I have some trouble about project euler problem 60.
The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property.
Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.
The code that I created is too slow to see the correct answer. And I don't even see that is it work correctly or not. The code is:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sat Aug 27 21:18:10 2022
#author: burak
"""
def is_prime(n, check_list_for_primes): #checks if value is prime
if check_list_for_primes.count(n) > 0: #checks if valu calculated before. if it were, it avoid loop.
return True
else:
if n == 1:
return False
if n == 2 or n == 3:
return True
i = 2
while i * i <= n:
if n % i == 0:
return False
exit(0)
i += 1
check_list_for_primes.append(n) # if it never calculated, stores the value to avoid loop at the beginning of function.
return True
def check_1(i, j): # checks the concanated calues if they are primes.
if is_prime(int(str(i)+str(j)), check_list_for_primes) == True and is_prime(int(str(j)+str(i)), check_list_for_primes) == True:
return True
else:
return False
def check_2(temp_list, n): # checks the final list that obtain the minimum summation.
if temp_list.count(n) == 0:
temp_list.append(n)
for i in temp_list:
for j in temp_list:
if len(temp_list) == 1:
return check_1(i, j)
elif i == j:
continue
elif len(temp_list) == 1:
return True
break
elif check_1(i, j) == False:
return False
return True
def func_(prime_list): # creates a dictionary summation of the five prime numbers in order to problem.
temp_list = []
result_dic = {}
k = 0
t = 0
for i in prime_list:
if i == 5:
continue
while k == 0:
t = k
for j in prime_list:
if i == j or j == 5:
continue
elif j < i:
continue
else:
temp_list.append(j)
if check_2(temp_list, i) == True:
continue
else:
temp_list.remove(j)
if t > 0 and len(temp_list) > 1:
t -= 1
temp_list.remove(max(temp_list))
continue
if len(temp_list) == 5:
result_dic[sum(temp_list)] = temp_list
elif len(temp_list) < 5:
k +=1
temp_list = []
return result_dic
if __name__ == "__main__":
dic_ = {}
prime_list = []
check_list_for_primes = []
for i in range(3, 9000, 1): #creates prime list between given range
if is_prime(i, check_list_for_primes) == True:
prime_list.append(i)
check_list_for_primes = prime_list.copy() #pseudo prime list to avoid calculating if the number is prime.
dic_ = func_(prime_list) #final dictionary to obtain minimum summation of five prime numbers.
x = min(list(dic_.keys()))
print(str(x) + " : " + str(dic_[x]))
I tried to type the examination of calculating order.
The main problem is at "func_" function. The for loop of "j" must be manipulated if the code not to get required list lenght. The "j" loop must be restart again after remove second element of "temp_list" and it must be start after shift to removed element of "prime_list".
Could you help me to see where I made mistakes and how can I improve calculation speed. Thanks so much.
I solved it. while loop changed and the "j" for loop determined by a list. The final code is;
def is_prime(n, check_list_for_primes): #checks if value is prime
if check_list_for_primes.count(n) > 0: #checks if valu calculated before. if it were, it avoid loop.
return True
else:
if n == 1:
return False
if n == 2 or n == 3:
return True
i = 2
while i * i <= n:
if n % i == 0:
return False
exit(0)
i += 1
check_list_for_primes.append(n) # if it never calculated, stores the value to avoid loop at the beginning of function.
return True
def check_1(i, j): # checks the concanated values if they are primes.
if is_prime(int(str(i)+str(j)), check_list_for_primes) == True and is_prime(int(str(j)+str(i)), check_list_for_primes) == True:
return True
else:
return False
def check_2(temp_list): # checks the final list that obtain the minimum summation.
for i in temp_list:
for j in temp_list:
if len(temp_list) == 1:
return check_1(i, j)
elif i == j:
continue
elif len(temp_list) == 1:
return True
break
elif check_1(i, j) == False:
return False
return True
def func_(prime_list): # creates a dictionary summation of the five prime numbers in order to problem.
