If I create a HyperLogLog per day to count unique visitors, and then the 1st of January I merge the last 365 ones will I get the same value than if I keep a single HyperLogLog for the whole 365 days?
I guess not. But how different would those values be?
Short answer: Yes, both solutions have the same accuracy.
The basis of the HyperLogLog algorithm is the observation that the cardinality of a multiset of uniformly distributed random numbers can be estimated by calculating the maximum number of leading zeros in the binary representation of each number in the set. If the maximum number of leading zeros observed is n, an estimate for the number of distinct elements in the set is 2**n.
Referred from wiki
No matter it's a single hyperloglog or 365 hyperloglogs, the numbers of leading zeros are the same. So they have the same accuracy, and that's why hyperloglog can be merged.
Related
Let's have a set of sequences with given start, end and score (not lenght). What is the best way to find a subset of sequences which gives the highest score? For better understanding here is the example, where we have a set of 3 sequences:
The output of algorithm should be this two subsequences with score 115:
Assume there will be milions of subsenqences and the time for completition should be maximum of tens of minutes on computer with 8GB ram.
what conclusion can be drawn from the resulting t-stats value When ttest_ind is applied on two independent series?
As you can read here, the scipy.stats.ttest_ind has two outputs
The calculated t-statistic.
The two-tailed p-value.
Very intuitively, you can read the t-statistic as a normalized difference of averages in both populations, considering their variances and sizes:
The larger are the samples, the more serious the difference of averages is because we have more evidence for that.
The larger are the variances, the less serious the difference of averages is because the absolute difference can be given by randomness only.
The higher is the value of the t-statistic, the more serious is the difference.
The p-value makes this intuition more explicit: it is the probability that the difference of averages can be considered as zero. If the p-value is bellow a threshold, e.g. 0.05, we say that the difference in not zero.
I'm writing a tone generator program for a microcontroller.
I use an hardware timer to trigger an interrupt and check if I need to set the signal to high or low in a particular moment for a given note.
I'm using pretty limited hardware, so the slower I run the timer the more time I have to do other stuff (serial communication, loading the next notes to generate, etc.).
I need to find the frequency at which I should run the timer to have an optimal result, which is, generate a frequency that is accurate enough and still have time to compute the other stuff.
To achieve this, I need to find an approximate (within some percent value, as the higher are the frequencies the more they need to be imprecise in value for a human ear to notice the error) LCM of all the frequencies I need to play: this value will be the frequency at which to run the hardware timer.
Is there a simple enough algorithm to compute such number? (EDIT, I shall clarify "simple enough": fast enough to run in a time t << 1 sec. for less than 50 values on a 8 bit AVR microcontroller and implementable in a few dozens of lines at worst.)
LCM(a,b,c) = LCM(LCM(a,b),c)
Thus you can compute LCMs in a loop, bringing in frequencies one at a time.
Furthermore,
LCM(a,b) = a*b/GCD(a,b)
and GCDs are easily computed without any factoring by using the Euclidean algorithm.
To make this an algorithm for approximate LCMs, do something like round lower frequencies to multiples of 10 Hz and higher frequencies to multiples of 50 Hz. Another idea that is a bit more principled would be to first convert the frequency to an octave (I think that the formula is f maps to log(f/16)/log(2)) This will give you a number between 0 and 10 (or slightly higher --but anything above 10 is almost beyond human hearing so you could perhaps round down). You could break 0-10 into say 50 intervals 0.0, 0.2, 0.4, ... and for each number compute ahead of time the frequency corresponding to that octave (which would be f = 16*2^o where o is the octave). For each of these -- go through by hand once and for all and find a nearby round number that has a number of smallish prime factors. For example, if o = 5.4 then f = 675.58 -- round to 675; if o = 5.8 then f = 891.44 -- round to 890. Assemble these 50 numbers into a sorted array, using binary search to replace each of your frequencies by the closest frequency in the array.
An idea:
project the frequency range to a smaller interval
Let's say your frequency range is from 20 to 20000 and you aim for a 2% accurary, you'll calculate for a 1-50 range. It has to be a non-linear transformation to keep the accurary for lower frequencies. The goal is both to compute the result faster and to have a smaller LCM.
