Can I easily skip pixels in Bresenham's line algorithm? - algorithm

I have a program which is using Bresenham's line algorithm to scan pixels in a line. This is reading pixels rather than writing them, and in my particular case, reading them is costly.
I can however determine that some spans of pixels do not need to be read. It looks something like this:
Normal scan of all pixels:
*start
\
\
\
\
\
*end
Scan without reading all pixels:
*start
\
\
- At this point I know I can skip (for example) the next 100 pixels
in the loop. Crucially, I can't know this until I reach the gap.
\
*end
The gap in the middle is much quicker because I can just iterate over the pixels without reading them.
However, can I modify the loop in any way to just jump directly forward 100 pixels within the loop, calculating directly the required values 100 steps ahead in the line algorithm?


Bresenhams middlepoint algorithm calculates 'distance' of point from a theoretical line going from (ax,ay)->(bx,by) by summing up digital differences delta_x = (by-ay), delta_y = (ax-bx).
Thus, if one want's to skip 7 pixels, one has to add accum += 7*delta_x; then dividing by delta_y one can check how many pixels should have been moved in y-direction and taking a remainder accum = accum % delta_y one should be able to continue at proper position.
The nice thing is that the algorithm is originated from the necessity of avoiding a division...
Disclaimer: whatever told may need to be adjusted by half delta.


Your main loop looks essentially something like:
while (cnt > 0) // cnt is 1 + the biggest of abs(x2-x1) and abs(y2-y1)
{
ReadOrWritePixel(x, y);
k += n; // n is the smallest of abs(x2-x1) and abs(y2-y1)
if (k < m) // m is the biggest of abs(x2-x1) and abs(y2-y1)
{
// continuing a horizontal/vertical segment
x += dx2; // dx2 = sgn(x2-x1) or 0
y += dy2; // dy2 = sgn(y2-y1) or 0
}
else
{
// beginning a new horizontal/vertical segment
k -= m;
x += dx1; // dx1 = sgn(x2-x1)
y += dy1; // dy1 = sgn(y2-y1)
}
cnt--;
}
So, skipping some q pixels is equivalent to the following adjustments (unless I made a mistake somewhere):
cntnew = cntold - q
knew = (kold + n * q) % m
xnew = xold + ((kold + n * q) / m) * dx1 + (q - ((kold + n * q) / m)) * dx2
ynew = yold + ((kold + n * q) / m) * dy1 + (q - ((kold + n * q) / m)) * dy2
Note that / and % are integer division and modulo operators.

Related

Given some rounded numbers, how to find the original fraction?

After asking this question on math.stackexchange.com I figured this might be a better place after all...
I have a small list of positive numbers rounded to (say) two decimals:
1.15 (can be 1.145 - 1.154999...)
1.92 (can be 1.915 - 1.924999...)
2.36 (can be 2.355 - 2.364999...)
2.63 (can be 2.625 - 2.634999...)
2.78 (can be 2.775 - 2.784999...)
3.14 (can be 3.135 - 3.144999...)
24.04 (can be 24.035 - 24.044999...)
I suspect that these numbers are fractions of integers and that all numerators or all denominators are equal. Choosing 100 as a common denominator would work in this case, that would leave the last value as 2404/100. But there could be a 'simpler' solution with much smaller integers.
How do I efficiently find the smallest common numerator and/or denominator? Or (if that is different) the one that would result in the smallest maximum denominator resp. numerator?
Of course I could brute force for small lists/numbers and few decimals. That would find 83/72, 138/72, 170/72, 189/72, 200/72, 226/72 and 1731/72 for this example.

Assuming the numbers don't have too many significant digits and aren't too big you can try increasing the denominator until you find a valid solution. It is not just brute-forcing. Additionally the following script is staying at the number violating the constraints as long as there is nothing found, in the hope of getting the denominator higher faster, without having to calculate for the non-problematic numbers.
It works based on the following formula:
x / y < a / b if x * b < a * y
This means a denominator d is valid if:
ceil(loNum * d / loDen) * hiDen < hiNum * d
The ceil(...) part calculates the smallest possible numerator satisfying the constraint of the low boundary and the rest is checking if it also satysfies the high boundary.
Better would be to work with real integer calculations, e.g. just longs in Java, then the ceil part becomes:
(loNum * d + loDen - 1) / loDen
function findRatios(arr) {
let lo = [], hi = [], consecutive = 0, d = 1
for (let i = 0; i < arr.length; i++) {
let x = '' + arr[i], len = x.length, dot = x.indexOf('.'),
num = parseInt(x.substr(0, dot) + x.substr(dot + 1)) * 10,
den = Math.pow(10, len - dot),
loGcd = gcd(num - 5, den), hiGcd = gcd(num + 5, den)
lo[i] = {num: (num - 5) / loGcd, den: den / loGcd}
hi[i] = {num: (num + 5) / hiGcd, den: den / hiGcd}
}
for (let index = 0; consecutive < arr.length; index = (index + 1) % arr.length) {
if (!valid(d, lo[index], hi[index])) {
consecutive = 1
d++
while (!valid(d, lo[index], hi[index]))
d++
} else {
consecutive++
}
}
for (let i = 0; i < arr.length; i++)
console.log(Math.ceil(lo[i].num * d / lo[i].den) + ' / ' + d)
}
function gcd(x, y) {
while(y) {
let t = y
y = x % y
x = t
}
return x
}
function valid(d, lo, hi) {
let n = Math.ceil(lo.num * d / lo.den)
return n * hi.den < hi.num * d
}
findRatios([1.15, 1.92, 2.36, 2.63, 2.78, 3.14, 24.04])

subscript indices must be either positiveintegers less than 2^31 or logicals

SOS i keep getting errors in the loop solving by finite difference method.
I either get the following error when i start with i = 2 : N :
diffusion: A(I,J): row index out of bounds; value 2 out of bound 1
error: called from
diffusion at line 37 column 10 % note line change due to edit!
or, I get the following error when i do i = 2 : N :
subscript indices must be either positive integers less than 2^31 or logicals
error: called from
diffusion at line 37 column 10 % note line change due to edit!
Please help
clear all; close all;
% mesh in space
dx = 0.1;
x = 0 : dx : 1;
% mesh in time
dt = 1 / 50;
t0 = 0;
tf = 10;
t = t0 : dt : tf;
% diffusivity
D = 0.5;
% number of nodes
N = 11;
% number of iterations
M = 10;
% initial conditions
if x <= .5 && x >= 0 % note, in octave, you don't need parentheses around the test expression
u0 = x;
elseif
u0 = 1-x;
endif
u = u0;
alpha = D * dt / (dx^2);
for j = 1 : M
for i = 1 : N
u(i, j+1) = u(i, j ) ...
+ alpha ...
* ( u(i-1, j) ...
+ u(i+1, j) ...
- 2 ...
* u(i, j) ...
) ;
end
u(N+1, j+1) = u(N+1, j) ...
+ alpha ...
* ( ...
u(N, j) ...
- 2 ...
* u(N+1, j) ...
+ u(N, j) ...
) ;
% boundary conditions
u(0, :) = u0;
u(1, :) = u1;
u1 = u0;
u0 = 0;
end
% exact solution with 14 terms
%k=14 % COMMENTED OUT
v = (4 / ((k * pi) .^ 2)) ...
* sin( (k * pi) / 2 ) ...
* sin( k * pi * x ) ...
* exp .^ (D * ((k * pi) ^ 2) * t) ;
exact = symsum( v, k, 1, 14 );
error = exact - u;
% plot stuff
plot( t, error );
xlabel( 'time' );
ylabel( 'error' );
legend( 't = 1 / 50' );

