I’m trying to build a program that plays draughts/checkers. At the moment I’m trying to make the function, that allows the computer to make and evaluate moves. My idea is to have the computer look at all it’s own possible moves and for each of these moves, look at the possible opponents moves and then for each of these moves, again look at it’s own possible moves.
With each ply it will evaluate if the move is good or bad for the player and assign points, at the end it picks the moves with the highest points.
So far I have managed to get a version of this working, but involves a lot of nested for loops. The code is a mess and not very readable at the moment, but this is a simple model of the same concept. Instead of evaluating and producing more lists, it just multiplies by two for the new list.
counter = 0
for x in list:
counter += 1
list_2 = [x * 2 for x in list]
print 'list_2', list_2, counter
for x in list_2:
counter += 1
list_3 = [x * 2 for x in list_2]
print 'list_3',list_3, counter
for x in list_3:
counter += 1
list_4 = [x * 2 for x in list_3]
print 'list_4', list_4, counter
If I run this code, I get what I want, except that I can't easily control the depth of the search without copying in more for loops. I thought recursion might be a way of doing this, but I can’t figure out how to stop the recursion after x levels of search depth.
Is there a better way of getting the same output form the code above, while getting rid of all the for loops? If I can get that to work, I think I can do the rest myself.
Here's an equivalent function that uses recursion. It controls the recursion with two parameters which track the current depth and maximum depth. If current depth exceeds the maximum depth it will return immediately thus stopping the recursion:
def evaluate(l, max_depth, cur_depth=0, counter=0):
if cur_depth > max_depth:
return counter
for x in l:
counter += 1
l2 = [x * 2 for x in l]
print cur_depth, l2, counter
counter = evaluate(l2, max_depth, cur_depth + 1, counter)
return counter
If called with max_depth=2 it will produce the same output except that instead of variable name the current depth is printed.
I thought recursion might be a way of doing this, but I can’t figure out how to stop the recursion after x levels of search depth.
Your intuition is correct, and a simple a way of doing this would be to have an incrementing number passed to each level. When the recursion gets the maximum value then the recursion is completed. A trivial example is below to demonstrate.
def countup(i=0):
print(i)
if i==MAX_COUNT: return
countup(i+1)
For your algorithm, you need a value to represent the board evaluation. For instance in the range [-1,1]. Player A could be said to be winning if the evaluation is -1 and Player B is winning if the evaluation is 1 for example. A recursive algorithm could be as follows.
def evaulate(board, player, depth=0):
if depth==MAX_DEPTH: return hueristicEvaluation(board)
bestMove = None
if player==PLAYER_A:
val=999 # some large value
for move in get_moves():
newboard = board.makeMove(move)
eval, _ = evaluate(newboard, PLAYER_B, depth+1)
if eval < val:
bestMove = move
val = eval
elif player==PLAYER_B:
val=-999 # some large negative value
for move in get_moves():
newboard = board.makeMove(move)
eval, _ = evaluate(newboard, PLAYER_A, depth+1)
if eval > val:
bestMove = move
val = eval
return val, bestMove
This is abstract, but the idea is there. Adjust depending on how your are representing the board or the players. The function hueristicEvaluation could be something as simple as counting the pieces on the board for each player and how close they are to the other side. Remember that this function needs to return a number between [-1,1]
Edge cases to consider, which I didn't take into account:
If all moves are winning and/or losing
If the are NO moves in the position, for example if your pieces are all blocked by your opponent's pieces
Many improvements exist to a simple search like this. Read if you're interested :)
For checkers, perhaps Memoization would speed things up a lot. I'm not sure, but I'd think it would be especially in the beginning. See python's way of doing this.
Pruning
Alpha-beta pruning
Branch and bound
Related
I'm so confused by backtracking because when the recursive call returns, won't you replace the solution found by replacing the grid back to zero. So even if you find the solution would it not be erased because after calling the solve function you are canceling what you did by replacing the value back to zero. I get the fact that you are backtracking but on the final recursive call that contains all the correct values are you not just replacing everything to 0?
# grid = ..... # defined as a global value,
# a list of 9 lists, 9-long each
def solve():
global grid
for y in range (0, 9):
for x in range (0, 9):
if grid[y][x] == 0:
for n in range(1,10):
if possible(y, x, n):
grid[y][x] = n
solve()
grid[y][x] = 0
return
# edit: missed this final line:
print (np.matrix(grid))
This was the code on Computerphile video by Prof. Thorsten Altenkirch.
