Question

been at problem set one for days now. i have a number of elements but cant get them to work together.

Answer 1

anyone wanna help?

Answer 2

what have you got so far?

Answer 3

haha well now that i look i feel like i have a lot less than i thought i did.
basically i have this:

for n in range(1000):
if n < 2: #number must be larger than two
pass
elif n%2==0: #number must not be even
pass
elif not n & 1:
pass
for x in range(3, int(n**0.5)+1, 2):
if n % x == 0:
pass

but i feel like i am doing many many many things wrong

Answer 4

for instance, should i be using a for loop?
and should i be using a range?
i get that i should get the program to count prime numbers 1000 times (or 999 not including 2) but i have no idea how to make that happen

Answer 5

well, let's try to describe the problem. What is the program going to do?

Answer 6

count 1000 prime numbers

Answer 7

ok. So if we can find prime numbers, it's easy to count them, right? that's just an int variable that gets one added every time we find a prime.

Answer 8

okay...

Answer 9

So we need to start at a number,
check to see if it's prime, and
If it is prime,
add one to the counter
otherwise
move to the next number

Answer 10

i see. so it will count to 1000 and show the prime number corresponding to the count?

Answer 11

right. the 1000th prime number

Answer 12

which is 7919, for when you need to check output

Answer 13

So say we have a number. We know we can get a number, we know we can increment a counter if it turns out to be prime, but for now we've got this number. What process can we use to tell if it's prime?

Answer 14

first make sure it's positive, then an integer, then odd, and then indivisible by... 2, 3, or 5?

Answer 15

how do we increment a counter if it turns out to be prime?

Answer 16

first, we assign a counter variable, as in:

ctr = 0

then, if the number is prime, we add to it:

if numberisprime:
ctr +=1

the += means 'equals itself plus,' so that's like saying ctr = ctr+1

Answer 17

okay. so then are my criteria for prime correct?

Answer 18

well, what if the number is divisible by 7? or 11? or a really high prime number that neither of us can think of?

One way to be absolutely sure is to just check. Try dividing the number by every number less than it.

Answer 19

We're using computers. We don't have to be clever, because we can do 2.9 million operations in a second.

Answer 20

so what function will divide by all integers below n?

Answer 21

well, a for loop would work. have you used those?

Answer 22

ive been playing around with them, went through this chapter last night - http://en.wikibooks.org/wiki/Python_Programming/Loops

Answer 23

So you know that with a for loop you can specify a range

Answer 24

what if you used a for loop with the range set to the number you wanted to test? That way it would iterate over every number less than the number you wanted to test.

Answer 25

but what number do i want to test? i could throw the 1000th prime in but wouldnt that defeat the purpose?

Answer 26

So if

numberToTest = 7

and your range is

for divisor in range (2,numberToTest)

then the for loop will start with x = 2, then x=3...up to 6.

Answer 27

oh, you want to start with a low number, and test every number going up until you collect 1000 primes

Answer 28

okay so i would start with a loop that lists primes and tell it to stop after it has counted 1000?

Answer 29

right:

while we haven't found as many primes as we need:
Check the next number
if the next number is prime
add one to the counter

When we have enough primes
Print the last prime we found.

Answer 30

while we haven't found as many primes as we need: # So we need a counter variable
Check the next number ##we need a number to test variable
if the next number is prime
add one to the counter

When we have enough primes ##we also need a goal variable
Print the last prime we found.

Answer 31

So we can start out saying

NumToTest = The number we're going to test for primeness
Divisor = The number we're going to divide NumToTest by to see if it's prime
Goal = the number of primes we want to find
ctr = the number of primes we've found

Answer 32

Now we can use those things as if they were actual numbers. So we can say

while ctr < Goal:

and that means "while we haven't found enough primes"

Answer 33

while ctr = goal: ##"while we haven't found enough primes
while Divisor < NumToTest ## while the number we're trying to divide by is smaller than the number we're testing

Answer 34

so ive got: NumToTest = 2
Divisor =
Goal = 1000
ctr = 0

while ctr < Goal:


but i do not know what to set divisor to

Answer 35

or should i start at 3?

Answer 36

while ctr = goal: ##"while we haven't found enough primes
while Divisor < NumToTest
<try to use an operation to see if the number is prime here>
<if the operation proves the number isn't prime>
<add something to the number to test to get the next one>
<reset the divisor>
<Otherwise>
<add something to the divisor>
<If we've made it this far, we've found a prime number>
<add one to our counter>
<reset the divisor>
<add something to the number to test to get the next one>
<if we've made it down here, we've found ENOUGH prime numbers>
<print the current NumberToTest minus whatever we added two lines up to get the next number>

Answer 37

We don't want to try to divide by 1--nothing would seem to be prime. We don't have to start at two if we only use odd numbers, and if we start out testing an odd number and always add two to it, we'll never have to test an even number. So the divisor should probably start at 3.

Answer 38

ahh thats nice.

Answer 39

I'm going to head home now. I'll be on later, but see what you can do to fill in that outline of a function. Make sure you start out with NumberToTest being odd, and make sure you increment things consistently--that is, whenever you reset the divisor, make sure you reset it to the same number, and whenever you add something to NumberToTest to get the next one, make sure you add the same number.