Expression 3, p15

Average Accuracy: 100.0% → 100.0%

Time: 2.4s

Precision: binary64

Cost: 13056

\[0 \leq x \land x \leq 2\]

Math

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

↓

\[\mathsf{fma}\left(x, x, {x}^{3}\right) \]

FPCore

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

↓

(FPCore (x) :precision binary64 (fma x x (pow x 3.0)))

C

double code(double x) {
	return (x * (x * x)) + (x * x);
}

↓

double code(double x) {
	return fma(x, x, pow(x, 3.0));
}

Julia

function code(x)
	return Float64(Float64(x * Float64(x * x)) + Float64(x * x))
end

↓

function code(x)
	return fma(x, x, (x ^ 3.0))
end

Wolfram

code[x_] := N[(N[(x * N[(x * x), $MachinePrecision]), $MachinePrecision] + N[(x * x), $MachinePrecision]), $MachinePrecision]

↓

code[x_] := N[(x * x + N[Power[x, 3.0], $MachinePrecision]), $MachinePrecision]

Target

Original	100.0%
Target	100.0%
Herbie	100.0%

\[\left(\left(1 + x\right) \cdot x\right) \cdot x \]

Derivation

1. Initial program 100.0%

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

2. Taylor expanded in x around 0 100.0%

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

3. Simplified 100.0%

   \[\leadsto \color{blue}{\mathsf{fma}\left(x, x, {x}^{3}\right)} \]
   Proof

   [Start] 100.0	\[ {x}^{2} + {x}^{3} \]
   unpow2 [=>] 100.0	\[ \color{blue}{x \cdot x} + {x}^{3} \]
   fma-udef [<=] 100.0	\[ \color{blue}{\mathsf{fma}\left(x, x, {x}^{3}\right)} \]

4. Final simplification 100.0%

   \[\leadsto \mathsf{fma}\left(x, x, {x}^{3}\right) \]

Alternatives

Alternative 1
Accuracy	100.0%
Cost	448

\[\left(x + 1\right) \cdot \left(x \cdot x\right) \]

Alternative 2
Accuracy	97.8%
Cost	192

\[x \cdot x \]

Reproduce

herbie shell --seed 2023152 
(FPCore (x)
  :name "Expression 3, p15"
  :precision binary64
  :pre (and (<= 0.0 x) (<= x 2.0))

  :herbie-target
  (* (* (+ 1.0 x) x) x)

  (+ (* x (* x x)) (* x x)))