I believe this is more of an algorithmic question but I also want to do this in C++.
Let me illustrate the question with an example.
Suppose I have N number of objects (not programming objects), each with different weights. And I have two vehicles to carry them. The vehicles are big enough to carry all the objects by each. These two vehicles have their own mileage and different levels of fuel in the tank. And also the mileage depends on the weight it carries.
The objective is to bring these N objects as far as possible. So I need to distribute the N objects in a certain way between the two vehicles. Note that I do not need to bring them the 'same' distance, but rather as far as possible. So example, I want the two vehicles to go 5km and 6 km, rather than one going 2km and other going 7km.
I cannot think of a theoretical closed-form calculation to determine which weights to be loaded in to each vehicle. because remember that I need to carry all the N objects which is a fixed value.
So as far as I can think, I need to try all the combinations.
Could someone advice of an efficient algorithm to try all the combinations?
For example I would have the following:
int weights[5] = {1,4,2,7,5}; // can be more values than 5
float vehicelONEMileage(int totalWeight);
float vehicleTWOMileage(int totalWeight);
How could I efficiently try all the combinations of weights[] with the two functions?
Thw two functions can be assumed as linear functions. I.e. the return value of the two mileage functions are linear functions with (different) negative slopes and (different) offsets.
So what I need to find is something like:
MAX(MIN(vehicleONEMileage(x), vehicleTWOMileage(sum(weights) - x)));
Thank you.
This should be on the cs or the math site.
Simplification: Instead of an array of objects, let's say we can distribute weight linearly.
The function we want to optimize is the minimum of both travel distances. Finding the maximum of the minimum is the same as finding the maximum of the product (Without proof. But to see this, think of the relationship between perimeter and area of rectangles. The rectangle with the biggest area given a perimeter is a square, which also happens to have the largest minimum side length).
In the following, we will scale the sum of all weights to 1. So, a distribution like (0.7, 0.3) means that 70% of all weights is loaded on vehicle 1. Let's call the load of vehicle 1 x and the load of vehicle 1-x.
Given the two linear functions f = a x + b and g = c x + d, where f is the mileage of vehicle 1 when loaded with weight x, and g the same for vehicle 2, we want to maximize
(a*x+b)*(c*(1-x)+d)
Let's ask Wolfram Alpha to do the hard work for us: www.wolframalpha.com/input/?i=derive+%28%28a*x%2Bb%29*%28c*%281-x%29%2Bd%29%29
It tells us that there is an extremum at
x_opt = (a * c + a * d - b * c) / (2 * a * c)
That's all you need to solve your problem efficiently.
The complete algorithm:
find a, b, c, d
b = vehicleONEMileage(0)
a = (vehicleONEMileage(1) - b) * sum_of_all_weights
same for c and d
calculate x_opt as above.
if x_opt < 0, load all weight onto vehicle 2
if x_opt > 1, load all weight onto vehicle 1
else, try to load tgt_load = x_opt*sum_of_all_weights onto vehicle 1, the rest onto vehicle 2.
The rest is a knapsack problem. See http://en.wikipedia.org/wiki/Knapsack_problem#0.2F1_Knapsack_Problem
How to apply this? Use the dynamic programming algorithm described there twice.
for maximizing a load up to tgt_load
for maximizing a load up to (sum_of_all_weights - tgt_load)
The first one, if loaded onto vehicle one, gives you a distribution with slightly less then expected on vehicle one.
The second one, if loaded onto vehicle two, gives you a distribution with slightly more than expected on vehicle two.
One of those is the best solution. Compare them and use the better one.
I leave the C++ part to you. ;-)
I can suggest the following solution:
The total number of combinations is 2^(number of weights). Using a bit logic we can loop through the all combinations and calculate maxDistance. Bits in the combination value show which weight goes to which vehicle.
Note that algorithm complexity is exponential and int has a limited number of bits!
float maxDistance = 0.f;
for (int combination = 0; combination < (1 << ARRAYSIZE(weights)); ++combination)
{
int weightForVehicleONE = 0;
int weightForVehicleTWO = 0;
for (int i = 0; i < ARRAYSIZE(weights); ++i)
{
if (combination & (1 << i)) // bit is set to 1 and goes to vechicleTWO
{
weightForVehicleTWO += weights[i];
}
else // bit is set to 0 and goes to vechicleONE
{
weightForVehicleONE += weights[i];
}
}
maxDistance = max(maxDistance, min(vehicelONEMileage(weightForVehicleONE), vehicleTWOMileage(weightForVehicleTWO)));
}
Related
I have an input of n unique points (X,Y) that are between 0 and 2^32 inclusive. The coordinates are integers.
