function [Fn,Ln] = fibonacci(n,modulus)
% fibonacci: vpi tool to efficiently compute the n'th Fibonacci number and the n'th Lucas number
% usage: [Fn,Ln] = fibonacci(n)
% usage: [Fn,Ln] = fibonacci(n,modulus)
%
% Compute the nth Fibonacci number as well as the nth Lucas
% Lucas number. In the event that all members of these
% sequences from 1 to n are desired, then a simple,
% direct loop would be more efficient, and more direct.
%
% Both the Fibonacci numbers and the Lucas numbers
% are defined by the same basic recursion:
%
% F(n) = F(n-1) + F(n-2)
% L(n) = L(n-1) + L(n-2)
%
% The difference is the starting point. The Fibonacci
% numbers start with F(1) = F(2) = 1, whereas the Lucas
% sequence starts with L(1) = 1, and L(2) = 3. The first
% few members of these sequences are:
%
% Fibonacci: [1 1 2 3 5 8 13 21 ... ]
% Lucas: [1 3 4 7 11 18 29 ... ]
%
% These sequences are also defined for n = 0 and for
% negative values of n.
%
% For efficiency, fibonacci uses a variety of tricks to
% maximize speed. While computation of fibonacci numbers
% is commonly done recursively, fibonacci does so using a
% direct iterative scheme given the binary representation
% of n. In addition, several Fibonacci and Lucas number
% identities are employed to maximize throughput.
%
% The methods employed by fibonacci will be O(log2(n)) in time.
%
%
% Arguments: (input)
% n - any non-negative integer, vpi or numeric.
% n must be less than 2^53 (in theory. Even that
% number would be impossibly huge to compute.
% Don't bother to try it.) In practice, recognize
% that F(1e6) is a number with 208995 digits.
%
% When n is a vector or array, fibonacci will
% generate each of the indicated Fibonacci and
% Lucas numbers.
%
% modulus - (OPTIONAL) - allows the computation of the
% indicated Fibonacci/Lucas numbers modulo a given
% modulus. This enables the computation of such
% numbers for truly immense index.
%
% Arguments: (output)
% Fn, Ln - scalar vpi number, containing the nth
% Fibonacci number and nth Lucas numbers in
% their respective sequences.
%
% Example:
% [Fn,Ln] = fibonacci(150)
%
% Fn =
% 9969216677189303386214405760200
% Ln =
% 22291846172619859445381409012498
%
% See also:
%
% Author: John D'Errico
% e-mail: woodchips@rochester.rr.com
% Release: 1.0
% Release date: 4/30/09
if (nargin < 1) || (nargin > 2)
error('fibonacci accepts only 1 or 2 arguments')
elseif any(n(:)~=round(n(:))) || any(abs(n) > 2^53)
error('n must be an integer, <= 2^53 in absolute value')
end
% was a modulus provided?
if (nargin < 2)
modulus = [];
end
% The first 15 Fibonacci and Lucas numbers to
% start things off efficiently.
Fseq = [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610];
Lseq = [2 1 3 4 7 11 18 29 47 76 123 199 322 521 843 1364];
if ~isempty(modulus)
Fseq = mod(Fseq,modulus);
Lseq = mod(Lseq,modulus);
end
% intialize Fn and Ln to the proper size,
% in case n is a vector or array
Fn = repmat(vpi(0),size(n));
Ln = Fn;
% much faster if n is not a vpi but a double.
% also, we don't need to worry about n being
% larger than 2^53, as this would have a vast
% number of digits.
n = double(n(:));
% catch any zeros in n first.
k = (n(:) == 0);
% Fn(k) is already zero
if any(k)
Ln(k) = vpi(2);
end
% Negative values for n will be inconvenient
% in a loop, so make them all positive. deal
% with any signs later.
nsign = 2*(n>=0) - 1;
if any(n<0)
neven = iseven(n);
n = abs(n);
end
if numel(n) > 1
% sort them and work upwards
[n,ntags] = sort(n);
flag = false;
for i = 1:numel(n)
if n(i) <= 15
% small values of n are pre-computed
Fn(i) = Fseq(n(i) + 1);
Ln(i) = Lseq(n(i) + 1);
elseif (i == 1) || ((n(i) - n(i-1)) > 15)
% the smallest value of n is > 15,
% or the difference from the last value
% computed was too large
[Fn(i),Ln(i)] = fibonacci(n(i),modulus);
flag = false;
elseif n(i) == n(i-1)
% will happen if there were positive and
% negative values in the list
Fn(i) = Fn(i-1);
Ln(i) = Ln(i-1);
flag = false;
elseif (n(i) == (n(i-1)+1))
% we can use an addition formula to
% get Fn & Ln. Which one?
