backtracking

Algorithm NQueens ( k, n) //Prints all Solution to the n-queens problem { for i := 1 to n do { if Place (k, i) then { x[k] := i; if ( k = n) then write ( x [1 : n] else NQueens ( k+1, n); } } } Algorithm Place (k, i) { for j := 1 to k-1 do if (( x[ j ] = // in the same column or (Abs( x [ j ] - i) =Abs ( j – k ))) // or in the same diagonal then return false; return true; } The above code is for solving N Queens problem using backtracking.I think that it can place the first 2 queens of two rows in respective columns and then when it comes to 3rd row queen it can't be placed as no queen needs

I'm stuck on the extended exercise 28.2 of How to Design Programs . I used a vector of true or false values to represent the board instead of using a list. This is what I've got which doesn't work: #lang Scheme (define-struct posn (i j)) ;takes in a position in i, j form and a board and ; returns a natural number that represents the position in index form ;example for board xxx ; xxx ; xxx ;(0, 1) -> 1 ;(2, 1) -> 7 (define (board-ref a-posn a-board) (+ (* (sqrt (vector-length a-board)) (posn-i a-posn)) (posn-j a-posn))) ;reverse of the above function ;1 -> (0, 1) ;7 -> (2, 1) (define (get-posn

The problem is next. I created an input field which has validation and this is valid data: 1-12, 14, 16, 22, 25-35, 41, 49, 55-90 1230-1992, 2001-2099, 9931 1-2 13 1,3,4,5,6,10 all Basically, any combination of those numbers (ranges, comma-separated ranges, comma-separated number, spaces after commas, no spaces after commas, word: 'all') My RegEx: /^(([0-9]{0,4},?\s{0,})+([0-9]{1,4}-[0-9]{1,4}){0,},?\s{0,})+$|^(all)$|^([0-9]{1,4}-[0-9]{1,4}){0,},?\s{0,}$/ It works almost fine, there is only 1 major problem. When I start typing, and after some comma-separated numbers I add something that is not

I'm getting quite interested in search algorithms and backtrack programming. For now, I have implemented Algorithm X (see my other post here: Determine conflict-free sets? ) to solve an exact cover problem. This works very well but I'm now interested in solving this with a more basic variant of backtracking. I just can't figure out how this can be done. The problem description is the same as before: Suppose you have a bunch of sets, whereas each set has a couple of subsets. Set1 = { (banana, pineapple, orange), (apple, kale, cucumber), (onion, garlic) } Set2 = { (banana, cucumber, garlic),

Hey guys, recently posted up about a problem with my algorithm. Finding the numbers from a set which give the minimum amount of waste Ive amended the code slightly, so it now backtracks to an extent, however the output is still flawed. Ive debugged this considerablychecking all the variable values and cant seem to find out the issue. Again advice as opposed to an outright solution would be of great help. I think there is only a couple of problems with my code, but i cant work out where. //from previous post: Basically a set is passed to this method below, and a length of a bar is also passed

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... side note: the string/regexes are not mine, I just took them from elsewhere on SE and saw this behavior

I have tried solving this problem with backtracking and it prints all possible solutions. Two questions came up: 1. Can i implement n queen using other techniques? 2. Is it possible to make the code below print only the first solution and then terminate? My current code uses backtracking: n = 8 x = [-1 for x in range(n)] def safe(k,i): for j in xrange(k): if x[j] == i or abs(x[j] - i) == abs(k - j) : return False return True def nqueen(k): for i in xrange(n): if safe(k,i): x[k] = i if k == n-1: print "SOLUTION", x else: nqueen(k+1) nqueen(0) Note : I am interested in techniques that do not