Have a look at the edited code I cleaned up for you above and study it.
Don't underestimate the importance of clean, readable code when hunting for bugs.
It will save you more time than it will cost. Especially a week from now when you will need to revisit this code and you will not remember at all what you were trying to do.
Now regarding your errors. (all line references are with respect to the cleaned up code above)
Scenario 1:
In line 29 you initialise u as a single value.
If you start your loop in line 35 starting with i = 2, then as soon as you try to do u(i, j+1), i.e. u(2,2) in the next line, octave will complain that you're trying to index the second row, in an array that so far only contains one row. (in fact, the same will apply for j at this point, since at this point you only have one column as well)
Scenario 2:
I assume the second scenario was a typo and you meant to say i = 1 : N.
If you start with i=1 in the loop, then have a look at line 38: you are trying to get element u(i-1, j), i.e. u(0,1). Therefore octave will complain that you're trying to get the zero element, but in octave arrays start from one and zero is not defined. Attempting to access any array with a zero will result in the error you see (try it in a terminal!).
UPDATE
Also, now that the code is clean, you can spot another bug, which octave helpfully warns you about if you try to run the code.
Look at line 26. There is NO condition in the elseif leg, so octave looks for the next statement as the test condition.
This means that the elseif condition will always succeed as long as the result of u0 = 1-x is non-zero.
This is clearly a bug. Either you forgot to put the condition for the elseif, or more likely, you probably just meant to say else, rather than elseif.

Improving the Efficiency Of This Code With Tracking Variable?

I have written the below code outline, basically to sum an array (a) where each element is multiplied by a value x^i:
y = a(0)
i = 0
{y = sum from i=0 to (n-1) a(i) * x^i AND 0 <= n <= a.length} //Invariant
while (i < (n-1))
{y = sum from i=0 to (n-1) a(i) * x^i AND 0 <= n <= a.length AND i < (n-1)}
y = y + a(i)*x^i
i = i + 1
end while
{y = sum from i=0 to (n-1) a(i) * x^i} //Postcondition
Note that I do not expect the code to compile - it's just a sensible outline of how the code should work. I need to improve the efficiency of the code by using a tracking variable, and thus, a linking invariant to bridge said variable with the rest of the code. This is where I am stuck. What would be useful to track in this case? I have thought about retaining sum values at each iteration, but I'm not sure if that does the trick. If I could figure out what to track, I'm pretty sure it would be trivial to link it to the space. Can anyone see how my algorithm might be improved via a tracking variable?

Your invariant logic has off-by-1 problems. Here is a corrected version that tracks partial power operations.
// Precondition: 1 <= n <= a.length
// Invariant:
{ 0 <= i < n AND xi = x^i AND y = sum(j = 0..i) . a(j) * x^j }
// Establish invariant at i = 0:
// xi = x^0 = 1 AND y = sum(j=0..0) . a(j) * x^j = a(0) * x^0 = a(0)
i = 0;
xi = 1;
y = a(0);
while (i < n - 1) {
i = i + 1; // Break the invariant
xi = xi * x; // Re-establish it
y = y + a(i) * xi
}
// Invariant was last established at i = n-1, so we have post condition:
{ y = sum(j = 0..n-1) . a(j) * x^j }
The more common and numerically stable way to calculate polynomials is with Horner's Rule
y = 0
for i = n-1 downto 0 do y = y * x + a(i)

So it seems like you're trying to end up with this:
(a(0)*x^0) + (a(1)*x^1) + ... + (a(n-1)*x^(n-1))
Is that right?
The only way I can see to improve performance would be if the ^ operation is more costly than the * operation. In that case, you could keep track of the x^n variable as you go, multiplying x by the value through each iteration.
In fact, in that case you could probably start at the end of the array and work your way backwards, multiplying by x each time, to produce:
(((...((a(n-1)*x+a(n-2))*x+...)+a(2))*x+a(1))*x)+a(0)
That would theoretically be slightly faster than recalculating x^i each time, but it's not going to be algorithmically faster. It probably wouldn't be an order of magnitude faster.

Calculating value for n when incrementing a value using a for loop

First of all, sorry for the bad title. I'm not really sure how to title this topic, so feel free to mod it where necessary.
I am drawing X rings inside my stage with given dimensions. To give this some sense of depth, each ring towards the screen boundaries is slightly wider:
The largest ring should be as wide as the largest dimension of the stage (note that in the picture i am drawing 3 extra rings which are drawn outside the stage boundaries). Also it should be twice as wide as the smallest ring. With ring i am refering to the space between 2 red circles.
After calculating an _innerRadius, which is the width of the smallest ring, i am drawing them using
const RINGS:Number = 10; //the amount of rings, note we will draw 3 extra rings to fill the screen
const DEPTH:Number = 7; //the amount of size difference between rings to create depth effect
var radius:Number = 0;
for (var i:uint = 0; i < RINGS + 3; i++) {
radius += _innerRadius + _innerRadius * ((i*DEPTH) / (RINGS - 1));
_graphics.lineStyle(1, 0xFF0000, 1);
_graphics.drawCircle(0, 0, radius * .5);
}
One of the sliders at the bottom goes from 0-100 being a percentage of the radius for the green ring which goes from the smallest to the largest ring.
I tried lerping between the smallest radius and the largest radius, which works fine if the DEPTH value is 1. However I don't want the distance between the rings to be the same for the sake of the illusion of depth.
Now I've been trying to figure this out for hours but it seems I've run into a wall.. it seems like I need some kind of non-linear formula here.. How would I calculate the radius based on the slider percentage value? Effectively for anywhere in between or on the red circles going from the smallest to the largest red circle?
thanks!
[edit]
Here's my example calculation for _innerRadius
//lets calculate _innerRadius for 10 rings
//inner ring width = X + 0/9 * X;
//ring 1 width = X + 1/9 * X;
//ring 2 width = X + 2/9 * X
//ring 3 width = X + 3/9 * X
//ring 4 width = X + 4/9 * X
//ring 5 width = X + 5/9 * X
//ring 6 width = X + 6/9 * X
//ring 8 width = X + 7/9 * X
//ring 9 width = X + 8/9 * X
//ring 10 width = X + 9/9 * X
//extent = Math.max(stage.stageWidth, stage.stageHeight);
//now we should solve extent = X + (X + 0/9 * X) + (X + 1/9 * X) + (X + 2/9 * X) + (X + 3/9 * X) + (X + 4/9 * X) + (X + 5/9 * X) + (X + 6/9 * X) + (X + 7/9 * X) + (X + 8/9 * X) + (X + 9/9 * X);
//lets add all X's
//extent = 10 * X + 45/9 * X
//extent = 15 * X;
//now reverse to solve for _innerRadius
//_innerRadius = extent / 15;

