ABCF->ab-angle a

Percentage Accurate: 18.6% → 50.3%

Time: 16.5s

Alternatives: 10

Speedup: 16.9×

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: 18.6% 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: 50.3% accurate, 1.3× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := C \cdot \left(A \cdot 4\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(C \cdot 2\right) \cdot \left(\left(F \cdot \left({B\_m}^{2} - t\_0\right)\right) \cdot 2\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* C (* A 4.0))))
   (if (<= (pow B_m 2.0) 5e-85)
     (/
      (sqrt (* (* C 2.0) (* (* F (- (pow B_m 2.0) t_0)) 2.0)))
      (- t_0 (pow B_m 2.0)))
     (/ (sqrt (* F 2.0)) (- (sqrt B_m))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = C * (A * 4.0);
	double tmp;
	if (pow(B_m, 2.0) <= 5e-85) {
		tmp = sqrt(((C * 2.0) * ((F * (pow(B_m, 2.0) - t_0)) * 2.0))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
	}
	return tmp;
}

Fortran

B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: tmp
    t_0 = c * (a * 4.0d0)
    if ((b_m ** 2.0d0) <= 5d-85) then
        tmp = sqrt(((c * 2.0d0) * ((f * ((b_m ** 2.0d0) - t_0)) * 2.0d0))) / (t_0 - (b_m ** 2.0d0))
    else
        tmp = sqrt((f * 2.0d0)) / -sqrt(b_m)
    end if
    code = tmp
end function

Java

B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = C * (A * 4.0);
	double tmp;
	if (Math.pow(B_m, 2.0) <= 5e-85) {
		tmp = Math.sqrt(((C * 2.0) * ((F * (Math.pow(B_m, 2.0) - t_0)) * 2.0))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((F * 2.0)) / -Math.sqrt(B_m);
	}
	return tmp;
}

Python

B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = C * (A * 4.0)
	tmp = 0
	if math.pow(B_m, 2.0) <= 5e-85:
		tmp = math.sqrt(((C * 2.0) * ((F * (math.pow(B_m, 2.0) - t_0)) * 2.0))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((F * 2.0)) / -math.sqrt(B_m)
	return tmp

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(C * Float64(A * 4.0))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e-85)
		tmp = Float64(sqrt(Float64(Float64(C * 2.0) * Float64(Float64(F * Float64((B_m ^ 2.0) - t_0)) * 2.0))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
	end
	return tmp
end

MATLAB

B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = C * (A * 4.0);
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 5e-85)
		tmp = sqrt(((C * 2.0) * ((F * ((B_m ^ 2.0) - t_0)) * 2.0))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
	end
	tmp_2 = tmp;
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(C * N[(A * 4.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-85], N[(N[Sqrt[N[(N[(C * 2.0), $MachinePrecision] * N[(N[(F * N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision]), $MachinePrecision] * 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 5.0000000000000002e-85

   1. Initial program 25.2%

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

   3. Taylor expanded in A around -inf

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

      1. *-commutative N/A

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

      2. lower-*.f64 32.8

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

   5. Applied rewrites 32.8%

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

   if 5.0000000000000002e-85 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 15.2%

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

   3. Taylor expanded in B around inf

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

      1. mul-1-neg N/A

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

      2. *-commutative N/A

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

      3. distribute-lft-neg-in N/A

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

      4. lower-*.f64 N/A

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

      5. lower-neg.f64 N/A

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

      6. lower-sqrt.f64 N/A

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

      7. lower-sqrt.f64 N/A

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

      8. lower-/.f64 21.3

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

   5. Applied rewrites 21.3%

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

   7. Applied rewrites 28.7%

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

   9. Applied rewrites 28.8%

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

4. Final simplification 30.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(C \cdot 2\right) \cdot \left(\left(F \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot 2\right)}}{C \cdot \left(A \cdot 4\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 50.2% accurate, 2.6× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(-4, C \cdot A, B\_m \cdot B\_m\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{-1}{t\_0} \cdot \sqrt{\left(t\_0 \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot 2\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (fma -4.0 (* C A) (* B_m B_m))))
   (if (<= (pow B_m 2.0) 5e-85)
     (* (/ -1.0 t_0) (sqrt (* (* t_0 (* F 2.0)) (* C 2.0))))
     (/ (sqrt (* F 2.0)) (- (sqrt B_m))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = fma(-4.0, (C * A), (B_m * B_m));
	double tmp;
	if (pow(B_m, 2.0) <= 5e-85) {
		tmp = (-1.0 / t_0) * sqrt(((t_0 * (F * 2.0)) * (C * 2.0)));
	} else {
		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
	}
	return tmp;
}

