dynamic-programming

I came across this question: There are two persons. There is an ordered sequence of n cities, and the distances between every pair of cities is given. You must partition the cities into two subsequences (not necessarily contiguous) such that person A visits all cities in the first subsequence (in order), person B visits all cities in the second subsequence (in order), and such that the sum of the total distances travelled by A and B is minimized. Assume that person A and person B start initially at the first city in their respective subsequences. I looked for its answer and the answer was: Let

I have a function that I want to write in tail recursive form. The function calculates the number of ways to get the sum of k by rolling an s sided die n times. I have seen the mathematical solution for this function on this answer . It is as follows: My reference recursive implementation in R is: sum_ways <- function(n_times, k_sum, s_side) { if (k_sum < n_times || k_sum > n_times * s_side) { return(0) } else if (n_times == 1) { return(1) } else { sigma_values <- sapply( 1:s_side, function(j) sum_ways(n_times - 1, k_sum - j, s_side) ) return(sum(sigma_values)) } } I have tried to re-write the

I'm a beginner, and I'm trying to write a working travelling salesman problem using dynamic programming approach. This is the code for my compute function: public static int compute(int[] unvisitedSet, int dest) { if (unvisitedSet.length == 1) return distMtx[dest][unvisitedSet[0]]; int[] newSet = new int[unvisitedSet.length-1]; int distMin = Integer.MAX_VALUE; for (int i = 0; i < unvisitedSet.length; i++) { for (int j = 0; j < newSet.length; j++) { if (j < i) newSet[j] = unvisitedSet[j]; else newSet[j] = unvisitedSet[j+1]; } int distCur; if (distMtx[dest][unvisitedSet[i]] != -1) { distCur =

The cost of a bond on each day is given in array prices of length n , and I need to find the maximum profit that I can make by buying and selling in exactly k transactions (buying and selling, in that order. not in the same day. but I can sell and then buy in the same day). I tried (Python): prices = [3, 1, 10] n = len(prices) def aux(i, j): if j == n - 1 or i == 0: return 0 s = [(prices[j + t] - prices[j]) + aux(i - 1, j + t) for t in range(1, n - j)] return max(aux(i, j + 1), max(s)) if s else aux(i, j + 1) def max_profit(k): return aux(k, 0) But for the given array in the code, and with k=2

There are n houses at locations a_1, a_2,..., a_n along a line. We want to set up porta potties along that same line so that every house is within distance R of at least one porta potty. These porta potties are restricted to the specified locations b_1, b_2,..., b_m. Let c_i be the cost of setting up a porta potty at location b_i. Find a dynamic programming algorithm that minimizes the total cost of setting up the porta potties. The algorithm should be able to detect if a solution does not exist. Assume that all a and b values are distinct. Inputs: A[1, 2,...n] holds the house locations B[1, 2

The subarray contains both positive and negative numbers. You have to find a maximum sum subarray such that the length of the sub-array is greater than or equal to k. Here is my code in c++ using Kadane's algorithm. #include <iostream> using namespace std; int main(){ int n,k; cin >> n >> k; int array[n]; int sum = 0; int maxsum = 0; int beststarts[n]; for(int i = 0;i < n; i++){ cin >> array[i]; } for(int i = 0;i < k-1;i ++){ sum = sum+array[i]; beststarts[i] = 0; } for(int i = k-1;i < n; i++){ //best end search with min length; sum = sum+array[i]; int testsum = sum; if(i > 0){ beststarts[i] =