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I have tried to implement the Naive and Heap Dijkstra as shown below but somehow my naive Dijkstra implementation is surprisingly faster. I debugged my code but couldn't understand where my problem in my implementations are.
Why is my heap-based implementation is slower than the naive implementation?
Raw data is stored here:
https://www.algorithmsilluminated.org/datasets/problem9.8.txt
Data Import and Manipulation:
import time
with open("DijkstraTest2.txt", 'r') as input:
lines = input.readlines()
lengths = {}
vertices = []
for line in lines:
contents = line.split("\t")
vertices.append(contents[0])
for content in contents:
content = content.replace('\n', '')
if ',' in content:
edge = contents[0] + '-' + content.split(',')[0]
lengths[edge] = int(content.split(',')[1])
Naive Dijkstra:
def NaiveDijkstra(vertices, start_point, lengths):
X = [start_point]
shortest_paths = {}
for vertex in vertices:
if vertex == start_point:
shortest_paths[vertex] = 0
else:
shortest_paths[vertex] = 999999999999
subset = [key for key in lengths.keys() if start_point == key.split('-')[0]
and key.split('-')[0] in X and key.split('-')[1] not in X]
while len(subset) > 0:
temp_min_dict = {}
for edge in subset:
temp_min = shortest_paths[edge.split('-')[0]] + lengths[edge]
temp_min_dict[edge] = temp_min
new_edge = min(temp_min_dict, key=temp_min_dict.get)
X.append(new_edge.split('-')[1])
shortest_paths[new_edge.split('-')[1]] = shortest_paths[new_edge.split('-')[0]] + lengths[new_edge]
subset = []
for key in lengths.keys():
if key.split('-')[0] in X and key.split('-')[1] not in X:
subset.append(key)
return shortest_paths
start_time = time.time()
print(NaiveDijkstra(vertices = vertices, start_point = '1', lengths = lengths)['197'])
print(time.time() - start_time, "seconds")
My Heap based Dijkstra code:
class Heap:
def __init__(self):
self.size = 0
self.lst = []
def swap(self, a):
if self.size == 1:
return self.lst
else:
if a == 1:
i = 1
else:
i = a // 2
while i > 0:
if i * 2 - 1 >= self.size:
break
elif self.lst[i - 1][1] > self.lst[i * 2 - 1][1]:
temp = self.lst[i - 1]
self.lst[i - 1] = self.lst[i * 2 - 1]
self.lst[i * 2 - 1] = temp
elif i * 2 >= self.size:
break
elif self.lst[i - 1][1] > self.lst[i * 2][1]:
temp = self.lst[i - 1]
self.lst[i - 1] = self.lst[i * 2]
self.lst[i * 2] = temp
i -= 1
# print(f"output: {self.lst}")
def insert(self, element):
# print(f"input: {self.lst}")
self.lst.append(element)
self.size += 1
self.swap(self.size)
def extractmin(self):
val = self.lst.pop(0)[0]
self.size -= 1
self.swap(self.size - 1)
return val
def delete(self, deleted):
ix = self.lst.index(deleted)
temp = self.lst[-1]
self.lst[ix] = temp
self.lst[-1] = deleted
self.lst.pop(-1)
self.size -= 1
#self.swap(self.size)
def FastDijkstra(vertices, start_point, lengths):
X = []
h = Heap()
width = {}
shortest_paths = {}
for vertex in vertices:
if vertex == start_point:
width[vertex] = 0
h.insert((vertex, width[vertex]))
else:
width[vertex] = 999999999999
h.insert((vertex, width[vertex]))
while h.size > 0:
w = h.extractmin()
X.append(w)
shortest_paths[w] = width[w]
Y = set(vertices).difference(X)
for x in X:
for y in Y:
key = f"{x}-{y}"
if lengths.get(key) is not None:
h.delete((y, width[y]))
if width[y] > shortest_paths[x] + lengths[key]:
width[y] = shortest_paths[x] + lengths[key]
h.insert((y, width[y]))
return shortest_paths
start_time = time.time()
print(FastDijkstra(vertices=vertices, start_point='1', lengths=lengths)['197'])
print(time.time() - start_time, "seconds")
The way you implemented the heap version is not efficient. Notably the following make it inefficient:
All nodes are put on the heap instead of only the direct neighbors of the visited nodes. This makes the heap large and slower than needed.
