Quadratic roots, medium range

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Percentage Accurate: 31.2% → 95.7%
Time: 19.8s
Precision: binary64
Cost: 40704

?

\[\left(\left(1.1102230246251565 \cdot 10^{-16} < a \land a < 9007199254740992\right) \land \left(1.1102230246251565 \cdot 10^{-16} < b \land b < 9007199254740992\right)\right) \land \left(1.1102230246251565 \cdot 10^{-16} < c \land c < 9007199254740992\right)\]
\[\frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a} \]
\[\mathsf{fma}\left(-0.25, {\left(c \cdot a\right)}^{4} \cdot \frac{20}{a \cdot {b}^{7}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{{b}^{3}}{c}} \]
(FPCore (a b c)
 :precision binary64
 (/ (+ (- b) (sqrt (- (* b b) (* (* 4.0 a) c)))) (* 2.0 a)))
(FPCore (a b c)
 :precision binary64
 (-
  (fma
   -0.25
   (* (pow (* c a) 4.0) (/ 20.0 (* a (pow b 7.0))))
   (- (/ (* -2.0 (pow c 3.0)) (/ (pow b 5.0) (* a a))) (/ c b)))
  (* a (/ c (/ (pow b 3.0) c)))))
double code(double a, double b, double c) {
	return (-b + sqrt(((b * b) - ((4.0 * a) * c)))) / (2.0 * a);
}
double code(double a, double b, double c) {
	return fma(-0.25, (pow((c * a), 4.0) * (20.0 / (a * pow(b, 7.0)))), (((-2.0 * pow(c, 3.0)) / (pow(b, 5.0) / (a * a))) - (c / b))) - (a * (c / (pow(b, 3.0) / c)));
}
function code(a, b, c)
	return Float64(Float64(Float64(-b) + sqrt(Float64(Float64(b * b) - Float64(Float64(4.0 * a) * c)))) / Float64(2.0 * a))
end
function code(a, b, c)
	return Float64(fma(-0.25, Float64((Float64(c * a) ^ 4.0) * Float64(20.0 / Float64(a * (b ^ 7.0)))), Float64(Float64(Float64(-2.0 * (c ^ 3.0)) / Float64((b ^ 5.0) / Float64(a * a))) - Float64(c / b))) - Float64(a * Float64(c / Float64((b ^ 3.0) / c))))
end
code[a_, b_, c_] := N[(N[((-b) + N[Sqrt[N[(N[(b * b), $MachinePrecision] - N[(N[(4.0 * a), $MachinePrecision] * c), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[(2.0 * a), $MachinePrecision]), $MachinePrecision]
code[a_, b_, c_] := N[(N[(-0.25 * N[(N[Power[N[(c * a), $MachinePrecision], 4.0], $MachinePrecision] * N[(20.0 / N[(a * N[Power[b, 7.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(N[(N[(-2.0 * N[Power[c, 3.0], $MachinePrecision]), $MachinePrecision] / N[(N[Power[b, 5.0], $MachinePrecision] / N[(a * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - N[(c / b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - N[(a * N[(c / N[(N[Power[b, 3.0], $MachinePrecision] / c), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a}
\mathsf{fma}\left(-0.25, {\left(c \cdot a\right)}^{4} \cdot \frac{20}{a \cdot {b}^{7}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{{b}^{3}}{c}}

Local Percentage Accuracy?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Derivation?

  1. Initial program 35.0%

    \[\frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a} \]
  2. Taylor expanded in b around inf 93.4%

    \[\leadsto \color{blue}{-1 \cdot \frac{{c}^{2} \cdot a}{{b}^{3}} + \left(-0.25 \cdot \frac{{\left(-2 \cdot \left({c}^{2} \cdot {a}^{2}\right)\right)}^{2} + 16 \cdot \left({c}^{4} \cdot {a}^{4}\right)}{a \cdot {b}^{7}} + \left(-1 \cdot \frac{c}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right)\right)} \]
  3. Simplified93.4%

    \[\leadsto \color{blue}{\mathsf{fma}\left(-0.25, \frac{\mathsf{fma}\left(16, {c}^{4} \cdot {a}^{4}, 4 \cdot \left({c}^{4} \cdot {a}^{4}\right)\right)}{a \cdot {b}^{7}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a} \]
    Step-by-step derivation

    [Start]93.4

    \[ -1 \cdot \frac{{c}^{2} \cdot a}{{b}^{3}} + \left(-0.25 \cdot \frac{{\left(-2 \cdot \left({c}^{2} \cdot {a}^{2}\right)\right)}^{2} + 16 \cdot \left({c}^{4} \cdot {a}^{4}\right)}{a \cdot {b}^{7}} + \left(-1 \cdot \frac{c}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right)\right) \]

