# Algorithm | : pencil2 : Useful Algorithm Implementation | Learning library

## kandi X-RAY | Algorithm Summary

## kandi X-RAY | Algorithm Summary

基于《算法 第四版》和《数据结构与算法分析 Java 语言描述》进行实现，包含 JUnit 测试代码。.

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### Top functions reviewed by kandi - BETA

Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of Algorithm

## Algorithm Key Features

## Algorithm Examples and Code Snippets

```
const bubbleSort = arr => {
let swapped = false;
const a = [...arr];
for (let i = 1; i < a.length; i++) {
swapped = false;
for (let j = 0; j < a.length - i; j++) {
if (a[j + 1] < a[j]) {
[a[j], a[j + 1]] = [a[j
```

```
const binarySearch = (arr, item) => {
let l = 0,
r = arr.length - 1;
while (l <= r) {
const mid = Math.floor((l + r) / 2);
const guess = arr[mid];
if (guess === item) return mid;
if (guess > item) r = mid - 1;
els
```

```
def pollard_rho(
num: int,
seed: int = 2,
step: int = 1,
attempts: int = 3,
) -> int | None:
"""
Use Pollard's Rho algorithm to return a nontrivial factor of ``num``.
The returned factor may be composite and require fur
```

```
def hill_climbing(
search_prob,
find_max: bool = True,
max_x: float = math.inf,
min_x: float = -math.inf,
max_y: float = math.inf,
min_y: float = -math.inf,
visualization: bool = False,
max_iter: int = 10000,
) -> S
```

```
def boruvka(self) -> None:
"""Performs Borůvka's algorithm to find MST."""
# Initialize additional lists required to algorithm.
component_size = []
mst_weight = 0
minimum_weight_edge: list[Any] = [-1] * se
```

## Community Discussions

Trending Discussions on Algorithm

QUESTION

I got this error when learning Next.js, using `npx create-next-app`

command according to site documentation here https://nextjs.org/docs/api-reference/create-next-app. Everything works until I start the server,

Error stack:

...ANSWER

Answered 2021-Nov-24 at 21:38I found this solution https://github.com/webpack/webpack/issues/14532

if using bash just run

`NODE_OPTIONS=--openssl-legacy-provider`

before any commandadding

`NODE_OPTIONS=--openssl-legacy-provider`

to package.json

QUESTION

I have an array of positive integers. For example:

...ANSWER

Answered 2022-Feb-27 at 22:44This problem has a fun O(n) solution.

If you draw a graph of cumulative sum vs index, then:

The average value in the subarray between any two indexes is the slope of the line between those points on the graph.

The first highest-average-prefix will end at the point that makes the highest angle from 0. The next highest-average-prefix must then have a *smaller* average, and it will end at the point that makes the highest angle from the first ending. Continuing to the end of the array, we find that...

These segments of highest average are exactly the segments in the **upper convex hull of the cumulative sum graph**.

Find these segments using the monotone chain algorithm. Since the points are already sorted, it takes O(n) time.

QUESTION

We have a test environment on a public site. There we use --disable-web-security flag on chrome for the testers to bypass CORS errors for public service calls during manual test phase. And also we have localhost requests on the agent machine. However today with Chrome 98 update we started struggling with the network requests targeting localhost.

The error we get is for the localhost requests from a public site:

`Access to XMLHttpRequest at 'https://localhost:3030/static/first.qjson' from origin 'https://....com' has been blocked by CORS policy: Request had no target IP address space, yet the resource is in address space `local`.`

The site on localhost is configured to return Access-Control-Allow-* CORS headers including "Access-Control-Allow-Private-Network: true".

And also I do not see any preflight request. Just one GET request with CORS error on it.

We suspect this might be a side effect caused when you disable web security by --disable-web-security. It might be preventing obtaining of the target IP address space. Our assumption is based on the CORS preflight section on https://wicg.github.io/private-network-access/

3.1.2. CORS preflight

The HTTP fetch algorithm should be adjusted to ensure that a preflight is triggered for all private network requests initiated from secure contexts.

The main issue here is again that the response’s IP address space is not known until a connection is obtained in HTTP-network fetch, which is layered under CORS-preflight fetch.

So does anyone know any workaround for Private Network Access with --disable-web-security flag ? Or maybe we are missing something. Thanks for the help.

...ANSWER

Answered 2022-Feb-09 at 04:20Below Steps can help to solve issue in chrome 98, for other browser like edge you need to do similar like chrome.

