function [Fn,Ln] = modfibonacci(n,modulus)
% fibonacci: compute the n'th Fibonacci number and the n'th Lucas number, all modulo a given value
% usage: [Fn,Ln] = modfibonacci(n,modulus)
% Compute the nth Fibonacci number as well as the nth Lucas
% Lucas number.
% 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.
% When n is a vector or array, fibonacci will
% generate the modulus for each of the indicated
% Fibonacci and Lucas numbers.
% modulus - scalar integer value, used to compute mod(n,modulus)
% Arguments: (output)
% Fn, Ln - scalar vpi number, containing the nth
% Fibonacci number and nth Lucas numbers in
% their respective sequences.
% % Find the 50 trailing digits of fibonacci(17^17) (an immensely huge number)
% tic,[Fn,Ln] = modfibonacci(vpi(17)^17,vpi(10)^50),toc
% Elapsed time is 0.853440 seconds.
% Fn =
% Ln =
% See also: fibonacci
% Author: John D'Errico
% e-mail: email@example.com
% Release: 1.0
% Release date: 10/13/2010
if (nargin ~= 2)
error('fibonacci accepts only 2 arguments')
error('n must be an integer')
% 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];
% 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
Ln(k) = vpi(2);
% 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;
neven = iseven(n);
n = abs(n);
for in = 1:numel(n)
ni = n(in);
% For each 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
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 = mod(vpi(Fseq(nhigh+1)),modulus);
Ln = mod(vpi(Lseq(nhigh+1)),modulus);
% 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');
% 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;
Fn = Fn + modulus;
Fn = mod(Fn/2,modulus);
Ln = 5 .*F2n + L2n;
Ln = Ln + modulus;
Ln = mod(Ln/2,modulus);
% the next bit was even. Use the 2*n
% rule to step up.
F2n = mod(Fn.*Ln,modulus);
Ln = mod(Ln.*Ln + 2*(-1)^(nhigh+1),modulus);
Fn = F2n;
% 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);
% End mainline, begin subfunctions.
function result = iseven(n)
% tests if a scalar value is an even integer, works
% for either numeric or vpi inputs
result = (mod(n,2) == 0);
% must have been a vpi
result = (mod(trailingdigit(n,1),2) == 0);
error('n must be either numeric or vpi')