This is weird code, but should work with some adaptations:
def solve():
global grid
for y in range(0, 9):
for x in range(0, 9):
if grid[y][x] == 0:
for n in range(1,10):
if possible(y, x, n):
grid[y][x] = n
if solve():
return True # return without reset
grid[y][x] = 0
return False # exhausted all options
return True # this is the deepest and last call with no more zeroes
Here is part of my code:
vlist = PossibleValueAtPosition(row,col) # find possible value at location (row, col)
for v in vlist: # try each possible value
puzzle[row][col] = v
if SolvePuzzle(n+1)==True: # n=81 means all filled then end loop
return True # if get a solution, you return True
puzzle[row][col] = 0 # if above return true, this line will never run
return False # return False for each fail attemp
Main program should like this
if SolvePuzzle(0)==True:
print(puzzle)
else:
print('No solution!')
It's not the final recursive call that contains all the correct values, but (each of) the deepest. Yes, this code enumerates all the solutions to the puzzle with the given board grid, not just the first solution.
For each (y,x) place, if it's empty, we try to place there each of the numbers from 1 through 9 in turn. If the placement was possible on the board as it is so far, we recurse with the changed grid board.
At the deepest level of recursion there were no empty (y,x) places on the board. Therefore we slide through to the print statement. (It could also be replaced by yield True for example, to turn it into a generator. On each next value we'd get from that generator, we'd have a complete solution -- in the changed grid. And when the generator would get exhausted, the grid would be again in its original state.)
When all the numbers from 1 through 9 have been tried, the current invocation has run its course. But the one above it in the recursion chain is waiting to continue its work trying to fill its (y,x) position. We must let it work on the same board it had before it invoked this invocation of solve(). And the only change on the board this invocation did was to change its (y,x) position's value from 0 to 1 through 9. So we must change it back to 0.
This means that the code could be restructured a little bit too, as
def solve():
global grid
for y in range (0, 9):
for x in range (0, 9): # for the first
if grid[y][x] == 0: # empty slot found:
for n in range(1,10): # try 1..9
if possible(y, x, n):
grid[y][x] = n
solve() # and recurse
# (move it here)
grid[y][x] = 0 # restore
return # and return
# no empty slots were found:
# we're at the deepest level of recursion and
# there are no more slots to fill:
yield True # was: print (np.matrix(grid))
Each invocation works only on one (y,x) location, the first empty position that it found by searching anew from the start on the changed board. This search is done by the first two nested loops on y and on x. That is a bit redundant; we know all the positions before this (y,x) are already filled. The code would be better restructured to pass the starting position (y,x) as a parameter to solve.
The paradigm of recursive backtracking is beautiful. Prolog is full of mystique, Haskell will dazzle you with cryptic monads talk (monads are actually just interpretable nestable data), but all it takes here are some nested loops, recursively created!
The paradigm is beautiful, but this code, not so much. A code's visual structure should reflect its true computational structure, but this code gives you an impression that the y-x- loops are the nested loops at work to create the backtracking structure, and they are not (they just implement a one-off linear search for the next empty space in the top-down left-to-right order).
That role is fulfilled by the n in range(1,10) loops. The y-x- loops should be stopped and exited explicitly when the empty space is found, to truly reflect in the code structure what is going on computationally, to make it apparent that the n in range(1,10) loop is not nested inside the y-x- loops, but comes in play after they finish their job.
Another problem is that it just assumes the validity of the numbers given to us in the grid before the very first call to solve(). That validity is never actually checked, only the validity of the numbers which we are placing in the empty cells is checked.
(note: previous versions of this answer were based on an erroneous reading of the code. there were some valid parts in them too. you can find them on the revisions list here).
I wrote code for a DFS after reading about what it is but not actually seeing the code. I did this to challenge myself (I always have believed that to learn something new you must always first challenge yourself). The thing is after I wrote my code, I compared my implementation to the one in the book I read it in (Introduction to the Design and Analysis of Algorithms - A. Levitin) and it is completely different. So now I am wondering well it works as intended... is it still a DFS?
I made the implementation to solve a maze. I will give a rundown on my code and also upload the code here (Some people hate reading other people's code while others do.)
Algorithm (What I understood and did):
Convert maze into a graph/map
Set start position as current node and run loop in which...
I choose one of the adjacent nodes as the next current node and do this until I stumble upon a dead end. Also I am adding each node I pass through into a list that acts as my stack.
Once I am at a dead end, I keep poping items from the stack and each time I pop, I check if it has adjacent nodes that have not been visited.
