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How I add to a list same number multiple times by count?

I've got 2 problems here.
my first problem is that the code shows me only one time a factor even though it's multiple x times by the same factor. I don't know how to add it to the factor list.
Another problem is I'm not sure in print - how the sep works and how can I write "*" only between elements of factor list.
I can't use any import functions here (intertools, maths etc.)
Please help me.
def factorize(n):
prvocisla = []
faktor = []
#prime numbers
for num in range(1, 2000):
if num > 1:
for i in range(2, num):
if (num % i) == 0:
break
else:
prvocisla.append(num)
count = 0
for i in prvocisla:
if n % i == 0:
count += 1
faktor.append(i)
print(n, " =", *faktor , sep=' *', end='\n')
factorize(360)
My result:
360 * = *2 *3 *5
The right result:
360 = 2 * 2 * 2 * 3 * 3 * 5
I try the count function with adding same factor to the list "count times" but it shows me an Error.

The problem is that in your second 'for' loop you evaluate if there is a prime number in your number, but not how many times it is present.
To do this you need to repeat the cycle every time you find a prime number and divide the initial number by the prime number. this way you will get to 1 and get all the factors in the array.
Here the right code:
def factorize(n):
prvocisla = []
faktor = []
#prime numbers
for num in range(1, 2000):
if num > 1:
for i in range(2, num):
if (num % i) == 0:
break
else:
prvocisla.append(num)
count = 0
t = n # <-- a temporary variable which get n value
while t>1:
for i in prvocisla:
if t % i == 0:
count += 1
faktor.append(i)
t = t/i <-- divide t every time you find a factor
break
print(f"{n!s} = {' * '.join(str(k) for k in faktor)}")
factorize(360)
For the print I use the #CreepyRaccoon suggestion.

Python: Code to find a number where first N digits are divisible by N (from 0-9)

I've been trying to write a recursive solution to a program to find a number where first N digits are divisible by N.
As an example: 3816547290, 3 is divisible by 1, 38 is divisible by 2, 381 is divisible by 3 and so on...
My recursive solution works fine while going "into" the recursion, but has issues when the stack unwinds (i.e. I don't specifically know how to backtrack or take steps on the way out
ARR = [0]*10
ARR[0] = 1 #dummy entry
def numSeq(pos, num):
if all(ARR):
print num
return True
if (pos>0) and (num%pos) != 0:
return False
for i in xrange(1,10):
if ARR[i] == 1:
continue
new_num = num*10 + i
if new_num%(pos+1) == 0:
ARR[i] = 1
numSeq(pos+1,new_num)
The problem with this code seems to be that it follows the number generation correctly while going into the recursion...so it correctly generates the number 123654 which is divisible by 6 and follows first N digits being divisible by N, but after it fails to find any further digits from 7-8 or 9 that divide 7, i don't get the next set of steps to "reset" the global ARR and begin from index 2, i.e. try 24xxxx,and eventually get to 3816547290
Thanks in Advance for your help!
EDIT: One condition I'd forgotten to mention is that each digit must be used exactly once (i.e. repetition of digits is disallowed)
2nd EDIT:
I was able to finally apply proper backtracking to solve the problem...this code works as is.
ARR = [0]*10
def numDivisibile(num,pos):
if all(ARR):
print num
return True
for i in xrange(0,10):
if ARR[i] == 1:
continue
new_num = num*10+i
#check for valid case
if new_num%(pos+1) == 0:
ARR[i] = 1
if numDivisibile(new_num, pos+1):
return True
#backtrack
ARR[i] = 0
return False
print numDivisibile(0, 0)

To generate all 10 digits integers where the first n digits are divisible by n for each n from 1 to 10 inclusive:
#!/usr/bin/env python3
def generate_ints_nth_digit_divisible_by_n(n=1, number=0):
number *= 10
if n == 10:
yield number # divisible by 10
else:
for digit in range(not number, 10):
candidate = number + digit
if candidate % n == 0: # divisible by n
yield from generate_ints_nth_digit_divisible_by_n(n + 1, candidate)
print("\n".join(map(str, generate_ints_nth_digit_divisible_by_n())))
Output
1020005640
1020061620
1020068010
...
9876062430
9876069630
9876545640
To get numbers where each digit occurs only once i.e., to find the permutations of the digits that satisfy the divisibility condition:
def divisibility_predicate(number):
digits = str(number)
for n in range(1, len(digits) + 1):
if int(digits[:n]) % n != 0:
return n - 1
return n
def generate_digits_permutation(n=1, number=0, digits=frozenset(range(1, 10))):
# precondition: number has n-1 digits
assert len(set(str(number))) == (n - 1) or (number == 0 and n == 1)
# and the divisibility condition holds for n-1
assert divisibility_predicate(number) == (n - 1) or (number == 0 and n == 1)
number *= 10
if n == 10:
assert not digits and divisibility_predicate(number) == 10
yield number # divisible by 10
else:
for digit in digits:
candidate = number + digit
if candidate % n == 0: # divisible by n
yield from generate_digits_permutation(n + 1, candidate, digits - {digit})
from string import digits
print([n for n in generate_ints_nth_digit_divisible_by_n()
if set(str(n)) == set(digits)])
print(list(generate_digits_permutation()))
Output
[3816547290]
[3816547290]

