ABCF->ab-angle b

Percentage Accurate: 19.5% → 47.7%

Time: 28.8s

Alternatives: 8

Speedup: 5.0×

Specification

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ \frac{-\sqrt{\left(2 \cdot \left(t_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t_0} \end{array} \end{array} \]

FPCore

(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (pow B 2.0) (* (* 4.0 A) C))))
   (/
    (-
     (sqrt
      (*
       (* 2.0 (* t_0 F))
       (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))))
    t_0)))

C

double code(double A, double B, double C, double F) {
	double t_0 = pow(B, 2.0) - ((4.0 * A) * C);
	return -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((pow((A - C), 2.0) + pow(B, 2.0)))))) / t_0;
}

Fortran

real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    t_0 = (b ** 2.0d0) - ((4.0d0 * a) * c)
    code = -sqrt(((2.0d0 * (t_0 * f)) * ((a + c) - sqrt((((a - c) ** 2.0d0) + (b ** 2.0d0)))))) / t_0
end function

Java

public static double code(double A, double B, double C, double F) {
	double t_0 = Math.pow(B, 2.0) - ((4.0 * A) * C);
	return -Math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0)))))) / t_0;
}

Python

def code(A, B, C, F):
	t_0 = math.pow(B, 2.0) - ((4.0 * A) * C)
	return -math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0)))))) / t_0

Julia

function code(A, B, C, F)
	t_0 = Float64((B ^ 2.0) - Float64(Float64(4.0 * A) * C))
	return Float64(Float64(-sqrt(Float64(Float64(2.0 * Float64(t_0 * F)) * Float64(Float64(A + C) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0))))))) / t_0)
end

MATLAB

function tmp = code(A, B, C, F)
	t_0 = (B ^ 2.0) - ((4.0 * A) * C);
	tmp = -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / t_0;
end

Wolfram

code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[Power[B, 2.0], $MachinePrecision] - N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]), $MachinePrecision]}, N[((-N[Sqrt[N[(N[(2.0 * N[(t$95$0 * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[B, 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]

Initial Program: 19.5% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ \frac{-\sqrt{\left(2 \cdot \left(t_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t_0} \end{array} \end{array} \]

FPCore

(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (pow B 2.0) (* (* 4.0 A) C))))
   (/
    (-
     (sqrt
      (*
       (* 2.0 (* t_0 F))
       (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))))
    t_0)))

C

double code(double A, double B, double C, double F) {
	double t_0 = pow(B, 2.0) - ((4.0 * A) * C);
	return -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((pow((A - C), 2.0) + pow(B, 2.0)))))) / t_0;
}

Fortran

real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    t_0 = (b ** 2.0d0) - ((4.0d0 * a) * c)
    code = -sqrt(((2.0d0 * (t_0 * f)) * ((a + c) - sqrt((((a - c) ** 2.0d0) + (b ** 2.0d0)))))) / t_0
end function

Java

public static double code(double A, double B, double C, double F) {
	double t_0 = Math.pow(B, 2.0) - ((4.0 * A) * C);
	return -Math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0)))))) / t_0;
}

Python

def code(A, B, C, F):
	t_0 = math.pow(B, 2.0) - ((4.0 * A) * C)
	return -math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0)))))) / t_0

Julia

function code(A, B, C, F)
	t_0 = Float64((B ^ 2.0) - Float64(Float64(4.0 * A) * C))
	return Float64(Float64(-sqrt(Float64(Float64(2.0 * Float64(t_0 * F)) * Float64(Float64(A + C) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0))))))) / t_0)
end

MATLAB

function tmp = code(A, B, C, F)
	t_0 = (B ^ 2.0) - ((4.0 * A) * C);
	tmp = -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / t_0;
end

Wolfram

code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[Power[B, 2.0], $MachinePrecision] - N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]), $MachinePrecision]}, N[((-N[Sqrt[N[(N[(2.0 * N[(t$95$0 * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[B, 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]

Alternative 1: 47.7% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 3.6 \cdot 10^{+27}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(t_0 \cdot F\right) \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* A C)))))
   (if (<= B 3.6e+27)
     (/ (- (sqrt (* 2.0 (* (* t_0 F) (+ A A))))) t_0)
     (- (/ (sqrt (* 2.0 (* F (- A (hypot B A))))) B)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 3.6e+27) {
		tmp = -sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
	} else {
		tmp = -(sqrt((2.0 * (F * (A - hypot(B, A))))) / B);
	}
	return tmp;
}