I need to create an algorithm that finds the number of pairs of points with a distance of exactly 2018.
I have thought of checking with every other point but it would be O(n^2) and I have to make it more efficient. I also thought of using a set or a vector and sort it using a comparator based on the distance with the origin point but it wouldn't help at all.
So how can I do it efficiently?
There is one Pythagorean triple with the hypotenuse of 2018: 11182+16802=20182.
Since all coordinates are integers, the only possible differences between the coordinates (both X an Y) of the two points are 0, 1118, 1680, and 2018.
Finding all pairs of points with a given difference between X (or Y) coordinates is a simple n log n operation.
Numbers other than 2018 might need a bit more work because they might be members of more than one Pythagorean triple (for example 2015 is a hypotenuse of 3 triples). If the number is not given as a constant, but provided at run time, you will have to generate all triples with this hypotenuse. This may require some sqrt(N) effort (N is the hypotenuse, not the number of points). One can find a recipe on the math stackexchange, e.g. here (there are many others).
You could try using a Quadtree. First you start sorting your points into the quadtree. You should specify a lower limit for the cell size of e.g. 2048 wich is a power of 2. Then iterate though the points and calculate distances to the points in the same cell and to the points in adjacent cells. That way you should be able to decrease the number of distance calculations drastically.
The main difficulty will probably be implementing the tree structure. You also have to find a way to find adjacent cells (you must include the possibility to traverse upwards in the tree)
The complexity of this is probably O(n*log(n)) in the best case but don't pin me down on that.
One additional word on the distance calculation: You are probably much faster if you don't do
dx = p1x - p2x;
dy = p1y - p2y;
if ( sqrt(dx*dx + dy*dy) == 2018 ) {
...
}
but
dx = p1x - p2x;
dy = p1y - p2y;
if ( dx*dx + dy*dy == 2018*2018 ) {
...
}
Squaring is faster than taking the sqare root. So just compare the square of the distance with the square of 2018.
I would like to write a C++ function which finds the median of an array of circular data.
For example, consider the reading from a compass where the readings are assumed to be in [0,360). Though 1 & 359 appears to be far away, they are very close due to the circular nature of the reading.
Finding median of N-elements in ordinary data is as follows.
1. sort the data of N-elements (ascending or descending order)
2. If N is odd, median is the (N+1)/2 th element in the sorted array.
3. If N is even, median is the average of the N/2 th and N/2+1 th elements in the sorted array.
However, the wrap around problem in the circular data takes the problem to a different dimension and the solution non-trivial.
A similar question to find mean from circular data is explained here How do you calculate the average of a set of circular data?
The suggestion in the above link is to find the unit vector corresponding to each angle and find the average. However, median requires sorting the data and sorting of vectors don't make any sense in this context. Hence I don't think we can use the proposed scheme to find median!
I've actually given this topic way more thought than is healthy so I'll share my thoughts and findings here. Maybe someone will have a similar problem and find this useful.
I haven't used C++ in many years so please forgive me if I write all the code in C#. I believe a fluent C++ speaker can pretty easily translate the algorithms.
Circular mean
First, let's define the circular mean. It's calculated by converting your points to radians, where your period (256, 360 or whatever - the value that is interpreted to be the same as zero) is scaled to 2*pi. You then calculate the sine and cosine of those radian values. Those are the y and x coordinates of your values on a unit circle. You then sum up all the sines and cosines and calculate atan2. This gives you the average angle, which can be easily converted back to your data point by dividing with the scaling factor.
var scalingFactor = 2 * Math.PI / period;
var sines = 0.0;
var cosines = 0.0;
foreach (var value in inputs)
{
var radians = value * scalingFactor;
sines += Math.Sin(radians);
cosines += Math.Cos(radians);
}
var circularMean = Math.Atan2(sines, cosines) / scalingFactor;
if (circularMean >= 0)
return circularMean;
else
return circularMean + period;
Marginal circular median
The simplest approach to a circular median is just a modified way of handling the circular mean.