Y = set(vertices).difference(X) is a slow operation, and makes Y unnecessary large.
The nested loop that tries every pair in the Cartesian product of X and Y to see if it is an edge. This point together with the previous should be replaced with a collection of edges starting from X, and then discarding edges that lead to already visited nodes.
For every found edge to delete the target node from the heap, and re-insert it, even if the width didn't change! Deletion is a costly operation (see next point). Only if the Heap implementation supports a decrease-key operation, this is an option, but otherwise the heap should just get an extra entry for the same vertex, knowing that the one with the lowest cost will come out of the heap first.
The heap's delete method has a bad time complexity due to the .index() call.
The heap's extractmin method has a bad time complexity, due to the .pop(0) call. This has O(n) time complexity.
The heap's extractmin does not give correct results (again due to that pop(0)). Here is a little script that shows a mistake:
h = Heap()
for value in 4, 3, 5, 2, 1:
h.insert((value, value))
print(h.extractmin()) # 1 = OK
print(h.extractmin()) # 2 = OK
print(h.extractmin()) # 4 = NOK. 3 expected.
The data structure lengths does not allow to quickly find the edges from a particular vertex. But this is a point that is also making the naive implementation slow. I would however suggest to turn that in a dict.
If this is done right it should run faster. Certainly when you would make use of the native heapq module you'll get good running times.
Here is a (much) faster implementation. It doesn't bother about unreachable vertices, and doesn't bother about possibly having multiple entries on the heap for the same node (with different distances). But it does start with only the starting node on the heap, and uses heapq:
from heapq import heappush, heappop
from collections import defaultdict
def FastDijkstra(vertices, start_point, lengths):
# Create a dictionary for the edges, keyed by source node
edges = defaultdict(list)
for key, length in lengths.items():
x, y = key.split("-")
edges[x].append((length, y))
heap = [(0, start_point)]
shortest_paths = {}
while heap:
cost, x = heappop(heap)
if x in shortest_paths:
continue # this vertex had already been on the heap before
shortest_paths[x] = cost
for length, y in edges[x]:
if y not in shortest_paths:
heappush(heap, (cost + length, y))
return shortest_paths
In my tests this ran hundreds times faster.
Thanks to the above answer (wonderful analysis) I adjusted my implementation which is way faster than the previous version. It is shown below.
class Heap:
def __init__(self):
self.size = 0
self.lst = []
def swap(self, a):
if self.size == 1:
return self.lst
else:
if a == 1:
i = 1
else:
i = a // 2
while i > 0:
if i * 2 - 1 >= self.size:
break
elif self.lst[i - 1][1] > self.lst[i * 2 - 1][1]:
temp = self.lst[i - 1]
self.lst[i - 1] = self.lst[i * 2 - 1]
self.lst[i * 2 - 1] = temp
elif i * 2 >= self.size:
break
elif self.lst[i - 1][1] > self.lst[i * 2][1]:
temp = self.lst[i - 1]
self.lst[i - 1] = self.lst[i * 2]
self.lst[i * 2] = temp
elif self.lst[2*i - 1][1] > self.lst[i * 2][1]:
temp = self.lst[2*i - 1]
self.lst[2*i - 1] = self.lst[i * 2]
self.lst[i * 2] = temp
i -= 1
#print(f"output: {self.lst}")
def insert(self, element):
#print(f"input: {self.lst}")
self.lst.append(element)
self.size += 1
self.swap(self.size)
def extractmin(self):
val = self.lst[0][0]
del self.lst[0]
self.size -= 1
self.swap(self.size-1)
return val
def delete(self, deleted):
ix = self.lst.index(deleted)
temp = self.lst[-1]
self.lst[ix] = temp
self.lst[-1] = deleted
del self.lst[-1]
self.size -= 1
#self.swap(self.size)
def FastDijkstra(vertices, start_point, lengths):
X = []
h = Heap()
width = {}
shortest_paths = {}
for vertex in vertices:
if vertex == start_point:
width[vertex] = 0
h.insert((vertex, width[vertex]))
else:
width[vertex] = 999999999999
h.insert((vertex, width[vertex]))
while h.size > 0:
w = h.extractmin()
X.append(w)
shortest_paths[w] = width[w]
Y = set(vertices).difference(X)
for y in Y:
key = f"{w}-{y}"
if lengths.get(key) is not None:
h.delete((y, width[y]))
if width[y] > shortest_paths[w] + lengths[key]:
width[y] = shortest_paths[w] + lengths[key]
h.insert((y, width[y]))
return shortest_paths
start_time = time.time()
print(FastDijkstra(vertices=vertices, start_point='1', lengths=lengths)['197'])
print(time.time() - start_time, "seconds")
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)
This question already has answers here:
Big O, how do you calculate/approximate it?
