I have a question to find a complexity class estimate of a algorithm. The question gives recorded times for an algorithm. So, do I just average out the times based on how it was computed?
Sorry, I missed a part.
ok, so it recorded time like N = 100, Algorithm = 300, next N = 200, Algorithm = 604, next N = 400 Algorithm = 1196, next N = 800 Algorithm 2395. So, do i calculate like 300/100, and 604/200 and find the average. Is that how I'm supposed to estimate the complexity class for the algorithm?
Try plotting running time vs. N and see if you get any insight. (e.g. if running time = f(N), is f(N) about equal to log(N), or sqrt(N), or... ?)
I don't think time will help figure out it's complexity class. Times can be very different even on exactly the same task (depends on the scheduler or other factors.)
Look at how many more steps it takes as your input get's larger. So if you had a sorting algorithm that took 100 steps to sort 10 items and 10000 steps to sort 100 items you'd say sorted in big O ( N^2 ) since
Input Steps
10 100 (which equals 10*10)
100 10000 (which equals 100*100)
It's not about averaging but looking for a function that maps the input to the number of steps and then finding what part of that function grows the fastest ( N^2 grows faster than N so if your function was N^2 + N you classify it as N^2).
At least that's how I remember it, but it's been ages!! :)
EDIT: Now that there are more details in your question, here is what I'd do, with the above in mind.
Don't think about averaging anything, just think about how f(100) = 300, f(200)=604, and f(400)=1196.
And it doesn't have to be exact, just in the ball park. So a simple linear function (such as f(x)=3*N ) where f(100)=300, f(200)=600, and f(400)=1200 that would describe the complexity of the algorithm you could say the complexity class of the algorithm was linear or big O(N).
Hope that helps!
Does it give you the inputs to the algorithm as well, which produce the recorded times? You can deduce the growth order according to the input size vs output running time.
i.e. input = 1, running time = 10
input = 100, running time = 100000
would appear to be O(N^2)
i.e.
with input = 1 and running time = 10, likely O(cn) where C = 10
with n = 1, N^2 and N are the same
with input = 10 and running time = 100000, likely O(cN^2) where C = 10
and N = 100*100 = 10000, * 10 = 100000
Hint: Calculate how much time the algorithm spent to process one single item.
How does these calculates time relate to each other?
Does the algorithm alway spent the same time to process one item, regardless how many items, is there a factor? maybe the time raises exponentially?
Related
I am seeing an increase in total within Sum of squares when I am using below code.Is this even possible or I am doing some mistake in code?
v<-foreach(i = 1:30,.combine = c) %dopar% {
iter <- kmeans (clustering_data,centers = i,iter.max = 1000)
iter$tot.withinss
}
K-means is a randomized algorithm. It does not guarantee to find the optimum.
So you simply had a bad random.
Yes. See Anony-Mousse's answer.
If you used the nstart = 25 argument of the kmeans() function, you would run the algorithm 25 times, let R collect the error measures from each run, and build averages internally. This way you do not need to construct a foreach-loop.
from the documentation of R's kmeans()
## random starts do help here with too many clusters
## (and are often recommended anyway!):
(cl <- kmeans(x, 5, nstart = 25))
You have to choose a reasonable value for nstart. Then, errors by different random initializations are more likely to have averaged out. (But there is no guarantee that tot.withinss is minimal after nstart runs. )
I have this piece of code
N=10^4;
for i = 1:N
[E,X,T] = fffun(); % Stochastic simulation. Returns every time three different vectors (whose length is 10^3).
X_(i,:)=X;
T_(i,:)=T;
GRID=[GRID T];
end
GRID=unique(GRID);
% Second part
for i=1:N
for j=1:(kmax)
f=find(GRID==T_(i,j) | GRID==T_(i,j+1));
s=f(1);
e=f(2)-1;
counter(X_(i,j), s:e)=counter(X_(i,j), s:e)+1;
end
end
The code performs N different simulations of a stochastic process (which consists of 10^3 events, occurring at discrete moments (T vector) that depends on the specific simulation.
