quad2m (problem 3.2.1, negative)

Average Accuracy: 52.1% → 84.6%

Time: 16.0s

Precision: binary64

Cost: 7432

Math

\[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a} \]

↓

\[\begin{array}{l} \mathbf{if}\;b_2 \leq -7.2 \cdot 10^{-16}:\\ \;\;\;\;\frac{-0.5 \cdot c}{b_2}\\ \mathbf{elif}\;b_2 \leq 10^{+132}:\\ \;\;\;\;\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - c \cdot a}}{a}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{b_2}{a}\\ \end{array} \]

FPCore

(FPCore (a b_2 c)
 :precision binary64
 (/ (- (- b_2) (sqrt (- (* b_2 b_2) (* a c)))) a))

↓

(FPCore (a b_2 c)
 :precision binary64
 (if (<= b_2 -7.2e-16)
   (/ (* -0.5 c) b_2)
   (if (<= b_2 1e+132)
     (/ (- (- b_2) (sqrt (- (* b_2 b_2) (* c a)))) a)
     (* -2.0 (/ b_2 a)))))

C

double code(double a, double b_2, double c) {
	return (-b_2 - sqrt(((b_2 * b_2) - (a * c)))) / a;
}

↓

double code(double a, double b_2, double c) {
	double tmp;
	if (b_2 <= -7.2e-16) {
		tmp = (-0.5 * c) / b_2;
	} else if (b_2 <= 1e+132) {
		tmp = (-b_2 - sqrt(((b_2 * b_2) - (c * a)))) / a;
	} else {
		tmp = -2.0 * (b_2 / a);
	}
	return tmp;
}

Fortran

real(8) function code(a, b_2, c)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_2
    real(8), intent (in) :: c
    code = (-b_2 - sqrt(((b_2 * b_2) - (a * c)))) / a
end function

↓

real(8) function code(a, b_2, c)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_2
    real(8), intent (in) :: c
    real(8) :: tmp
    if (b_2 <= (-7.2d-16)) then
        tmp = ((-0.5d0) * c) / b_2
    else if (b_2 <= 1d+132) then
        tmp = (-b_2 - sqrt(((b_2 * b_2) - (c * a)))) / a
    else
        tmp = (-2.0d0) * (b_2 / a)
    end if
    code = tmp
end function

Java

public static double code(double a, double b_2, double c) {
	return (-b_2 - Math.sqrt(((b_2 * b_2) - (a * c)))) / a;
}

↓

public static double code(double a, double b_2, double c) {
	double tmp;
	if (b_2 <= -7.2e-16) {
		tmp = (-0.5 * c) / b_2;
	} else if (b_2 <= 1e+132) {
		tmp = (-b_2 - Math.sqrt(((b_2 * b_2) - (c * a)))) / a;
	} else {
		tmp = -2.0 * (b_2 / a);
	}
	return tmp;
}

Python

def code(a, b_2, c):
	return (-b_2 - math.sqrt(((b_2 * b_2) - (a * c)))) / a

↓

def code(a, b_2, c):
	tmp = 0
	if b_2 <= -7.2e-16:
		tmp = (-0.5 * c) / b_2
	elif b_2 <= 1e+132:
		tmp = (-b_2 - math.sqrt(((b_2 * b_2) - (c * a)))) / a
	else:
		tmp = -2.0 * (b_2 / a)
	return tmp

Julia

function code(a, b_2, c)
	return Float64(Float64(Float64(-b_2) - sqrt(Float64(Float64(b_2 * b_2) - Float64(a * c)))) / a)
end

↓

function code(a, b_2, c)
	tmp = 0.0
	if (b_2 <= -7.2e-16)
		tmp = Float64(Float64(-0.5 * c) / b_2);
	elseif (b_2 <= 1e+132)
		tmp = Float64(Float64(Float64(-b_2) - sqrt(Float64(Float64(b_2 * b_2) - Float64(c * a)))) / a);
	else
		tmp = Float64(-2.0 * Float64(b_2 / a));
	end
	return tmp
end

MATLAB

function tmp = code(a, b_2, c)
	tmp = (-b_2 - sqrt(((b_2 * b_2) - (a * c)))) / a;
end

↓

function tmp_2 = code(a, b_2, c)
	tmp = 0.0;
	if (b_2 <= -7.2e-16)
		tmp = (-0.5 * c) / b_2;
	elseif (b_2 <= 1e+132)
		tmp = (-b_2 - sqrt(((b_2 * b_2) - (c * a)))) / a;
	else
		tmp = -2.0 * (b_2 / a);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b$95$2_, c_] := N[(N[((-b$95$2) - N[Sqrt[N[(N[(b$95$2 * b$95$2), $MachinePrecision] - N[(a * c), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / a), $MachinePrecision]

