10001st prime
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
Even though this problem is easy as it looks, it took me 1 day to find a better solution for this problem. I have started writing solution for this program yesterday. I have searched on internet a lot about prime numbers to find the perfect solution.
The first solution which gave me inspiration was this one: Jason Martin's answer to What are good ways to find nth prime number in the fastest way? - Quora
While answer by Jason Martin was pretty straight forward, I first thought that there would only be one prime in the range given by nln(n)+n(ln(ln(n))-1) < nth prime < nln(n)+nln(ln(n)). But what I later discovered was that my assumption was false and simply using the above formula was of no use.
As there was no formula for generating nth prime easily, my next option was to check if the given number is prime or not. And after I have written the function for Primality test, I will have to loop through each and every number till I have reached the 10001 prime.
This was the start for my solution(first program). I found this page on wiki which was really useful and I have learned a lot.- Primality test - Wikipedia, the free encyclopedia.
Accordingly there are few conditions for checking if the number is prime or not. They are as follows.(Let us assume the number is n)
Using the first condition have saved a lot of time.
Okay no more confusion, I will show you the first program I have written:
What is the 10 001st prime number?
Even though this problem is easy as it looks, it took me 1 day to find a better solution for this problem. I have started writing solution for this program yesterday. I have searched on internet a lot about prime numbers to find the perfect solution.
The first solution which gave me inspiration was this one: Jason Martin's answer to What are good ways to find nth prime number in the fastest way? - Quora
While answer by Jason Martin was pretty straight forward, I first thought that there would only be one prime in the range given by nln(n)+n(ln(ln(n))-1) < nth prime < nln(n)+nln(ln(n)). But what I later discovered was that my assumption was false and simply using the above formula was of no use.
As there was no formula for generating nth prime easily, my next option was to check if the given number is prime or not. And after I have written the function for Primality test, I will have to loop through each and every number till I have reached the 10001 prime.
This was the start for my solution(first program). I found this page on wiki which was really useful and I have learned a lot.- Primality test - Wikipedia, the free encyclopedia.
Accordingly there are few conditions for checking if the number is prime or not. They are as follows.(Let us assume the number is n)
- (n+1) or (n-1) should be evenly divisible by 6. i.e.
(n+1)%6 == 0 or (n-1)%6 == 0
- A prime number will be divisible by a number less than its square root.
Using the first condition have saved a lot of time.
Okay no more confusion, I will show you the first program I have written:
#http://radiusofcircle.blogspot.com #import time and math modules import time, math #Time at the start of execution start = time.time() #Function to check if the number is prime or not def isPrime(n): if (n+1)%6 == 0 or (n-1)%6 == 0: for i in range(2, int(round(math.sqrt(n)+1))): if n%i == 0: return False return True return False #Counter for counting the prime numbers #We took it equal to 2 because #we are starting with 4 and have already counted 2,3 counter = 2 #Numbers starting from 4 i = 4 #Assuming the nth-prime to be 0 prime = 0 #Starting a for loop while counter < 10001: if isPrime(i): counter = counter+1 prime = i i = i+1 i = i+1 #Printing the output print str(counter)+' prime is '+str(prime) #Time at the end of execution end = time.time() #Printing the time of execution print end-start
I think this program is pretty simple to understand, also I have added comments to this program so that it will be easy for everyone to understand.
You may be confused on why I have used
counter=2
and i=4
.
See we know that the prime numbers are 2,3,5,7...
when I have started iterating from 4,that means I have left 2 prime numbers already and so the counter which counts the prime number's rank is 2
Even though this program was efficient and had an execution time of almost 0.217seconds, I was not satisfied and wanted a faster program and so is the next program I will present it with you.
This program uses the concept of Sieve of Erotosthenes. And for writing the program you can see Sieve of Erotosthenes The python way and Finding Prime Numbers. The use of Sieve of Erotosthenes was suggested by Charles.
Program
Okay the program is as follows:
I am assuming that you have read about the sieve of Erotosthenes atleast from one of the links above and so I am not explaining the program.
But if you still have not understood the sieve concept then I would suggest you to take an example value(Don't take very large values) and start iterating manually, I am sure that you will understand the concept easily.If not comment below.
If you want to download the file you can download it from github gist
Output
Summary
Before solving this problem I didn't even have any idea about the tests with which we can find whether the given number is prime or not. After solving this problem now I know atleast few properties of prime numbers, primality tests etc. I am happy that I have written a program which has a good execution time using the sieve of Erotosthenes.
Eventhough I have written a program which had small execution time, there might be still better program with you which even has little execution time. I would be glad if you can share your code with me and this community.
I have not explained any code in this post because, to understand these programs you need to know some other concepts for which I have linked to. Even after reading if you are have any doubt or didn't understand anything then please do comment in the comment box below and I will be glad to help you.
Please don't hesitate to comment in the comment box if I have made any typo or have missed any description.
Thank you. Have a nice day☺️.