Now (second part) I want to know, as a function of time istant, how many simulations are in a particular state (X assumes value between 1 and 10). The idea I had: create a grid vector with all the moments at which something happens in any simulation. Then, looping over the simulations, loop over the timesteps in which something happens and incrementing all the counter indeces that corresponds to this particular slice of time.
However this second part is very heavy (I mean days of processing on a standard quad-core CPU). And it shouldn't.
Are there any ideas (maybe about comparing vectors in a more efficient way) to cut the CPU time?
This is a standalone 'second_part'
N=5000;
counter=zeros(11,length(GRID));
for i=1:N
disp(['Counting sim #' num2str(i)]);
for j=1:(kmax)
f=find(GRID==T_(i,j) | GRID==T_(i,j+1),2);
s=f(1);
e=f(2)-1;
counter(X_(i,j), s:e)=counter(X_(i,j), s:e)+1;
end
end
counter=counter/N;
stop=find(GRID==Tmin);
stop=stop-1;
plot(counter(:,(stop-500):stop)')
with associated dummy data ( filedropper.com/data_38 ). In the real context the matrix has 2x rows and 10x columns.
Here is what I understand:
T_ is a matrix of time steps from N simulations.
X_ is a matrix of simulation state at T_ in those simulations.
so if you do:
[ut,~,ic]= unique(T_(:));
you get ic which is a vector of indices for all unique elements in T_. Then you can write:
counter = accumarray([ic X_(:)],1);
and get counter with no. of rows as your unique timesteps, and no. of columns as the unique states in X_ (which are all, and must be, integers). Now you can say that for each timestep ut(k) the number of time that the simulation was in state m is counter(k,m).
In your data, the only combination of m and k that has a value greater than 1 is (1,1).
Edit:
From the comments below, I understand that you record all state changes, and the time steps when they occur. Then every time a simulation change a state you want to collect all the states from all simulations and count how many states are from each type.
The main problem here is that your time is continuous, so basically each element in T_ is unique, and you have over a million time steps to loop over. Fully vectorizing such a process will need about 80GB of memory which will probably stuck your computer.
So I looked for a combination of vectorizing and looping through the time steps. We start by finding all unique intervals, and preallocating counter:
ut = unique(T_(:));
stt = 11; % no. of states
counter = zeros(stt,numel(ut));r = 1:size(T_,1);
r = 1:size(T_,1); % we will need that also later
Then we loop over all element in ut, and each time look for the relevant timestep in T_ in all simulations in a vectorized way. And finally we use histcounts to count all the states:
for k = 1:numel(ut)
temp = T_<=ut(k); % mark all time steps before ut(k)
s = cumsum(temp,2); % count the columns
col_ind = s(:,end); % fins the column index for each simulation
% convert the coulmns to linear indices:
linind = sub2ind(size(T_),r,col_ind.');
% count the states:
counter(:,k) = histcounts(X_(linind),1:stt+1);
end
This takes about 4 seconds at my computer for 1000 simulations, so it adds to a little more than one hour for the whole process. Not very quick...
You can try also one or two of the tweaks below to squeeze run time a little bit more:
As you can read here, accumarray seems to work faster in small arrays then histcouns. So may want to switch to it.
Also, computing linear indices directly is a quicker method than sub2ind, so you may want to try that.
implementing these suggestions in the loop above, we get:
R = size(T_,1);
r = (1:R).';
for k = 1:K
temp = T_<=ut(k); % mark all time steps before ut(k)
s = cumsum(temp,2); % count the columns
col_ind = s(:,end); % fins the column index for each simulation
% convert the coulmns to linear indices:
linind = R*(col_ind-1)+r;
% count the states:
counter(:,k) = accumarray(X_(linind),1,[stt 1]);
end
In my computer switching to accumarray and or removing sub2ind gain a slight improvement but it was not consistent (using timeit for testing on 100 or 1K elements in ut), so you better test it yourself. However, this still remains very long.
One thing that you may want to consider is trying to discretize your timesteps, so you will have much less unique elements to loop over. In your data about 8% of the time intervals a smaller than 1. If you can assume that this is short enough to be treated as one time step, then you could round your T_ and get only ~12.5K unique elements, which take about a minute to loop over. You can do the same for 0.1 intervals (which are less than 1% of the time intervals), and get 122K elements to loop over, what will take about 8 hours...
