floyd | A raft consensus implementation | Architecture library
kandi X-RAY | floyd Summary
kandi X-RAY | floyd Summary
| type | API | Status | | --------- | --------------- | ------- | | Consensus | Read | support | | Consensus | Write | support | | Consensus | Delete | support | | Local | DirtyRead | support | | Local | DirtyWrite | support | | Query | GetLeader | support | | Query | GetServerStatus | support | | Debug | set_log_level | support |. | Language | Leader election + Log Replication | Membership Changes | Log Compaction | | -------- | --------------------------------- | ------------------ | -------------- | | C++ | Yes | No | No |.
Support
Quality
Security
License
Reuse
Top functions reviewed by kandi - BETA
Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of floyd
floyd Key Features
floyd Examples and Code Snippets
Lines of Code : 38License : Permissive (MIT License)
Copydef floyd_warshall(graph, v):
"""
:param graph: 2D array calculated from weight[edge[i, j]]
:type graph: List[List[float]]
:param v: number of vertices
:type v: int
:return: shortest distance between all vertex pairs
dista public static ArrayList floyd_warshall(ArrayList[] adjlist, int source, int destination){
boolean[] visited = new boolean[adjlist.length];
int[] distance = new int[adjlist.length];
int[] prev = new int[adjlist.length];
Arrays.fill(distance, I def floyd(n):
"""
Parameters:
n : size of pattern
"""
for i in range(0, n):
for j in range(0, n - i - 1): # printing spaces
print(" ", end="")
for k in range(0, i + 1): # printing stars
Community Discussions
Trending Discussions on floyd
QUESTION
I am trying to sort this file that has this information below
...ANSWER
Answered 2021-May-28 at 04:45Below part is problematic in some ways:
QUESTION
I am using a case statement and cursor together to perform a calculation on each row and then show the sum of the results.
I have a table that looks like this:
...ANSWER
Answered 2021-May-26 at 16:28Just to expand on Steve's comment.
Example or dbFiddle
QUESTION
I have been looking at the application of merge sort to linked lists. Some of the articles I have looked at tout that merge sort is the best algorithm for sorting a linked list. It makes sense for the conquer part in the divide and conquer strategy where you merge the two sorted linked lists as you end up saving on required memory (compared to an array). But, what I don't understand is the time complexity of the divide step in the algorithm.
For an array, this step is constant time by leveraging random access and splitting the array into smaller chunks. But, for a linked list isn't this going to take an additional O(n)? I have seen Floyd's algorithm (tortoise-hare) used to find the mid-point of a linked list and divide the problem into smaller chunks. I did some analysis on the divide step. Suppose the linked list is of size n, then the # of operations involved in just dividing the problem is as follows,
n/2 + n/4 * 2 + n/8 * 4 + ... = n/2 * log(n)
From the above, it looks like compared to the array case, an additional factor of "n" appears from Floyd's algorithm. So, the final time complexity would be O(n^2 * log(n)). Can someone please explain the discrepancy?
Edit: based on @Yves comment, I identified the mistake,
I multiplied the work while merging back the sorted blocks from bottom to top when it should be added. So, the net time would be: nlogn/2 + nlogn = O(nlogn),
This is probably is most valid answer to the above question; other answers are a bit indirect/ provide no explanation
...ANSWER
Answered 2021-May-22 at 21:29The issue with your question is that the additional O(n/2) time complexity for the scanning of half a sub-list for each level of recursion translates into an overall time complexity of O((0.5 n log(n) + 1.0 (n log(n)) = O(1.5 n log(n)), not O(n^2 (log(n))), and O(1.5 (n log(n))) translates into O(n log(n)), since time complexity ignores lower order terms or constants. However in my actual testing for a large list with scattered nodes, where most node accesses result in a cache miss, my benchmarks show an relative time complexity of recursive versus iterative to be O(1.4 n log(n)), using a count based scan to split lists, rather than tortoise-hare approach.
For recursive version, using tortoise-hare approach is relatively slow and can be improved by using a count of nodes, which may require a one time scan of n node if the linked list container doesn't maintain a count of nodes (for example C++ std::list::size()). The reduces the overhead to advancing a single pointer halfway (sub-count / 2) through a linked list run.