The circular median can be calculated in a similar way, by just finding the median of the sines and cosines instead of the sums, and calculating the atan2 of that. This way, you are finding the marginal median of the circle points and taking its angle as a result.
var scalingFactor = 2 * Math.PI / period;
var sines = new List<double>();
var cosines = new List<double>();
foreach (var value in inputs)
{
var radians = value * scalingFactor;
sines.Add(Math.Sin(radians));
cosines.Add(Math.Cos(radians));
}
var circularMedian = Math.Atan2(Median(sines), Median(cosines)) / scalingFactor;
if (circularMedian >= 0)
return circularMedian;
else
return circularMedian + period;
This approach is O(n), robust to outliers and very simple to implement. It may suit your purposes well enough, but it has a problem: rotating the input points will give you different results. Depending on the distribution of your input data, it may or may not be a problem.
Circular arc median
To understand this other approach, you need to stop thinking of means and medians in terms of "this is how it's calculated", but in terms of what the resulting values actually represent.
For non-cyclic data, you get the mean by summing up all the values and dividing by the number of elements. What this number represents, though, is the value with the minimal sum of all squared distances to data elements. (I hear statisticians call this value the L2 estimate of location, but a statistician should probably confirm or deny this.)
Likewise for median. You get it by finding the data element that would end up in the middle if all data were sorted (ideally, using an O(n) selection algorithm, like nth_element in C++). What this number is, though, is a value that has the minimal sum of all absolute (non-squared!) distances to data elements. (Supposedly, this value is called an L1 estimate of location.)
Sorting circular data doesn't help you find a middle, so the usual way of thinking about medians doesn't work, but you can still find this point that minimizes the sum of absolute distances from all data points. Here's the algorithm that I came up with, that runs in O(n) time assuming the input data is normalized to >= 0 and < period, and then sorted. (If you need to do this sorting as part of your calculation, then the runtime is O(n log n).)
It works by going through all the data points and keeping track of the sum of distances. When you shift to the right data point by a distance D, the sum of distances to all the left points increases by D*LeftCount and the sum of all distances to all the right points decreases by D*RightCount. Then, if some of the left points are now actually the right points, because their left distance is larger than period/2, you subtract their previous distance and add the new, correct distance.
For comparing the current sum to the best sum, I added a bit of tolerance to guard against inexact floating point arithmetic.
There may be multiple or infinitely many points that satisfy the minimum distances condition. With non-circular medians with even number of values, the median can be any value between the two central values. It's usually taken to be the average of those two central values, so I took the similar approach with this median algorithm. I find all data points that minimize the distances and then just calculate the circular mean of those points.
// Requires a sorted list with values normalized to [0,period).
// Doing an initialization pass:
// * candidate is the lowest number
// * finding the index where the circle with this candidate starts
// * calculating the score for this candidate - the sum of absolute distances
// * counting the number of values to the left of the candidate
int i;
var candidate = list[0];
var distanceSum = 0.0;
for (i = 1; i < list.Count; ++i)
{
if (list[i] >= candidate + period / 2)
break;
distanceSum += list[i] - candidate;
}
var leftCount = list.Count - i;
var circleStart = i;
if (circleStart == list.Count)
circleStart = 0;
else
for (; i < list.Count; ++i)
distanceSum += candidate + period - list[i];
var previousCandidate = candidate;
var bestCandidates = new List<double> { candidate };
var bestDistanceSum = distanceSum;
var equalityTolerance = period * 1e-10;
for (i = 1; i < list.Count; ++i)
{
candidate = list[i];
// A formula for correcting the distance given the movement to the right.
// It doesn't take into account that some values may have wrapped to the other side of the circle.
++leftCount;
distanceSum += (2 * leftCount - list.Count) * (candidate - previousCandidate);
// Counting all the values that wrapped to the other side of the circle
// and correcting the sum of distances from the candidate.
if (i <= circleStart)
while (list[circleStart] < candidate + period / 2)
{
--leftCount;
distanceSum += 2 * (list[circleStart] - candidate) - period;
++circleStart;
if (circleStart == list.Count)
{
circleStart = 0;
break; // Letting the next loop continue.