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = fma(-4.0, Float64(C * A), Float64(B_m * B_m))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e-85)
		tmp = Float64(Float64(-1.0 / t_0) * sqrt(Float64(Float64(t_0 * Float64(F * 2.0)) * Float64(C * 2.0))));
	else
		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
	end
	return tmp
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(-4.0 * N[(C * A), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-85], N[(N[(-1.0 / t$95$0), $MachinePrecision] * N[Sqrt[N[(N[(t$95$0 * N[(F * 2.0), $MachinePrecision]), $MachinePrecision] * N[(C * 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 5.0000000000000002e-85

   1. Initial program 25.2%

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

   4. Applied rewrites 25.2%

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

   5. Taylor expanded in C around inf

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

      1. distribute-lft1-in N/A

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

      2. metadata-eval N/A

         \[\leadsto \sqrt{\left(C \cdot \left(2 + \color{blue}{0} \cdot \frac{A}{C}\right)\right) \cdot \left(\left(2 \cdot F\right) \cdot \mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      3. mul0-lft N/A

         \[\leadsto \sqrt{\left(C \cdot \left(2 + \color{blue}{0}\right)\right) \cdot \left(\left(2 \cdot F\right) \cdot \mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      4. metadata-eval N/A

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

      5. lower-*.f64 32.7

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

   7. Applied rewrites 32.7%

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

   if 5.0000000000000002e-85 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 15.2%

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

   3. Taylor expanded in B around inf

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

      1. mul-1-neg N/A

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

      2. *-commutative N/A

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

      3. distribute-lft-neg-in N/A

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

      4. lower-*.f64 N/A

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

      5. lower-neg.f64 N/A

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

      6. lower-sqrt.f64 N/A

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

      7. lower-sqrt.f64 N/A

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

      8. lower-/.f64 21.3

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

   5. Applied rewrites 21.3%

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

   7. Applied rewrites 28.7%

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

   9. Applied rewrites 28.8%

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

4. Final simplification 30.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \cdot \sqrt{\left(\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right) \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot 2\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 50.3% accurate, 2.7× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(-4 \cdot C, A, B\_m \cdot B\_m\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(t\_0 \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot 2\right)}}{-t\_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (fma (* -4.0 C) A (* B_m B_m))))
   (if (<= (pow B_m 2.0) 5e-85)
     (/ (sqrt (* (* t_0 (* F 2.0)) (* C 2.0))) (- t_0))
     (/ (sqrt (* F 2.0)) (- (sqrt B_m))))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = fma((-4.0 * C), A, (B_m * B_m));
	double tmp;
	if (pow(B_m, 2.0) <= 5e-85) {
		tmp = sqrt(((t_0 * (F * 2.0)) * (C * 2.0))) / -t_0;
	} else {
		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
	}
	return tmp;
}

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = fma(Float64(-4.0 * C), A, Float64(B_m * B_m))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e-85)
		tmp = Float64(sqrt(Float64(Float64(t_0 * Float64(F * 2.0)) * Float64(C * 2.0))) / Float64(-t_0));
	else
		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
	end
	return tmp
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(-4.0 * C), $MachinePrecision] * A + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-85], N[(N[Sqrt[N[(N[(t$95$0 * N[(F * 2.0), $MachinePrecision]), $MachinePrecision] * N[(C * 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 5.0000000000000002e-85

   1. Initial program 25.2%

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

   3. Taylor expanded in A around -inf

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

      1. *-commutative N/A

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

      2. lower-*.f64 32.8

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

   5. Applied rewrites 32.8%

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

      1. lift-/.f64 N/A

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

      2. frac-2neg N/A

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

      3. lift-neg.f64 N/A

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

      4. remove-double-neg N/A

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

   7. Applied rewrites 32.7%

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

   if 5.0000000000000002e-85 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 15.2%

