Hacktoberfest: Added first solution to Project Euler problem 58 (#3599)

* Added solution to problem 58

* Update sol1.py
Co-authored-by: NJohn Law <johnlaw.po@gmail.com>

project_euler/problem_058/__init__.py

+#

project_euler/problem_058/sol1.py

+"""
+Project Euler Problem 58:https://projecteuler.net/problem=58
+
+
+Starting with 1 and spiralling anticlockwise in the following way,
+a square spiral with side length 7 is formed.
+
+37 36 35 34 33 32 31
+38 17 16 15 14 13 30
+39 18  5  4  3 12 29
+40 19  6  1  2 11 28
+41 20  7  8  9 10 27
+42 21 22 23 24 25 26
+43 44 45 46 47 48 49
+
+It is interesting to note that the odd squares lie along the bottom right
+diagonal ,but what is more interesting is that 8 out of the 13 numbers
+lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
+
+If one complete new layer is wrapped around the spiral above,
+a square spiral with side length 9 will be formed.
+If this process is continued,
+what is the side length of the square spiral for which
+the ratio of primes along both diagonals first falls below 10%?
+
+Solution: We have to find an odd length side for which square falls below
+10%. With every layer we add 4 elements are being added to the diagonals
+,lets say we have a square spiral of odd length with side length j,
+then if we move from j to j+2, we are adding j*j+j+1,j*j+2*(j+1),j*j+3*(j+1)
+j*j+4*(j+1). Out of these 4 only the first three can become prime
+because last one reduces to (j+2)*(j+2).
+So we check individually each one of these before incrementing our
+count of current primes.
+
+"""
+
+
+def isprime(d: int) -> int:
+    """
+    returns whether the given digit is prime or not
+    >>> isprime(1)
+    0
+    >>> isprime(17)
+    1
+    >>> isprime(10000)
+    0
+    """
+    if d == 1:
+        return 0
+
+    i = 2
+    while i * i <= d:
+        if d % i == 0:
+            return 0
+        i = i + 1
+    return 1
+
+
+def solution(ratio: float = 0.1) -> int:
+    """
+    returns the side length of the square spiral of odd length greater
+    than 1 for which the ratio of primes along both diagonals
+    first falls below the given ratio.
+    >>> solution(.5)
+    11
+    >>> solution(.2)
+    309
+    >>> solution(.111)
+    11317
+    """
+
+    j = 3
+    primes = 3
+
+    while primes / (2 * j - 1) >= ratio:
+        for i in range(j * j + j + 1, (j + 2) * (j + 2), j + 1):
+            primes = primes + isprime(i)
+
+        j = j + 2
+    return j
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()