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

Percentage Accurate: 100.0% → 100.0%

Time: 6.8s

Precision: binary64

Cost: 6720

Math

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

↓

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

FPCore

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

↓

(FPCore (x y z) :precision binary64 (fma x (+ y 0.5) 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(x, (y + 0.5), 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(x, Float64(y + 0.5), 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[(x * N[(y + 0.5), $MachinePrecision] + z), $MachinePrecision]

Derivation

1. Initial program 100.0%

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

2. Simplified 100.0%

   \[\leadsto \color{blue}{\frac{x}{2} + \left(x \cdot y + z\right)} \]
   Step-by-step derivation

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

3. Taylor expanded in x around 0 100.0%

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

4. Simplified 100.0%

   \[\leadsto \color{blue}{\mathsf{fma}\left(x, y + 0.5, z\right)} \]
   Step-by-step derivation

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

5. Final simplification 100.0%

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

Alternatives

Alternative 1
Accuracy	100.0%
Cost	6720

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

Alternative 2
Accuracy	60.2%
Cost	1116

\[\begin{array}{l} \mathbf{if}\;y \leq -0.5:\\ \;\;\;\;x \cdot y\\ \mathbf{elif}\;y \leq -2.16 \cdot 10^{-128}:\\ \;\;\;\;x \cdot 0.5\\ \mathbf{elif}\;y \leq -1.08 \cdot 10^{-250}:\\ \;\;\;\;z\\ \mathbf{elif}\;y \leq 7.4 \cdot 10^{-160}:\\ \;\;\;\;x \cdot 0.5\\ \mathbf{elif}\;y \leq 2.5 \cdot 10^{-123}:\\ \;\;\;\;z\\ \mathbf{elif}\;y \leq 1.4 \cdot 10^{-47}:\\ \;\;\;\;x \cdot 0.5\\ \mathbf{elif}\;y \leq 8.8 \cdot 10^{+22}:\\ \;\;\;\;z\\ \mathbf{else}:\\ \;\;\;\;x \cdot y\\ \end{array} \]

Alternative 3
Accuracy	73.4%
Cost	849

\[\begin{array}{l} \mathbf{if}\;z \leq -2.6 \cdot 10^{+120}:\\ \;\;\;\;z\\ \mathbf{elif}\;z \leq -2.7 \cdot 10^{+52} \lor \neg \left(z \leq -3.9 \cdot 10^{+27}\right) \land z \leq 6.8 \cdot 10^{+131}:\\ \;\;\;\;x \cdot \left(y + 0.5\right)\\ \mathbf{else}:\\ \;\;\;\;z\\ \end{array} \]

Alternative 4
Accuracy	83.6%
Cost	585

\[\begin{array}{l} \mathbf{if}\;z \leq -1.12 \cdot 10^{-8} \lor \neg \left(z \leq 3.2 \cdot 10^{+77}\right):\\ \;\;\;\;z + x \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(y + 0.5\right)\\ \end{array} \]

Alternative 5
Accuracy	98.9%
Cost	585

\[\begin{array}{l} \mathbf{if}\;y \leq -0.5 \lor \neg \left(y \leq 1.6 \cdot 10^{-8}\right):\\ \;\;\;\;z + x \cdot y\\ \mathbf{else}:\\ \;\;\;\;z + \frac{x}{2}\\ \end{array} \]

Alternative 6
Accuracy	48.9%
Cost	456

\[\begin{array}{l} \mathbf{if}\;z \leq -1.28 \cdot 10^{-8}:\\ \;\;\;\;z\\ \mathbf{elif}\;z \leq 1.7 \cdot 10^{+85}:\\ \;\;\;\;x \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;z\\ \end{array} \]

Alternative 7
Accuracy	100.0%
Cost	448

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

Alternative 8
Accuracy	41.1%
Cost	64

\[z \]

Reproduce

herbie shell --seed 2023272 
(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))