while True:
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yield next(lhs)+next(rhs)
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f=fib()
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print [next(f) for _ in range(9)]
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F = {0: 0, 1: 1, 2: 1}
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def fib(n):
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if n in F:
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return F[n]
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f1 = fib(n // 2 + 1)
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f2 = fib((n - 1) // 2)
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F[n] = (f1 * f1 + f2 * f2 if n & 1 else f1 * f1 - f2 * f2)
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return F[n]
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@ -1,11 +1,11 @@
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from itertools import islice
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def fib():
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yield 0
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"""Yield fib[n+1] + fib[n]"""
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yield 1
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a, b = fib(), fib()
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next(b)
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lhs, rhs = fib(), fib()
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# move lhs one iteration ahead
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yield next(lhs)
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while True:
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yield next(a)+next(b)
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yield next(lhs)+next(rhs)
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print(tuple(islice(fib(), 10)))
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f=fib()
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print [next(f) for _ in range(9)]
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@ -1,23 +1,11 @@
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'''Fibonacci accumulation'''
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from itertools import islice
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from itertools import accumulate
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def fib():
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yield 0
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yield 1
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a, b = fib(), fib()
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next(b)
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while True:
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yield next(a)+next(b)
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# fibs :: Integer :: [Integer]
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def fibs(n):
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'''An accumulation of the first n integers in
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the Fibonacci series. The accumulator is a
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pair of the two preceding numbers.
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'''
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return [
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a
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for a, b in accumulate(
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range(1, n), # we don't actually use these numbers
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lambda acc, _: (acc[1], sum(acc)),
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initial = (0, 1)
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)
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]
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# MAIN ---
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if __name__ == '__main__':
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print(f'First twenty: {fibs(20)}')
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print(tuple(islice(fib(), 10)))
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@ -1,18 +1,21 @@
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'''Nth Fibonacci term (by folding)'''
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def fibs(n):
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"""Fibonacci accumulation
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from functools import reduce
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An accumulation of the first n integers in the Fibonacci series. The accumulator is a
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pair of the two preceding numbers.
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"""
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# Local import is more efficient.
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from itertools import accumulate
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# nthFib :: Integer -> Integer
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def nthFib(n):
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'''Nth integer in the Fibonacci series.'''
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return reduce(
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lambda acc, _: (acc[1], sum(acc)),
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# Note: Numbers generated in range(1, n) [or range(n-1)] call will not be used.
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return [a for a, b in accumulate(
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range(1, n),
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(0, 1)
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)[0]
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lambda acc, _: (acc[1], sum(acc)),
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initial = (0, 1)
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)
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]
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# MAIN ---
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if __name__ == '__main__':
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n = 1000
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print(f'{n}th term: {nthFib(n)}')
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print(f'First twenty: {fibs(20)}')
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def fib(n):
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def nth_fib(n):
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"""Nth Fibonacci term (by folding)
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Nth integer in the Fibonacci series.
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"""
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from functools import reduce
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return reduce(lambda x, y: (x[1], x[0] + x[1]), range(n), (0, 1))[0]
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return reduce(
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lambda acc, _: (acc[1], sum(acc)),
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range(1, n),
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(0, 1)
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)[0]
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# MAIN ---
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if __name__ == '__main__':
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n = 1000
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print(f'{n}th term: {nth_fib(n)}')
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fibseq = [1,1,]
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fiblength = 21
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for x in range(1,fiblength-1):
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xcount = fibseq[x-1] + fibseq[x]
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fibseq.append(xcount)
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print(xcount)
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def fib(n):
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from functools import reduce
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return reduce(lambda x, y: (x[1], x[0] + x[1]), range(n), (0, 1))[0]
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def fibIter(n):
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if n < 2:
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return n
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fibPrev = 1
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fib = 1
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for _ in range(2, n):
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fibPrev, fib = fib, fib + fibPrev
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return fib
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def analytic_fibonacci91(m):
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"""
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Binet's algebraic formula for the nth Fibonacci number.
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Good for up to n=91
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Uses numpy longdoubles:
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See: https://artofproblemsolving.com/wiki/index.php/Binet%27s_Formula
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"""
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import numpy as np
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assert isinstance(m,int), "parameter must be an integer."
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assert 0<=m<=91 , "n must be in the range 0 .. 91 due to double precision floating point precision limitations."
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if m < 2: return m
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# Make sure that nothing causes conversion to single
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n=np.longdouble(m)
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C1=np.longdouble(1)
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C2=np.longdouble(2)
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C5=np.longdouble(5)
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Chalf=C1/C2
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Cfifth=C1/C5
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root5=C5**Chalf
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t1=(C1+root5)/C2
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t2=(C1-root5)/C2
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f=(t1**n-t2**n)/root5
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return int(f+0.1)
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# Usage
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print(f:=[[i,analytic_fibonacci91(i)] for i in range(92)])
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def fib(n,x=[0,1]):
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for i in range(abs(n)-1): x=[x[1],sum(x)]
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return x[1]*pow(-1,abs(n)-1) if n<0 else x[1] if n else 0
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def fib4k(n): # FIBonacci's numbers, Binet's Formula, Barron's Binomial expansion, Kra's code.
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# (c) 2025 David A. Kra GNUFDL1.3 and Copyleft Creative Commons CC BY-SA Attribution-ShareAlike
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#
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# ALL INTEGER. Tested up to Fib(4000).
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# NOTE: This implementation is NOT faster than simply progressing up to Fib(n) by addition, starting from Fib(1).
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# Thanks, Acknowledgement, and Appreciation to Barron https://stackexchange.com/users/9594318/barron
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# No floats were exploited in the production of this function.
