algorithms | : crystal_ball : Algorithm Visualizations | Data Visualization library
kandi X-RAY | algorithms Summary
kandi X-RAY | algorithms Summary
This repository is part of the project Algorithm Visualizer. algorithms contains algorithm visualizations shown on the side menu of algorithm-visualizer.org. The project currently supports JavaScript, C++, and Java.
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Top functions reviewed by kandi - BETA
- Build the Maze .
- Visit a vertex .
- Create a zarr
- fisher algorithm
- Match pattern with a pattern as a pattern .
- Clean up grid layout
- do merge merge
- Try to find the next element that is not found in the array .
- generate cell data
- Finishes the graph
algorithms Key Features
algorithms Examples and Code Snippets
Lines of Code : 67License : Permissive (MIT License)
Copydef find_optimal_binary_search_tree(nodes):
"""
This function calculates and prints the optimal binary search tree.
The dynamic programming algorithm below runs in O(n^2) time.
Implemented from CLRS (Introduction to Algorithms) book.
private void doGraphFrameAlgorithms(GraphFrame graph) {
graph.pageRank().maxIter(20).resetProbability(0.15).run().vertices().show();
graph.connectedComponents().run().show();
graph.triangleCount().run().show();
} public Stream getAllAlgorithms() {
return RandomGeneratorFactory.all().map(RandomGeneratorFactory::name);
} Community Discussions
Trending Discussions on algorithms
QUESTION
I am a computer science student; I am studying the Algorithms course independently.
During the course, I saw this question:
Given an n-bit integer N, find a polynomial (in n) time algorithm that decides whether N is a power (that is, there are integers a and k > 1 so that a^k = N).
I thought of a first option that is exponential in n: For all k , 1
For example, if N = 27, I will start with k = 2 , because 2 doesn't divide 27, I will go to next k =3. I will divide 27 / 3 to get 9, and divide it again until I will get 1. This is not a good solution because it is exponential in n.
My second option is using Modular arithmetic, using ak ≡ 1 mod (k+1) if gcd(a, k+1 ) = 1 (Euler's theorem). I don't know if a and k are relatively prime.
I am trying to write an algorithm, but I am struggling to do it:
...ANSWER
Answered 2022-Mar-15 at 10:07Ignoring the cases when N is 0 or 1, you want to know if N is representable as a^b for a>1, b>1.
If you knew b, you could find a in O(log(N)) arithmetic operations (using binary search). Each arithmetic operation including exponentiation runs in polynomial time in log(N), so that would be polynomial too.
It's possible to bound b: it can be at most log_2(N)+1, otherwise a will be less than 2.
So simply try each b from 2 to floor(log_2(N)+1). Each try is polynomial in n (n ~= log_2(N)), and there's O(n) trials, so the resulting time is polynomial in n.
QUESTION
For example, given a List of Integer List list = Arrays.asList(5,4,5,2,2), how can I get a maxHeap from this List in O(n) time complexity?
The naive method:
...ANSWER
Answered 2021-Aug-28 at 15:09According to the java doc of PriorityQueue(PriorityQueue)
Creates a PriorityQueue containing the elements in the specified priority queue. This priority queue will be ordered according to the same ordering as the given priority queue.
So we can extend PriorityQueue as CustomComparatorPriorityQueue to hold the desired comparator and the Collection we need to heapify. Then call new PriorityQueue(PriorityQueue) with an instance of CustomComparatorPriorityQueue.
Below is tested to work in Java 15.
QUESTION
I am learning how to write a Maximum Likelihood implementation in Julia and currently, I am following this material (highly recommended btw!).
So the thing is I do not fully understand what a closure is in Julia nor when should I actually use it. Even after reading the official documentation the concept still remain a bit obscure to me.
