backtracking

问题 I have read in Wikipedia and have also Googled it, but I cannot figure out what "Backtracking Algorithm" means. I saw this solution from "Cracking the Code Interviews" and wonder why is this a backtracking algorithm? 回答1: "Backtracking" is a term that occurs in enumerating algorithms. You built a "solution" (that is a structure where every variable is assigned a value). It is however possible that during construction, you realize that the solution is not successful (does not satisfy certain

问题 I have read in Wikipedia and have also Googled it, but I cannot figure out what "Backtracking Algorithm" means. I saw this solution from "Cracking the Code Interviews" and wonder why is this a backtracking algorithm? 回答1: "Backtracking" is a term that occurs in enumerating algorithms. You built a "solution" (that is a structure where every variable is assigned a value). It is however possible that during construction, you realize that the solution is not successful (does not satisfy certain

问题 I was testing two almost identical regexes against a string (on regex101.com), and I noticed that there was a huge difference in the number of steps that they were taking. Here are the two regexes: (Stake: £)(\d+(?:\.\d+)?) (winnings: £)(\d+(?:\.\d+)?) This is the string I was running them against (with modifiers g , i , m , u ): Start Game, Credit: £200.00game num: 1, Stake: £2.00Spinning Reels:NINE SEVEN KINGKING STAR ACEQUEEN JACK KINGtotal winnings: £0.00End Game, Credit: £198Start...

问题 We have a REST API for querying records in a MongoDB. Very simple, something along the following: GET /api/items?q=foo During development, it was convenient to allow regular expressions as the query q . We would simply pass the query parameter to a MongoDB $regex operator and not do any escaping: db.getCollection('items').find({ name: { $regex: req.query.q, $options: 'i' } }); Thus we have a very flexible and convenient way of querying our data. Now, that things are getting “serious” i.e.

问题 I'm trying to implement Unification, but having problems.. already got dozen of examples,but all they do is to muddy the water. I get more confused than enlightened : http://www.cs.trincoll.edu/~ram/cpsc352/notes/unification.html https://www.doc.ic.ac.uk/~sgc/teaching/pre2012/v231/lecture8.html [code below is based on this intro] http://www.cs.bham.ac.uk/research/projects/poplog/paradigms_lectures/lecture20.html#representing https://norvig.com/unify-bug.pdf How can I implement the unification

问题 I'm trying to implement the backtrack method on a new class named BacktrackingBST, which consists of the root node of a Binary Search Tree and a stack. the stack is used to store inserted or deleted nodes from the tree in order to allow us to backtrack said methods. i encounter an issue while trying to backtrack deletion , the problem was that while i was to restore the deleted node back to its original place i failed to come up with a solution on how to get the new node in its place like its

问题 I'm trying to implement the backtrack method on a new class named BacktrackingBST, which consists of the root node of a Binary Search Tree and a stack. the stack is used to store inserted or deleted nodes from the tree in order to allow us to backtrack said methods. i encounter an issue while trying to backtrack deletion , the problem was that while i was to restore the deleted node back to its original place i failed to come up with a solution on how to get the new node in its place like its

问题 I'm trying to implement the backtrack method on a new class named BacktrackingBST, which consists of the root node of a Binary Search Tree and a stack. the stack is used to store inserted or deleted nodes from the tree in order to allow us to backtrack said methods. i encounter an issue while trying to backtrack deletion , the problem was that while i was to restore the deleted node back to its original place i failed to come up with a solution on how to get the new node in its place like its

问题 I have some code here that solves the n queens problem using backtracking in python. When I run it, the odds always take much less time than the evens. This is especially evident when the n gets to around 20+. Does anyone know why this is? import time global N N = int(input("Enter N: ")) def display_solution(board): print('\n'.join(['\t'.join([str(cell) for cell in row]) for row in board])) def safe(board, row, col): for i in range(col): if board[row][i] == 1: return False for i, j in zip

问题 I have some code here that solves the n queens problem using backtracking in python. When I run it, the odds always take much less time than the evens. This is especially evident when the n gets to around 20+. Does anyone know why this is? import time global N N = int(input("Enter N: ")) def display_solution(board): print('\n'.join(['\t'.join([str(cell) for cell in row]) for row in board])) def safe(board, row, col): for i in range(col): if board[row][i] == 1: return False for i, j in zip