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foobar failing test case - a* with breakable wall

Working on the google foo-bar challenges and I am stuck on one failing test case (which is a hidden one - so I can't directly see what the problem is)
basically the test is an implementation of a maze solver with a single breakable wall.
I'm doing a modified a* search - with a Boolean flag for paths that contain a broken wall, so that when it reaches the second wall (along a path with a broken wall) it skips it for that path.
I've gone over this code and I can't seem to see the error (if there even is one - at this point I'm almost convinced the test case is somehow wrong)
the parameters:
given a grid of height x,y filled with zero or 1, representing spaces and walls respectively: find the shortest path if you can break exactly 1 wall (if needed)
an example:
[
[0,1,0,0,0,1,0,0,0,0,0]
[0,0,0,1,0,0,1,0,1,1,0]
[1,1,1,1,1,0,1,0,1,0,1]
[0,0,0,0,0,0,0,0,1,1,0]
]
22 is the shortest path breaking 1 wall.
I want a pointer in the right direction: I feel like whatever I'm missing is trivial.
below is the code.
from math import sqrt, ceil
cardinal_moves=[(0,1), (0,-1), (1,0), (-1,0)]
class Node:
def __init__(self, pos ,parent =None):
self.parent = parent
self.pos = pos
self.g = 0
self.h = 0
self.f = 0
self.wall_broken = False
def __eq__(self, other):
return ((self.pos == other.pos) and (self.wall_broken == other.wall_broken))
def update_heuristic(self,parent,end):
self.g = parent.g + 1
self.h = ceil(sqrt((end.pos[0] - self.pos[0])**2 + (end.pos[1] - self.pos[1])**2))
self.f = self.g + self.h
def not_inside_maze(pos,maze):
return pos[0] < 0 or pos[0] >= len(maze) or pos[1] < 0 or pos[1] >= len(maze[0])
def astar_with_1_breakable_wall(maze):
start_pos = (0,0)
end_pos = (len(maze)-1, len(maze[0])-1)
start = Node(start_pos)
end = Node(end_pos)
not_visted = []
visited = []
not_visted.append(start)
while not_visted:
current_node = not_visted[0]
for node in not_visted:
if current_node.f > node.f:
current_node = node
not_visted.remove(current_node)
visited.append(current_node)
if current_node.pos == end.pos:
temp = current_node
path = []
while temp:
path.append(temp.pos)
temp = temp.parent
return path[::-1]
children = []
for position in cardinal_moves:
new_pos = (current_node.pos[0] + position[0], current_node.pos[1] + position[1])
if not_inside_maze(new_pos,maze):
continue
if maze[new_pos[0]][new_pos[1]] == 1:
if current_node.wall_broken:
continue
else:
check = Node(new_pos,current_node)
check.wall_broken = True
already_visited = False
for node in visited:
if check == node:
already_visited = True
break
if not already_visited:
children.append(check)
continue
already_visited = False
check = Node(new_pos,current_node)
check.wall_broken = current_node.wall_broken
for node in visited:
if check == node:
already_visited = True
break
if already_visited:
continue
children.append(check)
for child in children:
child.update_heuristic(current_node,end)
for open_node in not_visted:
if open_node == child:
if open_node.g > child.g:
idx = not_visted.index(open_node)
not_visted[idx] = child
continue
else:
continue
not_visted.append(child)
def shortest_path(maze):
if (astar_with_1_breakable_wall(maze)):
return len(astar_with_1_breakable_wall(maze))
else:
return -1
but every check I make on my machine says this is correct:
#imports added
import sys
#then added this below the maze solver
test = [
[0,1,0,0,0,1,0,0,0,0,0],
[0,0,0,1,0,0,1,0,1,1,0],
[1,1,1,1,1,0,1,0,1,0,1],
[0,0,0,0,0,0,0,0,1,1,0],
]
test2=[
[0,1,0,0,0,0,1,0,0,0,0],
[0,1,0,1,1,0,1,0,1,0,0],
[0,0,0,1,0,0,1,0,1,0,0],
[1,1,1,1,0,1,1,0,1,0,1],
[0,0,0,0,0,0,0,0,1,0,1],
]
print("first test")
print(shortest_path(test))
print("second test")
print(shortest_path(test))
#both of these tests give the correct result
def generate_matrix(h,w,n):
sequence = "{0:b}".format(n).zfill(h*w)
maze =[]
if(sequence[0]=="1" or sequence[len(sequence)-1]=="1"):
return -1
i=0
for y in range(h):
maze.append([])
for x in range(w):
maze[y].append(int(sequence[i]))
i=i+1
return(maze)
for i in range(20):
for j in range(20):
if(i < 6 or j <6):
continue
shortest=j+i
