n can be arbitrarily large Well, n can’t be arbitrarily large – if n >= m, then n! ≡ 0 (mod m) (because m is one of the factors, by the definition of factorial). Assuming n << m and you need an exact value, your algorithm can’t get any faster, to my knowledge. However, if … Read more
The length of a year is (more or less) 365.242196 days. So we have to subtract, more or less, a quarter of a day to make it fit : 365.242196 – 0.25 = 364.992196 (by adding 1 day in 4 years) : but oops, now it’s too small!! lets add a hundreth of a day … Read more
You can use the matrix exponentiation method (linear recurrence method). You can find detailed explanation and procedure in this or this blog. Run time is O(log n). I don’t think there is a better way of doing this.
Wikipedia has a nice article on lexicographical order generation. It also describes an algorithm to generate the next permutation. Quoting: The following algorithm generates the next permutation lexicographically after a given permutation. It changes the given permutation in-place. Find the highest index i such that s[i] < s[i+1]. If no such index exists, the permutation … Read more
O(log log n) terms can show up in a variety of different places, but there are typically two main routes that will arrive at this runtime. Shrinking by a Square Root As mentioned in the answer to the linked question, a common way for an algorithm to have time complexity O(log n) is for that … Read more
Store IDs of the objects in a hash table mapping to the specific object. Enumerate through all the objects and find their parent if it exists and update its parent pointer accordingly. class MyObject { // The actual object public int ParentID { get; set; } public int ID { get; set; } } class … Read more
You could average the RGB values of random colors with those of a constant color: (example in Java) public Color generateRandomColor(Color mix) { Random random = new Random(); int red = random.nextInt(256); int green = random.nextInt(256); int blue = random.nextInt(256); // mix the color if (mix != null) { red = (red + mix.getRed()) / … Read more
According to Pippenger [1996], when comparing a Lisp system that is purely functional (and has strict evaluation semantics, not lazy) to one that can mutate data, an algorithm written for the impure Lisp that runs in O(n) can be translated to an algorithm in the pure Lisp that runs in O(n log n) time (based … Read more
you should also add the input in vector form so we can test on your data … I would approach like this: find center of axis aligned bbox O(n) compute max distance in each angle O(n) just create table for enough m angles (like 5 deg step so m = 360/5) where for each angle … Read more
Perhaps you should check out (PDF warning! Also note that, for some reason, the order of the pages is reversed) “Optimal Algorithms for Computing the Minimum Distance Between Two Finite Planar Sets” by Toussaint and Bhattacharya: It is shown in this paper that the minimum distance between two finite planar sets if [sic] n points … Read more