The way your drawing algorithm works, your radii are:
r[i + 1] = r[i] + (1 + i*a)*r0
where a is a constant that is depth / (rings - 1). This results in:
r0
r1 = r0 + (1 + a)*r0 = (2 + a)*r0
r2 = r1 + (1 + 2*a)*r0 = (3 + 3*a)*r0
r3 = r2 + (1 + 3*a)*r0 = (4 + 6*a)*r0
...
rn = (1 + n + a * sum(1 ... n))*r0
= (1 + n + a * n*(n - 1) / 2)*r0
Because you want ring n - 1 to correspond with your outer radius (let's forget about the three extra rings for now), you get:
r[n - 1] = (n + (n - 1)*(n - 2) / 2)*r0
a = 2 * (extent / r0 - n) / (n - 1) / (n - 2)
Then you can draw the rings:
for (var i = 0; i < rings; i++) {
r = (1 + i + 0.5 * a * (i - 1)*i) * r0;
// draw circle with radius r
}
You must have at least three rings in order not to have a division by zero when calculating a. Also note that this does not yield good results for all combinations: if the ratio of outer and inner circles is smaller than the number of rings, you get a negative a and have the depth effect reversed.
Another, maybe simpler, approach to create the depth effect is to make each circle's radius a constant multiple of the previous:
r[i + 1] = r[i] * c
or
r[i] = r0 * Math.pow(c, i)
and draw them like this:
c = Math.pow(extent / r0, 1 / (rings - 1))
r = r0
for (var i = 0; i < rings; i++) {
// draw circle with radius r
r *= c;
}
This will create a "positive" depth effect as long as the ratio of radii is positive. (And as long as there is more than one ring, of course.)

How do you calculate the average of a set of circular data?

I want to calculate the average of a set of circular data. For example, I might have several samples from the reading of a compass. The problem of course is how to deal with the wraparound. The same algorithm might be useful for a clockface.
The actual question is more complicated - what do statistics mean on a sphere or in an algebraic space which "wraps around", e.g. the additive group mod n. The answer may not be unique, e.g. the average of 359 degrees and 1 degree could be 0 degrees or 180, but statistically 0 looks better.
This is a real programming problem for me and I'm trying to make it not look like just a Math problem.

Compute unit vectors from the angles and take the angle of their average.

This question is examined in detail in the book:
"Statistics On Spheres", Geoffrey S. Watson, University of Arkansas Lecture
Notes in the Mathematical Sciences, 1983 John Wiley & Sons, Inc. as mentioned at http://catless.ncl.ac.uk/Risks/7.44.html#subj4 by Bruce Karsh.
A good way to estimate an average angle, A, from a set of angle measurements
a[i] 0<=i
sum_i_from_1_to_N sin(a[i])
a = arctangent ---------------------------
sum_i_from_1_to_N cos(a[i])
The method given by starblue is computationally equivalent, but his reasons are clearer and probably programmatically more efficient, and also work well in the zero case, so kudos to him.
The subject is now explored in more detail on Wikipedia, and with other uses, like fractional parts.

I see the problem - for example, if you have a 45' angle and a 315' angle, the "natural" average would be 180', but the value you want is actually 0'.
I think Starblue is onto something. Just calculate the (x, y) cartesian coordinates for each angle, and add those resulting vectors together. The angular offset of the final vector should be your required result.
x = y = 0
foreach angle {
x += cos(angle)
y += sin(angle)
}
average_angle = atan2(y, x)
I'm ignoring for now that a compass heading starts at north, and goes clockwise, whereas "normal" cartesian coordinates start with zero along the X axis, and then go anti-clockwise. The maths should work out the same way regardless.

FOR THE SPECIAL CASE OF TWO ANGLES:
The answer ( (a + b) mod 360 ) / 2 is WRONG. For angles 350 and 2, the closest point is 356, not 176.
The unit vector and trig solutions may be too expensive.
What I've got from a little tinkering is:
diff = ( ( a - b + 180 + 360 ) mod 360 ) - 180
angle = (360 + b + ( diff / 2 ) ) mod 360
0, 180 -> 90 (two answers for this: this equation takes the clockwise answer from a)
180, 0 -> 270 (see above)
180, 1 -> 90.5
1, 180 -> 90.5
20, 350 -> 5
350, 20 -> 5 (all following examples reverse properly too)
10, 20 -> 15
350, 2 -> 356
359, 0 -> 359.5
180, 180 -> 180

