Question

ps1.2 - logarithm of prime numbers - where i am stuck. plz help...

Answer 1

from math import *

n = 2
Goal = int(raw_input('Please input a prime number to count '))
TotLn = 0 ##sum of log(prime)s so far

def isprime(n): ## defines a prime number
if n == 2:
return 1 ## 2 is prime
if n % 2 == 0:
return 0 ## all other evens are not prime

max = n**0.5+1
i = 3

while i <= max: ## determines if remaining odd numbers are
if n % i == 0: ## composite
return 0
i += 2

return 1

while n < Goal: ## produces a list of prime numbers n, Goal

if isprime(n) == 1:
print n
NewLn = log(n) ##defines NewLn as log of current n
print NewLn
elif isprime(n) == 0:
pass
n += 1
TotLn += NewLn ##adds NewLn to the sum of previous logs
print TotLn
if n == Goal:
NewLn = log(n)
TotLn += NewLn
print 'n = ', n
print 'log(n) = ', log(n)
print 'sum of logs of primes up to n =', TotLn
print 'the ratio between the two =', n/TotLn

Answer 2

the problem i am having is that it adds too many times to TotLn

Please input a prime number to count 5
2
0.69314718056
0.69314718056
3
1.09861228867
1.79175946923
2.8903717579
n = 5
log(n) = 1.60943791243
sum of logs of primes up to n = 4.49980967033
the ratio between the two = 1.11115810808
>>> log(5)
1.6094379124341003

Answer 3

so you see there, it prints the prime number, then it's log, then TotLn, which works for 2, but when it gets to three it seems as if it's adding twice.

Answer 4

here is an update with some context in the prints
Please input a prime number to count 7
this is the current iteration of n, 2
this is the log of the current iteration of n, 0.69314718056
this is the total of all logs so far, 0.69314718056
this is the current iteration of n, 3
this is the log of the current iteration of n, 1.09861228867
this is the total of all logs so far, 1.79175946923
this is the total of all logs so far, 2.8903717579
this is the current iteration of n, 5
this is the log of the current iteration of n, 1.60943791243
this is the total of all logs so far, 4.49980967033
this is the total of all logs so far, 6.10924758276
n = 7
log(n) = 1.94591014906
sum of logs of primes up to n = 8.05515773182
the ratio between the two = 0.869008433236

Answer 5

so it seems as if it is adding NewLn to TotLn twice. and i have no idea why.

Answer 6

oops nevermind i figured it out. looks like placing it in the form of a question on here made me think of it in a new context. hahahah

Answer 7

I soooo identify with that last observation about posting a problem and then having the answer come to me. It's happend to me a ton of times. And if I may make a suggestion about style...

It's more "Pythonic" to return True or False from isPrime(). Then you can just say "if isPrime():". The way you have it looks like a C programmer would write it, since C has no such thing as True or False.

Answer 8

I feel like such the newb. :-)
I'll post my code - I used a list to do everything, so my log sums was pretty easy.

import math
somelist = []

for x in range(1,10000):
somelist.append(x)
somelist.remove(1)

for q in range(2,int(math.sqrt(10000)+1)):
for y in (somelist):
if y%q == 0:
if y >= 4 and y !=q:
somelist.remove(y)

while len(somelist) > 1000:
somelist.remove(somelist[-1])
print "1000th Prime is:", somelist[-1]

z = 0
for a in somelist:
#z = z + math.log(a,2) #Sum of logs, base 2.
z = z + math.log(a)
print "Sum of all those logs:",z

Answer 9

I am going to rewrite this code to accomplish it without the lists...wish me luck!

Answer 10

ps. Any constructive criticism of my code is welcomed!

Answer 11

One thing I noticed right off is the way you build somelist. if you just replace everything from "somelist = []" to "somelist.remove(1) with the single line "somelist = range(2, 10000), you'll get the same result in a much cleaner way.

Answer 12

Good point Radly...being completely green in the programming world, these are things I like to know. Sometimes it's hard to be critical of your own code, and I tend to miss the simple things.