Develop Reference

Calculate PI using sum of inverse squares

I need to calculate PI with predefined precision using this formula:
So I ended up with this solution.
private static double CalculatePIWithPrecision(int presicion)
{
if (presicion == 0)
{
return PI_ZERO_PRECISION;
}
double sum = 0;
double numberOfSumElements = Math.Pow(10, presicion + 2);
for (double i = 1; i < numberOfSumElements; i++)
{
sum += 1 / (i * i);
}
double pi = Math.Sqrt(sum * 6);
return pi;
}
So this works correct, but I faced the problem with efficiency. It's very slow with precision values 8 and higher.
Is there a better (and faster!) way to calculate PI using that formula?

double numberOfSumElements = Math.Pow(10, presicion + 2);
I'm going to talk about this strictly in practical software engineering terms, avoiding getting lost in the formal math. Just practical tips that any software engineer should know.
First observe the complexity of your code. How long it takes to execute is strictly determined by this expression. You've written an exponential algorithm, the value you calculate very rapidly goes up as presicion increases. You quote the uncomfortable number, 8 produces 10^10 or a loop that makes ten billion calculations. Yes, you notice this, that's when computers starts to take seconds to produce a result, no matter how fast they are.
Exponential algorithms are bad, they perform very poorly. You can only do worse with one that has factorial complexity, O(n!), that goes up even faster. Otherwise the complexity of many real-world problems.
Now, is that expression actually accurate? You can do this with an "elbow test", using a practical back-of-the-envelope example. Let's pick a precision of 5 digits as a target and write it out:
1.0000 + 0.2500 + 0.1111 + 0.0625 + 0.0400 + 0.0278 + ... = 1.6433
You can tell that the additions rapidly get smaller, it converges quickly. You can reason out that, once the next number you add gets small enough then it does very little to make the result more accurate. Let's say that when the next number is less than 0.00001 then it's time to stop trying to improve the result.
So you'll stop at 1 / (n * n) = 0.00001 => n * n = 100000 => n = sqrt(100000) => n ~= 316
Your expression says to stop at 10^(5+2) = 10,000,000
You can tell that you are way off, looping entirely too often and not improving the accuracy of the result with the last 9.999 million iterations.
Time to talk about the real problem, too bad that you didn't explain how you got to such a drastically wrong algorithm. But surely you discovered when testing your code that it just was not very good at calculating a more precise value for pi. So you figured that by iterating more often, you'd get a better result.
Do note that in this elbow-test, it is also very important that you are able to calculate the additions with sufficient precision. I intentionally rounded the numbers, as though it was calculated on a machine capable of performing additions with 5 digits of precision. Whatever you do, the result can never be more precise than 5 digits.
You are using the double type in your code. Directly supported by the processor, it does not have infinite precision. The one and only rule you ever need to keep in mind is that calculations with double are never more precise than 15 digits. Also memorize the rule for float, it is never more precise than 7 digits.
So no matter what value you pass for presicion, the result can never be more precise than 15 digits. That is not useful at all, you already have the value of pi accurate to 15 digits. It is Math.Pi
The one thing you need to do to fix this is using a type that has more precision than double. In fact, it needs to be a type that has arbitrary precision, it needs to be at least as accurate as the presicion value you pass. Such a type does not exist in the .NET framework. Finding a library that can provide you with one is a common question at SO.

Calculating algorithm time

I have an algorithm that works with extremely large numbers around the order of 2 raised to the power 4,500,000. I use the BigInteger class in .NET 4 to handle these numbers.
The algorithm is very simple in that it is a single loop that reduces a large initial number based on some predefined criteria. With each iteration, the number is reduced by around 10 exponents so 4,500,000 would become 4,499,990 in the next iteration.
I am currently getting 5.16 iterations per second or 0.193798 seconds per iteration. Based on that the total time for the algorithm should be roughly 22 hours to bring the exponent value down to 0.
The problem is, as the number is reduced, the time required to process the number in memory is reduced as well. Plus, as the exponent reduces to the 200,000 range, the iterations per second become huge and the reduction per iteration also increases exponentially.
Instead of letting the algo run for a whole day, is there a mathematical way to calculate how much time it would take based on an initial starting number and iterations per second?
This would be very helpful since I can measure improvements of optimization attempts quickly.
Consider the following psuedocode:
double e = 4500000; // 4,500,000.
Random r = new Random();
while (e > 0)
{
BigInteger b = BigInteger.Power(2, e);
double log = BigInteger.Log(b, 10);
e -= Math.Abs(log * r.Next(1, 10));
}

