Question

Having difficulty with PSet1 Prob2. Here is what I have thus far...


from math import *
prime_poss = 2
sum = 0
n = int(raw_input("Enter a number: "))
while prime_poss < n:
if prime_poss%2 != 0:
for divisor in range (2, (prime_poss/2)):
if prime_poss%divisor != 0:
logprime = log(prime_poss)
print logprime
sum = logprime + sum
prime_poss += 1

else:
prime_poss += 1
print sum

Ideas?

Answer 1

made numerous edits. my question is this ... when setting the range for the divisor to check a primes, what is the end limit... 'n'? n/2? our candidate prime?

Answer 2

Technically, the square root of the number

Answer 3

# is there a functional way to do this? in part 1a, where i wrote the 1000th prime finder, i used a FOR loop..
for divisor in range (2, (candidate/2)):
# and this was how i was able to test a range of divisors to identify a prime

is it the same concept (given that i am not using a square-root function?

Answer 4

I think you are trying too hard, let me help you understand. You have done problem 1, correct? Please show you your code for number 1 and, for the sake of simplicity and efficiency, replace your "n/2" by "math.sqrt(n)+1"

Then I'll build from that code to get the next problem done =D

Answer 5

Basically, you need a MINOR modification from code 1 to get code 2 working

Answer 6

## here is my code for part 1a:
prime_count = 1
candidate = 2
not_one_if_true = 0

while prime_count < 1000:
if candidate%2 == 0: # to eliminate even nnumbers
candidate += 1

else:
for divisor in range(2,(candidate/2)): # find divisor range
if candidate%divisor == 0: # find primes
not_one_if_true = 1 # if not prime

if not_one_if_true == 0: if prime, increment
prime_count += 1
candidate +=1

else: # if not a prime, increment, reset Boolean truth
candidate += 1
not_one_if_true = 0

print candidate - 1

Answer 7

That code seems to have a problem. Maybe use pastebin, because there seem to be problems.

Answer 8

# i forgot to comment out a note... i just ran this newly pasted code and got 7919
# don't know how to use pastebin...

prime_count = 1
candidate = 2
not_one_if_true = 0

while prime_count < 1000:
if candidate%2 == 0: # to eliminate even nnumbers
candidate += 1

else:
for divisor in range(2,(candidate/2)): # find divisor range
if candidate%divisor == 0: # find primes
not_one_if_true = 1 # if not prime

if not_one_if_true == 0: #if prime, increment
prime_count += 1
candidate +=1

else: # if not a prime, increment, reset Boolean truth
candidate += 1
not_one_if_true = 0

print candidate - 1

Answer 9

Okay, here is the easy part. I made your code work for problem 2, but I will not show it. Create a variable named Sum. Every time you find a prime number, add the math.log of that number to Sum.

Answer is 7803

Answer 10

implementing ....

Answer 11

If you come close to 7803, tell me. I will tell you why you may be a "little" off

Answer 12

7811.333

Answer 13

7811.59

Answer 14

Okay. Remember this.

1) You must NOT count the math.log of the last number ( 7919 ). This is said in the assignment. It has to be all numbers BELOW the nth prime.

Is your answer now exact?

Answer 15

7802.6

Answer 16

so i set my limit too high (1000), instead of 999...

thanks for the help!

Answer 17

Your answer is wrong. Do you count the math.log(2) somewhere in your code?

Answer 18

no, i couldn't figure that part out... my code can be broken down into
1. determine if odd; if even, increment by one
2. if odd, determine if there are factors within range (2 to n/2) [is it prime]
3. if prime, increment by one, and add

for part 1b, also add sum = math.log(candidate) + sum

but since i am eliminating the even in my initial code, i don't have the math.log(2) incorporated

Answer 19

Then add math.log(2) somewhere in the code. The exact answer is near 7803.3

Answer 20

will do, thanks!