for k in range(2**(i*j)):
if not k%2==0:
continue
if k> 2**(i*j-1):
continue
maze = generate_matrix(i,j,k)
if(not maze == -1):
path = astar_with_1_breakable_wall(maze)
res = shortest_path(maze)
if(res > shortest):
print("\n",res, "shortest was ", shortest)
for index,line in enumerate(maze):
for indx,cell in enumerate(line):
if ((index,indx) in path):
print("\033[94m"+str(cell),end="")
else:
print("\033[92m"+str(cell),end="")
print("\n")
for point in path:
print(point ," ", end="")
print("\n\n")
else:
sys.stdout.write("\r")
the above code makes every matrix possibility, and prints (with highlighting the path) the matrix if the path is longer than the base case.
every result is as I expect - returning the correct shortest path... I have not found the issue with why only the 3rd test case fails...

the answer was something specific to python 2.7.13 apparently - it took a bit of debugging to figure it out, but the not_visited.remove(current_node) was the thing that was failing in certain cases.

A* path finding algo gives path but it's not the shortest path

I am trying to implement a* pathfinding in my program, but it is returning the following output.
Here is the output image
In the image blue blocks are visited blocks, the one with circles are yet to visit blocks and the yellow blocks are path retuned to us.
I could not figure out the problem in my code here is the following code the I used.
class Node:
def __init__(self, parent=None, position=None):
self.parent = parent
self.position = position
self.g = 0
self.h = 0
self.f = 0
def __eq__(self, other):
return self.position == other.position
def search(maze, cost, start, end): # Hear maze is a 2D list with 1 means a wall and 0 is a clear block
start_node = Node(None, tuple(start))
start_node.g = start_node.h = start_node.f = 0
end_node = Node(None, tuple(end))
end_node.g = end_node.h = end_node.f = 0
yet_to_visit_list = []
visited_list = []
yet_to_visit_list.append(start_node)
outer_iterations = 0
max_iterations = (len(maze) // 2) ** 10
move = [[-1, 0], [0, -1], [1, 0], [0, 1]]
no_rows, no_columns = np.shape(maze)
while len(yet_to_visit_list) > 0:
outer_iterations += 1
current_node = yet_to_visit_list[0]
current_index = 0
for index, item in enumerate(yet_to_visit_list):
if item.f < current_node.f:
current_node = item
current_index = index
if outer_iterations > max_iterations:
print("Cann't find the path... too many iterations")
return return_path(current_node, maze)
yet_to_visit_list.pop(current_index)
visited_list.append(current_node)
maze[current_node.position[0]][current_node.position[1]] = 2 # 1 wall 2 visited 3 yet to visit
if current_node == end_node:
return return_path(current_node, maze)
children = []
for new_position in move:
node_position = (current_node.position[0]+ new_position[0], current_node.position[1]+ new_position[1])
if (node_position[0] > (no_rows - 1) or node_position[0] < 0 or node_position[1]>(no_columns-1) or node_position[1] < 0):
continue
if maze[node_position[0]][node_position[1]] != 0:
continue
new_node = Node(current_node, node_position)
children.append(new_node)
for child in children:
if len([visited_child for visited_child in visited_list if visited_child == child]) > 0:
continue
child.g = current_node.g + cost
child.h = (((child.position[0] - end_node.position[0]) ** 2 ) +
((child.position[0] - end_node.position[0])) ** 2)
child.f = child.g + child.h
if len([i for i in yet_to_visit_list if child == i and child.g>i.g]) > 0:
continue
yet_to_visit_list.append(child)
And the following function returns the path
def return_path(current_node, maze):
path = []
no_rows, no_columns = np.shape(maze)
result = maze
current = current_node
while current is not None:
path.append(current.position)
current = current.parent
path = path[::-1]
start_value = 0
for i in range(len(path)):
result[path[i][0]][path[i][1]] = start_value
start_value += 1
return result

In your for loop over move, you exclude neighbors with maze[.][.] != 0: i.e., neighbors that are either "wall" or "visited". The original A* algorithm requires you to reconsider visited nodes to see if the cost can be further reduced.
