[PYTHON] Project Euler 37

problem

3797 has an interesting property. First, it is a prime number, and when the digits are removed from left to right, it is all prime numbers (3797, 797, 97, 7). Similarly, the digits from right to left. All are prime numbers except for (3797, 379, 37, 3).

There are only 11 prime numbers that can be truncated from the right or from the left. Find the sum.

Note: We do not consider 2, 3, 5, 7 to be truncated prime numbers. http://odz.sakura.ne.jp/projecteuler/index.php?cmd=read&page=Problem%2037

Mathematical consideration

Such numbers have a first digit of 3 or 7. If the first digit is 1,4,6,8,9, the number itself is not a prime number, and if the first digit of two or more digits is 2,5, it is a multiple of 2 or a multiple of 5. Also, you don't need 2,4,5,6,8 in the middle digits. This is because if 2,4,5,6,8 is entered in the middle when rounded down from the right, the number is no longer a prime number. In addition, there is no number with 3 or more digits that has a maximum digit of 2.5. From the above consideration, it is 1,3,7,9 except for that digit, but if the most significant digit is 2 or 5 and there is one or more 1, 7 in the middle, it will always be a multiple of 3 in the middle. Also, if the most significant digit is 2 or 5 and the rest is 3 or 9, it will be a multiple of 3 when rounded down from the left. For two-digit numbers, 23 and 53 are prime numbers that satisfy the condition. For numbers other than these, it can be seen that numbers with each digit consisting of 1,3,5,7 should be considered.

Answer

For n digits, the number that is a prime number among the combined numbers of 1, 3, 5, and 7 from the right is held as a set type object nr.
For n digits, the number that is a prime number among the combined numbers of 1, 3, 5, and 7 from the right is held as a set type object nl.
Add the intersection of nr and nl to the answer set s.
Next, for n + 1 digits, attach any of 1,3,7,9 to the prime number truncated from the right or left from the appropriate direction to determine whether it is a prime number or not. By doing so, the two prime number sets in steps 1 and 2 above are created.

It ends when the number of answer columns reaches 11.

def conjoin(f, iter1, iter2):
  return set(f(e1,e2) for e1 in iter1 for e2 in iter2)

def connect_num(e1,e2):
  return str(e1)+str(e2)

def is_prime(num,pri):
  num = int(num)
  if num < len(pri['bool']):
    return pri['bool'][num]

  M = (num**0.5)+1
  #print num
  for p in pri['list']:
    if p > M:
      return True
    if (num % p) == 0:
      return False

  p = pri['list'][-1]+2
  while p<M:
    if (num % p) == 0:
      return False
    p += 2
  return True
    

import mymath

def get_pri(iter1, pri):
  ret = []
  for n in iter1:
    if is_prime(n,pri):
      ret.append(n)
  return set(ret)

def main():
  MAX = 10**7
  pri = mymath.get_primes(MAX)
  nr = [3,7]
  nl = [3,7]
  n2 = [1,3,7,9]
  s = set(['23', '53'])
  MAX_LEN = 11
  while len(s)<MAX_LEN:
    if len(nr) == 0:
      break
    nr = set(get_pri(conjoin(connect_num,nr,n2),pri))
    nl = set(get_pri(conjoin(connect_num,n2,nl),pri))
    s = s | (nr & nl)
  ans = 0
  for n in s:
    ans += int(n)
  print s
  print ans
main()

I wrote the code for 23,53 as well. In this case, the number to be combined includes not only 1,3,7,9 but also 2,5. Also, the set nr (N is the number of repetitions + 1) of N-digit numbers of prime numbers joined from the right is empty, or the set nr (N is) of N-digit numbers of prime numbers joined from the left. It ends when the number of repetitions + 1) becomes empty. This code also confirmed that the number above was only 11.

import copy
import mymath
def main():
  MAX = 10**7
  pri = mymath.get_primes(MAX)
  nr = set(get_pri(range(1,10),pri))
  nl = copy.deepcopy(nr)
  ns = [1,2,3,5,7,9]
  s = set([])
  while len(nr)>0 and len(ns)>0:
    nr = set(get_pri(conjoin(connect_num,nr,ns),pri))
    nl = set(get_pri(conjoin(connect_num,ns,nl),pri))
    s = s | (nr & nl)
  ans = 0
  for n in s:
    ans += int(n)
  print s
  print ans