backtracking

Is there a way that I can know how long the apps in the foreground has been running? I have three possible solutions in mind: I. use battery consumption and battery consumption rate (iOS 8 and later tell you the battery usage of the app, but the batter consumption will be difficult to handle) II. use system processes monitor III. use Apple's diagnostics logs. This approach is quite "backdoor." Plus I am not sure if Apple allows us to use the information or not. Can someone tell me if any of the above solution is realistic? If not, I want to know is it possible to find out the duration of a

def paren(n): lst = ['(' for x in range(n)] current_string = ''.join(lst) solutions = list() for i in range(len(current_string)+1): close(current_string, n, i, solutions) return solutions def close(current_string, num_close_parens, index, solutions): """close parentheses recursively""" if num_close_parens == 0: if current_string not in solutions: solutions.append(current_string) return new_str = current_string[:index] + ')' + current_string[index:] if num_close_parens and is_valid(new_str[:index+1]): return close(new_str, num_close_parens-1, index+1, solutions) else: return close(current

I have a problem that is seemingly solvable by enumerating all possible solutions and then finding the best. In order to do so, I devised a backtracking algorithm that enumerates and stores the best solution if found. It works fine so far. Now, I wanted to port this algorithm to CUDA. Therefore, I created a procedure that generates some distinct basic cases. These basic cases should be processed in parallel on the GPU. If one of the CUDA-threads finds an optimal solution, all the other threads can - of course - stop their work. So, I wanted kind of the following: The thread that finds the

def solve(n): #prepare a board board = [[0 for x in range(n)] for x in range(n)] #set initial positions place_queen(board, 0, 0) def place_queen(board, row, column): """place a queen that satisfies all the conditions""" #base case if row > len(board)-1: print board #check every column of the current row if its safe to place a queen while column < len(board): if is_safe(board, row, column): #place a queen board[row][column] = 1 #place the next queen with an updated board return place_queen(board, row+1, 0) else: column += 1 #there is no column that satisfies the conditions. Backtrack for c in

So I've been trying to solve this assignment whole day, just can't get it. The following function accepts 2 strings, the 2nd (not 1st) possibly containing * 's (asterisks). An * is a replacement for a string (empty, 1 char or more), it can appear appear (only in s2) once, twice, more or not at all, it cannot be adjacent to another * ( ab**c ), no need to check that. public static boolean samePattern(String s1, String s2) It returns true if strings are of the same pattern. It must be recursive , not use any loops, static & global variables. Can use local variables & method overloading. Can use

I'm reviewing a programming problem from a local programming contest. You can download the problem here (pdf). It's in dutch but the pictures will help to understand it. You receive a m*m grid as input with some pipes and some missing spots (the questionmarks). The remaining pipes have to be placed in the grid so that they connect with the others. Each pipe is represented as a letter (see picture on page 2). The letter 'A' has value 1, 'B' has value 2, .. I tried solving it with backtracking (it doesn't quite work yet). But since the grid can be 10x10 this will be too slow. Can someone suggest