In the beginning of the function I calculate the total weight of a protein sequence and define it as seq_weight.
After that I calculate the weight of several fragments and make combinations of those weights that sum to the total weight of the first proteins sequence.
The first print statement prints the total weight correctly, but near the end of the function it seems to forget that value when I want to define it as the result of the sum.
When I type the value manually I get the result I want:
def fragmentcombinations(sequence, fragments):
for seq_rec in sequence:
seq_weight = 0.0
for i in seq_rec.seq:
seq_weight += SeqUtils.molecular_weight(i, "protein")
print("The protein sequence: " + seq_rec.seq)
print("The molecular weight: " + str(round(seq_weight, 2)) + " Da.")
nums = []
for a in fragments:
fragment_weights = 0.0
for aa in a.seq:
fragment_weights += SeqUtils.molecular_weight(aa, 'protein')
nums.append(round(fragment_weights, 2))
print(nums)
weights_array = []
combs = []
if len(nums) > 0:
for r in range(0,len(nums)+1):
weights_array += list(combinations(nums, r))
for item in weights_array:
if sum(item) == 4364.85: #Sequence weight needs to inserted manually -> not ideal
combs.append(item)
print(" ")
print("The possible combinations of fragment weights that may cover the protein sequence without overlap are: ")
for row in combs:
print(*row, sep = ", ")
fragmentcombinations(seq_list3, seq_list4)
This is the result:
The protein sequence: IEEATHMTPCYELHGLRWVQIQDYAINVMQCL
The molecular weight: 4364.85 Da.
[3611.86, 2269.63, 469.53, 556.56, 1198.41, 2609.88, 547.69, 1976.23, 2306.48, 938.01, 1613.87, 789.87, 737.75, 2498.71, 2064.25, 1184.39, 1671.87]
The possible combinations of fragment weights that may cover the protein sequence without overlap are:
556.56, 1198.41, 2609.88
469.53, 2609.88, 547.69, 737.75
556.56, 1198.41, 938.01, 1671.87
469.53, 547.69, 938.01, 737.75, 1671.87
If I write
if sum(item) == seq_weight:
the result doesn't print the combination of weights like I intended.
Sorry if the code is kind of messy, I'm still a beginner.
Thanks in advance!
The problem is not that your variable is not remembered anymore. The problem is that you perform an exact comparison between floating point numbers. In programming floating point numbers are the "decimal" numbers, but they are not the exact presentation of your numbers. They only are up to an arbitrary precision.
Let's do some basic maths Python.
>>> a = 0.2 + 0.1
>>> a
0.30000000000000004
>>> a == 0.3
False
As you can see, there is clearly happening something weird here. But this is just how floating point arithmetic works.
Now we have explained that. What should you do to make your program work? There are multiple solutions.
One way to deal with it, is to compare your numbers to some fixed difference. ie
if abs(sum(item) - seq_weight) < 0.00001
Another way to deal with this is using fixed precision decimal objects, but that can be more difficult than you think it is. https://docs.python.org/3/library/decimal.html
I'm adding a pair of unsigned 32bit binary integers (including overflow). The addition is expressive rather than actually computed, so there's no need for an efficient algorithm, but since each component is manually specified in terms of individual bits, I need one with a compact representation. Any suggestions?
Edit: In terms of boolean operators. So I'm thinking that carry = a & b; sum = a ^ b; for the first bit, but the other 31?
Oh, and subtraction!
You can not perform addition with simple boolean operators, you need an adder. (Of course the adder can be built using some more complex boolean operators.)
The adder adds two bits plus carry, and passes carry out to next bit.
Pseudocode:
carry = 0
for i = 31 to 0
sum = a[i] + b[i] + carry
result[i] = sum & 1
carry = sum >> 1
next i
Here is an implementation using the macro language of VEDIT text editor.
The two numbers to be added are given as ASCII strings, one on each line.
The results are inserted on the third line.
