bfs | distributed file system ( small file storage

 by   Terry-Mao Go Version: v2.1.0 License: No License

kandi X-RAY | bfs Summary

kandi X-RAY | bfs Summary

bfs is a Go library. bfs has no bugs, it has no vulnerabilities and it has medium support. You can download it from GitHub.

distributed file system(small file storage) writen in golang.
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            kandi-support Support

              bfs has a medium active ecosystem.
              It has 1561 star(s) with 315 fork(s). There are 122 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 4 open issues and 22 have been closed. On average issues are closed in 178 days. There are 2 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of bfs is v2.1.0

            kandi-Quality Quality

              bfs has 0 bugs and 0 code smells.

            kandi-Security Security

              bfs has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              bfs code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              bfs does not have a standard license declared.
              Check the repository for any license declaration and review the terms closely.
              OutlinedDot
              Without a license, all rights are reserved, and you cannot use the library in your applications.

            kandi-Reuse Reuse

              bfs releases are available to install and integrate.
              Installation instructions are not available. Examples and code snippets are available.

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            bfs Key Features

            No Key Features are available at this moment for bfs.

            bfs Examples and Code Snippets

            Prints the BFS tree .
            javadot img1Lines of Code : 21dot img1License : Non-SPDX (GNU General Public License v3.0)
            copy iconCopy
            public static void main(String[] args) {
                    BinaryTree bt = new BinaryTree<>();
                    bt.put(1);
                    bt.put(2);
                    bt.put(3);
                    bt.put(4);
                    bt.put(5);
                    bt.put(6);
                    bt.put(7);
                    bt.put(8);
            
                    
            Fills a bfs image .
            javadot img2Lines of Code : 19dot img2no licencesLicense : No License
            copy iconCopy
            public static int[][] floodFill_bfs(int[][] image, int sr, int sc, int newColor) {
                    int rows = image.length, cols = image[0].length;
                    Set visited = new HashSet<>();
                    Queue queue = new LinkedList<>();
                    queue.add  
            BFS breadth - first search .
            pythondot img3Lines of Code : 16dot img3License : Permissive (MIT License)
            copy iconCopy
            def BFS(graph, s, t, parent):
                # Return True if there is node that has not iterated.
                visited = [False] * len(graph)
                queue = []
                queue.append(s)
                visited[s] = True
            
                while queue:
                    u = queue.pop(0)
                    for ind in range(le  

            Community Discussions

            QUESTION

            Count nodes within k distance of marked nodes in grid
            Asked 2022-Feb-25 at 09:45

            I am attempting to solve a coding challenge however my solution is not very performant, I'm looking for advice or suggestions on how I can improve my algorithm.

            The puzzle is as follows:

            You are given a grid of cells that represents an orchard, each cell can be either an empty spot (0) or a fruit tree (1). A farmer wishes to know how many empty spots there are within the orchard that are within k distance from all fruit trees.

            Distance is counted using taxicab geometry, for example:

            ...

            ANSWER

            Answered 2021-Sep-07 at 01:11

            This wouldn't be easy to implement but could be sublinear for many cases, and at most linear. Consider representing the perimeter of each tree as four corners (they mark a square rotated 45 degrees). For each tree compute it's perimeter intersection with the current intersection. The difficulty comes with managing the corners of the intersection, which could include more than one point because of the diagonal alignments. Run inside the final intersection to count how many empty spots are within it.

            Source https://stackoverflow.com/questions/69075779

            QUESTION

            Why can Dijkstra's Algorithm be modified to find K shortest paths?
            Asked 2022-Feb-17 at 19:54

            I am trying to find an intuitive explanation as to why we can generalize Dijkstra's Algorithm to find the K shortest (simple) paths from a single source in a directed, weighted graph with no negative edges. According to Wikipedia, the pseudocode for the modified Dijkstra is as follows:

            ...

            ANSWER

            Answered 2022-Feb-17 at 19:54

            The Wiki article doesn't specify, but that code will only solve the 'loopy' version of k-shortest-paths, where paths are not required to be simple.

            The simple path version of the problem is harder: you'll want to look at something like Yen's algorithm, which does clever filtering to avoid repeated points when generating paths. Yen's algorithm can use Dijkstra's algorithm as a subroutine, but any other shortest-path algorithm can also be used instead.

            There is no obvious way to modify Dijkstra's algorithm to solve the k-shortest-simple-paths problem. You'd need to track the paths in the priority queue (which is already done in your posted code), but there's an exponential upper bound on the number of times each vertex can be explored.