Example C / C++ code:
Time taken to sort numbers in Linked List
However, in such a case (large list, scattered nodes), it is faster to copy the data from the list into an array, sort the array, then create a new sorted list from the sorted array. This is because elements in an array are merged sequentially (not via random linked list next pointers), which is cache friendly.
QUESTION
How to maintain an adding a new edge to the graph and efficiently update Floyd-Warshall all-pairs distances? Floyd-Warshall algorithm uses distances matrix, how we can update it after adding a new edge to the graph?
We can rerun Floyd-Warshall Algorithm, and it will take O(V^3). Can we make it faster?
...ANSWER
Answered 2021-Apr-24 at 12:30Let's say the edge goes from vertex v to vertex w and has cost c:
If the distance matrix already has a shorter path from v to w, then adding the edge has no effect, so there's nothing to do.
Otherwise, the new edge becomes the shortest path from v to w, so enter it into the distance matrix, and then, for every other pair of vertices a and b, see if it can be made shorter by using the new edge. From the distance matrix you can easily find the cost of a-v-w-b and a-w-v-b. Note that one a or b might by the same as v or w.
Since you have to check every pair of vertices, this takes O(V2) time.
QUESTION
import requests
from bs4 import BeautifulSoup
def getPage(url):
try:
req = requests.get(url)
except requests.exceptions.RequestException:
return None
return BeautifulSoup(req.text, 'html.parser')
bs = getPage('https://www.reuters.com/world/us/us-launch-probe-minneapolis-police-after-george-floyd-murder'
'-report-2021-04-21/')
bs.select_one('div.ArticleHeader__container___3rO4Ad h1')
ANSWER
Answered 2021-Apr-22 at 03:59In BS you can use find('span').next_sibling
QUESTION
How to modify my script, so in result there is no key "author" (I want to have only song and place)? The songs' list looks like that (obviously I do not copy whole list)
...ANSWER
Answered 2021-Apr-13 at 12:56Using Array.reduce() or lodash's _.reduce() you can destructure each object, and take out author, and then group the objects:
QUESTION
I was trying to code Floyd's triangle in MIPS and for some reason, I cannot seem to figure out the problem but since my QTspim is crashing while I am running the code therefore I am guessing it is stuck in an infinite loop. I need help in getting out of the infinte loop. This is the part of code with the loops and everything :
...ANSWER
Answered 2021-Mar-31 at 15:51It seems like you are not incrementing the inner loop i.e. loop 2 anywhere and therefore its having issues and is getting stuck in an infinite loop. Apart from that it looks fine. Here is my code: main:
QUESTION
I'm doing research on two algorithms for my bachelor thesis: Floyd-Warshall and A* algorithm. Time complexity is an important part in comparison of both algorithms for my work. But due to heuristics in A*, there is no constant time complexity of algorithm. Only information that I've found is that in worst case time complexity can be exponentially difficult.
What is an average, and best possible time complexity of A* algorithm in normal practice?
...ANSWER
Answered 2021-Mar-25 at 16:01first I suggest you look on A* as some kind of Dijkstra with heuristic approximation for the solution, so on the worst case your solution would be the time complexity of Dijkstra (which is O(|V|+|E|) * log|V|) in the original algorithm) so surely isn't exponential, of course if you consider that the heuristic function isn't bigger than the actual distance itself). for understanding sake you can look on my comparators of the "relax" function codes here when I implemented Dijkstra and A* on the same algorithm:
QUESTION
I'm trying to apply the map method to a dynamically created table in order to select the span elements with class = book_info. The results array contains duplicate data because of the nested table, but I'm not sure why - can someone please explain? What is the correct way to apply the map method to this table to get the desired data? Or should I use the each method instead?
...ANSWER
Answered 2021-Mar-17 at 16:58Thank you to @charlietfl
QUESTION
- This is normal text.Hi, I need to solve a problem with a list of dictionaries and conditional statements, however my code gives me no output - but no error message either. This is the problem:
Define a variable named ‘dataset’ that contains the following information:
...ANSWER
Answered 2021-Mar-14 at 00:52There is an error in your code, instead of asking if there is a 'Plays' property on dataset, you should access the x variable you declared on the iteration.
Try this, it should work:
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install floyd
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