↓

code[a_, b$95$2_, c_] := If[LessEqual[b$95$2, -7.2e-16], N[(N[(-0.5 * c), $MachinePrecision] / b$95$2), $MachinePrecision], If[LessEqual[b$95$2, 1e+132], N[(N[((-b$95$2) - N[Sqrt[N[(N[(b$95$2 * b$95$2), $MachinePrecision] - N[(c * a), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / a), $MachinePrecision], N[(-2.0 * N[(b$95$2 / a), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if b_2 < -7.19999999999999965e-16

   1. Initial program 17.2%

      \[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a} \]

   2. Taylor expanded in b_2 around -inf 88.1%

      \[\leadsto \color{blue}{-0.5 \cdot \frac{c}{b_2}} \]

   3. Simplified 88.1%

      \[\leadsto \color{blue}{\frac{-0.5 \cdot c}{b_2}} \]
      Step-by-step derivation

      [Start] 88.1	\[ -0.5 \cdot \frac{c}{b_2} \]
      associate-*r/ [=>] 88.1	\[ \color{blue}{\frac{-0.5 \cdot c}{b_2}} \]

   if -7.19999999999999965e-16 < b_2 < 9.99999999999999991e131

   1. Initial program 85.4%

      \[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a} \]

   if 9.99999999999999991e131 < b_2

   1. Initial program 44.0%

      \[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a} \]

   2. Taylor expanded in b_2 around inf 100.0%

      \[\leadsto \color{blue}{-2 \cdot \frac{b_2}{a}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 89.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \leq -7.2 \cdot 10^{-16}:\\ \;\;\;\;\frac{-0.5 \cdot c}{b_2}\\ \mathbf{elif}\;b_2 \leq 10^{+132}:\\ \;\;\;\;\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - c \cdot a}}{a}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{b_2}{a}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	79.4%
Cost	7240

\[\begin{array}{l} \mathbf{if}\;b_2 \leq -4.3 \cdot 10^{-11}:\\ \;\;\;\;\frac{-0.5 \cdot c}{b_2}\\ \mathbf{elif}\;b_2 \leq 2.9 \cdot 10^{-69}:\\ \;\;\;\;\frac{\left(-b_2\right) - \sqrt{a \cdot \left(-c\right)}}{a}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{b_2}{a} + 0.5 \cdot \frac{c}{b_2}\\ \end{array} \]

Alternative 2
Accuracy	78.8%
Cost	7112

\[\begin{array}{l} \mathbf{if}\;b_2 \leq -9.5 \cdot 10^{-18}:\\ \;\;\;\;\frac{-0.5 \cdot c}{b_2}\\ \mathbf{elif}\;b_2 \leq 5.5 \cdot 10^{-120}:\\ \;\;\;\;\frac{-\sqrt{a \cdot \left(-c\right)}}{a}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{b_2}{a} + 0.5 \cdot \frac{c}{b_2}\\ \end{array} \]

Alternative 3
Accuracy	66.9%
Cost	836

\[\begin{array}{l} \mathbf{if}\;b_2 \leq -1 \cdot 10^{-310}:\\ \;\;\;\;\frac{-0.5 \cdot c}{b_2}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{b_2}{a} + 0.5 \cdot \frac{c}{b_2}\\ \end{array} \]

Alternative 4
Accuracy	42.4%
Cost	452

\[\begin{array}{l} \mathbf{if}\;b_2 \leq -1 \cdot 10^{-310}:\\ \;\;\;\;\frac{0}{a}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{b_2}{a}\\ \end{array} \]

Alternative 5
Accuracy	66.6%
Cost	452

\[\begin{array}{l} \mathbf{if}\;b_2 \leq -4.6 \cdot 10^{-289}:\\ \;\;\;\;c \cdot \frac{-0.5}{b_2}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{b_2}{a}\\ \end{array} \]

Alternative 6
Accuracy	66.3%
Cost	452

\[\begin{array}{l} \mathbf{if}\;b_2 \leq -1.9 \cdot 10^{-298}:\\ \;\;\;\;\frac{-0.5}{\frac{b_2}{c}}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{b_2}{a}\\ \end{array} \]

Alternative 7
Accuracy	66.7%
Cost	452

\[\begin{array}{l} \mathbf{if}\;b_2 \leq -2.2 \cdot 10^{-298}:\\ \;\;\;\;\frac{-0.5 \cdot c}{b_2}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{b_2}{a}\\ \end{array} \]

Alternative 8
Accuracy	23.0%
Cost	388

\[\begin{array}{l} \mathbf{if}\;b_2 \leq -1.9 \cdot 10^{-298}:\\ \;\;\;\;\frac{0}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-b_2}{a}\\ \end{array} \]

Alternative 9
Accuracy	10.6%
Cost	192

\[\frac{0}{a} \]

Reproduce

herbie shell --seed 2023164 
(FPCore (a b_2 c)
  :name "quad2m (problem 3.2.1, negative)"
  :precision binary64
  (/ (- (- b_2) (sqrt (- (* b_2 b_2) (* a c)))) a))