Of course, all the timing above are rough estimates using the same algorithm. If you do choose to round the times there may be even better ways to solve this.
I have an EMG signal of a subject walking on a treadmill.
We used footswitches to be able to see when the subject is placing his foot, so we can see how many periods (steps) there are in time.
We would like to cut the signal in periods (steps) and get one average signal (100% step cycle).
I tried the reshape function but it does not work
when I count 38 steps:
nwaves = 38;
sig2 = reshape(sig,[numel(sig)/nwaves nwaves])';
avgSig = mean(sig2,1);
plot(avgSig);
the error displayed is this: Size arguments must be real integers.
Can anyone help me with this? Thanks!
First of all, reshaping the array is a bad approach to the problem. In real world one cannot assume that the person on the treadmill will step rhythmically with millisecond-precision (i.e. for the same amount of samples).
A more realistic approach is to use the footswitch signal: assume is really a switch on a single foot (1=foot on, 0=foot off), and its actions are filtered to avoid noise (Schmidt trigger, for example), you can get the samples index when the foot is removed from the treadmill with:
foot_off = find(diff(footswitch) < 0);
then you can transform your signal in a cell array (variable lengths) of vectors of data between consecutive steps:
step_len = diff([0, foot_off, numel(footswitch)]);
sig2 = mat2cell(sig(:), step_len, 1);
The problem now is you can't apply mean() to the signal slices in order to get an "average step": you must process each step first, then average the results.
It's probably because numel(sig)/nwaves isn't an integer. You need to round it to the nearest integer with round(numel(sig)/nwaves).
EDIT based on comments:
Your problem is you can't divide 51116 by 38 (it's 1345.2), so you can't reshape your signal in chunks of 38 long. You need a signal whose length is exactly a multiple of 38 if you want to be able to reshape it in chunks of 38. Either that, or remove the last (or first) 6 values from your signal to have an exact multiple of 38 (1345 * 38 = 51110):
nwaves = 38;
n_chunks = round(numel(sig)/nwaves);
max_sig_length = n_chunks * nwaves;
sig2 = reshape(sig(1:max_sig_length),[n_chunks nwaves])';
avgSig = mean(sig2,1);
plot(avgSig);
Having read carefully the previous question
Random numbers that add to 100: Matlab
I am struggling to solve a similar but slightly more complex problem.
I would like to create an array of n elements that sums to 1, however I want an added constraint that the minimum increment (or if you like number of significant figures) for each element is fixed.
For example if I want 10 numbers that sum to 1 without any constraint the following works perfectly:
num_stocks=10;
num_simulations=100000;
temp = [zeros(num_simulations,1),sort(rand(num_simulations,num_stocks-1),2),ones(num_simulations,1)];
weights = diff(temp,[],2);
I foolishly thought that by scaling this I could add the constraint as follows
num_stocks=10;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp2 = [zeros(num_simulations,1),sort(round(rand(num_simulations,num_stocks-1)*scaling)/scaling,2),ones(num_simulations,1)];
weights2 = diff(temp2,[],2);
However though this works for small values of n & small values of increment, if for example n=1,000 & the increment is 0.1% then over a large number of trials the first and last numbers have a mean which is consistently below 0.1%.
I am sure there is a logical explanation/solution to this but I have been tearing my hair out to try & find it & wondered anybody would be so kind as to point me in the right direction. To put the problem into context create random stock portfolios (hence the sum to 1).
Thanks in advance
Thank you for the responses so far, just to clarify (as I think my initial question was perhaps badly phrased), it is the weights that have a fixed increment of 0.1% so 0%, 0.1%, 0.2% etc.
I did try using integers initially
num_stocks=1000;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp = [zeros(num_simulations,1),sort(randi([0 scaling],num_simulations,num_stocks-1),2),ones(num_simulations,1)*scaling];
weights = (diff(temp,[],2)/scaling);
test=mean(weights);
but this was worse, the mean for the 1st & last weights is well below 0.1%.....
