Add Python solution for problem 808

Python/p808.py

@@ -0,0 +1,47 @@
+# Both 169 and 961 are the square of a prime. 169 is the reverse of 961.
+#
+# We call a number a reversible prime square if:
+#
+# 1. It is not a palindrome, and
+# 2. It is the square of a prime, and
+# 3. Its reverse is also the square of a prime.
+#
+# 169 and 961 are not palindromes, so both are reversible prime squares.
+#
+# Find the sum of the first 50 reversible prime squares.
+
+from math import sqrt
+
+from projecteuler import is_palindrome, sieve, timing
+
+
+@timing
+def p808():
+    n = 1
+    primes = sieve(50000000)
+    reversible_prime_squares = []
+
+    for n, p in enumerate(primes):
+        if not p:
+            continue
+
+        n_squared = n * n
+
+        if not is_palindrome(n_squared):
+            n_reverse = int(str(n_squared)[::-1])
+            n_reverse_sqrt = sqrt(n_reverse)
+
+            if n_reverse_sqrt.is_integer() and primes[int(n_reverse_sqrt)]:
+                reversible_prime_squares.append(n_squared)
+
+        if len(reversible_prime_squares) == 50:
+            break
+
+    res = sum(reversible_prime_squares)
+
+    print('Project Euler, Problem 808')
+    print(f'Answer: {res}')
+
+
+if __name__ == '__main__':
+    p808()

Python/projecteuler.py

@@ -10,20 +10,19 @@ from numpy import zeros
 def is_prime(num):
     if num < 4:
         # If num is 2 or 3 then it's prime.
-        return num == 2 or num == 3
+        return num in (2, 3)
 
     # If num is divisible by 2 or 3 then it's not prime.
     if num % 2 == 0 or num % 3 == 0:
         return False
+
     # Any number can have only one prime factor greater than its
     # square root. If we reach the square root and we haven't found
     # any smaller prime factors, then the number is prime.
     limit = floor(sqrt(num)) + 1
 
-    # Every prime other than 2 and 3 is in the form 6k+1 or 6k-1.
-    # If I check all those value no prime factors of the number
-    # will be missed. If a factor is found, the number is not prime
-    # and the function returns 0.
+    # Every prime other than 2 and 3 is in the form 6k+1 or 6k-1. If I check all those values no prime factors of the number
+    # will be missed. If a factor is found, the number is not prime and the function returns False.
     for i in range(5, limit, 6):
         if num % i == 0 or num % (i + 2) == 0:
             return False
@@ -32,7 +31,7 @@ def is_prime(num):
     return True
 
 
-def is_palindrome(num, base):
+def is_palindrome(num, base=10):
     reverse = 0
 
     tmp = num
@@ -426,6 +425,7 @@ def partition_fn(n, partitions, mod=-1):
     # The partition function for zero is 1 by definition.
     if n == 0:
         partitions[n] = 1
+
         return 1
 
     # If the partition for the current n has already been calculated, return the value.

README.md

@@ -4,4 +4,4 @@ These are my solutions in C and Python, not necessarily the best solutions. I've
 # Notes
 - Solutions for problems 82, 86, 95, and 145 in Python are quite slow.
 - Solutions for problems 84, 89, 102, and 124 have been implemented in C but not in Python.
-- Solutions for problems 88, 90 and 91 have been implemented in Python but not in C.
+- Solutions for problems 88, 90, 91, and 808 have been implemented in Python but not in C.