            Here, if count[u] <= K puts an upper bound of K+1 on the number of times a vertex can be explored, which works for the non-simple path case. On the other hand, a direct modification of Dijkstra's algorithm for simple paths would, in the worst case, require you to explore a node once for each of the 2^(V-1) possibilities of which nodes had been previously visited (or possibly a slightly smaller exponential).

            Source https://stackoverflow.com/questions/71140260

            QUESTION

            How to print a Binary Tree diagram (vertical) and ensure fairly unbalanced trees are not improperly printed?
            Asked 2022-Jan-17 at 07:49

            I'm trying to write functionality to print a vertical binary tree diagram,

            I've got the correct breadth-first search algorithm written, and it outputs the BFS-ordered tree traversal to an integer vector. The code can be seen below:

            ...

            ANSWER

            Answered 2022-Jan-17 at 07:49

            Here's a great answer, thanks to @NicoSchertler:

            "You can push prev and next to travQueue even if they are nullptr. When you reach a nullptr in your iteration, add the dummy value to the result and two more nullptr to travQueue for the non-existing children."

            And here's my code for it:

            Source https://stackoverflow.com/questions/70727970

            QUESTION

            How to find if there is a simple path from vertex x to vertex y that includes the edge e
            Asked 2022-Jan-04 at 09:14

            so I faced this question and I hope that someone can help me in it.

            Given an undirected graph G = (V, E), 2 vertices x,y and an edge e = (v,u).

            Suggest an algorithm to find if there's a simple path from x to y that includes the edge e.

            So the focus here is on the simple path and not a regular path, for a regular path it's an easy problem using the BFS to search a path from x to v and a path from u to y.

            I know that the problem can be solved using the max-flow approach but I just don't recognize how to build a new graph that can implement a max-flow algorithm on it so it tells whether the above criterion is achieved or not, I hope for help.

            Thanks in advance.

            ...

            ANSWER

            Answered 2022-Jan-04 at 09:14
            Without sharing edges (edge-independent)

            You could solve max flow with +1 sources at x and y, and -1 sinks at u and v.

            Remove the edge e, and set all other edges to capacity 1.

            A simple path from x to y via edge e exists if and only if you can find a flow of 2 in this new max flow problem.

            Without sharing vertices (vertex-independent i.e. simple path)

            Split each vertex v[i] in the original graph into two vertices, a[i] and b[i].

            For each undirected edge between v[i] and v[j] in the original, add directed edges b[j] to a[i] and b[i] to a[j] with capacity 1.

            Also add a directed edge from a[i] to b[i] with capacity 1 for each vertex v[i].

            The idea is that flow must always arrive at an a[i] vertex, and leave from a b[i] vertex, after passing through the capacity 1 bottleneck from a[i] to b[i]. This ensures that each vertex can only be used once.

            With this new graph, proceed as for the edge-independent case.

            Source https://stackoverflow.com/questions/70571584

            QUESTION

            Store Grid N*N into an Adjacency Graph? Position and Neighbors
            Asked 2021-Dec-09 at 18:08

            Updating this post, once again.

            This time to make things clearer. I am trying, to parse in a Grid of size 9x9, but this size can change overtime, its not fixed. This is a board game called Quoridor. What I have at my disposal, is a Board class. This provides me with the following, horizontal bool[,], and vertical bool[,], I can loop over each and print out x, y position. But these differ, depending on if its horizontal direction, or vertical direction and position.

            The player can move one step ONLY either north, south, west, or east in terms of direction. The other player (human) can put a wall (obstacle) which covers two blocks horizontally or vertically. My automated player has to build a Graph of Nodes from the board, and refresh the Graph based on the changes on the board and its own position. For example, if the player cannot go to left, from current position, then an edge which connects two nodes will be deleted between them, only if caused by an obstacle. And then the BFS will run again against the Graph and return a new position (x, y) which this (automated) player uses and performs its move.

            Every single block on the 9x9 grid, will represent one Node in the Graph. Meaning the number of vertices or nodes List in the Graph will be 9x9=81. Each of the Nodes hold a list or 2D array of size 4 for representing North, South, West, and East which could be of a bool type.

            Now, I have provided a sample code of the Graph class I wrote along with a Node class. I hope the latest information herein makes it clear. I have already implemented the BFS algorithm. But this part is what I cannot understand correctly. I watched this video for some ideas: https://www.youtube.com/watch?v=KiCBXu4P-2Y

            Code

            ...