Also, I get the sense that you should be removing items from the visited_list at some point in your program but I don't see this happening.
It would be easier to tell exactly what's wrong once you make the program input available, since it would then become possible to step through your code with a debugger.
Another problem is that you are using the squared distance for your heuristic cost, which likely overestimates the cost. With up-down-left-right neighbors, use the Manhattan distance function (see the warning about using squared Euclidean distance in the link). I don't think this is at the heart of what's wrong with your program, though.

How can I implement IDA* algorithm in Python for 15-Puzzle problem?

I'm trying to solve the 15-Puzzle problem using IDA* algorithm and Manhattan heuristic.
I already implemented the algorithm from the pseudocode in this Wikipedia page (link).
Here's my code so far :
def IDA(initial_state, goal_state):
initial_node = Node(initial_state)
goal_node = Node(goal_state)
threshold = manhattan_heuristic(initial_state, goal_state)
path = [initial_node]
while 1:
tmp = search(path, goal_state, 0, threshold)
if tmp == True:
return path, threshold
elif tmp == float('inf'):
return False
else:
threshold = tmp
def search(path, goal_state, g, threshold):
node = path[-1]
f = g + manhattan_heuristic(node.state, goal_state)
if f > threshold:
return f
if np.array_equal(node.state, goal_state):
return True
minimum = float('inf')
for n in node.nextnodes():
if n not in path:
path.append(n)
tmp = search(path, goal_state, g + 1, threshold)
if tmp == True:
return True
if tmp < minimum:
minimum = tmp
path.pop()
return minimum
def manhattan_heuristic(state1, state2):
size = range(1, len(state1) ** 2)
distances = [count_distance(num, state1, state2) for num in size]
return sum(distances)
def count_distance(number, state1, state2):
position1 = np.where(state1 == number)
position2 = np.where(state2 == number)
return manhattan_distance(position1, position2)
def manhattan_distance(a, b):
return abs(b[0] - a[0]) + abs(b[1] - a[1])
class Node():
def __init__(self, state):
self.state = state
def nextnodes(self):
zero = np.where(self.state == 0)
y,x = zero
y = int(y)
x = int(x)
up = (y - 1, x)
down = (y + 1, x)
right = (y, x + 1)
left = (y, x - 1)
arr = []
for direction in (up, down, right, left):
if len(self.state) - 1 >= direction[0] >= 0 and len(self.state) - 1 >= direction[1] >= 0:
tmp = np.copy(self.state)
tmp[direction[0], direction[1]], tmp[zero] = tmp[zero], tmp[direction[0], direction[1]]
arr.append(Node(tmp))
return arr
I'm testing this code with a 3x3 Puzzle and here's the infinite loop! Due to the recursion I have some trouble testing my code...
I think the error might be here : tmp = search(path, goal_state, g + 1, threshold) (in the search function). I'm adding only one to the g cost value. It should be correct though, because I can only move a tile 1 place away.
Here's how to call the IDA() function:
initial_state = np.array([8 7 3],[4 1 2],[0 5 6])
goal_state = np.array([1 2 3],[8 0 4],[7 6 5])
IDA(initial_state, goal_state)
Can someone help me on this ?

There are couple of issues in your IDA* implementation. First, what is the purpose of the variable path? I found two purposes of path in your code:
Use as a flag/map to keep the board-states that is already been visited.
Use as a stack to manage recursion states.
But, it is not possible to do both of them by using a single data structure. So, the first modification that your code requires:
Fix-1: Pass current node as a parameter to the search method.