}
}
if (i > circleStart)
while (list[circleStart] < candidate - period / 2)
{
--leftCount;
distanceSum += 2 * (list[circleStart] - candidate) + period;
++circleStart;
}
// Comparing current sum to the best one, using the given tolerance.
if (distanceSum <= bestDistanceSum + equalityTolerance)
{
if (distanceSum >= bestDistanceSum - equalityTolerance)
{
// The numbers are close, so using their average as the next best.
bestDistanceSum = (bestCandidates.Count * bestDistanceSum + distanceSum) / (bestCandidates.Count + 1);
}
else
{
// The new number is significantly better, clearing.
bestDistanceSum = distanceSum;
bestCandidates.Clear();
}
bestCandidates.Add(candidate);
}
previousCandidate = candidate;
}
if (bestCandidates.Count == 1)
return bestCandidates[0];
else
return CircularMean(bestCandidates, period);
Geometric circular median
There is an inconsistency in the previous algorithm, in the way the median is defined in relation to the circular mean. The circular mean minimizes the sum of squared euclidian distances between points on a circle. In other words, it looks at the straight lines connecting points on a circle, cutting through the circle.
The arc median, as I calculate it above, looks at the arc distances: how far the points are to each other by moving on the perimeter of the circle, not by taking a straight line between them.
I have thought about how to address this issue, if it bothers you, but I haven't really done any experiments so I can't claim the following method works. In short, I believe you could use a modification of the Iteratively reweighted least squares algorithm (IRLS), which is what is usually used to calculate geometric medians.
The idea is to pick a starting value (for instance, the circular mean or the arc median presented above), and calculate the euclidean distance to each point: Di = sqrt(dxi^2 + dyi^2). Circular mean will minimize the squares of those distances, so the weights of each point should cancel out the square and reset to just D: Wi = Di / Di^2, which is just Wi = 1 / Di.
With these weights, calculate the weighted circular mean (same as the circular mean, but multiply each sine and cosine by the weight of that point before summing them up) and repeat the process. Repeat until enough iterations have passed or until the result stops changing much.
The problem with this algorithm is that it has a division by zero if the current solution falls exactly on a data point. Even if the distance isn't exactly zero, the solution will stop moving if you hit close enough to the point because the weight will become enormous compared to all the other ones. This can be fixed by adding a small fixed offset to the distance before dividing by it. This will make the solution suboptimal, but at least it won't stop on a wrong point.
It will still take some number of iterations to dig itself out of that wrong point unless the offset is relatively large, and the final solution is worse the bigger the offset is. So the best way would probably be to start with a fairly large offset and then progressively making it smaller for each next iteration.
Two properties of median allow inventing two distinct algorithms for median finding.
1) Median minimizes sum of absolute distance to all other elements -- O(n^2) algo:
for (i = 0; i < N; i++)
{
sum = 0;
for (j = 0; j < N; j++)
sum += abs(item[i] - item[j]) % 360;
if (sum < best_so_far) { best_so_far = sum; index = i; }
}
2) Median satisfies that half of items are less and half are greater
sort the items
locate the first set of items (i=0...I), satisfying either that
I <= N/2, OR item[I] > i + 180
if the condition for median is not satisfied, advance either i, or I.
requires O(N*log N) for sorting and O(N) for the next scan
Of course in cyclical data all items (and all items inbetween data points) can be a proper candidate for the median.
For definition and discussion of circular median see
N.I. Fisher's 'Statistical Analysis of Circular Data', Cambridge Univ. Press 1993
and the discussion surrounding equations 2.32 and 2.33. For multi-modal or isotropic data a unique median may not exist.
Find an axis that divides the data into 2 equal groups and choose the end of the axis at the smaller value of the angle. If the sample size is odd the median will be a data point, otherwise it will be the midpoint of 2 data points.
There are packages in other languages (e.g. R, MatLab) that would help provide test values for any function you write.
e.g.
https://www.rdocumentation.org/packages/circular/versions/0.4-93
See in particular median.circular and medianHL.circular
or
Berens, Philipp. ‘CircStat: A MATLAB Toolbox for Circular Statistics’. Journal of Statistical Software 31, no. 1 (23 September 2009): 1–21. https://doi.org/10.18637/jss.v031.i10.
and see circ_median
With your vector of angular datapoints (i.e. vector of numbers from 0 to 259), create two new vectors, I'll call them x and y. These two new vectors are the sine and cosine respectively of your angular datapoints.