For instance, in the tutorial, I mentioned the author defines the log-likelihood function as:
...ANSWER
Answered 2022-Feb-03 at 18:34In the context you ask about you can think that closure is a function that references to some variables that are defined in its outer scope (for other cases see the answer by @phipsgabler). Here is a minimal example:
QUESTION
I have a Python 3 application running on CentOS Linux 7.7 executing SSH commands against remote hosts. It works properly but today I encountered an odd error executing a command against a "new" remote server (server based on RHEL 6.10):
encountered RSA key, expected OPENSSH key
Executing the same command from the system shell (using the same private key of course) works perfectly fine.
On the remote server I discovered in /var/log/secure that when SSH connection and commands are issued from the source server with Python (using Paramiko) sshd complains about unsupported public key algorithm:
userauth_pubkey: unsupported public key algorithm: rsa-sha2-512
Note that target servers with higher RHEL/CentOS like 7.x don't encounter the issue.
It seems like Paramiko picks/offers the wrong algorithm when negotiating with the remote server when on the contrary SSH shell performs the negotiation properly in the context of this "old" target server. How to get the Python program to work as expected?
Python code
...ANSWER
Answered 2022-Jan-13 at 14:49Imo, it's a bug in Paramiko. It does not handle correctly absence of server-sig-algs extension on the server side.
Try disabling rsa-sha2-* on Paramiko side altogether:
QUESTION
I'm trying to detect angle difference between two circular objects, which be shown as 2 image below.
I'm thinking about rotate one of image with some small angle. Every time one image rotated, SSIM between rotated image and the another image will be calculated. The angle with maximum SSIM will be the angle difference.
But, finding the extremes is never an easy problem. So my question is: Are there another algorithms (opencv) can be used is this case?
EDIT:
Thanks @Micka, I just do the same way he suggest and remove black region like @Yves Daoust said to improve processing time. Here is my final result:
ORIGINAL IMAGE ROTATED + SHIFTED IMAGE
...ANSWER
Answered 2021-Dec-15 at 09:19Here's a way to do it:
- detect circles (for the example I assume circle is in the image center and radius is 50% of the image width)
- unroll circle images by polar coordinates
- make sure that the second image is fully visible in the first image, without a "circle end overflow"
- simple template matching
Result for the following code:
QUESTION
(Solution has been found, please avoid reading on.)
I am creating a pixel art editor for Android, and as for all pixel art editors, a paint bucket (fill tool) is a must need.
To do this, I did some research on flood fill algorithms online.
I stumbled across the following video which explained how to implement an iterative flood fill algorithm in your code. The code used in the video was JavaScript, but I was easily able to convert the code from the video to Kotlin:
https://www.youtube.com/watch?v=5Bochyn8MMI&t=72s&ab_channel=crayoncode
Here is an excerpt of the JavaScript code from the video:
Converted code:
...ANSWER
Answered 2021-Dec-29 at 08:28I think the performance issue is because of expandToNeighbors method generates 4 points all the time. It becomes crucial on the border, where you'd better generate 3 (or even 2 on corner) points, so extra point is current position again. So first border point doubles following points count, second one doubles it again (now it's x4) and so on.
If I'm right, you saw not the slow method work, but it was called too often.
QUESTION
Imagine there's have an array of integers but you aren't allowed to access any of the values (so no Arr[i] > Arr[i+1] or whatever). The only way to discern the integers from one another is by using a query() function: this function takes a subset of elements as inputs and returns the number of unique integers in this subset. The goal is to partition the integers into groups based on their values — integers in the same group should have the same value, while integers in different groups have different values.
The catch - the code has to be O(nlog(n)), or in other words the query() function can only be called O(nlog(n)) times.
I've spent hours optimizing different algorithms in Python, but all of them have been O(n^2). For reference, here's the code I start out with:
...ANSWER
Answered 2021-Nov-04 at 14:49Let's say you have an element x and an array of distinct elements, A = [x0, x1, ..., x_{k-1}] and want to know if the x is equivalent to some element in the array and if yes, to which element.
