I am trying to create a fast prime generator in Java. It is (more or less) accepted that the fastest way for this is the segmented sieve of Eratosthenes: https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes. Lots of optimizations can be further implemented to make it faster. As of now, my implementation generates 50847534 primes below 10^9 in about 1.6 seconds, but I am looking to make it faster and at least break the 1 second barrier. To increase the chance of getting good replies, I will include a walkthrough of the algorithm as well as the code.
Still, as a TL;DR, I am looking to include multi-threading into the code
For the purposes of this question, I want to separate between the 'segmented' and the 'traditional' sieves of Eratosthenes. The traditional sieve requires O(n) space and therefore is very limited in range of the input (the limit of it). The segmented sieve however only requires O(n^0.5) space and can operate on much larger limits. (A main speed-up is using a cache-friendly segmentation, taking into account the L1 & L2 cache sizes of the specific computer). Finally, the main difference that concerns my question is that the traditional sieve is sequential, meaning it can only continue once the previous steps are completed. The segmented sieve however, is not. Each segment is independent, and is 'processed' individually against the sieving primes (the primes not larger than n^0.5). This means that theoretically, once I have the sieving primes, I can divide the work between multiple computers, each processing a different segment. The work of eachother is independent of the others. Assuming (wrongly) that each segment requires the same amount of time t to complete, and there are k segments, One computer would require total time of T = k * t, whereas k computers, each working on a different segment would require a total amount of time T = t to complete the entire process. (Practically, this is wrong, but for the sake of simplicity of the example).
This brought me to reading about multithreading - dividing the work to a few threads each processing a smaller amount of work for better usage of CPU. To my understanding, the traditional sieve cannot be multithreaded exactly because it is sequential. Each thread would depend on the previous, rendering the entire idea unfeasible. But a segmented sieve may indeed (I think) be multithreaded.
Instead of jumping straight into my question, I think it is important to introduce my code first, so I am hereby including my current fastest implementation of the segmented sieve. I have worked quite hard on it. It took quite some time, slowly tweaking and adding optimizations to it. The code is not simple. It is rather complex, I would say. I therefore assume the reader is familiar with the concepts I am introducing, such as wheel factorization, prime numbers, segmentation and more. I have included notes to make it easier to follow.
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
public class primeGen {
public static long x = (long)Math.pow(10, 9); //limit
public static int sqrtx;
public static boolean [] sievingPrimes; //the sieving primes, <= sqrtx
public static int [] wheels = new int [] {2,3,5,7,11,13,17,19}; // base wheel primes
public static int [] gaps; //the gaps, according to the wheel. will enable skipping multiples of the wheel primes
public static int nextp; // the first prime > wheel primes
public static int l; // the amount of gaps in the wheel
public static void main(String[] args)
{
long startTime = System.currentTimeMillis();
preCalc(); // creating the sieving primes and calculating the list of gaps
int segSize = Math.max(sqrtx, 32768*8); //size of each segment
long u = nextp; // 'u' is the running index of the program. will continue from one segment to the next
int wh = 0; // the will be the gap index, indicating by how much we increment 'u' each time, skipping the multiples of the wheel primes
long pi = pisqrtx(); // the primes count. initialize with the number of primes <= sqrtx
for (long low = 0 ; low < x ; low += segSize) //the heart of the code. enumerating the primes through segmentation. enumeration will begin at p > sqrtx
{
long high = Math.min(x, low + segSize);
boolean [] segment = new boolean [(int) (high - low + 1)];
int g = -1;
for (int i = nextp ; i <= sqrtx ; i += gaps[g])
{
if (sievingPrimes[(i + 1) / 2])
{
long firstMultiple = (long) (low / i * i);
if (firstMultiple < low)
firstMultiple += i;
if (firstMultiple % 2 == 0) //start with the first odd multiple of the current prime in the segment
firstMultiple += i;
for (long j = firstMultiple ; j < high ; j += i * 2)
segment[(int) (j - low)] = true;
}
g++;
//if (g == l) //due to segment size, the full list of gaps is never used **within just one segment** , and therefore this check is redundant.
