I am a beginner at learning python 3 and I am just writing basic programs. I wrote this simple program which would take a number in and divide it by numbers starting from 1 to the square root of the number and find the remainders and add it to a list and print it.
import math
def prime_checker(num):
n=1
list_of_remainder=[]
while n == math.floor(num**0.5):
var=int(num % n)
list_of_remainder.append(var)
n += 1
return list_of_remainder
var=prime_checker(10)
print(var)
Please tell me what I did wrong. I would like to point out here that I did try to research a bit and find error but I couldn't and only then have I posted this question.
The problem that I faced was that it printed out an empty list.
to start with, your while loop is not executed even once. The condition for your while loop is
while n == math.floor(num**0.5):
The argument num you are passing to the function prime_checker is equal to 10. In this case your condition test is:
while 1 == math.floor(10**0.5)
which is
while 1 == 3 which is obviously not true and as a result the loop is not executed even once.
import math
def prime_checker(num):
list_of_remainder = []
number=num;
n=1
x=math.floor(number**0.5)
while n <= x:
v=int(number % n)
list_of_remainder.append(v)
n += 1
return list_of_remainder
var=prime_checker(10)
print(var)
Related
I was solving the 3rd Question on Project Euler (Largest Prime Factor) and I'm a beginner at Python 3.
This is the solution I came up with, it works but not with very large numbers
x=int(input("Enter a number:"))
a=[]
for i in range(1,x+1):
cnt=0
if x%i==0:
for j in range(1,i+1):
if i%j==0:
cnt=cnt+1
if cnt==2:
a.append(i)
print(a[len(a)-1])
I understand its very basic, and its too slow to run large inputs, but is there any way a compiler could give me the output for this input - 600851475143. I tried using pypy3, it was taking too long as well.
Its my first time I'm using stackoverflow, so let me know if I'm doing anything wrong too.
I know you said you don't want to change the code but you would have to, if you want to solve it efficiently.
There actually a lib just for this eulerlib but the built-in math module can do it too.
If you want to use python with no modules you could try this but it is probably just as slow for large numbers
def Largest_Prime_Factor(n):
prime_factor = 1
i = 2
while i <= n / i:
if n % i == 0:
prime_factor = i
n /= i
else:
i += 1
prime_factor = max(prime_factor, n)
return prime_factor
The built-in math module can also do this and is far quicker. Since it is built-in you don't need any external libs like eulerlib
import math
# Getting input from user
n = int(input("Enter the number : "))
maxPrimeFactor = 0
# Checking and converting the number to odd
while n % 2 == 0:
maxPrimeFactor = 2
n = n/2
# Finding and dividing the number by all
# prime factors and replacing maxPrimeFactor
for i in range(3, int(math.sqrt(n)) + 1, 2):
while n % i == 0:
maxPrimeFactor = i
n = n / i
if n > 2:
maxPrimeFactor = n
print("The largest prime Factor of the number is ",int(maxPrimeFactor))
sum = 2
x=3
y=5000
for i in range (x,y):
for j in range (2,i):
if i%j==0:
break
elif i%j!=0 and j==i-1:
sum += i
if i==y-1 and y<2000000:
x=y
y+=5000
else:
continue
print(sum)
**I am not getting what is wrong in this code. By running this I came to know that the Last If and Else statement are not running **
Given your code, there are a couple of things wrong. First, sum is a python function name and should never be used as a variable name. It will get you into trouble in more ways than I care to think about. Second, the last else statement is not needed, because whether the if clause above it is or is not executed executed, the for loop will be executed again. Third, I don't understand the purpose of y and the magical value 5000, unless you are trying to provide an end value for your loop. The problem with this approach is you seem to try and extend it's range in increments of 5000. The problem is that once the initial for loop is executed, it creates a local iterable from x to 5000, and subsequent changes to y do not affect the for loops range.
I would approach the problem differently, by creating a list of primes and then use the python sum method to add all the values. Here is the code:
def sum_primes(max_prime):
""" Return Sum of primes Less than max_prime"""
primes = [2]
indx_num = 3
while primes[-1] <= max_prime:
update = True
for prime in primes:
if indx_num == prime or indx_num % prime == 0:
update = False
break
if update:
primes.append(indx_num)
indx_num += 2
#return summ of all values except the last
return sum(primes[:-1])
Executing sum_primes(2000000)
yields 1709600813
So first of all, I realise there are much easier ways to get a list of prime numbers, but I'm just doing this to learn. I have a very poor understanding of a lot of this (as you'll see) so sorry if this is a dumb question. I'm trying to learn.
#make an empty list to store primes in
primes = list()
#make a variable to easily change the amount of numbers I test for primality
high_val = 15
#Allocate a range that I will test all numbers in for primality
for n in range(2, high_val):
#Within the previous for loop, start another for loop to test every integer against every
#value inside the primes list
for p in primes:
if n % p == 0:
print(%s is not prime" % n)
else:
#If n is prime, I add it to the list and print that it is prime
primes.append(n)
print("%s is a prime" % n)
I don't know if those comments make it harder to read, but that's my logic. There is no print output for the function. So I figured, there's just no value in primes, I need to give it something to compare to. So I added a primes.append(2) at the start immediately after the first line, and changed the range to (3, high_val)...
