Bouland and Aaronson, Equation (26)

Average Accuracy: 99.9% → 100.0%

Time: 5.2s

Precision: binary64

Cost: 19776

• 61.6% of points produce a very large (infinite) output. You may want to add a precondition.

Math

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

↓

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

FPCore

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

↓

(FPCore (a b)
 :precision binary64
 (+ (pow (hypot a b) 4.0) (fma b (* b 4.0) -1.0)))

C

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

↓

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

Julia

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

↓

function code(a, b)
	return Float64((hypot(a, b) ^ 4.0) + fma(b, Float64(b * 4.0), -1.0))
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[(b * b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - 1.0), $MachinePrecision]

↓

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

Derivation

1. Initial program 99.9%

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

2. Simplified 100.0%

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

   [Start] 99.9	\[ \left({\left(a \cdot a + b \cdot b\right)}^{2} + 4 \cdot \left(b \cdot b\right)\right) - 1 \]
   associate--l+ [=>] 99.9	\[ \color{blue}{{\left(a \cdot a + b \cdot b\right)}^{2} + \left(4 \cdot \left(b \cdot b\right) - 1\right)} \]
   unpow2 [=>] 99.9	\[ \color{blue}{\left(a \cdot a + b \cdot b\right) \cdot \left(a \cdot a + b \cdot b\right)} + \left(4 \cdot \left(b \cdot b\right) - 1\right) \]
   unpow1 [<=] 99.9	\[ \left(a \cdot a + b \cdot b\right) \cdot \color{blue}{{\left(a \cdot a + b \cdot b\right)}^{1}} + \left(4 \cdot \left(b \cdot b\right) - 1\right) \]
   sqr-pow [=>] 99.9	\[ \left(a \cdot a + b \cdot b\right) \cdot \color{blue}{\left({\left(a \cdot a + b \cdot b\right)}^{\left(\frac{1}{2}\right)} \cdot {\left(a \cdot a + b \cdot b\right)}^{\left(\frac{1}{2}\right)}\right)} + \left(4 \cdot \left(b \cdot b\right) - 1\right) \]
   associate-*r* [=>] 99.9	\[ \color{blue}{\left(\left(a \cdot a + b \cdot b\right) \cdot {\left(a \cdot a + b \cdot b\right)}^{\left(\frac{1}{2}\right)}\right) \cdot {\left(a \cdot a + b \cdot b\right)}^{\left(\frac{1}{2}\right)}} + \left(4 \cdot \left(b \cdot b\right) - 1\right) \]

3. Final simplification 100.0%

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

Alternatives

Alternative 1
Accuracy	99.9%
Cost	7424

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

Alternative 2
Accuracy	94.1%
Cost	7304

\[\begin{array}{l} \mathbf{if}\;b \leq -1.26 \cdot 10^{+39}:\\ \;\;\;\;\left(b \cdot b\right) \cdot \left(4 + b \cdot b\right) + -1\\ \mathbf{elif}\;b \leq 6.2:\\ \;\;\;\;{a}^{4} + -1\\ \mathbf{else}:\\ \;\;\;\;\left(4 \cdot \left(b \cdot b\right) + {b}^{4}\right) + -1\\ \end{array} \]

Alternative 3
Accuracy	94.2%
Cost	6920

\[\begin{array}{l} \mathbf{if}\;b \leq -2.25 \cdot 10^{+39}:\\ \;\;\;\;\left(b \cdot b\right) \cdot \left(4 + b \cdot b\right) + -1\\ \mathbf{elif}\;b \leq 250000000000:\\ \;\;\;\;{a}^{4} + -1\\ \mathbf{else}:\\ \;\;\;\;{b}^{4}\\ \end{array} \]

Alternative 4
Accuracy	93.9%
Cost	6792

\[\begin{array}{l} \mathbf{if}\;a \leq -2.8 \cdot 10^{+64}:\\ \;\;\;\;{a}^{4}\\ \mathbf{elif}\;a \leq 2.05 \cdot 10^{+16}:\\ \;\;\;\;\left(b \cdot b\right) \cdot \left(4 + b \cdot b\right) + -1\\ \mathbf{else}:\\ \;\;\;\;{a}^{4}\\ \end{array} \]

Alternative 5
Accuracy	68.9%
Cost	704

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

Alternative 6
Accuracy	50.1%
Cost	585

\[\begin{array}{l} \mathbf{if}\;b \leq -1.8 \cdot 10^{-11} \lor \neg \left(b \leq 0.0125\right):\\ \;\;\;\;4 \cdot \left(b \cdot b\right)\\ \mathbf{else}:\\ \;\;\;\;-1\\ \end{array} \]

Alternative 7
Accuracy	50.5%
Cost	448

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

Alternative 8
Accuracy	24.1%
Cost	64

\[-1 \]

Reproduce

herbie shell --seed 2023159 
(FPCore (a b)
  :name "Bouland and Aaronson, Equation (26)"
  :precision binary64
  (- (+ (pow (+ (* a a) (* b b)) 2.0) (* 4.0 (* b b))) 1.0))