Fix-2: flag should be a data structure that can perform a not in query efficiently.
Now, fix-1 is easy as we can just pass the current visiting node as the parameter in the search method. For fix-2, we need to change the type of flag from list to set as:
list's average case complexity for x in s is: O(n)
set's
Average case complexity for x in s is: O(1)
Worst case complexity for x in s is: O(n)
You can check more details about performance for testing memberships: list vs sets for more details.
Now, to keep the Node information into a set, you need to implement __eq__ and __hash__ in your Node class. In the following, I have attached the modified code.
import timeit
import numpy as np
def IDA(initial_state, goal_state):
initial_node = Node(initial_state)
goal_node = Node(goal_state)
threshold = manhattan_heuristic(initial_state, goal_state)
#print("heuristic threshold: {}".format(threshold))
loop_counter = 0
while 1:
path = set([initial_node])
tmp = search(initial_node, goal_state, 0, threshold, path)
#print("tmp: {}".format(tmp))
if tmp == True:
return True, threshold
elif tmp == float('inf'):
return False, float('inf')
else:
threshold = tmp
def search(node, goal_state, g, threshold, path):
#print("node-state: {}".format(node.state))
f = g + manhattan_heuristic(node.state, goal_state)
if f > threshold:
return f
if np.array_equal(node.state, goal_state):
return True
minimum = float('inf')
for n in node.nextnodes():
if n not in path:
path.add(n)
tmp = search(n, goal_state, g + 1, threshold, path)
if tmp == True:
return True
if tmp < minimum:
minimum = tmp
return minimum
def manhattan_heuristic(state1, state2):
size = range(1, len(state1) ** 2)
distances = [count_distance(num, state1, state2) for num in size]
return sum(distances)
def count_distance(number, state1, state2):
position1 = np.where(state1 == number)
position2 = np.where(state2 == number)
return manhattan_distance(position1, position2)
def manhattan_distance(a, b):
return abs(b[0] - a[0]) + abs(b[1] - a[1])
class Node():
def __init__(self, state):
self.state = state
def __repr__(self):
return np.array_str(self.state.flatten())
def __hash__(self):
return hash(self.__repr__())
def __eq__(self, other):
return self.__hash__() == other.__hash__()
def nextnodes(self):
zero = np.where(self.state == 0)
y,x = zero
y = int(y)
x = int(x)
up = (y - 1, x)
down = (y + 1, x)
right = (y, x + 1)
left = (y, x - 1)
arr = []
for direction in (up, down, right, left):
if len(self.state) - 1 >= direction[0] >= 0 and len(self.state) - 1 >= direction[1] >= 0:
tmp = np.copy(self.state)
tmp[direction[0], direction[1]], tmp[zero] = tmp[zero], tmp[direction[0], direction[1]]
arr.append(Node(tmp))
return arr
initial_state = np.array([[8, 7, 3],[4, 1, 2],[0, 5, 6]])
goal_state = np.array([[1, 2, 3],[8, 0, 4],[7, 6, 5]])
start = timeit.default_timer()
is_found, th = IDA(initial_state, goal_state)
stop = timeit.default_timer()
print('Time: {} seconds'.format(stop - start))
if is_found is True:
print("Solution found with heuristic-upperbound: {}".format(th))
else:
print("Solution not found!")
Node: Please double check your Node.nextnodes() and manhattan_heuristic() methods as I did not pay much attention in those areas. You can check this GitHub repository for other algorithmic implementations (i.e., A*, IDS, DLS) to solve this problem.
References:
Python Wiki: Time Complexity
TechnoBeans: Performance for testing memberships: list vs tuples vs sets
GitHub: Puzzle Solver (by using problem solving techniques)

My A-star implementation seems very slow, need advice and help on what I am doing wrong

My tests of my implementations of Dijkstra and A-Star have revealed that my A-star implementation is approximately 2 times SLOWER. Usually equivalent implementations of Dijkstra and A-star should see A-star beating out Dijkstra. But that isn't the case here and so it has led me to question my implementation of A-star. So I want someone to tell me what I am doing wrong in my implementation of A-star.