ackb is right that these vector based solutions cannot be considered true averages of angles, they are only an average of the unit vector counterparts. However, ackb's suggested solution does not appear to mathematically sound.
The following is a solution that is mathematically derived from the goal of minimising (angle[i] - avgAngle)^2 (where the difference is corrected if necessary), which makes it a true arithmetic mean of the angles.
First, we need to look at exactly which cases the difference between angles is different to the difference between their normal number counterparts. Consider angles x and y, if y >= x - 180 and y <= x + 180, then we can use the difference (x-y) directly. Otherwise, if the first condition is not met then we must use (y+360) in the calculation instead of y. Corresponding, if the second condition is not met then we must use (y-360) instead of y. Since the equation of the curve we are minimising only changes at the points where these inequalities change from true to false or vice versa, we can separate the full [0,360) range into a set of segments, separated by these points. Then, we only need to find the minimum of each of these segments, and then the minimum of each segment's minimum, which is the average.
Here's an image demonstrating where the problems occur in calculating angle differences. If x lies in the gray area then there will be a problem.
To minimise a variable, depending on the curve, we can take the derivative of what we want to minimise and then we find the turning point (which is where the derivative = 0).
Here we will apply the idea of minimise the squared difference to derive the common arithmetic mean formula: sum(a[i])/n. The curve y = sum((a[i]-x)^2) can be minimised in this way:
y = sum((a[i]-x)^2)
= sum(a[i]^2 - 2*a[i]*x + x^2)
= sum(a[i]^2) - 2*x*sum(a[i]) + n*x^2
dy\dx = -2*sum(a[i]) + 2*n*x
for dy/dx = 0:
-2*sum(a[i]) + 2*n*x = 0
-> n*x = sum(a[i])
-> x = sum(a[i])/n
Now applying it to curves with our adjusted differences:
b = subset of a where the correct (angular) difference a[i]-x
c = subset of a where the correct (angular) difference (a[i]-360)-x
cn = size of c
d = subset of a where the correct (angular) difference (a[i]+360)-x
dn = size of d
y = sum((b[i]-x)^2) + sum(((c[i]-360)-b)^2) + sum(((d[i]+360)-c)^2)
= sum(b[i]^2 - 2*b[i]*x + x^2)
+ sum((c[i]-360)^2 - 2*(c[i]-360)*x + x^2)
+ sum((d[i]+360)^2 - 2*(d[i]+360)*x + x^2)
= sum(b[i]^2) - 2*x*sum(b[i])
+ sum((c[i]-360)^2) - 2*x*(sum(c[i]) - 360*cn)
+ sum((d[i]+360)^2) - 2*x*(sum(d[i]) + 360*dn)
+ n*x^2
= sum(b[i]^2) + sum((c[i]-360)^2) + sum((d[i]+360)^2)
- 2*x*(sum(b[i]) + sum(c[i]) + sum(d[i]))
- 2*x*(360*dn - 360*cn)
+ n*x^2
= sum(b[i]^2) + sum((c[i]-360)^2) + sum((d[i]+360)^2)
- 2*x*sum(x[i])
- 2*x*360*(dn - cn)
+ n*x^2
dy/dx = 2*n*x - 2*sum(x[i]) - 2*360*(dn - cn)
for dy/dx = 0:
2*n*x - 2*sum(x[i]) - 2*360*(dn - cn) = 0
n*x = sum(x[i]) + 360*(dn - cn)
x = (sum(x[i]) + 360*(dn - cn))/n
This alone is not quite enough to get the minimum, while it works for normal values, that has an unbounded set, so the result will definitely lie within set's range and is therefore valid. We need the minimum within a range (defined by the segment). If the minimum is less than our segment's lower bound then the minimum of that segment must be at the lower bound (because quadratic curves only have 1 turning point) and if the minimum is greater than our segment's upper bound then the segment's minimum is at the upper bound. After we have the minimum for each segment, we simply find the one that has the lowest value for what we're minimising (sum((b[i]-x)^2) + sum(((c[i]-360)-b)^2) + sum(((d[i]+360)-c)^2)).
Here is an image to the curve, which shows how it changes at the points where x=(a[i]+180)%360. The data set is in question is {65,92,230,320,250}.
Here is an implementation of the algorithm in Java, including some optimisations, its complexity is O(nlogn). It can be reduced to O(n) if you replace the comparison based sort with a non comparison based sort, such as radix sort.
static double varnc(double _mean, int _n, double _sumX, double _sumSqrX)
{
return _mean*(_n*_mean - 2*_sumX) + _sumSqrX;
}
//with lower correction
static double varlc(double _mean, int _n, double _sumX, double _sumSqrX, int _nc, double _sumC)
{
return _mean*(_n*_mean - 2*_sumX) + _sumSqrX
+ 2*360*_sumC + _nc*(-2*360*_mean + 360*360);
}
//with upper correction
static double varuc(double _mean, int _n, double _sumX, double _sumSqrX, int _nc, double _sumC)
{
return _mean*(_n*_mean - 2*_sumX) + _sumSqrX
- 2*360*_sumC + _nc*(2*360*_mean + 360*360);
}
static double[] averageAngles(double[] _angles)
{
double sumAngles;
double sumSqrAngles;
double[] lowerAngles;
double[] upperAngles;
{
List<Double> lowerAngles_ = new LinkedList<Double>();
List<Double> upperAngles_ = new LinkedList<Double>();
sumAngles = 0;
sumSqrAngles = 0;
for(double angle : _angles)
{
sumAngles += angle;
sumSqrAngles += angle*angle;
if(angle < 180)
lowerAngles_.add(angle);
else if(angle > 180)
upperAngles_.add(angle);
}
Collections.sort(lowerAngles_);
Collections.sort(upperAngles_,Collections.reverseOrder());
lowerAngles = new double[lowerAngles_.size()];
Iterator<Double> lowerAnglesIter = lowerAngles_.iterator();
for(int i = 0; i < lowerAngles_.size(); i++)
lowerAngles[i] = lowerAnglesIter.next();
upperAngles = new double[upperAngles_.size()];
Iterator<Double> upperAnglesIter = upperAngles_.iterator();
for(int i = 0; i < upperAngles_.size(); i++)
upperAngles[i] = upperAnglesIter.next();
}
List<Double> averageAngles = new LinkedList<Double>();
averageAngles.add(180d);
double variance = varnc(180,_angles.length,sumAngles,sumSqrAngles);
double lowerBound = 180;
double sumLC = 0;
for(int i = 0; i < lowerAngles.length; i++)
{
//get average for a segment based on minimum
double testAverageAngle = (sumAngles + 360*i)/_angles.length;
//minimum is outside segment range (therefore not directly relevant)
//since it is greater than lowerAngles[i], the minimum for the segment
//must lie on the boundary lowerAngles[i]
if(testAverageAngle > lowerAngles[i]+180)
testAverageAngle = lowerAngles[i];
if(testAverageAngle > lowerBound)
{
double testVariance = varlc(testAverageAngle,_angles.length,sumAngles,sumSqrAngles,i,sumLC);
if(testVariance < variance)
{
averageAngles.clear();
averageAngles.add(testAverageAngle);
variance = testVariance;
}
else if(testVariance == variance)
averageAngles.add(testAverageAngle);
}
lowerBound = lowerAngles[i];
sumLC += lowerAngles[i];
}
//Test last segment
{
//get average for a segment based on minimum
double testAverageAngle = (sumAngles + 360*lowerAngles.length)/_angles.length;
//minimum is inside segment range
//we will test average 0 (360) later
if(testAverageAngle < 360 && testAverageAngle > lowerBound)
{
double testVariance = varlc(testAverageAngle,_angles.length,sumAngles,sumSqrAngles,lowerAngles.length,sumLC);
if(testVariance < variance)
{
averageAngles.clear();
averageAngles.add(testAverageAngle);
variance = testVariance;
}
else if(testVariance == variance)
averageAngles.add(testAverageAngle);
}
}
double upperBound = 180;
double sumUC = 0;
for(int i = 0; i < upperAngles.length; i++)
{
//get average for a segment based on minimum
double testAverageAngle = (sumAngles - 360*i)/_angles.length;
//minimum is outside segment range (therefore not directly relevant)
//since it is greater than lowerAngles[i], the minimum for the segment
//must lie on the boundary lowerAngles[i]
if(testAverageAngle < upperAngles[i]-180)
testAverageAngle = upperAngles[i];
if(testAverageAngle < upperBound)
{
double testVariance = varuc(testAverageAngle,_angles.length,sumAngles,sumSqrAngles,i,sumUC);
if(testVariance < variance)
{
averageAngles.clear();
averageAngles.add(testAverageAngle);
variance = testVariance;
}
else if(testVariance == variance)
averageAngles.add(testAverageAngle);
}
upperBound = upperAngles[i];
sumUC += upperBound;
}
//Test last segment
{
//get average for a segment based on minimum
double testAverageAngle = (sumAngles - 360*upperAngles.length)/_angles.length;
//minimum is inside segment range
//we test average 0 (360) now
if(testAverageAngle < 0)
testAverageAngle = 0;
if(testAverageAngle < upperBound)
{
double testVariance = varuc(testAverageAngle,_angles.length,sumAngles,sumSqrAngles,upperAngles.length,sumUC);
if(testVariance < variance)
{
averageAngles.clear();
averageAngles.add(testAverageAngle);
variance = testVariance;
}
else if(testVariance == variance)
averageAngles.add(testAverageAngle);
}
}
double[] averageAngles_ = new double[averageAngles.size()];
Iterator<Double> averageAnglesIter = averageAngles.iterator();
for(int i = 0; i < averageAngles_.length; i++)
averageAngles_[i] = averageAnglesIter.next();
return averageAngles_;
}
The arithmetic mean of a set of angles may not agree with your intuitive idea of what the average should be. For example, the arithmetic mean of the set {179,179,0,181,181} is 216 (and 144). The answer you immediately think of is probably 180, however it is well known that the arithmetic mean is heavily affected by edge values. You should also remember that angles are not vectors, as appealing as that may seem when dealing with angles sometimes.
This algorithm does of course also apply to all quantities that obey modular arithmetic (with minimal adjustment), such as the time of day.
I would also like to stress that even though this is a true average of angles, unlike the vector solutions, that does not necessarily mean it is the solution you should be using, the average of the corresponding unit vectors may well be the value you actually should to be using.