That is, x[n] = cos(data[n]) and y[n] = sin(data[n]) where data is your angular data vector and n is however many datapoints there are.
Next, add up all the values in the x vector to get a single value, call it say sum_x and add up all the values in the y vector to get a another single value, call it sum_y.
Now you can do tangent-inverse (e.g. atan(sum_y/sum_x)) to get a new value. And this value is very meaningful. This value is basically telling you which direction your data is "pointing", i.e. where the majority of your data exists. NOTE: You must be careful of dividing by 0 (when sum_x=0) and when the indeterminate forms occurs (when both sum_x=0 and sum_y=0). The indeterminate form just means your data is evenly distributed, in which case the median is meaningless, and when sum_x=0 but sum_y!=0, then it is effectively atan(inf) or atan(-inf), both of which are known.
EDIT:
My previous answer needed some tweaking after this point.
From here, it is easy. Take the value you got in the previous step (atan(sum_y/sum_x)) and add 180 degrees to that value. This is your reference point of where your data starts and ends. From here, you can sort your angular data with this reference point as both the starting and ending point, and find the median of that data.
It is not possible to canonically extend the concept of median to circular data. For the sake of simplicity lets consider numbers in [0 10) and as an example the (already ordered) set { 1 3 5 7 8 }. Depending on how you rotate the array you get different values for the median:
1 3 5 7 8 -> 5
3 5 7 8 1 -> 7
5 7 8 1 3 -> 8
...etc...
and any is as good as the other.
I am not claiming that it is not possible to define a median on circular data. I am just claiming that the "normal" median cannot be extended to that case in a meaningful way without adding additional constraints or making an arbitrary choice.
There are n points with each having two attributes:
1. Position (from axis)
2. Attraction value (integer)
Attraction force between two points A & B is given by:
Attraction_force(A, B) = (distance between them) * Max(Attraction_val_A, Attraction_val_B);
Find the summation of all the forces between all possible pairs of points?
I tried by calculating and adding forces between all the pairs
for(int i=0; i<n-1; i++) {
for(int j=i+1; j<n; j++) {
force += abs(P[i].pos - P[j].pos) * max(P[i].attraction_val, P[j].attraction_val);
}
}
Example:
Points P1 P2 P3
Points distance: 2 3 4
Attraction Val: 4 5 6
Force = abs(2 - 3) * max(4, 5) + abs(2 - 4) * max(4, 6) + abs(3 - 4) * max(5, 6) = 23
But this takes O(n^2) time, I can't think of a way to reduce it further!!
Scheme of a solution:
Sort all points by their attraction value and process them one-by-one, starting with the one with lowest attraction.
For each point you have to quickly calculate sum of distances to all previously added points. That can be done using any online Range Sum Query problem solution, like segment tree or BIT. Key idea is that all points to the left are really not different and sum of their coordinates is enough to calculate sum of distances to them.
For each newly added point you just multiply that sum of distances (obtained on step 2) by point's attraction value and add that to the answer.
Intuitive observations that I made in order to invent this solution:
We have two "bad" functions here (somewhat "discrete"): max and modulo (in distance).
We can get rid of max by sorting our points and processing them in a specific order.
We can get rid of modulo if we process points to the left and to the right separately.
After all these transformations, we have to calculate something which, after some simple algebraic transformations, converts to an online RSQ problem.
An algorithm of:
O(N2)
is optimal, because you need the actual distance between all possible pairs.
At the moment i'm working with a camera to detect a marker. I use opencv and the Aruco Libary.
Only I'm stuck with a problem right now. I need to detect if the distance between 2 marker is less than a specific value. I have a function to calculate the distance, I can compare everything. But I'm looking for the most efficient way to keep track of all the markers (around 5/6) and how close they are together.
There is a list with markers but I cant find a efficient way to compare all of them.
I have a
Vector <Marker>
I also have a function called getDistance.
double getDistance(cv::Point2f punt1, cv::Point2f punt2)
{
float xd = punt2.x-punt1.x;
float yd = punt2.y-punt1.y;
double Distance = sqrtf(xd*xd + yd*yd);
return Distance;
}
The Markers contain a Point2f, so i can compare them easily.