Here is my code:
from copy import deepcopy
from math import inf, sqrt
import maze_builderV2 as mb
if __name__ == '__main__':
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)][z] = [sqrt((pos[0]-z[0])**2+(pos[1]-z[1])**2),
sqrt((finalpos[0]-z[0])**2+(finalpos[1]-z[1])**2)] # Euclidean distance to destination (Heuristic)
scan()
unvisited = deepcopy(mazemap)
distances = {}
paths = {}
# Initialization of distances:
for node in unvisited:
if node == pos:
distances[node] = [0, sqrt((finalpos[0]-node[0])**2+(finalpos[1]-node[1])**2)]
else:
distances[node] = [inf, inf]
while unvisited != {}:
curnode = None
for node in unvisited:
if curnode == None:
curnode = node
elif (distances[node][0]+distances[node][1]) < (distances[curnode][0]+distances[curnode][1]):
curnode = node
else:
pass
for childnode, lengths in mazemap[curnode].items():
# Length to nearby childnode - G length, Euclidean (Heuristic) length from curnode to finalpos - H length
# G length + H length < Euclidean length to reach that childnode directly + Euclidean length to finalpos from that childnode = Better path found, update known distance and paths
if lengths[0] + lengths[1] < distances[childnode][0] + distances[childnode][1]:
distances[childnode] = [lengths[0], lengths[1]]
paths[childnode] = curnode
unvisited.pop(curnode)
def shortestroute(paths, start, end):
shortestpath = []
try:
def rec(start, end):
if end == start:
shortestpath.append(end)
return shortestpath[::-1]
else:
shortestpath.append(end)
return rec(start, paths[end])
return rec(start, end)
except KeyError:
return False
finalpath = shortestroute(paths, pos, finalpos)
if finalpath:
for x in finalpath:
if x == pos or x == finalpos:
pass
else:
maze[x[0]][x[1]] = 'W'
else:
print("This maze not solvable, Blyat!")
print()
spit()
For those who find my code too messy and can't bother to read the comments I added to help with the reading... Here is a gist of my code:
Creates a mazemap (all the coordinates and its connected neighbors along with their euclidean distances from that neighboring point to the start position (G Cost) as well as to the final position (H Cost)... in a dictionary)
start position is selected as the current node. All distances to other nodes is initialised as infinity.
For every node we compare the total path cost i.e is the G cost + H cost. The one with least total cost is selected as then next current node. Each time we select new current node, we add that node to a dictionary that keeps track of through which node it was reached, so that it is easier to backtrack and find our path.
Process continues until current node is the final position.
If anyone can help me out on this, that would be great!
EDIT: On account of people asking for the maze building algorithm, here it is:
# Maze generator - v2: Generates mazes that look like city streets (more or less...)
from copy import deepcopy
from random import randint, choice
if __name__ == "__main__":
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)])
pos = (1, 1)
finalpos = (order, order)
maze[pos[0]][pos[1]] = 'S' # Initializing a start position
maze[finalpos[1]][finalpos[1]] = 'O' # Initializing a end position
def spit():
for x in maze:
print(x)
blocks = []
freespaces = [(x, y) for x in range(1, order+1) for y in range(1, order+1)]
def blockbuilder(kind):
param1 = param2 = 0
double = randint(0, 1)
if kind == 0:
param2 = randint(3, 5)
if double:
param1 = 2
else:
param1 = 1
else:
param1 = randint(3, 5)
if double:
param2 = 2
else:
param2 = 1
for a in range(blockstarter[0], blockstarter[0]+param2):
for b in range(blockstarter[1], blockstarter[1]+param1):
if (a+1, b) in blocks or (a-1, b) in blocks or (a, b+1) in blocks or (a, b-1) in blocks or (a, b) in blocks or (a+1, b+1) in blocks or (a-1, b+1) in blocks or (a+1, b-1) in blocks or (a-1, b-1) in blocks:
pass
else:
if a > order+1 or b > order+1:
pass
else:
if maze[a][b] == 'X':
blocks.append((a, b))
else:
spaces = [(a+1, b), (a-1, b), (a, b+1), (a, b-1)]
for c in spaces:
if maze[c[0]][c[1]] == 'X':
break
else:
maze[a][b] = 'X'
blocks.append((a, b))
for x in range(1, order+1):
for y in range(1, order+1):
if (x, y) in freespaces:
t = [(x+1, y), (x-1, y), (x, y+1), (x, y-1)]
i = 0
while i < len(t):
if maze[t[i][0]][t[i][1]] == 'X' or (t[i][0], t[i][1]) == pos or (t[i][0], t[i][1]) == finalpos:
del t[i]
else:
i += 1
if len(t) > 2:
blockstarter = t[randint(0, len(t)-1)]
kind = randint(0, 1) # 0 - vertical, 1 - horizontal
blockbuilder(kind)
else:
pass
# rch = choice(['d', 'u', 'r', 'l'])
b = 0
while b < len(blocks):
block = blocks[b]
t = {'d': (block[0]+2, block[1]), 'u': (block[0]-2, block[1]),
'r': (block[0], block[1]+2), 'l': (block[0], block[1]-2)}
rch = choice(['d', 'u', 'r', 'l'])
z = t[rch]