Edit to reflect excellent answer by Floris & clarify
The original code I was using to solve this problem (before finding this forum) was
function x = monkey_weights_original(simulations,stocks)
stockmatrix=1:stocks;
base_weight=1/stocks;
r=randi(stocks,stocks,simulations);
x=histc(r,stockmatrix)*base_weight;
end
This runs very fast, which was important considering I want to run a total of 10,000,000 simulations, 10,000 simulations on 1,000 stocks takes just over 2 seconds with a single core & I am running the whole code on an 8 core machine using the parallel toolbox.
It also gives exactly the distribution I was looking for in terms of means, and I think that it is just as likely to get a portfolio that is 100% in 1 stock as it is to geta portfolio that is 0.1% in every stock (though I'm happy to be corrected).
My issue issue is that although it works for 1,000 stocks & an increment of 0.1% and I guess it works for 100 stocks & an increment of 1%, as the number of stocks decreases then each pick becomes a very large percentage (in the extreme with 2 stocks you will always get a 50/50 portfolio).
In effect I think this solution is like the binomial solution Floris suggests (but more limited)
However my question has arrisen because I would like to make my approach more flexible & have the possibility of say 3 stocks & an increment of 1% which my current code will not handle correctly, hence how I stumbled accross the original question on stackoverflow
Floris's recursive approach will get to the right answer, but the speed will be a major issue considering the scale of the problem.
An example of the original research is here
http://www.huffingtonpost.com/2013/04/05/monkeys-stocks-study_n_3021285.html
I am currently working on extending it with more flexibility on portfolio weights & numbers of stock in the index, but it appears my programming & probability theory ability are a limiting factor.......
One problem I can see is that your formula allows for numbers to be zero - when the rounding operation results in two consecutive numbers to be the same after sorting. Not sure if you consider that a problem - but I suggest you think about it (it would mean your model portfolio has fewer than N stocks in it since the contribution of one of the stocks would be zero).
The other thing to note is that the probability of getting the extreme values in your distribution is half of what you want them to be: If you have uniformly distributed numbers from 0 to 1000, and you round them, the numbers that round to 0 were in the interval [0 0.5>; the ones that round to 1 came from [0.5 1.5> - twice as big. The last number (rounding to 1000) is again from a smaller interval: [999.5 1000]. Thus you will not get the first and last number as often as you think. If instead of round you use floor I think you will get the answer you expect.
EDIT
I thought about this some more, and came up with a slow but (I think) accurate method for doing this. The basic idea is this:
Think in terms of integers; rather than dividing the interval 0 - 1 in steps of 0.001, divide the interval 0 - 1000 in integer steps
If we try to divide N into m intervals, the mean size of a step should be N / m; but being integer, we would expect the intervals to be binomially distributed
This suggests an algorithm in which we choose the first interval as a binomially distributed variate with mean (N/m) - call the first value v1; then divide the remaining interval N - v1 into m-1 steps; we can do so recursively.
The following code implements this:
% random integers adding up to a definite sum
function r = randomInt(n, limit)
% returns an array of n random integers
% whose sum is limit
% calls itself recursively; slow but accurate
if n>1
v = binomialRandom(limit, 1 / n);
r = [v randomInt(n-1, limit - v)];
else
r = limit;
end
function b = binomialRandom(N, p)
b = sum(rand(1,N)<p); % slow but direct
To get 10000 instances, you run this as follows:
tic
portfolio = zeros(10000, 10);
for ii = 1:10000
portfolio(ii,:) = randomInt(10, 1000);
end
toc
This ran in 3.8 seconds on a modest machine (single thread) - of course the method for obtaining a binomially distributed random variate is the thing slowing it down; there are statistical toolboxes with more efficient functions but I don't have one. If you increase the granularity (for example, by setting limit=10000) it will slow down more since you increase the number of random number samples that are generated; with limit = 10000 the above loop took 13.3 seconds to complete.
As a test, I found mean(portfolio)' and std(portfolio)' as follows (with limit=1000):
100.20 9.446
99.90 9.547
100.09 9.456
100.00 9.548
100.01 9.356
100.00 9.484
99.69 9.639
100.06 9.493
99.94 9.599
100.11 9.453
This looks like a pretty convincing "flat" distribution to me. We would expect the numbers to be binomially distributed with a mean of 100, and standard deviation of sqrt(p*(1-p)*n). In this case, p=0.1 so we expect s = 9.4868. The values I actually got were again quite close.
