Bouland and Aaronson, Equation (24)

Average Accuracy: 73.7% → 98.2%

Time: 11.1s

Precision: binary64

Cost: 41284

• 60.6% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[\left({\left(a \cdot a + b \cdot b\right)}^{2} + 4 \cdot \left(\left(a \cdot a\right) \cdot \left(1 - a\right) + \left(b \cdot b\right) \cdot \left(3 + a\right)\right)\right) - 1 \]

↓

\[\begin{array}{l} \mathbf{if}\;{\left(a \cdot a + b \cdot b\right)}^{2} + 4 \cdot \left(\left(a \cdot a\right) \cdot \left(1 - a\right) + \left(b \cdot b\right) \cdot \left(a + 3\right)\right) \leq \infty:\\ \;\;\;\;{\left(\mathsf{hypot}\left(a, b\right)\right)}^{4} + \mathsf{fma}\left(4, \mathsf{fma}\left(b \cdot b, a + 3, a \cdot a\right) - {a}^{3}, -1\right)\\ \mathbf{else}:\\ \;\;\;\;{a}^{4}\\ \end{array} \]

FPCore

(FPCore (a b)
 :precision binary64
 (-
  (+
   (pow (+ (* a a) (* b b)) 2.0)
   (* 4.0 (+ (* (* a a) (- 1.0 a)) (* (* b b) (+ 3.0 a)))))
  1.0))

↓

(FPCore (a b)
 :precision binary64
 (if (<=
      (+
       (pow (+ (* a a) (* b b)) 2.0)
       (* 4.0 (+ (* (* a a) (- 1.0 a)) (* (* b b) (+ a 3.0)))))
      INFINITY)
   (+
    (pow (hypot a b) 4.0)
    (fma 4.0 (- (fma (* b b) (+ a 3.0) (* a a)) (pow a 3.0)) -1.0))
   (pow a 4.0)))

C

double code(double a, double b) {
	return (pow(((a * a) + (b * b)), 2.0) + (4.0 * (((a * a) * (1.0 - a)) + ((b * b) * (3.0 + a))))) - 1.0;
}

↓

double code(double a, double b) {
	double tmp;
	if ((pow(((a * a) + (b * b)), 2.0) + (4.0 * (((a * a) * (1.0 - a)) + ((b * b) * (a + 3.0))))) <= ((double) INFINITY)) {
		tmp = pow(hypot(a, b), 4.0) + fma(4.0, (fma((b * b), (a + 3.0), (a * a)) - pow(a, 3.0)), -1.0);
	} else {
		tmp = pow(a, 4.0);
	}
	return tmp;
}

Julia

function code(a, b)
	return Float64(Float64((Float64(Float64(a * a) + Float64(b * b)) ^ 2.0) + Float64(4.0 * Float64(Float64(Float64(a * a) * Float64(1.0 - a)) + Float64(Float64(b * b) * Float64(3.0 + a))))) - 1.0)
end

↓

function code(a, b)
	tmp = 0.0
	if (Float64((Float64(Float64(a * a) + Float64(b * b)) ^ 2.0) + Float64(4.0 * Float64(Float64(Float64(a * a) * Float64(1.0 - a)) + Float64(Float64(b * b) * Float64(a + 3.0))))) <= Inf)
		tmp = Float64((hypot(a, b) ^ 4.0) + fma(4.0, Float64(fma(Float64(b * b), Float64(a + 3.0), Float64(a * a)) - (a ^ 3.0)), -1.0));
	else
		tmp = a ^ 4.0;
	end
	return tmp
end

Wolfram

code[a_, b_] := N[(N[(N[Power[N[(N[(a * a), $MachinePrecision] + N[(b * b), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[(4.0 * N[(N[(N[(a * a), $MachinePrecision] * N[(1.0 - a), $MachinePrecision]), $MachinePrecision] + N[(N[(b * b), $MachinePrecision] * N[(3.0 + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - 1.0), $MachinePrecision]

↓

code[a_, b_] := If[LessEqual[N[(N[Power[N[(N[(a * a), $MachinePrecision] + N[(b * b), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[(4.0 * N[(N[(N[(a * a), $MachinePrecision] * N[(1.0 - a), $MachinePrecision]), $MachinePrecision] + N[(N[(b * b), $MachinePrecision] * N[(a + 3.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], Infinity], N[(N[Power[N[Sqrt[a ^ 2 + b ^ 2], $MachinePrecision], 4.0], $MachinePrecision] + N[(4.0 * N[(N[(N[(b * b), $MachinePrecision] * N[(a + 3.0), $MachinePrecision] + N[(a * a), $MachinePrecision]), $MachinePrecision] - N[Power[a, 3.0], $MachinePrecision]), $MachinePrecision] + -1.0), $MachinePrecision]), $MachinePrecision], N[Power[a, 4.0], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (+.f64 (pow.f64 (+.f64 (*.f64 a a) (*.f64 b b)) 2) (*.f64 4 (+.f64 (*.f64 (*.f64 a a) (-.f64 1 a)) (*.f64 (*.f64 b b) (+.f64 3 a))))) < +inf.0