//should be used with bigger segment sizes or smaller lists of gaps
//g = 0;
}
while (u <= high)
{
if (!segment[(int) (u - low)])
pi++;
u += gaps[wh];
wh++;
if (wh == l)
wh = 0;
}
}
System.out.println(pi);
long endTime = System.currentTimeMillis();
System.out.println("Solution took "+(endTime - startTime) + " ms");
}
public static boolean [] simpleSieve (int l)
{
long sqrtl = (long)Math.sqrt(l);
boolean [] primes = new boolean [l/2+2];
Arrays.fill(primes, true);
int g = -1;
for (int i = nextp ; i <= sqrtl ; i += gaps[g])
{
if (primes[(i + 1) / 2])
for (int j = i * i ; j <= l ; j += i * 2)
primes[(j + 1) / 2]=false;
g++;
if (g == l)
g=0;
}
return primes;
}
public static long pisqrtx ()
{
int pi = wheels.length;
if (x < wheels[wheels.length-1])
{
if (x < 2)
return 0;
int k = 0;
while (wheels[k] <= x)
k++;
return k;
}
int g = -1;
for (int i = nextp ; i <= sqrtx ; i += gaps[g])
{
if(sievingPrimes[( i + 1 ) / 2])
pi++;
g++;
if (g == l)
g=0;
}
return pi;
}
public static void preCalc ()
{
sqrtx = (int) Math.sqrt(x);
int prod = 1;
for (long p : wheels)
prod *= p; // primorial
nextp = BigInteger.valueOf(wheels[wheels.length-1]).nextProbablePrime().intValue(); //the first prime that comes after the wheel
int lim = prod + nextp; // circumference of the wheel
boolean [] marks = new boolean [lim + 1];
Arrays.fill(marks, true);
for (int j = 2 * 2 ;j <= lim ; j += 2)
marks[j] = false;
for (int i = 1 ; i < wheels.length ; i++)
{
int p = wheels[i];
for (int j = p * p ; j <= lim ; j += 2 * p)
marks[j]=false; // removing all integers that are NOT comprime with the base wheel primes
}
ArrayList <Integer> gs = new ArrayList <Integer>(); //list of the gaps between the integers that are coprime with the base wheel primes
int d = nextp;
for (int p = d + 2 ; p < marks.length ; p += 2)
{
if (marks[p]) //d is prime. if p is also prime, then a gap is identified, and is noted.
{
gs.add(p - d);
d = p;
}
}
gaps = new int [gs.size()];
for (int i = 0 ; i < gs.size() ; i++)
gaps[i] = gs.get(i); // Arrays are faster than lists, so moving the list of gaps to an array
l = gaps.length;
sievingPrimes = simpleSieve(sqrtx); //initializing the sieving primes
}
}
Currently, it produces 50847534 primes below 10^9 in about 1.6 seconds. This is very impressive, at least by my standards, but I am looking to make it faster, possibly break the 1 second barrier. Even then, I believe it can be made much faster still.
The whole program is based on wheel factorization: https://en.wikipedia.org/wiki/Wheel_factorization. I have noticed I am getting the fastest results using a wheel of all primes up to 19.
public static int [] wheels = new int [] {2,3,5,7,11,13,17,19}; // base wheel primes
This means that the multiples of those primes are skipped, resulting in a much smaller searching range. The gaps between numbers which we need to take are then calculated in the preCalc method. If we make those jumps between the the numbers in the searching range we skip the multiples of the base primes.
