Kruskal algorithm in python - python-3.x

import heapq
from collections import defaultdict
a = list(map(int, input().split()))
nodes = a[0]
disjoint_set = [-1]*(nodes+1)
rank_set = [0]*(nodes+1)
edges = a[1]
heap = []
def get_parent(u):
if disjoint_set[u] == -1:
return u
return get_parent(disjoint_set[u])
def make_union(x, y):
x_parent = get_parent(x)
y_parent = get_parent(y)
if rank_set[x_parent] == rank_set[y_parent]:
disjoint_set[x_parent] = y_parent
rank_set[x_parent] +=1
elif rank_set[x_parent] > rank_set[y_parent]:
disjoint_set[x_parent] = y_parent
else:
disjoint_set[y_parent] = x_parent
def not_cycle(*item):
x_parent = get_parent(item[1])
y_parent = get_parent(item[2])
if x_parent == y_parent:
return False;
make_union(x_parent, y_parent)
return True
while(edges!=0):
edge = list(map(int, input().split()))
heapq.heappush(heap, [edge[2], edge[0], edge[1]])
edges-=1
cnt = 0
total = 0
while(cnt!=nodes-1):
item = heapq.heappop(heap)
if(not_cycle(*item) is True):
total+= item[0]
cnt+=1
print(total)
I implemented the kruskal algorthm in python. I am getting RecursionError:maximum recursion depth exceeded in comparison error. make_union and get_parent are method of disjoint set algorithm. I am getting the error in get_parent method. How to solve this?

In not_cycle you are passing the parents to make_union but then in make_union you are trying to get the parents again. After the first change the parents will no longer be -1 and you will recurse "forever"[1]
[1] "forever" in this case is until the maximum depth of your stack.

-
aa,bb=list(map(int,input().split()))
c=[] for i in range(bb):
z=list(map(int,input().split()))
c.append(z) c.sort(key=lambda x: x[2])
a=[]
b=[]
for i in c:
a.append((i[0]-1,i[1]-1))
b.append(i[2])
arr=[]
size=[]
for i in range(len(b)):
arr.append(i)
size.append(1)
def root(i):
while arr[i]!=i:
i=arr[i]
return i
def unions(arr,size,p,q):
root_a=root(p)
root_b=root(q)
if size[root_a]>size[root_b]:
arr[root_b]=arr[root_a]
size[root_a]+=size[root_b]
else:
arr[root_a]=arr[root_b]
size[root_b]+=size[root_a]
def kruskals(b,a,aa):
te=[]
i=0
while (len(te))<aa-1:
(p,q)=a[i]
if root(p)!=root(q):
te.append(b[i])
unions(arr,size,p,q)
i+=1
return sum(te)
print(kruskals(b,a,aa))