# if z[0] > order+1 or z[1] > order+1 or z[0] < 1 or z[1] < 1:
# Decreased chance of having non solvable maze being generated...
if z[0] > order-2 or z[1] > order-2 or z[0] < 2+2 or z[1] < 2+2:
pass
else:
if maze[z[0]][z[1]] == 'X':
if randint(0, 1):
set = None
if rch == 'u':
set = (z[0]+1, z[1])
elif rch == 'd':
set = (z[0]-1, z[1])
elif rch == 'r':
set = (z[0], z[1]-1)
elif rch == 'l':
set = (z[0], z[1]+1)
else:
pass
if maze[set[0]][set[1]] == '_':
# Checks so that no walls that block the entire way are formed
# Makes sure maze is solvable
sets, count = [
(set[0]+1, set[1]), (set[0]-1, set[1]), (set[0], set[1]+1), (set[0], set[1]-1)], 0
for blyat in sets:
while blyat[0] != 0 and blyat[1] != 0 and blyat[0] != order+1 and blyat[1] != order+1:
ch = [(blyat[0]+1, blyat[1]), (blyat[0]-1, blyat[1]),
(blyat[0], blyat[1]+1), (blyat[0], blyat[1]-1)]
suka = []
for i in ch:
if ch not in suka:
if maze[i[0]][i[1]] == 'X':
blyat = i
break
else:
pass
suka.append(ch)
else:
pass
else:
blyat = None
if blyat == None:
break
else:
pass
else:
count += 1
if count < 1:
maze[set[0]][set[1]] = 'X'
blocks.append(set)
else:
pass
else:
pass
else:
pass
b += 1
mazebuilder(maze, order)
spit()
Sorry for leaving this out!

Just at a quick glance, it looks like you don't have a closed set at all?? Your unvisited structure appears to contain every node in the map. This algorithm is not A* at all.
Once you fix that, make sure to change unvisited from a list to a priority queue also.

Slow performance in agent based model python

I originally posted this on code-review (hence the lengthy code) but failed to get an answer.
My model is based on this game https://en.wikipedia.org/wiki/Ultimatum_game . I won't go into the intuition behind it but generally speaking it functions as follows:
The game consists of a n x n lattice on which an agent is placed at each node.
During each time step, each player on each node plays against a random neighbour by playing a particular strategy.
Each of their strategies (a value between 1-9) has a propensity attached to it (which is randomly assigned and is just some number). The propensity then in turn determines the probability of playing that strategy. The probability is calculated as the propensity of that strategy over the sum of the propensities of all strategies.
If a game results in a positive payoff, then the payoffs from that game get added to the propensities for those strategies.
These propensities then determine the probabilities for their strategies in the next time step, and so on.
The simulation ends after time step N is reached.
For games with large lattices and large time steps, my code runs really really slowly. I ran cProfiler to check where the bottleneck(s) are, and as I suspected the update_probabilitiesand play_rounds functions seem to be taking up a lot time. I want to be able to run the game with gridsize of about 40x40 for about 100000+ time steps, but right now that is not happening.