In your function, you never do return numSeq(...), this seems like causing the issue.
If you want to have a iterative solution, you can check the following:
def getN(number):
strNum = str(number)
for i in range(1, len(strNum)+1):
if int(strNum[:i]) % i != 0:
return i-1
return i
print getN(3816)
print getN(3817)
print getN(38165)
Output:
4
3
5

We can modify your recursive function a little to try different possibilities. Rather than have a global record (ARR) of used positions, each thread of the recursion will have its own hash of used digits:
def numSeq(pos, num, hash):
if pos != 1 and num % (pos - 1) != 0: # number does not pass the test
return
elif pos == 11: # number passed all the tests
print num
elif pos == 5:
numSeq(pos + 1,10 * num + 5,hash) # digit is 5 at position 5
elif pos == 10:
numSeq(pos + 1,10 * num,hash) # digit is 0 at position 10
else:
k = 2 if pos % 2 == 0 else 1 # digit is even at even positions
for i in xrange(k,10,2):
if hash & (1 << i): # digit has already been used, skip it
continue
numSeq(pos + 1,10 * num + i,hash | (1 << i))
numSeq(1,0,0) # 3816547290

How to generate the 1000th prime in python?

count = 0
i = 11
while count <= 1000 and i <= 10000:
if i%2 != 0:
if (i%3 == 0 or i%4 == 0 or i%5 == 0 or i%6 == 0 or i%7 == 0 or i%9 == 0):
continue
else:
print i,'is prime.'
count += 1
i+=1
I'm trying to generate the 1000th prime number only through the use of loops. I generate the primes correctly but the last prime i get is not the 1000th prime. How can i modify my code to do so. Thank in advance for the help.
EDIT: I understand how to do this problem now. But can someone please explain why the following code does not work ? This is the code I wrote before I posted the second one on here.
count = 1
i = 3
while count != 1000:
if i%2 != 0:
for k in range(2,i):
if i%k == 0:
print(i)
count += 1
break
i += 1

Let's see.
count = 1
i = 3
while count != 1000:
if i%2 != 0:
for k in range(2,i):
if i%k == 0: # 'i' is _not_ a prime!
print(i) # ??
count += 1 # ??
break
i += 1 # should be one space to the left,
# for proper indentation
If i%k==0, then i is not a prime. If we detect that it's not a prime, we should (a) not print it out, (b) not increment the counter of found primes and (c) we indeed should break out from the for loop - no need to test any more numbers.
Also, instead of testing i%2, we can just increment by 2, starting from 3 - they will all be odd then, by construction.
So, we now have
count = 1
i = 3
while count != 1000:
for k in range(2,i):
if i%k == 0:
break
else:
print(i)
count += 1
i += 2
The else after for gets executed if the for loop was not broken out of prematurely.
It works, but it works too hard, so is much slower than necessary. It tests a number by all the numbers below it, but it's enough to test it just up to its square root. Why? Because if a number n == p*q, with p and q between 1 and n, then at least one of p or q will be not greater than the square root of n: if they both were greater, their product would be greater than n.
So the improved code is:
from math import sqrt
count = 1
i = 1
while count < 1000:
i += 2
for k in range(2, 1+int(sqrt(i+1))):
if i%k == 0:
break
else:
# print(i) ,
count += 1
# if count%20==0: print ""
print i
Just try running it with range(2,i) (as in the previous code), and see how slow it gets. For 1000 primes it takes 1.16 secs, and for 2000 – 4.89 secs (3000 – 12.15 ses). But with the sqrt it takes just 0.21 secs to produce 3000 primes, 0.84 secs for 10,000 and 2.44 secs for 20,000 (orders of growth of ~ n2.1...2.2 vs. ~ n1.5).
The algorithm used above is known as trial division. There's one more improvement needed to make it an optimal trial division, i.e. testing by primes only. An example can be seen here, which runs about 3x faster, and at better empirical complexity of ~ n1.3.
Then there's the sieve of Eratosthenes, which is quite faster (for 20,000 primes, 12x faster than "improved code" above, and much faster yet after that: its empirical order of growth is ~ n1.1, for producing n primes, measured up to n = 1,000,000 primes):
from math import log
count = 1 ; i = 1 ; D = {}
n = 100000 # 20k:0.20s
m = int(n*(log(n)+log(log(n)))) # 100k:1.15s 200k:2.36s-7.8M
while count < n: # 400k:5.26s-8.7M
i += 2 # 800k:11.21-7.8M
if i not in D: # 1mln:13.20-7.8M (n^1.1)
count += 1
k = i*i
if k > m: break # break, when all is already marked
while k <= m:
D[k] = 0
k += 2*i
while count < n:
i += 2
if i not in D: count += 1
if i >= m: print "invalid: top value estimate too small",i,m ; error
print i,m
The truly unbounded, incremental, "sliding" sieve of Eratosthenes is about 1.5x faster yet, in this range as tested here.