public static void preCalc ()
{
sqrtx = (int) Math.sqrt(x);
int prod = 1;
for (long p : wheels)
prod *= p; // primorial
nextp = BigInteger.valueOf(wheels[wheels.length-1]).nextProbablePrime().intValue(); //the first prime that comes after the wheel
int lim = prod + nextp; // circumference of the wheel
boolean [] marks = new boolean [lim + 1];
Arrays.fill(marks, true);
for (int j = 2 * 2 ;j <= lim ; j += 2)
marks[j] = false;
for (int i = 1 ; i < wheels.length ; i++)
{
int p = wheels[i];
for (int j = p * p ; j <= lim ; j += 2 * p)
marks[j]=false; // removing all integers that are NOT comprime with the base wheel primes
}
ArrayList <Integer> gs = new ArrayList <Integer>(); //list of the gaps between the integers that are coprime with the base wheel primes
int d = nextp;
for (int p = d + 2 ; p < marks.length ; p += 2)
{
if (marks[p]) //d is prime. if p is also prime, then a gap is identified, and is noted.
{
gs.add(p - d);
d = p;
}
}
gaps = new int [gs.size()];
for (int i = 0 ; i < gs.size() ; i++)
gaps[i] = gs.get(i); // Arrays are faster than lists, so moving the list of gaps to an array
l = gaps.length;
sievingPrimes = simpleSieve(sqrtx); //initializing the sieving primes
}
At the end of the preCalc method, the simpleSieve method is called, efficiently sieving all the sieving primes mentioned before, the primes <= sqrtx. This is a simple Eratosthenes sieve, rather than segmented, but it is still based on wheel factorization, perviously computed.
public static boolean [] simpleSieve (int l)
{
long sqrtl = (long)Math.sqrt(l);
boolean [] primes = new boolean [l/2+2];
Arrays.fill(primes, true);
int g = -1;
for (int i = nextp ; i <= sqrtl ; i += gaps[g])
{
if (primes[(i + 1) / 2])
for (int j = i * i ; j <= l ; j += i * 2)
primes[(j + 1) / 2]=false;
g++;
if (g == l)
g=0;
}
return primes;
}
Finally, we reach the heart of the algorithm. We start by enumerating all primes <= sqrtx, with the following call:
long pi = pisqrtx();`
which used the following method:
public static long pisqrtx ()
{
int pi = wheels.length;
if (x < wheels[wheels.length-1])
{
if (x < 2)
return 0;
int k = 0;
while (wheels[k] <= x)
k++;
return k;
}
int g = -1;
for (int i = nextp ; i <= sqrtx ; i += gaps[g])
{
if(sievingPrimes[( i + 1 ) / 2])
pi++;
g++;
if (g == l)
g=0;
}
return pi;
}
Then, after initializing the pi variable which keeps track of the enumeration of primes, we perform the mentioned segmentation, starting the enumeration from the first prime > sqrtx:
int segSize = Math.max(sqrtx, 32768*8); //size of each segment
long u = nextp; // 'u' is the running index of the program. will continue from one segment to the next
int wh = 0; // the will be the gap index, indicating by how much we increment 'u' each time, skipping the multiples of the wheel primes
long pi = pisqrtx(); // the primes count. initialize with the number of primes <= sqrtx
for (long low = 0 ; low < x ; low += segSize) //the heart of the code. enumerating the primes through segmentation. enumeration will begin at p > sqrtx
{
long high = Math.min(x, low + segSize);
boolean [] segment = new boolean [(int) (high - low + 1)];
int g = -1;
for (int i = nextp ; i <= sqrtx ; i += gaps[g])
{
if (sievingPrimes[(i + 1) / 2])
{
long firstMultiple = (long) (low / i * i);
if (firstMultiple < low)
firstMultiple += i;
if (firstMultiple % 2 == 0) //start with the first odd multiple of the current prime in the segment
firstMultiple += i;
for (long j = firstMultiple ; j < high ; j += i * 2)
segment[(int) (j - low)] = true;
}
g++;
//if (g == l) //due to segment size, the full list of gaps is never used **within just one segment** , and therefore this check is redundant.