(24 answers)
Closed 2 years ago.
def is_prime(x):
'''
Function to check if a number is prime
'''
if x == 2:
return True
if x%2 != 0: #Check if number is even since all primes are odd except 2
a = [x % i for i in range(2,x+1)]
b = [i for i in a if i == 0] # Checks to make sure there's only one modulus of 0
if len(b) == 1:
return True
else:
return False
else:
return False
So like yeah please what is the time complexity (all those 0/n things) and how do i find that, a good resource link would be helpful (:
Your complexity is O(x) as you run one loop from 2 to x+1.
You can just check upto sqrt(x). That will bring down the complexity to O(sqrt(x)). (You can break once you find one factor, even though it won't bring down the worst time complexity)
Just change this line -
a = [x % i for i in range(2,math.sqrt(x+1))]
Why up to square root?
You can just google, there are many proofs. The simple one being, if
a = b.c, then the at least there is one divisor of a which is less than sqrt(a), or equal if a is square number a = b*b.
There are many fast heuristic and probabilistic (Miller-Rabin is very famous and frequently used) algorithms for faster prime detection.
Here's one which is deterministic:
The Adleman–Pomerance–Rumely primality test is an algorithm for determining whether a number is prime. Unlike other, more efficient algorithms for this purpose, it avoids the use of random numbers, so it is a deterministic primality test.
It's complexity is log(x)^O(logloglogx)
import copy
import time
from math import gcd # version >= 3.5
# primality test by trial division
def isprime_slow(n):
if n<2:
return False
elif n==2 or n==3:
return True
elif n%2==0:
return False
else:
i = 3
while i*i <= n:
if n%i == 0:
return False
i+=2
return True
# v_q(t): how many time is t divided by q
def v(q, t):
ans = 0
while(t % q == 0):
ans +=1
t//= q
return ans
def prime_factorize(n):
ret = []
p=2
while p*p <= n:
if n%p==0:
num = 0
while n%p==0:
num+=1
n//= p
ret.append((p,num))
p+= 1
if n!=1:
ret.append((n,1))
return ret
# calculate e(t)
def e(t):
s = 1
q_list = []
for q in range(2, t+2):
if t % (q-1) == 0 and isprime_slow(q):
s *= q ** (1+v(q,t))
q_list.append(q)
return 2*s, q_list
# Jacobi sum
class JacobiSum(object):
def __init__(self, p, k, q):
self.p = p
self.k = k
self.q = q
self.m = (p-1)*p**(k-1)
self.pk = p**k
self.coef = [0]*self.m
# 1
def one(self):
self.coef[0] = 1
for i in range(1,self.m):
self.coef[i] = 0
return self
# product of JacobiSum
# jac : JacobiSum
def mul(self, jac):
m = self.m
pk = self.pk
j_ret=JacobiSum(self.p, self.k, self.q)
for i in range(m):
for j in range(m):
if (i+j)% pk < m:
j_ret.coef[(i+j)% pk] += self.coef[i] * jac.coef[j]
else:
r = (i+j) % pk - self.p ** (self.k-1)
while r>=0:
j_ret.coef[r] -= self.coef[i] * jac.coef[j]
r-= self.p ** (self.k-1)
return j_ret
def __mul__(self, right):
