does eight queens backtracking solution require any sorting at all?
By : Mostafa Amer
Date : March 29 2020, 07:55 AM
I wish this help you I think in this case you're misinterpreting what "sorting" means. In order for backtracking to work you need to be analyzing positions in some sort of predictable ordering. If your algorithm is not analyzing positions in a set order, then when you "prune" a set of positions, you may have pruned a valid position. Without this "ordering" or tree like structure of positions backtracking does not work. You do not, however, need to presort a set of positions or anything like that. This would, in fact, defeat the purpose of backtracking. The idea is that, some combination of positions never even have to be built. Once a conflict is found ALL iterations involving that combination are no longer even considered. It is the ordering in which these combinations are built that is a concern, not sorting them ahead of time. All combinations must be built and considered in proper oder. This allows u to know when we give up on a "branch" that all combinations built on this branch would have been equally(or even worse) incorrect as the option we just rejected, otherwise you can "over prune" your result set, and miss a proper combination. But no NlogN sorting algorithms are required. At least not in the example of the nqueens problem. In fact, if you prebuilt all positions and sorted them, you are completely ignoring the dynamic programming elements that allow us to speed up the computation of this problem considerably.

N Queens: backtracking solution implemented by Python generator
By : harold
Date : March 29 2020, 07:55 AM
With these it helps The algorithm works exactly the same way as the one using a list comprehension, the only difference really is that it uses a generator expression  in fact, it creates a generator with more generators nested inside! So instead of listing all the possible solutions at once, it lazily generates more solutions as needed. You can easily verify this by increasing some globally defined counter variable each time under_attack is called, and seeing how many times this counter has been increased. code :
gen = solve(8) # just create the generators, under_attack called 0 times
next(gen) # first solution, under_attack called 876 times
list(gen) # all remaining solutions, under_attack called 15720
def under_attack(col, queens):
return col in queens or any(abs(col  x) == len(queens)i # (1)
for i,x in enumerate(queens)) # (2)
def solve(n):
solutions = [[]]
for row in range(n): # (3)
solutions = [solution+[i] for solution in solutions # (4)
for i in range(n) # (5)
if not under_attack(i, solution)] # (6)
print solutions # remove this when using generator! # (7)
return solutions
[[0], [1], [2], [3]] # 1st queen
[[0, 2], [0, 3], [1, 3], [2, 0], [3, 0], [3, 1]] # 2nd queen
[[0, 3, 1], [1, 3, 0], [2, 0, 3], [3, 0, 2]] # 3rd queen
[ [1, 3, 0, 2], [2, 0, 3, 1]] # 4th queen

Appending the list with each unique solution to the Queens Puzzle. I don't know why the last solution is the only soluti
By : Raquel Ribeiro
Date : March 29 2020, 07:55 AM
like below fixes the issue I am trying to save each individual solution to the NQueens Puzzle to a list. However, when I try to add each list as a sublist, only the last solution is added to the list (10 times), as opposed to the 10 individual solutions. My goal is to ensure that each time I run the board only new solutions, which have not been found already, are printed and added to the list. , Just change code :
solutions.append(bd)
solutions.append(list(bd))
if not has_clashes(bd):
if not has_clashes(bd) and bd not in solutions:

Solution to variation of the N Queens puzzle
By : Omar Naji
Date : March 29 2020, 07:55 AM
This might help you With these super queens, you will no longer be able to fit N queens on an NxN board, other than the trivial 1x1 board. One way to see this is that there are 2N1 diagonals (let's use lower left to upper right) on an NxN board. Each queen will be attacking 3 diagonals, except if they are in the corner they will attack 2 diagonals. Let's say we have one queen in the corner, occupying 2 diagonals. Then e.g. on an 8x8 board we have 13 diagonals left which can be used by floor(13/3) queens or 4 queens. So at most we could have 5 queens on an 8x8 board. I don't know if this is a tight upper bound.

NQueens puzzle solution  cannot understand
By : wang jie
Date : March 29 2020, 07:55 AM
wish of those help First of all, it doesn't matter whether you use columns or rows  the output will be the same, because the problem is symmetric. This symmetry will create some confusion; be prepared for this. Without regard to your specific questions, the idea here is to do recursion. The problem talks about an arrangement of 8 queens. If you have placed k1 queens, you got a "position". From each position, you can get several "extended" positions, in which you have placed one more queen (so there are k queens). So for each set of positions with k1 queens, there is a set of positions with k queens.

