Big-O is an upper bound. Big-Theta is a tight bound, i.e. upper and lower bound. When people only worry about what’s the worst that can happen, big-O is sufficient; i.e. it says that “it can’t get much worse than this”. The tighter the bound the better, of course, but a tight bound isn’t always easy … Read more
This paper has some analysis. Also, from Wikipedia: The most direct competitor of quicksort is heapsort. Heapsort is typically somewhat slower than quicksort, but the worst-case running time is always Θ(nlogn). Quicksort is usually faster, though there remains the chance of worst case performance except in the introsort variant, which switches to heapsort when a bad … Read more
You comment says you are looking for shortest path. This problem is a variation of TSP on a planar graph, and thus is NP-Hard. Possible heuristics function for A* that can solve the problem but is not admissible [thus the path found is not guaranteed to be optimal]: Sum of manhattan distances from all fruits to the agent. You can also use an … Read more
(via Introduction to Programming in Java)
std::string doesn’t contain such function but you could use stand-alone replace function from algorithm header.
Tail-call optimization is where you are able to avoid allocating a new stack frame for a function because the calling function will simply return the value that it gets from the called function. The most common use is tail-recursion, where a recursive function written to take advantage of tail-call optimization can use constant stack space. … Read more
If you were using JGraph, you should give a try to JGraphT which is designed for algorithms. One of its features is visualization using the JGraph library. It’s still developed, but pretty stable. I analyzed the complexity of JGraphT algorithms some time ago. Some of them aren’t the quickest, but if you’re going to implement them on … Read more
and so on you can substitute the value of T(n-1) and T(n-2) in T(n) to get a general idea of the pattern. For base case: => k = n – 1substitute in (1) Which you can see is of Order n2 => O(n2).
The perceptron’s output is the hard limit of the dot product between the instance and the weight. Let’s see how this changes after the update. Since w(t + 1) = w(t) + y(t)x(t), then x(t) ⋅ w(t + 1) = x(t) ⋅ w(t) + x(t) ⋅ (y(t) x(t)) = x(t) ⋅ w(t) + y(t) [x(t) … Read more
From http://www.astrolog.org/labyrnth/algrithm.htm Recursive backtracker: This is somewhat related to the recursive backtracker solving method described below, and requires stack up to the size of the Maze. When carving, be as greedy as possible, and always carve into an unmade section if one is next to the current cell. Each time you move to a new cell, … Read more