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finding a very big prime number in Python [duplicate]

I want to generate two really large prime numbers using an algorithm I found online and changed slightly.
I get this error on line 5:
Python OverflowError: cannot fit 'long' into an index=sized integer
My code:
import math
def atkin(end):
if end < 2: return []
lng = ((end/2)-1+end%2)
**sieve = [True]*(lng+1)**
for i in range(int(math.sqrt(end)) >> 1):
if not sieve[i]: continue
for j in range( (i*(i + 3) << 1) + 3, lng, (i << 1) + 3):
sieve[j] = False
primes = [2]
primes.extend([(i << 1) + 3 for i in range(lng) if sieve[i]])
return primes
How can I fix my error?
If you know a better way to generate large primes, that would be helpful also.

The following code demonstrates the problem that you are running into:
import sys
x = [True]*(sys.maxint+1)
which yields an OverflowError. If you instead do:
x = [True]*(sys.maxint)
then you should get a MemoryError.
Here is what is going on. Python can handle arbitrarily large integers with its own extendible data type. However, when you try to make a list like above, Python tries to convert the number of times the small list is repeated, which is a Python integer, to a C integer of type Py_ssize_t. Py_ssize_t is defined differently depending on your build but can be a ssize_t, long, or int. Essentially, Python checks if the Python integer can fit in the C integer type before doing the conversion and raises the OverflowError if it won't work.

Line 5 trues to allocate a really long list full of True values. Probably your lng is too large to fit that list in memory?
I was not able to exactly reproduce your error; in the worst case I ended up with just a MemoryError instead.
Probably the algorithm is ok (though I can't bet), just try a smaller number.

python multiplying python float with numpy float

So I'm trying to do the following:
self.cashflows["Balance"] * self.assumptions["Default Rates"][current_period - 1]
where cashflows is a list with python floats and assumptions is a list with numpy floats.
I used numpy to fill the assumption vector and I am getting the following error when I try to multiply the two:
can't multiply sequence by non-int of type 'numpy.float64
I get the error but what would be my best course of action here?
thx

Depends on what you want to do. Do you want to multiply every item in the list self.cashflow["Balance"] with self.assumptions["Default Rates"][current_period - 1]? Then you can use some list comprehension:
result = [q * self.assumptions["Default Rates"][current_period - 1] for q in self.cashflow["Balance"]]
or convert your second argument to np.float:
result = self.assumptions["Default Rates"][current_period - 1]* np.asarray(self.cashflow["Balance"])
Otherwise, multiplying a whole list by N repeats that list N times. If thats what you want, cast your np.float64 to int.
EDIT: Added missing multiplication sign

Python input multiple lines and spaces

Im trying to solve one of the a2oj problems "given three numbers a , b and c. print the total sum of the three numbers added to itself."
I came with this
import sys
numbers = [int(x) for x in sys.stdin.read().split()]
print(numbers[0] + numbers[1] + numbers[2])
I saw many topics but I cant figure out how to read just 3 values from input. I know I can stop this procces by typing CTRL+D, but is there any possibility to make it automatic (after reaching third value)?
Thanks
// Thanks for very quick answers, I made mistake and posted only Problem Statement without Input Format: "three numbers separated by bunch of spaces and/or new lines"
So for example input should look like this:
2
1 4
// Ok thanks to you guys finally I made this:
n = []
while len(n) < 3:
s=input()
i = s.split()
[n.append(int(j)) for j in i]
print(2 * sum(n))
It's working but when I sent my results I got Runtime Error. I have no idea why:
Link: https://a2oj.com/p?ID=346

You could just use:
sys.argv
import sys
numbers = [int(x) for x in sys.argv[1:4]]
print(numbers)
print(sum(numbers))

When inputs are given line by line.
from sys import stdin
sum = 0
for num in stdin.readline(4):
sum = sum + int(num)
print(sum)
When inputs are given on CLI.
from sys import argv
sum = 0
for num in argv[1:4]:
sum = sum + int(num)
print(sum)
Use Python strip() and split() functions as per your usecases

I am not sure what you are looking for, but it seems that you are looking for is the input function, from python's builtins:
x=input()
This reads any input from the user, as a string. You have then to convert it to a number if needed.
You can read three values:
x=input("First value:")
y=input("Second value:")
z=input("Third value:")
As you have now specified more precisely the problem statement, I edit my answer:
In your case, this is not very complicated. I am not going to give you the answer straight away, as it would defeat the point, but the idea is to wrap the input inside a while loop. Something like:
numbers=[]
while (you have less than 3 numbers):
(input one line and add the numbers to your list)
(print the sum of your numbers)
That way you are waiting for as many inputs as you need until you reach 3 numbers. By the way, depending on your input, you might have to check whether you do not get more than 3 numbers.

