I'm wondering how you can quantify the results of the Needleman-Wunsch algorithm (typically used for aligning nucleotide/protein sequences).
Consider some fixed scoring scheme and two sequences of varying length S1 and S2. Say we calculate every possible alignment of S1 and S2 by brute force, and the highest scoring alignment has a score x. And of course, this has considerably higher complexity than the Needleman-Wunsch approach.
When using the Needleman-Wunsch algorithm to find a sequence alignment, say that it has a score y.
Consider r to be the score generated via Needleman-Wunsch for two random sequences R1 and R2.
How does x compare to y? Is y always greater than r for two sequences of known homology?
In general, I do understand that we use the Needleman-Wunsch algorithm to significantly speed up sequence alignment (vs a brute-force approach), but don't understand the cost in accuracy (if any) that comes with it. I had a go at reading the original paper (Needleman & Wunsch, 1970) but am still left with this question.
Needlman-Wunsch always produces an optimal answer - it's much faster than brute force and doesn't sacrifice accuracy in the process. The key insight it uses is that it's not actually necessary to generate all possible alignments, since most of them contain bad sub-alignments and couldn't possibly be optimal. The Needleman-Wunsch algorithm works by instead slowly building up optimal alignments for fragments of the original strands and then slowly growing those smaller alignments into larger alignments using the guarantee that any optimal alignment must contain an optimal alignment for a slightly smaller case.
I think your question boils down to whether dynamic programming finds the optimal solution ie, garantees that y >= x. For a discussion on this I would refer to people who are likely smarter than me:
https://cs.stackexchange.com/questions/23599/how-is-dynamic-programming-different-from-brute-force
Basically, it says that dynamic programming will likely produce optimal result ie, same as brute force, but only for particular problems that satisfy the Bellman principle of optimality.
According to Wikipedia page for Needleman-Wunsch, the problem does satisfy Bellman principle of optimality:
https://en.wikipedia.org/wiki/Needleman%E2%80%93Wunsch_algorithm
Specifically:
The Needleman–Wunsch algorithm is still widely used for optimal global
alignment, particularly when the quality of the global alignment is of
the utmost importance. However, the algorithm is expensive with
respect to time and space, proportional to the product of the length
of two sequences and hence is not suitable for long sequences.
There is also mention of optimality elsewhere in the same Wikipedia page.
Related
I am learning statistics, and have some basic yet core questions on SD:
s = sample size
n = total number of observations
xi = ith observation
μ = arithmetic mean of all observations
σ = the usual definition of SD, i.e. ((1/(n-1))*sum([(xi-μ)**2 for xi in s])**(1/2) in Python lingo
f = frequency of an observation value
I do understand that (1/n)*sum([xi-μ for xi in s]) would be useless (= 0), but would not (1/n)*sum([abs(xi-μ) for xi in s]) have been a measure of variation?
Why stop at power of 1 or 2? Would ((1/(n-1))*sum([abs((xi-μ)**3) for xi in s])**(1/3) or ((1/(n-1))*sum([(xi-μ)**4 for xi in s])**(1/4) and so on have made any sense?
My notion of squaring is that it 'amplifies' the measure of variation from the arithmetic mean while the simple absolute difference is somewhat a linear scale notionally. Would it not amplify it even more if I cubed it (and made absolute value of course) or quad it?
I do agree computationally cubes and quads would have been more expensive. But with the same argument, the absolute values would have been less expensive... So why squares?
Why is the Normal Distribution like it is, i.e. f = (1/(σ*math.sqrt(2*pi)))*e**((-1/2)*((xi-μ)/σ))?
What impact would it have on the normal distribution formula above if I calculated SD as described in (1) and (2) above?
Is it only a matter of our 'getting used to the squares', it could well have been linear, cubed or quad, and we would have trained our minds likewise?
(I may not have been 100% accurate in my number of opening and closing brackets above, but you will get the idea.)