Once I have found an unvisited adjacent node, we continue the entire process from step 3.
We do this until current node is the end position.
Then I just retrace my way back through the stack.
Here is my code:
# Depth First Search implementation for maze...
# from random import choice
from copy import deepcopy
import maze_builderV2 as mb
order = 10
space = ['X']+['_' for x in range(order)]+['X']
maze = [deepcopy(space) for x in range(order)]
maze.append(['X' for x in range(order+2)])
maze.insert(0, ['X' for x in range(order+2)])
finalpos = (order, order)
pos = (1, 1)
maze[pos[0]][pos[1]] = 'S' # Initializing a start position
maze[finalpos[0]][finalpos[1]] = 'O' # Initializing a end position
mb.mazebuilder(maze=maze)
def spit():
for x in maze:
print(x)
spit()
print()
mazemap = {}
def scan(): # Converts raw map/maze into a suitable datastructure.
for x in range(1, order+1):
for y in range(1, order+1):
mazemap[(x, y)] = []
t = [(x-1, y), (x+1, y), (x, y-1), (x, y+1)]
for z in t:
if maze[z[0]][z[1]] == 'X':
pass
else:
mazemap[(x, y)].append(z)
scan()
path = [pos] # stack
impossible = False
while path[-1] != finalpos:
curpos = path[-1]
i = 0
while i < len(mazemap[curpos]):
if mazemap[curpos][i] in path:
del mazemap[curpos][i]
else:
i += 1
nextpos = None
if mazemap[curpos] == []:
while nextpos == None:
try:
wrongpos = path.pop(-1)
if mazemap[wrongpos] == []:
pass
else:
path.append(wrongpos)
# nextpos = choice(mazemap[wrongpos])
nextpos = mazemap[wrongpos][-1]
mazemap[wrongpos].remove(nextpos)
except IndexError:
impossible = True
break
else:
# nextpos = choice(mazemap[curpos])
nextpos = mazemap[curpos][-1]
if impossible:
break
path.append(nextpos)
if not impossible:
for x in path:
if x == pos or x == finalpos:
pass
else:
maze[x[0]][x[1]] = 'W'
else:
print("This maze not solvable, Blyat!")
print()
spit()
As always, I greatly appreciate your suggestions!
Your algorithm looks DFS to me. DFS means exploring the path as deep as possible, backtrack to the previous node only if there is no solution and your algorithm works in a similar way by popping nodes from the stack. You just mimic the recursion stack using your own stack so it looks quite different from the standard solution.
Essentially, all recursive algorithms can be simulated using stack and loop. But most of the time doing this will make the algorithm much less readable. To tackle a difficult problem, I think the usual way to do it is to first come up with the recursive solution. After making sure the recursive solution is bug-free, then start implementing the iterative version using stack if you care a lot about the efficiency.
Other Suggestion:
if mazemap[curpos][i] in path: is a O(n) operation since path is a normal list. Consider using a separate hash set to store visited nodes and use the set to check repetition instead to make it O(1).
I am currently trying to implement dynamic programming in Python, but I don't know how to setup the backtracking portion so that it does not repeat permutations.
For example, an input would be (6, [1,5]) and the expected output should be 2 because there are 2 possible ways to arrange 1 and 5 so that their sum is equivalent to 6. Those combinations are {1,1,1,1,1,1} and {1,5} but the way my program currently works, it accounts for the combinations displayed above and the combination {5,1}. This causes the output to be 3 which is not what I wanted. So my question is "How do I prevent from repeating permutations?". My current code is shown below.
import collections as c
class DynamicProgram(object):
def __init__(self):
self.fib_memo = {}
# nested dictionary, collections.defaultdict works better than a regular nested dictionary
self.coin_change_memo = c.defaultdict(dict)
self.__dict__.update({x:k for x, k in locals().items() if x != 'self'})
def coin_change(self, n, coin_array):
# check cache
if n in self.coin_change_memo:
if len(coin_array) in self.coin_change_memo[n]:
return [n][len(coin_array)]
# base cases
if n < 0: return 0
elif n == 1 or n == 0: return 1
result = 0
i = 0
# backtracking (the backbone of how this function works)
while i <= n and i < len(coin_array):
result += self.coin_change(n-coin_array[i], coin_array)
i += 1
# append to cache
self.coin_change_memo[n][len(coin_array)] = result
# return result
return result
One of the way of avoiding permutation is to use the numbers in "non-decreasing" order. By doing so you will never add answer for [5 1] because it is not in "non-decreasing" order.And [1 5] will be added as it is in "non-decreasing" order.
