Question

Don't know what I am missing with problem set 1 b

Answer 1

here's what i have:
#!/usr/bin/env python


# print out prime numbers up to 1000

from math import *

max_prime = 1000 # number of prime numbers to output

isPrime = 1 # boolean testing for prime numbers

prime_candidates = []
prime_numbers = [2]
x = 2

while x < max_prime:

if (x/2)* 2 == x:
x += 1
else:
prime_candidates.append(x)
x += 1


index = 0 # index of prime_candidates list

while index < len(prime_candidates) -1:
isPrime = 1

for i in range (2, prime_candidates[index]):
if prime_candidates[index] % i ==0:
isPrime == 0
index += 1

if isPrime == 1:
prime_numbers.append(prime_candidates[index])
index += 1

sum = 0
n = 80


for i in range (0, n):
sum = sum + log(prime_numbers[i])

print "sum is" , sum , " n is ", n , "ratio is" , sum/n

according to the problem the sum of the logs should be less than n, and as n increases the ratio should be close to 1, which are not the case.

Answer 2

for one thing, isPrime == 0 is a comparison, not an assignment.

Answer 3

Also, n should probably equal max_prime, based on the problem statement:
In this case, the conjecture above reduces to the claim that the sum of the logarithms of all the primes less than n is less than n, and that as n grows, the ratio of this sum to n gets close to 1.

Answer 4

Hi Somnamniac, thanks for the reply! Shouldn't n be less than or equal to len(prime_numbers), as this is the actual number of primes up until max_prime?

I changed my code in part b to:
n = 20
for i in range (0,n):
if n > prime_numbers[i]:
sum = sum + log(prime_numbers[i])
print "sum is" , sum , " n is ", n , "ratio is" , sum/n

When I calculate by hand for n = 20 ( log(2) + log(3) + log(5) + log (7) + log (11) + log (13) + log (17) + log(19) ) I believe I get the correct answer. However when I plugin large values for n, the ratio exceeds 1.0.

Answer 5

I think we have different interpretations of the problem statement. I'll explain mine (using 1000 as n), and you can see if it makes sense:

the sum of the logarithms of all the primes less than n

translation: "the sum of the logarithms of all the primes less than 1000"


is less than n
"is less than 1000"

Answer 6

I generally copy everything before I post, in case it gets stuck

Answer 7

This is my interpretation:
"the sum of the logarithms of all the primes less than n" (with n =1000)
This would be a total of 191 numbers. [2,3,5,7,11,13,..........967,971,977,983,991,997]

so then the sum would be the log of each of these individual numbers added together, right?

Answer 8

right. and what is that less than?

Answer 9

It's less than n, 1000.
I assume then my is error that I am using the number of primes (191) to calculate the ratio? If I use n = 1000 it looks good. However if I test higher values this way I see the ratio is around 0.176 - 0.18. Like the problem states it does not increase monotonically, but it doesn't seem to get close to 1.

Answer 10

Can you paste your current version?

Answer 11

I found a flaw in my code when I calculated my last post. I attempted to fix it, but now the sum is greater than n and the ratio is greater than 1, when n =1000 :(

#!/usr/bin/env python
# print out prime numbers up to 1000

from math import *

n = 1000 # number until to calculate prime numbers

isPrime = 1 # boolean testing for prime number

prime_candidates = []
prime_numbers = [2]
x = 2

while x < n:

if (x/2)* 2 == x:
x += 1
else:
prime_candidates.append(x)
x += 1


index = 0 # index of prime_candidates list

while index < len(prime_candidates) -1:
isPrime = 1

for i in range (2, prime_candidates[index]):
if prime_candidates[index] % i ==0:
isPrime == 0
index += 1

if isPrime == 1:
prime_numbers.append(prime_candidates[index])
index += 1


sum = 0

#print prime_numbers

num_of_primes = len(prime_numbers)

#print num_of_primes

for i in range (0,num_of_primes):
if n > prime_numbers[i]:
sum = sum + log(prime_numbers[i])

print "sum is" , sum , " n is ",n , "ratio is" , sum/n

Answer 12

You've still got the

isPrime == 0

problem. That should be an assignment, not a comparison

Answer 13

D'oh!
Changing that fixes it! Thank you so much for your patience, I really appreciate it :)

Answer 14

No problem. I'm glad you got it!