# How could I neaten this code up.

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Scragmore on 24 Oct 2011
Hi
Currently working through some of Project Eulers problems as mentioned in the forum discussion 'Best way(s) to master MATLAB'. I have created this code, it works but there is something about it I find messy. For a start I don't like all the declared variables.
function eusol2 = euler2(euin)
% for No's in fibonacci sequence < 4mil, sum all even No terms.
x = [1 2];
y = 1;
xout = 2;
fibMax = 0;
while fibMax < euin
fibMax = x(1) + x(2);
if fibMax < euin
if mod(fibMax, 2) == 0;
xout = xout + fibMax;
else
end
else
end
x(y) = fibMax;
y = 3 - y;
eusol2 = xout;
end
Any suggestions or remarks
Jan on 26 Oct 2011
You started the sequence with [1,2] accidently instead of [1,1]. Using xout=2 instead of 0 compensates this such that the correct answer is calculated.
A very nice example for a "stealth bug": No effect for standard usage, generalising the function for any Fibonacci sequence [a,b] fails.

Jan on 25 Oct 2011
Your function looks fine already. Most of all it is very fast, e.g. twice as fast as Daniel's program.
The result is wrong, because you start with xout=2. The empty ELSE branch wastes time. The "if fibMax < euin" check can be omited, if fibMax is increased after adding the new term. I prefer meaningful names instead of x and y.
function out = euler2(limit)
accum = [1, 1]; % [EDITED], was [1,2] as in the original question
index = 1;
out = 0;
fibMax = 0;
while fibMax < limit
if mod(fibMax, 2) == 0
out = out + fibMax;
end
fibMax = accum(1) + accum(2);
accum(index) = fibMax;
index = 3 - index;
end
Of course Binet's formula would be more sophisticated. But calling this function 10'000 times with the input 4e6 needs 0.15438 seconds only.
The MOD(fibMax, 2) fails after 75 terms due to the limited precision. Exploiting the simple pattern of even terms might be interesting, but a further "improvement" of the function is of accademical interest only. For a practical use this function is eitehr fast enough, of a lookup table with the 75 values is faster brute-force approach.
Scragmore on 26 Oct 2011
Thanks for the answers and explanation. I don't blame matlab, my maths is not good enough to accurately go beyond the size of those numbers either.

bym on 24 Oct 2011
Have a look at Binet's formula and look for a pattern in the distribution of even Fibonnaci numbers...I bet a one liner could be written

Daniel Shub on 25 Oct 2011
x = [1 2];
eusol2 = 0;
while sum(x) < euin
if mod(sum(x), 2) == 0;
eusol2 = eusol2 + sum(x);
end
x = [x(2), sum(x)];
% k = sum(x);
end
This will likely be a little slower since it calculates the sum(x) 4 times per loop. If you add another declared variable (k) you can make it so sum(x) is calculated only once. I eliminated y, because it doesn't seem to do anything.
Scragmore on 25 Oct 2011
Thanks,
I didn't know you could sum a complete matrix like that. I presume as its matlap and matrices are native you can sum in a hole bunch of ingenious ways.
My y is a switch, it flips between 1 & 2, in this case cell positions of x(y), allowing me to alternate storing the latest number in the series.
From reading your code I can use my function output, in this case eusol2 as a declared variable. I was worried that every time I wrote to this variable it would return it from the function. I will test/try this.
Thanks,

bym on 26 Oct 2011
just to let you know where I was headed:
clc;clear;tic
phi = (1+sqrt(5))/2;
i = 1:11; % 11 even terms < 4e6
fib = (phi.^(i.*3)-(-1/phi).^(i.*3))/sqrt(5);
fprintf('%6.0f\n',sum(fib))
toc
as Jan suspected, it runs slower than yours (about .8 msec vs .07 msec). +1 vote for the question