# Program to find common fraction between min and max using given constraint in Python

Suppose we have two long integer values maximum and minimum. We have to find a common fraction n/d such that min <= d <= max. And |n/d - pi| is smallest. Here pi = 3.14159265... and if there are more than one fractions holding this condition, then return the fraction with smallest denominator.

So, if the input is like minimum = 1 maximum = 10, then the output will be 22/7.

To solve this, we will follow these steps −

• P := a fraction (5706674932067741 / 1816491048114374) - 3
• a := 0, b := 1, c := 1, d := 1
• farey := an array of pairs, it has two pairs initially (a, b) and (c, d)
• Loop through the following unconditionally -
• f := b + d
• if f > maximum - minimum, then
• come out from the loop
• e := a + c
• insert pair (e, f) at the end of farey
• if P < the value of (e/f), then
• c := e and d := f
• therwise,
• a := e and b := f
• p_min := floor of (P * minimum)
• while minimum <= maximum, do
• c := 0, d := 0
• for each pair (a, b) in farey, do
• if minimum + b > maximum, then
• come out from the loop
• if |(p_min + a)/ (minimum + b) - P| <|p_min / minimum - P|, then
• c := a, d := b
• come out from the loop
• if d is same as 0, then
• come out from the loop
• p_min := p_min + c
• o minimum := minimum + d
• o return fraction (p_min + 3 * minimum) / minimum

## Example

Let us see the following implementation to get better understanding −

from fractions import Fraction

def solve(minimum, maximum):
P = Fraction(5706674932067741, 1816491048114374) - 3

a, b, c, d = 0, 1, 1, 1
farey = [(a,b),(c,d)]

while True:
f = b + d
if f > maximum - minimum:
break

e = a + c
farey.append((e, f))
if P < Fraction(e, f):
c, d = e, f
else:
a, b = e, f

p_min = int(P * minimum)

while minimum <= maximum:
c, d = 0, 0
for a, b in farey:
if minimum + b > maximum:
break
if abs(Fraction(p_min + a, minimum + b).real - P) < abs(Fraction(p_min, minimum).real - P):
c, d = a, b
break
if d == 0:
break
p_min += c
minimum += d
return ("{}/{}".format(p_min + 3 * minimum, minimum))

minimum = 1
maximum = 10
print(solve(minimum, maximum))

## Input

4, 27


## Output

22/7