Use a prime factors table to easily compute the LCM on that reduced range
Store the pre-calculated prime factors powers in an array (size about 50x7 for range 1-50), and then use it for the LCM: the LCM of a number is the product of multiplying the highest power of each prime factor of the number together. It's easy to code and blazingly fast to run.
Do the first step in reverse to get the final number.
I'v got some problem to understand the difference between Logarithmic(Lcc) and Uniform(Ucc) cost criteria and also how to use it in calculations.
Could someone please explain the difference between the two and perhaps show how to calculate the complexity for a problem like A+B*C
(Yes this is part of an assignment =) )
Thx for any help!
/Marthin
Uniform Cost Criteria assigns a constant cost to every machine operation regardless of the number of bits involved WHILE Logarithm Cost Criteria assigns a cost to every machine operation proportional to the number of bits involved
Problem size influence complexity
Since complexity depends on the size of the
problem we define complexity to be a function
of problem size
Definition: Let T(n) denote the complexity for
an algorithm that is applied to a problem of
size n.
The size (n) of a problem instance (I) is the
number of (binary) bits used to represent the
instance. So problem size is the length of the
binary description of the instance.
This is called Logarithmic cost criteria
Unit Cost Criteria
If you assume that:
- every computer instruction takes one time
unit,
- every register is one storage unit
- and that a number always fits in a register
then you can use the number of inputs as
problem size since the length of input (in bits)
will be a constant times the number of inputs.
Uniform cost criteria assume that every instruction takes a single unit of time and that every register requires a single unit of space.
Logarithmic cost criteria assume that every instruction takes a logarithmic number of time units (with respect to the length of the operands) and that every register requires a logarithmic number of units of space.
In simpler terms, what this means is that uniform cost criteria count the number of operations, and logarithmic cost criteria count the number of bit operations.
For example, suppose we have an 8-bit adder.
If we're using uniform cost criteria to analyze the run-time of the adder, we would say that addition takes a single time unit; i.e., T(N)=1.
If we're using logarithmic cost criteria to analyze the run-time of the adder, we would say that addition takes lgn time units; i.e., T(N)=lgn, where n is the worst case number you would have to add in terms of time complexity (in this example, n would be 256). Thus, T(N)=8.
More specifically, say we're adding 256 to 32. To perform the addition, we have to add the binary bits together in the 1s column, the 2s column, the 4s column, etc (columns meaning the bit locations). The number 256 requires 8 bits. This is where logarithms come into our analysis. lg256=8. So to add the two numbers, we have to perform addition on 8 columns. Logarithmic cost criteria say that each of these 8 addition calculations takes a single unit of time. Uniform cost criteria say that the entire set of 8 addition calculations takes a single unit of time.
Similar analysis can be made in terms of space as well. Registers either take up a constant amount of space (under uniform cost criteria) or a logarithmic amount of space (under uniform cost criteria).
I think you should do some research on Big O notation... http://en.wikipedia.org/wiki/Big_O_notation#Orders_of_common_functions
If there is a part of the description you find difficult edit your question.
I searched the site but did not find exactly what I was looking for... I wanted to generate a discrete random number from normal distribution.
For example, if I have a range from a minimum of 4 and a maximum of 10 and an average of 7. What code or function call ( Objective C preferred ) would I need to return a number in that range. Naturally, due to normal distribution more numbers returned would center round the average of 7.
As a second example, can the bell curve/distribution be skewed toward one end of the other? Lets say I need to generate a random number with a range of minimum of 4 and maximum of 10, and I want the majority of the numbers returned to center around the number 8 with a natural fall of based on a skewed bell curve.
Any help is greatly appreciated....
Anthony
What do you need this for? Can you do it the craps player's way?
Generate two random integers in the range of 2 to 5 (inclusive, of course) and add them together. Or flip a coin (0,1) six times and add 4 to the result.
Summing multiple dice produces a normal distribution (a "bell curve"), while eliminating high or low throws can be used to skew the distribution in various ways.
The key is you are going for discrete numbers (and I hope you mean integers by that). Multiple dice throws famously generate a normal distribution. In fact, I think that's how we were first introduced to the Gaussian curve in school.
