dynamic-programming

How to convert recursive “divide and conquer” function to a dynamic programming function using decorators?

浪子不回头ぞ 提交于 2020-01-16 18:54:37
问题 I am trying to write a decorator function which converts a pure recursive function with two arguments that uses a "divide and conquer" strategy to an equivalent but more efficient one using dynamic programming. Note: it is designated to decorate two input functions. So I am trying to memoize the values but I am not sure how to correctly implement it in the form of a decorator? Also how can it decorate two input functions? EDIT: This is what I have managed to do: profile_results = {} t = {} ''

How to convert recursive “divide and conquer” function to a dynamic programming function using decorators?

邮差的信 提交于 2020-01-16 18:53:13
问题 I am trying to write a decorator function which converts a pure recursive function with two arguments that uses a "divide and conquer" strategy to an equivalent but more efficient one using dynamic programming. Note: it is designated to decorate two input functions. So I am trying to memoize the values but I am not sure how to correctly implement it in the form of a decorator? Also how can it decorate two input functions? EDIT: This is what I have managed to do: profile_results = {} t = {} ''

How to convert recursive “divide and conquer” function to a dynamic programming function using decorators?

谁都会走 提交于 2020-01-16 18:53:05
问题 I am trying to write a decorator function which converts a pure recursive function with two arguments that uses a "divide and conquer" strategy to an equivalent but more efficient one using dynamic programming. Note: it is designated to decorate two input functions. So I am trying to memoize the values but I am not sure how to correctly implement it in the form of a decorator? Also how can it decorate two input functions? EDIT: This is what I have managed to do: profile_results = {} t = {} ''

Why is the time complexity of this problem only consider the previous recursive call and not the entire problem?

我怕爱的太早我们不能终老 提交于 2020-01-16 09:07:10
问题 Here we have a box that is 4 * 7 and it can be filled with rectangles that are either 1 * 2 or 2 * 1. This depiction is from the book Competitive Programmer's Handbook . To solve this problem most efficiently, the book mentions using the parts that can be in a particular row: Since there are 4 things in this set, the maximum unique rows we can have is 4^m, where m is the number of columns. From each constructed row, we construct the next row such that it is valid. Valid means we cannot have

Parallelize or vectorize all-against-all operation on a large number of matrices?

放肆的年华 提交于 2020-01-16 04:08:12
问题 I have approximately 5,000 matrices with the same number of rows and varying numbers of columns (20 x ~200). Each of these matrices must be compared against every other in a dynamic programming algorithm. In this question, I asked how to perform the comparison quickly and was given an excellent answer involving a 2D convolution. Serially, iteratively applying that method, like so list = who('data_matrix_prefix*') H = cell(numel(list),numel(list)); for i=1:numel(list) for j=1:numel(list) if i

Perfect Square in Leetcode

て烟熏妆下的殇ゞ 提交于 2020-01-14 03:24:20
问题 I am having trouble understanding one of a Leetcode Problem. Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n. For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9. Solution: int numSquares(int n) { static vector<int> dp {0}; while (dp.size() <= n) { int m = dp.size(), squares = INT_MAX; for (int i=1; i*i<=m; ++i) squares = min(squares, dp[m-i*i] + 1); dp.push_back

Can a convolution function written in tail recursive form?

旧巷老猫 提交于 2020-01-13 18:23:05
问题 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)

Minimum sum partition of an array

。_饼干妹妹 提交于 2020-01-13 05:50:36
问题 Problem Statement: Given an array, the task is to divide it into two sets S1 and S2 such that the absolute difference between their sums is minimum. Sample Inputs , [1,6,5,11] => 1 . The 2 subsets are {1,5,6} and {11} with sums being 12 and 11 . Hence answer is 1 . [36,7,46,40] => 23 . The 2 subsets are {7,46} and {36,40} with sums being 53 and 76 . Hence answer is 23 . Constraints 1 <= size of array <= 50 1 <= a[i] <= 50 My Effort: int someFunction(int n, int *arr) { qsort(arr, n, sizeof(int

Longest K Sequential Increasing Subsequences

最后都变了- 提交于 2020-01-12 05:26:09
问题 Why I created a duplicate thread I created this thread after reading Longest increasing subsequence with K exceptions allowed. I realised that the person who was asking the question hadn't really understood the problem, because he was referring to a link which solves the "Longest Increasing sub-array with one change allowed" problem. So the answers he got were actually irrelevant to LIS problem. Description of the problem Suppose that an array A is given with length N . Find the longest

8-queen problem using Dynamic programming

こ雲淡風輕ζ 提交于 2020-01-12 01:43:07
问题 I am quite confused with idea of implementing 8-queen problem using dynamic programming. It seems it is not possible at one end as for DP " if the problem was broken up into a series of subproblems and the optimal solution for each subproblem was found, then the resulting solution would be realized through the solution to these subproblems. A problem that does not have this structure cannot be solved with dynamic programming" (Reference). By taking this in account, the optimal solution for