Challenge:
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?
Programming Language:
LISP
Solution:
(defun isprime-p(n)
(defparameter isprime NIL)
(if (and (oddp n) (> n 2))
(loop for x from 2 below n
do (if (eql 0 (mod n x))
(progn (setq isprime NIL)
(return NIL))
(setq isprime T))))
(if (eq n 2)
(setq isprime T))
(print isprime))
(defun nth-prime(num)
(defparameter primes NIL)
(defparameter p 2)
(loop while (< (list-length primes) num)
do (if (eql T (isprime-p p))
(push p primes))
(setq p (+ 1 p)))
(print (car primes)))Overview:
Coming soon…
Programming Language:
x86 Assembly
Solution:
; receives input (n), prints all primes greater than 1 up to n.
format PE console
entry start
include 'win32a.inc'
; ============================================
section '.text' code readable executable
start: ; the program begins here:
call read_hex
mov ebx, eax ; setup counter in ebx
prime_loop:
cmp ebx, 1 ; check that counter is
jbe exit ; below or equal to 1. if so, exit.
; otherwise, continue.
mov eax, ebx ; move counter's value into eax
; (input of 'start_check_prime' loop.
start_check_prime:
mov esi, eax ; store input in esi for later.
cmp eax, 2 ; if input is 2 (special case prime)
je prime ; num is prime. jump to 'prime' label.
cmp eax, 3 ; if input is 3
je prime ; also jump to 'prime' label.
mov edx, 0 ; initialize edx to 0 for div with 32-bit integers [EDX:EAX].
mov ecx, 2 ; setup divsor (2)
div ecx ; check if input is even and
; get half its value for prime check.
cmp edx, 0 ; if edx is 0, remainder of 'div 2' = 0
je n_prime ; input is even.
; Otherwise input is odd. Check for prime.
mov ecx, eax ; setup ecx (counter) with half of input
; and check divisbility by nums up to ecx.
check_prime: ; check for prime loop
mov eax, esi ; restore eax with original input for check_prime.
mov edx, 0 ; initialize edx to 0 for div with 32-bit integers [EDX:EAX].
div ecx ; divide by current ecx (counter) value
cmp edx, 0 ; if input num is evenly divisible by ecx (counter)
je n_prime ; then not prime.
dec ecx ; decrease counter by 1
cmp ecx, 1 ; but make sure counter is above 1 (because all nums evenly
ja check_prime ; divisible by 1). if counter is above 1, check next ecx num.
; once all ecx nums are checked, and no jumps have been made
; from 'check_prime' loop, then number is prime.
; continue to following line.
prime: ; number is prime.
mov eax, esi
call print_eax
dec ebx
jmp prime_loop
n_prime: ; number is not prime.
dec ebx
jmp prime_loop
exit: ; exit the process:
push 0
call [ExitProcess]
include 'training.inc'
Overview:
Coming soon…