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

   3. Taylor expanded in B around inf

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

      1. mul-1-neg N/A

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

      2. *-commutative N/A

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

      3. distribute-lft-neg-in N/A

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

      4. lower-*.f64 N/A

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

      5. lower-neg.f64 N/A

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

      6. lower-sqrt.f64 N/A

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

      7. lower-sqrt.f64 N/A

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

      8. lower-/.f64 21.3

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

   5. Applied rewrites 21.3%

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

   7. Applied rewrites 28.7%

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

   9. Applied rewrites 28.8%

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

4. Final simplification 30.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(\mathsf{fma}\left(-4 \cdot C, A, B \cdot B\right) \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot 2\right)}}{-\mathsf{fma}\left(-4 \cdot C, A, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 49.1% accurate, 2.9× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 4 \cdot 10^{-138}:\\ \;\;\;\;\sqrt{\left(\left(\left(C \cdot A\right) \cdot F\right) \cdot C\right) \cdot -16} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B\_m \cdot B\_m\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 4e-138)
   (*
    (sqrt (* (* (* (* C A) F) C) -16.0))
    (/ -1.0 (fma -4.0 (* C A) (* B_m B_m))))
   (/ (sqrt (* F 2.0)) (- (sqrt B_m)))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 4e-138) {
		tmp = sqrt(((((C * A) * F) * C) * -16.0)) * (-1.0 / fma(-4.0, (C * A), (B_m * B_m)));
	} else {
		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
	}
	return tmp;
}

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 4e-138)
		tmp = Float64(sqrt(Float64(Float64(Float64(Float64(C * A) * F) * C) * -16.0)) * Float64(-1.0 / fma(-4.0, Float64(C * A), Float64(B_m * B_m))));
	else
		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
	end
	return tmp
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 4e-138], N[(N[Sqrt[N[(N[(N[(N[(C * A), $MachinePrecision] * F), $MachinePrecision] * C), $MachinePrecision] * -16.0), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / N[(-4.0 * N[(C * A), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 4.00000000000000027e-138

   1. Initial program 23.9%

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

   4. Applied rewrites 23.9%

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

   5. Taylor expanded in C around inf

      \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(A \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
   6. Step-by-step derivation

      1. lower-*.f64 N/A

         \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(A \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      2. associate-*r* N/A

         \[\leadsto \sqrt{-16 \cdot \color{blue}{\left(\left(A \cdot {C}^{2}\right) \cdot F\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      3. lower-*.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \color{blue}{\left(\left(A \cdot {C}^{2}\right) \cdot F\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      4. lower-*.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\color{blue}{\left(A \cdot {C}^{2}\right)} \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      5. unpow2 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \color{blue}{\left(C \cdot C\right)}\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      6. lower-*.f64 23.3

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \color{blue}{\left(C \cdot C\right)}\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

   7. Applied rewrites 23.3%

      \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
   8. Step-by-step derivation

   9. Applied rewrites 32.1%

      \[\leadsto \sqrt{-16 \cdot \left(\left(F \cdot \left(C \cdot A\right)\right) \cdot \color{blue}{C}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

   if 4.00000000000000027e-138 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 16.6%

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

   3. Taylor expanded in B around inf

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

      1. mul-1-neg N/A

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

      2. *-commutative N/A

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

      3. distribute-lft-neg-in N/A

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

      4. lower-*.f64 N/A

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

      5. lower-neg.f64 N/A

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

      6. lower-sqrt.f64 N/A

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

      7. lower-sqrt.f64 N/A

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

      8. lower-/.f64 20.4

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

   5. Applied rewrites 20.4%

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

   7. Applied rewrites 27.5%

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

   9. Applied rewrites 27.5%

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

4. Final simplification 29.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 4 \cdot 10^{-138}:\\ \;\;\;\;\sqrt{\left(\left(\left(C \cdot A\right) \cdot F\right) \cdot C\right) \cdot -16} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 49.2% accurate, 2.9× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 4 \cdot 10^{-138}:\\ \;\;\;\;\sqrt{\left(\left(F \cdot C\right) \cdot \left(C \cdot A\right)\right) \cdot -16} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B\_m \cdot B\_m\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 4e-138)
   (*
    (sqrt (* (* (* F C) (* C A)) -16.0))
    (/ -1.0 (fma -4.0 (* C A) (* B_m B_m))))
   (/ (sqrt (* F 2.0)) (- (sqrt B_m)))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 4e-138) {
		tmp = sqrt((((F * C) * (C * A)) * -16.0)) * (-1.0 / fma(-4.0, (C * A), (B_m * B_m)));
	} else {
		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
	}
	return tmp;
}