   1. Initial program 99.8%

      \[\left({\left(a \cdot a + b \cdot b\right)}^{2} + 4 \cdot \left(\left(a \cdot a\right) \cdot \left(1 - a\right) + \left(b \cdot b\right) \cdot \left(3 + a\right)\right)\right) - 1 \]

   2. Simplified 100.0%

      \[\leadsto \color{blue}{{\left(\mathsf{hypot}\left(a, b\right)\right)}^{4} + \mathsf{fma}\left(4, \mathsf{fma}\left(b \cdot b, a + 3, a \cdot a\right) - {a}^{3}, -1\right)} \]
      Proof

      [Start] 99.8	\[ \left({\left(a \cdot a + b \cdot b\right)}^{2} + 4 \cdot \left(\left(a \cdot a\right) \cdot \left(1 - a\right) + \left(b \cdot b\right) \cdot \left(3 + a\right)\right)\right) - 1 \]
      associate--l+ [=>] 99.8	\[ \color{blue}{{\left(a \cdot a + b \cdot b\right)}^{2} + \left(4 \cdot \left(\left(a \cdot a\right) \cdot \left(1 - a\right) + \left(b \cdot b\right) \cdot \left(3 + a\right)\right) - 1\right)} \]

   if +inf.0 < (+.f64 (pow.f64 (+.f64 (*.f64 a a) (*.f64 b b)) 2) (*.f64 4 (+.f64 (*.f64 (*.f64 a a) (-.f64 1 a)) (*.f64 (*.f64 b b) (+.f64 3 a)))))

   1. Initial program 0.0%

      \[\left({\left(a \cdot a + b \cdot b\right)}^{2} + 4 \cdot \left(\left(a \cdot a\right) \cdot \left(1 - a\right) + \left(b \cdot b\right) \cdot \left(3 + a\right)\right)\right) - 1 \]

   2. Simplified 14.8%

      \[\leadsto \color{blue}{{\left(\mathsf{hypot}\left(a, b\right)\right)}^{4} + \mathsf{fma}\left(4, \mathsf{fma}\left(b \cdot b, a + 3, a \cdot a\right) - {a}^{3}, -1\right)} \]
      Proof

      [Start] 0.0	\[ \left({\left(a \cdot a + b \cdot b\right)}^{2} + 4 \cdot \left(\left(a \cdot a\right) \cdot \left(1 - a\right) + \left(b \cdot b\right) \cdot \left(3 + a\right)\right)\right) - 1 \]
      associate--l+ [=>] 0.0	\[ \color{blue}{{\left(a \cdot a + b \cdot b\right)}^{2} + \left(4 \cdot \left(\left(a \cdot a\right) \cdot \left(1 - a\right) + \left(b \cdot b\right) \cdot \left(3 + a\right)\right) - 1\right)} \]

   3. Taylor expanded in a around inf 96.9%

      \[\leadsto \color{blue}{{a}^{4}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 99.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{\left(a \cdot a + b \cdot b\right)}^{2} + 4 \cdot \left(\left(a \cdot a\right) \cdot \left(1 - a\right) + \left(b \cdot b\right) \cdot \left(a + 3\right)\right) \leq \infty:\\ \;\;\;\;{\left(\mathsf{hypot}\left(a, b\right)\right)}^{4} + \mathsf{fma}\left(4, \mathsf{fma}\left(b \cdot b, a + 3, a \cdot a\right) - {a}^{3}, -1\right)\\ \mathbf{else}:\\ \;\;\;\;{a}^{4}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	97.9%
Cost	8457

\[\begin{array}{l} \mathbf{if}\;a \leq -1.65 \cdot 10^{+70} \lor \neg \left(a \leq 2.2 \cdot 10^{+73}\right):\\ \;\;\;\;{a}^{4}\\ \mathbf{else}:\\ \;\;\;\;\left({\left(a \cdot a + b \cdot b\right)}^{2} + 4 \cdot \left(\left(a \cdot a\right) \cdot \left(1 - a\right) + \left(b \cdot b\right) \cdot \left(a + 3\right)\right)\right) + -1\\ \end{array} \]