//should be used with bigger segment sizes or smaller lists of gaps
//g = 0;
}
while (u <= high)
{
if (!segment[(int) (u - low)])
pi++;
u += gaps[wh];
wh++;
if (wh == l)
wh = 0;
}
}
I have also included it as a note, but will explain as well. Because the segment size is relatively small, we will not go through the entire list of gaps within just one segment, and checking it - is redundant. (Assuming we use a 19-wheel). But in a broader scope overview of the program, we will make use of the entire array of gaps, so the variable u has to follow it and not accidentally surpass it:
while (u <= high)
{
if (!segment[(int) (u - low)])
pi++;
u += gaps[wh];
wh++;
if (wh == l)
wh = 0;
}
Using higher limits will eventually render a bigger segment, which might result in a neccessity of checking we don't surpass the gaps list even within the segment. This, or tweaking the wheel primes base might have this effect on the program. Switching to bit-sieving can largely improve the segment limit though.
As an important side-note, I am aware that efficient segmentation is
one that takes the L1 & L2 cache-sizes into account. I get the
fastest results using a segment size of 32,768 * 8 = 262,144 = 2^18. I am not sure what the cache-size of my computer is, but I do
not think it can be that big, as I see most cache sizes <= 32,768.
Still, this produces the fastest run time on my computer, so this is
why it's the chosen segment size.
As I mentioned, I am still looking to improve this by a lot. I
believe, according to my introduction, that multithreading can result
in a speed-up factor of 4, using 4 threads (corresponding to 4
cores). The idea is that each thread will still use the idea of the
segmented sieve, but work on different portions. Divide the n
into 4 equal portions - threads, each in turn performing the
segmentation on the n/4 elements it is responsible for, using the
above program. My question is how do I do that? Reading about
multithreading and examples, unfortunately, did not bring to me any
insight on how to implement it in the case above efficiently. It
seems to me, as opposed to the logic behind it, that the threads were
running sequentially, rather than simultaneously. This is why I
excluded it from the code to make it more readable. I will really
appreciate a code sample on how to do it in this specific code, but a
good explanation and reference will maybe do the trick too.
Additionally, I would like to hear about more ways of speeding-up
this program even more, any ideas you have, I would love to hear!
Really want to make it very fast and efficient. Thank you!
An example like this should help you get started.
An outline of a solution:
Define a data structure ("Task") that encompasses a specific segment; you can put all the immutable shared data into it for extra neatness, too. If you're careful enough, you can pass a common mutable array to all tasks, along with the segment limits, and only update the part of the array within these limits. This is more error-prone, but can simplify the step of joining the results (AFAICT; YMMV).
Define a data structure ("Result") that stores the result of a Task computation. Even if you just update a shared resulting structure, you may need to signal which part of that structure has been updated so far.
Create a Runnable that accepts a Task, runs a computation, and puts the results into a given result queue.
Create a blocking input queue for Tasks, and a queue for Results.
Create a ThreadPoolExecutor with the number of threads close to the number of machine cores.
Submit all your Tasks to the thread pool executor. They will be scheduled to run on the threads from the pool, and will put their results into the output queue, not necessarily in order.
Wait for all the tasks in the thread pool to finish.
Drain the output queue and join the partial results into the final result.
Extra speedup may (or may not) be achieved by joining the results in a separate task that reads the output queue, or even by updating a mutable shared output structure under synchronized, depending on how much work the joining step involves.
Hope this helps.
Are you familiar with the work of Tomas Oliveira e Silva? He has a very fast implementation of the Sieve of Eratosthenes.
How interested in speed are you? Would you consider using c++?
$ time ../c_code/segmented_bit_sieve 1000000000
50847534 primes found.
real 0m0.875s
user 0m0.813s
sys 0m0.016s
$ time ../c_code/segmented_bit_isprime 1000000000
50847534 primes found.
real 0m0.816s
user 0m0.797s
sys 0m0.000s
(on my newish laptop with an i5)
The first is from #Kim Walisch using a bit array of odd prime candidates.
https://github.com/kimwalisch/primesieve/wiki/Segmented-sieve-of-Eratosthenes
The second is my tweak to Kim's with IsPrime[] also implemented as bit array, which is slightly less clear to read, although a little faster for big N due to the reduced memory footprint.