So, if you are looking for an index of dispersion, you actually don't have to use the standard deviation. You can indeed report mean absolute deviation, the summary statistic you suggested. You merely need to be aware of how each summary statistic behaves, for example the SD assigns more weight to outlying variables. You should also consider how each one can be interpreted. For example, with a normal distribution, we know how much of the distribution lies between ±2SD from the mean. For some discussion of mean absolute deviation (and other measures of average absolute deviation, such as the median average deviation) and their uses see here.
Beyond its use as a measure of spread though, SD is related to variance and this is related to some of the other reasons it's popular, because the variance has some nice mathematical properties. A mathematician or statistician would be able to provide a more informed answer here, but squared difference is a smooth function and is differentiable everywhere, allowing one to analytically identify a minimum, which helps when fitting functions to data using least squares estimation. For more detail and for a comparison with least absolute deviations see here. Another major area where variance shines is that it can be easily decomposed and summed, which is useful for example in ANOVA and regression models generally. See here for a discussion.
As to your questions about raising to higher powers, they actually do have uses in statistics! In general, the mean (which is related to average absolute mean), the variance (related to standard deviation), skewness (related to the third power) and kurtosis (related to the fourth power) are all related to the moments of a distribution. Taking differences raised to those powers and standardizing them provides useful information about the shape of a distribution. The video I linked provides some easy intuition.
For some other answers and a larger discussion of why SD is so popular, See here.
Regarding the relationship of sigma and the normal distribution, sigma is simply a parameter that stretches the standard normal distribution, just like the mean changes its location. This is simply a result of the way the standard normal distribution (a normal distribution with mean=0 and SD=variance=1) is mathematically defined, and note that all normal distributions can be derived from the standard normal distribution. This answer illustrates this. Now, you can parameterize a normal distribution in other ways as well, but I believe you do need to provide sigma, whether using the SD or precisions. I don't think you can even parametrize a normal distribution using just the mean and the mean absolute difference. Now, a deeper question is why normal distributions are so incredibly useful in representing widely different phenomena and crop up everywhere. I think this is related to the Central Limit Theorem, but I do not understand the proofs of the theorem well enough to comment further.
I am reading the Rabin-Karb algorithm from Sedgewick. The book says:
We use a random prime Q taking as large a value as possible while
avoiding overflow
At first reading I didn't notice the significance of random and when I saw that in the code a long is used my first thoughts were:
a) Use Eratosthene's sieve to find a big prime that fits a long
or
b) look up from a list of primes any prime large enough that is greater than int and use it as a constant.
But then the rest of the explanation says:
We will use a long value greater than 10^20 making the probability
that a collision happens less than 10^-20
This part got me confused since a long can not fit 10^20 let alone a value greater than that.
Then when I checked the calculation for the prime the book defers to an exercise that has just the following hint:
A random n-digit number is prime with probability proportional to 1/n
What does that mean?
So basically what I don't get is:
a) what is the meaning of using a random prime? Why can't we just pre-calculate it and use it as a constant?
b) why is the 10^20 mentioned since it is out of range for long?
c) How is that hint helpful? What does it mean exactly?
Once again, Sedgewick has tried to simplify an algorithm and gotten the details slightly wrong. First, as you observe, 1020 cannot be represented in 64 bits. Even taking a prime close to 263 − 1, however, you probably would want a bit of room to multiply the normal way without overflowing so that the subsequent modulo is correct. The answer uses a 31-bit prime, which makes this easy but only offers collision probabilities in the 10−9 range.
The original version uses Rabin fingerprints and a random irreducible polynomial over 𝔽2[x], which from the perspective of algebraic number theory behaves a lot like a random prime over the integers. If we choose the polynomial to be degree 32 or 64, then the fingerprints fit perfectly into a computer word of the appropriate length, and polynomial addition and subtraction both work out to bitwise XOR, so there is no overflow.
Now, Sedgewick presumably didn't want to explain how polynomial rings work. Fine. If I had to implement this approach in practice, I'd choose a prime p close to the max that was easy to mod by with cheap instructions (I'm partial to 231 − 227 + 1; EDIT actually 231 − 1 works even better since we don't need a smooth prime here) and then choose a random number in [1, p−1] to evaluate the polynomials at (this is how Wikipedia explains it). The reason that we need some randomness is that otherwise the oblivious adversary could choose an input that would be guaranteed to have a lot of hash collisions, which would severely degrade the running time.