One way to increase performance is to keep all the distances squared and avoid using the square root function. If you square the specific value you are checking against then this should work fine.
There isn't really a lot to recommend. If I understand the question and I'm counting the pairs correctly, you'll need to calculate 10 distances when you have 5 points, and 15 distances when you have 6 points. If you need to determine all of the distances, then you have no choice but to calculate all of the distances. I don't see any way around that. The only advice I can give is to make sure you calculate the distance between each pair only once (e.g., once you know the distance between points A and B, you don't need to calculate the distance between B and A).
It might be possible to sort the vector in such a way that you can short circuit your loop. For instance, if you sort it correctly and the distance between point A and point B is larger than your threshold, then the distances between A and C and A and D will also be larger than the threshold. But keep in mind that sorting isn't free, and it's likely that for small sets of points it would be faster to just calculate all distances ("Fancy algorithms are slow when n is small, and n is usually small. Fancy algorithms have big constants. Until you know that n is frequently going to be big, don't get fancy. ... For example, binary trees are always faster than splay trees for workaday problems.").
Newer versions of the C and C++ standard library have a hypot function for calculating distance between points:
#include <cmath>
double getDistance(cv::Point2f punt1, cv::Point2f punt2)
{
return std::hypot(punt2.x - punt1.x, punt2.y - punt1.y);
}
It's not necessarily faster, but it should be implemented in a way that avoids overflow when the points are far apart.
One minor optimization is to simply check if the change in X or change in Y exceeds the threshold. If it does, you can ignore the distance between those two points because the overall distance will also exceed the threshold:
const double threshold = ...;
std::vector<cv::Point2f> points;
// populate points
...
for (auto i = points.begin(); i != points.end(); ++i) {
for (auto j = i + 1; j != points.end(); ++j) {
double dx = std::abs(i->x - j->x), dy = std::abs(i->y - j->y);
if (dx > threshold || dy > threshold) {
continue;
}
double distance = std::hypot(dx, dy);
if (distance > threshold) {
continue;
}
...
}
}
If you're dealing with large amounts of data inside your vector you may want to consider some multithreading using future.
Vector <Marker> could be chunked into X chunks which are asynchronously computed together and stored inside std::future<>, putting to use #Sesame's suggestion will also increase your speed as well.
Is there a way to do a quick and dirty 3D distance check where the results are rough, but it is very very fast? I need to do depth sorting. I use STL sort like this:
bool sortfunc(CBox* a, CBox* b)
{
return a->Get3dDistance(Player.center,a->center) <
b->Get3dDistance(Player.center,b->center);
}
float CBox::Get3dDistance( Vec3 c1, Vec3 c2 )
{
//(Dx*Dx+Dy*Dy+Dz*Dz)^.5
float dx = c2.x - c1.x;
float dy = c2.y - c1.y;
float dz = c2.z - c1.z;
return sqrt((float)(dx * dx + dy * dy + dz * dz));
}
Is there possibly a way to do it without a square root or possibly without multiplication?
You can leave out the square root because for all positive (or really, non-negative) numbers x and y, if sqrt(x) < sqrt(y) then x < y. Since you're summing squares of real numbers, the square of every real number is non-negative, and the sum of any positive numbers is positive, the square root condition holds.
You cannot eliminate the multiplication, however, without changing the algorithm. Here's a counterexample: if x is (3, 1, 1) and y is (4, 0, 0), |x| < |y| because sqrt(1*1+1*1+3*3) < sqrt(4*4+0*0+0*0) and 1*1+1*1+3*3 < 4*4+0*0+0*0, but 1+1+3 > 4+0+0.
Since modern CPUs can compute a dot product faster than they can actually load the operands from memory, it's unlikely that you would have anything to gain by eliminating the multiply anyway (I think the newest CPUs have a special instruction that can compute a dot product every 3 cycles!).
I would not consider changing the algorithm without doing some profiling first. Your choice of algorithm will heavily depend on the size of your dataset (does it fit in cache?), how often you have to run it, and what you do with the results (collision detection? proximity? occlusion?).
