Data.Histogram.Bin.BinF:$cfromIndex from histogram-fill-0.8.4.1

Average Accuracy: 100.0% → 100.0%

Time: 6.9s

Precision: binary64

Cost: 6720

Math

\[\left(\frac{x}{2} + y \cdot x\right) + z \]

↓

\[\mathsf{fma}\left(0.5 + y, x, z\right) \]

FPCore

(FPCore (x y z) :precision binary64 (+ (+ (/ x 2.0) (* y x)) z))

↓

(FPCore (x y z) :precision binary64 (fma (+ 0.5 y) x z))

C

double code(double x, double y, double z) {
	return ((x / 2.0) + (y * x)) + z;
}

↓

double code(double x, double y, double z) {
	return fma((0.5 + y), x, z);
}

Julia

function code(x, y, z)
	return Float64(Float64(Float64(x / 2.0) + Float64(y * x)) + z)
end

↓

function code(x, y, z)
	return fma(Float64(0.5 + y), x, z)
end

Wolfram

code[x_, y_, z_] := N[(N[(N[(x / 2.0), $MachinePrecision] + N[(y * x), $MachinePrecision]), $MachinePrecision] + z), $MachinePrecision]

↓

code[x_, y_, z_] := N[(N[(0.5 + y), $MachinePrecision] * x + z), $MachinePrecision]

Derivation

1. Initial program 100.0%

   \[\left(\frac{x}{2} + y \cdot x\right) + z \]

2. Simplified 100.0%

   \[\leadsto \color{blue}{\frac{x}{2} + \mathsf{fma}\left(x, y, z\right)} \]
   Proof

   [Start] 100.0	\[ \left(\frac{x}{2} + y \cdot x\right) + z \]
   associate-+l+ [=>] 100.0	\[ \color{blue}{\frac{x}{2} + \left(y \cdot x + z\right)} \]
   *-commutative [=>] 100.0	\[ \frac{x}{2} + \left(\color{blue}{x \cdot y} + z\right) \]
   fma-def [=>] 100.0	\[ \frac{x}{2} + \color{blue}{\mathsf{fma}\left(x, y, z\right)} \]

3. Taylor expanded in x around 0 100.0%

   \[\leadsto \color{blue}{z + \left(0.5 + y\right) \cdot x} \]

4. Simplified 100.0%

   \[\leadsto \color{blue}{\mathsf{fma}\left(0.5 + y, x, z\right)} \]
   Proof

   [Start] 100.0	\[ z + \left(0.5 + y\right) \cdot x \]
   +-commutative [=>] 100.0	\[ \color{blue}{\left(0.5 + y\right) \cdot x + z} \]
   fma-def [=>] 100.0	\[ \color{blue}{\mathsf{fma}\left(0.5 + y, x, z\right)} \]

5. Final simplification 100.0%

   \[\leadsto \mathsf{fma}\left(0.5 + y, x, z\right) \]

Alternatives

Alternative 1
Accuracy	57.0%
Cost	1380

\[\begin{array}{l} \mathbf{if}\;z \leq -380000000000:\\ \;\;\;\;z\\ \mathbf{elif}\;z \leq -5 \cdot 10^{-239}:\\ \;\;\;\;0.5 \cdot x\\ \mathbf{elif}\;z \leq -1.45 \cdot 10^{-297}:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;z \leq 6 \cdot 10^{-223}:\\ \;\;\;\;0.5 \cdot x\\ \mathbf{elif}\;z \leq 1.25 \cdot 10^{-170}:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;z \leq 5.9 \cdot 10^{-37}:\\ \;\;\;\;0.5 \cdot x\\ \mathbf{elif}\;z \leq 1.2:\\ \;\;\;\;y \cdot x\\ \mathbf{elif}\;z \leq 7.5 \cdot 10^{+142}:\\ \;\;\;\;z\\ \mathbf{elif}\;z \leq 8.5 \cdot 10^{+158}:\\ \;\;\;\;0.5 \cdot x\\ \mathbf{else}:\\ \;\;\;\;z\\ \end{array} \]

Alternative 2
Accuracy	74.2%
Cost	849

\[\begin{array}{l} \mathbf{if}\;z \leq -76000000000000:\\ \;\;\;\;z\\ \mathbf{elif}\;z \leq 1.25 \cdot 10^{+53} \lor \neg \left(z \leq 6.2 \cdot 10^{+101}\right) \land z \leq 8.5 \cdot 10^{+158}:\\ \;\;\;\;\left(0.5 + y\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;z\\ \end{array} \]

Alternative 3
Accuracy	57.7%
Cost	720

\[\begin{array}{l} \mathbf{if}\;z \leq -740000000000:\\ \;\;\;\;z\\ \mathbf{elif}\;z \leq 7.6 \cdot 10^{-41}:\\ \;\;\;\;0.5 \cdot x\\ \mathbf{elif}\;z \leq 7.5 \cdot 10^{+142}:\\ \;\;\;\;z\\ \mathbf{elif}\;z \leq 8.5 \cdot 10^{+158}:\\ \;\;\;\;0.5 \cdot x\\ \mathbf{else}:\\ \;\;\;\;z\\ \end{array} \]

Alternative 4
Accuracy	83.0%
Cost	585

\[\begin{array}{l} \mathbf{if}\;x \leq -3.8 \cdot 10^{+23} \lor \neg \left(x \leq 86000000000\right):\\ \;\;\;\;\left(0.5 + y\right) \cdot x\\ \mathbf{else}:\\ \;\;\;\;z + y \cdot x\\ \end{array} \]

Alternative 5
Accuracy	98.7%
Cost	585

\[\begin{array}{l} \mathbf{if}\;y \leq -35 \lor \neg \left(y \leq 6.2 \cdot 10^{-9}\right):\\ \;\;\;\;z + y \cdot x\\ \mathbf{else}:\\ \;\;\;\;z + \frac{x}{2}\\ \end{array} \]

Alternative 6
Accuracy	100.0%
Cost	448

\[z + \left(0.5 + y\right) \cdot x \]

Alternative 7
Accuracy	46.6%
Cost	64

\[z \]

Reproduce

herbie shell --seed 2023122 
(FPCore (x y z)
  :name "Data.Histogram.Bin.BinF:$cfromIndex from histogram-fill-0.8.4.1"
  :precision binary64
  (+ (+ (/ x 2.0) (* y x)) z))