if flag
% the last two values of n were
% separated by 1, so just use the
% two term recurrence
Fn(i) = (Fn(i-1) + Fn(i-2));
Ln(i) = (Ln(i-1) + Ln(i-2));
if ~isempty(modulus)
Fn(i) = mod(Fn(i),modulus);
Ln(i) = mod(Ln(i),modulus);
end
else
% we must use the addition formula
Fn(i) = (Fn(i-1) + Ln(i-1))./2;
Ln(i) = (5 .*Fn(i-1) + Ln(i-1))./2;
if ~isempty(modulus)
Fn(i) = mod(Fn(i),modulus);
Ln(i) = mod(Ln(i),modulus);
end
% flag indicates whether the last two
% numbers were consecutive
flag = true;
end
else
% 1 < n(i) - n(i-1) <= 15
m = n(i) - n(i-1);
Fm = Fseq(m+1);
Lm = Lseq(m+1);
% use an addition formula
Fn(i) = (Fm.*Ln(i-1) + Lm.*Fn(i-1))./2;
Ln(i) = (5 .*Fm.*Fn(i-1) + Lm.*Ln(i-1))./2;
if ~isempty(modulus)
Fn(i) = mod(Fn(i),modulus);
Ln(i) = mod(Ln(i),modulus);
end
flag = false;
end
end % for i = 1:numel(n)
% shuffle Fn and Ln to reflect the sort
Fn(ntags) = Fn;
Ln(ntags) = Ln;
else
% For only one value of n, compute it individually.
% Uses an efficient iterative (recursive, but not
% really so) scheme.
% get the binary representation of n. Thus
% nbin is a character vector, of length
% ceil(log2(n)).
nbin = dec2bin(n);
% get the 4 highest order bits from nbin
k = min(numel(nbin),4);
% start the sequence from the top
% few bits of n.
nhigh = bin2dec(nbin(1:k));
Fn = vpi(Fseq(nhigh+1));
Ln = vpi(Lseq(nhigh+1));
% We need to loop forwards. Essentially, we started
% with the highest order bit(s) of the binary representation
% for n. Look at each successively lower order bit.
% If the next bit is 0, then we are essentially doubling
% the index at this step. If the next bit is odd, then
% we are moving to 2*n+1.
for k = 5:numel(nbin)
bit = (nbin(k) == '1');
if bit
% the next bit was odd. Use
% a 2*n+1 rule to step up.
F2n = Fn.*Ln;
% we want to do this...
% L2n = (5 .*Fn.*Fn + Ln.*Ln)./2;
% Instead use the identity that
% 5F(n)^2 + L(n)^2 = 2*L(n)^2 + 4*(-1)^(n+1)
% to make that expression more efficiently
% computed. See that the form used below
% has only a single multiplication between a
% pair of large integers, whereas the prior
% form for L2n would have had several multiples
% as well as a divide.
L2n = Ln.*Ln + 2*(-1)^(nhigh+1);
Fn = (L2n + F2n)./2;
Ln = (5 .*F2n + L2n)./2;
if ~isempty(modulus)
Fn = mod(Fn,modulus);
Ln = mod(Ln,modulus);
end
else
% the next bit was even. Use the 2*n
% rule to step up.
F2n = Fn.*Ln;
Ln = Ln.*Ln + 2*(-1)^(nhigh+1);
Fn = F2n;
if ~isempty(modulus)
Fn = mod(Fn,modulus);
Ln = mod(Ln,modulus);
end
end
% update the top bits of n
nhigh = 2*nhigh + bit;
end % for k = 5:numel(nbin)
end % if numel(n) > 1
% if n was negative, then we may need to apply a
% sign change to Fn and Ln.
if any(nsign < 0)
k = neven & (nsign < 0);
Fn(k) = -Fn(k);
k = (~neven) & (nsign < 0);
Ln(k) = -Ln(k);
if ~isempty(modulus)
Fn(k) = mod(Fn(k),modulus);
Ln(k) = mod(Ln(k),modulus);
end
end
% ==================================
% End mainline, begin subfunctions.
% ==================================
function result = iseven(n)
% tests if a scalar value is an even integer, works
% for either numeric or vpi inputs
if isnumeric(n)
result = (mod(n,2) == 0);
elseif isa(n,'vpi')
% must have been a vpi
result = (mod(trailingdigit(n,1),2) == 0);
else
error('n must be either numeric or vpi')
end