You have to define average more accurately. For the specific case of two angles, I can think of two different scenarios:
The "true" average, i.e. (a + b) / 2 % 360.
The angle that points "between" the two others while staying in the same semicircle, e.g. for 355 and 5, this would be 0, not 180. To do this, you need to check if the difference between the two angles is larger than 180 or not. If so, increment the smaller angle by 360 before using the above formula.
I don't see how the second alternative can be generalized for the case of more than two angles, though.

I'd like to share an method I used with a microcontroller which did not have floating point or trigonometry capabilities. I still needed to "average" 10 raw bearing readings in order to smooth out variations.
Check whether the first bearing is the range 270-360 or 0-90 degrees (northern two quadrants)
If it is, rotate this and all subsequent readings by 180 degrees, keeping all values in the range 0 <= bearing < 360. Otherwise take the readings as they come.
Once 10 readings have been taken calculate the numerical average assuming that there has been no wraparound
If the 180 degree rotation had been in effect then rotate the calculated average by 180 degrees to get back to a "true" bearing.
It's not ideal; it can break. I got away with it in this case because the device only rotates very slowly. I'll put it out there in case anyone else finds themselves working under similar restrictions.

Like all averages, the answer depends upon the choice of metric. For a given metric M, the average of some angles a_k in [-pi,pi] for k in [1,N] is that angle a_M which minimizes the sum of squared distances d^2_M(a_M,a_k). For a weighted mean, one simply includes in the sum the weights w_k (such that sum_k w_k = 1). That is,
a_M = arg min_x sum_k w_k d^2_M(x,a_k)
Two common choices of metric are the Frobenius and the Riemann metrics. For the Frobenius metric, a direct formula exists that corresponds to the usual notion of average bearing in circular statistics. See "Means and Averaging in the Group of Rotations", Maher Moakher, SIAM Journal on Matrix Analysis and Applications, Volume 24, Issue 1, 2002, for details.
http://link.aip.org/link/?SJMAEL/24/1/1
Here's a function for GNU Octave 3.2.4 that does the computation:
function ma=meanangleoct(a,w,hp,ntype)
% ma=meanangleoct(a,w,hp,ntype) returns the average of angles a
% given weights w and half-period hp using norm type ntype
% Ref: "Means and Averaging in the Group of Rotations",
% Maher Moakher, SIAM Journal on Matrix Analysis and Applications,
% Volume 24, Issue 1, 2002.
if (nargin<1) | (nargin>4), help meanangleoct, return, end
if isempty(a), error('no measurement angles'), end
la=length(a); sa=size(a);
if prod(sa)~=la, error('a must be a vector'); end
if (nargin<4) || isempty(ntype), ntype='F'; end
if ~sum(ntype==['F' 'R']), error('ntype must be F or R'), end
if (nargin<3) || isempty(hp), hp=pi; end
if (nargin<2) || isempty(w), w=1/la+0*a; end
lw=length(w); sw=size(w);
if prod(sw)~=lw, error('w must be a vector'); end
if lw~=la, error('length of w must equal length of a'), end
if sum(w)~=1, warning('resumming weights to unity'), w=w/sum(w); end
a=a(:); % make column vector
w=w(:); % make column vector
a=mod(a+hp,2*hp)-hp; % reduce to central period
a=a/hp*pi; % scale to half period pi
z=exp(i*a); % U(1) elements
% % NOTA BENE:
% % fminbnd can get hung up near the boundaries.
% % If that happens, shift the input angles a
% % forward by one half period, then shift the
% % resulting mean ma back by one half period.
% X=fminbnd(#meritfcn,-pi,pi,[],z,w,ntype);
% % seems to work better
x0=imag(log(sum(w.*z)));
X=fminbnd(#meritfcn,x0-pi,x0+pi,[],z,w,ntype);
% X=real(X); % truncate some roundoff
X=mod(X+pi,2*pi)-pi; % reduce to central period
ma=X*hp/pi; % scale to half period hp
return
%%%%%%
function d2=meritfcn(x,z,w,ntype)
x=exp(i*x);
if ntype=='F'
y=x-z;
else % ntype=='R'
y=log(x'*z);
end
d2=y'*diag(w)*y;
return
%%%%%%
% % test script
% %
% % NOTA BENE: meanangleoct(a,[],[],'R') will equal mean(a)
% % when all abs(a-b) < pi/2 for some value b
% %
% na=3, a=sort(mod(randn(1,na)+1,2)-1)*pi;
% da=diff([a a(1)+2*pi]); [mda,ndx]=min(da);
% a=circshift(a,[0 2-ndx]) % so that diff(a(2:3)) is smallest
% A=exp(i*a), B1=expm(a(1)*[0 -1; 1 0]),
% B2=expm(a(2)*[0 -1; 1 0]), B3=expm(a(3)*[0 -1; 1 0]),
% masimpl=[angle(mean(exp(i*a))) mean(a)]
% Bsum=B1+B2+B3; BmeanF=Bsum/sqrt(det(Bsum));
% % this expression for BmeanR should be correct for ordering of a above
% BmeanR=B1*(B1'*B2*(B2'*B3)^(1/2))^(2/3);
% mamtrx=real([[0 1]*logm(BmeanF)*[1 0]' [0 1]*logm(BmeanR)*[1 0]'])
% manorm=[meanangleoct(a,[],[],'F') meanangleoct(a,[],[],'R')]
% polar(a,1+0*a,'b*'), axis square, hold on
% polar(manorm(1),1,'rs'), polar(manorm(2),1,'gd'), hold off
% Meanangleoct Version 1.0
% Copyright (C) 2011 Alphawave Research, robjohnson#alphawaveresearch.com
% Released under GNU GPLv3 -- see file COPYING for more info.
%
% Meanangle is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or (at
% your option) any later version.
%
% Meanangle is distributed in the hope that it will be useful, but
% WITHOUT ANY WARRANTY; without even the implied warranty of
% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
% General Public License for more details.
%
% You should have received a copy of the GNU General Public License
% along with this program. If not, see `http://www.gnu.org/licenses/'.