I will read your post carefully as I am interested in primes and performance no matter what language is used. I hope this isn't too far off topic or premature. But I noticed I was already beyond your performance goal.
I'm quite new to coding, as you all can see from the clumsy code below. However, looking at this code you can see what I'm getting at. The code basically does what its supposed to, but I would like to write it as a loop to make it more efficient. Could someone maybe point me in the right direction? I have done some digging and thought about recursion, but I haven't been able to figure out how to apply it here.
public static void main(String[] args) {
double a = 10;
double b = 2;
double c = 3;
double avg = (a + b + c)/3;
double avg1 = (avg + b + c)/3;
double avg2 = (avg1 + b + c)/3;
double avg3 = (avg2 + b + c)/3;
System.out.println(avg+ "\n" + avg1+ "\n"+ avg2 + "\n"+ avg3);
}
Functionally, this would be equivalent to what you have done:
public static void main(String[] args) {
double a = 10;
double b = 2;
double c = 3;
double avg = (a + b + c)/3;
System.out.println(avg);
for (int i=0; i<3; i++) {
avg = (avg + b + c)/3;
System.out.println(avg);
}
}
But also you should know that shorter code does not always mean efficient code. The solution may be more concise, but I doubt there will be any change in performance.
If you mean shorter code with efficieny you can do it like this.
public static void main(String[] args) {
double a = 10;
double b = 2;
double c = 3;
for (int i = 0; i < 4; i++) {
a = (a + b + c) / 3;
System.out.println(a);
}
}
I have no idea what this calculation represents (some sort of specialised weighted average?) but rather than use repetition and loops, you can reach the exact same calculation by using a bit of algebra and refactoring the terms:
public static double directlyCalculateWeightedAverage(double a, double b,
double c) {
return a / 81 + 40 * b / 81 + 40 * c / 81;
}
This reformulation is reached because the factor a appears just once in the mix and is then divided by 34 which is 81. Then each of b and c appear at various levels of division, so that b sums to this:
b/81 + b/27 + b/9 + b/3
== b/81 + 3b/81 + 9b/81 + 27b/81
== 40b/81
and c is treated exactly the same.
Which gives the direct calculation
a/81 + 40b/81 + 40c/81
Assuming your formula does not change, I'd recommend using this direct approach rather than resorting to repeated calculations and loops.
Your problem can be solved by 2 approaches: iterative (with a loop) or recursive (with a recursive function).
Iterative approach : for loop
The for loop allow you to repeat a group of instructions au given number of times.
In your case, you could write the following :
double a = 10, b = 2, c = 3;
double avg = a;
for (int i = 0; i < 4; i++) {
avg = (avg + b + c) / 3;
System.out.println(avg);
}
This will print the 4 first results of your calculation.
In my example, I overwrite the variable avg to only keep the last result, which might not be what you want. To keep the result of each loop iteration, you may store the result in an array.
Recursive approach
In Java, there is no such thing as a standalone function. In order to use recursion, you have to use static methods :
private static double recursiveAvg(double avg, int count) {
// Always check that the condition for your recursion is valid !
if (count == 0) {
return avg;
}
// Apply your formula
avg = (avg + 2 + 3) / 3;
// Call the same function with the new avg value, and decrease the iteration count.
return recursiveAvg(avg, count - 1);
}
public static void main(String[] args) {
// Start from a = 10, and repeat the operation 4 times.
double avg = recursiveAvg(10, 4);
System.out.println(avg);
}
Always check for a condition that will end the recursion. In our example, it's the number of times the operation should be performed.
Note that most programmers prefer the iterative approach : easier to write and read, and less error prone.