A couple of problems are obvious. First, since you're starting at 11, you've already skipped over the first 5 primes, so count should start at 5.
More importantly, your prime detection algorithm just isn't going to work. You have to keep track of all the primes smaller than i for this kind of simplistic "sieve of Eratosthanes"-like prime detection. For example, your algorithm will think 11 * 13 = 143 is prime, but obviously it isn't.
PGsimple1 here is a correct implementatioin of what the prime detection you're trying to do here, but the other algorithms there are much faster.

Are you sure you are checking for primes correctly? A typical solution is to have a separate "isPrime" function you know that works.
def isPrime(num):
i = 0
for factor in xrange(2, num):
if num%factor == 0:
return False
return True
(There are ways to make the above function more effective, such as only checking odds, and only numbers below the square root, etc.)
Then, to find the n'th prime, count all the primes until you have found it:
def nthPrime(n):
found = 0
guess = 1
while found < n:
guess = guess + 1
if isPrime(guess):
found = found + 1
return guess

your logic is not so correct.
while :
if i%2 != 0:
if (i%3 == 0 or i%4 == 0 or i%5 == 0 or i%6 == 0 or i%7 == 0 or i%9 == 0):
this cannot judge if a number is prime or not .
i think you should check if all numbers below sqrt(i) divide i .

Here's a is_prime function I ran across somewhere, probably on SO.
def is_prime(n):
return all((n%j > 0) for j in xrange(2, n))
primes = []
n = 1
while len(primes) <= 1000:
if is_prime(n):
primes.append(n)
n += 1
Or if you want it all in the loop, just use the return of the is_prime function.
primes = []
n = 1
while len(primes) <= 1000:
if all((n%j > 0) for j in xrange(2, n)):
primes.append(n)
n += 1

This is probably faster: try to devide the num from 2 to sqrt(num)+1 instead of range(2,num).
from math import sqrt
i = 2
count = 1
while True:
i += 1
prime = True
div = 2
limit = sqrt(i) + 1
while div < limit:
if not (i % div):
prime = False
break
else:
div += 1
if prime:
count += 1
if count == 1000:
print "The 1000th prime number is %s" %i
break

Try this:
def isprime(num):
count = num//2 + 1
while count > 1:
if num %count == 0:
return False
count -= 1
else:
return True
num = 0
count = 0
while count < 1000:
num += 1
if isprime(num):
count += 1
if count == 1000:
prime = num
Problems with your code:
No need to check if i <= 10000.
You are doing this
if i%2 != 0:
if (i%3 == 0 or i%4 == 0 or i%5 == 0 or i%6 == 0 or i%7 == 0 or i%9 == 0):
Here, you are not checking if the number is divisible by a prime number greater than 7.
Thus your result: most probably divisible by 11
Because of 2. your algorithm says 17 * 13 * 11 is a prime(which it is not)

How about this:
#!/usr/bin/python
from math import sqrt
def is_prime(n):
if n == 2:
return True
if (n < 2) or (n % 2 == 0):
return False
return all(n % i for i in xrange(3, int(sqrt(n)) + 1, 2))
def which_prime(N):
n = 2
p = 1
while True:
x = is_prime(n)
if x:
if p == N:
return n
else:
p += 1
n += 1
print which_prime(1000)

n=2 ## the first prime no.
prime=1 ## we already know 2 is the first prime no.
while prime!=1000: ## to get 1000th prime no.
n+=1 ## increase number by 1
pon=1 ## sets prime_or_not(pon) counter to 1
for i in range(2,n): ## i varies from 2 to n-1
if (n%i)==0: ## if n is divisible by i, n is not prime
pon+=1 ## increases prime_or_not counter if n is not prime
if pon==1: ## checks if n is prime or not at the end of for loop
prime+=1 ## if n is prime, increase prime counter by 1
print n ## prints the thousandth prime no.