What you can do is a simple recursion (let's call it check-eq):
- Check if
query([x, A]) == k + 1. If yes, then you know thatxis distinct from every element inA. - Otherwise, you know that
xis equivalent to some element ofA. LetA1 = A[:k/2], A2 = A[k/2+1:]. Ifquery([x, A1]) == len(A1), then you know thatxis equivalent to some element inA1, so recurse inA1. Otherwise recurse inA2.
This recursion takes at most O(logk) steps. Now, let our initial array be T = [x0, x1, ..., x_{n-1}]. A will be an array of "representative" of the groups of elements. What you do is first take A = [x0] and x = x1. Now use check-eq to see if x1 is in the same group as x0. If no, then let A = [x0, x1]. Otherwise do nothing. Proceed with x = x2. You can see how it goes.
Complexity is of course O(nlogn), because check-eq is called exactly n-1 times and each call take O(logn) time.
QUESTION
Given a sorted array like [1,2,4,4,4]. Each element in the array should occur exactly same number of times as element. For Example 1 should occur 1 time, 2 should occur 2 times, n should occur n times, After minimum number of moves, the array will be rendered as [1,4,4,4,4]
ANSWER
Answered 2021-Sep-24 at 19:58The error is in this comparison:
QUESTION
In C++20 we got concepts for iterator categories, e.g. std::forward_iterator corresponds to named requirement ForwardIterator.
They are not equivalent. In some (but not all) regards, the concepts are weaker:
(1) "Unlike the [ForwardIterator or stricter] requirements, the [corresponding] concept does not require dereference to return an lvalue."
(2) "Unlike the [InputIterator] requirements, the
input_iteratorconcept does not requireequality_comparable, since input iterators are typically compared with sentinels."...
The new std::ranges algorithms seem to use the concepts to check iterator requirements.
But what about the other standard algorithms? (that existed pre-C++20)
Do they still use the same iterator requirements in C++20 as they were pre-C++20, or did the requirements for them get relaxed to match the concepts?
...ANSWER
Answered 2021-Sep-24 at 18:17Do they still use the same iterator requirements in C++20 as they were pre-C++20, or did the requirements for them get relaxed to match the concepts?
The existing std:: algorithms still use the named requirements. For instance, [alg.find] has both:
QUESTION
Given is a set S of n axis-aligned cubes. The task is to find the volume of the union of all cubes in S. This means that every volume-overlap of 2 or more cubes is only counted once. The set specifically contains all the coordinates of all the cubes.
I have found several papers regarding the subject, presenting algorithms to complete the task. This paper for example generalizes the problem to any dimension d rather than the trivial d=3, and to boxes rather than cubes. This paper as well as a few others solve the problem in time O(n^1.5) or slightly better. Another paper which looks promising and is specific to 3d-cubes is this one which solves the task in O(n^4/3 log n). But the papers seem rather complex, at least for me, and I cannot follow them clearly.
Is there any relatively simple pseudocode or procedure that I can follow to implement this idea? I am looking for a set of instructions, what exactly to do with the cubes. Any implementation will be also excellent. And O(n^2) or even O(n^3) are fine.
Currently, my approach was to compute the total volume, i.e the sum of all volumes of all the cubes, and then compute the overlap of every 2 cubes, and reduce it from the total volume. The problem is that every such overlap may (or may not) be of a different pair of cubes, meaning an overlap can be for example shared by 5 cubes. In this approach the overlap will be counted 10 times rather just once. So I was thinking maybe an inclusion-exclusion principle may prove itself useful, but I don't know exactly how it may be implemented specifically. Computing every overlap (naively) is already O(n^2), but doable. Is there any better way to compute all such overlaps?
Anyways, I don't assume this is a useful approach, to continue along these lines.
ANSWER
Answered 2021-Sep-10 at 21:41Here's some Python (sorry, didn't notice the Java tag) implementing user3386109's suggestion. This algorithm is O(n³ log n). We could get down to O(n³) by sorting the events for all cubes once and extracting the sorted sub-sequence that we need each time, but perhaps this is good enough.
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