prime_list.remove(5)
temp_list = []
result_dic = {}
# k = 0
copy_primes = prime_list.copy()
for i in prime_list:
for z in prime_list:
if z <= i:
copy_primes.remove(z)
else:
break
if i == max(prime_list):
break
elif i == 5:
continue
# for z in range(len(prime_list) - 1, 0, -1):
# if check_1(i,prime_list[z]) == True:
# max_prime_of_i = prime_list[z]
# break
while len(temp_list) < 5:
# if temp_list[1] == max_prime_of_i:
# break
if len(temp_list) == 1:
break
if len(copy_primes) == 0 and len(temp_list) > 1:
copy_primes = prime_list.copy()
for z in prime_list:
if z <= temp_list[1]:
copy_primes.remove(z)
else:
break
temp_list = []
for j in prime_list:
if len(copy_primes) > 0:
j = copy_primes[0]
copy_primes.remove(j)
else:
break
if temp_list.count(i) == 0:
temp_list.append(i)
continue
temp_list.append(j)
temp_list.sort()
if check_2(temp_list) == True and len(temp_list) > 1:
continue
elif check_2(temp_list) == False and len(temp_list) > 1:
temp_list.remove(j)
print(i)
print(temp_list)
if len(temp_list) < 5 and len(copy_primes) == 0:
continue
elif len(temp_list) == 5:
break
copy_primes = prime_list.copy()
if len(temp_list) == 5:
result_dic[sum(temp_list)] = temp_list
print(str(min(list(result_dic.keys()))) + " : " + str(result_dic[min(list(result_dic.keys()))]))
weight_ = 0
check_weight = 0
for p in temp_list:
weight_ = weight_ + len(str(p))
if weight_ < check_weight or check_weight == 0:
check_weight = weight_
elif check_weight < weight_ and len(temp_list) == 5:
return result_dic
temp_list = []
return result_dic
if __name__ == "__main__":
dic_ = {}
prime_list = []
check_list_for_primes = []
for i in range(3, 9000, 1): #creates prime list between given range
if is_prime(i, check_list_for_primes) == True:
prime_list.append(i)
check_list_for_primes = prime_list.copy() #pseudo prime list to avoid calculating if the number is prime.
dic_ = func_(prime_list) #final dictionary to obtain minimum summation of five prime numbers.
x = min(list(dic_.keys()))
print(str(x) + " : " + str(dic_[x]))
Related
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
I have these functions. They are working perfectly, but is there a way to speed them up? I tried to split the dataset, but it takes the same or more time than the original functions. I'm working with big arrays (1Mill+X2504X2). create_needed_pos takes arount 350sec for 1.2millX2054X2 array, but my biggest is around 10bilionx2054x2.
#nb.njit
def create_needed_pos(chr_pos, pos):
needed_pos = nb.typed.List.empty_list(nb.int32)
for i in range(len(chr_pos)):
for k in range(len(pos)):
if chr_pos[i] == pos[k]:
if i == k == 1:
needed_pos = nb.typed.List([pos[k]])
else:
needed_pos.append(pos[k])
return needed_pos
#nb.njit
def create_mat(geno):
# create matrix as np.uint8 (1 byte) instead of list of python integers (8 byte)
# also no need to dynamically resize / increase list size
geno_mat = np.zeros((len(geno[:, 0]), len(geno[1, :])), dtype=np.uint8)
for i in np.arange(len(geno[:, 0])):
for k in np.arange(len(geno[1, :])):
g = geno[i, k]
# nested ifs to avoid duplicate comparisons
if g[0] == 0:
if g[1] == 0:
geno_mat[i, k] = 2
elif g[1] == 1:
geno_mat[i, k] = 1
else:
geno_mat[i, k] = 9
elif g[0] == 1:
if g[1] == 0:
geno_mat[i, k] = 1
elif g[1] == 1:
geno_mat[i, k] = 0
else:
geno_mat[i, k] = 9
else:
geno_mat[i, k] = 9
return geno_mat
I'm trying to make and anagram checker without any built-in functions. So far, I've managed this:
def isa1(s1, s2):
a = s1.lower()
b = s2.lower()
c = list(a)
d = list(b)
l = len(s1)
counter = 0
for i in range(l):
if c[i] == d[0]:
del d[0]
counter += 1
elif c[i] == d[1]:
del d[1]
counter += 1
elif c[i] == d[2]:
del d[2]
counter += 1
elif c[i] == d[3]:
del d[3]
counter += 1
elif c[i] == d[4]:
del d[4]
counter += 1
elif c[i] == d[5]:
del d[5]
counter += 1
else:
pass
if counter == len(s1):
return True
else:
return False
I'm happy with the start, bar the assignment naming, but I cant figure out how to iterate through my second string, s2, without the for-loop being ridiculous. Plus this code will only work for a string/list 6 characters long.