I realize that this is inefficient for large values of limit, and I made no attempt at efficiency. I find that clarity trumps speed when you develop something new. But for instance you could pre-compute the cumulative binomial distributions for p=1./(1:10), then do a random lookup; but if you are just going to do this once, for 100,000 instances, it will run in under a minute; unless you intend to do it many times, I wouldn't bother. But if anyone wants to improve this code I'd be happy to hear from them.
Eventually I have solved this problem!
I found a paper by 2 academics at John Hopkins University "Sampling Uniformly From The Unit Simplex"
http://www.cs.cmu.edu/~nasmith/papers/smith+tromble.tr04.pdf
In the paper they outline how naive algorthms don't work, in a way very similar to woodchips answer to the Random numbers that add to 100 question. They then go on to show that the method suggested by David Schwartz can also be slightly biased and propose a modified algorithm which appear to work.
If you want x numbers that sum to y
Sample uniformly x-1 random numbers from the range 1 to x+y-1 without replacement
Sort them
Add a zero at the beginning & x+y at the end
difference them & subtract 1 from each value
If you want to scale them as I do, then divide by y
It took me a while to realise why this works when the original approach didn't and it come down to the probability of getting a zero weight (as highlighted by Floris in his answer). To get a zero weight in the original version for all but the 1st or last weights your random numbers had to have 2 values the same but for the 1st & last ones then a random number of zero or the maximum number would result in a zero weight which is more likely.
In the revised algorithm, zero & the maximum number are not in the set of random choices & a zero weight occurs only if you select two consecutive numbers which is equally likely for every position.
I coded it up in Matlab as follows
function weights = unbiased_monkey_weights(num_simulations,num_stocks,min_increment)
scaling=1/min_increment;
sample=NaN(num_simulations,num_stocks-1);
for i=1:num_simulations
allcomb=randperm(scaling+num_stocks-1);
sample(i,:)=allcomb(1:num_stocks-1);
end
temp = [zeros(num_simulations,1),sort(sample,2),ones(num_simulations,1)*(scaling+num_stocks)];
weights = (diff(temp,[],2)-1)/scaling;
end
Obviously the loop is a bit clunky and as I'm using the 2009 version the randperm function only allows you to generate permutations of the whole set, however despite this I can run 10,000 simulations for 1,000 numbers in 5 seconds on my clunky laptop which is fast enough.
The mean weights are now correct & as a quick test I replicated woodchips generating 3 numbers that sum to 1 with the minimum increment being 0.01% & it also look right
Thank you all for your help and I hope this solution is useful to somebody else in the future
The simple answer is to use the schemes that work well with NO minimum increment, then transform the problem. As always, be careful. Some methods do NOT yield uniform sets of numbers.
Thus, suppose I want 11 numbers that sum to 100, with a constraint of a minimum increment of 5. I would first find 11 numbers that sum to 45, with no lower bound on the samples (other than zero.) I could use a tool from the file exchange for this. Simplest is to simply sample 10 numbers in the interval [0,45]. Sort them, then find the differences.
X = diff([0,sort(rand(1,10)),1]*45);
The vector X is a sample of numbers that sums to 45. But the vector Y sums to 100, with a minimum value of 5.
Y = X + 5;
Of course, this is trivially vectorized if you wish to find multiple sets of numbers with the given constraint.
I have production (q) values from 4 different methods stored in the 4 matrices. Each of the 4 matrices contains q values from a different method as:
Matrix_1 = 1 row x 20 column
Matrix_2 = 100 rows x 20 columns
Matrix_3 = 100 rows x 20 columns
Matrix_4 = 100 rows x 20 columns
The number of columns indicate the number of years. 1 row would contain the production values corresponding to the 20 years. Other 99 rows for matrix 2, 3 and 4 are just the different realizations (or simulation runs). So basically the other 99 rows for matrix 2,3 and 4 are repeat cases (but not with exact values because of random numbers).