Related

AttibuteError when trying to make configs

AttributeError: module 'collections' has no attribute 'Sequence'
i get this error everything i try to run my code but there isn't any information about how to use Mothur except for the the documentation.
`# python3
import sys
import queue
import itertools
from collections import deque
from mothur_py import Mothur
import collections.abc as collections
class KmerIdMgmt:
def __init__(self):
self.id = 0
self.ids_map = {}
self.kmers = {}
def insert(self, kmer):
if kmer not in self.ids_map:
self.ids_map[kmer] = self.id
self.kmers[self.id] = kmer
self.id += 1
return self.ids_map[kmer]
class DeBruijnGraph(object):
def __init__(self, k, reads):
self.k = k
self.threshold = self.k + 1
self.kmer_ids = KmerIdMgmt()
self.coverage = {}
self.graph = {}
self.outgoing_num = lambda k: len(self.graph[k][0])
self.incoming_num = lambda k: self.graph[k][1]
self.make_deBruijn_graph(self.break_reads_into_kmers(reads))
def break_reads_into_kmers(self, reads):
break_read = lambda read: [ read[j:j + self.k] for j in range(len(read) - self.k + 1) ]
return [ kmer for read in reads for kmer in break_read(read) ]
def make_deBruijn_graph(self, kmers):
def add_edge(graph, coverage, left, right):
graph.setdefault(left, [set(), 0])
graph.setdefault(right, [set(), 0])
coverage.setdefault((left, right), 0)
coverage[(left, right)] += 1
if right not in graph[left][0]:
graph[left][0].add(right)
graph[right][1] += 1
for kmer in kmers:
left = self.kmer_ids.insert(kmer[:-1])
right = self.kmer_ids.insert(kmer[1:])
if left != right:
add_edge(self.graph, self.coverage, left, right)
def remove_leaves(self):
removable = [ k for k, v in self.graph.items() if len(v[0]) == 0 ]
for k in removable:
del self.graph[k]
def print_graph(self):
for k, v in self.graph.items():
print(k, v)
class TipRemoval(DeBruijnGraph):
def __init__(self, k, reads):
DeBruijnGraph.__init__(self, k, reads)
def remove_tips(self):
for k, v in self.graph.items():
find_and_remove = None
if self.outgoing_num(k) == 1 and self.incoming_num(k) == 0:
find_and_remove = self.find_and_remove_incoming
elif self.outgoing_num(k) > 1:
find_and_remove = self.find_and_remove_outgoing
else: continue
condition = True
while condition:
condition = False
for edge in v[0]:
if find_and_remove(edge, 0):
v[0].remove(edge)
condition = True
break
def find_and_remove_outgoing(self, current, depth):
if self.outgoing_num(current) > 1 or self.incoming_num(current) > 1:
return False
if depth == self.threshold:
return False
if self.outgoing_num(current) == 0:
return True
if self.find_and_remove_outgoing(next(iter(self.graph[current][0])), depth + 1):
to = next(iter(self.graph[current][0]))
self.graph[current][0].pop()
self.graph[to][1] -= 1
return True
return False
def find_and_remove_incoming(self, current, depth):
if self.outgoing_num(current) == 0 or self.incoming_num(current) > 1:
return True
if depth == self.threshold:
return False
if self.find_and_remove_incoming(next(iter(self.graph[current][0])), depth + 1):
to = next(iter(self.graph[current][0]))
self.graph[current][0].pop()
self.graph[to][1] -= 1
return True
return False
class BubbleRemoval(TipRemoval):
def __init__(self, k, reads):
TipRemoval.__init__(self, k, reads)
self.paths = {}
def remove_bubbles(self):
for k, v in self.graph.items():
if self.outgoing_num(k) > 1:
self.dfs(path=[k], current=k, depth=0)
for pair, candidates_list in self.paths.items():
source, target = pair[0], pair[1]
best_path = max(candidates_list, key=lambda item: item[1])[0]
for path, _ in candidates_list:
if best_path == path or not self.bubble_possible(source, target):
continue
if self.paths_disjoint(best_path, path) and self.path_exists(path):
self.remove_path(path)
def bubble_possible(self, source, target):
return len(self.graph[source][0]) > 1 and self.graph[target][1] > 1
def path_exists(self, path):
for j in range(len(path) -1):
if path[j +1] not in self.graph[path[j]][0]:
return False
return True
def remove_path(self, path):
for j in range(len(path) -1):
self.graph[path[j]][0].remove(path[j +1])
self.graph[path[j +1]][1] -= 1
del self.coverage[(path[j], path[j +1])]
def paths_disjoint(self, a, b):
return len(set(a) & set(b)) == 2
def dfs(self, path, current, depth):
if current != path[0] and self.incoming_num(current) > 1:
weight = sum(self.coverage[(path[i], path[i+1])] for i in range(len(path)-1)) / len(path)
self.paths.setdefault((path[0], current), list()).append((path[:], weight))
if depth == self.threshold:
return
for next_ in self.graph[current][0]:
if next_ not in path:
path.append(next_)
self.dfs(path, next_, depth + 1)
path.remove(next_)
class PhiX174GenomeAssembler(BubbleRemoval):
def __init__(self, k, reads):
BubbleRemoval.__init__(self, k, reads)
def make_Euler_cycle(self):
verteces = deque()
path = []
# line 191
current = next(iter(self.graph))
verteces.append(current)
while verteces:
current = verteces[0]
if len(self.graph[current][0]) != 0:
t = next(iter(self.graph[current][0]))
verteces.append(t)
self.graph[current][0].remove(t)
continue
path.append(current)
verteces.popleft()
return path
def assemble(self):
self.remove_tips()
self.remove_leaves()
self.remove_bubbles()
cycle = self.make_Euler_cycle()
circular_genome = self.kmer_ids.kmers[cycle[0]]
for i in range(1, len(cycle) - (self.k - 1)):
circular_genome += self.kmer_ids.kmers[cycle[i]][-1]
return circular_genome
if __name__ == "__main__":
n_kmers = int(input())
for _ in range(n_kmers):
reads = list(input())
reads = str(reads)
with open('reads.fasta', 'w') as read:
read.write(reads)
k = 100
m = Mothur()
contig = m.make.contigs(ffasta = read)
for x in range(n_kmers):
print(">CONTIG", x)
print(contig)
`