if type(right) is int:
# product with integer
j_ret=JacobiSum(self.p, self.k, self.q)
for i in range(self.m):
j_ret.coef[i] = self.coef[i] * right
return j_ret
else:
# product with JacobiSum
return self.mul(right)
# power of JacobiSum（x-th power mod n）
def modpow(self, x, n):
j_ret=JacobiSum(self.p, self.k, self.q)
j_ret.coef[0]=1
j_a = copy.deepcopy(self)
while x>0:
if x%2==1:
j_ret = (j_ret * j_a).mod(n)
j_a = j_a*j_a
j_a.mod(n)
x //= 2
return j_ret
# applying "mod n" to coefficient of self
def mod(self, n):
for i in range(self.m):
self.coef[i] %= n
return self
# operate sigma_x
# verification for sigma_inv
def sigma(self, x):
m = self.m
pk = self.pk
j_ret=JacobiSum(self.p, self.k, self.q)
for i in range(m):
if (i*x) % pk < m:
j_ret.coef[(i*x) % pk] += self.coef[i]
else:
r = (i*x) % pk - self.p ** (self.k-1)
while r>=0:
j_ret.coef[r] -= self.coef[i]
r-= self.p ** (self.k-1)
return j_ret
# operate sigma_x^(-1)
def sigma_inv(self, x):
m = self.m
pk = self.pk
j_ret=JacobiSum(self.p, self.k, self.q)
for i in range(pk):
if i<m:
if (i*x)%pk < m:
j_ret.coef[i] += self.coef[(i*x)%pk]
else:
r = i - self.p ** (self.k-1)
while r>=0:
if (i*x)%pk < m:
j_ret.coef[r] -= self.coef[(i*x)%pk]
r-= self.p ** (self.k-1)
return j_ret
# Is self p^k-th root of unity (mod N)
# if so, return h where self is zeta^h
def is_root_of_unity(self, N):
m = self.m
p = self.p
k = self.k
# case of zeta^h (h<m)
one = 0
for i in range(m):
if self.coef[i]==1:
one += 1
h = i
elif self.coef[i] == 0:
continue
elif (self.coef[i] - (-1)) %N != 0:
return False, None
if one == 1:
return True, h
# case of zeta^h (h>=m)
for i in range(m):
if self.coef[i]!=0:
break
r = i % (p**(k-1))
for i in range(m):
if i % (p**(k-1)) == r:
if (self.coef[i] - (-1))%N != 0:
return False, None
else:
if self.coef[i] !=0:
return False, None
return True, (p-1)*p**(k-1)+ r
# find primitive root
def smallest_primitive_root(q):
for r in range(2, q):
s = set({})
m = 1
for i in range(1, q):
m = (m*r) % q
s.add(m)
if len(s) == q-1:
return r
return None # error
# calculate f_q(x)
def calc_f(q):
g = smallest_primitive_root(q)
m = {}
for x in range(1,q-1):
m[pow(g,x,q)] = x
f = {}
for x in range(1,q-1):
f[x] = m[ (1-pow(g,x,q))%q ]
return f
# sum zeta^(a*x+b*f(x))
def calc_J_ab(p, k, q, a, b):
j_ret = JacobiSum(p,k,q)
f = calc_f(q)
for x in range(1,q-1):
pk = p**k
if (a*x+b*f[x]) % pk < j_ret.m:
j_ret.coef[(a*x+b*f[x]) % pk] += 1
else:
r = (a*x+b*f[x]) % pk - p**(k-1)
while r>=0:
j_ret.coef[r] -= 1
r-= p**(k-1)
return j_ret
# calculate J(p,q)（p>=3 or p,q=2,2）
def calc_J(p, k, q):
return calc_J_ab(p, k, q, 1, 1)
# calculate J_3(q)（p=2 and k>=3）
def calc_J3(p, k, q):
j2q = calc_J(p, k, q)
j21 = calc_J_ab(p, k, q, 2, 1)
j_ret = j2q * j21
return j_ret
# calculate J_2(q)（p=2 and k>=3）
def calc_J2(p, k, q):
j31 = calc_J_ab(2, 3, q, 3, 1)
j_conv = JacobiSum(p, k, q)
for i in range(j31.m):