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 3.6e+27) {
		tmp = -Math.sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
	} else {
		tmp = -(Math.sqrt((2.0 * (F * (A - Math.hypot(B, A))))) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	tmp = 0
	if B <= 3.6e+27:
		tmp = -math.sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0
	else:
		tmp = -(math.sqrt((2.0 * (F * (A - math.hypot(B, A))))) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 3.6e+27)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(t_0 * F) * Float64(A + A))))) / t_0);
	else
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(F * Float64(A - hypot(B, A))))) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (A * C));
	tmp = 0.0;
	if (B <= 3.6e+27)
		tmp = -sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
	else
		tmp = -(sqrt((2.0 * (F * (A - hypot(B, A))))) / B);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 3.6e+27], N[((-N[Sqrt[N[(2.0 * N[(N[(t$95$0 * F), $MachinePrecision] * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], (-N[(N[Sqrt[N[(2.0 * N[(F * N[(A - N[Sqrt[B ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision])]]

Derivation

1. Split input into 2 regimes

2. if B < 3.59999999999999983e27

   1. Initial program 19.9%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

   3. Simplified 19.9%

      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

   4. Taylor expanded in C around inf 16.7%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
   5. Step-by-step derivation

      1. cancel-sign-sub-inv 16.7%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(--1\right) \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

      2. metadata-eval 16.7%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{1} \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

      3. *-lft-identity 16.7%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{A}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   6. Simplified 16.7%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   if 3.59999999999999983e27 < B

   1. Initial program 13.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

   3. Simplified 13.0%

      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

   4. Taylor expanded in C around 0 27.8%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
   5. Step-by-step derivation

      1. mul-1-neg 27.8%

         \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]

      2. associate-*l/ 27.8%

         \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]

      3. distribute-neg-frac 27.8%

         \[\leadsto \color{blue}{\frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]

      4. unpow2 27.8%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)}}{B} \]

      5. unpow2 27.8%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)}}{B} \]

      6. hypot-def 57.2%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)}}{B} \]

   6. Simplified 57.2%

      \[\leadsto \color{blue}{\frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}}{B}} \]
   7. Step-by-step derivation

      1. div-inv 57.2%

         \[\leadsto \color{blue}{\left(-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}\right) \cdot \frac{1}{B}} \]

      2. sqrt-unprod 57.3%

         \[\leadsto \left(-\color{blue}{\sqrt{2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}\right) \cdot \frac{1}{B} \]

      3. *-commutative 57.3%

         \[\leadsto \left(-\sqrt{2 \cdot \color{blue}{\left(\left(A - \mathsf{hypot}\left(A, B\right)\right) \cdot F\right)}}\right) \cdot \frac{1}{B} \]

   8. Applied egg-rr 57.3%

      \[\leadsto \color{blue}{\left(-\sqrt{2 \cdot \left(\left(A - \mathsf{hypot}\left(A, B\right)\right) \cdot F\right)}\right) \cdot \frac{1}{B}} \]
   9. Step-by-step derivation

      1. associate-*r/ 57.3%

         \[\leadsto \color{blue}{\frac{\left(-\sqrt{2 \cdot \left(\left(A - \mathsf{hypot}\left(A, B\right)\right) \cdot F\right)}\right) \cdot 1}{B}} \]

   10. Simplified 57.3%

       \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}{B}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 25.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 3.6 \cdot 10^{+27}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}{B}\\ \end{array} \]