For MACRequestly with chrome version 98. You need to follow following steps :- Run this command on terminal

`defaults write com.google.Chrome InsecurePrivateNetworkRequestsAllowed -bool true`

Restart your Browser, Not work then restart your machine

- Run 'regedit' to open windows registry (If permission issue came then run that command with Admin command prompt)
- Go to Computer\HKEY_LOCAL_MACHINE\SOFTWARE\Policies\Google\Chrome
- Create new DWORD value with "InsecurePrivateNetworkRequestsAllowed" Name
- Change Value to "1"
- Restart your Browser

QUESTION

I was looking for the canonical implementation of MergeSort on Haskell to port to HOVM, and I found this StackOverflow answer. When porting the algorithm, I realized something looked silly: the algorithm has a "halve" function that does nothing but split a list in two, using half of the length, before recursing and merging. So I thought: why not make a better use of this pass, and use a pivot, to make each half respectively smaller and bigger than that pivot? That would increase the odds that recursive merge calls are applied to already-sorted lists, which might speed up the algorithm!

I've done this change, resulting in the following code:

...ANSWER

Answered 2022-Jan-27 at 19:15Your `split`

splits the list in two *ordered* halves, so `merge`

consumes its first argument first and then just produces the second half in full. In other words it is equivalent to `++`

, doing redundant comparisons on the first half which always turn out to be `True`

.

In the true mergesort the merge actually does twice the work on random data because the two parts are not ordered.

The `split`

though spends some work on the partitioning whereas an online bottom-up mergesort would spend no work there at all. But the built-in sort tries to detect ordered runs in the input, and apparently that extra work is not negligible.

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 using a string Encryption/Decryption class similar to the one provided here as a solution.

This worked well for me in .Net 5.

Now I wanted to update my project to .Net 6.

**When using .Net 6, the decrypted string does get cut off a certain point depending on the length of the input string.**

▶️ To make it easy to debug/reproduce my issue, I created a public repro Repository here.

- The encryption code is on purpose in a Standard 2.0 Project.
- Referencing this project are both a .Net 6 as well as a .Net 5 Console project.

Both are calling the encryption methods with the exact same input of `"12345678901234567890"`

with the path phrase of `"nzv86ri4H2qYHqc&m6rL"`

.

.Net 5 output: `"12345678901234567890"`

.Net 6 output: `"1234567890123456"`

The difference in length is `4`

.

I also looked at the breaking changes for .Net 6, but could not find something which guided me to a solution.

I'm glad for any suggestions regarding my issue, thanks!

Encryption Class

...ANSWER

Answered 2021-Nov-10 at 10:25The reason is this breaking change:

DeflateStream, GZipStream, and CryptoStream diverged from typical Stream.Read and Stream.ReadAsync behavior in two ways:

They didn't complete the read operation until either the buffer passed to the read operation was completely filled or the end of the stream was reached.

And the new behaviour is:

Starting in .NET 6, when Stream.Read or Stream.ReadAsync is called on one of the affected stream types with a buffer of length N, the operation completes when:

At least one byte has been read from the stream, or The underlying stream they wrap returns 0 from a call to its read, indicating no more data is available.

In your case you are affected because of this code in `Decrypt`

method:

QUESTION

I am trying to implement a reasonably fast version of Floyd-Warshall algorithm in Rust. This algorithm finds a shortest paths between all vertices in a directed weighted graph.

The main part of the algorithm could be written like this:

...ANSWER

Answered 2021-Nov-21 at 19:55At first blush, one would hope this would be enough:

QUESTION

I am trying to find a more efficient solution to a combinatorics problem than the solution I have already found.

Suppose I have a set of **N** objects (indexed **0..N-1**) and wish to consider each subset of size **K** (**0<=K<=N**). There are **S=C(N,K)** (i.e., "N choose K") such subsets. I wish to map (or "encode") each such subset to a unique integer in the range **0..S-1**.

Using **N=7** (i.e., indexes are **0..6**) and **K=4** (**S=35**) as an example, the following mapping is the goal:

**0 1 2 3 --> 0
0 1 2 4 --> 1
...
2 4 5 6 --> 33
3 4 5 6 --> 34**

**N** and **K** were chosen small for the purposes of illustration. However, in my actual application, **C(N,K)** is far too large to obtain these mappings from a lookup table. They must be computed on-the-fly.