In python, with angles between [-180, 180)
def add_angles(a, b):
return (a + b + 180) % 360 - 180
def average_angles(a, b):
return add_angles(a, add_angles(-a, b)/2)
Details:
For the average of two angles there are two averages 180° apart, but we may want the closer average.
Visually, the average of the blue (b) and green (a) the yields the teal point:
Angles 'wrap around' (e.g. 355 + 10 = 5), but standard arithmetic will ignore this branch point.
However if angle b is opposite to the branch point, then (b + g)/2 gives the closest average: the teal point.
For any two angles, we can rotate the problem so one of the angles is opposite to the branch point, perform standard averaging, then rotate back.

Here is the full solution:
(the input is an array of bearing in degrees (0-360)
public static int getAvarageBearing(int[] arr)
{
double sunSin = 0;
double sunCos = 0;
int counter = 0;
for (double bearing : arr)
{
bearing *= Math.PI/180;
sunSin += Math.sin(bearing);
sunCos += Math.cos(bearing);
counter++;
}
int avBearing = INVALID_ANGLE_VALUE;
if (counter > 0)
{
double bearingInRad = Math.atan2(sunSin/counter, sunCos/counter);
avBearing = (int) (bearingInRad*180f/Math.PI);
if (avBearing<0)
avBearing += 360;
}
return avBearing;
}

In english:
Make a second data set with all angles shifted by 180.
Take the variance of both data sets.
Take the average of the data set with the smallest variance.
If this average is from the shifted set then shift the answer again by 180.
In python:
A #numpy NX1 array of angles
if np.var(A) < np.var((A-180)%360):
average = np.average(A)
else:
average = (np.average((A-180)%360)+180)%360

If anyone is looking for a JavaScript solution to this, I've translated the example given in the wikipedia page Mean of circular quantities (which was also referred to in Nick's answer) into JavaScript/NodeJS code, with help from the mathjs library.
If your angles are in degrees:
const maths = require('mathjs');
getAverageDegrees = (array) => {
let arrayLength = array.length;
let sinTotal = 0;
let cosTotal = 0;
for (let i = 0; i < arrayLength; i++) {
sinTotal += maths.sin(array[i] * (maths.pi / 180));
cosTotal += maths.cos(array[i] * (maths.pi / 180));
}
let averageDirection = maths.atan(sinTotal / cosTotal) * (180 / maths.pi);
if (cosTotal < 0) {
averageDirection += 180;
} else if (sinTotal < 0) {
averageDirection += 360;
}
return averageDirection;
}
This solution worked really well for me in order to find the average direction from a set of compass directions. I've tested this on a large range of directional data (0-360 degrees) and it seems very robust.
Alternatively, if your angles are in radians:
const maths = require('mathjs');
getAverageRadians = (array) => {
let arrayLength = array.length;
let sinTotal = 0;
let cosTotal = 0;
for (let i = 0; i < arrayLength; i++) {
sinTotal += maths.sin(array[i]);
cosTotal += maths.cos(array[i]);
}
let averageDirection = maths.atan(sinTotal / cosTotal);
if (cosTotal < 0) {
averageDirection += 180;
} else if (sinTotal < 0) {
averageDirection += 360;
}
return averageDirection;
}
Hopefully these solutions are helpful to someone facing a similar programming challenge to me.

I would go the vector way using complex numbers. My example is in Python, which has built-in complex numbers:
import cmath # complex math
def average_angle(list_of_angles):
# make a new list of vectors
vectors= [cmath.rect(1, angle) # length 1 for each vector
for angle in list_of_angles]
vector_sum= sum(vectors)
# no need to average, we don't care for the modulus
return cmath.phase(vector_sum)
Note that Python does not need to build a temporary new list of vectors, all of the above can be done in one step; I just chose this way to approximate pseudo-code applicable to other languages too.

Here's a complete C++ solution:
#include <vector>
#include <cmath>
double dAngleAvg(const vector<double>& angles) {
auto avgSin = double{ 0.0 };
auto avgCos = double{ 0.0 };
static const auto conv = double{ 0.01745329251994 }; // PI / 180
static const auto i_conv = double{ 57.2957795130823 }; // 180 / PI
for (const auto& theta : angles) {
avgSin += sin(theta*conv);
avgCos += cos(theta*conv);
}
avgSin /= (double)angles.size();
avgCos /= (double)angles.size();
auto ret = double{ 90.0 - atan2(avgCos, avgSin) * i_conv };
if (ret<0.0) ret += 360.0;
return fmod(ret, 360.0);
}
It takes the angles in the form of a vector of doubles, and returns the average simply as a double. The angles must be in degrees, and of course the average is in degrees as well.

Based on Alnitak's answer, I've written a Java method for calculating the average of multiple angles:
If your angles are in radians:
public static double averageAngleRadians(double... angles) {
double x = 0;
double y = 0;
for (double a : angles) {
x += Math.cos(a);
y += Math.sin(a);
}
return Math.atan2(y, x);
}
If your angles are in degrees:
public static double averageAngleDegrees(double... angles) {
double x = 0;
double y = 0;
for (double a : angles) {
x += Math.cos(Math.toRadians(a));
y += Math.sin(Math.toRadians(a));
}
return Math.toDegrees(Math.atan2(y, x));
}

Here's an idea: build the average iteratively by always calculating the average of the angles that are closest together, keeping a weight.
Another idea: find the largest gap between the given angles. Find the point that bisects it, and then pick the opposite point on the circle as the reference zero to calculate the average from.