Alternative 2: 44.6% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 8.8 \cdot 10^{+36}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(t_0 \cdot F\right) \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot \left(A - B\right)} \cdot \left(-\sqrt{2}\right)}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* A C)))))
   (if (<= B 8.8e+36)
     (/ (- (sqrt (* 2.0 (* (* t_0 F) (+ A A))))) t_0)
     (/ (* (sqrt (* F (- A B))) (- (sqrt 2.0))) B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 8.8e+36) {
		tmp = -sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
	} else {
		tmp = (sqrt((F * (A - B))) * -sqrt(2.0)) / B;
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (a * c))
    if (b <= 8.8d+36) then
        tmp = -sqrt((2.0d0 * ((t_0 * f) * (a + a)))) / t_0
    else
        tmp = (sqrt((f * (a - b))) * -sqrt(2.0d0)) / b
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 8.8e+36) {
		tmp = -Math.sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
	} else {
		tmp = (Math.sqrt((F * (A - B))) * -Math.sqrt(2.0)) / B;
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	tmp = 0
	if B <= 8.8e+36:
		tmp = -math.sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0
	else:
		tmp = (math.sqrt((F * (A - B))) * -math.sqrt(2.0)) / B
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 8.8e+36)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(t_0 * F) * Float64(A + A))))) / t_0);
	else
		tmp = Float64(Float64(sqrt(Float64(F * Float64(A - B))) * Float64(-sqrt(2.0))) / B);
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (A * C));
	tmp = 0.0;
	if (B <= 8.8e+36)
		tmp = -sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
	else
		tmp = (sqrt((F * (A - B))) * -sqrt(2.0)) / B;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 8.8e+36], N[((-N[Sqrt[N[(2.0 * N[(N[(t$95$0 * F), $MachinePrecision] * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[N[(F * N[(A - B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision] / B), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 8.80000000000000002e36

   1. Initial program 20.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

   3. Simplified 20.2%

      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

   4. Taylor expanded in C around inf 16.6%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
   5. Step-by-step derivation

      1. cancel-sign-sub-inv 16.6%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(--1\right) \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

      2. metadata-eval 16.6%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{1} \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

      3. *-lft-identity 16.6%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{A}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   6. Simplified 16.6%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   if 8.80000000000000002e36 < B

   1. Initial program 11.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

   3. Simplified 11.6%

      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

   4. Taylor expanded in C around 0 28.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
   5. Step-by-step derivation

      1. mul-1-neg 28.6%

         \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]

      2. associate-*l/ 28.6%

         \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]

      3. distribute-neg-frac 28.6%

         \[\leadsto \color{blue}{\frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]

      4. unpow2 28.6%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)}}{B} \]

      5. unpow2 28.6%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)}}{B} \]

      6. hypot-def 59.1%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)}}{B} \]

   6. Simplified 59.1%

      \[\leadsto \color{blue}{\frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}}{B}} \]

   7. Taylor expanded in A around 0 51.7%

      \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \color{blue}{\left(A + -1 \cdot B\right)}}}{B} \]
   8. Step-by-step derivation

      1. mul-1-neg 51.7%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A + \color{blue}{\left(-B\right)}\right)}}{B} \]

      2. unsub-neg 51.7%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \color{blue}{\left(A - B\right)}}}{B} \]

   9. Simplified 51.7%

      \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \color{blue}{\left(A - B\right)}}}{B} \]
3. Recombined 2 regimes into one program.

4. Final simplification 24.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 8.8 \cdot 10^{+36}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot \left(A - B\right)} \cdot \left(-\sqrt{2}\right)}{B}\\ \end{array} \]

Alternative 3: 44.2% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 7 \cdot 10^{+36}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(t_0 \cdot F\right) \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{B \cdot \left(-F\right)} \cdot \left(-\sqrt{2}\right)}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* A C)))))
   (if (<= B 7e+36)
     (/ (- (sqrt (* 2.0 (* (* t_0 F) (+ A A))))) t_0)
     (/ (* (sqrt (* B (- F))) (- (sqrt 2.0))) B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 7e+36) {
		tmp = -sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
	} else {
		tmp = (sqrt((B * -F)) * -sqrt(2.0)) / B;
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (a * c))
    if (b <= 7d+36) then
        tmp = -sqrt((2.0d0 * ((t_0 * f) * (a + a)))) / t_0
    else
        tmp = (sqrt((b * -f)) * -sqrt(2.0d0)) / b
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 7e+36) {
		tmp = -Math.sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
	} else {
		tmp = (Math.sqrt((B * -F)) * -Math.sqrt(2.0)) / B;
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	tmp = 0
	if B <= 7e+36:
		tmp = -math.sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0
	else:
		tmp = (math.sqrt((B * -F)) * -math.sqrt(2.0)) / B
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 7e+36)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(t_0 * F) * Float64(A + A))))) / t_0);
	else
		tmp = Float64(Float64(sqrt(Float64(B * Float64(-F))) * Float64(-sqrt(2.0))) / B);
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (A * C));
	tmp = 0.0;
	if (B <= 7e+36)
		tmp = -sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
	else
		tmp = (sqrt((B * -F)) * -sqrt(2.0)) / B;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 7e+36], N[((-N[Sqrt[N[(2.0 * N[(N[(t$95$0 * F), $MachinePrecision] * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[N[(B * (-F)), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision] / B), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 6.9999999999999996e36