So the change in your code will be if you fix to use the ith number in sorted order than you will never ever use the number which is strictly lower than this.
The code change will be as described in Suparshva's answer with initial list of numbers sorted.
Quick fix would be:
result += self.coin_change(n-coin_array[i], coin_array[i:]) # notice coin_array[i:] instead of coin_array
But you want to avoid this as each time you will be creating a new list.
Better fix would be:
Simply add a parameter lastUsedCoinIndex in the function. Then always use coins with index >= lastUsedCoinIndex from coin array. This will ensure that the solutions are distinct.
Also you will have to make changes in your memo state. You are presently storing sum n and size of array(size of array is not changing in your provided implementation unlike the quick fix I provided, so its of no use there!!) together as a state for memo. Now you will have n and lastUsedCoinIndex, together determining a memo state.
EDIT:
Your function would look like:
def coin_change(self,coin_array,n,lastUsedCoinIndex):
Here, the only variables changing will be n and lastUsedCoinIndex. So you can also modify your constructor such that it takes coin_array as input and then you will access the coin_array initialized by constructor through self.coin_array. Then the function would become simply:
def coin_change(self,n,lastUsedCoinIndex):
Does anyone understand the following iterative algorithm for producing all permutations of a list of numbers?
I do not understand the logic within the while len(stack) loop. Can someone please explain how it works?
# Non-Recursion
#param nums: A list of Integers.
#return: A list of permutations.
def permute(self, nums):
if nums is None:
return []
nums = sorted(nums)
permutation = []
stack = [-1]
permutations = []
while len(stack):
index = stack.pop()
index += 1
while index < len(nums):
if nums[index] not in permutation:
break
index += 1
else:
if len(permutation):
permutation.pop()
continue
stack.append(index)
stack.append(-1)
permutation.append(nums[index])
if len(permutation) == len(nums):
permutations.append(list(permutation))
return permutations
I'm just trying to understand the code above.
As mentioned in the comments section to your question, debugging may provide a helpful way to understand what the code does. However, let me provide a high-level perspective of what your code does.
First of all, although there are no recursive calls to the function permute, the code your provided is effectively recursive, as all it does is keeping its own stack, instead of using the one provided by the memory manager of your OS. Specifically, the variable stack is keeping the recursive state, so to speak, that is passed from one recursive call to another. You could, and perhaps should, consider each iteration of the outer while loop in the permute function as a recursive call. If you do so, you will see that the outer while loop helps 'recursively' traverse each permutation of nums in a depth-first manner.
Noticing this, it's fairly easy to figure out what each 'recursive call' does. Basically, the variable permutation keeps the current permutation of nums which is being formed as while loop progresses. Variable permutations store all the permutations of nums that are found. As you may observe, permutations are updated only when len(permutation) is equal to len(nums) which can be considered as the base case of the recurrence relation that is being implemented using a custom stack. Finally, the inner while loop picks which element of nums to add to the current permutation(i.e. stored in variable permutation) being formed.
So that is about it, really. You can figure out what is exactly being done on the lines relevant to the maintenance of stack using a debugger, as suggested. As a final note, let me repeat that I, personally, would not consider this implementation to be non-recursive. It just so happens that, instead of using the abstraction provided by the OS, this recursive solution keeps its own stack. To provide a better understanding of how a proper non-recursive solution would be, you may observe the difference in recursive and iterative solutions to the problem of finding nth Fibonacci number provided below. As you can see, the non-recursive solution keeps no stack, and instead of dividing the problem into smaller instances of it(recursion) it builds up the solution from smaller solutions. (dynamic programming)
def recursive_fib(n):
if n == 0:
return 0
elif n == 1:
return 1
return recursive_fib(n-1) + recursive_fib(n-2)
def iterative_fib(n):
f_0 = 0
f_1 = 1
for i in range(3, n):
f_2 = f_1 + f_0
f_0 = f_1
f_1 = f_2
return f_1
The answer from #ilim is correct and should be the accepted answer but I just wanted to add another point that wouldn't fit as a comment. Whilst I imagine you are studying this algorithm as an exercise it should be pointed out that a better way to proceed, depending on the size of the list, may be to user itertools's permutations() function:
print [x for x in itertools.permutations([1, 2, 3])]
Testing on my machine with a list of 11 items (39m permutations) took 1.7secs with itertools.permutations(x) but took 76secs using the custom solution above. Note however that with 12 items (479m permutations) the itertools solution blows up with a memory error. If you need to generate permutations of such size efficiently you may be better dropping to native code.
