complexity-theory

procedure matrixvector(n:integer); var i,j:integer; begin for i<-1 to n do begin B[i] = 0; C[i] = 0; for j<-1 to i do B[i]<- B[i]+ A[i,j]; for j<-n down to i+1 do C[i]<-C[i] + A[i,j] end end; O(n^2), if I read it right. Why you need two inner loops is beyond me. Why not sum B and C in the same loop? sven worst case is O(n²). there are indeed three loops, but not all inside each other, thus giving O(n²). also, you can clearly see that the inner loops won't go from 1 to n (like the outer loop does). But because this would only change the time complexity by some constant, we can ignore this and

So I am trying to write a simple method which takes in set of four coordinates and decide whether they form a square or not.My approach is start with a point and calculate the distance between the other three points and the base point.From this we can get the two sides which have same value and the one which is a diagonal.Then I use Pythagoras theorem to find if the sides square is equal to the diagonal.If it is the isSquare method return true else false.The thing I want to find out is there some cases I might be missing out on or if something is wrong with the approach.Thanks for the all the

This is a short problem from edx's course Introduction to Computer Science and Programming using Python. def program1(x): total = 0 for i in range(1000): total += i while x > 0: x -= 1 total += x return total Question : What is the number of steps it will take to run Program 1 in the best case? Express your answer in terms of n, the size of the input x Answer : Best case : 3003 and Worst Case : 5n+3003 I am confused with the answer 3003 because according to me the problem's best case would be if x =-1 and other statements execute which have a constant amount of time .hence statement total =0 /

I am refreshing on Master Theorem a bit and I am trying to figure out the running time of an algorithm that solves a problem of size n by recursively solving 2 subproblems of size n-1 and combine solutions in constant time. So the formula is: T(N) = 2T(N - 1) + O(1) But I am not sure how can I formulate the condition of master theorem. I mean we don't have T(N/b) so is b of the Master Theorem formula in this case b=N/(N-1) ? If yes since obviously a > b^k since k=0 and is O(N^z) where z=log2 with base of (N/N-1) how can I make sense out of this? Assuming I am right so far? ah, enough with the

I use SonarLint with Eclipse , and I'm codding an application using AngularJS . I had a problem with a controller so I was trying to clean it a bit to see clearer, and then SonarLint popped me up an issue : Function has a complexity of 11 which is greater than 10 authorized. And here's the code of my controller : app.controller('LauncherCtrl', function ($scope, $http) { $scope.genStatus = "stopped"; $scope.startgenerator = function() { $http.get('/start').success(function () { $scope.updateStatus(); }); }; $scope.resumegenerator = function() { $http.get('/resume').success(function () { $scope