   1. Initial program 20.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

   3. Simplified 20.2%

      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

   4. Taylor expanded in C around inf 16.6%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
   5. Step-by-step derivation

      1. cancel-sign-sub-inv 16.6%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(--1\right) \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

      2. metadata-eval 16.6%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{1} \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

      3. *-lft-identity 16.6%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{A}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   6. Simplified 16.6%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   if 6.9999999999999996e36 < B

   1. Initial program 11.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

   3. Simplified 11.6%

      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

   4. Taylor expanded in C around 0 28.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}\right)} \]
   5. Step-by-step derivation

      1. mul-1-neg 28.6%

         \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}} \]

      2. associate-*l/ 28.6%

         \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]

      3. distribute-neg-frac 28.6%

         \[\leadsto \color{blue}{\frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{{A}^{2} + {B}^{2}}\right)}}{B}} \]

      4. unpow2 28.6%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)}}{B} \]

      5. unpow2 28.6%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)}}{B} \]

      6. hypot-def 59.1%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)}}{B} \]

   6. Simplified 59.1%

      \[\leadsto \color{blue}{\frac{-\sqrt{2} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}}{B}} \]

   7. Taylor expanded in A around 0 51.3%

      \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \color{blue}{\left(-1 \cdot B\right)}}}{B} \]
   8. Step-by-step derivation

      1. mul-1-neg 51.3%

         \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \color{blue}{\left(-B\right)}}}{B} \]

   9. Simplified 51.3%

      \[\leadsto \frac{-\sqrt{2} \cdot \sqrt{F \cdot \color{blue}{\left(-B\right)}}}{B} \]
3. Recombined 2 regimes into one program.

4. Final simplification 24.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 7 \cdot 10^{+36}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{B \cdot \left(-F\right)} \cdot \left(-\sqrt{2}\right)}{B}\\ \end{array} \]

Alternative 4: 29.0% accurate, 4.8× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := t_0 \cdot F\\ \mathbf{if}\;B \leq 1.1 \cdot 10^{+37}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + \left(C - B\right)\right)\right)}}{t_0}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* A C)))) (t_1 (* t_0 F)))
   (if (<= B 1.1e+37)
     (/ (- (sqrt (* 2.0 (* t_1 (+ A A))))) t_0)
     (/ (- (sqrt (* 2.0 (* t_1 (+ A (- C B)))))) t_0))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = t_0 * F;
	double tmp;
	if (B <= 1.1e+37) {
		tmp = -sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	} else {
		tmp = -sqrt((2.0 * (t_1 * (A + (C - B))))) / t_0;
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (a * c))
    t_1 = t_0 * f
    if (b <= 1.1d+37) then
        tmp = -sqrt((2.0d0 * (t_1 * (a + a)))) / t_0
    else
        tmp = -sqrt((2.0d0 * (t_1 * (a + (c - b))))) / t_0
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = t_0 * F;
	double tmp;
	if (B <= 1.1e+37) {
		tmp = -Math.sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	} else {
		tmp = -Math.sqrt((2.0 * (t_1 * (A + (C - B))))) / t_0;
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	t_1 = t_0 * F
	tmp = 0
	if B <= 1.1e+37:
		tmp = -math.sqrt((2.0 * (t_1 * (A + A)))) / t_0
	else:
		tmp = -math.sqrt((2.0 * (t_1 * (A + (C - B))))) / t_0
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	t_1 = Float64(t_0 * F)
	tmp = 0.0
	if (B <= 1.1e+37)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A + A))))) / t_0);
	else
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A + Float64(C - B)))))) / t_0);
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (A * C));
	t_1 = t_0 * F;
	tmp = 0.0;
	if (B <= 1.1e+37)
		tmp = -sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	else
		tmp = -sqrt((2.0 * (t_1 * (A + (C - B))))) / t_0;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 * F), $MachinePrecision]}, If[LessEqual[B, 1.1e+37], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + N[(C - B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if B < 1.1e37

   1. Initial program 20.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

   3. Simplified 20.2%

      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

   4. Taylor expanded in C around inf 16.6%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
   5. Step-by-step derivation

