## LAB: Sieve of Eratosthenes

The end goal is to verify the included program is getting all the primes it should be.

To this end, primes_file.txt contains the first 10,000 prime numbers, up to over 104,000.

First, lets see how we read in the file and turn it into a simple Python list, of primes only. This published list came from The Prime Pages.

``````

In [1]:

``````
``````

Out[1]:

``````
``````

In [2]:

with open("primes_file.txt", "r") as primes:
output = []
for line in primes.readlines()[3:]: # skip first 4 lines
if line.strip() == 'end.':
break
for column in line.split():
num = int(column.strip())
output.append(num)

``````

output will be global since the above cell is run top level, not inside the scope of a function. Checking the last few entries below, to confirm everything worked:

``````

In [3]:

# verify we have the data we want
print(output[0:10]) # first 10
print(output[-10: -1]) # last 10

``````
``````

[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
[104677, 104681, 104683, 104693, 104701, 104707, 104711, 104717, 104723]

``````

Lets check on primes to 1000. Up to 100,000 is left as an exercise to the more committed.

``````

In [4]:

from math import sqrt as root2, floor

def eratosthenes(n):
the_max = floor(root2(n)) # upper limit of eliminator
sieve = list(range(0, n+1))
eliminator = 2
while True:
if eliminator > the_max:
break
print("Eliminating multiples of:", eliminator)
for k in range(2 * eliminator, n + 1, eliminator):
sieve[k] = 0
while sieve[eliminator + 1] == 0:
eliminator += 1
else:
eliminator = sieve[eliminator + 1]

# shrink me down (compact the sieve)
sieve = [n for n in sieve if n != 0][1:]  # list comprehension!
return sieve

# apply fancy formatting to output
output = eratosthenes(1000)
for row in range(0, len(output), 10):
print(", ".join(map(lambda s: str.format("{:3d}", s), output[row:row+10])))

``````
``````

Eliminating multiples of: 2
Eliminating multiples of: 3
Eliminating multiples of: 5
Eliminating multiples of: 7
Eliminating multiples of: 11
Eliminating multiples of: 13
Eliminating multiples of: 17
Eliminating multiples of: 19
Eliminating multiples of: 23
Eliminating multiples of: 29
Eliminating multiples of: 31
2,   3,   5,   7,  11,  13,  17,  19,  23,  29
31,  37,  41,  43,  47,  53,  59,  61,  67,  71
73,  79,  83,  89,  97, 101, 103, 107, 109, 113
127, 131, 137, 139, 149, 151, 157, 163, 167, 173
179, 181, 191, 193, 197, 199, 211, 223, 227, 229
233, 239, 241, 251, 257, 263, 269, 271, 277, 281
283, 293, 307, 311, 313, 317, 331, 337, 347, 349
353, 359, 367, 373, 379, 383, 389, 397, 401, 409
419, 421, 431, 433, 439, 443, 449, 457, 461, 463
467, 479, 487, 491, 499, 503, 509, 521, 523, 541
547, 557, 563, 569, 571, 577, 587, 593, 599, 601
607, 613, 617, 619, 631, 641, 643, 647, 653, 659
661, 673, 677, 683, 691, 701, 709, 719, 727, 733
739, 743, 751, 757, 761, 769, 773, 787, 797, 809
811, 821, 823, 827, 829, 839, 853, 857, 859, 863
877, 881, 883, 887, 907, 911, 919, 929, 937, 941
947, 953, 967, 971, 977, 983, 991, 997

``````

Now we're interested in checking our list against the elements obtained from primes_file.txt, up to and including our largest prime.

We may simply count the primes we have, chop off that many from output and perform and equality test...

``````

In [5]:

how_many = len(output)
check_list = output[:how_many]  # get as many from the published pool as are in the sieve
print(output == check_list)     # check for equality

``````
``````

True

``````

Below is a screen shot of what the sieve algorithm looks like on the Codesters development platform, used with 5th and 6th graders in some schools. You may actually run the code in Codesters by visiting this link.