divide-and-conquer

问题 enter image description here “⊕” is the bitwise XOR operation. I think Karatsuba’s algorithm may be used to solve the problem, but when I try to use XOR instead of "+" in the Karatsuba’s algorithm, it is tough to get the sub-problem. 回答1: The convolution theorem gives you F(C) = F(A) . F(B) where F is a Fourier-related transform, in this case the Hadamard transform, and . is point-wise multiplication. Using the fast Walsh–Hadamard transform, you can compute F(A) , F(B) , and finally C (using

问题 There are two key attributes that a problem must have in order for dynamic programming to be applicable: optimal substructure and overlapping subproblems [1]. For this question, we going to focus on the latter property only. There are various definitions for overlapping subproblems , two of which are: A problem is said to have overlapping subproblems if the problem can be broken down into subproblems which are reused several times OR a recursive algorithm for the problem solves the same

问题 I have a homework problem which is really messing with me right now, and I could use some help in how to implement it in DrRacket. I do not wish for code, just guidance, as I am very new to DrRacket. The assignment is to implement this phrase: "If n = 0, return the empty tiling (list of tile structs). Otherwise, place a tromino (L-shaped domino) in the center of the chessboard so that it covers the three quadrants of the chessboard that have no missing tile in them, and tile each of the

问题 Problem statement: // m is the number, n is upto-length of subsequences // m = 20125, n =3 should print 201, 202, 205, 212, 215, 225, 012, 015, 125 // m = 20125, n =2 should print 20, 21, 22, 25, 01, 02, 05, 12, 15, 25 // m = 20125, n =1 should print 2, 0, 1, 2, 5 // m = 20125, n =4 should print 2012, 2015, 2125, 0125, 2025 // m = 20125, n =5 should print 20125 Below is the recursive solution implemented in GoLang: package recursion import ( "fmt" "strconv" ) // m is the number, n is upto

问题 Problem statement: // m is the number, n is upto-length of subsequences // m = 20125, n =3 should print 201, 202, 205, 212, 215, 225, 012, 015, 125 // m = 20125, n =2 should print 20, 21, 22, 25, 01, 02, 05, 12, 15, 25 // m = 20125, n =1 should print 2, 0, 1, 2, 5 // m = 20125, n =4 should print 2012, 2015, 2125, 0125, 2025 // m = 20125, n =5 should print 20125 Below is the recursive solution implemented in GoLang: package recursion import ( "fmt" "strconv" ) // m is the number, n is upto

问题 I'm trying to implement in C++ the divide and conquer algorithm of finding the convex hull from a set of two dimensional points. For simplicity let's assume that all the points are described with integers. The most important part of the algorithm is merging the two convex hulls that you have computed from previous recursive calls. This part involves finding the lower and upper tangents of the two convex hulls and then proceeding with the merging. The merging is trivial, if you have found the

问题 For master's theorem T(n) = a*T(n/b) + f(n) I am using 3 cases: If a*f(n/b) = c*f(n) for some constant c > 1 then T(n) = (n^log(b) a) If a*f(n/b) = f(n) then T(n) = (f(n) log(b) n) If a*f(n/b) = c*f(n) for some constant c < 1 then T(n) = (f(n)) But when f(n) = log n or n*log n , the value of c is dependent on value of n. How do I solve the recursive function using master's theorem? 回答1: You might find these three cases from the Wikipedia article on the Master theorem a bit more useful: Case 1

问题 I'm trying to solve the famous skyline problem (see gif): Input (1,11,5), (2,6,7), (3,13,9), (12,7,16), (14,3,25), (19,18,22), (23,13,29), (24,4,28) Should return, the points that are behind other buildings should be gone and the coordinates of changes in the Y-axis should be in the returning skyline: (1, 11), (3, 13), (9, 0), (12, 7), (16, 3), (19, 18), (22, 3), (23, 13), (29, 0) I'm trying to do so by using a divide and conquer approach to the algorithm as to achieve a running time of O(n

问题 Suppose we have two arrays A[n] and b[n], the goal is to compare every element in A to elements in B. Then return a list result[n] that records the number of each element in A that is larger than the elements in B. For example, A = [38, 24, 43, 3], B = [9, 82, 10, 11] Since 38 is larger than 9, 10 and 11, so result[0] is 3. Then result is [3, 3, 3, 0]. It will be the best if you can provide some pseudocode. Thank you. 回答1: You can perform the above algorithm in O(nlogn) complexity where n is

问题 I have been following Coursera's course on Algorithms and came up with a thought about the divide/conquer algorithm for the closest pair problem, that I want clarified. As per Prof Roughgarden's algorithm (which you can see here if you're interested): For a given set of points P, of which we have two copies - sorted in X and Y direction - Px and Py, the algorithm can be given as closestPair(Px,Py): Divide points into left half - Q, and right half - R, and form sorted copies of both halves