Main:bigenough2 from A

Percentage Accurate: 100.0% → 100.0%

Time: 2.5s

Precision: binary64

Cost: 6720

• 20.6% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[x + y \cdot \left(z + x\right) \]

↓

\[\mathsf{fma}\left(y, x + z, x\right) \]

FPCore

(FPCore (x y z) :precision binary64 (+ x (* y (+ z x))))

↓

(FPCore (x y z) :precision binary64 (fma y (+ x z) x))

C

double code(double x, double y, double z) {
	return x + (y * (z + x));
}

↓

double code(double x, double y, double z) {
	return fma(y, (x + z), x);
}

Julia

function code(x, y, z)
	return Float64(x + Float64(y * Float64(z + x)))
end

↓

function code(x, y, z)
	return fma(y, Float64(x + z), x)
end

Wolfram

code[x_, y_, z_] := N[(x + N[(y * N[(z + x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

↓

code[x_, y_, z_] := N[(y * N[(x + z), $MachinePrecision] + x), $MachinePrecision]

Derivation

1. Initial program 100.0%

   \[x + y \cdot \left(z + x\right) \]

2. Simplified 100.0%

   \[\leadsto \color{blue}{\mathsf{fma}\left(y, x + z, x\right)} \]
   Step-by-step derivation

   [Start] 100.0	\[ x + y \cdot \left(z + x\right) \]
   +-commutative [=>] 100.0	\[ \color{blue}{y \cdot \left(z + x\right) + x} \]
   fma-def [=>] 100.0	\[ \color{blue}{\mathsf{fma}\left(y, z + x, x\right)} \]
   +-commutative [=>] 100.0	\[ \mathsf{fma}\left(y, \color{blue}{x + z}, x\right) \]

3. Final simplification 100.0%

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

Alternatives

Alternative 1
Accuracy	61.4%
Cost	1116

\[\begin{array}{l} \mathbf{if}\;y \leq -2 \cdot 10^{+274}:\\ \;\;\;\;y \cdot z\\ \mathbf{elif}\;y \leq -9 \cdot 10^{+234}:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;y \leq -1.7 \cdot 10^{+149}:\\ \;\;\;\;y \cdot z\\ \mathbf{elif}\;y \leq -1.3 \cdot 10^{+73}:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;y \leq -8.5 \cdot 10^{+45}:\\ \;\;\;\;y \cdot z\\ \mathbf{elif}\;y \leq -1:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;y \leq 2.3 \cdot 10^{-47}:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;y \cdot z\\ \end{array} \]

Alternative 2
Accuracy	84.7%
Cost	585

\[\begin{array}{l} \mathbf{if}\;y \leq -4.1 \cdot 10^{-10} \lor \neg \left(y \leq 5.8 \cdot 10^{-54}\right):\\ \;\;\;\;y \cdot \left(x + z\right)\\ \mathbf{else}:\\ \;\;\;\;x\\ \end{array} \]

Alternative 3
Accuracy	99.0%
Cost	585

\[\begin{array}{l} \mathbf{if}\;y \leq -1 \lor \neg \left(y \leq 1\right):\\ \;\;\;\;y \cdot \left(x + z\right)\\ \mathbf{else}:\\ \;\;\;\;x + y \cdot z\\ \end{array} \]

Alternative 4
Accuracy	60.3%
Cost	456

\[\begin{array}{l} \mathbf{if}\;y \leq -1:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;y \leq 6.1 \cdot 10^{+29}:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;y \cdot x\\ \end{array} \]

Alternative 5
Accuracy	100.0%
Cost	448

\[x + y \cdot \left(x + z\right) \]

Alternative 6
Accuracy	37.3%
Cost	64

\[x \]

Reproduce

herbie shell --seed 2023160 
(FPCore (x y z)
  :name "Main:bigenough2 from A"
  :precision binary64
  (+ x (* y (+ z x))))