Find all the primes from 2 to n, using the Sieve of Eratosthenes

Question

I have a word problem I am trying to solve but am getting stuck on a key part.

Convert the following English description into Python code.

Initialize n to be 100. Initialize numbers to be a list of numbers from 2 to n, but not including n. With results starting as the empty list, repeat the following as long as numbers contains any numbers.

Add the first number in numbers to the end of results.

Remove every number in numbers that is evenly divisible by (has no remainder when divided by) the number that you had just added to results.

How long is result?

When n is 100, the length of results is 25.

So far I have understood to set n = 100, and a range(2, 100), results = [] and that the result will be an append situation as in results.append(numbers[]), but I am having a mental block figuring the key of "Remove every number in numbers that is divisible by the number that was added to results".

======after a couple minutes=======

As Michael0x2a said - this is a problem when i have to implement an algorithm that finds all the primes from 2 to n, using the Sieve of Eratosthenes.

I think I can continue to deal with this problem.

Thank's a lot for your answers guys.

If it helps, it looks like the question is asking you to implement an algorithm that finds all the primes from 2 to n, using the Sieve of Eratosthenes. — Michael0x2a, Nov 23 '13 at 21:33
What is the question? Are you asking how to go through the list of numbers or how to figure out that a number is divisible by another a number? — Lasse V. Karlsen, Nov 23 '13 at 21:35
Reopen please. I've edited with clear explanation what i needed in my question. — Matt, Nov 24 '13 at 17:51

score 3 · Accepted Answer · answered Nov 23 '13 at 22:52

3

n = 100
numbers = range(2,100)
results = []
while len(numbers) > 0:
    results.append(numbers[0])
    numbers = [number for number in numbers if number % results[-1] != 0]
print len(results)

answered Nov 23 '13 at 22:52

Kyle Neary

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6
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Dear Kyle, Can you please write a full implementation of your solution? I just trying to figure out how exactly it works. Thanks. – Matt Nov 24 '13 at 17:28
Thank's for help Kyle. I've done with your help right that way: `codedef long_implementation(n): numbers = range(2,n) results = [] while len(numbers) > 0: results.append(numbers[0]) for number in numbers: if number % results[-1] == 0: # numbers.remove(number)` – Matt Nov 24 '13 at 18:03
@Matvey after `results.append(numbers[0])`, `results[-1]` is just `== numbers[0]`; also, `for number in numbers: if number % results[-1] == 0: numbers.remove(number)` is ***trial division***; but it is the same as `for number in range(numbers[0],100,numbers[0]): if number in numbers: numbers.remove(number)` and *that* ***is*** the sieve of Eratosthenes. – Will Ness Nov 25 '13 at 17:30

erewok · Answer 2 · 2013-11-23T21:42:59.043

The following:

Add the first number in numbers to the end of results.

...is not the same as:

results.append(numbers[])

To access the first number in a list, you can use the first index.

Thus:

numbers[0] ## would be the first "number" in numbers

Further, to do the following:

But I am having a mental block figuring the key of "Remove every number in numbers that is divisible by the number that was added to results".

...you will need some way of testing what a number is divisible by, which means you'll need to use modulo:

some_number % some_other_number

If there's no remainder, the modulo will result in a zero.

Find all the primes from 2 to n, using the Sieve of Eratosthenes

2 Answers2