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# References:
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# https://math.stackexchange.com/questions/674570/prove-that-binets-formula-gives-an-integer-using-the-binomial-theorem
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# https://math.stackexchange.com/questions/2002702/fibonacci-identity-with-binomial-coefficients
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# https://artofproblemsolving.com/wiki/index.php/Binet%27s_Formula
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# https://latex.artofproblemsolving.com/8/6/d/86d486c560727727342090b432e23ba85ac098b1.png
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# https://www.geeksforgeeks.org/find-nth-fibonacci-number-using-binets-formula/
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# https://discuss.geeksforgeeks.org/comment/540dc728-8cfc-41e3-853e-e920e0a85101/gfg
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# # Fn=(1/(2**(n-1))) * SUM(j=0,n//2,((5**j)*comb(n,2*j+1))
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#
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# qc == Quick Combination. Instead of using comb, with its loop on each call,
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# derive the next ( comb(n,2*j+1) ) by building up from what had already been calculated,
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# Given comb(n,2*j+1), then
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# comb(n,2*(j+1)+1) = comb(n,2*j+1) * (n-(2*j+1))*(n-2*j+1)-1) // ( (2*j+1)+1)*(2*j+1)+2) )
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# qp == Quick Power. Instead of using 5**j, derive the next fttj as fttj*=5
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#
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f=0
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fttj=1
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qc=n # = comb(n,1)
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for j in range(0,int(n/2+1)):
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f+=fttj*qc # (5**k)*qc # qc == comb(n,2*k+1)
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j2p1=j*2+1
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# calculate the 5**k and the combinations for the next iteration,
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# but for now, k and k2p1 have this iteration's values.
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fttj*=5
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qc=qc* (n-j2p1)*(n-j2p1-1) // ( (j2p1+1)*(j2p1+2) ) # for the next iteration
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for i in range(-30,31): print fib(i),
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f=f//(2 ** (n-1) )
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return f
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</syntaxhighlight lang="python">
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<pre>
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Test
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print([[i,fib4k(i)] for i in [4,40,400,4000]])
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output
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[[4, 3], [40, 102334155], [400, 176023680645013966468226945392411250770384383304492191886725992896575345044216019675], [4000, 39909473435004422792081248094960912600792570982820257852628876326523051818641373433549136769424132442293969306537520118273879628025443235370362250955435654171592897966790864814458223141914272590897468472180370639695334449662650312874735560926298246249404168309064214351044459077749425236777660809226095151852052781352975449482565838369809183771787439660825140502824343131911711296392457138867486593923544177893735428602238212249156564631452507658603400012003685322984838488962351492632577755354452904049241294565662519417235020049873873878602731379207893212335423484873469083054556329894167262818692599815209582517277965059068235543139459375028276851221435815957374273143824422909416395375178739268544368126894240979135322176080374780998010657710775625856041594078495411724236560242597759185543824798332467919613598667003025993715274875]]
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</pre>
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===Iterative===
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<syntaxhighlight lang="python">def fib_iter(n):
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if n < 2:
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return n
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fib_prev = 1
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fib = 1
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for _ in range(2, n):
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fib_prev, fib = fib, fib + fib_prev
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return fib
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def fibRec(n):
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if n < 2:
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return n
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else:
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return fibRec(n-1) + fibRec(n-2)
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def fib(n,x=[0,1]):
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for i in range(abs(n)-1): x=[x[1],sum(x)]
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return x[1]*pow(-1,abs(n)-1) if n<0 else x[1] if n else 0
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for i in range(-30,31): print fib(i),
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def fibMemo():
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pad = {0:0, 1:1}
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def func(n):
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if n not in pad:
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pad[n] = func(n-1) + func(n-2)
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return pad[n]
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return func
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fm = fibMemo()
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for i in range(1,31):
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print fm(i),
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def fib_rec(n):
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if n < 2:
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return n
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return fib_rec(n-1) + fib_rec(n-2)
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def fibFastRec(n):
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def fib(prvprv, prv, c):
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if c < 1:
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return prvprv
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else:
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return fib(prv, prvprv + prv, c - 1)
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return fib(0, 1, n)
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def fib_memo():
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pad = {0:0, 1:1}
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def sub_func(n):
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if not n in pad:
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pad[n] = sub_func(n-1) + sub_func(n-2)
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return pad[n]
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return sub_func
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fm = fib_memo()
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for i in range(1,31):
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print(fm(i))
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def fibGen(n):
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a, b = 0, 1
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while n>0:
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yield a
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a, b, n = b, a+b, n-1
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def fib_fast_rec(n):
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def inner_fib(prvprv, prv, c):
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if c < 1:
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return prvprv
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return inner_fib(prv, prvprv + prv, c - 1)
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return inner_fib(0, 1, n)
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>>> [i for i in fibGen(11)]
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[0,1,1,2,3,5,8,13,21,34,55]
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def fib_gen(n):
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a, b = 0, 1
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while n>0:
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yield a
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a, b, n = b, a+b, n-1
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def prevPowTwo(n):
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'Gets the power of two that is less than or equal to the given input'
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if ((n & -n) == n):
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return n
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else:
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n -= 1
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n |= n >> 1
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n |= n >> 2
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n |= n >> 4
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n |= n >> 8
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n |= n >> 16
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n += 1
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return (n/2)
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>>> [i for i in fib_gen(11)]
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def crazyFib(n):
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'Crazy fast fibonacci number calculation'
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powTwo = prevPowTwo(n)
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q = r = i = 1
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s = 0
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while(i < powTwo):
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i *= 2
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q, r, s = q*q + r*r, r * (q + s), (r*r + s*s)
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while(i < n):
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i += 1
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q, r, s = q+r, q, r
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return q
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[0,1,1,2,3,5,8,13,21,34,55]
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