What I usually do is first filter by Manhattan distance
float CBox::Within3DManhattanDistance( Vec3 c1, Vec3 c2, float distance )
{
float dx = abs(c2.x - c1.x);
float dy = abs(c2.y - c1.y);
float dz = abs(c2.z - c1.z);
if (dx > distance) return 0; // too far in x direction
if (dy > distance) return 0; // too far in y direction
if (dz > distance) return 0; // too far in z direction
return 1; // we're within the cube
}
Actually you can optimize this further if you know more about your environment. For example, in an environment where there is a ground like a flight simulator or a first person shooter game, the horizontal axis is very much larger than the vertical axis. In such an environment, if two objects are far apart they are very likely separated more by the x and y axis rather than the z axis (in a first person shooter most objects share the same z axis). So if you first compare x and y you can return early from the function and avoid doing extra calculations:
float CBox::Within3DManhattanDistance( Vec3 c1, Vec3 c2, float distance )
{
float dx = abs(c2.x - c1.x);
if (dx > distance) return 0; // too far in x direction
float dy = abs(c2.y - c1.y);
if (dy > distance) return 0; // too far in y direction
// since x and y distance are likely to be larger than
// z distance most of the time we don't need to execute
// the code below:
float dz = abs(c2.z - c1.z);
if (dz > distance) return 0; // too far in z direction
return 1; // we're within the cube
}
Sorry, I didn't realize the function is used for sorting. You can still use Manhattan distance to get a very rough first sort:
float CBox::ManhattanDistance( Vec3 c1, Vec3 c2 )
{
float dx = abs(c2.x - c1.x);
float dy = abs(c2.y - c1.y);
float dz = abs(c2.z - c1.z);
return dx+dy+dz;
}
After the rough first sort you can then take the topmost results, say the top 10 closest players, and re-sort using proper distance calculations.
Here's an equation that might help you get rid of both sqrt and multiply:
max(|dx|, |dy|, |dz|) <= distance(dx,dy,dz) <= |dx| + |dy| + |dz|
This gets you a range estimate for the distance which pins it down to within a factor of 3 (the upper and lower bounds can differ by at most 3x). You can then sort on, say, the lower number. You then need to process the array until you reach an object which is 3x farther away than the first obscuring object. You are then guaranteed to not find any object that is closer later in the array.
By the way, sorting is overkill here. A more efficient way would be to make a series of buckets with different distance estimates, say [1-3], [3-9], [9-27], .... Then put each element in a bucket. Process the buckets from smallest to largest until you reach an obscuring object. Process 1 additional bucket just to be sure.
By the way, floating point multiply is pretty fast nowadays. I'm not sure you gain much by converting it to absolute value.
I'm disappointed that the great old mathematical tricks seem to be getting lost. Here is the answer you're asking for. Source is Paul Hsieh's excellent web site: http://www.azillionmonkeys.com/qed/sqroot.html . Note that you don't care about distance; you will do fine for your sort with square of distance, which will be much faster.
In 2D, we can get a crude approximation of the distance metric without a square root with the formula:
distanceapprox (x, y) =
which will deviate from the true answer by at most about 8%. A similar derivation for 3 dimensions leads to:
distanceapprox (x, y, z) =
with a maximum error of about 16%.
However, something that should be pointed out, is that often the distance is only required for comparison purposes. For example, in the classical mandelbrot set (z←z2+c) calculation, the magnitude of a complex number is typically compared to a boundary radius length of 2. In these cases, one can simply drop the square root, by essentially squaring both sides of the comparison (since distances are always non-negative). That is to say:
√(Δx2+Δy2) < d is equivalent to Δx2+Δy2 < d2, if d ≥ 0
I should also mention that Chapter 13.2 of Richard G. Lyons's "Understanding Digital Signal Processing" has an incredible collection of 2D distance algorithms (a.k.a complex number magnitude approximations). As one example:
Max = x > y ? x : y;
Min = x < y ? x : y;
if ( Min < 0.04142135Max )
|V| = 0.99 * Max + 0.197 * Min;
else
|V| = 0.84 * Max + 0.561 * Min;
which has a maximum error of 1.0% from the actual distance. The penalty of course is that you're doing a couple branches; but even the "most accepted" answer to this question has at least three branches in it.
