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 std.stdio, std.regex, std.conv, std.string, std.range,
       arithmetic_rational;
 
struct Pair { int x; Rational r; }
 
Pair[][] parseEquations(in string text) /*pure nothrow*/ {
    auto r = regex(r"\s*(?P<sign>[+-])?\s*(?:(?P<mul>\d+)\s*\*)?\s*" ~
                   r"arctan\((?P<num>\d+)/(?P<denom>\d+)\)");
    Pair[][] machins;
    foreach (const line; text.splitLines) {
        Pair[] formula;
        foreach (part; line.split("=")[1].matchAll(r)) {
            immutable mul = part["mul"],
                      num = part["num"],
                      denom = part["denom"];
            formula ~= Pair((part["sign"] == "-" ? -1 : 1) *
                            (mul.empty ? 1 : mul.to!int),
                            Rational(num.to!int,
                                     denom.empty ? 1 : denom.to!int));
        }
        machins ~= formula;
    }
    return machins;
}
 
 
Rational tans(in Pair[] xs) pure nothrow {
    static Rational tanEval(in int coef, in Rational f)
    pure nothrow {
        if (coef == 1)
            return f;
        if (coef < 0)
            return -tanEval(-coef, f);
        immutable a = tanEval(coef / 2, f),
                  b = tanEval(coef - coef / 2, f);
        return (a + b) / (1 - a * b);
    }
 
    if (xs.length == 1)
        return tanEval(xs[0].tupleof);
    immutable a = xs[0 .. $ / 2].tans,
              b = xs[$ / 2 .. $].tans;
    return (a + b) / (1 - a * b);
}
 
void main() {
    immutable 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)";
 
    const machins = equationText.parseEquations;
    foreach (const machin, const eqn; machins.zip(equationText.splitLines)) {
        immutable ans = machin.tans;
        writefln("%5s: %s", ans == 1 ? "OK" : "ERROR", eqn);
    }
}