Reg_Empty(10) // result as ASCII string
#0 = 0 // carry bit
for (#9=31; #9>=0; #9--) {
#1 = CC(#9)-'0' // a bit from first number
#2 = CC(#9+34)-'0' // a bit from second number
#3 = #0+#1+#2 // add with carry
#4 = #3 & 1 // resulting bit
#0 = #3 >> 1 // new carry
Num_Str(#4, 11, LEFT) // convert bit to ASCII
Reg_Set(10, #11, INSERT) // insert bit to start of string
}
Line(2)
Reg_Ins(10) IN
Return
Example input and output:
00010011011111110101000111100001
00110110111010101100101101110111
01001010011010100001110101011000
Edit:
Here is pseudocode where the adder has been implemented with boolean operations:
carry = 0
for i = 31 to 0
sum[i] = a[i] ^ b[i] ^ carry
carry = (a[i] & b[i]) | (a[i] & carry) | (b[i] & carry)
next i
Perhaps you can begin by stating addition for two 1-bit numbers, with overflow (=carry):
A | B | SUM | CARRY
===================
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
To generalize this further, you need a "full adder" which also takes a carry as an input, from the preceding stage. Then you can express the 32-bit addition as a chain of 32 such full adders (with the first stage's carry input tied to 0).
Regarding data structure part to represent these numbers. There are 4 ways
1) Bit Array
A bit array is an array data structure that compactly stores individual bits.
They are also known as bitmap, bitset or bitstring.
2) Bit Field
A bit field is a common idiom used in computer programming to compactly store multiple logical values as a short series of bits where each of the single bits can be addressed separately.
3) Bit Plane
A bit plane of a digital discrete signal (such as image or sound) is a set of bits corresponding to a given bit position in each of the binary numbers representing the signal.
4) Bit Board
A bitboard or bit field is a format that stuffs a whole group of related boolean variables into the same integer, typically representing positions on a board game.
Regarding implementation, you can check that at each step, we have following
S = a xor b xor c
S is result of sum of current bits a an b
c is input carry
Cout - output carry is (a & b) xor (c & (a xor b))
This isn't necessarily a programming question but i'm sure you folks know how to do it. How would i convert floating point numbers into binary.
The number i am looking at is 27.625.
27 would be 11011, but what do i do with the .625?
On paper, a good algorithm to convert the fractional part of a decimal number is the "repeated multiplication by 2" algorithm (see details at http://www.exploringbinary.com/base-conversion-in-php-using-bcmath/, under the heading "dec2bin_f()"). For example, 0.8125 converts to binary as follows:
1. 0.8125 * 2 = 1.625
2. 0.625 * 2 = 1.25
3. 0.25 * 2 = 0.5
4. 0.5 * 2 = 1.0
The integer parts are stripped off and saved at each step, forming the binary result: 0.1101.
If you want a tool to do these kinds of conversions automatically, see my decimal/binary converter.
Assuming you are not thinking about inside a PC, just thinking about binary vs decimal as physically represented on a piece of paper:
You know .1 in binary should be .5 in decimal, so the .1's place is worth .5 (1/2)
the .01 is worth .25 (1/4) (half of the previous one)
the .001 is worth (1/8) (Half of 1/4)
Notice how the denominator is progressing just like the whole numbers to the left of the decimal do--standard ^2 pattern? The next should be 1/16...
So you start with your .625, is it higher than .5? Yes, so set the first bit and subtract the .5
.1 binary with a decimal remainder of .125
Now you have the next spot, it's worth .25dec, is that less than your current remainder of .125? No, so you don't have enough decimal "Money" to buy that second spot, it has to be a 0
.10 binary, still .125 remainder.
Now go to the third position, etc. (Hint: I don't think there will be too much etc.)
There are several different ways to encode a non-integral number in binary. By far the most common type are floating point representations, especially the one codified in IEEE 754.
the code works for me is as below , you can use this code to convert any type of dobule values:
private static String doubleToBinaryString( double n ) {
String val = Integer.toBinaryString((int)n)+"."; // Setting up string for result
String newN ="0" + (""+n).substring((""+n).indexOf("."));
n = Double.parseDouble(newN);
while ( n > 0 ) { // While the fraction is greater than zero (not equal or less than zero)
double r = n * 2; // Multiply current fraction (n) by 2
if( r >= 1 ) { // If the ones-place digit >= 1
val += "1"; // Concat a "1" to the end of the result string (val)
n = r - 1; // Remove the 1 from the current fraction (n)
}else{ // If the ones-place digit == 0
val += "0"; // Concat a "0" to the end of the result string (val)
n = r; // Set the current fraction (n) to the new fraction
}
}
return val; // return the string result with all appended binary values
}
Edit: Since it appears nobody is reading the original question this links to, let me bring in a synopsis of it here.
The original problem, as asked by someone else, was that, given a large number of values, where the sum would exceed what a data type of Double would hold, how can one calculate the average of those values.