First rewrite
double log = BigInteger.Log(b, 10);
as
double log = log(2)/log(10) * e; // approx 0.3 * e
Then you notice that the algorithm terminates after O(1) iterations (~70% termination chance on each iteration), you can probably neglect the cost of everything apart from the first iteration.
The total cost of your algo is about 1 to 2 times as expensive as Math.Pow(2, e) for the initial exponent e. For base=2 this is a trivial bitshift, for others you'll need square-and-multiply

There is no way to estimate the time of the unknown since you are using Random!

generating completely random even numbers with C#

is there any good method to generate random even numbers with C#? I came up with this code:
Random RandString = new Random(0);
code:
private void button6_Click(object sender, EventArgs e)
{
int min = 0;
int max = 50;
int rand = RandString.Next(min, max-1);
while (rand % 2 == 1) // odd
{
rand = RandString.Next(min, max-1);
}
textBox4.Text = "" + rand;
}
Problems with this code:
my current solution might not be the fastest of all
very often, numbers that are generated by this method are repeated with the same sequence! For example after 5 iterations I can get: 8, 10, 20, 30, 16. After more clicking (for like 15 seconds) I can get 10, 30, 20 in a row or 30, 20, 10.. I'm not sure, but that's NOT what I call "random".
so is there any way to deal with the problem that I've described, or is there any other method?

textBox4.Text = (2 * rand.Next(min / 2, max / 2)).ToString();

even easier would be to take your result and multiply it by 2
private void button6_Click(object sender, EventArgs e)
{
int min = 0;
int max = 25;
int rand = RandString.Next(min, max-1)*2;
textBox4.Text = rand.ToString();
}

Random random = new Random();
int nextnum = random.Next(min/2,max/2) * 2

You could use the RNGCryptoServiceProvider to get a "more random" (which is a silly term I admit) number. There is a nice eample on MSDN's documentation page.
As others recommended, you can just multiply the result with 2 to get an even number.

There are a number of flaws in your approach.
First of all, Random from .NET isn't completely random in the cryptographic sense AFAIK, but it should be much much better than the results you report. Most likely, you are creating a new Random object on every iteration of your loop; since the current system time is used to seed the RNG, you will get the same sequences of pseudo-random numbers if you do runs very short after one another. Instead, create one instance of Random at startup, and keep it around for as long as possible.
Then, creating an even number from a random integer is much easier than looping until you find one that happens to be even: just generate one random number and clear the least significant bit: myRand.Next(min, max-1) & ~1;. A decent RNG has a uniform distribution over all of its bits, so clearing any bit shouldn't reduce entropy by more than, well, one bit.
Getting back to the 'completely random' part: Random provides a pseudo-random number generator. It is seeded once, based on a value that is somewhat random-ish (the least-significant parts of the current system time), but given the same seed, the RNG will reliably and deterministically produce the same numbers on every run. If you want true randomness (a.k.a. 'entropy'), you'd be surprised how hard it is to produce it on a machine that was built for deterministic calculations. UNIX and Unix-like systems typically offer a pool of entropy through a special kernel-generated file (/dev/random), using things like hard disk access timing, network noise, and whatever other sources of actual randomness they can find, and distillate those into a uniform distribution using fairly complicated calculations. Windows can probably do the same, but I am unaware of any API for this in either .NET or the classic win32 API.

factorial method resulting in error

I'm trying to get the factorial value of number 66, but my method resulting me an output 0. But whenever I try to get the factorial of 5, it is resulting me an output 120. Could anyone please tell me why?
public static int factorial(int n)
{
if (n == 1)
return n;
return n * factorial(n - 1);
}

Sure - factorials get very big, very fast. You're overflowing the bounds of int very quickly... and at some point you'll have multiplied by enough factors to get an overflow to 0, which will then keep the value at 0 forever.
According to a quick Google search, 66 factorial is 5.44344939 × 1092 - which is considerably more than int can handle, or even long or decimal. You could get double to handle it - you'd lose a huge amount of precision, and that would accumulate really quickly too, but at least it wouldn't overflow...