Here is yet another submission:
ans = 0;
primeCounter = 0;
while primeCounter < 1000:
ans += 1;
if ans % 2 != 0:
# we have an odd number
# start testing for prime
divisor = 2;
isPrime = True;
while divisor < ans:
if ans % divisor == 0:
isPrime = False;
break;
divisor += 1;
if isPrime:
print str(ans) + ' is the ' + str(primeCounter) + ' prime';
primeCounter += 1;
print 'the 1000th prime is ' + str(ans);

Here's a method using only if & while loops. This will print out only the 1000th prime number. It skips 2. I did this as problem set 1 for MIT's OCW 6.00 course & therefore only includes commands taught up to the second lecture.
prime_counter = 0
number = 3
while(prime_counter < 999):
divisor = 2
divcounter = 0
while(divisor < number):
if(number%divisor == 0):
divcounter = 1
divisor += 1
if(divcounter == 0):
prime_counter+=1
if(prime_counter == 999):
print '1000th prime number: ', number
number+=2

I just wrote this one. It will ask you how many prime number user wants to see, in this case it will be 1000. Feel free to use it :).
# p is the sequence number of prime series
# n is the sequence of natural numbers to be tested if prime or not
# i is the sequence of natural numbers which will be used to devide n for testing
# L is the sequence limit of prime series user wants to see
p=2;n=3
L=int(input('Enter the how many prime numbers you want to see: '))
print ('# 1 prime is 2')
while(p<=L):
i=2
while i<n:
if n%i==0:break
i+=1
else:print('#',p,' prime is',n); p+=1
n+=1 #Line X
#when it breaks it doesn't execute the else and goes to the line 'X'

This will be the optimized code with less number of executions, it can calculate and display 10000 prime numbers within a second.
it will display all the prime numbers, if want only nth prime number, just set while condition and print the prime number after you come out of the loop. if you want to check a number is prime or not just assign number to n, and remove while loop..
it uses the prime number property that
* if a number is not divisible by the numbers which are less than its square root then it is prime number.
* instead of checking till the end(Means 1000 iteration to figure out 1000 is prime or not) we can end the loop within 35 iterations,
* break the loop if it is divided by any number at the beginning(if it is even loop will break on first iteration, if it is divisible by 3 then 2 iteration) so we iterate till the end only for the prime numbers
remember one thing you can still optimize the iterations by using the property *if a number is not divisible with the prime numbers less than that then it is prime number but the code will be too large, we have to keep track of the calculated prime numbers, also it is difficult to find a particular number is a prime or not, so this will be the Best logic or code
import math
number=1
count = 0
while(count<10000):
isprime=1
number+=1
for j in range(2,int(math.sqrt(number))+1):
if(number%j==0):
isprime=0
break
if(isprime==1):
print(number,end=" ")
count+=1

Project Euler #3, infinite loop on factorization

So I'm doing Project Euler because dear gods do I need to practice writing code, and also my math skills are rusty as something very rusty. Thusly; Project Euler. I'm sure most here have already seen or heard of the problem, but I'll put it here just for completeness:
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
For this, I've written two functions:
from math import sqrt
def isprime(n):
if n == 1:
return False
elif n == 2:
return True
elif n % 2 == 0:
return False
for x in range(3, round(sqrt(n))+1, 2):
if n % x == 0:
return False
else:
return True
This just checks any fed number for primality. It's working as intended (as far as I know), but now that I've said that I grow unsure. Anyway, it checks for special cases first: 1 (never prime), 2 (prime) or if it's divisible by 2 (not prime). If none of the special cases happen, it runs a general primality test.
This is my factorization code:
def factorization(n):
factor = 2
x = 3
while True:
if n % x == 0:
if isprime(x):
factor = x
n = n // x
if n == 1:
return factor
else:
return factor
x += 2
And this is definitely not working as intended. It is, sadly, working for the particular value of the Project Euler problem, but it doesn't work for, say, 100. I'm unsure what I need to do to fix this: what happens is that if it's a number like 100, it will correctly find the first 5 (2*2*5), but after that will loop around and set x = 7, which will make the entire thing loop infinitely because the answer is 2*2*5*5. Would recursion help here? I tried it, but it didn't get any prettier (it would still go into an endless loop for some numbers). I'm unsure how to solve this now.

