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I am trying to figure out the time complexity of the below problem.
import math
def prime(n):
for i in range(2,n+1):
for j in range(2, int(math.sqrt(i))+1):
if i%j == 0:
break
else:
print(i)
prime(36)
This problem prints the prime numbers until 36.
My understanding of the above program:
for every n the inner loop runs for sqrt(n) times so on until n.
so the Big-o-Notation is O(n sqrt(n)).
Does my understanding is right? Please correct me if I am wrong...
Time complexity measures the increase in number or steps (basic operations) as the input scales up:
O(1) : constant (hash look-up)
O(log n) : logarithmic in base 2 (binary search)
O(n) : linear (search for an element in unsorted list)
O(n^2) : quadratic (bubble sort)
To determine the exact complexity of an algorithm requires a lot of math and algorithms knowledge. You can find a detailed description of them here: time complexity
Also keep in mind that these values are considered for very large values of n, so as a rule of thumb, whenever you see nested for loops, think O(n^2).
You can add a steps counter inside your inner for loop and record its value for different values of n, then print the relation in a graph. Then you can compare your graph with the graphs of n, log n, n * sqrt(n) and n^2 to determine exactly where your algorithm is placed.
What is the correct way to generate exact value from 0 to 999999 randomly since 1000000 is not a power of 2?
This is my approach:
use crypto.randomBytes to generate 3 bytes and convert to hex
use the first 5 characters to convert to integer (max is fffff == 1048575 > 999999)
if the result > 999999, start from step 1 again
It will somehow create a recursive function. Is it logically correct and will it cause a concern of performance?
There are several way to extract random numbers in a range from random bits. Some common ones are described in NIST Special Publication 800-90A revision 1: Recommendation for Random Number Generation Using Deterministic Random Bit Generators
Although this standard is about deterministic random bit generations there is a helpful appendix called A.5 Converting Random Bits into a Random Number which describes three useful methods.
The methods described are:
A.5.1 The Simple Discard Method
A.5.2 The Complex Discard Method
A.5.3 The Simple Modular Method
The first two of them are not deterministic with regards to running time but generate a number with no bias at all. They are based on rejection sampling.
The complex discard method discusses a more optimal scheme for generating large quantities of random numbers in a range. I think it is too complex for almost any normal use; I would look at the Optimized Simple Discard method described below if you require additional efficiency instead.
The Simple Modular Method is time constant and deterministic but has non-zero (but negligible) bias. It requires a relatively large amount of additional randomness to achieve the negligible bias though; basically to have a bias of one out of 2^128 you need 128 bits on top of the bit size of the range required. This is probably not the method to choose for smaller numbers.
Your algorithm is clearly a version of the Simple Discard Method (more generally called "rejection sampling"), so it is fine.
I've myself thought of a very efficient algorithm based on the Simple Discard Method called the "Optimized Simple Discard Method" or RNG-BC where "BC" stands for "binary compare". It is based on the observation that comparison only looks at the most significant bits, which means that the least significant bits should still be considered random and can therefore be reused. Beware that this method has not been officially peer reviewed; I do present an informal proof of equivalence with the Simple Discard Method.
Of course you should rather use a generic method that is efficient given any value of N. In that case the Complex Discard Method or Simple Modular Method should be considered over the Simple Discard Method. There are other, much more complex algorithms that are even more efficient, but generally you're fine when using either of these two.
Note that it is often beneficial to first check if N is a power of two when generating a random in the range [0, N). If N is a power of two then there is no need to use any of these possibly expensive computations; just use the bits you need from the random bit or byte generator.
It's a correct algorithm (https://en.wikipedia.org/wiki/Rejection_sampling), though you could consider using bitwise operations instead of converting to hex. It can run forever if the random number generator is malfunctioning -- you could consider trying a fixed number of times and then throwing an exception instead of looping forever.
The main possible performance problem is that on some platforms, crypto.randomBytes can block if it runs out of entropy. So you don't want to waste any randomness if you're using it.
