Those are not actual errors, but warnings. They are there because you are trying to divide something by zero. Namely, you are setting M[0] = 0 and then dividing by M[0] (at first iteration, where m = 0) and same for P[0]. The question is, what do you want the first values to be? Maybe a solution would be initialize the … Read more
This link might be helpful to you, as it details the use of the Haversine formula to calculate the distance. Excerpt: This script [in Javascript] calculates great-circle distances between the two points – that is, the shortest distance over the earth’s surface – using the ‘Haversine’ formula.
Because math. 10e1 is 100 and 10e-1 is 1. 10e1 * 10e-1 is 100. 10e2 * 10e-2 is 100. 10e3 * 10e-3 is 100. You can very easily extend this to figure out that 10eN * 10e-N is always going to be 100. If you want actual scientific notation, as in 1 * 10 ^ … Read more
As long as your input is within the half-open interval (0, 1] (not including 0), you are fine. You can’t be too close to zero: So simply checking for exactly zero (maybe as the result of an underflow) should be enough, or alternatively catch the exception and handle it. EDIT: This also holds for the … Read more
Try this: See Ruby’s Enumerable Documentation (note: the 0 base case is needed so that 0 will be returned on an empty array instead of nil)
Let’s imagine we have a truly random hash function that hashes from strings to n-bit numbers. This means that there are 2n possible hash codes, and each string’s hash code is chosen uniformly at random from all of those possibilities. The birthday paradox specifically says that once you’ve seen roughly √(2k) items, there’s a 50% chance … Read more
Initialize your variable with an expression that evaluates to a double rather than an int: Otherwise the expression is done using integer arithmetic (which ends up truncating the result). That truncated result then gets converted to a double.
Let’s forget the math and try to solve this intuitively. First, if we want to map input numbers in the range [0, x] to output range [0, y], we just need to scale by an appropriate amount. 0 goes to 0, x goes to y, and a number t will go to (y/x)*t. So, let’s reduce your problem to the above simpler problem. … Read more
Remember that You can get the upper bound by And you can get the lower bound by doing a similar thing after throwing away the first half of the sum:
If you have N nodes, there are N – 1 directed edges than can lead from it (going to every other node). Therefore, the maximum number of edges is N * (N – 1).