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

PythonServer Side ProgrammingProgramming

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