The following algorithm can me implemented with O(N+M) complexity. Take any vertex u. Use flood fill algorithm to reach other vertices. If any vertex is not reachable, return NOK. Now do the same, but go to the opposite direction. If any vertex is not reachable, return NOK. Return OK. (Here we know that there is … Read more
As you only return h, you don’t actually need t. It is also not needed to use randrange — you can achieve the same with random. For the actual counting you can use sum, and the boolean result can be converted to 0 or 1 with int: def prob(times): return sum(int(rd.random() < 0.7) for _ … Read more
From the modulo laws on DAle’s linked Wikipedia page (on your previous question), we can obtain two formulas: From the first formula it is clear that we can iteratively calculate the modulo for n from the result for n / 2. This is done by the lines x = x * x; x = x … Read more
Yes. Algorithms that do that exist. You need to use kernel or mask on an image in order to transform it: http://en.wikipedia.org/wiki/Kernel_%28image_processing%29 Many libraries have that built in, for example openCV. solved algorithms to detect lines in an image?
The problem is in this line: if(p2 !== p1) { In your use of the function, p1 and p2 are arrays with x and y coordinates. Such a comparison is only true when p1 and p2 are the same array reference. This is never the case in your use cases. You need this (as is … Read more
It sounds like you’re having trouble reading two numbers in at once (the weight and destination). Are you ready to have your mind blown? You can chain the input operator! That means that you can do: int dest, weight; while( is >> dest >> weight ) { edge++; E[edge].beg = row; E[edge].end = dest; E[edge].weight … Read more
You can find the least common multiple using prime factorization. Here, lcm(a, b) has to contain all the prime factors of a and b in their highest multiple they appear in in any of the two numbers. For example, 8 = 2^3 and 12 = 2^2 * 3, so lcm(8, 12) = 2^3 * 3 … Read more
Face detection topic is really wide and the problem is quite complex, but in a few words I will try to describe it. In digital image processing you have to treat image as an two-dimensional array of pixels. If you proceed RGB image each pixel contains 3 values (from r,g,b channels). For the simple solution … Read more
Set L = sum of lengths While L > 0 Get largest box B where size S <= L OR smallest box set numB = Floor division L / B OR 1 store numB of B Subtract numB * B from L you’ll need some adjustments for a case where you might have waste (ie … Read more
Construct a directed graph of containing all characters as vertices. Create an edge from each character to each character directly following that character in any word. For example, if you have a word @ % ^, you’d have edges @ -> % and % -> ^. Run a topological sort on the graph to get … Read more
You only need to carefully read the section B. Proposed methodology. It says something like this: Now our aim is to find the best set of values for these four parameters which can produce the optimal result and to perform this work PSO is used. P number of particles are initialized, each with four parameters … Read more
What is length? length does not exist, but Node.length does. The reason return(1+length(self.next)) doesn’t work is because you’re calling length(), not Node.length(). Instead, try this: return(1+LinkedList.length(self.next)) Or this (my preference): return(1+self.next.length()) solved Why isn’t length(self.next) a valid expression in python? [closed]
You can impose a tree structure on an array. The leaves of the tree are the array itself. The next level up has one node for each pair of array elements. The next level up has four descendants for each node, and so on… 15 13 14 9 10 11 12 1 2 3 4 … Read more
You will obviously compute the partial sums X.n and stop when X.n<4 and X.n+1>4. To compute the partial sums, keep an accumulator variable and add the fractions one after the other n= 0 S= 0 // Repeat the following instructions n+= 1 S+= 1/(n+1) // Now, S = X.n Remains to find the stopping condition. … Read more
Clearly, O(n) time-complexity, as i iterates from 0 to n-1. solved What is the running time of this algorithm? [closed]