The question is: print out all numbers from 1 to 1000 that satisfy the two conditions:

1. Those are prime numbers
2. The numbers after being reversed are also prime numbers. e.g., 13 satisfies (as 13 and 31 are prime numbers), but 19 does not satisfy (19 is a prime number, while 91 is not). My codes:

def prime(n):
  if n<2:
    return False
  for i in range(1, n):
    if n%i == 0:
      return False
  else:
    return True

def reverse(n):
  List = []
  while n>0:
    List.append(n)
    n = n//10

  string = [str(integer) for integer in List]
  a_string = "".join(string)
  result = int(a_string)
  print(result)

L = []
for i in range (1, 1000):
  if prime(i) == prime(reverse(i)) == True:
    L.append(i)
print(L)

Mine seems to contain some errors as the outcome is not as expected, either it shows none or still shows 19 in the list. Thank you in advance!

CodePudding user response：

First your prime method is wrong, because the loops starts at 1, and every numbre satisfies n%1 == 0 , it needs to starts at 2

def prime(n):
    if n < 2:
        return False
    for i in range(2, n):
        if n % i == 0:
            return False
    return True

Then your reverse method returns nothing so reverse(5) gives None, you have tried it manually.

def reverse(n):
    values = []
    while n > 0:
        values.append(n % 10)
        n = n // 10
    return int("".join(map(str, values)))

Then simplify the condition to be

for i in range(1, 20):
    if prime(i) and prime(reverse(i)):
        L.append(i)

The reverse process can be done with string also, and so short that it can be inlined

L = []
for i in range(1, 20):
    if prime(i) and prime(int(str(i)[::-1])):
        L.append(i)

CodePudding user response：

It would seem running prime twice is very inefficient. I would recommend the following:

def prime(n):
  if n<2:
    return False
  for i in range(2, n):
    if n%i == 0:
      return False
  return True

l = [i for i in range(1, 1001) if prime(i)]
print(*(x for x in l if int(str(x)[::-1]) in l))