Solution to ps1

[jetamors]

The solution I wrote is behind the cut. Please let me know if you have any questions about the math or the code itself, or feel free to share your own solution in the comments.

# ps1 for MIT OpenCourse 6.00 Introduction to Computing
# Written by Jetamors, January 2010

# Problem 1: compute and print the 1000th prime number

# Remember that a prime number is an integer (whole number) that is not evenly divisible by any integer
# other than itself and 1.  Therefore, my strategy is to divide every number bigger than 2 by all the integers smaller
# than it, except 1. For example, the number 5 would be divided by 4, 3, and 2.
# If every division gives me a remainder, then the number is prime; if even one remainder is zero, then the
# number cannot be prime.
# 
# There are two things I did to simplify this problem:
# 1. Every even number bigger than 2 cannot be a prime number; therefore I stripped out all the prime numbers.
# 2. A number can't be evenly divisible by any integer more than half its size.  For example, the biggest number you can
#    evenly divide 10 by is 5. (I'm not really sure how to explain the math behind this, but if you test it it'll be
#    true.)  Therefore, I don't bother dividing a potential prime number by any number more than half its size.
# You don't have to do either of these things, but your code will take much longer to run if you don't do them. This code
# could be made to run even faster by immediately going to the next candidate as soon as you get a zero remainder,
# but... I didn't do that, oh well.

print 'Problem 1:'
primeiter = 1 # This counts the number of primes. We initialize this to 1 to account for the prime number 2.
candidate = 2 # This is the variable we use for each candidate for prime number. This is initialized to 2 so the first number
              # the while loop evaluates will be 3.

while (primeiter<1000):
    candidate = candidate + 1                               # increment at the beginning so the last number won't be incremented
    # print 'Current iteration is',candidate
    remaindercheck = 1                                      # this is the variable we use when evaluating remainders
    if (candidate/2)*2 != candidate:                        # this strips out all the even numbers
        for i in range(3,candidate/2):                      # this strips out all the numbers bigger than 1/2.
            remaindercheck = remaindercheck * candidate%i   # This number is only non-zero if every remainder is non-zero.
            # print remaindercheck
        if remaindercheck != 0:
            # print 'Prime number is',candidate
            primeiter = primeiter + 1                       # increment this every time we see a prime number

# printing the output
print 'The 1000th prime number is', candidate
print ' '





# Problem 2: compute the sum of the logarithms of all the primes from 2 to some number n.
#            Print out the sum of the logs of the primes, the number n, and the ratio of these two quantities.

# Here, I modified the code from problem 1. The major difference is in the while loop.  Before, the
# while loop continued until we reached the 1000th prime number.  Now, we want the loop to continue until
# n is reached.  Notice also that we add the log of a number only if it's a prime number.

print 'Problem 2:'
from math import *      # so we can use the log function
candidate = 2 
logprimesum = log(2)    # this initalizes the sum of the logs with the log of the prime number 2.
n = int(raw_input('Please enter a number! '))

while (candidate<=n):   # note that the while statement has changed; now it's based on n, not the number of primes.
    candidate = candidate + 1
    remaindercheck = 1
    if (candidate/2)*2 != candidate:
        for i in range(3,candidate/2):
            remaindercheck = remaindercheck * candidate%i
        if remaindercheck != 0:
            logprimesum = logprimesum + log(candidate)
            
# printing the output
print ' '
print 'The sum of the logs of the prime numbers from 2 to',n,'is',logprimesum
print 'The ratio of that sum to',n,'is',logprimesum/n

Comments

[owl]

Here's my version. I decided to store the primes I'd already calculated in a list and use them as the divisors for the next candidate, rather than determining the divisors each time through the loop. Python's high-level, I imagine its list implementation is efficient enough for me not to worry about it :)
A number can't have a factor that's greater than its square root, so if it has reached that point, it's a prime and the control can break out of the loop.

#Problem 1
primes = [2] #list of primes. We know that 2 is the first prime
integer = 3 #so start checking at 3

while len(primes) < 1000:
   for p in primes: #we want to divide the candidate integer by all the primes we already have
       is_prime = 1 #we'll assume it's prime until proven otherwise
           if (integer % p == 0):
	       is_prime = 0 #if it can be divided equally by a prime, it's not prime
	       break        #so break out of the loop here
	   if p*p >= integer:
		break #don't keep checking with divisors greater than the square root
    if is_prime: #stopped dividing; is it prime?
        primes.append(integer) #add it to the array of primes
    integer = integer + 1 #and then move on to the next integer
#we're out of the while loop, so print the last prime, 
#and let's check how many we have while we're at it
print "The %dth prime is: %d" %(len(primes), primes[len(primes)-1]) #Don't forget the list is 
#zero-indexed, or we'll get a lovely index out of bounds error here

And the slightly adapted version for the second problem:

from math import *
n = int(raw_input("Enter a number: ")) #ask the user what is n
primes = [2] #list of primes. We know that 2 is the first prime
integer = 3 #so start checking at 3
sum_logs = log(2) #the log of the primes we have so far

while len(primes) < n:
    for p in primes: #we want to divide the candidate integer by all the primes we already have
        is_prime = 1 #we'll assume it's prime until proven otherwise
            if (integer % p == 0):
        is_prime = 0 #if it can be divided equally by a prime, it's not prime
        break        #so break out of the loop here
        if p*p >= integer:
            break #don't keep checking with divisors greater than the square root
     if is_prime: #stopped dividing; is it prime?
        primes.append(integer) #add it to the array of primes
        sum_logs = sum_logs + log(integer) #find the log and add it to the total
    integer = integer + 1 #and then move on to the next integer
#print out the sum of the logs, and the ratio
print "The sum of the logs of the first",len(primes), "primes is",sum_logs
print "The ratio of the sum to",n,"equals",(sum_logs / n)

I don't think I've put so many comments in code since I was in first year of uni ;)
You might also notice that I like to put double quotes around strings, and have a space on each side of the operators. I just think it's easier to read that way.

[jetamors]

This one also runs pretty fast! I do see an error in your part 2, though; the question asks you to find all the primes between 1 and n, not n number of primes. If you do it that way, the ratio you get should always be between 0 and 1.

[owl]

Ooops! Requirements reading fail on my part, you'd think I'd know better by now. If I change while len(primes) < n: to while integer <= n: that should do the trick.

[aranthe]

Very elegant! Since I used the same basic logic outline, I merged our approaches: Used your square root limit with my increment of 2 and knocked about another 4000 tests off. Cool!