After seeing the update from the question author and linked the online judge question description, the tweak to his code needed is below. It's worth noting that the expected output is in float and has precision set to 6 and the output is 2 * sum of all inputs, not just sum. There is no description on this in the online judge question and you've to understand from the input vs output.
n = []
while len(n) < 3:
s = input()
i = s.split()
n.extend(float(j) for j in i)
print(format(2 * sum(n), '.6f'))
Screenshot below
But the first version of this answer is still valid to the first version of this question. Keeping them if anyone else is looking for the following scenarios.
To separate inputs by enter aka New lines:
numbers_List = []
for i in range(3):
number = int(input())
numbers_List.append(number)
print("Sum of all numbers: ", sum(numbers_List))
Screenshot:
To separate inputs by space aka Bunch of spaces:
Use map before taking input. I'd suggest using input as well instead of sys.stdin.read() to get input from users, separated by space, and ended by pressing Enter key.
Very easy implementation below for any number of inputs and to add using sum function on a list:
numbers = list(map(int, input("Numbers: ").split()))
print("Sum of all numbers: ", sum(numbers))
The screenshot below and link to the program is here
Read Python's Built-in Functions documentation to know more about all the functions I used above.

summing over a list of int overflow(?) python

Let's consider a list of large integers, for example one given by:
def primesfrom2to(n):
# http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
""" Input n>=6, Returns a array of primes, 2 <= p < n """
sieve = np.ones(n/3 + (n%6==2), dtype=np.bool)
sieve[0] = False
for i in xrange(int(n**0.5)/3+1):
if sieve[i]:
k=3*i+1|1
sieve[ ((k*k)/3) ::2*k] = False
sieve[(k*k+4*k-2*k*(i&1))/3::2*k] = False
return np.r_[2,3,((3*np.nonzero(sieve)[0]+1)|1)]
primesfrom2to(2000000)
I want to calculate the sum of that, and the expected result is 142913828922.
But if I do:
sum(primesfrom2to(2000000))
I get 1179908154, which is clearly wrong. The problem is that I have an int overflow, but I don't understand why. Let's me explain.Consider this testing code:
a=primesfrom2to(2000000)
b=[float(i) for i in a]
c=[long(i) for i in a]
sumI=0
sumF=0
sumL=0
m=0
for i,j,k in zip(a,b,c):
m=m+1
sumI=sumI+i
sumF=sumF+j
sumL=sumL+k
print sumI,sumF,sumL
if sumI<0:
print i,m
break
I found out that the first integer overflow is happening at a[i=20444]=225289
If I do:
>>> sum(a[:20043])+225289
-2147310677
But if I do:
>>> sum(a[:20043])
2147431330
>>> 2147431330+225289
2147656619L
What's happening? Why such a different behaviour? Why can't sum switch automatically to long type and give the correct result?

Look at the types of your results. You are summing a numpy array, which is using numpy datatypes, which can overflow. When you do sum(a[:20043]), you get a numpy object back (some sort of int32 or the like), which overflows when added to another number. When you manually type in the same number, you're creating a Python builtin int, which can auto-promote to long. Numpy arrays cannot autopromote like Python builtin types, because the array type (and its memory layout) have to be fixed when the array is created. This makes operations much faster at the expense of type flexibility.
You may be able to get around the problem by using a different datatype (like np.int64) instead of np.bool. However, it depends how big your numbers are. A simple example:
# Python types ok
>>> 2**62
4611686018427387904L
>>> 2**63
9223372036854775808L
# numpy types overflow
>>> np.int64(2)**62
4611686018427387904
>>> np.int64(2)**63
-9223372036854775808
Your example works correctly for me on 64-bit Python, so I guess you're using 32-bit Python. If you can use 64-bit types you will be able to get past the limit you found, but as my example shows you will eventually overflow 64-bit ints too if your numbers get super huge.

Python: Number ranges that are extremely large?

val = long(raw_input("Please enter the maximum value of the range:")) + 1
start_time = time.time()
numbers = range(0, val)
shuffle(numbers)
I cannot find a simple way to make this work with extremely large inputs - can anyone help?
I saw a question like this - but I could not implement the range function they described in a way that works with shuffle. Thanks.

To get a random permutation of the range [0, n) in a memory efficient manner; you could use numpy.random.permutation():
import numpy as np
numbers = np.random.permutation(n)
If you need only small fraction of values from the range e.g., to get k random values from [0, n) range:
import random
from functools import partial
def sample(n, k):
# assume n is much larger than k
randbelow = partial(random.randrange, n)
# from random.py
result = [None] * k
selected = set()
selected_add = selected.add
for i in range(k):
j = randbelow()
while j in selected:
j = randbelow()
selected_add(j)
result[i] = j
return result
print(sample(10**100, 10))

If you don't need the full list of numbers (and if you are getting billions, its hard to imagine why you would need them all), you might be better off taking a random.sample of your number range, rather than shuffling them all. In Python 3, random.sample can work on a range object too, so your memory use can be quite modest.
For example, here's code that will sample ten thousand random numbers from a range up to whatever maximum value you specify. It should require only a relatively small amount of memory beyond the 10000 result values, even if your maximum is 100 billion (or whatever enormous number you want):
import random
def get10kRandomNumbers(maximum):
pop = range(1, maximum+1) # this is memory efficient in Python 3
sample = random.sample(pop, 10000)
return sample
Alas, this doesn't work as nicely in Python 2, since xrange objects don't allow maximum values greater than the system's integer type can hold.

An important point to note is that it will be impossible for a computer to have the list of numbers in memory if it is larger than a few billion elements: its memory footprint becomes larger than the typical RAM size (as it takes about 4 GB for 1 billion 32-bit numbers).
In the question, val is a long integer, which seems to indicate that you are indeed using more than a billion integer, so this cannot be done conveniently in memory (i.e., shuffling will be slow, as the operating system will swap).
That said, if the number of elements is small enough (let's say smaller than 0.5 billion), then a list of elements can fit in memory thanks to the compact representation offered by the array module, and be shuffled. This can be done with the standard module array:
import array, random
numbers = array.array('I', xrange(10**8)) # or 'L', if the number of bytes per item (numbers.itemsize) is too small with 'I'
random.shuffle(numbers)