Let's represent these angles with points on the circumference of the circle.
Can we assume that all these points fall on the same half of the circle? (Otherwise, there is no obvious way to define the "average angle". Think of two points on the diameter, e.g. 0 deg and 180 deg --- is the average 90 deg or 270 deg? What happens when we have 3 or more evenly spread out points?)
With this assumption, we pick an arbitrary point on that semicircle as the "origin", and measure the given set of angles with respect to this origin (call this the "relative angle"). Note that the relative angle has an absolute value strictly less than 180 deg. Finally, take the mean of these relative angles to get the desired average angle (relative to our origin of course).

There's no single "right answer". I recommend reading the book,
K. V. Mardia and P. E. Jupp, "Directional Statistics", (Wiley, 1999),
for a thorough analysis.

(Just want to share my viewpoint from Estimation Theory or Statistical Inference)
Nimble's trial is to get the MMSE^ estimate of a set of angles, but it's one of choices to find an "averaged" direction; one can also find an MMAE^ estimate, or some other estimate to be the "averaged" direction, and it depends on your metric quantifying error of direction; or more generally in estimation theory, the definition of cost function.
^ MMSE/MMAE corresponds to minimum mean squared/absolute error.
ackb said "The average angle phi_avg should have the property that sum_i|phi_avg-phi_i|^2 becomes minimal...they average something, but not angles"
---- you quantify errors in mean-squared sense and it's one of the mostly common way, however, not the only way. The answer favored by most people here (i.e., sum of the unit vectors and get the angle of the result) is actually one of the reasonable solutions. It is (can be proved) the ML estimator that serves as the "averaged" direction we want, if the directions of the vectors are modeled as von Mises distribution. This distribution is not fancy, and is just a periodically sampled distribution from a 2D Guassian. See Eqn. (2.179) in Bishop's book "Pattern Recognition and Machine Learning". Again, by no means it's the only best one to represent "average" direction, however, it is quite reasonable one that have both good theoretical justification and simple implementation.
Nimble said "ackb is right that these vector based solutions cannot be considered true averages of angles, they are only an average of the unit vector counterparts"
----this is not true. The "unit vector counterparts" reveals the information of the direction of a vector. The angle is a quantity without considering the length of the vector, and the unit vector is something with additional information that the length is 1. You can define your "unit" vector to be of length 2, it does not really matter.

You can see a solution and a little explanation in the following link, for ANY programming language:
https://rosettacode.org/wiki/Averages/Mean_angle
For instance, C++ solution:
#include<math.h>
#include<stdio.h>
double
meanAngle (double *angles, int size)
{
double y_part = 0, x_part = 0;
int i;
for (i = 0; i < size; i++)
{
x_part += cos (angles[i] * M_PI / 180);
y_part += sin (angles[i] * M_PI / 180);
}
return atan2 (y_part / size, x_part / size) * 180 / M_PI;
}
int
main ()
{
double angleSet1[] = { 350, 10 };
double angleSet2[] = { 90, 180, 270, 360};
double angleSet3[] = { 10, 20, 30};
printf ("\nMean Angle for 1st set : %lf degrees", meanAngle (angleSet1, 2));
printf ("\nMean Angle for 2nd set : %lf degrees", meanAngle (angleSet2, 4));
printf ("\nMean Angle for 3rd set : %lf degrees\n", meanAngle (angleSet3, 3));
return 0;
}
Output:
Mean Angle for 1st set : -0.000000 degrees
Mean Angle for 2nd set : -90.000000 degrees
Mean Angle for 3rd set : 20.000000 degrees
Or Matlab solution:
function u = mean_angle(phi)
u = angle(mean(exp(i*pi*phi/180)))*180/pi;
end
mean_angle([350, 10])
ans = -2.7452e-14
mean_angle([90, 180, 270, 360])
ans = -90
mean_angle([10, 20, 30])
ans = 20.000

Here is a completely arithmetic solution using moving averages and taking care to normalize values. It is fast and delivers correct answers if all angles are on one side of the circle (within 180° of each other).
It is mathimatically equivalent to adding the offset which shifts the values into the range (0, 180), calulating the mean and then subtracting the offset.
The comments describe what range a specific value can take on at any given time
// angles have to be in the range [0, 360) and within 180° of each other.
// n >= 1
// returns the circular average of the angles int the range [0, 360).
double meanAngle(double* angles, int n)
{
double average = angles[0];
for (int i = 1; i<n; i++)
{
// average: (0, 360)
double diff = angles[i]-average;
// diff: (-540, 540)
if (diff < -180)
diff += 360;
else if (diff >= 180)
diff -= 360;
// diff: (-180, 180)
average += diff/(i+1);
// average: (-180, 540)
if (average < 0)
average += 360;
else if (average >= 360)
average -= 360;
// average: (0, 360)
}
return average;
}

Well I'm hugely late to the party but thought I'd add my 2 cents worth as I couldn't really find any definitive answer. In the end I implemented the following Java version of the Mitsuta method which, I hope, provides a simple and robust solution. Particularly as the Standard Deviation provides both a measure dispersion and, if sd == 90, indicates that the input angles result in an ambiguous mean.
EDIT: Actually I realised that my original implementation can be even further simplified, in fact worryingly simple considering all the conversation and trigonometry going on in the other answers.
/**
* The Mitsuta method
*
* #param angles Angles from 0 - 360
* #return double array containing
* 0 - mean
* 1 - sd: a measure of angular dispersion, in the range [0..360], similar to standard deviation.
* Note if sd == 90 then the mean can also be its inverse, i.e. 360 == 0, 300 == 60.
*/
public static double[] getAngleStatsMitsuta(double... angles) {
double sum = 0;
double sumsq = 0;
for (double angle : angles) {
if (angle >= 180) {
angle -= 360;
}
sum += angle;
sumsq += angle * angle;
}
double mean = sum / angles.length;
return new double[]{mean <= 0 ? 360 + mean: mean, Math.sqrt(sumsq / angles.length - (mean * mean))};
}
... and for all you (Java) geeks out there, you can use the above approach to get the mean angle in one line.
Arrays.stream(angles).map(angle -> angle<180 ? angle: (angle-360)).sum() / angles.length;

Alnitak has the right solution. Nick Fortescue's solution is functionally the same.
For the special case of where
( sum(x_component) = 0.0 && sum(y_component) = 0.0 ) // e.g. 2 angles of 10. and 190. degrees ea.
use 0.0 degrees as the sum
Computationally you have to test for this case since atan2(0. , 0.) is undefined and will generate an error.