Therefore instead of your string comparison I would use the following integer operation.
if (random_bytes < 16700000) {
return random_bytes = random_bytes - 100000 * Math.floor(random_bytes/100000);
}
This has about a 99.54% chance of producing an answer from the first 3 bytes, as opposed to around 76% odds with your approach.
I would suggest the following approach:
private generateCode(): string {
let code: string = "";
do {
code += randomBytes(3).readUIntBE(0, 3);
// code += Number.parseInt(randomBytes(3).toString("hex"), 16);
} while (code.length < 6);
return code.slice(0, 6);
}
This returns the numeric code as string, but if it is necessary to get it as a number, then change to return Number.parseInt(code.slice(0, 6))
I call it the random_6d algo. Worst case just a single additional loop.
var random_6d = function(n2){
var n1 = crypto.randomBytes(3).readUIntLE(0, 3) >>> 4;
if(n1 < 1000000)
return n1;
if(typeof n2 === 'undefined')
return random_6d(n1);
return Math.abs(n1 - n2);
};
loop version:
var random_6d = function(){
var n1, n2;
while(true){
n1 = crypto.randomBytes(3).readUIntLE(0, 3) >>> 4;
if(n1 < 1000000)
return n1;
if(typeof n2 === 'undefined')
n2 = n1;
else
return Math.abs(n1 - n2);
};
};
I am trying to evaluate points in a large piecewise polynomial, which is obtained from a cubic-spline. This takes a long time to do and I would like to speed it up.
As such, I would like to evaluate a points on a piecewise polynomial with parallel processes, rather than sequentially.
Code:
z = zeros(1e6, 1) ; % preallocate some memory for speed
Y = rand(11220,161) ; %some data, rand for generating a working example
X = 0 : 0.0125 : 2 ; % vector of data sites
pp = spline(X, Y) ; % get the piecewise polynomial form of the cubic spline.
The resulting structure is large.
for t = 1 : 1e6 % big number
hcurrent = ppval(pp,t); %evaluate the piecewise polynomial at t
z(t) = sum(x(t:t+M-1).*hcurrent,1) ; % do some operation of the interpolated value. Most likely not relevant to this question.
end
Unfortunately, with matrix form and using:
hcurrent = flipud(ppval(pp, 1: 1e6 ))
requires too much memory to process, so cannot be done. Is there a way that I can batch process this code to speed it up?
For scalar second arguments, as in your example, you're dealing with two issues. First, there's a good amount of function call overhead and redundant computation (e.g., unmkpp(pp) is called every loop iteration). Second, ppval is written to be general so it's not fully vectorized and does a lot of things that aren't necessary in your case.
Below is vectorized code code that take advantage of some of the structure of your problem (e.g., t is an integer greater than 0), avoids function call overhead, move some calculations outside of your main for loop (at the cost of a bit of extra memory), and gets rid of a for loop inside of ppval:
n = 1e6;
z = zeros(n,1);
X = 0:0.0125:2;
Y = rand(11220,numel(X));
pp = spline(X,Y);
[b,c,l,k,dd] = unmkpp(pp);
T = 1:n;
idx = discretize(T,[-Inf b(2:l) Inf]); % Or: [~,idx] = histc(T,[-Inf b(2:l) Inf]);
x = bsxfun(#power,T-b(idx),(k-1:-1:0).').';
idx = dd*idx;
d = 1-dd:0;
for t = T
hcurrent = sum(bsxfun(#times,c(idx(t)+d,:),x(t,:)),2);
z(t) = ...;
end
The resultant code takes ~34% of the time of your example for n=1e6. Note that because of the vectorization, calculations are performed in a different order. This will result in slight differences between outputs from ppval and my optimized version due to the nature of floating point math. Any differences should be on the order of a few times eps(hcurrent). You can still try using parfor to further speed up the calculation (with four already running workers, my system took just 12% of your code's original time).
I consider the above a proof of concept. I may have over-optmized the code above if your example doesn't correspond well to your actual code and data. In that case, I suggest creating your own optimized version. You can start by looking at the code for ppval by typing edit ppval in your Command Window. You may be able to implement further optimizations by looking at the structure of your problem and what you ultimately want in your z vector.