      1. cancel-sign-sub-inv 16.6%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(--1\right) \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

      2. metadata-eval 16.6%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{1} \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

      3. *-lft-identity 16.6%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{A}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   6. Simplified 16.6%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   if 1.1e37 < B

   1. Initial program 11.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

   3. Simplified 11.6%

      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

   4. Taylor expanded in B around inf 10.1%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(C + -1 \cdot B\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
   5. Step-by-step derivation

      1. mul-1-neg 10.1%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(C + \color{blue}{\left(-B\right)}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

      2. unsub-neg 10.1%

         \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{\left(C - B\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   6. Simplified 10.1%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(C - B\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 15.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.1 \cdot 10^{+37}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(C - B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

Alternative 5: 26.8% accurate, 5.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \frac{-\sqrt{2 \cdot \left(\left(t_0 \cdot F\right) \cdot \left(A + A\right)\right)}}{t_0} \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* A C)))))
   (/ (- (sqrt (* 2.0 (* (* t_0 F) (+ A A))))) t_0)))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	return -sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    t_0 = (b * b) - (4.0d0 * (a * c))
    code = -sqrt((2.0d0 * ((t_0 * f) * (a + a)))) / t_0
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	return -Math.sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	return -math.sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	return Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(t_0 * F) * Float64(A + A))))) / t_0)
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (A * C));
	tmp = -sqrt((2.0 * ((t_0 * F) * (A + A)))) / t_0;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[((-N[Sqrt[N[(2.0 * N[(N[(t$95$0 * F), $MachinePrecision] * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]

Derivation

1. Initial program 18.4%

   \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

3. Simplified 18.4%

   \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

4. Taylor expanded in C around inf 13.2%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
5. Step-by-step derivation

   1. cancel-sign-sub-inv 13.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(--1\right) \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   2. metadata-eval 13.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{1} \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   3. *-lft-identity 13.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{A}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

6. Simplified 13.2%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

7. Final simplification 13.2%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

Alternative 6: 2.2% accurate, 5.1× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \frac{-\sqrt{2 \cdot \left(\left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right) \cdot \left(C + C\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (/
  (- (sqrt (* 2.0 (* (* -4.0 (* A (* C F))) (+ C C)))))
  (- (* B B) (* 4.0 (* A C)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	return -sqrt((2.0 * ((-4.0 * (A * (C * F))) * (C + C)))) / ((B * B) - (4.0 * (A * C)));
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = -sqrt((2.0d0 * (((-4.0d0) * (a * (c * f))) * (c + c)))) / ((b * b) - (4.0d0 * (a * c)))
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	return -Math.sqrt((2.0 * ((-4.0 * (A * (C * F))) * (C + C)))) / ((B * B) - (4.0 * (A * C)));
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	return -math.sqrt((2.0 * ((-4.0 * (A * (C * F))) * (C + C)))) / ((B * B) - (4.0 * (A * C)))

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	return Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(-4.0 * Float64(A * Float64(C * F))) * Float64(C + C))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))))
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp = code(A, B, C, F)
	tmp = -sqrt((2.0 * ((-4.0 * (A * (C * F))) * (C + C)))) / ((B * B) - (4.0 * (A * C)));
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := N[((-N[Sqrt[N[(2.0 * N[(N[(-4.0 * N[(A * N[(C * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(C + C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 18.4%

   \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

3. Simplified 18.4%

   \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

4. Taylor expanded in A around inf 17.2%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C - -1 \cdot C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
5. Step-by-step derivation

   1. cancel-sign-sub-inv 17.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C + \left(--1\right) \cdot C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   2. metadata-eval 17.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(C + \color{blue}{1} \cdot C\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   3. *-lft-identity 17.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(C + \color{blue}{C}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

6. Simplified 17.2%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C + C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

7. Taylor expanded in B around 0 14.3%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)} \cdot \left(C + C\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

8. Final simplification 14.3%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right) \cdot \left(C + C\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

Alternative 7: 1.4% accurate, 5.2× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \frac{-\sqrt{2 \cdot \left(\left(C + C\right) \cdot \left(\left(B \cdot B\right) \cdot F\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (/ (- (sqrt (* 2.0 (* (+ C C) (* (* B B) F))))) (- (* B B) (* 4.0 (* A C)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	return -sqrt((2.0 * ((C + C) * ((B * B) * F)))) / ((B * B) - (4.0 * (A * C)));
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = -sqrt((2.0d0 * ((c + c) * ((b * b) * f)))) / ((b * b) - (4.0d0 * (a * c)))
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	return -Math.sqrt((2.0 * ((C + C) * ((B * B) * F)))) / ((B * B) - (4.0 * (A * C)));
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	return -math.sqrt((2.0 * ((C + C) * ((B * B) * F)))) / ((B * B) - (4.0 * (A * C)))

