Hi Dev,
I am currently on this Project Euler Challenge journey. And it is my 7^th day! Yay~💪🏻
Today's problem 7 is particularly easy until the test case; to search for the 10001^th prime number.
I will share with you my thought process and two things I learnt today about prime numbers.
The problem
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
nth prime number?
Given num a prime number?
We need to find out any given number is a prime number; we build this function similar to the one used for solution 3.
function is_prime_num(num) {
if (num % 2 === 0 || num % 3 === 0){
return true;
}
var i = 5;
while (i <= Math.ceil(Math.sqrt(num))) {
if (num % i === 0)
return false;
if (num % (i + 2) === 0)
return false;
i+= 6;
}
return true;
}
Today I learnt that #1
prime numbers always satisfy the
6n+1or6n-1conditions, but that doesn't mean that all numbers of that meet the conditions are prime numbers [source]
Lets visualise this with examples:
// 6n – 1: → 5, 11, 17, 23, 29, 35, 41, ...
// 6n + 1: → 7, 13, 19, 25, 31, 37, 43, ...
This means that we can save a lot of computation, by extracting the numbers with this formula until we get the number of prime numbers we need.
function generate_prime_numbers(n){
var list_prime_numbers = [2, 3];
for(var i=1;i<n;i++){
current_prime = 6*i - 1;
if(is_prime_num(current_prime)){
list_prime_numbers.push(current_prime);
}
current_prime = 6*i + 1;
if(is_prime_num(current_prime)){
list_prime_numbers.push(current_prime);
}
}
}
Unfortunately, it did not return the correct result for 1000 and 10001. The results are as follows:
Test value: 6
Output: 13
Took 0.22499999613501132 ms
Test value: 10
Output: 29
Took 0.04499999340623617 ms
Test value: 100
Output: 541
Took 0.2049999893642962 ms
Test value: 1000
Output: 5987
Took 4.014999984065071 ms
Test value: 10001
Output: 59999
Took 32.11999998893589 ms
The expected output for 1000 and 10001:
nthPrime(1000) should return 7919.
nthPrime(10001) should return 104743.
Today I learnt that #2
There is an upper bound to n^th prime number, and we can calculate the upper bound with this formula: n * (Math.log(n) + Math.log(Math.log(n))) [source]
function upper_bound_for_n_prime(n){
if(n<6){
return 13;
}
return n * (Math.log(n) + Math.log(Math.log(n)))
}
With this knowledge, we can safely use while loop to generate prime numbers and stop once we reached this upper bound number.
function generate_prime_numbers(n){
var list_prime_numbers = [2, 3];
var upper_bound = upper_bound_for_n_prime(n);
var current_prime = 3;
var i = 0;
while(current_prime < upper_bound){
i+=1;
current_prime = 6*i - 1;
if(is_prime_num(current_prime)){
list_prime_numbers.push(current_prime);
}
current_prime = 6*i + 1;
if(is_prime_num(current_prime)){
list_prime_numbers.push(current_prime);
}
}
return list_prime_numbers.slice(0, n);
}
All together now
// list of numbers we wanna test
var test_values = [6, 10, 100, 1000, 10001];
// this function execute the code and records the time to execute
function run_function(func, test_values) {
for(var i in test_values){
console.log('Test value:', test_values[i]);
var t0 = performance.now();
console.log('Output:', func(test_values[i]));
var t1 = performance.now();
console.log("Took " + (t1 - t0) + " ms");
console.log();
}
}
function nthPrime(n) {
var prime_numbers = generate_prime_numbers(n);
return prime_numbers[prime_numbers.length-1];
}
// today i learnt that, the upper bound for nth prime number cannot be larger than
function upper_bound_for_n_prime(n){
if(n<6){
return 13;
}
return n * (Math.log(n) + Math.log(Math.log(n)))
}
function is_prime_num(num) {
if (num % 2 === 0 || num % 3 === 0){
return true;
}
var i = 5;
while (i <= Math.ceil(Math.sqrt(num))) {
if (num % i === 0)
return false;
if (num % (i + 2) === 0)
return false;
i+= 6;
}
return true;
}
// lets generate the prime numbers give that all prime numbers > 3 are fulfils `6n – 1` or `6n + 1`:
// 6n – 1: → 5, 11, 17, 23, 29, 35, 41, ...
// 6n + 1: → 7, 13, 19, 25, 31, 37, 43, ...
function generate_prime_numbers(n){
var list_prime_numbers = [2, 3];
var upper_bound = upper_bound_for_n_prime(n);
var current_prime = 3;
var i = 0;
while(current_prime < upper_bound){
i+=1;
current_prime = 6*i - 1;
if(is_prime_num(current_prime)){
list_prime_numbers.push(current_prime);
}
current_prime = 6*i + 1;
if(is_prime_num(current_prime)){
list_prime_numbers.push(current_prime);
}
}
return list_prime_numbers.slice(0, n);
}
run_function(nthPrime, test_values);
Output:
Test value: 6
Output: 13
Took 0.2099999983329326 ms
Test value: 10
Output: 29
Took 0.040000013541430235 ms
Test value: 100
Output: 541
Took 0.2500000118743628 ms
Test value: 1000
Output: 7919
Took 5.414999992353842 ms
Test value: 10001
Output: 104743
Took 73.96999999764375 ms
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