There was several answers that said to calculate in sets, like taking 50 and 50 numbers, and calculating the average inside those sets, and then finally take the average of all those sets and combine those to get the final average value.
My position was that unless you can guarantee that all those values can be split into a number of equally sized sets, you cannot use this approach. Someone dared me to ask the question here, in order to provide the answer, so here it is.
Basically, given an arbitrary number of values, where:
I know the number of values beforehand (but again, how would your answer change if you didn't?`)
I cannot gather up all the numbers, nor can I sum them (the sum will be too big for a normal data type in your programming language)
how can I calculate the average?
The rest of the question here outlines how, and the problems with, the approach to split into equally sized sets, but I'd really just like to know how you can do it.
Note that I know perfectly well enough math to know that in math theory terms, calculating the sum of A[1..N]/N will give me the average, let's assume that there are reasons that it isn't just as simple, and I need to split up the workload, and that the number of values isn't necessarily going to be divisable by 3, 7, 50, 1000 or whatever.
In other words, the solution I'm after will have to be general.
From this question:
What is a good solution for calculating an average where the sum of all values exceeds a double’s limits?
my position was that splitting the workload up into sets is no good, unless you can ensure that the size of those sets are equal.
Edit: The original question was about the upper limit that a particular data type could hold, and since he was summing up a lot of numbers (count that was given as example was 10^9), the data type could not hold the sum. Since this was a problem in the original solution, I'm assuming (and this is a prerequisite for my question, sorry for missing that) that the numbers are too big to give any meaningful answers.
So, dividing by the total number of values directly is out. The original reason for why a normal SUM/COUNT solution was out was that SUM would overflow, but let's assume, for this question that SET-SET/SET-SIZE will underflow, or whatever.
The important part is that I cannot simply sum, I cannot simply divide by the number of total values. If I cannot do that, will my approach work, or not, and what can I do to fix it?
Let me outline the problem.
Let's assume you're going to calculate the average of the numbers 1 through 6, but you cannot (for whatever reason) do so by summing the numbers, counting the numbers, and then dividing the sum by the count. In other words, you cannot simply do (1+2+3+4+5+6)/6.
In other words, SUM(1..6)/COUNT(1..6) is out. We're not considering NULL's (as in database NULL's) here.
Several of the answers to that question alluded to being able to split the numbers being averaged into sets, say 3 or 50 or 1000 numbers, then calculating some number for that, and then finally combining those values to get the final average.
My position is that this is not possible in the general case, since this will make some numbers, the ones appearing in the final set, more or less valuable than all the ones in the previous sets, unless you can split all the numbers into equally sized sets.
For instance, to calculate the average of 1-6, you can split it up into sets of 3 numbers like this:
/ 1 2 3 \ / 4 5 6 \
| - + - + - | + | - + - + - |
\ 3 3 3 / \ 3 3 3 / <-- 3 because 3 numbers in the set
---------- -----------
2 2 <-- 2 because 2 equally sized groups
Which gives you this:
2 5
- + - = 3.5
2 2
(note: (1+2+3+4+5+6)/6 = 3.5, so this is correct here)
However, my point is that once the number of values cannot be split into a number of equally sized sets, this method falls apart. For instance, what about the sequence 1-7, which contains a prime number of values.
Can a similar approach, that won't sum all the values, and count all the values, in one go, work?
So, is there such an approach? How do I calculate the average of an arbitrary number of values in which the following holds true:
I cannot do a normal sum/count approach, for whatever reason
I know the number of values beforehand (what if I don't, will that change the answer?)
Well, suppose you added three numbers and divided by three, and then added two numbers and divided by two. Can you get the average from these?
x = (a + b + c) / 3
y = (d + e) / 2
z = (f + g) / 2
And you want
r = (a + b + c + d + e + f + g) / 7
That is equal to
r = (3 * (a + b + c) / 3 + 2 * (d + e) / 2 + 2 * (f + g) / 2) / 7
r = (3 * x + 2 * y + 2 * z) / 7
Both lines above overflow, of course, but since division is distributive, we do
r = (3.0 / 7.0) * x + (2.0 / 7.0) * y + (2.0 / 7.0) * z
Which guarantees that you won't overflow, as I'm multiplying x, y and z by fractions less than one.
This is the fundamental point here. Neither I'm dividing all numbers beforehand by the total count, nor am I ever exceeding the overflow.
