Machin-like formulas   are useful for efficiently computing numerical approximations for 



π


{\displaystyle \pi }





Task

Verify the following Machin-like formulas are correct by calculating the value of tan   (right hand side) for each equation using exact arithmetic and showing they equal 1:

 





π
4


=
arctan
⁡


1
2


+
arctan
⁡


1
3




{\displaystyle {\pi \over 4}=\arctan {1 \over 2}+\arctan {1 \over 3}}

 
 





π
4


=
2
arctan
⁡


1
3


+
arctan
⁡


1
7




{\displaystyle {\pi \over 4}=2\arctan {1 \over 3}+\arctan {1 \over 7}}


 





π
4


=
4
arctan
⁡


1
5


−
arctan
⁡


1
239




{\displaystyle {\pi \over 4}=4\arctan {1 \over 5}-\arctan {1 \over 239}}


 





π
4


=
5
arctan
⁡


1
7


+
2
arctan
⁡


3
79




{\displaystyle {\pi \over 4}=5\arctan {1 \over 7}+2\arctan {3 \over 79}}


 





π
4


=
5
arctan
⁡


29
278


+
7
arctan
⁡


3
79




{\displaystyle {\pi \over 4}=5\arctan {29 \over 278}+7\arctan {3 \over 79}}


 





π
4


=
arctan
⁡


1
2


+
arctan
⁡


1
5


+
arctan
⁡


1
8




{\displaystyle {\pi \over 4}=\arctan {1 \over 2}+\arctan {1 \over 5}+\arctan {1 \over 8}}

 
 





π
4


=
4
arctan
⁡


1
5


−
arctan
⁡


1
70


+
arctan
⁡


1
99




{\displaystyle {\pi \over 4}=4\arctan {1 \over 5}-\arctan {1 \over 70}+\arctan {1 \over 99}}

 
 





π
4


=
5
arctan
⁡


1
7


+
4
arctan
⁡


1
53


+
2
arctan
⁡


1
4443




{\displaystyle {\pi \over 4}=5\arctan {1 \over 7}+4\arctan {1 \over 53}+2\arctan {1 \over 4443}}


 





π
4


=
6
arctan
⁡


1
8


+
2
arctan
⁡


1
57


+
arctan
⁡


1
239




{\displaystyle {\pi \over 4}=6\arctan {1 \over 8}+2\arctan {1 \over 57}+\arctan {1 \over 239}}


 





π
4


=
8
arctan
⁡


1
10


−
arctan
⁡


1
239


−
4
arctan
⁡


1
515




{\displaystyle {\pi \over 4}=8\arctan {1 \over 10}-\arctan {1 \over 239}-4\arctan {1 \over 515}}


 





π
4


=
12
arctan
⁡


1
18


+
8
arctan
⁡


1
57


−
5
arctan
⁡


1
239




{\displaystyle {\pi \over 4}=12\arctan {1 \over 18}+8\arctan {1 \over 57}-5\arctan {1 \over 239}}


 





π
4


=
16
arctan
⁡


1
21


+
3
arctan
⁡


1
239


+
4
arctan
⁡


3
1042




{\displaystyle {\pi \over 4}=16\arctan {1 \over 21}+3\arctan {1 \over 239}+4\arctan {3 \over 1042}}


 





π
4


=
22
arctan
⁡


1
28


+
2
arctan
⁡


1
443


−
5
arctan
⁡


1
1393


−
10
arctan
⁡


1
11018




{\displaystyle {\pi \over 4}=22\arctan {1 \over 28}+2\arctan {1 \over 443}-5\arctan {1 \over 1393}-10\arctan {1 \over 11018}}


 





π
4


=
22
arctan
⁡


1
38


+
17
arctan
⁡


7
601


+
10
arctan
⁡


7
8149




{\displaystyle {\pi \over 4}=22\arctan {1 \over 38}+17\arctan {7 \over 601}+10\arctan {7 \over 8149}}


 





π
4


=
44
arctan
⁡


1
57


+
7
arctan
⁡


1
239


−
12
arctan
⁡


1
682


+
24
arctan
⁡


1
12943




{\displaystyle {\pi \over 4}=44\arctan {1 \over 57}+7\arctan {1 \over 239}-12\arctan {1 \over 682}+24\arctan {1 \over 12943}}


 





π
4


=
88
arctan
⁡


1
172


+
51
arctan
⁡


1
239


+
32
arctan
⁡


1
682


+
44
arctan
⁡


1
5357


+
68
arctan
⁡


1
12943




{\displaystyle {\pi \over 4}=88\arctan {1 \over 172}+51\arctan {1 \over 239}+32\arctan {1 \over 682}+44\arctan {1 \over 5357}+68\arctan {1 \over 12943}}


and confirm that the following formula is incorrect by showing   tan   (right hand side)   is not   1:

 





π
4


=
88
arctan
⁡


1
172


+
51
arctan
⁡


1
239


+
32
arctan
⁡


1
682


+
44
arctan
⁡


1
5357


+
68
arctan
⁡


1
12944




{\displaystyle {\pi \over 4}=88\arctan {1 \over 172}+51\arctan {1 \over 239}+32\arctan {1 \over 682}+44\arctan {1 \over 5357}+68\arctan {1 \over 12944}}


These identities are useful in calculating the values:

 



tan
⁡
(
a
+
b
)
=



tan
⁡
(
a
)
+
tan
⁡
(
b
)


1
−
tan
⁡
(
a
)
tan
⁡
(
b
)





{\displaystyle \tan(a+b)={\tan(a)+\tan(b) \over 1-\tan(a)\tan(b)}}


 



tan
⁡

(
arctan
⁡


a
b


)

=


a
b




{\displaystyle \tan \left(\arctan {a \over b}\right)={a \over b}}


 



tan
⁡
(
−
a
)
=
−
tan
⁡
(
a
)


{\displaystyle \tan(-a)=-\tan(a)}




You can store the equations in any convenient data structure, but for extra credit parse them from human-readable text input.

Note: to formally prove the formula correct, it would have to be shown that 






−
3
p
i

4




{\displaystyle {-3pi \over 4}}

 < right hand side < 






5
p
i

4




{\displaystyle {5pi \over 4}}

 due to 



tan
⁡
(
)


{\displaystyle \tan()}

 periodicity.



import re
from fractions import Fraction
from pprint import pprint as pp
 
 
equationtext = '''\
  pi/4 = arctan(1/2) + arctan(1/3) 
  pi/4 = 2*arctan(1/3) + arctan(1/7)
  pi/4 = 4*arctan(1/5) - arctan(1/239)
  pi/4 = 5*arctan(1/7) + 2*arctan(3/79)
  pi/4 = 5*arctan(29/278) + 7*arctan(3/79)
  pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/8) 
  pi/4 = 4*arctan(1/5) - arctan(1/70) + arctan(1/99) 
  pi/4 = 5*arctan(1/7) + 4*arctan(1/53) + 2*arctan(1/4443)
  pi/4 = 6*arctan(1/8) + 2*arctan(1/57) + arctan(1/239)
  pi/4 = 8*arctan(1/10) - arctan(1/239) - 4*arctan(1/515)
  pi/4 = 12*arctan(1/18) + 8*arctan(1/57) - 5*arctan(1/239)
  pi/4 = 16*arctan(1/21) + 3*arctan(1/239) + 4*arctan(3/1042)
  pi/4 = 22*arctan(1/28) + 2*arctan(1/443) - 5*arctan(1/1393) - 10*arctan(1/11018)
  pi/4 = 22*arctan(1/38) + 17*arctan(7/601) + 10*arctan(7/8149)
  pi/4 = 44*arctan(1/57) + 7*arctan(1/239) - 12*arctan(1/682) + 24*arctan(1/12943)
  pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12943)
  pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12944)
'''
 
def parse_eqn(equationtext=equationtext):
    eqn_re = re.compile(r"""(?mx)
    (?P<lhs> ^ \s* pi/4 \s* = \s*)?             # LHS of equation
    (?:                                         # RHS
        \s* (?P<sign> [+-])? \s* 
        (?: (?P<mult> \d+) \s* \*)? 
        \s* arctan\( (?P<numer> \d+) / (?P<denom> \d+)
    )""")
 
    found = eqn_re.findall(equationtext)
    machins, part = [], []
    for lhs, sign, mult, numer, denom in eqn_re.findall(equationtext):
        if lhs and part:
            machins.append(part)
            part = []
        part.append( ( (-1 if sign == '-' else 1) * ( int(mult) if mult else 1),
                       Fraction(int(numer), (int(denom) if denom else 1)) ) )
    machins.append(part)
    return machins
 
 
def tans(xs):
    xslen = len(xs)
    if xslen == 1:
        return tanEval(*xs[0])
    aa, bb = xs[:xslen//2], xs[xslen//2:]
    a, b = tans(aa), tans(bb)
    return (a + b) / (1 - a * b)
 
def tanEval(coef, f):
    if coef == 1:
        return f
    if coef < 0:
        return -tanEval(-coef, f)
    ca = coef // 2
    cb = coef - ca
    a, b = tanEval(ca, f), tanEval(cb, f)
    return (a + b) / (1 - a * b)
 
 
if __name__ == '__main__':
    machins = parse_eqn()
    #pp(machins, width=160)
    for machin, eqn in zip(machins, equationtext.split('\n')):
        ans = tans(machin)
        print('%5s: %s' % ( ('OK' if ans == 1 else 'ERROR'), eqn))