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 4e-138)
		tmp = Float64(sqrt(Float64(Float64(Float64(F * C) * Float64(C * A)) * -16.0)) * Float64(-1.0 / fma(-4.0, Float64(C * A), Float64(B_m * B_m))));
	else
		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
	end
	return tmp
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 4e-138], N[(N[Sqrt[N[(N[(N[(F * C), $MachinePrecision] * N[(C * A), $MachinePrecision]), $MachinePrecision] * -16.0), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / N[(-4.0 * N[(C * A), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 4.00000000000000027e-138

   1. Initial program 23.9%

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

   4. Applied rewrites 23.9%

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

   5. Taylor expanded in C around inf

      \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(A \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
   6. Step-by-step derivation

      1. lower-*.f64 N/A

         \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(A \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      2. associate-*r* N/A

         \[\leadsto \sqrt{-16 \cdot \color{blue}{\left(\left(A \cdot {C}^{2}\right) \cdot F\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      3. lower-*.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \color{blue}{\left(\left(A \cdot {C}^{2}\right) \cdot F\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      4. lower-*.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\color{blue}{\left(A \cdot {C}^{2}\right)} \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      5. unpow2 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \color{blue}{\left(C \cdot C\right)}\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      6. lower-*.f64 23.3

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \color{blue}{\left(C \cdot C\right)}\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

   7. Applied rewrites 23.3%

      \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
   8. Step-by-step derivation

   9. Applied rewrites 32.8%

      \[\leadsto \sqrt{-16 \cdot \left(\left(C \cdot A\right) \cdot \color{blue}{\left(C \cdot F\right)}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

   if 4.00000000000000027e-138 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 16.6%

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

   3. Taylor expanded in B around inf

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

      1. mul-1-neg N/A

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

      2. *-commutative N/A

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

      3. distribute-lft-neg-in N/A

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

      4. lower-*.f64 N/A

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

      5. lower-neg.f64 N/A

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

      6. lower-sqrt.f64 N/A

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

      7. lower-sqrt.f64 N/A

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

      8. lower-/.f64 20.4

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

   5. Applied rewrites 20.4%

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

   7. Applied rewrites 27.5%

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

   9. Applied rewrites 27.5%

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

4. Final simplification 29.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 4 \cdot 10^{-138}:\\ \;\;\;\;\sqrt{\left(\left(F \cdot C\right) \cdot \left(C \cdot A\right)\right) \cdot -16} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 44.0% accurate, 2.9× speedup

Math

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 4 \cdot 10^{-138}:\\ \;\;\;\;\frac{\sqrt{\left(\left(\left(C \cdot C\right) \cdot A\right) \cdot F\right) \cdot -16}}{-\mathsf{fma}\left(-4 \cdot C, A, B\_m \cdot B\_m\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (if (<= (pow B_m 2.0) 4e-138)
   (/ (sqrt (* (* (* (* C C) A) F) -16.0)) (- (fma (* -4.0 C) A (* B_m B_m))))
   (/ (sqrt (* F 2.0)) (- (sqrt B_m)))))

C

B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double tmp;
	if (pow(B_m, 2.0) <= 4e-138) {
		tmp = sqrt(((((C * C) * A) * F) * -16.0)) / -fma((-4.0 * C), A, (B_m * B_m));
	} else {
		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
	}
	return tmp;
}

Julia

B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 4e-138)
		tmp = Float64(sqrt(Float64(Float64(Float64(Float64(C * C) * A) * F) * -16.0)) / Float64(-fma(Float64(-4.0 * C), A, Float64(B_m * B_m))));
	else
		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
	end
	return tmp
end