j_conv.coef[i*(p**k)//8] = j31.coef[i]
j_ret = j_conv * j_conv
return j_ret
# in case of p>=3
def APRtest_step4a(p, k, q, N):
print("Step 4a. (p^k, q = {0}^{1}, {2})".format(p,k,q))
J = calc_J(p, k, q)
# initialize s1=1
s1 = JacobiSum(p,k,q).one()
# J^Theta
for x in range(p**k):
if x % p == 0:
continue
t = J.sigma_inv(x)
t = t.modpow(x, N)
s1 = s1 * t
s1.mod(N)
# r = N mod p^k
r = N % (p**k)
# s2 = s1 ^ (N/p^k)
s2 = s1.modpow(N//(p**k), N)
# J^alpha
J_alpha = JacobiSum(p,k,q).one()
for x in range(p**k):
if x % p == 0:
continue
t = J.sigma_inv(x)
t = t.modpow((r*x)//(p**k), N)
J_alpha = J_alpha * t
J_alpha.mod(N)
# S = s2 * J_alpha
S = (s2 * J_alpha).mod(N)
# Is S root of unity
exist, h = S.is_root_of_unity(N)
if not exist:
# composite!
return False, None
else:
# possible prime
if h%p!=0:
l_p = 1
else:
l_p = 0
return True, l_p
# in case of p=2 and k>=3
def APRtest_step4b(p, k, q, N):
print("Step 4b. (p^k, q = {0}^{1}, {2})".format(p,k,q))
J = calc_J3(p, k, q)
# initialize s1=1
s1 = JacobiSum(p,k,q).one()
# J3^Theta
for x in range(p**k):
if x % 8 not in [1,3]:
continue
t = J.sigma_inv(x)
t = t.modpow(x, N)
s1 = s1 * t
s1.mod(N)
# r = N mod p^k
r = N % (p**k)
# s2 = s1 ^ (N/p^k)
s2 = s1.modpow(N//(p**k), N)
# J3^alpha
J_alpha = JacobiSum(p,k,q).one()
for x in range(p**k):
if x % 8 not in [1,3]:
continue
t = J.sigma_inv(x)
t = t.modpow((r*x)//(p**k), N)
J_alpha = J_alpha * t
J_alpha.mod(N)
# S = s2 * J_alpha * J2^delta
if N%8 in [1,3]:
S = (s2 * J_alpha ).mod(N)
else:
J2_delta = calc_J2(p,k,q)
S = (s2 * J_alpha * J2_delta).mod(N)
# Is S root of unity
exist, h = S.is_root_of_unity(N)
if not exist:
# composite
return False, None
else:
# possible prime
if h%p!=0 and (pow(q,(N-1)//2,N) + 1)%N==0:
l_p = 1
else:
l_p = 0
return True, l_p
# in case of p=2 and k=2
def APRtest_step4c(p, k, q, N):
print("Step 4c. (p^k, q = {0}^{1}, {2})".format(p,k,q))
J2q = calc_J(p, k, q)
# s1 = J(2,q)^2 * q (mod N)
s1 = (J2q * J2q * q).mod(N)
# s2 = s1 ^ (N/4)
s2 = s1.modpow(N//4, N)
if N%4 == 1:
S = s2
elif N%4 == 3:
S = (s2 * J2q * J2q).mod(N)
else:
print("Error")
# Is S root of unity
exist, h = S.is_root_of_unity(N)
if not exist:
# composite
return False, None
else:
# possible prime
if h%p!=0 and (pow(q,(N-1)//2,N) + 1)%N==0:
l_p = 1
else:
l_p = 0
return True, l_p
# in case of p=2 and k=1
def APRtest_step4d(p, k, q, N):
print("Step 4d. (p^k, q = {0}^{1}, {2})".format(p,k,q))
S2q = pow(-q, (N-1)//2, N)
if (S2q-1)%N != 0 and (S2q+1)%N != 0:
# composite
return False, None
else:
# possible prime
if (S2q + 1)%N == 0 and (N-1)%4==0:
l_p=1
else:
l_p=0
return True, l_p
# Step 4
def APRtest_step4(p, k, q, N):
if p>=3:
result, l_p = APRtest_step4a(p, k, q, N)
elif p==2 and k>=3:
result, l_p = APRtest_step4b(p, k, q, N)
elif p==2 and k==2:
result, l_p = APRtest_step4c(p, k, q, N)
elif p==2 and k==1:
result, l_p = APRtest_step4d(p, k, q, N)