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	return Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(C + C) * Float64(Float64(B * B) * F))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))))
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp = code(A, B, C, F)
	tmp = -sqrt((2.0 * ((C + C) * ((B * B) * F)))) / ((B * B) - (4.0 * (A * C)));
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := N[((-N[Sqrt[N[(2.0 * N[(N[(C + C), $MachinePrecision] * N[(N[(B * B), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 18.4%

   \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

3. Simplified 18.4%

   \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

4. Taylor expanded in A around inf 17.2%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C - -1 \cdot C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
5. Step-by-step derivation

   1. cancel-sign-sub-inv 17.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C + \left(--1\right) \cdot C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   2. metadata-eval 17.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(C + \color{blue}{1} \cdot C\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   3. *-lft-identity 17.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(C + \color{blue}{C}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

6. Simplified 17.2%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C + C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

7. Taylor expanded in B around inf 4.0%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\color{blue}{{B}^{2}} \cdot F\right) \cdot \left(C + C\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
8. Step-by-step derivation

   1. unpow2 4.0%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\color{blue}{\left(B \cdot B\right)} \cdot F\right) \cdot \left(C + C\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

9. Simplified 4.0%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\color{blue}{\left(B \cdot B\right)} \cdot F\right) \cdot \left(C + C\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

10. Final simplification 4.0%

    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(C + C\right) \cdot \left(\left(B \cdot B\right) \cdot F\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

Alternative 8: 1.0% accurate, 5.9× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ -2 \cdot \frac{\sqrt{C \cdot F}}{B} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F) :precision binary64 (* -2.0 (/ (sqrt (* C F)) B)))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	return -2.0 * (sqrt((C * F)) / B);
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    code = (-2.0d0) * (sqrt((c * f)) / b)
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	return -2.0 * (Math.sqrt((C * F)) / B);
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	return -2.0 * (math.sqrt((C * F)) / B)

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	return Float64(-2.0 * Float64(sqrt(Float64(C * F)) / B))
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp = code(A, B, C, F)
	tmp = -2.0 * (sqrt((C * F)) / B);
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := N[(-2.0 * N[(N[Sqrt[N[(C * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 18.4%

   \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]

3. Simplified 18.4%

   \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]

4. Taylor expanded in A around inf 17.2%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C - -1 \cdot C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
5. Step-by-step derivation

   1. cancel-sign-sub-inv 17.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C + \left(--1\right) \cdot C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   2. metadata-eval 17.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(C + \color{blue}{1} \cdot C\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

   3. *-lft-identity 17.2%

      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(C + \color{blue}{C}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

6. Simplified 17.2%

   \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C + C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

7. Taylor expanded in B around inf 4.1%

   \[\leadsto \color{blue}{-2 \cdot \left(\frac{1}{B} \cdot \sqrt{C \cdot F}\right)} \]
8. Step-by-step derivation

   1. associate-*l/ 4.1%

      \[\leadsto -2 \cdot \color{blue}{\frac{1 \cdot \sqrt{C \cdot F}}{B}} \]

   2. *-lft-identity 4.1%

      \[\leadsto -2 \cdot \frac{\color{blue}{\sqrt{C \cdot F}}}{B} \]

   3. *-commutative 4.1%

      \[\leadsto -2 \cdot \frac{\sqrt{\color{blue}{F \cdot C}}}{B} \]

9. Simplified 4.1%

   \[\leadsto \color{blue}{-2 \cdot \frac{\sqrt{F \cdot C}}{B}} \]

10. Final simplification 4.1%

    \[\leadsto -2 \cdot \frac{\sqrt{C \cdot F}}{B} \]

Reproduce

herbie shell --seed 2023283 
(FPCore (A B C F)
  :name "ABCF->ab-angle b"
  :precision binary64
  (/ (- (sqrt (* (* 2.0 (* (- (pow B 2.0) (* (* 4.0 A) C)) F)) (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))) (- (pow B 2.0) (* (* 4.0 A) C))))