intro_to_cs | Solution to ps1

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

Flat | Top-Level Comments Only

Since primes require so much computing power, I tried to close down the number of candidates and tests as much as I could. To do that, I did three things:

Since 2 is the first prime, generated only odd number candidates.
Stored the primes and used a slice based on the prime nearest 1/3 of the candidate value as test divisors. (Since all the even numbers are eliminated, 1/3 becomes the upper limit.)
Used the break keyword to exit the loop as soon as a candidate number was weeded out.

It runs pretty fast; here are the stats:

Total candidates tested: 3959
Total tests: 203,514
Average tests per candidate: 51

# Initialize state variables.
primes = [2]
count = 1
test_slice = primes[0:]

# Do this until we have 1000 primes
while len(primes) < 1000:

    # Increment counter by 2. (Eliminates all divisible by 2.)
    count += 2
    
    # Set prime switch true.
    is_prime = True

    # Loop through the testing slice.
    for prime in test_slice:

        # Check to see if count is prime.
        if count % prime == 0:
            
            # Not a prime, set switch to false.
            is_prime = False

            # If divisible by 3, figure new test slice upper limit.
            if prime == 3:

                # Get the divisor limit.
                divisor_limit = count / 3

                # Check to see if it's in the prime array.
                # (If divisor_limit is a power of 3, it won't be.)
                if divisor_limit in primes:

                    # If it's a prime, find its index and add 1.
                    i = 1 + primes.index(divisor_limit)
                    # set the new test slice range from 1 to i.
                    test_slice = primes[1:i]

            # Don't waste any more time on this number.
            break

    # Check the switch; if it's still true...
    if is_prime:
        #...count is prime. Add it to the primes array.
        primes.append(count)

        # This is only for the first loop.
        if count == 3:
            test_slice = [count]

# print
ordinal = str(len(primes)) + 'th'
print 'The', ordinal, 'prime is', count

I'd love to see if I could improve on that, but I'm really out of time this week.

Haha, that runs a lot faster than mine--good catch on the 1/3rd and using the break command.

Solution to ps1

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