In the code that follows, `combinations_table`

is a pre-computed two-dimensional array for fast lookup of **C(N,K)** values.

All code given is compliant with the **C++14** standard.

If the objects in a subset are ordered by increasing order of their indexes, the following code will compute that subset's encoding:

...ANSWER

Answered 2021-Oct-21 at 02:18Take a look at the recursive formula for combinations:

Suppose you have a combination space `C(n,k)`

. You can divide that space into two subspaces:

`C(n-1,k-1)`

all combinations, where the first element of the original set (of length`n`

) is present`C(n-1, k)`

where first element is not preset

If you have an index X that corresponds to a combination from `C(n,k)`

, you can identify whether the first element of your original set belongs to the subset (which corresponds to `X`

), if you check whether `X`

belongs to either subspace:

`X < C(n-1, k-1)`

: belongs`X >= C(n-1, k-1)`

: doesn't belong

Then you can recursively apply the same approach for `C(n-1, ...)`

and so on, until you've found the answer for all `n`

elements of the original set.

Python code to illustrate this approach:

QUESTION

Some answer originally had this sorting algorithm:

...ANSWER

Answered 2021-Oct-24 at 16:59To prove that it's correct, you have to find some sort of invariant. Something that's true during every pass of the loop.

Looking at it, after the very first pass of the inner loop, the *largest* element of the list will actually be in the *first* position.

Now in the second pass of the inner loop, `i = 1`

, and the very first comparison is between `i = 1`

and `j = 0`

. So, the largest element was in position 0, and after this comparison, it will be swapped to position 1.

In general, then it's not hard to see that after each step of the outer loop, the largest element will have moved one to the right. So after the full steps, we know at least the largest element will be in the correct position.

What about all the rest? Let's say the second-largest element sits at position `i`

of the current loop. We know that the *largest* element sits at position `i-1`

as per the previous discussion. Counter `j`

starts at 0. So now we're looking for the first `A[j]`

such that it's `A[j] > A[i]`

. Well, the `A[i]`

is the second largest element, so the first time that happens is when `j = i-1`

, at the first largest element. Thus, they're adjacent and get swapped, and are now in the "right" order. Now `A[i]`

again points to the largest element, and hence for the rest of the inner loop no more swaps are performed.

So we can say: Once the outer loop index has moved past the location of the second largest element, the second and first largest elements will be in the right order. They will now slide up together, in every iteration of the outer loop, so we know that at the end of the algorithm both the first and second-largest elements will be in the right position.

What about the third-largest element? Well, we can use the same logic again: Once the outer loop counter `i`

is at the position of the third-largest element, it'll be swapped such that it'll be just below the second largest element (if we have found that one already!) or otherwise just below the first largest element.

Ah. And here we now have our invariant: After `k`

iterations of the outer loop, the k-length sequence of elements, ending at position `k-1`

, will be in sorted order:

After the 1st iteration, the 1-length sequence, at position 0, will be in the correct order. That's trivial.

After the 2nd iteration, we know the largest element is at position 1, so obviously the sequence `A[0]`

, `A[1]`

is in the correct order.

Now let's assume we're at step `k`

, so all the elements up to position `k-1`

will be in order. Now `i = k`

and we iterate over `j`

. What this does is basically find the position at which the new element needs to be slotted into the existing sorted sequence so that it'll be properly sorted. Once that happens, the rest of the elements "bubble one up" until now the largest element sits at position `i = k`

and no further swaps happen.

Thus finally at the end of step `N`

, all the elements up to position `N-1`

are in the correct order, QED.

QUESTION

Given a list of 20 float numbers, I want to find a largest subset where any two of the candidates are different from each other larger than a `mindiff = 1.`

. Right now I am using a brute-force method to search from largest to smallest subsets using `itertools.combinations`

. As shown below, the code finds a subset after 4 s for a list of 20 numbers.

ANSWER

Answered 2021-Jun-18 at 18:48I probably don't fully understand the question, because right now the solution is quite trivial. EDIT: yes, I misunderstood after all, the OP does not just want an optimal solution, but wishes to **randomly sample from the set of optimal solutions**. This answer is not incorrect but it also is an answer to a different question than what OP is interested in.

Simply sort the numbers and greedily construct the subset:

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

## Vulnerabilities

No vulnerabilities reported

## Install Algorithm

You can use Algorithm like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the Algorithm component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .

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