I am having a really hard time understanding these so please help me so I may get an example, so I can get a chance to solve more of those. I have tried many different way but none seem to be able to come to a result, my guess is that it is wrong and it is O(n) but I cannot prove it.
//Is this statement true and why?
square_root(n5) log(square_root(n5)) = O(n3)
Following the steps below would help you deduce the result discussed in the question.
Because:
With an appendix:
Related
I'm debugging the code of an api and I found a cypher instruction that takes 6 minutes to return the data.
I ran the neo4j code in smaller chunks and found that this snippet is causing the problem: MATCH(copart:CopartOperadora) WHERE NOT (copart)-[:FROM_TO]->(:Coexistence)
I'm new to neo4j so I still haven't figured out how I can optimize this instruction.
Thanks to everyone who contributed.
Optimizations of this kind, usually depend on the schema, of your graph database, without that it's very hard to provide any insights. But you can try this:
MATCH (copart:CopartOperadora)-[:FROM_TO]->(:Coexistence)
WITH collect(id(copart)) AS connectedNodesIds
MATCH (copart:CopartOperadora) WHERE id(copart) NOT IN connectedNodesIds
We can't create any index as such, unfortunately. But if the relationship FROM_TO is only present from CopartOperadora to Coexistence nodes. Then you can remove the node label for Coexistence, all together, which will be optimal. Something like this:
MATCH(copart:CopartOperadora) WHERE NOT (copart)-[:FROM_TO]->()
I have a dataset with 3.3M rows and 8k unique products.
I wanted to apply apriori algorithm to find association rules and connections between products.
Well, I did it before on a much smaller database with 50k rows and maybe 200 unique products..
Someone knows how can I do it effectively with larger scales of data? How can I still make it work for me maybe there are tricks to reduce the scale of the data but still get effective results.
Any help would be amazing! Reach me out if you experienced with this algorithm.
The trick is: Don't use Apriori.
Use LCM or the top-down version of FP-Growth.
You can find my implementations here:
command line programs: https://borgelt.net/fim.html (eclat with option -o gives LCM)
Python: https://borgelt.net/pyfim.html
R: https://borgelt.net/fim4r.html
In the code examples from the SAT4J documentation, calling the solver multiple times on the same SAT problem always yields the same solution, even if multiple possible solutions exist - that is, the result is deterministic.
I'm looking for a way to get different solutions on multiple runs, that is, a nondeterministic/random result. For each possible solution, there should be a non-zero probability for the solution to be picked. Ideally, every solution should be picked with the same probability, but that's not a strict requirement.
I'm aware of the possibility to (deterministically) iterate over all solutions and just take a random one, but that's not a feasible solution in my case since there are too many solutions to begin with, and calculating them all takes too long.
Yes, Sat4j is by default deterministic: it will always find the same solution if you run it several times on the same problem from the command line.
The way to add some non determinism in the heuristics is to use the RandomWalkDecorator, as found for instance in the GreedySolver in org.sat4j.minisat.SolverFactory.
Note however that if you several times such solver from the command line :
java -jar org.sat4j.core.jar GreedySolver file.cnf
you will still be deterministic, since the pseudo random numbers generator is seeded by a constant.
Thus you need to ask several models within your Java code.
As mentioned in your question, you can use a ModelIterator decorator with a bound for that:
ISolver solver = SolverFactory.newGreedySolver();
ModelIterator mi = new ModelIterator(solver,10); // look for 10 models
Given a of 6sided pair of dices, in how many ways can I get sum K.
Can someone give a detailed explanation to me how to use dynamic programming to solve this problem.
I have done reading on it but still hard to understand link
This is something simply to ease my curiosity, if someone would feel like answering it though that would be fantastic.
With if statements, is the time taken to calculate the result affected by the way it's written?
So what I mean is (if that wasn't overly clear) would the following two statements take the same amount of time to process?
if 1 < 2 and 3 = 3 then
//do something
end if
compared to
if 1 < 2 then
if 3 = 3 then
//Do something
end if
end if
If we consider that the compiler will not optimize these two calls, then the second statement will require two branching instructions instead of one. And branching requires some extra work for the CPU because of pipelining. So, technically, the second version will require more work, but it should not matter here.
This is just another case of premature-optimization. You are not going to gain anything by thinking a lot about this.
What you should be focusing on is how to make your code more readable.