If I do that, it ends up printing about 5 times for every number that it is prime and 5 more messages saying it is not prime. Clearly I'm doing something massively wrong, if anyone knows where I went wrong and/or how to fix this that would be greatly appreciated. Thanks!
The thing is that you are missing some flags: you are printing if the number is prime every time you check it against another number. You should print it only after iterating over all primes.
About this, consider using all:
primes = [2]
high_val = 15
for n in range(3, high_val, 2):
if all(n % p == 0 for p in primes):
print(f"{n} is prime")
primes.append(n)
else:
print(f"{n} is not prime")
ps: I didn't test it.
I am a novice trying to learn python from Automate The Boring Stuff with Python by Al Sweigart and I came across his block of code to answer a math problem: "What is the sum of all the numbers from 0 to 100?" Apparently, this was a question Gauss got when his teacher wanted to keep him busy.
Sweigart used a for loop and range() function to get the answer:
total = 0
for num in range(101):
total=total+num
print(total)
A page later he states that "you can actually use a while loop to do the same thing as a for loop; for loops are more concise."
How would this statement be rendered in a while loop?
I tried replacing the for with while but got an error: "name 'num' is not defined." I also tried to set up a summation math equation using another block of code from another forum, but completely got lost.
print('Gauss was presented with a math problem: add up all the numbers from 0 to 100. What was the total?')
a=[1,2,3,4,5,...,100]
i=0
while i< len(a)-1:
result=(a[i]+a[i+1])/2
print(result)
i +=1
Then, I tried to setup i in an equation that would loop until each number was added, but got stuck.
print('Gauss was presented with a math problem: add up all the numbers from 0 to 100. What was the total?')
i=0
while i<101:
i=i+1
a=i
Would the while statement be too complex to warrant the effort?
Your last example comes close.
A for loop of this form:
for x in range(N):
# ...
can be replaced by a while loop like this:
x = 0
while x < N:
# ...
x += 1 # equivalent to x = x + 1
Just make sure you leave the rest of the code unchanged!
The for loop is more concise. Notice how we need a "counter" variable , in this case i with the while loop. That's not to say we don't need them in a for loop, however they're integrated nicely into the syntax to make for cleaner code.
i = 0
total = 0
while i < 101:
total += i
i += 1
print(total)
Python's for loop syntax is also a foreach equivalent:
for eachItem in list:
# Do something
I'm using python 3 for a small extra credit assignment to write an RSA cracker. The teacher has given us a fairly large (large enough to require more than 32 bits) int and the public key. My code works for primes < 32 bits. One of the reasons I chose python 3 is because I heard it can handle arbitrarily large integers. In the python terminal I tested this by doing small things such as 2**35 and factorial(70). This stuff worked fine.
Now that I've written the code, I'm running in to problems with overflow errors etc. Why is it that operations on large numbers seem to work in the terminal but won't work in my actual code? The errors state that they cannot be converted to their C types, so my first guess would be that for some reason the stuff in the python interpreter is not being converter to C types while the coded stuff is. Is there anyway to get this working?
As a first attempt, I tried calculating a list of all primes between 1 and n (the large number). This sort of worked until I realized that the list indexers [ ] only accept ints and explode if the number is higher than int. Also, creating an array that is n in length won't work if n > 2**32. (not to mention the memory this would take up)
Because of this, I switched to using a function I found that could give a very accurate guess as to whether or not a number was prime. These methods are pasted below.
As you can see, I am only doing , *, /, and % operations. All of these seem to work in the interpreter but I get "cannot convert to c-type" errors when used with this code.
def power_mod(a,b,n):
if b < 0:
return 0
elif b == 0:
return 1
elif b % 2 == 0:
return power_mod(a*a, b/2, n) % n
else:
return (a * power_mod(a,b-1,n)) % n
Those last 3 lines are where the cannot convert to c-type appears.
The below function estimates with a very high degree of certainty that a number is prime. As mentioned above, I used this to avoid creating massive arrays.
def rabin_miller(n, tries = 7):
if n == 2:
return True
if n % 2 == 0 or n < 2:
return False
p = primes(tries**2)
if n in p:
return True
s = n - 1
r = 0
while s % 2 == 0:
r = r+1
s = s/2
for i in range(tries):
a = p[i]
if power_mod(a,s,n) == 1:
continue
else:
for j in range(0,r):
if power_mod(a, (2**j)*s, n) == n - 1:
break
else:
return False
continue
return True
Perhaps I should be more specific by pasting the error:
line 19, in power_mod
return (a * power_mod(a,b-1,n)) % n
OverflowError: Python int too large to convert to C double
This is the type of error I get when performing arithmetic. Int errors occur when trying to create incredibly large lists, sets etc
Your problem (I think) is that you are converting to floating point by using the / operator. Change it to // and you should stay in the int domain.
Many C routines still have C int limitations. Do your work using Python routines instead.