Consider Matrix_1 as the reference truth (or base case ). Now I want to compare the other 3 matrices with Matrix_1 to see which one among those three matrices (each with 100 repeats) compares best, or closely imitates, with Matrix_1.
How can this be done in Matlab?
I know, manually, that we use confidence interval (CI) by plotting the mean of Matrix_1, and drawing each distribution of mean of Matrix_2, mean of Matrix_3 and mean of Matrix_4. The largest CI among matrix 2, 3 and 4 which contains the reference truth (or mean of Matrix_1) will be the answer.
mean of Matrix_1 = (1 row x 1 column)
mean of Matrix_2 = (100 rows x 1 column)
mean of Matrix_3 = (100 rows x 1 column)
mean of Matrix_4 = (100 rows x 1 column)
I hope the question is clear and relevant to SO. Otherwise please feel free to edit/suggest anything in question. Thanks!
EDIT: My three methods I talked about are a1, a2 and a3 respectively. Here's my result:
ci_a1 =
1.0e+008 *
4.084733001497999
4.097677503988565
ci_a2 =
1.0e+008 *
5.424396063219890
5.586301025525149
ci_a3 =
1.0e+008 *
2.429145282593182
2.838897116739112
p_a1 =
8.094614835195452e-130
p_a2 =
2.824626709966993e-072
p_a3 =
3.054667629953656e-012
h_a1 = 1; h_a2 = 1; h_a3 = 1
None of my CI, from the three methods, includes the mean ( = 3.454992884900722e+008) inside it. So do we still consider p-value to choose the best result?
If I understand correctly the calculation in MATLAB is pretty strait-forward.
Steps 1-2 (mean calculation):
k1_mean = mean(k1);
k2_mean = mean(k2);
k3_mean = mean(k3);
k4_mean = mean(k4);
Step 3, use HIST to plot distribution histograms:
hist([k2_mean; k3_mean; k4_mean]')
Step 4. You can do t-test comparing your vectors 2, 3 and 4 against normal distribution with mean k1_mean and unknown variance. See TTEST for details.
[h,p,ci] = ttest(k2_mean,k1_mean);
EDIT : I misinterpreted your question. See the answer of Yuk and following comments. My answer is what you need if you want to compare distributions of two vectors instead of a vector against a single value. Apparently, the latter is the case here.
Regarding your t-tests, you should keep in mind that they test against a "true" mean. Given the number of values for each matrix and the confidence intervals it's not too difficult to guess the standard deviation on your results. This is a measure of the "spread" of your results. Now the error on your mean is calculated as the standard deviation of your results divided by the number of observations. And the confidence interval is calculated by multiplying that standard error with appx. 2.
This confidence interval contains the true mean in 95% of the cases. So if the true mean is exactly at the border of that interval, the p-value is 0.05 the further away the mean, the lower the p-value. This can be interpreted as the chance that the values you have in matrix 2, 3 or 4 come from a population with a mean as in matrix 1. If you see your p-values, these chances can be said to be non-existent.
So you see that when the number of values get high, the confidence interval becomes smaller and the t-test becomes very sensitive. What this tells you, is nothing more that the three matrices differ significantly from the mean. If you have to choose one, I'd take a look at the distributions anyway. Otherwise the one with the closest mean seems a good guess. If you want to get deeper into this, you could also ask on stats.stackexchange.com
Your question and your method aren't really clear :
Is the distribution equal in all columns? This is important, as two distributions can have the same mean, but differ significantly :
is there a reason why you don't use the Central Limit Theorem? This seems to me like a very complex way of obtaining a result that can easily be found using the fact that the distribution of a mean approaches a normal distribution where sd(mean) = sd(observations)/number of observations. Saves you quite some work -if the distributions are alike! -
Now if the question is really the comparison of distributions, you should consider looking at a qqplot for a general idea, and at a 2-sample kolmogorov-smirnov test for formal testing. But please read in on this test, as you have to understand what it does in order to interprete the results correctly.
On a sidenote : if you do this test on multiple cases, make sure you understand the problem of multiple comparisons and use the appropriate correction, eg. Bonferroni or Dunn-Sidak.