Your method overflows. See the following example:
static void Main(string[] args)
{
Console.WriteLine(factorial(66));
}
public static int factorial(int n)
{
if (n == 1)
return n;
var result = n * factorial(n - 1);
Console.WriteLine("{0} : {1}", n, result);
return result;
}
With this example, the results of each iteration is printed.
You see that at one point, result becomes 0 and this means that every iteration from that point on becomes n * 0.
You can try using BigInteger. This will give the correct result. Calculate factorials in C# contains more information on this.

66! does not fit into an int. Use BigInteger.

The problem is that the factorial of 66 is way to large to fit into an int. I think it will also we way to large to fit into a long.
As an example, factorial(20) will return 2432902008176640000

The factorial of 50 is 3.0414093202×1064 which exeeds already what a int can contain.
Use long or BigInteger for this.

You get numeric overflow, 66! ~= 5e92 which is way larger than an int can handle. Also, factorials are better calculated using a for loop.

Approximately 13 or 14 is the largest number whose factorial fits in an int... If you switch to long, it'll be aroung 18 or 19 if I recall correctly. If you wish arbitraty big numbers you'd have to write your own big arithmetic library or use an existing one :)

You need to use the appropriate data type.
In this case the Big Integer data type is probably best seeing as how fast the numbers get real big.
Here is the way you use this data type.
Right click on your project, choose the add reference menu.
Look for the system.numerics library and add it.
You then add the using clause in your code.
And then you can initialise variable with the keyword as usual.

How do I find the average in a LARGE set of numbers?

I have a large set of numbers, probably in the multiple gigabytes range. First issue is that I can't store all of these in memory. Second is that any attempt at addition of these will result in an overflow. I was thinking of using more of a rolling average, but it needs to be accurate. Any ideas?
These are all floating point numbers.
This is not read from a database, it is a CSV file collected from multiple sources. It has to be accurate as it is stored as parts of a second (e.g; 0.293482888929) and a rolling average can be the difference between .2 and .3
It is a set of #'s representing how long users took to respond to certain form actions. For example when showing a messagebox, how long did it take them to press OK or Cancel. The data was sent to me stored as seconds.portions of a second; 1.2347 seconds for example. Converting it to milliseconds and I overflow int, long, etc.. rather quickly. Even if I don't convert it, I still overflow it rather quickly. I guess the one answer below is correct, that maybe I don't have to be 100% accurate, just look within a certain range inside of a sepcific StdDev and I would be close enough.

You can sample randomly from your set ("population") to get an average ("mean"). The accuracy will be determined by how much your samples vary (as determined by "standard deviation" or variance).
The advantage is that you have billions of observations, and you only have to sample a fraction of them to get a decent accuracy or the "confidence range" of your choice. If the conditions are right, this cuts down the amount of work you will be doing.
Here's a numerical library for C# that includes a random sequence generator. Just make a random sequence of numbers that reference indices in your array of elements (from 1 to x, the number of elements in your array). Dereference to get the values, and then calculate your mean and standard deviation.
If you want to test the distribution of your data, consider using the Chi-Squared Fit test or the K-S test, which you'll find in many spreadsheet and statistical packages (e.g., R). That will help confirm whether this approach is usable or not.

Integers or floats?
If they're integers, you need to accumulate a frequency distribution by reading the numbers and recording how many of each value you see. That can be averaged easily.
For floating point, this is a bit of a problem. Given the overall range of the floats, and the actual distribution, you have to work out a bin-size that preserves the accuracy you want without preserving all of the numbers.
Edit
First, you need to sample your data to get a mean and a standard deviation. A few thousand points should be good enough.
Then, you need to determine a respectable range. Folks pick things like ±6σ (standard deviations) around the mean. You'll divide this range into as many buckets as you can stand.
In effect, the number of buckets determines the number of significant digits in your average. So, pick 10,000 or 100,000 buckets to get 4 or 5 digits of precision. Since it's a measurement, odds are good that your measurements only have two or three digits.
Edit
What you'll discover is that the mean of your initial sample is very close to the mean of any other sample. And any sample mean is close to the population mean. You'll note that most (but not all) of your means are with 1 standard deviation of each other.
You should find that your measurement errors and inaccuracies are larger than your standard deviation.
This means that a sample mean is as useful as a population mean.