Internally, ppval still uses histc, which has been deprecated. My code above uses discretize to perform the same task, as suggested by the documentation.
Use parfor command for parallel loops. see here, also precompute z vector as z(j) = x(j:j+M-1) and hcurrent in parfor for speed up.
The Spline Parameters estimation can be written in Matrix form.
Once you write it in Matrix form and solve it you can use the Model Matrix to evaluate the Spline on all data point using Matrix Multiplication which is probably the most tuned operation in MATLAB.
I want to plot the frequency spectrum of a music file (like they do for example in Audacity). Hence I want the frequency in Hertz on the x-axis and the amplitude (or desibel) on the y-axis.
I devide the song (about 20 million samples) into blocks of 4096 samples at a time. These blocks will result in 2049 (N/2 + 1) complex numbers (sine and cosine -> real and imaginary part). So now I have these thousands of individual 2049-arrays, how do I combine them?
Lets say I do the FFT 5000 times resulting in 5000 2049-arrays of complex numbers. Do I plus all the values of the 5000 arrays and then take the magnitude of the combined 2049-array? Do I then sacle the x-axis with the songs sample rate / 2 (eg: 22050 for a 44100hz file)?
Any information will be appriciated
What application are you using for this? I assume you are not doing this by hand, so here is a Matlab example:
>> fbins = fs/N * (0:(N/2 - 1)); % Where N is the number of fft samples
now you can perform
>> plot(fbins, abs(fftOfSignal(1:N/2)))
Stolen
edit: check this out http://www.codeproject.com/Articles/9388/How-to-implement-the-FFT-algorithm
Wow I've written a load about this just recently.
I even turned it into a blog post available here.
My explanation is leaning towards spectrograms but its just as easy to render a chart like you describe!
I might not be correct on this one, but as far as I'm aware, you have 2 ways to get the spectrum of the whole song.
1) Do a single FFT on the whole song, which will give you an extremely good frequency resolution, but is in practice not efficient, and you don't need this kind of resolution anyway.
2) Divide it into small chunks (like 4096 samples blocks, as you said), get the FFT for each of those and average the spectra. You will compromise on the frequency resolution, but make the calculation more manageable (and also decrease the variance of the spectrum). Wilhelmsen link's describes how to compute an FFT in C++, and I think some library already exists to do that, like FFTW (but I never managed to compile it, to be fair =) ).
To obtain the magnitude spectrum, average the energy (square of the magnitude) accross all you chunks for every single bins. To get the result in dB, just 10 * log10 the results. That is of course assuming that you are not interested in the phase spectrum. I think this is known as the Barlett's method.
I would do something like this:
// At this point you have the FFT chunks
float sum[N/2+1];
// For each bin
for (int binIndex = 0; binIndex < N/2 + 1; binIndex++)
{
for (int chunkIndex = 0; chunkIndex < chunkNb; chunkIndex++)
{
// Get the magnitude of the complex number
float magnitude = FFTChunk[chunkIndex].bins[binIndex].real * FFTChunk[chunkIndex].bins[binIndex].real
+ FFTChunk[chunkIndex].bins[binIndex].im * FFTChunk[chunkIndex].bins[binIndex].im;
magnitude = sqrt(magnitude);
// Add the energy
sum[binIndex] += magnitude * magnitude;
}
// Average the energy;
sum[binIndex] /= chunkNb;
}
// Then get the values in decibel
for (int binIndex = 0; binIndex < N/2 + 1; binIndex++)
{
sum[binIndex] = 10 * log10f(sum[binIndex]);
}
Hope this answers your question.
Edit: Goz's post will give you plenty of information on the matter =)
Commonly, you would take just one of the arrays, corresponding to the point in time of the music in which you are interested. The you would calculate the log of the magnitude of each complex array element. Plot the N/2 results as Y values, and scale the X axis from 0 to Fs/2 (where Fs is the sampling rate).
I have a corpus of 900,000 strings. They vary in length, but have an average character count of about 4,500. I need to find the most efficient way of computing the Dice coefficient of every string as it relates to every other string. Unfortunately, this results in the Dice coefficient algorithm being used some 810,000,000,000 times.