So... if you you keep adding to an accumulator, keep track of how many numbers you have added, and always test if the next number will cause an overflow, you can then get partial averages, and compute the final average.
And no, if you don't know the values beforehand, it doesn't change anything (provided that you can count them as you sum them).
Here is a Scala function that does it. It's not idiomatic Scala, so that it can be more easily understood:
def avg(input: List[Double]): Double = {
var partialAverages: List[(Double, Int)] = Nil
var inputLength = 0
var currentSum = 0.0
var currentCount = 0
var numbers = input
while (numbers.nonEmpty) {
val number = numbers.head
val rest = numbers.tail
if (number > 0 && currentSum > 0 && Double.MaxValue - currentSum < number) {
partialAverages = (currentSum / currentCount, currentCount) :: partialAverages
currentSum = 0
currentCount = 0
} else if (number < 0 && currentSum < 0 && Double.MinValue - currentSum > number) {
partialAverages = (currentSum / currentCount, currentCount) :: partialAverages
currentSum = 0
currentCount = 0
}
currentSum += number
currentCount += 1
inputLength += 1
numbers = rest
}
partialAverages = (currentSum / currentCount, currentCount) :: partialAverages
var result = 0.0
while (partialAverages.nonEmpty) {
val ((partialSum, partialCount) :: rest) = partialAverages
result += partialSum * (partialCount.toDouble / inputLength)
partialAverages = rest
}
result
}
EDIT:
Won't multiplying with 2, and 3, get me back into the range of "not supporter by the data type?"
No. If you were diving by 7 at the end, absolutely. But here you are dividing at each step of the sum. Even in your real case the weights (2/7 and 3/7) would be in the range of manageble numbers (e.g. 1/10 ~ 1/10000) which wouldn't make a big difference compared to your weight (i.e. 1).
PS: I wonder why I'm working on this answer instead of writing mine where I can earn my rep :-)
If you know the number of values beforehand (say it's N), you just add 1/N + 2/N + 3/N etc, supposing that you had values 1, 2, 3. You can split this into as many calculations as you like, and just add up your results. It may lead to a slight loss of precision, but this shouldn't be an issue unless you also need a super-accurate result.
If you don't know the number of items ahead of time, you might have to be more creative. But you can, again, do it progressively. Say the list is 1, 2, 3, 4. Start with mean = 1. Then mean = mean*(1/2) + 2*(1/2). Then mean = mean*(2/3) + 3*(1/3). Then mean = mean*(3/4) + 4*(1/4) etc. It's easy to generalize, and you just have to make sure the bracketed quantities are calculated in advance, to prevent overflow.
Of course, if you want extreme accuracy (say, more than 0.001% accuracy), you may need to be a bit more careful than this, but otherwise you should be fine.
Let X be your sample set. Partition it into two sets A and B in any way that you like. Define delta = m_B - m_A where m_S denotes the mean of a set S. Then
m_X = m_A + delta * |B| / |X|
where |S| denotes the cardinality of a set S. Now you can repeatedly apply this to partition and calculate the mean.
Why is this true? Let s = 1 / |A| and t = 1 / |B| and u = 1 / |X| (for convenience of notation) and let aSigma and bSigma denote the sum of the elements in A and B respectively so that:
m_A + delta * |B| / |X|
= s * aSigma + u * |B| * (t * bSigma - s * aSigma)
= s * aSigma + u * (bSigma - |B| * s * aSigma)
= s * aSigma + u * bSigma - u * |B| * s * aSigma
= s * aSigma * (1 - u * |B|) + u * bSigma
= s * aSigma * (u * |X| - u * |B|) + u * bSigma
= s * u * aSigma * (|X| - |B|) + u * bSigma
= s * u * aSigma * |A| + u * bSigma
= u * aSigma + u * bSigma
= u * (aSigma + bSigma)
= u * (xSigma)
= xSigma / |X|
= m_X
The proof is complete.
From here it is obvious how to use this to either recursively compute a mean (say by repeatedly splitting a set in half) or how to use this to parallelize the computation of the mean of a set.
The well-known on-line algorithm for calculating the mean is just a special case of this. This is the algorithm that if m is the mean of {x_1, x_2, ... , x_n} then the mean of {x_1, x_2, ..., x_n, x_(n+1)} is m + ((x_(n+1) - m)) / (n + 1). So with X = {x_1, x_2, ..., x_(n+1)}, A = {x_(n+1)}, and B = {x_1, x_2, ..., x_n} we recover the on-line algorithm.
