I am trying trying to find the number of circular primes from a given limit. The prime(x) will return whether a number is a prime or not. The rotations() will return a list of rotated numbers. Lastly, prime_count() will output the total amount of circular primes based on the given limit. Both prime() and rotations() gave me the correct output; however, prime_count() is not incrementing like it should. Any ideas on what i did wrong?
def prime(number): #return true or false
return all(number% i for i in range(2,number))
def rotations(num): #rotating number and return list
list = []
m = str(num)
counter = 0
while counter < len(str(num)):
m=m[1:] + m[0]
list.append(int(m))
counter+=1
list1=sorted(list,key=int)
return list1
def prime_count(limit): #return numbers of circular primes from given limit
counter = 0
for i in range(1,limit+1):
a=rotations(i)
for j in a:
if j == prime(j):
counter+=1
return counter
print(prime_count(100))
There are a few problems with your code:
Your prime function has a bug:
In [8]: prime(1)
Out[8]: True
It erroneously returns True for any number less than 2 due to range(2, n) being empty and any([]) == True.
prime_count should be counting the total number of circular primes below limit. prime(j) returns a boolean, but you check j == prime(j), which can only be true if j is zero or one, which definitely isn't what you want. Try creating an is_circular_prime function that takes in an integer n and returns whether or not the prime is circular. Then, prime_count becomes easy to write.
This is the heart of the problem:
a=rotations(i)
for j in a:
if j == prime(j):
counter+=1
You're counting the wrong thing (e.g. 13 counts twice independent of 31 counting twice) and you're comparing the wrong thing (numbers against booleans.) The problem is simpler than you're making it. Rearranging your code:
def prime(number):
return number > 1 and all(number % i != 0 for i in range(2, number))
def rotations(num):
rotated = []
m = str(num)
for _ in m:
rotated.append(int(m))
m = m[1:] + m[0]
return rotated
def prime_count(limit):
counter = 0
for number in range(1, limit + 1):
if all(prime(rotation) for rotation in rotations(number)):
counter += 1
return counter
print(prime_count(100))
Note that you don't need to sort the rotations for this purpose. Also list, or any other Python built-in function, is a bad name for a variable.
Problem may be here:
for i in range(1,limit+1):
a=rotations(i)
for j in a:
if j == prime(j): # Prime will return True or False, comapring with J will cause it False ,except when J = 1
counter+=1
Changing it to prime(j)
for i in range(2,number//2):
if(number%i==0):
return False
return True
def rotations(num):
count=0
rotation_lst=[]
num=str(num)
while(count<len(num)):
num=num[1:]+num[0]
count+=1
rotation_lst.append(int(num))
rotation_lst1=sorted(rotation_lst,key=int)
return rotation_lst1
def get_circular_prime_count(limit):
count=0
for i in range(1,limit+1):
a=rotations(i)
for j in a:
if(check_prime(j)):
count+=1
return count
print(get_circular_prime_count(1000) ```
Related
I am pretty new to Python and just got started in Leet and I am doing the Happy Number question, only half of the test cases have been passed. I would appreciate any help. Thanks
A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
My test cases were 19, 100 where it output True correctly but when I do 7, it is wrong
def isHappy(self, n: int) -> bool:
if (n == 1):
return True
sum = 0
flag = False
for j in range(1,100):
x = str(n)
a = list(x)
for i in range(0,len(a)):
sum += int(a[i])*int(a[i])
if sum == 1:
return True
break
else:
x = sum
return False
Here is an implementation using a set to keep track of numbers that we have already seen. (I've removed the self argument here for sake of something that can be run outside of your test class.)
def isHappy(n: int) -> bool:
seen = set()
while True:
if n == 1:
return True
if n in seen:
return False
seen.add(n)
n = sum(int(c) ** 2 for c in str(n))
Your code has various issues aside from the fact that the number 100 is arbitrary.