else:
print("error")
if not result:
print("Composite")
return result, l_p
def APRtest(N):
t_list = [
2,
12,
60,
180,
840,
1260,
1680,
2520,
5040,
15120,
55440,
110880,
720720,
1441440,
4324320,
24504480,
73513440
]
print("N=", N)
if N<=3:
print("input should be greater than 3")
return False
# Select t
for t in t_list:
et, q_list = e(t)
if N < et*et:
break
else:
print("t not found")
return False
print("t=", t)
print("e(t)=", et, q_list)
# Step 1
print("=== Step 1 ===")
g = gcd(t*et, N)
if g > 1:
print("Composite")
return False
# Step 2
print("=== Step 2 ===")
l = {}
fac_t = prime_factorize(t)
for p, k in fac_t:
if p>=3 and pow(N, p-1, p*p)!=1:
l[p] = 1
else:
l[p] = 0
print("l_p=", l)
# Step 3 & Step 4
print("=== Step 3&4 ===")
for q in q_list:
if q == 2:
continue
fac = prime_factorize(q-1)
for p,k in fac:
# Step 4
result, l_p = APRtest_step4(p, k, q, N)
if not result:
# composite
print("Composite")
return False
elif l_p==1:
l[p] = 1
# Step 5
print("=== Step 5 ===")
print("l_p=", l)
for p, value in l.items():
if value==0:
# try other pair of (p,q)
print("Try other (p,q). p={}".format(p))
count = 0
i = 1
found = False
# try maximum 30 times
while count < 30:
q = p*i+1
if N%q != 0 and isprime_slow(q) and (q not in q_list):
count += 1
k = v(p, q-1)
# Step 4
result, l_p = APRtest_step4(p, k, q, N)
if not result:
# composite
print("Composite")
return False
elif l_p == 1:
found = True
break
i += 1
if not found:
print("error in Step 5")
return False
# Step 6
print("=== Step 6 ===")
r = 1
for t in range(t-1):
r = (r*N) % et
if r!=1 and r!= N and N % r == 0:
print("Composite", r)
return False
# prime!!
print("Prime!")
return True
if __name__ == '__main__':
start_time = time.time()
APRtest(2**521-1) # 157 digit, 18 sec
# APRtest(2**1279-1) # 386 digit, 355 sec
# APRtest(2074722246773485207821695222107608587480996474721117292752992589912196684750549658310084416732550077)
end_time = time.time()
print(end_time - start_time, "sec")
credit: https://github.com/wacchoz/APR_CL/blob/master/APR_CL.py
You only need to test odd numbers, except for 2 which you can test specially before the loop. This doesn't change the complexity, since it's a constant factor, but it reduces the time in half.
You should only test numbers up to math.sqrt(x). This changes the worst case complexity from O(n) to O(sqrt(n)).
You should stop as soon as you find a factor, rather than creating a list of all the x % i. This improves the best case complexity.
import math
def is_prime(x):
'''
Function to check if a number is prime
'''
if x == 2:
return True
if x % 2 == 0:
return False
for i in range(3, int(math.sqrt(x))+1, 2):
if x % i == 0:
return False
return True
Even better than checking all odd numbers is to check only prime numbers, using the Sieve of Eratosthenes.
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).
