How can I estimate the mean of the normal distribution, while it is an increasing function?
In other word, ascending the mean is a key requirement.
Related
I know that the algorithm for live variable analysis can finally terminate and give a solution. However, I'd like to know whether the iteration number of the algorithm is determined(i.e., can I calculate the iteration number of the algorithm with some parameters, I guess the parameters may be related to the program to be analyzed).
Although I still don't know how to calculate the accurate iteration number, it's easy to compute the maximum iteration number we need. It can be solved by applying the lattice theorems.
In terms of simplicity, assume the height of the lattice corresponding to the analysis is i and the number of the nodes in CFG is k, then the maximum iteration number is i*k.
I was thinking about sorting algorithm that starts with categorizing numbers by their length (i.e group with number 3 and 4, and another with 100 and 101) and then sorting them in their collective groups which i think would decrease sorting time (maybe even adding first number grouping to even further decrease the comparing time). Does that make sense? Do you have any recommendation how one would start that?
For simplicity, assume that you split the dataset in k subsets of equal size n. The sorting workload would become k.n.log(n) instead of k.n.log(k.n) = k.n.log(n) + k.log(k).n.
So from a theoretical standpoint, this makes no difference because both asymptotic complexities are O(n.log(n)). From a practical standpoint, you have to trade the gain on the number of key comparisons, to the loss due to extra element classification by length and pre/post-sort moves.
Only an actual implementation can tell you.
I have a large pool of short strings and a custom distance function on them (let's say Damerau–Levenshtein distance).
Q: What is the state-of-the-art solution for getting top N strings from the pool according to the custom distance?
I am looking for both a theoretical approach to this problem as well as coded implementation (Java, Python, etc).
The straight forward approach is to iterate over all strings, calculate the distance for each and keep only the best N while you iterate.
If you need to do this task a lot, you should think if you can come up with a upper-bound / lower bound estimation for the costs that can be calculated much faster than your real cost function. E.g. pre-calculate all n-grams (e.g. 3-grams) for your strings. or maybe comparing the length difference can already give a lower bound for the distance. than you can skip the calculation of the distance for all strings which have a lower bound distance higher than your current distance of the n-th best match.
Problem description:
We have a lot of bitstrings of the same size. The number and the size of bitstrings is huge. E.g.: 10100101 and 00001111.
Now there is a distance function that just counts the number of same bit positions. In this example: distance is 2 - because the third last and last bits are set by both bitstrings.
-> Now we can make a tour through the bitstrings with the maximal distance, because every bitstring can be converted to a vertex which is connected to all other bitstrings (by the distance function).
Goal
However, this has the complexity of O(N²). My idea is to use a hashing function, that preserves the similarities and than do a simple sort on the hash values. This should result in a near maximum tour. Of course it is not the best result, but it should be a somewhat good result.
Current Problem
My own hashing function rates the left bits higher than the left bits. So that they have a more significant effect to the sort.
Actual Question
Does such an algorithm exists?
Is it possible to use Locality-Sensitive Hashing (LSH) for that aim and if so, can you formulate the respective algorithm. (I didn't understood the algorithm right now)
Thank you, guys!
Is it possible to use dynamic programming in the calculation of power set of a string (ie, all possible subsequences of that string) to reduce the number of computations significantly?
No. If you are calculating a powerset, you are calculating a powerset, which always has the same number of elements.
You can never reduce complexity below linear with the size of the output, because you need go through each of the output bits some way or another. This is true for all problems, regardless of algorithm used. So 2^n is the lower bound for computation of the power set, because you need to output 2^n strings (and every string is multiple characters, which depends on n on average, so even higher).