The average angle phi_avg should have the property that sum_i|phi_avg-phi_i|^2 becomes minimal, where the difference has to be in [-Pi, Pi) (because it might be shorter to go the other way around!). This is easily achieved by normalizing all input values to [0, 2Pi), keeping a running average phi_run and choosing normalizing |phi_i-phi_run| to [-Pi,Pi)
(by adding or subtractin 2Pi). Most suggestions above do something else that does not
have that minimal property, i.e., they average something, but not angles.

I solved the problem with the help of the answer from #David_Hanak.
As he states:
The angle that points "between" the two others while staying in the same semicircle, e.g. for 355 and 5, this would be 0, not 180. To do this, you need to check if the difference between the two angles is larger than 180 or not. If so, increment the smaller angle by 360 before using the above formula.
So what I did was calculate the average of all the angles. And then all the angles that are less than this, increase them by 360. Then recalculate the average by adding them all and dividing them by their length.
float angleY = 0f;
int count = eulerAngles.Count;
for (byte i = 0; i < count; i++)
angleY += eulerAngles[i].y;
float averageAngle = angleY / count;
angleY = 0f;
for (byte i = 0; i < count; i++)
{
float angle = eulerAngles[i].y;
if (angle < averageAngle)
angle += 360f;
angleY += angle;
}
angleY = angleY / count;
Works perfectly.

Python function:
from math import sin,cos,atan2,pi
import numpy as np
def meanangle(angles,weights=0,setting='degrees'):
'''computes the mean angle'''
if weights==0:
weights=np.ones(len(angles))
sumsin=0
sumcos=0
if setting=='degrees':
angles=np.array(angles)*pi/180
for i in range(len(angles)):
sumsin+=weights[i]/sum(weights)*sin(angles[i])
sumcos+=weights[i]/sum(weights)*cos(angles[i])
average=atan2(sumsin,sumcos)
if setting=='degrees':
average=average*180/pi
return average

You can use this function in Matlab:
function retVal=DegreeAngleMean(x)
len=length(x);
sum1=0;
sum2=0;
count1=0;
count2=0;
for i=1:len
if x(i)<180
sum1=sum1+x(i);
count1=count1+1;
else
sum2=sum2+x(i);
count2=count2+1;
end
end
if (count1>0)
k1=sum1/count1;
end
if (count2>0)
k2=sum2/count2;
end
if count1>0 && count2>0
if(k2-k1 >= 180)
retVal = ((sum1+sum2)-count2*360)/len;
else
retVal = (sum1+sum2)/len;
end
elseif count1>0
retVal = k1;
else
retVal = k2;
end

While starblue's answer gives the angle of the average unit vector, it is possible to extend the concept of the arithmetic mean to angles if you accept that there may be more than one answer in the range of 0 to 2*pi (or 0° to 360°). For example, the average of 0° and 180° may be either 90° or 270°.
The arithmetic mean has the property of being the single value with the minimum sum of squared distances to the input values. The distance along the unit circle between two unit vectors can be easily calculated as the inverse cosine of their dot product. If we choose a unit vector by minimizing the sum of the squared inverse cosine of the dot product of our vector and each input unit vector then we have an equivalent average. Again, keep in mind that there may be two or more minimums in exceptional cases.
This concept could be extended to any number of dimensions, since the distance along the unit sphere can be calculated in the exact same way as the distance along the unit circle--the inverse cosine of the dot product of two unit vectors.
For circles we could solve for this average in a number of ways, but I propose the following O(n^2) algorithm (angles are in radians, and I avoid calculating the unit vectors):
var bestAverage = -1
double minimumSquareDistance
for each a1 in input
var sumA = 0;
for each a2 in input
var a = (a2 - a1) mod (2*pi) + a1
sumA += a
end for
var averageHere = sumA / input.count
var sumSqDistHere = 0
for each a2 in input
var dist = (a2 - averageHere + pi) mod (2*pi) - pi // keep within range of -pi to pi
sumSqDistHere += dist * dist
end for
if (bestAverage < 0 OR sumSqDistHere < minimumSquareDistance) // for exceptional cases, sumSqDistHere may be equal to minimumSquareDistance at least once. In these cases we will only find one of the averages
minimumSquareDistance = sumSqDistHere
bestAverage = averageHere
end if
end for
return bestAverage
If all the angles are within 180° of each other, then we could use a simpler O(n)+O(sort) algorithm (again using radians and avoiding use of unit vectors):
sort(input)
var largestGapEnd = input[0]
var largestGapSize = (input[0] - input[input.count-1]) mod (2*pi)
for (int i = 1; i < input.count; ++i)
var gapSize = (input[i] - input[i - 1]) mod (2*pi)
if (largestGapEnd < 0 OR gapSize > largestGapSize)
largestGapSize = gapSize
largestGapEnd = input[i]
end if
end for
double sum = 0
for each angle in input
var a2 = (angle - largestGapEnd) mod (2*pi) + largestGapEnd
sum += a2
end for
return sum / input.count
To use degrees, simply replace pi with 180. If you plan to use more dimensions then you will most likely have to use an iterative method to solve for the average.

The problem is extremely simple.
1. Make sure all angles are between -180 and 180 degrees.
2. a Add all non-negative angles, take their average, and COUNT how many
2. b.Add all negative angles, take their average and COUNT how many.
3. Take the difference of pos_average minus neg_average
If difference is greater than 180 then change difference to 360 minus difference. Otherwise just change the sign of difference. Note that difference is always non-negative.
The Average_Angle equals the pos_average plus difference times the "weight", negative count divided by the sum of negative and positive count

Here is some java code to average angles, I think it's reasonably robust.
public static double getAverageAngle(List<Double> angles)
{
// r = right (0 to 180 degrees)
// l = left (180 to 360 degrees)
double rTotal = 0;
double lTotal = 0;
double rCtr = 0;
double lCtr = 0;
for (Double angle : angles)
{
double norm = normalize(angle);
if (norm >= 180)
{
lTotal += norm;
lCtr++;
} else
{
rTotal += norm;
rCtr++;
}
}
double rAvg = rTotal / Math.max(rCtr, 1.0);
double lAvg = lTotal / Math.max(lCtr, 1.0);
if (rAvg > lAvg + 180)
{
lAvg += 360;
}
if (lAvg > rAvg + 180)
{
rAvg += 360;
}
double rPortion = rAvg * (rCtr / (rCtr + lCtr));
double lPortion = lAvg * (lCtr / (lCtr + rCtr));
return normalize(rPortion + lPortion);
}
public static double normalize(double angle)
{
double result = angle;
if (angle >= 360)
{
result = angle % 360;
}
if (angle < 0)
{
result = 360 + (angle % 360);
}
return result;
}

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