Thinking outside the box: Use the median instead. It's much easier to calculate - there are tons of algorithms out there (e.g. using queues), you can often construct good arguments as to why it's more meaningful for data sets (less swayed by extreme values; etc) and you will have zero problems with numerical accuracy. It will be fast and efficient. Plus, for large data sets (which it sounds like you have), unless the distributions are truly weird, the values for the mean and median will be similar.
When you split the numbers into sets you're just dividing by the total number or am I missing something?
You have written it as
/ 1 2 3 \ / 4 5 6 \
| - + - + - | + | - + - + - |
\ 3 3 3 / \ 3 3 3 /
---------- -----------
2 2
but that's just
/ 1 2 3 \ / 4 5 6 \
| - + - + - | + | - + - + - |
\ 6 6 6 / \ 6 6 6 /
so for the numbers from 1 to 7 one possible grouping is just
/ 1 2 3 \ / 4 5 6 \ / 7 \
| - + - + - | + | - + - + - | + | - |
\ 7 7 7 / \ 7 7 7 / \ 7 /
Average of x_1 .. x_N
= (Sum(i=1,N,x_i)) / N
= (Sum(i=1,M,x_i) + Sum(i=M+1,N,x_i)) / N
= (Sum(i=1,M,x_i)) / N + (Sum(i=M+1,N,x_i)) / N
This can be repeatedly applied, and is true regardless of whether the summations are of equal size. So:
Keep adding terms until both:
adding another one will overflow (or otherwise lose precision)
dividing by N will not underflow
Divide the sum by N
Add the result to the average-so-far
There's one obvious awkward case, which is that there are some very small terms at the end of the sequence, such that you run out of values before you satisfy the condition "dividing by N will not underflow". In which case just discard those values - if their contribution to the average cannot be represented in your floating type, then it is in particular smaller than the precision of your average. So it doesn't make any difference to the result whether you include those terms or not.
There are also some less obvious awkward cases to do with loss of precision on individual summations. For example, what's the average of the values:
10^100, 1, -10^100
Mathematics says it's 1, but floating-point arithmetic says it depends what order you add up the terms, and in 4 of the 6 possibilities it's 0, because (10^100) + 1 = 10^100. But I think that the non-commutativity of floating-point arithmetic is a different and more general problem than this question. If sorting the input is out of the question, I think there are things you can do where you maintain lots of accumulators of different magnitudes, and add each new value to whichever one of them will give best precision. But I don't really know.
Here's another approach. You're 'receiving' numbers one-by-one from some source, but you can keep track of the mean at each step.
First, I will write out the formula for mean at step n+1:
mean[n+1] = mean[n] - (mean[n] - x[n+1]) / (n+1)
with the initial condition:
mean[0] = x[0]
(the index starts at zero).
The first equation can be simplified to:
mean[n+1] = n * mean[n] / (n+1) + x[n+1]/(n+1)
The idea is that you keep track of the mean, and when you 'receive' the next value in your sequence, you figure out its offset from the current mean, and divide it equally between the n+1 samples seen so far, and adjust your mean accordingly. If your numbers don't have a lot of variance, your running mean will need to be adjusted very slightly with the new numbers as n becomes large.
Obviously, this method works even if you don't know the total number of values when you start. It has an additional advantage that you know the value of the current mean at all times. One disadvantage that I can think of is the it probably gives more 'weight' to the numbers seen in the beginning (not in a strict mathematical sense, but because of floating point representations).
Finally, all such calculations are bound to run into floating-point 'errors' if one is not careful enough. See my answer to another question for some of the problems with floating point calculations and how to test for potential problems.
As a test, I generated N=100000 normally distributed random numbers with mean zero and variance 1. Then I calculated their mean by three methods.
sum(numbers) / N, call it m1,
my method above, call it m2,
sort the numbers, and then use my method above, call it m3.
Here's what I found: m1 − m2 ∼ −4.6×10−17, m1 − m3 ∼ −3×10−15, m2 − m3 ∼ −3×10−15. So, if your numbers are sorted, the error might not be small enough for you. (Note however that even the worst error is 10−15 parts in 1 for 100000 numbers, so it might be good enough anyway.)
Some of the mathematical solutions here are very good. Here's a simple technical solution.