Trying a method to find the power of a number using a for loop and without using Math.pow. For this result I just got 2.0 as it doesn't seem to go back round through the loop. Help please.
public void test() {
{
double t = 1;
double b = 2; // base number
double exponent = 2;
for (int i = 1; i<=exponent; i++);
t = t*b;
System.out.println(t);
Try this.
double t = 1;
double b = 2; // base number
double exponent = 2;
for (int i = 1; i<=exponent; i++) t = t*b;
System.out.println(t);
It's because the first iteration around you are setting T equal to B. You aren't multiplying by the exponent just yet. So it needs to iterate 1 further time then you are expecting. just decrease the I value in your for loop. EG
for(int i = 1; i <= exponent; i++)
t=t*b;
Hope this helps!
I am using the "think java" book and I am stuck on exercise 7.6. The goal here is to write a function that can find . It gives you a couple hints:
One way to evaluate is
to use the infinite series expansion:
In other words, we need to add up a series of terms where the ith term
is equal to
Here is the code I came up with, but it is horribly wrong (when compared to Math.exp) for anything other than a power of 1. I don't understand why, as far as I can tell the code is correct with the formula from the book. I'm not sure if this is more of a math question or something related to how big of a number double and int can hold, but I am just trying to understand why this doesn't work.
public static void main(String[] args) {
System.out.println("Find exp(-x^2)");
double x = inDouble("Enter x: ");
System.out.println("myexp(" + -x*x + ") = " + gauss(x, 20));
System.out.println("Math.exp(" + -x*x + ") = " + Math.exp(-x*x));
}
public static double gauss(double x, int n) {
x = -x*x;
System.out.println(x);
double exp = 1;
double prevnum = 1;
int prevdenom = 1;
int i = 1;
while (i < n) {
exp = exp + (prevnum*x)/(prevdenom*i);
prevnum = prevnum*x;
prevdenom = prevdenom*i;
i++;
}
return exp;
} // I can't figure out why this is so inacurate, as far as I can tell the math is accurate to what the book says the formula is
public static double inDouble(String string) {
Scanner in = new Scanner (System.in);
System.out.print(string);
return in.nextDouble();
}
I am about to add to the comment on your question. I do this because I feel I have a slightly better implementation.
Your approach
Your approach is to have the function accept two arguments, where the second argument is the number of iterations. This isn't bad, but as #JamesKPolk pointed out, you might have to do some manual searching for an int (or long) that won't overflow
My approach
My approach would use something called the machine epsilon for a data type. The machine epsilon is the smallest number of that type (in your case, double) that is representable as that number. There exists algorithm for determining what that machine epsilon is, if you are not "allowed" to access machine epsilon in the Double class.
There is math behind this:
The series representation for your function is
Since it is alternating series, the error term is the absolute value of the first term you choose not to include (I leave the proof to you).
What this means is that we can have an error-based implementation that doesn't use iterations! The best part is that you could implement it for floats, and data types that are "more" than doubles! I present thus:
public static double gauss(double x)
{
x = -x*x;
double exp = 0, error = 1, numerator = 1, denominator = 1;
double machineEpsilon = 1.0;
// calculate machineEpsilon
while ((1.0 + 0.5 * machineEpsilon) != 1.0)
machineEpsilon = 0.5 * machineEpsilon;
int n = 0; //
// while the error is large enough to be representable in terms of the current data type
while ((error >= machineEpsilon) || (-error <= -machineEpsilon))
{
exp += error;
// calculate the numerator (it is 1 if we just start, but -x times its past value otherwise)
numerator = ((n == 0) ? 1 : -numerator * x);
// calculate the denominator (denominator gets multiplied by n)
denominator *= (n++);
// calculate error
error = numerator/denominator;
}
return exp;
}
Let me know how this works!
I'm trying to calculate the Mean Difference average of a set of data. I have two (supposedly equivalent) formulas which calculate this, with one being more efficient (O^n) than the other (O^n2).
The problem is that while the inefficient formula gives correct output, the efficient one does not. Just by looking at both formulas I had a hunch that they weren't equivalent, but wrote it off because the derivation was made by a statician in a scientific journal. So i'm assuming the problem is my translation. Can anyone help me translate the efficient function properly?
