backtracking

问题 So I came up with this implementation for solving knights tour for a 8*8 chess board. But seems like it is taking a long time running (so long that I have to stop it). But if I replace the dx, dy arrays with the one written in comments and the program works like magic and gives the output. They say that it is cleverly choosing the array, so that the bruteforce algo is lucky to find the solution quickly. But how does one come up with this array at first place, this array (dx,dy) I got from

问题 There are three different solutions to implement regular expression matching: DFA, NFA and Backtracking. I am looking for examples: a regexp, which can be solved with a DFA and the reason, why a DFA is sufficient. a regexp, which requires a NFA and the reason why a NFA is necessary. a regexp, which requires backtracking and the reason why backtracking is necessary. A recommendation for some good literature about this topic would be nice, too. 回答1: i guess there is more than 1 meaning to the

问题 I am trying to solve the following question using BackTracking in C but I don't know how to continue from here... The question is: Chris is planning to travel in a country with N cities. He will get help from a matrix NxN that the cell (I,J) represents the length of the road from city I to City J. The length of the road from City A to city B is not the same when compared to the road from City B to City A. The Road From the same source city back to it (directly) is 0. Chris noticed that the

问题 I want to print all the paths from the root to the leaf nodes in an N-ary tree in python. I have an idea to print it in binary tree but doing this in N-ary doesn't give me the correct result. I'm poping and Visiting each node from the child node list here but not sure how to print the paths separately for each leaf node. class createnode: def __init__(self,val): self.data=val self.child=[] def traverse(root): global path if root.child: while(len(root.child)>0): x=root.child.pop(0) path.append

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 I have this homework assignment to make a recursive method to crack a password of a given length, n (unlimited and unknown!) made of small English letters, a-z ONLY. Here's the class "Password" that creates a random password: import java.util.Random; public class Password { private String _password = ""; public Password(int length) { Random generator = new Random(); for (int i = 0; i < length; ++i) { this._password = this._password + (char) (generator.nextInt(26) + 97); } } public boolean

问题 I'm trying to write code for the Knight's Tour: A knight's tour is a sequence of moves of a knight on a chessboard such that the knight visits every square exactly once. I've been trying to alter someone else's code, but the backtracking seems to not work properly - it never finds the solution. It works perfectly fine when the knight starts at 0, 0 but if it starts at any other spot on the 2D grid, the program goes on forever. Where is the bug in this code? #include <iostream> #include <ctime

问题 This predicate should print a list of size N containing possible permutations of 0 and 1 . My question is : does the value of H get carried over with each recursion or does the creation of the list with values of bit(H) take place in the backtracking phase? bit(0). bit(1). gen(0,[]). gen(N,[H|T]) :- N > 0, bit(H), N1 is N - 1, gen(N1,T). 回答1: Prolog execution is all about choice points. Here a choice point is left at each recursion step by the bit/1 predicate. When you ask Prolog to give you

问题 The following expression: ^(#ifdef FEATURE)+?\s*$((\r\n.*?)*^(#endif)+\s*[\/\/]*\s*(end of)*\s*FEATURE)+?$ Overrides the matching buffer when running my compiled .Jar file. The matching string can be similar to: this is a junk line #ifdef FEATURE #endif // end of FEATURE this is a junk line #ifdef FEATURE this is a junk line that should be matched: HOLasduiqwhei & // FEATURE fjfefj #endif // h #endif FEATURE this is a junk line So, the bold strings should match. The error is as follows: at

问题 "Programming Challenges (The Programming Contest Training Manual)" is probably one of the nicest exercises book on algorithms. I've resolved the first 11 exercises, but now I am stuck with "Crypt Kicker" problem: Crypt Kicker A common but insecure method of encrypting text is to permute the letters of the alphabet. In other words, each letter of the alphabet is consistently replaced in the text by some other letter. To ensure that the encryption is reversible, no two letters are replaced by