So what would be a more efficient way to calculate and update the probabilities/propensities of each player in the grid? I've considered implementing NumPy arrays but I am not sure if it would be worth the hassle here?
import numpy as np
import random
from random import randint
from numpy.random import choice
from numpy.random import multinomial
import cProfile
mew = 0.001
error = 0.05
def create_grid(row, col):
return [[0 for j in range(col)] for i in range(row)]
def create_random_propensities():
propensities = {}
pre_propensities = [random.uniform(0, 1) for i in range(9)]
a = np.sum(pre_propensities)
for i in range(1, 10):
propensities[i] = (pre_propensities[i - 1]/a) * 10 # normalize sum of propensities to 10
return propensities
class Proposer:
def __init__(self):
self.propensities = create_random_propensities()
self.probabilites = []
self.demand = 0 # the amount the proposer demands for themselves
def pick_strat(self, n_trials): # gets strategy, an integer in the interval [1, 9]
results = multinomial(n_trials, self.probabilites)
i, = np.where(results == max(results))
if len(i) > 1:
return choice(i) + 1
else:
return i[0] + 1
def calculate_probability(self, dict_data, index, total_sum): # calculates probability for particular strat, taking propensity
return dict_data[index]/total_sum # of that strat as input
def calculate_sum(self, dict_data):
return sum(dict_data.values())
def initialize(self):
init_sum = self.calculate_sum(self.propensities)
for strategy in range(1, 10):
self.probabilites.append(self.calculate_probability(self.propensities, strategy, init_sum))
self.demand = self.pick_strat(1)
def update_strategy(self):
self.demand = self.pick_strat(1)
def update_probablities(self):
for i in range(9):
self.propensities[1 + i] *= 1 - mew
pensity_sum = self.calculate_sum(self.propensities)
for i in range(9):
self.probabilites[i] = self.calculate_probability(self.propensities, 1 + i, pensity_sum)
def update(self):
self.update_probablities()
self.update_strategy()
class Responder: # methods same as proposer class, can skip-over
def __init__(self):
self.propensities = create_random_propensities()
self.probabilites = []
self.max_thresh = 0 # the maximum demand they are willing to accept
def pick_strat(self, n_trials):
results = multinomial(n_trials, self.probabilites)
i, = np.where(results == max(results))
if len(i) > 1:
return choice(i) + 1
else:
return i[0] + 1
def calculate_probability(self, dict_data, index, total_sum):
return dict_data[index]/total_sum
def calculate_sum(self, dict_data):
return sum(dict_data.values())
def initialize(self):
init_sum = self.calculate_sum(self.propensities)
for strategy in range(1, 10):
self.probabilites.append(self.calculate_probability(self.propensities, strategy, init_sum))
self.max_thresh = self.pick_strat(1)
def update_strategy(self):
self.max_thresh = self.pick_strat(1)
def update_probablities(self):
for i in range(9):
self.propensities[1 + i] *= 1 - mew # stops sum of propensites from growing without bound
pensity_sum = self.calculate_sum(self.propensities)
for i in range(9):
self.probabilites[i] = self.calculate_probability(self.propensities, 1 + i, pensity_sum)
def update(self):
self.update_probablities()
self.update_strategy()
class Agent:
def __init__(self):
self.prop_side = Proposer()
self.resp_side = Responder()
self.prop_side.initialize()
self.resp_side.initialize()
def update_all(self):
self.prop_side.update()
self.resp_side.update()
class Grid:
def __init__(self, rowsize, colsize):
self.rowsize = rowsize
self.colsize = colsize
def make_lattice(self):
return [[Agent() for j in range(self.colsize)] for i in range(self.rowsize)]
#staticmethod
def von_neumann_neighbourhood(array, row, col, wrapped=True): # gets up, bottom, left, right neighbours of some node
neighbours = set([])
if row + 1 <= len(array) - 1:
neighbours.add(array[row + 1][col])
if row - 1 >= 0:
neighbours.add(array[row - 1][col])
if col + 1 <= len(array[0]) - 1:
neighbours.add(array[row][col + 1])
if col - 1 >= 0:
neighbours.add(array[row][col - 1])
#if wrapped is on, conditions for out of bound points
if row - 1 < 0 and wrapped == True:
neighbours.add(array[-1][col])
if col - 1 < 0 and wrapped == True:
neighbours.add(array[row][-1])
if row + 1 > len(array) - 1 and wrapped == True:
neighbours.add(array[0][col])
if col + 1 > len(array[0]) - 1 and wrapped == True:
neighbours.add(array[row][0])
return neighbours
def get_error_term(pay, strategy):
index_strat_2, index_strat_8 = 2, 8
if strategy == 1:
return (1 - (error/2)) * pay, error/2 * pay, index_strat_2
if strategy == 9:
return (1 - (error/2)) * pay, error/2 * pay, index_strat_8
else:
return (1 - error) * pay, error/2 * pay, 0
class Games:
def __init__(self, n_rows, n_cols, n_rounds):
self.rounds = n_rounds
self.rows = n_rows
self.cols = n_cols
self.lattice = Grid(self.rows, self.cols).make_lattice()
self.lookup_table = np.full((self.rows, self.cols), False, dtype=bool) # if player on grid has updated their strat, set to True
def reset_look_tab(self):
self.lookup_table = np.full((self.rows, self.cols), False, dtype=bool)
def run_game(self):
n = 0
while n < self.rounds:
for r in range(self.rows):
for c in range(self.cols):
if n != 0:
self.lattice[r][c].update_all()
self.lookup_table[r][c] = True
self.play_rounds(self.lattice, r, c)
self.reset_look_tab()
n += 1
def play_rounds(self, grid, row, col):
neighbours = Grid.von_neumann_neighbourhood(grid, row, col)
neighbour = random.sample(neighbours, 1).pop()
neighbour_index = [(ix, iy) for ix, row in enumerate(self.lattice) for iy, i in enumerate(row) if i == neighbour]
if self.lookup_table[neighbour_index[0][0]][neighbour_index[0][1]] == False: # see if neighbour has already updated their strat
neighbour.update_all()
player = grid[row][col]
coin_toss = randint(0, 1) # which player acts as proposer or responder in game
if coin_toss == 1:
if player.prop_side.demand <= neighbour.resp_side.max_thresh: # postive payoff
payoff, adjacent_payoff, index = get_error_term(player.prop_side.demand, player.prop_side.demand)
if player.prop_side.demand == 1 or player.prop_side.demand == 9: # extreme strategies get bonus payoffs
player.prop_side.propensities[player.prop_side.demand] += payoff
player.prop_side.propensities[index] += adjacent_payoff
else:
player.prop_side.propensities[player.prop_side.demand] += payoff
player.prop_side.propensities[player.prop_side.demand - 1] += adjacent_payoff
player.prop_side.propensities[player.prop_side.demand + 1] += adjacent_payoff
else:
return 0 # if demand > max thresh -> both get zero
if coin_toss != 1:
if neighbour.prop_side.demand <= player.resp_side.max_thresh:
payoff, adjacent_payoff, index = get_error_term(10 - neighbour.prop_side.demand, player.resp_side.max_thresh)
if player.resp_side.max_thresh == 1 or player.resp_side.max_thresh == 9:
player.resp_side.propensities[player.resp_side.max_thresh] += payoff
player.resp_side.propensities[index] += adjacent_payoff
else:
player.resp_side.propensities[player.resp_side.max_thresh] += payoff
player.resp_side.propensities[player.resp_side.max_thresh - 1] += adjacent_payoff
player.resp_side.propensities[player.resp_side.max_thresh + 1] += adjacent_payoff
else:
return 0
#pr = cProfile.Profile()
#pr.enable()
my_game = Games(10, 10, 2000) # (rowsize, colsize, n_steps)
my_game.run_game()
#pr.disable()
#pr.print_stats(sort='time')
(For those who might be wondering, the get_error_term just returns the propensities for strategies that are next to strategies that receive a positive payoff, for example if the strategy 8 works, then 7 and 9's propensities also get adjusted upwards and this is calculated by said function. And the first for loop inside update_probabilities just makes sure that the sum of propensities don't grow without bound).