Sedgewick wanted to follow the original a little more closely than that, however, which in essence evaluates the polynomials at a fixed value of x (literally x in the original version that uses polynomial rings). He needs a random prime so that the oblivious adversary can't engineer collisions. Sieving numbers big enough is quite inefficient, so he turns to the Prime Number Theorem (which is the math behind his hint, but it holds only asymptotically, which makes a big mess theoretically) and a fast primality test (which can be probabilistic; the cases where it fails won't influence the correctness of the algorithm, and they are rare enough that they won't affect the expected running time).
I'm not sure how he proves a formal bound on the collision probability. My rough idea is basically, show that there are enough primes in the window of interest, use the Chinese Remainder Theorem to show that it's impossible for there to be a collision for too many primes at once, conclude that the collision probability is bounded by the probability of picking a bad prime, which is low. But the Prime Number Theorem holds only asymptotically, so we have to rely on computer experiments regarding the density of primes in machine word ranges. Not great.
I'm trying to invent programming exercise on Suffix Arrays. I learned O(n*log(n)^2) algorithm for constructing it and then started playing with random input strings of varying length in order to find out when naive approach becomes too slow. E.g. I wanted to choose string length so that people will need to implement "advanced" algorithm.
Suddenly I found that naive algorithm (with using logarithmic sort on all suffixes) is not as slow as O(n^2 * log(n)) means. After thinking a bit, I understand that comparison of suffixes of a randomly generated string is not O(n) amortized. Really, we usually only compare few first characters before we come to difference and there we return from comparison function. This of course depends on the size of the alphabet, but anyway it does not depend much on the length of suffixes.
I tried simple implementation in PHP processing 50000-characters string in 2 seconds (despite slowness of scripting language). If it will work at least as O(n^2) we'll expect it to work at least several minutes (with 1e7 operations per second and ~1e9 operations total).
So I understand that even if it is O(n^2 * log(n)) then the constant factor is a very small fraction of 1, really something close to 0. Or we should say about such complexity as worst-case only, right?
But what is the amortized time complexity of the naive approach? I'm bit bewildered about how to assess it.
You seem to be confusing amortized and expected complexity. In this case you are talking about expected complexity. And yes the stated complexity is computed assuming that the suffix comparison takes O(n). This will be the worst case for suffix comparison and for random generated input you will only perform constant number of comparisons in most cases. Thus O(n^2*log(n)) is worst case complexity.
One more note - on a modern computer you can perform a few billion elementary instructions in a second and it is possible that you execute in the order of 50000^2 in 2 seconds. The correct way to benchmark complexity of an algorithm is to measure the time it takes to complete e.g. for input of size N, N*2, N*4,...(as many as you can go) and then to interpolate the function that would describe the computational complexity
I'm testing string search algorithms from this site: EXACT STRING MATCHING ALGORITHMS. Christian Charras, Thierry Lecroq. Test text is a random sequence of DNA bases (ACGT) of 1 GByte size. Test patterns are a list of random sequences of random size (1kB max). Test system is a AMD Phenom II x4 955 at 3.2 GHz, 4 GB of RAM and Windows 7 64 bits. Code witten in C and compiled with MinGW with -O3 flag.
Naive search algorithm takes 4 seconds for short patterns to 8 seconds for 1kB patterns. Deterministic finite state machine takes 2 seconds for short patterns to 4 seconds for 1kB patterns. Boyer-Moore algorithm takes 4 seconds for very short patters, about 1/2 second for short pattherns and 2 seconds for 1kB patterns. The remaining algorithm performance is worst than naive search algorithm.
How can be naive search algorithm search algorithm faster than most other algorithms?
How can a deterministic finite state machine implemented with a transition table (O(n) execution time always) be 2 to 8 times slower than Boyer-Moore algorithm? Yes, BM best case is O(n/m), but his average case is O(n) and worst case is O(nm).
There is no perfect string matching algorithm which is best for all circumstances.