Mainly that you never update n in your loop. Also you do not wait for the completion of the for loop before testing sum (in fact your break is never reached), you initialise sum only once, and you return False prematurely. Here is a minimally corrected version of your code, although still subject to the fact that there is no rule about 100 maximum iterations.
def isHappy(n: int) -> bool:
if (n == 1):
return True
for j in range(1,100):
x = str(n)
a = list(x)
sum = 0
for i in range(0,len(a)):
sum += int(a[i])*int(a[i])
if sum == 1:
return True
else:
n = sum
return False
I have a hackkerank coding challenge to print first n non prime numbers, i have the working code but the problem is that they have a locked code which prints numbers from 1 to n along with the output, in order to pass the test i need to print only the non prime numbers not 1...n numbers along with it. I cant comment the printing part of 1...n as it is blocked. please let me know the idea to print only 1st n non prime numbers:
here is my solution:
def manipulate_generator(generator, n):
if n>1:
ls=[1]
for elm in generator:
if elm>3 and len(ls)<n:
for k in range(2,elm):
if elm%k==0 and elm not in ls:
ls.append(elm)
print(elm)
if len(ls)==n:
return ls
That's the code I added but here is the code that's locked on which I have to write the code above to make it print the number one at a time
def positive_integers_generator():
n = 1
while True:
x = yield n
if x is not None:
n = x
else:
n += 1
k = int(input())
g = positive_integers_generator()
for _ in range(k):
n = next(g)
print(n)
manipulate_generator(g, n)
the point is the for _ in range(k): already prints out number which includes numbers I don't want printed out: This is the desired kind of output I want:for n=10 I want it to print out:
output:
1
4
6
8
9
10
12
14
15
16
I can't change this code but the one above is what I wrote and can be changed... Pleae help me out... Thanks in anticipation
Why not to throw away the numbers which we don't need? Look at this solution which I implemented...
def is_prime(n):
for i in range(2, n):
if n%i == 0:
return False
return True
def manipulate_generator(generator, n):
if is_prime(n+1):
next(generator)
manipulate_generator(generator, n+1)
Note: I understand that the logic can be improved to make it more efficient. But, its the idea of skipping unnecessary number printing which is important here !
You can print all the numbers from 1 up to the first prime number and then from that first number to the next one until you reach n.
I'm not sure of your hackerrank situation, but printing the first N non-prime numbers efficiently can be done this way.
def non_prime_numbers_till_n(n):
primes = set()
for num in range(2,number + 1):
if num > 1:
for i in range(2, math.sqrt(num)):
if (num % i) == 0:
break
else:
primes.add(num)
result = []
for i in range(1, n):
if i not in primes:
result.append(i)
return result
Depending on what your online editor expects you can either print them, or store them in a list and return the list.
Also bear in mind, you can only check upto the sqrt of the number to determine if its a prime or not.
I eventually came up with this answer which I believe should solve it but id there;s a better way to solve it please add your answers:
def manipulate_generator(generator, n):
for num in range(3,100000):
for q in range(2,num):
if num%q==0 and num>n:
generator.send(num-1)
return
this link python generator helped me to understand python generator
I just solved that right now. Like Swapnil Godse said you need to deal with all special cases to optimize computations. This link might be helpful: click here.
Here is the solution:
from math import sqrt
def is_prime(n):
if (n <= 1):
return False
if (n == 2):
return True
if (n % 2 == 0):
return False
i = 3
while i <= sqrt(n):
if n % i == 0:
return False
i = i + 2
return True
def manipulate_generator(g, n):
if is_prime(n+1):
next(g)
manipulate_generator(g, n+1)
prime = int(input('Please enter the range: '))
prime_number = []
for num in range(prime):
if num > 1:
for i in range(2,num):
if num % i == 0:
break
else:
prime_number.append(num)
print(f'Prime numbers in range {prime} is {prime_number}')
all_number = []
for i in range(2,prime+1):
all_number.append(i)
Non_prime_number = []
for element in all_number:
if element not in prime_number:
Non_prime_number.append(element)
print(f'Non Prime numbers in range {prime} is {Non_prime_number}')
I have a recursive function to count the number of way 8 queens can fit on an 8x8 chess board without intersecting each other (for a class). It works well and gives the correct permutations, the interesting thing happens when I have the program try to count the # of answers -it constantly returns my counter to zero. When I manually count the permutations it's 92 (which is correct).
