sol1 Algorithm
The sol1 Algorithm, also known as "Squaring the Circle" algorithm, is a mathematical technique that aims to solve the ancient geometrical problem of constructing a square with the same area as a given circle using only compass and straightedge. This algorithm is based on the approximation of the value of Pi (π), which is the ratio of the circumference of a circle to its diameter. The main idea behind the sol1 Algorithm is to find the side length of a square that, when multiplied by itself, gives the same area as that of a circle with a given radius.
The sol1 Algorithm begins by drawing a circle with the desired radius, followed by constructing an inscribed square within the circle. The next step involves dividing the circle's circumference into a number of equal segments, which are then used to create a polygon that approximates the circle. The area of this polygon can be easily calculated using basic trigonometry, and as the number of segments increases, the approximation of the circle's area becomes more accurate. Finally, the side length of the square is determined by finding the square root of the approximated circle's area, and a square with this side length is constructed using a compass and straightedge. Although the sol1 Algorithm provides an approximation to the problem of squaring the circle, it has been proven mathematically impossible to achieve an exact solution using only compass and straightedge due to the transcendental nature of the number π.
"""
Problem:
The fraction 49/98 is a curious fraction, as an inexperienced
mathematician in attempting to simplify it may incorrectly believe
that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.
We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
There are exactly four non-trivial examples of this type of fraction,
less than one in value, and containing two digits in the numerator
and denominator.
If the product of these four fractions is given in its lowest common
terms, find the value of the denominator.
"""
def isDigitCancelling(num, den):
if num != den:
if num % 10 == den // 10:
if (num // 10) / (den % 10) == num / den:
return True
def solve(digit_len: int) -> str:
"""
>>> solve(2)
'16/64 , 19/95 , 26/65 , 49/98'
>>> solve(3)
'16/64 , 19/95 , 26/65 , 49/98'
>>> solve(4)
'16/64 , 19/95 , 26/65 , 49/98'
>>> solve(0)
''
>>> solve(5)
'16/64 , 19/95 , 26/65 , 49/98'
"""
solutions = []
den = 11
last_digit = int("1" + "0" * digit_len)
for num in range(den, last_digit):
while den <= 99:
if (num != den) and (num % 10 == den // 10) and (den % 10 != 0):
if isDigitCancelling(num, den):
solutions.append(f"{num}/{den}")
den += 1
num += 1
den = 10
solutions = " , ".join(solutions)
return solutions
if __name__ == "__main__":
print(solve(2))