I'm currently working with a recursive function in Python, and I've run into a wall. As titled, the problem is to return the maximum depth of an arbitrarily nested list.
Here is what I have so far:
def depthCount(lst):
'takes an arbitrarily nested list as a parameter and returns the maximum depth to which the list has nested sub-lists.'
var = 0
if len(lst) > 0:
if type(lst[0]) == list:
var += 1
depthCount(lst[1:])
else:
depthCount(lst[1:])
else:
return var
I feel that the problem is with my recursive calls (this may be obvious). It will indeed return var when the list has reached the end, but when I have a nonempty list, things go awry. Nothing is returned at all.
Am I slicing wrong? Should I be doing something before the slice in the recursive call?
The problem may also be with my base case.
If they are just nested lists, e.g., [[[], []], [], [[]]], here's a nice solution:
def depthCount(lst):
return 1 + max(map(depthCount, lst), default=0)
Here's a slight variation you could use if you don't use Python 3.4, where the default argument was introduced:
def depthCount(lst):
return len(lst) and 1 + max(map(depthCount, lst))
They also differ by how they count. The first considers the empty list to be depth 1, the second to be depth 0. The first one is easy to adapt, though, just make the default -1.
If they're not just nested lists, e.g., [[[1], 'a', [-5.5]], [(6,3)], [['hi']]]), here are adaptions to that:
def depthCount(x):
return 1 + max(map(depthCount, x)) if x and isinstance(x, list) else 0
def depthCount(x):
return int(isinstance(x, list)) and len(x) and 1 + max(map(depthCount, x))
Make sure you understand how the latter one works. If you don't know it yet, it'll teach you how and works in Python :-)
Taking the "purely recursive" challenge:
def depthCount(x, depth=0):
if not x or not isinstance(x, list):
return depth
return max(depthCount(x[0], depth+1),
depthCount(x[1:], depth))
Granted, the extra argument is slightly ugly, but I think it's ok.
It will indeed return var when the list has reached the end, but when I have a nonempty list, things go awry. Nothing is returned at all.
That's because you have no return statement, except in the else base case for an empty list. And if you fall off the end of the function without hitting a return, that means the function returns None.
But you have another problem on top of that. You're starting var = 0, then possibly doing var += 1… but you're not passing that down into the recursive calls, or using any result from the recursive calls. So the recursive calls have no useful effect at all.
What you probably meant is something like this:
def depthCount(lst):
'takes an arbitrarily nested list as a parameter and returns the maximum depth to which the list has nested sub-lists.'
if len(lst) > 0:
if type(lst[0]) == list:
return 1 + depthCount(lst[1:])
else:
return depthCount(lst[1:])
else:
return 0
But this still isn't actually right. The depth count of a list is 1 more than the depth count of its deepest element. Just checking its second element won't do you any good; you need to check all of them. So, what you really want is something like this:
def depthCount(lst):
'takes an arbitrarily nested list as a parameter and returns the maximum depth to which the list has nested sub-lists.'
if isinstance(lst, list):
return 1 + max(depthCount(x) for x in lst)
else:
return 0
If you want to replace that iterative for x in lst with a second layer of recursion, of course you can, but I can't see any good reason to do so; it just makes the code more complicated for no reason. For example:
def max_child_count(lst):
if lst:
return max(depth_count(lst[0]), max_child_count(lst[1:]))
else:
return 0
def depth_count(lst):
if isinstance(lst, list):
return 1 + max_child_count(lst)
else:
return 0
This may still not be right. It definitely does the right thing for, e.g., [1, [2,3], [4, [5]]]. But what should it do for, say, []? I can't tell from your question. If it should return 0 or 1, you'll obviously need to change the if appropriately. If that's illegal input, then it's already doing the right thing. (And that should also answer the question of what it should do for, e.g., [[[], []], [], [[]]], but make sure you think through that case as well.)
So, essentially, the data structure that you're referring to is a k-ary tree, also known as n-ary tree, with arbitrary branching. Here's the code for determining the max. depth of a n-ary tree with arbitrary branching.
def maxdepth(tree):
if isleaf(tree):
return 1
maximum = 0
for child in children(tree):
depth = maxdepth(child)
if depth > maximum:
maximum = depth
return maximum + 1
You can see the code in action with different test inputs here.