def can_be_extended_to_solution(perm):
i = len(perm) - 1
for j in range(i):
if i - j == abs(perm[i] - perm[j]):
return False
return True
def extend(perm,count, n):
if len(perm)==n:
count=count+1
print "cycle counter= ",count
print(perm)
for k in range(n):
if k not in perm:
perm.append(k)
if can_be_extended_to_solution(perm): # if it works
extend(perm, count, n)
perm.pop()
extend(perm = [], count=0, n = 8)
The issue is that you never allow the recursive call to modify the value of count. You pass the count value in to the function, but then when the count = count + 1 line is called, it only modifies the local value of count for that function call, and does not modify the value in the call that recursively called it.
The following modification works just fine (the return value of extend is 92).
def can_be_extended_to_solution(perm):
i = len(perm) - 1
for j in range(i):
if i - j == abs(perm[i] - perm[j]):
return False
return True
def extend(perm, count, n):
if len(perm) == n:
count = count + 1
print("cycle counter= " + str(count))
print(perm)
for k in range(n):
if k not in perm:
perm.append(k)
if can_be_extended_to_solution(perm): # if it works
count = extend(perm, count, n)
perm.pop()
return count
print(extend(perm=[], count=0, n=8))
I was wondering if there is any easier way to achieve what this code is achieving. Now the code creates all 4-digits number in a list (if the number starts with a 0 is doesn't count as a 4-digit, for example 0123) and no digit is repeated within the number. So for example 1231 is not in the list. Preferable I want a code that does what this one is doing but a depending on what argument N is given to the function upon calling it creates this kind of list with all numbers with N digits. I hope this wasn't impossible to understand since Im new to programing.
def guessables():
'''creates a list of all 4 digit numbers wherest every
element has no repeating digits inside of that number+
it doesn't count as a 4 digit number if it starts with a 0'''
guesses=[]
for a in range(1,10):
for b in range(0,10):
if a!=b:
for c in range(0,10):
if b!=c and a!=c:
for d in range(0,10):
if c!=d and d!=b and d!=a:
guesses.append(str(a)+str(b)+str(c)+str(d))
return guesses
This can be expressed more easily.
def all_digits_unique(number):
# `set` will only record unique elements.
# We convert to a string and check if the unique
# elements are the same number as the original elements.
return len(str(number)) == len(set(str(number)))
Edit:
def guesses(N):
return filter(all_digits_unique, range(10**(N-1), 10**N))
print guesses(4)
I'd use itertools for this which is in my opinion the simplest generic answer:
import itertools
def guessables(num):
guesses = []
for p in itertools.permutations(xrange(10),num):
if p[0] != 0:
guesses.append(''.join(map(str, p)))
return guesses
Simply call this function with guessables(4) and get a list with all the numbers you want.
You can do in one line:
print([str(a)+str(b)+str(c)+str(d) for a in range(1,10) for b in range(0,10) if a!=b for c in range(0,10) if b!=c and a!=c for d in range(0,10) if c!=d and d!=b and d!=a])
Try the following:
def guessables(n):
''' Returns an array with the combination of different digits of size "n" '''
if n > 10:
raise ValueError("The maximum number of different digits is 10.")
elif n < 1:
raise ValueError("The minimum number of digits is 1.")
else:
results = []
for i in range(1, 10):
_recursiveDigit([i], n, results)
return results
def _formatDigit(l):
''' Return a formated number from a list of its digits. '''
return "".join(map(str, l))
def _recursiveDigit(l, n, results):
''' Recursive function to calculate the following digit. '''
if len(l) < n:
for i in range(0, 10):
if i not in l:
_recursiveDigit(l + [i], n, results)
else:
results.append(_formatDigit(l))
The functions that are prefixed with an underscore(_) should not be called from outside of this script. If you prefer to have the result as something different than an array of strings, such as an array of ints for example, you can change the _formatDigit() function as follows:
def _formatDigit(l):
''' Return a formated number from a list of its digits. '''
return int("".join(map(str, l)))
c=list(range(10))
print c
def fun(n,k,i,getnum): # function , result in getnum
if n==0:
if k not in getnum and len(set(list(k)))==len(k) and k[0]!='0':
getnum.append(k)
return
if i>=len(c):
return
fun(n-1,k+str(c[i]),0,getnum)
fun(n,k,i+1,getnum)
getnum=[]
d=fun(4,"",0,getnum)
print getnum
These types of problems are easily solved with recursion.
