Question

Hi guys,
I'm doing the 2008 course and I think it's going well but I have no way to check my assignments. So I'm attaching my solution to the second problem on set 1 - Product of the primes.

I would love it if someone would give me feedback. My output is pretty close to 1, but I'm not sure if it's close enough.
This is my solution:

```
from math import *

def prime_compute(n):
is_it_prime = 3
# prime_list = [2] #put 2 inside. 2 is a Prime
prime_sum = log1p(2)
remains = 1
while is_it_prime < n:
for i in range(3, (is_it_prime/2)):
remains *= is_it_prime%i
if remains == 0:
remains = 1
else:
# prime_list.append(is_it_prime)
prime_sum += log1p(is_it_prime)
remains = 1
is_it_prime += 2
print "the sum of all primes is ", prime_sum
print "the prime number is ", n
print "the ratio is ", float(n/prime_sum)

n = 9973
prime_compute(n)

```


Thanks in advance!
Michelle [Hakabuk]

Answer 1

There is code quoting here and code paste sites that make sharing code easier. To code quote here, use ``` (the one on the ~ key) above and below the code block.

```
from math import *
def prime_compute(n):
is_it_prime = 3
```
As you can see, this does code highlighting.

I have attached what I typed so you can see what the input looked like to get that output.

Answer 2

Thanks for the tip, my bad. Here's the code again:
```
from math import *

def prime_compute(n):
is_it_prime = 3
# prime_list = [2] #put 2 inside. 2 is a Prime
prime_sum = log1p(2)
remains = 1
while is_it_prime < n:
for i in range(3, (is_it_prime/2)):
remains *= is_it_prime%i
if remains == 0:
remains = 1
else:
# prime_list.append(is_it_prime)
prime_sum += log1p(is_it_prime)
remains = 1
is_it_prime += 2
print "the sum of all primes is ", prime_sum
print "the prime number is ", n
print "the ratio is ", float(n/prime_sum)

n = 9973
prime_compute(n)

```

Answer 3

Hey, first off, lots of thought you've put into that! You're good!!

I have a couple of questions, if you're interested in style comments. But, first, I wanted to make sure that I'm correctly understanding your algorithm.

Just to make sure I'm understanding your intent, I've added comments saying what I think your algorithm is doing. Does it look like I'm understanding your intent correctly?
```
def prime_compute(n):

# IS_IT_PRIME is a number that MAY be a prime. We can start it at
# 3, skipping 2, because we initialize our PRIME_SUM with
# the log of 2, in the statement that follows this one. So, we know we'll
# take care of 2- which we KNOW is a prime- by using it to start our sum.
is_it_prime = 3


# take care of 2, the first prime, by starting with its log as our sum.
prime_sum = log1p(2)

# we must initialize REMAINS, because we multiply by it in the loop below
remains = 1

while is_it_prime < n:

# In the following FOR, we check to see if IS_IT_PRIME can be evenly
# divided- i.e. can be divided without leaving any remainder.
#
# We can start the FOR loop at 3 because we KNOW that IS_IT_PRIME isn't
# divisible by 2: we only give set it to the odd values 3, 5, 7...
for i in range(3, (is_it_prime/2)):

# When remainder (%) operation in next statement returns 0, we have
# REMAINS = REMAINS * 0, which means REMAINS will become 0.
# After that, REMAINS will STAY 0, since it's one of the operands
# in the multiplication, and multiplying with 0 always gives 0.
remains *= is_it_prime%i

# if REMAINS is now 0, then the remainder at some point must have
# been 0. So, IS_IT_PRIME is NOT a prime, since there was a number
# that divided it evenly, because we had some remainder that was 0.
if remains == 0:

# IS_IT_PRIME was NOT a prime.
# Set REMAINS back to 1 to make ready for the next FOR iteration.
remains = 1

else:

# IS_IT_PRIME indeed WAS a prime, so add its log into our sum
prime_sum += log1p(is_it_prime)

# then, as we did in the other clause of this IF statement,
# set REMAINS back to 1 to make ready for the next FOR iteration.
remains = 1

# increment IS_IT_PRIME by 2, to make it the next odd number: 3, 5, 7...
is_it_prime += 2

print "the sum of all primes is ", prime_sum
print "the prime number is ", n
print "the ratio is ", float(n/prime_sum)

n = 9973
prime_compute(n)


```

Answer 4

you can compare your answer with mine: https://gist.github.com/jhonny111s/10928777

Answer 5

Here's mine (python 3 code):

from math import *

n = 100000
log_primes = log(2)

possible_primes = range(3, n, 2)

for i in possible_primes:
for j in range(2, i//2):
if i%j == 0:
break
else:
log_primes += log(i)

print("The sum of the log of primes less than %d is %f. The ratio is %f" %
(n, log_primes, log_primes/n))