Use a larger data type. This breaks down into two possibilities:
Use a high-precision floating point library. One who encounters a need to average a billion numbers probably has the resources to purchase, or the brain power to write, a 128-bit (or longer) floating point library.
I understand the drawbacks here. It would certainly be slower than using intrinsic types. You still might over/underflow if the number of values grows too high. Yada yada.
If your values are integers or can be easily scaled to integers, keep your sum in a list of integers. When you overflow, simply add another integer. This is essentially a simplified implementation of the first option. A simple (untested) example in C# follows
class BigMeanSet{
List<uint> list = new List<uint>();
public double GetAverage(IEnumerable<uint> values){
list.Clear();
list.Add(0);
uint count = 0;
foreach(uint value in values){
Add(0, value);
count++;
}
return DivideBy(count);
}
void Add(int listIndex, uint value){
if((list[listIndex] += value) < value){ // then overflow has ocurred
if(list.Count == listIndex + 1)
list.Add(0);
Add(listIndex + 1, 1);
}
}
double DivideBy(uint count){
const double shift = 4.0 * 1024 * 1024 * 1024;
double rtn = 0;
long remainder = 0;
for(int i = list.Count - 1; i >= 0; i--){
rtn *= shift;
remainder <<= 32;
rtn += Math.DivRem(remainder + list[i], count, out remainder);
}
rtn += remainder / (double)count;
return rtn;
}
}
Like I said, this is untested—I don't have a billion values I really want to average—so I've probably made a mistake or two, especially in the DivideBy function, but it should demonstrate the general idea.
This should provide as much accuracy as a double can represent and should work for any number of 32-bit elements, up to 232 - 1. If more elements are needed, then the count variable will need be expanded and the DivideBy function will increase in complexity, but I'll leave that as an exercise for the reader.
In terms of efficiency, it should be as fast or faster than any other technique here, as it only requires iterating through the list once, only performs one division operation (well, one set of them), and does most of its work with integers. I didn't optimize it, though, and I'm pretty certain it could be made slightly faster still if necessary. Ditching the recursive function call and list indexing would be a good start. Again, an exercise for the reader. The code is intended to be easy to understand.
If anybody more motivated than I am at the moment feels like verifying the correctness of the code, and fixing whatever problems there might be, please be my guest.
I've now tested this code, and made a couple of small corrections (a missing pair of parentheses in the List<uint> constructor call, and an incorrect divisor in the final division of the DivideBy function).
I tested it by first running it through 1000 sets of random length (ranging between 1 and 1000) filled with random integers (ranging between 0 and 232 - 1). These were sets for which I could easily and quickly verify accuracy by also running a canonical mean on them.
I then tested with 100* large series, with random length between 105 and 109. The lower and upper bounds of these series were also chosen at random, constrained so that the series would fit within the range of a 32-bit integer. For any series, the results are easily verifiable as (lowerbound + upperbound) / 2.
*Okay, that's a little white lie. I aborted the large-series test after about 20 or 30 successful runs. A series of length 109 takes just under a minute and a half to run on my machine, so half an hour or so of testing this routine was enough for my tastes.
For those interested, my test code is below:
static IEnumerable<uint> GetSeries(uint lowerbound, uint upperbound){
for(uint i = lowerbound; i <= upperbound; i++)
yield return i;
}
static void Test(){
Console.BufferHeight = 1200;
Random rnd = new Random();
for(int i = 0; i < 1000; i++){
uint[] numbers = new uint[rnd.Next(1, 1000)];
for(int j = 0; j < numbers.Length; j++)
numbers[j] = (uint)rnd.Next();
double sum = 0;
foreach(uint n in numbers)
sum += n;
double avg = sum / numbers.Length;
double ans = new BigMeanSet().GetAverage(numbers);
Console.WriteLine("{0}: {1} - {2} = {3}", numbers.Length, avg, ans, avg - ans);
if(avg != ans)
Debugger.Break();
}
for(int i = 0; i < 100; i++){
uint length = (uint)rnd.Next(100000, 1000000001);
uint lowerbound = (uint)rnd.Next(int.MaxValue - (int)length);
uint upperbound = lowerbound + length;
double avg = ((double)lowerbound + upperbound) / 2;
double ans = new BigMeanSet().GetAverage(GetSeries(lowerbound, upperbound));
Console.WriteLine("{0}: {1} - {2} = {3}", length, avg, ans, avg - ans);
if(avg != ans)
Debugger.Break();
}
}