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 4e-138], N[(N[Sqrt[N[(N[(N[(N[(C * C), $MachinePrecision] * A), $MachinePrecision] * F), $MachinePrecision] * -16.0), $MachinePrecision]], $MachinePrecision] / (-N[(N[(-4.0 * C), $MachinePrecision] * A + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision])), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (pow.f64 B #s(literal 2 binary64)) < 4.00000000000000027e-138

   1. Initial program 23.9%

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

   4. Applied rewrites 23.9%

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

   5. Taylor expanded in C around inf

      \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(A \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
   6. Step-by-step derivation

      1. lower-*.f64 N/A

         \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(A \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      2. associate-*r* N/A

         \[\leadsto \sqrt{-16 \cdot \color{blue}{\left(\left(A \cdot {C}^{2}\right) \cdot F\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      3. lower-*.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \color{blue}{\left(\left(A \cdot {C}^{2}\right) \cdot F\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      4. lower-*.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\color{blue}{\left(A \cdot {C}^{2}\right)} \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      5. unpow2 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \color{blue}{\left(C \cdot C\right)}\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

      6. lower-*.f64 23.3

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \color{blue}{\left(C \cdot C\right)}\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

   7. Applied rewrites 23.3%

      \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
   8. Step-by-step derivation

      1. lift-*.f64 N/A

         \[\leadsto \color{blue}{\sqrt{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)}} \]

      2. lift-/.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)} \cdot \color{blue}{\frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)}} \]

      3. frac-2neg N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)} \cdot \color{blue}{\frac{\mathsf{neg}\left(-1\right)}{\mathsf{neg}\left(\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)}} \]

      4. metadata-eval N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)} \cdot \frac{\color{blue}{1}}{\mathsf{neg}\left(\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \]

      5. lift-fma.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)} \cdot \frac{1}{\mathsf{neg}\left(\color{blue}{\left(-4 \cdot \left(C \cdot A\right) + B \cdot B\right)}\right)} \]

      6. lift-*.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)} \cdot \frac{1}{\mathsf{neg}\left(\left(-4 \cdot \color{blue}{\left(C \cdot A\right)} + B \cdot B\right)\right)} \]

      7. associate-*r* N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)} \cdot \frac{1}{\mathsf{neg}\left(\left(\color{blue}{\left(-4 \cdot C\right) \cdot A} + B \cdot B\right)\right)} \]

      8. lift-*.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)} \cdot \frac{1}{\mathsf{neg}\left(\left(\color{blue}{\left(-4 \cdot C\right)} \cdot A + B \cdot B\right)\right)} \]

      9. lift-fma.f64 N/A

         \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)} \cdot \frac{1}{\mathsf{neg}\left(\color{blue}{\mathsf{fma}\left(-4 \cdot C, A, B \cdot B\right)}\right)} \]

      10. un-div-inv N/A

          \[\leadsto \color{blue}{\frac{\sqrt{-16 \cdot \left(\left(A \cdot \left(C \cdot C\right)\right) \cdot F\right)}}{\mathsf{neg}\left(\mathsf{fma}\left(-4 \cdot C, A, B \cdot B\right)\right)}} \]

   9. Applied rewrites 23.3%

      \[\leadsto \color{blue}{\frac{\sqrt{\left(\left(\left(C \cdot C\right) \cdot A\right) \cdot F\right) \cdot -16}}{-\mathsf{fma}\left(-4 \cdot C, A, B \cdot B\right)}} \]

   if 4.00000000000000027e-138 < (pow.f64 B #s(literal 2 binary64))

   1. Initial program 16.6%

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

   3. Taylor expanded in B around inf

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

      1. mul-1-neg N/A

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

      2. *-commutative N/A

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

      3. distribute-lft-neg-in N/A

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

      4. lower-*.f64 N/A

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

      5. lower-neg.f64 N/A

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

      6. lower-sqrt.f64 N/A

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

      7. lower-sqrt.f64 N/A

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

      8. lower-/.f64 20.4

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

   5. Applied rewrites 20.4%

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

   7. Applied rewrites 27.5%

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

   9. Applied rewrites 27.5%

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

4. Final simplification 25.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 4 \cdot 10^{-138}:\\ \;\;\;\;\frac{\sqrt{\left(\left(\left(C \cdot C\right) \cdot A\right) \cdot F\right) \cdot -16}}{-\mathsf{fma}\left(-4 \cdot C, A, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 35.3% accurate, 12.6× speedup