Inefficient formula:
Inefficient formula translation (Java):
public static double calculateMeanDifference(ArrayList<Integer> valuesArrayList)
{
int valuesArrayListSize = valuesArrayList.size();
int sum = 0;
for(int i = 0; i < valuesArrayListSize; i++)
{
for(int j = 0; j < valuesArrayListSize; j++)
sum += (i != j ? Math.abs(valuesArrayList.get(i) - valuesArrayList.get(j)) : 0);
}
return new Double( (sum * 1.0)/ (valuesArrayListSize * (valuesArrayListSize - 1)));
}
Efficient derived formula:
where (sorry, don't know how to use MathML on here):
x(subscript i) = the ith order statistic of the data set
x(bar) = the mean of the data set
Efficient derived formula translation (Java):
public static double calculateMean(ArrayList<Integer> valuesArrayList)
{
double sum = 0;
int valuesArrayListSize = valuesArrayList.size();
for(int i = 0; i < valuesArrayListSize; i++)
sum += valuesArrayList.get(i);
return sum / (valuesArrayListSize * 1.0);
}
public static double calculateMeanDifference(ArrayList<Integer> valuesArrayList)
{
double sum = 0;
double mean = calculateMean(valuesArrayList);
int size = valuesArrayList.size();
double rightHandTerm = mean * size * (size + 1);
double denominator = (size * (size - 1)) / 2.0;
Collections.sort(valuesArrayList);
for(int i = 0; i < size; i++)
sum += (i * valuesArrayList.get(i) - rightHandTerm);
double meanDifference = (2 * sum) / denominator;
return meanDifference;
}
My data set consists of a collection of integers each having a value bounded by the set [0,5].
Randomly generating such sets and using the two functions on them gives different results. The inefficient one seems to be the one producing results in line with what is being measured: the absolute average difference between any two values in the set.
Can anyone tell me what's wrong with my translation?
EDIT: I created a simpler implementation that is O(N) provided the all your data has values limited to a relatively small set.The formula sticks to the methodology of the first method and thus, gives identical results to it (unlike the derived formula). If it fits your use case, I suggest people use this instead of the derived efficient formula, especially since the latter seems to give negative values when N is small).
Efficient, non-derived translation (Java):
public static double calculateMeanDifference3(ArrayList<Integer> valuesArrayList)
{
HashMap<Integer, Double> valueCountsHashMap = new HashMap<Integer, Double>();
double size = valuesArrayList.size();
for(int i = 0; i < size; i++)
{
int currentValue = valuesArrayList.get(i);
if(!valueCountsHashMap.containsKey(currentValue))
valueCountsHashMap.put(currentValue, new Double(1));
else
valueCountsHashMap.put(currentValue, valueCountsHashMap.get(currentValue)+ 1);
}
double sum = 0;
for(Map.Entry<Integer, Double> valueCountKeyValuePair : valueCountsHashMap.entrySet())
{
int currentValue = valueCountKeyValuePair.getKey();
Double currentCount = valueCountKeyValuePair.getValue();
for(Map.Entry<Integer, Double> valueCountKeyValuePair1 : valueCountsHashMap.entrySet())
{
int loopValue = valueCountKeyValuePair1.getKey();
Double loopCount = valueCountKeyValuePair1.getValue();
sum += (currentValue != loopValue ? Math.abs(currentValue - loopValue) * loopCount * currentCount : 0);
}
}
return new Double( sum/ (size * (size - 1)));
}
Your interpretation of sum += (i * valuesArrayList.get(i) - rightHandTerm); is wrong, it should be sum += i * valuesArrayList.get(i);, then after your for, double meanDifference = ((2 * sum) - rightHandTerm) / denominator;
Both equations yields about the same value, but they are not equal. Still, this should help you a little.
You subtract rightHandTerm on each iteration, so it gets [over]multiplied to N.
The big Sigma in the nominator touches only (i x_i), not the right hand term.
One more note: mean * size == sum. You don't have to divide sum by N and then remultiply it back.