Stack in python with O(1) time complexity

I'm trying to write a code in python that imitates a stack, it has 3 properties:
push(x): append x to the end of the list
pop: pops rightmost value and return it
increase(k): increases the top k elements of list by 1, meaning if the list is [1,2,3] with a k value of 2; the updated list is [1,3,4].
I was able to do this easily, but i'm required to do it with a O(1) time complexity, here is my code:
from collections import deque
class Stack:
def __init__(self):
self.s = []
self.new = []
def push(self, x):
self.s.append(x)
self.new.append(0)
def pop(self):
return self.new.pop() + self.s.pop()
def increase(self, k):
b = self.new[-k:]
a = [1] * k
self.new = self.new[0:-k]
self.new += map(sum, zip(a,b))
if __name__ == "__main__":
s = Stack()
s.push(1)
s.push(2)
s.push(3)
s.increase(2)
print(s.pop()) # 4
s.push(1)
s.increase(2)
print(s.pop()) # 2
print(s.pop()) # 4
print(s.pop()) # 1
The list itself does not have to change, but these are the desired answers when calling the pop() function. Meaning only the illusion of the list being manipulated is enough.
Edit:
I found the fastest way to do this, here is the code:
class Stack:
def __init__(self):
self.s = []
self.new = [0]
def push(self, x):
self.s.append(x)
self.new.append(0)
def pop(self):
self.new[-2] += self.new[-1]
return self.new.pop() + self.s.pop()
def increase(self, k):
self.new[-1] += 1
self.new[-1-k] -= 1

How to find the shortest path

In the function of find_shortest_func, i think if now position isn't "T" which is also known as the terminal or exit, then i will try to find for direction and see if it is "T", if not, check if it is space and i can go there. Besides, tell the next state function now output and dic to tell the place where i visited. But some errors occur and I don't know why.
I think the problem may occur where I tried to deepcopy the output list
import copy
def set_symbol(symbol_name):
def set_symbol_decorator(func):
def wrapper(self, symbol):
setattr(self, symbol_name, symbol)
return wrapper
return set_symbol_decorator
class Maze:
space_symbol = " "
obstacle_symbol = "X"
path_symbol = "•"
output = []
dis = 0
def __init__(self, input_string):
self.maze = []
if input_string.endswith("txt"):
with open(input_string) as f:
count = 0
for line in f.readlines():
self.maze.append([])
for j in line:
if j != '\n':
self.maze[count].append(j)
count += 1
else:
count = 0
for i in input_string.split("\n"):
self.maze.append([])
for j in i:
self.maze[count].append(j)
count += 1
def __str__(self):
output_string = ""
for i in range(20):
for j in range(20):
output_string += self.maze[i][j]
output_string += "\n"
return output_string
#set_symbol("space_symbol")
def set_space_symbol(self, change):
pass
#set_symbol("obstacle_symbol")
def set_obstacle_symbol(self, change):
pass
#set_symbol("path_symbol")
def set_path_symbol(self, change):
pass
def find_shortest_func(self, position: tuple, d: dict, out: list, dis: int):
dic = copy.deepcopy(d)
output = copy.deepcopy(out)
dic[(position[0], position[1])] = 1
output.append((position[0], (position[1])))
dis += 1
if self.maze[position[0]][position[1]] != "T":
if position[0]+1 < 20 and self.maze[position[0]+1][position[1]] == self.space_symbol and (position[0]+1, position[1]) not in dic:
self.find_shortest_func(
(position[0]+1, position[1]), dic, output, dis)
if position[1]+1 < 20 and self.maze[position[0]][position[1]+1] == self.space_symbol and (position[0], position[1]+1) not in dic:
self.find_shortest_func(
(position[0], position[1]+1), dic, output, dis)
if position[0]-1 >= 0 and self.maze[position[0]-1][position[1]] == self.space_symbol and (position[0]-1, position[1]) not in dic:
self.find_shortest_func(
(position[0]-1, position[1]), dic, output, dis)
if position[1]-1 >= 0 and self.maze[position[0]][position[1]-1] == self.space_symbol and (position[0], position[1]-1) not in dic:
self.find_shortest_func(
(position[0], position[1]-1), dic, output, dis)
if self.maze[position[0]][position[1]] == "T":
if dis < self.dis:
self.output = copy.deepcopy(output)
self.dis = dis
return
def find_shortest_path(self):
d = dict()
output = []
dis = -1
self.find_shortest_func((1, 0), d, output, dis)
return self.output, self.dis

