recursion

问题 I wrote some code that calculates the determinant of a given nxn matrix using Leibniz formula for determinants. I am trying to figure out it's complexity in O-notation. I think it should be something like: O(n!) * O(n^2) + O(n) = O(n!*n^2) or O((n+2)!) Reasoning: I think O(n!) is the complexity of permutations. and O(n) the complexity of perm_parity, and O(n^2) is the multiplication of n items each iteration. this is my code: def determinant_leibnitz(self): assert self.dim()[0] == self.dim()

问题 In the class I'm taking my teacher is discouraging the students from calling functions recursively. For example, if I want to call the main function in the following code: if (x > 5) { printf("Your number's too high. Enter a number below 5."); main(); } I'm instead encouraged to use: int x; while (x > 5) { printf("\nYour number's too high. Enter a number below 5."); scanf("%d", &x); } While I understand that the while function is a simpler way of going about this particular problem, we're

问题 I have a recursive function living in a module called test_module import requests def send_msg(msg, retries=0): try: # send the message here, e.g. a http request response = requests.get("http://www.doesnotexist98734.com") # if url does not exist raise an exception except Exception as e: if retries == 0: raise e else: return send_msg(msg, retries=retries-1) My question is how can I write a unittest that checks the send_msg function is called n times when I set retries = n. I was playing around

问题 My data structure will look like this: var tree = [ { id: 1, children: [] }, { id: 2, children: [ { id: 3, children: [] } ] } ]; There can be any number of nodes or children on one branch. My goal is to build a path to every node. For example id: 3 will have a path of 1 > 2 > 3 id: 2 will have a path of 1 > 2 I want to run my tree through the algorithm so it will be modified like this: var tree = [ { id: 1, path: [1], children: [] }, { id: 2, path: [2], children: [ { id: 3, path: [2, 3],

问题 I'm trying to understand what exactly is recursion and have not been able to find an answer to the following. My current understanding of recursion is that it is anytime a method calls itself. I.E Menu() { if(i<2) {Console.WriteLine();} else {Menu();} } The above is an example of recursion a method calling itself. What I'm not sure about is a scenario like: Menu() { if(i<2) {Console.WriteLine();} else {Console.WriteLine("Something Went Wrong!"); MenuError();} } MenuError() { Console.WriteLine

问题 I have a program written recursively which, due to its nature, will fluctuate dramatically and thus, go over Python's limit. The shape of the program is also quite important, so I'd rather not change that. I've come across increasing the recursion limit like so: sys.setrecursionlimit(1500) My question is whether or not it is advisable in my situation to increase the recursion limit like this? I've seen people advise against it for numerous reasons. 回答1: sys.setrecursionlimit(1500) is

问题 My goal is a slideshow of background images with HTML/CSS/JS. Many solutions that I've found promote something like this: my_recursion(); function my_recursion () { // cycle the Background image ... setTimeout(my_recursion, 3000); } Am I wrong to assume that this is bad style? I would expect that at e.g. cycle 1000 all the other 999 instances of my_recursion are still open / on the stack? Doesn't this create and infinite stack which consumes more and more memory? Or is there some sort of

问题 Hi I have a number of different files that need to be renamed to something else. I got this far but I want to have it so that I can have many items to replace and their corresponding replacements rather than type each one out, run the code then retype it again. UPDATE* Also I need the rename to only change part of the file not the whole thing so if there was a "Cat5e_1mBend1bottom50m2mBend2top-Aqeoiu31" it would just change it to "'Cat5e50m1mBED_50m2mBE2U-Aqeoiu31" import os, glob #searches

问题 So I want to write a Fibonacci function in scala that outputs a tree like so: fib(3) | fib(2) | | fib(1) | | = 1 | | fib(0) | | = 0 | = 1 | fib(1) | = 1 = 2 and my current code is as follows: var depth: Int = 0 def depthFibonacci(n:Int, depth: Int): Int={ def fibonnaciTailRec(t: Int,i: Int, j: Int): Int = { println(("| " * depth) + "fib(" + t + ")") if (t==0) { println(("| " * depth) + "=" + j) return j } else if (t==1) { println (("| " * depth) + "=" + i) return i } else { depthFibonacci(t-1

问题 #include <iostream> using namespace std; void convertToBinary(unsigned int n) { if (n>0) { convertToBinary(n / 2); cout << n % 2; } } int main(){ unsigned int n; cin >> n; convertToBinary(n); } Here is a function recursion that coverts from decimal to binary, for example If I give this function n equals to 10 the output is 1010 , How come 1 is at the rightmost digit? It would be 0 since 10 % 2 = 0 , then print 0 , so the expected output would be 101, since there's leading a 0 . 回答1: The data