Math

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

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F) :precision binary64 (/ (sqrt (* F 2.0)) (- (sqrt B_m))))

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]

Derivation

1. Initial program 19.6%

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

3. Taylor expanded in B around inf

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

   1. mul-1-neg N/A

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

   2. *-commutative N/A

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

   3. distribute-lft-neg-in N/A

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

   4. lower-*.f64 N/A

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

   5. lower-neg.f64 N/A

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

   6. lower-sqrt.f64 N/A

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

   7. lower-sqrt.f64 N/A

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

   8. lower-/.f64 12.8

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

5. Applied rewrites 12.8%

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

7. Applied rewrites 17.0%

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

9. Applied rewrites 17.0%

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

10. Final simplification 17.0%

    \[\leadsto \frac{\sqrt{F \cdot 2}}{-\sqrt{B}} \]
11. Add Preprocessing

Alternative 8: 35.3% accurate, 12.6× speedup

Math

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

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F) :precision binary64 (* (- (sqrt F)) (sqrt (/ 2.0 B_m))))

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := N[((-N[Sqrt[F], $MachinePrecision]) * N[Sqrt[N[(2.0 / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 19.6%

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

3. Taylor expanded in B around inf

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

   1. mul-1-neg N/A

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

   2. *-commutative N/A

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

   3. distribute-lft-neg-in N/A

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

   4. lower-*.f64 N/A

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

   5. lower-neg.f64 N/A

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

   6. lower-sqrt.f64 N/A

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

   7. lower-sqrt.f64 N/A

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

   8. lower-/.f64 12.8

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

5. Applied rewrites 12.8%

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

7. Applied rewrites 12.9%

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

9. Applied rewrites 17.0%

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

10. Final simplification 17.0%

    \[\leadsto \left(-\sqrt{F}\right) \cdot \sqrt{\frac{2}{B}} \]
11. Add Preprocessing

Alternative 9: 27.3% accurate, 16.9× speedup

Math

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

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F) :precision binary64 (- (sqrt (* (/ F B_m) 2.0))))

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := (-N[Sqrt[N[(N[(F / B$95$m), $MachinePrecision] * 2.0), $MachinePrecision]], $MachinePrecision])

Derivation

1. Initial program 19.6%

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

3. Taylor expanded in B around inf

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

   1. mul-1-neg N/A

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

   2. *-commutative N/A

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

   3. distribute-lft-neg-in N/A

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

   4. lower-*.f64 N/A

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

   5. lower-neg.f64 N/A

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

   6. lower-sqrt.f64 N/A

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

   7. lower-sqrt.f64 N/A

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

   8. lower-/.f64 12.8

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

5. Applied rewrites 12.8%

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

7. Applied rewrites 12.9%

   \[\leadsto \color{blue}{-\sqrt{\frac{F}{B} \cdot 2}} \]
8. Add Preprocessing

Alternative 10: 27.3% accurate, 16.9× speedup

Math

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

FPCore

B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F) :precision binary64 (- (sqrt (* (/ 2.0 B_m) F))))

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := (-N[Sqrt[N[(N[(2.0 / B$95$m), $MachinePrecision] * F), $MachinePrecision]], $MachinePrecision])

Derivation

1. Initial program 19.6%

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

3. Taylor expanded in B around inf

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

   1. mul-1-neg N/A

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

   2. *-commutative N/A

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

   3. distribute-lft-neg-in N/A

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

   4. lower-*.f64 N/A

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

   5. lower-neg.f64 N/A

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

   6. lower-sqrt.f64 N/A

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

   7. lower-sqrt.f64 N/A

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

   8. lower-/.f64 12.8

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

5. Applied rewrites 12.8%

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

7. Applied rewrites 12.9%

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

9. Applied rewrites 12.9%

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

10. Final simplification 12.9%

    \[\leadsto -\sqrt{\frac{2}{B} \cdot F} \]
11. Add Preprocessing

Reproduce

herbie shell --seed 2024235 
(FPCore (A B C F)
  :name "ABCF->ab-angle a"
  :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))))