You're on a good track, but you need to take account of the possibility of repeating factors. You can do that with something like this:
factors = []
while num % 2 == 0:
factors.append(2)
num /= 2
The idea here being that you're going to continue adding 2's to the factor list until the number you're testing becomes odd. You can use similar logic for other factors as well to enhance your factorization method.

I think you have made the problem more complicated than necessary
Here is some pseudo code that you should be able to turn into Python code
from itertools import count
n=600851475143
for x in count(2):
while x divides n:
divide n by x
if n==1:
print x # largest factor will be the last one
break

For the repeating (odd) factors, just increment x when a divisor has not been found:
def factorization(n):
factor = 2
x = 3
while True:
if n % x == 0:
if isprime(x):
factor = x
n = n // x
if n == 1:
return factor
else:
return factor
else:
x += 2
OTOS, it seems that you miss always the "2" factors. Stick them on top and then do the main loop
EDIT (after comment)
You can do a much simpler:
def factorization(n):
factors = []
x = 2
while True:
while n % x == 0:
factors.push(x)
n /= x
if n == 1:
return factors
if x == 2:
x = 3
else:
x += 2

Here is another optimized solution:
import math
def find_prime(num):
if num <= 1:
return False
elif(num == 2):
return True
elif( num % 2 == 0):
return False
for i in range(3, int(math.sqrt(num))+1, 2):
if num%i == 0:
return False
return True
def prime_factor(number):
pf = number;
divList = [];
for i in range(2, int(math.sqrt(number))):
if number % i == 0 :
divList.append(i)
for n in divList:
if(find_prime(n)):
pf = n;
return pf
num = 600851475143
print("Max prime factor :", prime_factor(num))

euler challenge 12, why does this python code fail?

The Following code keeps telling me a wrong number and I can't see why, know it's brute force but it should still work... also the number it returns has indeed over 500 divisors, 512 to be exact, help would be much appreciated
Number = 1
Count = 2
Found = False
while Found == False:
Divisors = 0
if (Number % 2) != 0:
for i in range(1, int(Number**(1/2)), 2):
if Number % i == 0:
Divisors += 1
else:
for i in range(1, int(Number**(1/2))):
if Number % i == 0:
Divisors += 1
if Divisors >= 500:
print (Number)
Found = True
else:
Number += Count
Count += 1
For reference: Problem 12 from the Euler Challange

The number of divisors of an integer is just the product of (1 + exponent) for each pure power in the factor decomposition of an integer.
As an example: 28 = 2^2 * 7
The powers are 2 and 1, so the number of divisors is (2+1)*(1+1) = 3*2 = 6. Easy one
Bigger one: 2047 * 2048 / 2 = 2^10 * 23 * 89
The powers are 10, 1 and 1, so the number of divisors is 11*2*2 = 44
Easier: 100 = 2^2 * 5^2
The powers are 2, 2 so there are 3*3=9 divisors. The same applies to 36=2^2*3^2. The only interesting part is the exponents.
So, use any prime factor decomposition (use a sieve, you don't need a primality test) it would be much faster and more reliable than trying each of the possible numbers.
def factorize(i):
# returns an array of prime factors
whatever
def number_of_divisors(i):
n = 1
for v in Counter(factorize(i)).values():
n *= v + 1
return n

I'm not sure what Euler Challenge 12 is, but one obvious issue is the (1/2). If you try typing that in a Python prompt, you'll get 0. The reason why is that it will try to do integer math. I suggest just putting (0.5), or alternatively you could do (1/2.0).

Your divisor counting method is wrong. 12 has 6 divisors, but your code only counts 2.
Problems:
a number often has divisors larger than its square root
range doesn't include its upper bound, so you're stopping too early

the code you have been write is searching till number**0.5 and it is wrong you must search until number/2
so the corrected answer is like below:
Note : I add some extra code to show the progress. and they are not affect the solution.
Another Note: since the Nubmer itself is not counted like in the problem example, I add once to perform that.
Number = 1
Count = 2
Found = False
big_Devisor = 0
print "Number Count Divisors"
while Found == False:
Divisors = 1 # because the Number is itself Devisor
if (Number % 2) != 0:
for i in range(1, int(Number/2), 2):
if Number % i == 0:
Divisors += 1
else:
for i in range(1, int(Number/2)):
if Number % i == 0:
Divisors += 1
if Divisors >= 500:
print (Number)
Found = True
else:
if Divisors > big_Devisor:
big_Devisor = Divisors
print Number,'\t', Count, '\t', Divisors
Number += Count
Count += 1