mirror prime generator python

Question

I am currently enrolled in a Python programming course and last week we got a homework problem which was to Develop a program to generate all prime numbers less or equal to n whose mirror is also prime, I can't see where I am going wrong, please help!

import math

def mirror_prime(n):
    answer = True
    # Test 0 and 1
    if n==0 or n==1:
        answer = False
    # End if

    # Test even numbers
    if n != 2 and n%2==0:
        answer= False
    # End if

    # Test if there is a proper odd divisor
        for d in range (3, int(math.sqrt(n))+1, 2):
            if n%d==0:
                answer=False
            # End if
        # End for



    #Reverse n
    mirror_n = int(str(n)[::-1])
    mirror_answer = True

    # Test 0 and 1
    if mirror_n==0 or mirror_n==1:
        mirror_answer = False
    # End if


def mirror_prime_generator(n):
    for i in range(3, n+1):
        print (mirror_prime(i))

I am expecting to get a list of all prime numbers whose mirror is also prime less or equal to n

The result i get when i put mirror_prime_generator(n) into the shell, it justs prints none however many times n is, so if n is 23 it will print none 23 times

Answer 1

I think the shortest way to define the function you described is this:

def mirrorPrimes(limit):
    primes = [n for n in range(2,limit+1) if all(n%d!=0 for d in range(2,n))]
    return [n for n in primes if int(str(n)[::-1]) in primes]

The first function line puts in primes all prime numbers from 2 to limit and the second line simply returns elements in primes only if there is a mirror in primes.

Simple and clean!

Answer 2

This is my code in the end after editing, it now does what I want it to do, thanks everyone for the help

import math

def mirror_prime(n):
    answer = True
    # Test 0 and 1
    if n==0 or n==1:
        answer = False
    # End if

    # Test even numbers
    if n != 2 and n%2==0:
        answer= False
    # End if

    # Test if there is a proper odd divisor
    for d in range (3, int(math.sqrt(n))+1, 2):
        if n%d==0:
            answer=False
        # End if

    # End for





    #Reverse n
    mirror_n = int(str(n)[::-1])
    mirror_answer = True

    # Test 0 and 1
    if mirror_n==0 or mirror_n==1:
        mirror_answer = False
    # End if
    # Test even numbers
    if n != 2 and n%2==0:
        mirror_answer= False
    # End if

    # Test if there is a proper odd divisor
    for d in range (3, int(math.sqrt(mirror_n))+1, 2):
        if mirror_n%d==0:
            mirror_answer=False
        # End if

        # End for

    if answer and mirror_answer==True:
        return n, mirror_n



def mirror_prime_generator(n):
    for i in range(3, n+1):
        if mirror_prime(i):
            print(i)

Answer 3

There were still some bugs. The indentation of for loops was wrong (they were inside the if) and you sometimes used n instead of mirror_n.

Your code

Here's a working code with the least amount of change:

import math

def mirror_prime(n):
    answer = True
    # Test 0 and 1
    if n==0 or n==1:
        answer = False
    # End if

    # Test even numbers
    if n != 2 and n%2==0:
        answer= False
    # End if

    # Test if there is a proper odd divisor
    for d in range (3, int(math.sqrt(n))+1, 2):
        if n%d==0:
            answer=False
        # End if

    # End for

    #Reverse n
    mirror_n = int(str(n)[::-1])
    mirror_answer = True

    # Test 0 and 1
    if mirror_n==0 or mirror_n==1:
        mirror_answer = False
    # End if
    # Test even numbers
    if mirror_n != 2 and mirror_n%2==0:
        mirror_answer= False
    # End if

    # Test if there is a proper odd divisor
    for d in range (3, int(math.sqrt(mirror_n))+1, 2):
        if mirror_n%d==0:
            mirror_answer=False
        # End if

    # End for

    if answer and mirror_answer==True:
        return n, mirror_n



def mirror_prime_generator(n):
    for i in range(3, n+1):
        if mirror_prime(i):
            print(i)

mirror_prime_generator(100)
# 3
# 5
# 7
# 11
# 13
# 17
# 31
# 37
# 71
# 73
# 79
# 97

Shorter version

You should try to avoid using duplicate code. The test for n and mirror_n is exactly the same, so you could put it inside a function:

def is_prime(n):
    if n == 2:
        return True
    if n < 2 or n % 2 == 0:
        return False
    for d in range(3, int(n**0.5) + 1, 2):
        if n % d == 0:
            return False
    return True


def is_mirror_prime(n):
    mirror_n = int(str(n)[::-1])
    return mirror_n != n and is_prime(n) and is_prime(mirror_n)

print([n for n in range(1000) if is_mirror_prime(n)])
# [13, 17, 31, 37, 71, 73, 79, 97, 107, 113, 149, 157, 167, 179, 199, 311, 337, 347, 359, 389, 701, 709, 733, 739, 743, 751, 761, 769, 907, 937, 941, 953, 967, 971, 983, 991]

Answer 4

You have to consider what you want your functions to do. As far as I can tell, the function mirror_prime(n) answers the question "are both n and its mirror prime?", which is a binary question (either true or false). If that is the case, which is entirely reasonable, restructure your mirror_prime_generator loop like this:

def mirror_prime_generator(n):
    for i in range(3, n+1):
        if mirror_prime(i):
            print(i)

This is, of course, assuming that mirror_prime actually returns a correct result, but that seems like your whole task so I won't give you that.