Sorry if this seems simply, I'm just starting Python and programming in general
Thanks!
if you are okay with using for in side of for you can do:
def isa1(s1, s2):
a = s1.lower()
b = s2.lower()
c = list(a)
d = list(b)
l = len(s1)
counter = 0
for i in range(l):
for j in range(len(d)):
if c[i] == d[j]:
del d[j]
counter += 1
break # to continue to the next letter
if counter == len(s1):
return True
else:
return False
this solution will check against each letter in the second list, and if it finds a match it will break the inner loop going to the next letter.
The i+=1 isn't working, it should have increased the i value but it isn't
n = int(input())
for j in range(n):
a = input()
pair = 0
for i in range(len(a)-1):
print(i)
if a[i] == "x" and a[i+1] == "y":
pair += 1
print("*")
elif a[i] == "y" and a[i+1] =="x":
pair += 1
print('**')
else:
continue
i+=1
print(pair)
print("****")
print(pair)strong text
```
You are trying to modify a parameter that is implicitly set by the for loop.
This isn't C-Code, increasing the counter variable will not skip the next iteration.
The reason for that is quite simple: for itself does not increase i in every iteration, it just steps through the given iteratable. In this case the iteratable is a range, which behaves like for would increase i every iteration, but it really just takes the next value from the range.
So i+=1 doesn't have an effect on the next iteration, as it doesn't modify the next value in the range.
Your for-loop already increments i on every iteration. But if you want to increment by more than 1 when certain condition is satisfied, then you can use while loop as follows:
n = int(input())
for j in range(n):
a = input()
pair = 0
i = 0
while i < len(a)-1:
print(i)
if a[i] == "x" and a[i+1] == "y":
pair += 1
print("*")
elif a[i] == "y" and a[i+1] =="x":
pair += 1
print('**')
else:
i += 1
continue
i+=2
print(pair)
print("****")
print(pair)strong text
You could explicitely skip the next iteration when a pair is found.
Here is your code with the changes pointed at by # <---.
n = int(input())
for j in range(n):
a = input()
pair = 0
skip_next = False # <---
for i in range(len(a)-1):
if skip_next is True: # <---
skip_next = False # <---
continue # <---
print(i)
if a[i] == "x" and a[i+1] == "y":
pair += 1
print("*")
skip_next = True # <---
elif a[i] == "y" and a[i+1] =="x":
pair += 1
print('**')
skip_next = True # <---
else:
continue
print(pair)
print("****")
print(pair)
I am trying to write a function that returns the number of prime numbers that exist up to and including a given number.
Initially this was my code:
def count_primes(num):
prime = [2]
x = 3
if num < 2:
return 0
while x <= num:
for y in prime:
if x%y == 0:
print('not prime')
x+=2
break
else:
prime.append(x)
x += 2
return len(prime)
How ever I realise this code will run forever because of the following line of code:
for y in prime:
if x%y == 0:
print('not prime')
x+=2
break
else:
prime.append(x)
x += 2
Can anyone help to explain to me why will this end up with an infinite loop compared to the following code?
for y in prime:
if x%y == 0:
print('not prime')
x+=2
break
else:
prime.append(x)
x += 2