I'm trying to recreate the n-queens problem and solve it with simulated annealing, although the board object from my Object class is throwing an error when I try to add the temperature using len(board)**2. Any help would be very appreciated!! I have included the source code and the output. Thanks!
import time
import random
import math
class Board(object):
"""An N-queens solution attempt."""
def __init__(self, queens):
"""Instances differ by their queen placements."""
self.queens = queens.copy()
def display(self):
"""Print the board."""
for r in range(len(self.queens)):
for c in range(len(self.queens)):
if self.queens[c] == r:
print 'Q',
else:
print '-',
print
print
def moves(self):
"""Return a list of possible moves given the current placements."""
bestMoves = []
optimalHeuristic = heuristic(board)
for a, b in moves.iteritems():
if b < optimalHeuristic:
optimalHeuristic = b
for a, b in moves.iteritems():
if b == optimalHeuristic:
bestMoves.append(a)
return bestMoves
def neighbor(self, move):
"""Return a Board instance like this one but with one move made."""
if len(bestMoves) > 0:
pick = random.randint(0, len(bestMoves) - 1)
column = bestMoves[pick][0]
row = bestMoves[pick][1]
board[column] = row
return board
def heuristic(self):
"""Compute the cost of this solution."""
h = 0
##checking columns
for i in range(1, n):
##checking rows
for j in range(i+1, n):
if board[i] == board[j]:
h += 1
x = j - i
##checking the diagonals
if board[i] == board[j] - x or board[i] == board[j] + x:
h += 1
return h
class Agent(object):
"""Knows how to solve an n-queens problem with simulated annealing."""
def anneal(self, board):
"""Return a list of moves to adjust queen placements."""
temperature = len(board)**2
annealRate = 0.95
newHeuristic = heuristic(board)
while newHeuristic > 0:
board = makeMove(board, newHeuristic, temperature)
newHeuristic = heuristic(board)
newTemperature = max(temperature * annealRate, 0.01)
temperature = newTemperature
##steps cap is here to avoid the algorithm getting stuck
if steps >= 10000:
break
boardCopy = list(board)
foundMove = False
while not foundMove:
boardCopy = list(board)
newRow = random.randint(0, len(board)-1)
newColumn = random.randint(0, len(board)-1)
boardCopy[newColumn] = newRow
newHeuristic = heuristic(boardCopy)
if newHeuristic < optimalHeuristic:
foundMove = True
else:
delta_e = optimalHeuristic - newHeuristic
##aceptance prob equation min(1, e**(delta e/temp))
acceptProbability = min(1, math.exp(delta_e/temperature))
foundMove = random.random() <= acceptProbability
return boardCopy
def main():
"""Create a problem, solve it with simulated anealing, and console-animate."""
print("Enter the number of queens")
n = input()
queens = dict()
for column in range(n):
row = random.choice(range(n))
queens[column] = row
board = Board(queens)
board.display()
agent = Agent()
path = agent.anneal(board)
while path:
move = path.pop(0)
board = board.neighbor(move)
time.sleep(0.1)
board.display()
if __name__ == '__main__':
main()
Output:
Enter the number of queens
8
- - - - - - Q -
- - - - - Q - -
- - - - - - - Q
- - - - - - - -
- Q - - - - - -
- - - Q Q - - -
- - - - - - - -
Q - Q - - - - -
Traceback (most recent call last):
File "F:\Intelligent Systems\annealingnqueens.py", line 119, in <module>
main()
File "F:\Intelligent Systems\annealingnqueens.py", line 110, in main
path = agent.anneal(board)
File "F:\Intelligent Systems\annealingnqueens.py", line 66, in anneal
temperature = len(board)**2
TypeError: object of type 'Board' has no len()
You need to define a __len__ method for your Board class, otherwise the call to len(board) will not be defined. This and many other special double-underscore methods are mentioned in the Python 3 documentation (or, if you prefer, the Python 2 documentation). Presuming that the length of queens is the quantity you want:
class Board(object):
"""An N-queens solution attempt."""
def __init__(self, queens):
"""Instances differ by their queen placements."""
self.queens = queens.copy()
def __len__(self):
return len(self.queens)