If you're serious about doing a super fast distance estimate to a specific precision, you could do so by writing your own simplified fsqrt() estimate using the same basic method as the compiler vendors do, but at a lower precision, by doing a fixed number of iterations. For example, you can eliminate the special case handling for extremely small or large numbers, and/or also reduce the number of Newton-Rapheson iterations. This was the key strategy underlying the so-called "Quake 3" fast inverse square root implementation -- it's the classic Newton algorithm with exactly one iteration.
Do not assume that your fsqrt() implementation is slow without benchmarking it and/or reading the sources. Most modern fsqrt() library implementations are branchless and really damned fast. Here for example is an old IBM floating point fsqrt implementation. Premature optimization is, and always will be, the root of all evil.
Note that for 2 (non-negative) distances A and B, if sqrt(A) < sqrt(B), then A < B. Create a specialized version of Get3DDistance() (GetSqrOf3DDistance()) that does not call sqrt() that would be used only for the sortfunc().
If you worry about performance, you should also take care of the way you send your arguments:
float Get3dDistance( Vec3 c1, Vec3 c2 );
implies two copies of Vec3 structure. Use references instead:
float Get3dDistance( Vec3 const & c1, Vec3 const & c2 );
You could compare squares of distances instead of the actual distances, since d2 = (x1-x2)2 + (y1-y2)2+ (z1-z2)2. It doesn't get rid of the multiplication, but it does eliminate the square root operation.
How often are the input vectors updated and how often are they sorted? Depending on your design, it might be quite efficient to extend the "Vec3" class with a pre-calculated distance and sort on that instead. Especially relevant if your implementation allows you to use vectorized operations.
Other than that, see the flipcode.com article on approximating distance functions for a discussion on yet another approach.
Depending slightly on the number of points that you are being used to compare with, what is below is pretty much guaranteed to be the get the list of points in approximate order assuming all points change at all iteration.
1) Rewrite the array into a single list of Manhattan distances with
out[ i ] = abs( posn[ i ].x - player.x ) + abs( posn[ i ].y - player.y ) + abs( posn[ i ].z - player.z );
2) Now you can use radix sort on floating point numbers to order them.
Note that in practice this is going to be a lot faster than sorting the list of 3d positions because it significantly reduces the memory bandwidth requirements in the sort operation which all of the time is going to be spend and in which unpredictable accesses and writes are going to occur. This will run on O(N) time.
If many of the points are stationary at each direction there are far faster algorithms like using KD-Trees, although implementation is quite a bit more complex and it is much harder to get good memory access patterns.
If this is simply a value for sorting, then you can swap the sqrt() for a abs(). If you need to compare distances against set values, get the square of that value.
E.g. instead of checking sqrt(...) against a, you can compare abs(...) against a*a.
You may want to consider caching the distance between the player and the object as you calculate it, and then use that in your sortfunc. This would depend upon how many times your sort function looks at each object, so you might have to profile to be sure.
What I'm getting at is that your sort function might do something like this:
compare(a,b);
compare(a,c);
compare(a,d);
and you would calculate the distance between the player and 'a' every time.
As others have mentioned, you can leave out the sqrt in this case.
If you could center your coordinates around the player, use spherical coordinates? Then you could sort by the radius.
That's a big if, though.
If your operation happens a lot, it might be worth to put it into some 3D data structure. You probably need the distance sorting to decide which object is visible, or some similar task. In order of complexity you can use:
Uniform (cubic) subdivision
Divide the used space into cells, and assign the objects to the cells. Fast access to element, neighbours are trivial, but empty cells take up a lot of space.
Quadtree
Given a threshold, divide used space recursively into four quads until less then threshold number of object is inside. Logarithmic access element if objects don't stack upon each other, neighbours are not hard to find, space efficient solution.
Octree
Same as Quadtree, but divides into 8, optimal even if objects are above each other.
Kd tree
Given some heuristic cost function, and a threshold, split space into two halves with a plane where the cost function is minimal. (Eg.: same amount of objects at each side.) Repeat recursively until threshold reached. Always logarithmic, neighbours are harder to get, space efficient (and works in all dimensions).
Using any of the above data structures, you can start from a position, and go from neighbour to neighbour to list the objects in increasing distance. You can stop at desired cut distance. You can also skip cells that cannot be seen from the camera.
For the distance check, you can do one of the above mentioned routines, but ultimately they wont scale well with increasing number of objects. These can be used to display data that takes hundreds of gigabytes of hard disc space.