Wouldn't a rolling average be as accurate as anything else (discounting rounding errors, I mean)? It might be kind of slow because of all the dividing.
You could group batches of numbers and average them recursively. Like average 100 numbers 100 times, then average the result. This would be less thrashing and mostly addition.
In fact, if you added 256 or 512 at once you might be able to bit-shift the result by either 8 or 9, (I believe you could do this in a double by simply changing the floating point mantissa)--this would make your program extremely quick and it could be written recursively in just a few lines of code (not counting the unsafe operation of the mantissa shift).
Perhaps dividing by 256 would already use this optimization? I may have to speed test dividing by 255 vs 256 and see if there is some massive improvement. I'm guessing not.

You mean of 32-bit and 64-bit numbers. But why not just use a proper Rational Big Num library? If you have so much data and you want an exact mean, then just code it.
class RationalBignum {
public Bignum Numerator { get; set; }
public Bignum Denominator { get; set; }
}
class BigMeanr {
public static int Main(string[] argv) {
var sum = new RationalBignum(0);
var n = new Bignum(0);
using (var s = new FileStream(argv[0])) {
using (var r = new BinaryReader(s)) {
try {
while (true) {
var flt = r.ReadSingle();
rat = new RationalBignum(flt);
sum += rat;
n++;
}
}
catch (EndOfStreamException) {
break;
}
}
}
Console.WriteLine("The mean is: {0}", sum / n);
}
}
Just remember, there are more numeric types out there than the ones your compiler offers you.

You could break the data into sets of, say, 1000 numbers, average these, and then average the averages.

This is a classic divide-and-conquer type problem.
The issue is that the average of a large set of numbers is the same
as the average of the first-half of the set, averaged with the average of the second-half of the set.
In other words:
AVG(A[1..N]) == AVG( AVG(A[1..N/2]), AVG(A[N/2..N]) )
Here is a simple, C#, recursive solution.
Its passed my tests, and should be completely correct.
public struct SubAverage
{
public float Average;
public int Count;
};
static SubAverage AverageMegaList(List<float> aList)
{
if (aList.Count <= 500) // Brute-force average 500 numbers or less.
{
SubAverage avg;
avg.Average = 0;
avg.Count = aList.Count;
foreach(float f in aList)
{
avg.Average += f;
}
avg.Average /= avg.Count;
return avg;
}
// For more than 500 numbers, break the list into two sub-lists.
SubAverage subAvg_A = AverageMegaList(aList.GetRange(0, aList.Count/2));
SubAverage subAvg_B = AverageMegaList(aList.GetRange(aList.Count/2, aList.Count-aList.Count/2));
SubAverage finalAnswer;
finalAnswer.Average = subAvg_A.Average * subAvg_A.Count/aList.Count +
subAvg_B.Average * subAvg_B.Count/aList.Count;
finalAnswer.Count = aList.Count;
Console.WriteLine("The average of {0} numbers is {1}",
finalAnswer.Count, finalAnswer.Average);
return finalAnswer;
}

The trick is that you're worried about an overflow. In that case, it all comes down to order of execution. The basic formula is like this:
Given:
A = current avg
C = count of items
V = next value in the sequence
The next average (A1) is:
(C * A) + V
A1 = ———————————
C + 1
The danger is over the course of evaulating the sequence, while A should stay relatively manageable C will become very large.
Eventually C * A will overflow the integer or double types.
One thing we can try is to re-write it like this, to reduce the chance of an overflow:
A1 = C/(C+1) * A/(C+1) + V/(C+1)
In this way, we never multiply C * A and only deal with smaller numbers. But the concern now is the result of the division operations. If C is very large, C/C+1 (for example) may not be meaningful when constrained to normal floating point representations. The best I can suggest is to use the largest type possible for C here.

