?

Average Error: 0.01% → 0%
Time: 1.8s
Precision: binary64
Cost: 6720

?

\[ \begin{array}{c}[x, y] = \mathsf{sort}([x, y])\\ \end{array} \]
\[x \cdot x + y \cdot y \]
\[\mathsf{fma}\left(y, y, x \cdot x\right) \]
(FPCore (x y) :precision binary64 (+ (* x x) (* y y)))
(FPCore (x y) :precision binary64 (fma y y (* x x)))
double code(double x, double y) {
	return (x * x) + (y * y);
}
double code(double x, double y) {
	return fma(y, y, (x * x));
}
function code(x, y)
	return Float64(Float64(x * x) + Float64(y * y))
end
function code(x, y)
	return fma(y, y, Float64(x * x))
end
code[x_, y_] := N[(N[(x * x), $MachinePrecision] + N[(y * y), $MachinePrecision]), $MachinePrecision]
code[x_, y_] := N[(y * y + N[(x * x), $MachinePrecision]), $MachinePrecision]
x \cdot x + y \cdot y
\mathsf{fma}\left(y, y, x \cdot x\right)

Error?

Derivation?

  1. Initial program 0.01

    \[x \cdot x + y \cdot y \]
  2. Taylor expanded in x around 0 0.01

    \[\leadsto \color{blue}{{y}^{2} + {x}^{2}} \]
  3. Simplified0

    \[\leadsto \color{blue}{\mathsf{fma}\left(y, y, x \cdot x\right)} \]
    Proof

    [Start]0.01

    \[ {y}^{2} + {x}^{2} \]

    unpow2 [=>]0.01

    \[ \color{blue}{y \cdot y} + {x}^{2} \]

    unpow2 [=>]0.01

    \[ y \cdot y + \color{blue}{x \cdot x} \]

    fma-udef [<=]0

    \[ \color{blue}{\mathsf{fma}\left(y, y, x \cdot x\right)} \]
  4. Final simplification0

    \[\leadsto \mathsf{fma}\left(y, y, x \cdot x\right) \]

Alternatives

Alternative 1
Error12.74%
Cost589
\[\begin{array}{l} \mathbf{if}\;y \leq 1.08 \cdot 10^{-160} \lor \neg \left(y \leq 5.2 \cdot 10^{-71}\right) \land y \leq 2.25 \cdot 10^{-57}:\\ \;\;\;\;x \cdot x\\ \mathbf{else}:\\ \;\;\;\;y \cdot y\\ \end{array} \]
Alternative 2
Error0.01%
Cost448
\[x \cdot x + y \cdot y \]
Alternative 3
Error43.05%
Cost192
\[x \cdot x \]

Error

Reproduce?

herbie shell --seed 2023090 
(FPCore (x y)
  :name "Graphics.Rasterific.Linear:$cquadrance from Rasterific-0.6.1"
  :precision binary64
  (+ (* x x) (* y y)))