Boyer-Moore (and Horspool, Sunday etc.) work by creating jump tables ('How far can I move the search pointer when the characters do not match? The more distinct letters in the strings, the better the positive impact. You can imagine, that a string with only 4 distinct letters creates a jump table with a maximum of 3 shifts per mismatch. Whereas searching an english word with case sensitive may result in a jumptable with (A-Z + a-z + punctiation) max. approx 55 shifts per mismatch.
On the other hand, there is a negative impact on both preparation (i.e. calculating the jump tables) and looping itself. So these algorithms perform poor on short strings (preparation creates an overhead) and strings with only a few distict letters (as mentioned before)
The naive search algorithm is very compact and there are very little operations inside the loop, so loop runs fast. As there is no overhead it performs better when searching short strings.
The (compared to the naive search) quite complex loop operations of a BM algorithm take much longer per loop run. This (partly) compensates for the positive performance impact of the jump tables.
So although you are using long strings, the small alphabet (=small jump tables) makes BM perform poorly. A KMP has less overhead in the loop (the jump table is smaller in general, but is similar to the BM with small alphabets) and so the KMP performs so well.
Theoretically good algorithms (lower time complexity) often have high bookkeeping costs that can overwhelm that of a naive algorithm for small problem sizes. Also implementation details matter. By optimizing an implementation you can sometimes improve runtime by factors of 2 or more.
The naive implementation actually has a linear expected running time (same as BM/KMP, etc) for random input data. I could not write a full proof here but it's accessible from Algorithms Design Techniques and Analysis.
Most exact matching algorithms are optimized version of the naive implementation to prevent being slowed down by certain patterns. For instance, suppose we are searching for:
aaaaaaaaaaaaaaaaaaaaaaaab
on a stream of:
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab
It fails at the b for lots of times. KMP/BM implementations are contrived to prevent repeatedly comparing the as. However, if the sequence is random by itself, such conditions are almost impossible to appear and the naive implementation is likely to work better due to its lower overhead in bookkeeping or possibly better spatial/temporal locality.
And, yeah, I'm not sure DNA sequences are random. Or alternatively are repetitions common in them. Anyway there's no way to examine this carefully without representative data.
I have n points in R^3 that I want to cover with k ellipsoids or cylinders (I don't really care; whichever is easier). I want to approximately minimize the union of the volumes. Let's say n is tens of thousands and k is a handful. Development time (i.e. simplicity) is more important than runtime.
Obviously I can run k-means and use perfect balls for my ellipsoids. Or I can run k-means, then use minimum enclosing ellipsoids per cluster rather than covering with balls, though in the worst case that's no better. I've seen talk of handling anisotropy with k-means but the links I saw seemed to think I had a tensor in hand; I don't, I just know the data will be a union of ellipsoids. Any suggestions?
[Edit: There's a couple votes for fitting a mixture of multivariate Gaussians, which seems like a viable thing to try. Firing up an EM code to do that won't minimize the volume of the union, but of course k-means doesn't minimize volume either.]
So you likely know k-means is NP-hard, and this problem is even more general (harder). Because you want to do ellipsoids it might make a lot of sense to fit a mixture of k multivariate gaussian distributions. You would probably want to try and find a maximum likelihood solution, which is a non-convex optimization, but at least it's easy to formulate and there is likely code available.
Other than that you're likely to have to write your own heuristic search algorithm from scratch, this is just a huge undertaking.
I did something similar with multi-variate gaussians using this method. The authors use kurtosis as the split measure, and I found it to be a satisfactory method for my application, clustering points obtained from a laser range finder (i.e. computer vision).
If the ellipsoids can overlap a lot,
then methods like k-means that try to assign points to single clusters
won't work very well.
Part of each ellipsoid has to fit the surface of your object,
but the rest may be inside it, don't-cares.
That is, covering algorithms
seem to me quite different from clustering / splitting algorithms;
unions are not splits.
Gaussian mixtures with lots of overlaps ?
No idea, but see the picture and code on Numerical Recipes p. 845.
Coverings are hard even in 2d, see
find-near-minimal-covering-set-of-discs-on-a-2-d-plane.