backtracking

问题 I am programming a Sudoku solver in Java for a 9x9 grid. I have methods for: printing the grid initializing the board with given values testing for conflicts (if same number is in same line or 3x3 sub-grid) a method to place the digits, one by one, which requires the most work. Before I go into detail with that method, keep in mind that I have to use recursion to solve it, as well as backtracking (watch the applet here as an example http://www.heimetli.ch/ffh/simplifiedsudoku.html ) Also, I

问题 The Problem I'm using the following regular expression to check for valid file paths: ^(?:[a-zA-Z]\:\\|\\\\)([^\\\/\:\*\?\<\>\"\|]+(\\){0,1})+$ Using the test string V:\Sample Names\Libraries\DeveloperLib\DeveloperComDlgs\res is recognized as valid. I can even add invalid characters to the beginning of the string without issues. However, when I add an invalid character towards the end of the string, the webpage freezes up from catastrophic backtracking. What is causing this in this regex

问题 The Problem I'm using the following regular expression to check for valid file paths: ^(?:[a-zA-Z]\:\\|\\\\)([^\\\/\:\*\?\<\>\"\|]+(\\){0,1})+$ Using the test string V:\Sample Names\Libraries\DeveloperLib\DeveloperComDlgs\res is recognized as valid. I can even add invalid characters to the beginning of the string without issues. However, when I add an invalid character towards the end of the string, the webpage freezes up from catastrophic backtracking. What is causing this in this regex

问题 I'm making a peg solitaire resolver with backtracking in java. This is the method I've done: private void solve(Board board, ArrayList<Movement> solution, Boolean result) { ArrayList<Movement> movs = board.getMovements(); for (Movement movement : movs) { if (board.isMovementValid(movement)) { board.doMovement(movement); if(!board.isSolution()) { solution.add(movement); solve(board, solution, result); result.setValue(false); } else { result.setValue(true); } } } result.setValue(false); } The

问题 I am solving Word Search question on LeetCode.com: Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once. The solution which I wrote with online help is as follows: class Solution { public: //compare this with Max Area of Island: //they 'look' similar, but this one uses a backtracking

问题 I found this sudoku solver which use backtracking when it is trying to solve the puzzle and for better understanding I would like to delay the process so I can analyse the backtracking. But I don't really know how to do that. I tried to use Thread.sleep(100); but I don't really know where exactly to put the delay. abstract class SudoKiller { private SudokuBoard sb; // Puzzle to solve; public SudoKiller(SudokuBoard sb) { this.sb = sb; } private boolean check(int num, int row, int col) { int r

问题 The problem is counting the number of paths of length n from a given vertex in a graph like the used to unlock Android devices.I'm trying to use backtracking to solve it but I don't get right, I'm I'm still learning how to use it. So here is some code I've been trying G = { 'a': set('bed'), 'b': set('cfeda'), 'c': set('feb'), 'd': set('abehg'), 'e': set('bcfihgda'), 'f': set('ciheb'), 'g': set('deh'), 'h': set('efigd'), 'i': set('fhe') } result = 0 def count_patterns(node, length): if length

问题 Using a findall in Prolog how can I perform operations inside the goal without affecting backtracking? The following example explains what I'm trying to achieve: value('M1', 11, 3). value('M2', 11, 3). connection('M1',1, 'A', 'B'). connection('M1',1, 'B', 'C'). connection('M1',2, 'C', 'D'). connection('M1',2, 'D', 'E'). connection('M2',1, 'D', 'F'). run :- bbR('C',[(0,'X',['A'])],_,_). run2 :- bbR2('C',[(0,['A'])],_,_). bbR(Destination,[(Cost,_,[Destination|T])|_],Result,Cost):- reverse(

问题 I don't understand why my predicate doesnt backtrack and find all solutions. person(john). person(erik). allExceptSpider(person(Spider),T ):- setof(person(X),person(X),S), subtract(S,[person(Spider) ],T). If I call this predicate with Two variables: allExceptSpider(person(Z),Q) Then it will only give me the answer Z = john, Q = [person(erik)] but it won't backtrack to find Z = erik ,Q = [person(john)] why? 回答1: TL;DR: If you use subtract/3 , your code may lose logical-purity. person(john).

问题 Given a matrix of characters and a string, find whether the string can be obtained from the matrix. From each character in the matrix, we can move up/down/right/left. For example, if the matrix[3][4] is: o f a s l l q w z o w k and the string is follow , then the function should return true. The only approach I can think of is a backtracking algorithm that searches whether the word is possible or not. Is there any other faster algorithm to approach this problem? And suppose I have a lot of