What is the best way to structure this program for increased efficiency? Obviously, I can prevent computing the Dice of sections A and B, and then B and A--but this only halves the work required. Should I consider taking some shortcuts or creating some sort of binary tree?
I'm using the following implementation of the Dice coefficient algorithm in Java:
public static double diceCoefficient(String s1, String s2) {
Set<String> nx = new HashSet<String>();
Set<String> ny = new HashSet<String>();
for (int i = 0; i < s1.length() - 1; i++) {
char x1 = s1.charAt(i);
char x2 = s1.charAt(i + 1);
String tmp = "" + x1 + x2;
nx.add(tmp);
}
for (int j = 0; j < s2.length() - 1; j++) {
char y1 = s2.charAt(j);
char y2 = s2.charAt(j + 1);
String tmp = "" + y1 + y2;
ny.add(tmp);
}
Set<String> intersection = new HashSet<String>(nx);
intersection.retainAll(ny);
double totcombigrams = intersection.size();
return (2 * totcombigrams) / (nx.size() + ny.size());
}
My ultimate goal is to output an ID for every section that has a Dice coefficient of greater than 0.9 with another section.
Thanks for any advice that you can provide!
Make a single pass over all the Strings, and build up a HashMap which maps each bigram to a set of the indexes of the Strings which contain that bigram. (Currently you are building the bigram set 900,000 times, redundantly, for each String.)
Then make a pass over all the sets, and build a HashMap of [index,index] pairs to common-bigram counts. (The latter Map should not contain redundant pairs of keys, like [1,2] and [2,1] -- just store one or the other.)
Both of these steps can easily be parallelized. If you need some sample code, please let me know.
NOTE one thing, though: from the 26 letters of the English alphabet, a total of 26x26 = 676 bigrams can be formed. Many of these will never or almost never be found, because they don't conform to the rules of English spelling. Since you are building up sets of bigrams for each String, and the Strings are so long, you will probably find almost the same bigrams in each String. If you were to build up lists of bigrams for each String (in other words, if the frequency of each bigram counted), it's more likely that you would actually be able to measure the degree of similarity between Strings, but then the calculation of Dice's coefficient as given in the Wikipedia article wouldn't work; you'd have to find a new formula.
I suggest you continue researching algorithms for determining similarity between Strings, try implementing a few of them, and run them on a smaller set of Strings to see how well they work.
You should come up with some kind of inequality like: D(X1,X2) > 1-p, D(X1,X3) < 1-q and p D(X2,X3) < 1-q+p . Or something like that. Now, if 1-q+p < 0.9, then probably you don't have to evaluate D(X2,X3).
PS: I am not sure about this exact inequality, but I have a gut feeling that this might be right (but I do not have enough time to actually do the derivations now). Look for some of the inequalities with other similarity measures and see if any of them are valid for Dice co-efficient.
=== Also ===
If there are a elements in set A, and if your threshold is r (=0.9), then set B should have number of elements b should be such that: r*a/(2-r) <= b <= (2-r)*a/r . This should eliminate need for lots of comparisons IMHO. You can probably sort the strings according to length and use the window describe above to limit comparisons.
Disclaimer first: This will not reduce the number of comparisons you'll have to make. But this should make a Dice comparison faster.
1) Don't build your HashSets every time you do a diceCoefficient() call! It should speed things up considerably if you just do it once for each string and keep the result around.
2) Since you only care about if a particular bigram is present in the string, you could get away with a BitSet with a bit for each possible bigram, rather than a full HashMap. Coefficient calculation would then be simplified to ANDing two bit sets and counting the number of set bits in the result.
3) Or, if you have a huge number of possible bigrams (Unicode, perhaps?) - or monotonous strings with only a handful of bigrams each - a sorted Array of bigrams might provide faster, more space-efficent comparisons.
Is their charset limited somehow? If it is, you can compute character counts by their code in each string and compare these numbers. After such pre-computation (it will occupy 2*900K*S bytes of memory [if we assume no character is found more then 65K time in the same string], where S is different character count). Then computing the coefficent would take O(S) time. Sure, this would be helpful if S<4500.