blog-code | Supporting code snippets for my blog posts | Blog library
kandi X-RAY | blog-code Summary
kandi X-RAY | blog-code Summary
Supporting code snippets for my blog posts.
Support
Quality
Security
License
Reuse
Top functions reviewed by kandi - BETA
- Fetch all pages from a URL .
- Schedules a task
- Show the index .
- Run a task
- Invoked when a return is received .
- hello message
blog-code Key Features
blog-code Examples and Code Snippets
Community Discussions
Trending Discussions on blog-code
QUESTION
We have Boost.PFR and we have the tuple iterator. If we combine the two, we might have a way of applying std algorithms over structs. Does a solution already exist? What I'm looking for is:
...ANSWER
Answered 2021-Apr-10 at 15:38You can already have the existing functionality with pfr::structure_tie, which results in a tuple of references.
This would allow us to use the newer features to process structs out of sequence or in parallel.
That would only make sense if processing per element has considerable execution cost. In that case, you're probably already fine with a bespoke
QUESTION
i want to know what are the chances for this unique id generator to collide https://github.com/vejuhust/blog-code/blob/master/python-short-id-generator/short_id_v5.py
i wanted to generate a unique url for my Django project. is it going to be safe to use. i am a beginner in python and Django.
...ANSWER
Answered 2019-Sep-02 at 08:18This is a base62 encoded 8 byte random number. The encoding does not matter however, since every random number is encoded differently. It thus boils down on what are the odds that two 8 byte random numbers are the same.
We can generate a total of 28×8=18'446'744'073'709'552'000 values with 8 bytes. So the odds of generating a second value that is the same is 1/18'446'744'073'709'552'000 or 0.000000000000000000542%.
If you generate k items, the odds of generating a duplicate is:
1 - (28×8)!/((28×8-k)!×28×8×k).
As k grows larger, the odds of a collision increase. If you mark the field unique, and thus have a retry mechanism, the odds that it will collide a second time (or a third time) are quite small.
QUESTION
I am trying to implement a Union-Find/Disjoint-Set data structure in C, using weighted Union and path compression in Find. I have some questions as to how the weighted union should be implemented to reduce time complexity when compared to the non weighted Union.
I have already found several solutions to this problem online and have implemented my own. In every solution, the root of each separate tree (representing a set) holds the number of nodes of the tree at all times. When uniting the sets of two random objects that belong to a different set, the roots are first found (path compression is used here) and then we compare the sizes of these roots. The root of the biggest tree is set as the parent of the of the root of the smallest tree.
In my understanding however, what we are trying to achieve with a weighted union is to reduce the height of the resulting tree (which is also what we are trying to achieve with path compression). Hence, it is no the tree with the lowest number of Nodes that should be connected to the other tree, but the tree with the lowest height. This keeps the height to a minimum.
Is this correct? Is checking the height and the size somehow equivalent given the rest of the implementation (we always start with a number of single (one node) sets)?
Supposing that it is the height that needs to be checked, keeping track of the height of a tree if path compression is not used is fairly straightforward. I have not however found a way to keep track of the height of the tree when path compression is used (at least not without traversing the whole tree, which increases the time complexity of the "find" algorithm.
Here is an example of an implementation I have found and uses what I described (very similar to what I have coded) in c++: https://github.com/kartikkukreja/blog-codes/blob/master/src/Union%20Find%20(Disjoint%20Set)%20Data%20Structure.cpp
...ANSWER
Answered 2019-Apr-16 at 12:51It looks like you've pretty much figured this all out yourself.
Union-by-height is the obvious way to make the shortest tree, but it's hard to keep track of the height when you use path compression...
So union-by-rank is commonly used instead. The 'rank' of a set is what it's height would be if we didn't do any path compression, so when you use union-by-rank with path compression it's like starting with union-by-height and then applying path compression as an optimization, ensuring that the path compression doesn't change how the merges work.
A lot of people (myself included) use union-by-size, however, because the size is often useful, and it can be shown that union-by-size produces the same worst-case complexities as union-by-rank. Again in this case, path compression doesn't affect the merges since it doesn't change the size of any roots.
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install blog-code
You can use blog-code like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.
Support
Reuse Trending Solutions
Find, review, and download reusable Libraries, Code Snippets, Cloud APIs from over 650 million Knowledge Items
Find more librariesStay Updated
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page