I've tried running the code but it says list index is out of range

from typing import List
# You are given an integer n, denoting the no of people who needs to be seated, and a list of m integer seats, where 0 represents a vacant seat. Find whether all people can be seated, provided that no two people can sit together
When I run this code in geeks for geeks for submission I get a error that List index is out of range.
but seems to work fine when I run it as a script.
class Solution:
def is_possible_to_get_seats(self, n: int, m: int, seats: List[int]) -> bool:
vacant_seats = 0
if len(seats) == 2:
if seats[0] or seats[1] == 1:
print(seats)
return False
else:
print(seats)
return True
else:
for x in range(len(seats)):
if x == 0:
if seats[x] == 0 and seats[x+1] == 0:
seats[x] = 1
vacant_seats += 1
elif x == len(seats)-1:
if seats[x] == 0 and seats[x-1] == 0:
seats[x] = 1
vacant_seats += 1
else:
if seats[x] == 0:
if seats[x+1] == 0 and seats[x-1] == 0:
seats[x] = 1
vacant_seats += 1
if vacant_seats < n:
return False
else:
return True
# {
# Driver Code Starts
class IntArray:
def __init__(self) -> None:
pass
def Input(self, n):
arr = [int(i) for i in input().strip().split()] # array input
return arr
def Print(self, arr):
for i in arr:
print(i, end=" ")
print()
if __name__ == "__main__":
t = int(input())
for _ in range(t):
n = int(input())
m = int(input())
seats = IntArray().Input(m)
obj = Solution()
res = obj.is_possible_to_get_seats(n, m, seats)
result_val = "Yes" if res else "No"
print(result_val)
# } Driver Code Ends

Python: Roll a dice for 12 times, calculate the probability if each number equally shows up twice

I've drafted the below code for the captioned question, but the return result is always 0. Could anyone please help me figure out what's the problem here?
Thanks a lot!
import random
dice_sides = 6
frequency_list = []
def roll_dice(times):
results = []
for roll_num in range(times):
result = random.randint(1,dice_sides)
results.append(result)
for i in range(dice_sides):
if results.count(i) != 2:
frequency = 0
break
else:
frequency = 1
return frequency
def occurrence(N,times):
for j in range(N):
frequency_list.append(roll_dice(times))
prob = frequency_list.count(1)
return prob
print(occurrence(10000,12))
You can try something like this
Code
import random
from collections import Counter
def roll_dice(n_sides, times):
if n_sides % times:
return 0
results = []
for roll_num in range(times):
result = random.randint(1, n_sides)
results.append(result)
# I'm using set here and will check its length,
# Counter(results) returns a dict of items (item, count)
# and if every item has the same count it should have length 1.
# More generic statement not only for (2 in this case)
res_dict = set(Counter(results).values())
if len(res_dict) == 1:
return 1
return 0
def mean(ar):
return sum(ar)/len(ar)
def occurrence(N, n_sides, times):
frequency_list = []
for j in range(N):
frequency_list.append(roll_dice(n_sides, times))
prob = mean(frequency_list)
return prob
if __name__ == '__main__':
N = 100000 # I intentionally made it 100k
n_sides = 6
times = 12
res_prob = occurrence(N, times)
print(res_prob)
Output
0.00604
[Finished in 3.6s]

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