N-Queens Part 1: Steepest Hill Climbing

The n-queens problem was first invented in the mid 1800s as a puzzle for people to solve in their spare time, but now serves as a good tool for discussing computer search algorithms. In chess, a queen is the only piece that can attack in any direction. The puzzle is to place a number of queens on a board in such a way that no queen is attacking any other. For example:A diagram of a solved 8-queens problemOne way we can describe this board is to say it has a heuristic cost of 0, because there are 0 pairs of queens attacking each other. We can then generalize this to say the heuristic cost of a given n-queens board is equal to the number of queens directly or indirectly attacking one another. Consider this 5-queens puzzle. There are 5 pairs of queens attacking each other therefore the heuristic cost of this board is 5.An unsolved 5-queens problem. We can calculate the heuristic cost easily if we represent a board as an array where the index is the column and the value is the row. The board above is [0,0,1,2,4].

def get_h_cost(board):
	h = 0
	for i in range(len(board)):
		#Check every column we haven't already checked
		for j in range(i + 1,len(board)):
			#Queens are in the same row
			if board[i] == board[j]:
				h += 1
			#Get the difference between the current column
			#and the check column
			offset = j - i
			#To be a diagonal, the check column value has to be 
			#equal to the current column value +/- the offset
			if board[i] == board[j] - offset 
				or board[i] == board[j] + offset:
				h += 1
	return h

To solve this puzzle, we need to take steps to reduce the heuristic cost to zero.

In order to evaluate which moves are best, we can calculate the heuristic cost of the board after one move. This diagram shows the heuristic costs of all possible moves from the current board. For simplicity, we will only move queens up or down in their rows.Complete heuristic costs of a given 5-queens puzzle. If you would choose the move with the lowest heuristic cost and then repeat the process, then you would be using the steepest hill climbing algorithm.

The hill climbing algorithm gets its name from the metaphor of climbing a hill where the peak is h=0. But there is more than one way to climb a hill. If we always choose the path with the best improvement in heuristic cost then we are using the steepest hill variety.

Steepest hill climbing can be implemented in Python as follows:

def make_move_steepest_hill(board):
	moves = {}
	for col in range(len(board)):
		best_move = board[col]
		for row in range(len(board)):
			if board[col] == row:
				#We don't need to evaluate the current
				#position, we already know the h-value
			board_copy = list(board)
			#Move the queen to the new row
			board_copy[col] = row
			moves[(col,row)] = get_h_cost(board_copy)
	best_moves = []
	h_to_beat = get_h_cost(board)
	for k,v in moves.iteritems():
		if v < h_to_beat:
			h_to_beat = v
	for k,v in moves.iteritems():
		if v == h_to_beat:
	#Pick a random best move
	if len(best_moves) > 0:
		pick = random.randint(0,len(best_moves) - 1)
		col = best_moves[pick][0]
		row = best_moves[pick][1]
		board[col] = row
	return board

But as I mentioned above, there are multiple ways to climb a hill! Next time we’ll look at some additional ways to solve n-queens problems.

8 thoughts on “N-Queens Part 1: Steepest Hill Climbing

  1. Phillip

    Thanks, I’ve updated the diagram to show a correct solution. Thankfully, this was just a diagram and doesn’t have any bearing on the code in the post.

    1. Phillip Johnson Post author

      Sorry, I don’t have it anymore. But as long as you have a function that checks if the board is solved, you can wrap that in a while loop and keep calling make_move_steepest_hill until a solution is found.

      1. Arafat

        Sir, actually I am confused with the board. It is supposed to be a 2D list. How do we configure it first? And the get_h_cost is the number of threats isn’t it?

        1. Shivangi

          Hi Arfat,

          So this is modeled like a constraint problem such that each column can only have one queen. So it is assumed that there will be no column clashes.

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