Here's one way to do it in pseudocode:
average=first
count=1
while more:
count+=1
diff=next-average
average+=diff/count
return average

Sorry for the late comment, but isn't it the formula above provided by Joel Coehoorn rewritten wrongly?
I mean, the basic formula is right:
Given:
A = current avg
C = count of items
V = next value in the sequence
The next average (A1) is:
A1 = ( (C * A) + V ) / ( C + 1 )
But instead of:
A1 = C/(C+1) * A/(C+1) + V/(C+1)
shouldn't we have:
A1 = C/(C+1) * A + V/(C+1)
That would explain kastermester's post:
"My math ticks off here - You have C, which you say "go towards infinity" or at least, a really big number, then: C/(C+1) goes towards 1. A /(C+1) goes towards 0. V/(C+1) goes towards 0. All in all: A1 = 1 * 0 + 0 So put shortly A1 goes towards 0 - seems a bit off. – kastermester"
Because we would have A1 = 1 * A + 0, i.e., A1 goes towards A, which it's right.
I've been using such method for calculating averages for a long time and the aforementioned precision problems have never been an issue for me.

With floating point numbers the problem is not overflow, but loss of precision when the accumulated value gets large. Adding a small number to a huge accumulated value will result in losing most of the bits of the small number.
There is a clever solution by the author of the IEEE floating point standard himself, the Kahan summation algorithm, which deals exactly with this kind of problems by checking the error at each step and keeping a running compensation term that prevents losing the small values.

If the numbers are int's, accumulate the total in a long. If the numbers are long's ... what language are you using? In Java you could accumulate the total in a BigInteger, which is an integer which will grow as large as it needs to be. You could always write your own class to reproduce this functionality. The gist of it is just to make an array of integers to hold each "big number". When you add two numbers, loop through starting with the low-order value. If the result of the addition sets the high order bit, clear this bit and carry the one to the next column.
Another option would be to find the average of, say, 1000 numbers at a time. Hold these intermediate results, then when you're done average them all together.

Why is a sum of floating point numbers overflowing? In order for that to happen, you would need to have values near the max float value, which sounds odd.
If you were dealing with integers I'd suggest using a BigInteger, or breaking the set into multiple subsets, recursively averaging the subsets, then averaging the averages.
If you're dealing with floats, it gets a bit weird. A rolling average could become very inaccurate. I suggest using a rolling average which is only updated when you hit an overflow exception or the end of the set. So effectively dividing the set into non-overflowing sets.

Two ideas from me:
If the numbers are ints, use an arbitrary precision library like IntX - this could be too slow, though
If the numbers are floats and you know the total amount, you can divide each entry by that number and add up the result. If you use double, the precision should be sufficient.

Why not just scale the numbers (down) before computing the average?

If I were to find the mean of billions of doubles as accurately as possible, I would take the following approach (NOT TESTED):
Find out 'M', an upper bound for log2(nb_of_input_data). If there are billions of data, 50 may be a good candidate (> 1 000 000 billions capacity). Create an L1 array of M double elements. If you're not sure about M, creating an extensible list will solve the issue, but it is slower.
Also create an associated L2 boolean array (all cells set to false by default).
For each incoming data D:
int i = 0;
double localMean = D;
while (L2[i]) {
L2[i] = false;
localMean = (localMean + L1[i]) / 2;
i++;
}
L1[i] = localMean;
L2[i] = true;
And your final mean will be:
double sum = 0;
double totalWeight = 0;
for (int i = 0; i < 50) {
if (L2[i]) {
long weight = 1 << i;
sum += L1[i] * weight;
totalWeight += weight;
}
}
return sum / totalWeight;
Notes:
Many proposed solutions in this thread miss the point of lost precision.
Using binary instead of 100-group-or-whatever provides better precision, and doubles can be safely doubled or halved without losing precision!

Try this
Iterate through the numbers incrementing a counter, and adding each number to a total, until adding the next number would result in an overflow, or you run out of numbers.
( It makes no difference if the inputs are integers or floats - use the largest precision float you can and convert each input to that type)
Divide the total by the counter to get a mean ( a floating point), and add it to a temp array
If you had run out of numbers, and there is only one element in temp, that's your result.
Start over using the temp array as input, ie iteratively recurse until you reached the end condition described earlier.

depending on the range of numbers it might be a good idea to have an array where the subscript is your number and the value is the quantity of that number, you could then do your calculation from this