    +-commutative [=>]93.4

    \[ \color{blue}{\left(-0.25 \cdot \frac{{\left(-2 \cdot \left({c}^{2} \cdot {a}^{2}\right)\right)}^{2} + 16 \cdot \left({c}^{4} \cdot {a}^{4}\right)}{a \cdot {b}^{7}} + \left(-1 \cdot \frac{c}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right)\right) + -1 \cdot \frac{{c}^{2} \cdot a}{{b}^{3}}} \]

    mul-1-neg [=>]93.4

    \[ \left(-0.25 \cdot \frac{{\left(-2 \cdot \left({c}^{2} \cdot {a}^{2}\right)\right)}^{2} + 16 \cdot \left({c}^{4} \cdot {a}^{4}\right)}{a \cdot {b}^{7}} + \left(-1 \cdot \frac{c}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right)\right) + \color{blue}{\left(-\frac{{c}^{2} \cdot a}{{b}^{3}}\right)} \]

    unsub-neg [=>]93.4

    \[ \color{blue}{\left(-0.25 \cdot \frac{{\left(-2 \cdot \left({c}^{2} \cdot {a}^{2}\right)\right)}^{2} + 16 \cdot \left({c}^{4} \cdot {a}^{4}\right)}{a \cdot {b}^{7}} + \left(-1 \cdot \frac{c}{b} + -2 \cdot \frac{{c}^{3} \cdot {a}^{2}}{{b}^{5}}\right)\right) - \frac{{c}^{2} \cdot a}{{b}^{3}}} \]
  4. Taylor expanded in c around 0 93.4%

    \[\leadsto \mathsf{fma}\left(-0.25, \color{blue}{\frac{{c}^{4} \cdot \left(4 \cdot {a}^{4} + 16 \cdot {a}^{4}\right)}{a \cdot {b}^{7}}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]
  5. Simplified93.4%

    \[\leadsto \mathsf{fma}\left(-0.25, \color{blue}{{\left(c \cdot a\right)}^{4} \cdot \frac{20}{a \cdot {b}^{7}}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]
    Step-by-step derivation

    [Start]93.4

    \[ \mathsf{fma}\left(-0.25, \frac{{c}^{4} \cdot \left(4 \cdot {a}^{4} + 16 \cdot {a}^{4}\right)}{a \cdot {b}^{7}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]

    distribute-rgt-out [=>]93.4

    \[ \mathsf{fma}\left(-0.25, \frac{{c}^{4} \cdot \color{blue}{\left({a}^{4} \cdot \left(4 + 16\right)\right)}}{a \cdot {b}^{7}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]

    metadata-eval [=>]93.4

    \[ \mathsf{fma}\left(-0.25, \frac{{c}^{4} \cdot \left({a}^{4} \cdot \color{blue}{20}\right)}{a \cdot {b}^{7}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]

    associate-*r* [=>]93.4

    \[ \mathsf{fma}\left(-0.25, \frac{\color{blue}{\left({c}^{4} \cdot {a}^{4}\right) \cdot 20}}{a \cdot {b}^{7}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]

    times-frac [=>]93.4

    \[ \mathsf{fma}\left(-0.25, \color{blue}{\frac{{c}^{4} \cdot {a}^{4}}{a} \cdot \frac{20}{{b}^{7}}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - \frac{c}{\frac{{b}^{3}}{c}} \cdot a \]
  6. Final simplification93.4%

    \[\leadsto \mathsf{fma}\left(-0.25, {\left(c \cdot a\right)}^{4} \cdot \frac{20}{a \cdot {b}^{7}}, \frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{{b}^{3}}{c}} \]

Alternatives

Alternative 1
Accuracy91.1%
Cost21060
\[\begin{array}{l} \mathbf{if}\;\frac{\sqrt{b \cdot b - c \cdot \left(a \cdot 4\right)} - b}{a \cdot 2} \leq -5000:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(b, b, \left(c \cdot a\right) \cdot -4\right)} - b}{a \cdot 2}\\ \mathbf{else}:\\ \;\;\;\;\frac{-c}{b} - \frac{c \cdot \left(c \cdot a\right)}{{b}^{3}}\\ \end{array} \]
Alternative 2
Accuracy94.2%
Cost20736
\[\left(\frac{-2 \cdot {c}^{3}}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{{b}^{3}}{c}} \]
Alternative 3
Accuracy91.1%
Cost14788
\[\begin{array}{l} \mathbf{if}\;\frac{\sqrt{b \cdot b - c \cdot \left(a \cdot 4\right)} - b}{a \cdot 2} \leq -5000:\\ \;\;\;\;\frac{\sqrt{b \cdot b + a \cdot \left(c \cdot -4\right)} - b}{a \cdot 2}\\ \mathbf{else}:\\ \;\;\;\;\frac{-c}{b} - \frac{c \cdot \left(c \cdot a\right)}{{b}^{3}}\\ \end{array} \]
Alternative 4
Accuracy91.1%
Cost7232
\[\frac{-c}{b} - \frac{c \cdot \left(c \cdot a\right)}{{b}^{3}} \]
Alternative 5
Accuracy90.7%
Cost1600
\[\left(2 \cdot \left(\frac{c}{\frac{b}{a}} + \frac{c \cdot c}{\frac{b}{a} \cdot \frac{b \cdot b}{a}}\right)\right) \cdot \frac{-0.5}{a} \]
Alternative 6
Accuracy81.5%
Cost256
\[\frac{-c}{b} \]
Alternative 7
Accuracy3.2%
Cost192
\[\frac{0}{a} \]

Error

Reproduce?

herbie shell --seed 2023159 
(FPCore (a b c)
  :name "Quadratic roots, medium range"
  :precision binary64
  :pre (and (and (and (< 1.1102230246251565e-16 a) (< a 9007199254740992.0)) (and (< 1.1102230246251565e-16 b) (< b 9007199254740992.0))) (and (< 1.1102230246251565e-16 c) (< c 9007199254740992.0)))
  (/ (+ (- b) (sqrt (- (* b b) (* (* 4.0 a) c)))) (* 2.0 a)))