def gen(num, n, saveto):
if len(num) == 1 and num[0] == '0':
return
if len(num) == n:
saveto.append(int(''.join(num)))
return
for i in range(0, 10):
i= str(i)
if i not in num:
gen(num+[i], n, saveto)
saveto= []
# generate 4 digit numbers
gen([], 4, saveto)
print(saveto)
Here I'm using the list num to construct the numbers by placing one digit at each call. When there are four digits added, it stores the number to the saveto list.
Edit: Here's a version of the above function that returns the list of numbers instead of appending them to a list.
def gen(num, n):
if len(num) == 1 and num[0] == '0':
return []
if len(num) == n:
return [int(''.join(num))]
ans = []
for i in range(0, 10):
i= str(i)
if i not in num:
ans.extend(gen(num+[i], n))
return ans
saveto= gen([], 4)
print(saveto)
numPool = []
for i in range(0, 10):
for j in range(0, 10):
for k in range(0,10):
for l in range(0,10):
if i != j and i != k and i != l and j != k and j != l and k != l :
numPool.append(str(i) + str(j) + str(k) + str(l))
This works, but keep in mind that this will also add "0123" or "0234" to the list as well. If you do not want the numbers that are starting with zero, you might want to add "i != 0" to the if query. Hope it helps.
I try to write it clear for absolute beginers ^^ Ofcourse it is possible to make it faster and shorter if you use combianations and advance array methods.
def f(n)
s = list(range(10**(n-1), 10**n))
number_list = []
for ss in s:
test_list = []
a = ss
while ss:
if ss % 10 in test_list:
break
test_list.append(ss % 10)
ss = ss // 10
if len(test_list) == n:
number_list.append(a)
return number_list
print(f(4))
This would sovlve the problem, without repeating digits:
from itertools import permutations
myperm = permutations([0,1,2,3,4,5,6,7,8,9],4)
for digits in list(myperm):
print(digits)
How about this?
def check_count(num):
if isinstance(num, str) == False:
num = str(num) # Convert to str
if len(num) == 1: # If total length number is single, return False
return False
return any(num.count(x) > 1 for x in num) # Check here
Return False if numbers all different, else return True
Usage:
# Get all numbers has all different. (1000-9999)
[x for x in range(1000, 10000) if check_count(x) == False]
I've created a function which, hopefully, creates a list of numbers that are both pentagonal and square.
Here is what i've got so far:
def sqpent(n):
i = 0
list = []
while n >= 0:
if n == 0:
list.append(0)
elif n == 1:
list.append(1)
elif (i*i == (i*(3*i-1)//2)):
list.append(i)
n -= 1
i += 1
But when it gets past the first two numbers it seems to be taking a while to do so...
You have two issues: the first is that the special-casing for n==0 and n==1 doesn't decrease n, so it goes into an infinite loop. The special-casing isn't really needed and can be dropped.
The second, and more significant one, is that in the test i*i == (i*(3*i-1)//2) you are assuming that the index i will be the same for the square and pentagonal number. But this will only happen for i==0 and i==1, so you won't find values past that.
I suggest:
Iterate over i instead of n to make things simpler.
Take the ith pentagonal number and check if it is a square number (e.g. int(sqrt(x))**2 == x).
Stop when you've reached n numbers.
Thanks to #interjay's advice, I came up with this answer which works perfectly:
import math
def sqpent(n):
counter = 0
i = 0
l = []
while counter < n:
x = (i*(3*i-1)//2)
#print(x)
if(int(math.sqrt(x))**2 == x):
#print("APPENDED: " + str(x))
l.append(x)
counter += 1
i += 1
return l
For an explanation:
It iterates through a value i, and gets the ith pentagonal number. Then it checks if it is a square and if so it appends it to a list which i ultimately return.
It does this until a final point when the counter reaches the number of items in the list you want.