Of course the more throws, the more closely you approximate the bell curve. Rolling a single die gives a flat line. Rolling two dice just creates a ramp up and down that isn't terribly close to a bell. Six coin flips gets you closer.
So consider this...
If I understand your question correctly, you only have seven possible outcomes--the integers (4,5,6,7,8,9,10). You can set up an array of seven probabilities to approximate any distribution you like.
Many frameworks and libraries have this built-in.
Also, just like TokenMacGuy said a normal distribution isn't characterized by the interval it's defined on, but rather by two parameters: Mean μ and standard deviation σ. With both these parameters you can confine a certain quantile of the distribution to a certain interval, so that 95 % of all points fall in that interval. But resticting it completely to any interval other than (−∞, ∞) is impossible.
There are several methods to generate normal-distributed values from uniform random values (which is what most random or pseudorandom number generators are generating:
The Box-Muller transform is probably the easiest although not exactly fast to compute. Depending on the number of numbers you need, it should be sufficient, though and definitely very easy to write.
Another option is Marsaglia's Polar method which is usually faster1.
A third method is the Ziggurat algorithm which is considerably faster to compute but much more complex to program. In applications that really use a lot of random numbers it may be the best choice, though.
As a general advice, though: Don't write it yourself if you have access to a library that generates normal-distributed random numbers for you already.
For skewing your distribution I'd just use a regular normal distribution, choosing μ and σ appropriately for one side of your curve and then determine on which side of your wanted mean a point fell, stretching it appropriately to fit your desired distribution.
For generating only integers I'd suggest you just round towards the nearest integer when the random number happens to fall within your desired interval and reject it if it doesn't (drawing a new random number then). This way you won't artificially skew the distribution (such as you would if you were clamping the values at 4 or 10, respectively).
1 In testing with deliberately bad random number generators (yes, worse than RANDU) I've noticed that the polar method results in an endless loop, rejecting every sample. Won't happen with random numbers that fulfill the usual statistic expectations to them, though.
Yes, there are sophisticated mathematical solutions, but for "simple but practical" I'd go with Nosredna's comment. For a simple Java solution:
Random random=new Random();
public int bell7()
{
int n=4;
for (int x=0;x<6;++x)
n+=random.nextInt(2);
return n;
}
If you're not a Java person, Random.nextInt(n) returns a random integer between 0 and n-1. I think the rest should be similar to what you'd see in any programming language.
If the range was large, then instead of nextInt(2)'s I'd use a bigger number in there so there would be fewer iterations through the loop, depending on frequency of call and performance requirements.
Dan Dyer and Jay are exactly right. What you really want is a binomial distribution, not a normal distribution. The shape of a binomial distribution looks a lot like a normal distribution, but it is discrete and bounded whereas a normal distribution is continuous and unbounded.
Jay's code generates a binomial distribution with 6 trials and a 50% probability of success on each trial. If you want to "skew" your distribution, simply change the line that decides whether to add 1 to n so that the probability is something other than 50%.
The normal distribution is not described by its endpoints. Normally it's described by it's mean (which you have given to be 7) and its standard deviation. An important feature of this is that it is possible to get a value far outside the expected range from this distribution, although that will be vanishingly rare, the further you get from the mean.
The usual means for getting a value from a distribution is to generate a random value from a uniform distribution, which is quite easily done with, for example, rand(), and then use that as an argument to a cumulative distribution function, which maps probabilities to upper bounds. For the standard distribution, this function is
F(x) = 0.5 - 0.5*erf( (x-μ)/(σ * sqrt(2.0)))
where erf() is the error function which may be described by a taylor series:
erf(z) = 2.0/sqrt(2.0) * Σ∞n=0 ((-1)nz2n + 1)/(n!(2n + 1))
I'll leave it as an excercise to translate this into C.
If you prefer not to engage in the exercise, you might consider using the Gnu Scientific Library, which among many other features, has a technique to generate random numbers in one of many common distributions, of which the Gaussian Distribution (hint) is one.
Obviously, all of these functions return floating point values. You will have to use some rounding strategy to convert to a discrete value. A useful (but naive) approach is to simply downcast to integer.