            ANSWER

            Answered 2021-Nov-24 at 22:11

            Based on what you said I understand your question as follows: How the to handle Nodes that on the edges like (x=0,y=0), (x=9,y=5) or (x=9.y=9) ..... you should handle 8 cases

            for case of left top corner the node only have 2 neighbors so set the top and left neighbors Null

            Source https://stackoverflow.com/questions/69972021

            QUESTION

            How to generate and plot all spanning trees?
            Asked 2021-Nov-28 at 17:26

            I have a toy graph g, then I have found the number of spanning trees by cofactor of the Laplacian. The number is 11.

            ...

            ANSWER

            Answered 2021-Nov-04 at 11:14

            First of all, I would say, my solution below is a brute-force method, thus only working well for graphs of small size, i.e., not many vertices or arcs.

            If you have large networks, you should refer to some more advanced algorithms, e.g., https://link.springer.com/article/10.1007/s40747-018-0079-7

            Since you have 6 arcs and 5 vertices, you only need to remove 2 arcs out of 6 to find the spanning tree. There would be combn(6,2) options, and you can delete those edge combinations one by one to check if a spanning tree remains

            Source https://stackoverflow.com/questions/69794497

            QUESTION

            When should us use normal BFS over bidirectional BFS?
            Asked 2021-Nov-21 at 15:10

            I understand that Bidirectional BFS has a lot of advantage over using normal BFS, as it theoretically halves the time to discover the shortest path between two nodes and the time to find if a node is reachable from another node.

            Also I understand that we should use Bidirectional only if we have Uniquely defined both the nodes.

            Is there any situation when we should prefer a normal BFS over bidirectional BFS?

            ...

            ANSWER

            Answered 2021-Nov-21 at 15:10

            I understand that bidirectional BFS consists of, given start and goal nodes, alternately expanding layers of nodes from start and goal, until a node in the middle has been reached from both ends. The shortest path from start to goal is then understood to be the shortest from start to middle node, continued by the shortest from middle to goal. I can see that less nodes may need to be expanded as compared with a standard BFS approach. However,

            • It is easier to implement standard BFS (sBFS) than to implement a bidirectional BFS (bBFS for short). Simple is often good, as it is easier to code and to later verify its correctness.
            • If the graph is directed and unweighted, sBFS guarantees that it will find the shortest path from start to goal in minimal steps; bBFS is not guaranteed to work with directed graphs.
            • After running sBFS, you can reconstruct shortest paths from the start to all nodes that are at most 1 step before the goal node (that is, before the search was stopped). This may be valuable in and of itself. Running bBFS does not generate such a list.

            With this in mind, I would argue the bBFS is only useful for a very narrow case (where, depending on the graph, it is expected to perform better than sBFS), and sBFS is both simpler and useful in a larger range of scenarios.

            Source https://stackoverflow.com/questions/70052625

            QUESTION

            Find nodes in directed graph that is reachable from all other nodes
            Asked 2021-Nov-11 at 16:27

            Questions:

            Given a directed graph of N nodes and M edges (M <= 2.N). Find all the nodes that is reachable from all other nodes.

            Example:

            The below graph has 4 nodes and 4 edges:

            Answer: Node (2) and (3) is reachable from all other nodes.

            P/S:

            The only solution I came up with is to revert the graph, BFS all nodes and check if they reach all other nodes. But it would take O(n^2).

            Is there any other approach that takes O(n.logn) or less?

            Here's my take on the O(n^2) approach:

            ...

            ANSWER

            Answered 2021-Nov-11 at 13:18

            Can You share that code which take O(n^2) ?

            Anyway, This approach only take O(n), Which is better then O(n log n)!

            Source https://stackoverflow.com/questions/69927348

            QUESTION

            create table of contents in python
            Asked 2021-Oct-25 at 14:04

            Assuming the following text:

            ...

            ANSWER

            Answered 2021-Oct-25 at 10:56

            You could do something like this:

            Source https://stackoverflow.com/questions/69704038

            QUESTION

            padding article elements in html
            Asked 2021-Aug-19 at 05:44

            I'm using an article header to divide up some projects I wish to display:

            ...

            ANSWER

            Answered 2021-Aug-18 at 16:53

            You might need margin-bottom instead of padding-bottom;

            By using padding you don't seperate them, padding works kind of from-inside.

            You may read about box-model here to understand this.

            Source https://stackoverflow.com/questions/68836316

            Community Discussions, Code Snippets contain sources that include Stack Exchange Network

            Vulnerabilities

            No vulnerabilities reported

            Install bfs

            You can download it from GitHub.

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