Alternative 2
Accuracy	93.8%
Cost	7560

\[\begin{array}{l} \mathbf{if}\;a \leq -0.000205:\\ \;\;\;\;\left({a}^{4} + \left(1 - a\right) \cdot \left(\left(a \cdot a\right) \cdot 4\right)\right) + -1\\ \mathbf{elif}\;a \leq 4.9 \cdot 10^{+48}:\\ \;\;\;\;\left({b}^{4} + \left(b \cdot b\right) \cdot \left(12 + a \cdot 4\right)\right) + -1\\ \mathbf{else}:\\ \;\;\;\;{a}^{4}\\ \end{array} \]

Alternative 3
Accuracy	93.8%
Cost	7428

\[\begin{array}{l} \mathbf{if}\;a \leq -7.5 \cdot 10^{-5}:\\ \;\;\;\;\left({a}^{4} + \left(1 - a\right) \cdot \left(\left(a \cdot a\right) \cdot 4\right)\right) + -1\\ \mathbf{elif}\;a \leq 4.1 \cdot 10^{+48}:\\ \;\;\;\;\left({b}^{4} + \left(b \cdot b\right) \cdot 12\right) + -1\\ \mathbf{else}:\\ \;\;\;\;{a}^{4}\\ \end{array} \]

Alternative 4
Accuracy	93.8%
Cost	7304

\[\begin{array}{l} \mathbf{if}\;a \leq -8.5 \cdot 10^{+20}:\\ \;\;\;\;{a}^{4}\\ \mathbf{elif}\;a \leq 4.1 \cdot 10^{+48}:\\ \;\;\;\;\left({b}^{4} + \left(b \cdot b\right) \cdot 12\right) + -1\\ \mathbf{else}:\\ \;\;\;\;{a}^{4}\\ \end{array} \]

Alternative 5
Accuracy	93.8%
Cost	6792

\[\begin{array}{l} \mathbf{if}\;a \leq -6.6 \cdot 10^{+20}:\\ \;\;\;\;{a}^{4}\\ \mathbf{elif}\;a \leq 5.4 \cdot 10^{+48}:\\ \;\;\;\;\left(b \cdot b\right) \cdot \left(b \cdot b + 12\right) + -1\\ \mathbf{else}:\\ \;\;\;\;{a}^{4}\\ \end{array} \]

Alternative 6
Accuracy	89.9%
Cost	968

\[\begin{array}{l} \mathbf{if}\;a \leq -4.3 \cdot 10^{+101}:\\ \;\;\;\;a \cdot \left(\left(a \cdot a\right) \cdot -4\right) + -1\\ \mathbf{elif}\;a \leq 3.5 \cdot 10^{+102}:\\ \;\;\;\;\left(b \cdot b\right) \cdot \left(b \cdot b + 12\right) + -1\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(a \cdot \left(a \cdot 4\right)\right) + -1\\ \end{array} \]

Alternative 7
Accuracy	59.7%
Cost	836

\[\begin{array}{l} \mathbf{if}\;a \leq 2.4:\\ \;\;\;\;a \cdot \left(4 \cdot \left(a - a \cdot a\right)\right) + -1\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(a \cdot \left(a \cdot 4\right)\right) + -1\\ \end{array} \]

Alternative 8
Accuracy	55.7%
Cost	708

\[\begin{array}{l} \mathbf{if}\;a \leq -1:\\ \;\;\;\;a \cdot \left(\left(a \cdot a\right) \cdot -4\right) + -1\\ \mathbf{else}:\\ \;\;\;\;\left(a \cdot a\right) \cdot 4 + -1\\ \end{array} \]

Alternative 9
Accuracy	59.5%
Cost	708

\[\begin{array}{l} \mathbf{if}\;a \leq 10^{-295}:\\ \;\;\;\;a \cdot \left(\left(a \cdot a\right) \cdot -4\right) + -1\\ \mathbf{else}:\\ \;\;\;\;a \cdot \left(a \cdot \left(a \cdot 4\right)\right) + -1\\ \end{array} \]

Alternative 10
Accuracy	51.6%
Cost	448

\[\left(a \cdot a\right) \cdot 4 + -1 \]

Reproduce

herbie shell --seed 2023160 
(FPCore (a b)
  :name "Bouland and Aaronson, Equation (24)"
  :precision binary64
  (- (+ (pow (+ (* a a) (* b b)) 2.0) (* 4.0 (+ (* (* a a) (- 1.0 a)) (* (* b b) (+ 3.0 a))))) 1.0))