Difference of squares

Percentage Accurate: 93.8% → 96.8%

Time: 2.9s

Precision: binary64

Cost: 7044

• 43.8% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[a \cdot a - b \cdot b \]

↓

\[\begin{array}{l} \mathbf{if}\;a \cdot a \leq 5 \cdot 10^{+304}:\\ \;\;\;\;\mathsf{fma}\left(a, a, b \cdot \left(-b\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

FPCore

(FPCore (a b) :precision binary64 (- (* a a) (* b b)))

↓

(FPCore (a b)
 :precision binary64
 (if (<= (* a a) 5e+304) (fma a a (* b (- b))) (* a a)))

C

double code(double a, double b) {
	return (a * a) - (b * b);
}

↓

double code(double a, double b) {
	double tmp;
	if ((a * a) <= 5e+304) {
		tmp = fma(a, a, (b * -b));
	} else {
		tmp = a * a;
	}
	return tmp;
}

Julia

function code(a, b)
	return Float64(Float64(a * a) - Float64(b * b))
end

↓

function code(a, b)
	tmp = 0.0
	if (Float64(a * a) <= 5e+304)
		tmp = fma(a, a, Float64(b * Float64(-b)));
	else
		tmp = Float64(a * a);
	end
	return tmp
end

Wolfram

code[a_, b_] := N[(N[(a * a), $MachinePrecision] - N[(b * b), $MachinePrecision]), $MachinePrecision]

↓

code[a_, b_] := If[LessEqual[N[(a * a), $MachinePrecision], 5e+304], N[(a * a + N[(b * (-b)), $MachinePrecision]), $MachinePrecision], N[(a * a), $MachinePrecision]]

Target

Original	93.8%
Target	100.0%
Herbie	96.8%

\[\left(a + b\right) \cdot \left(a - b\right) \]

Derivation

1. Split input into 2 regimes

2. if (*.f64 a a) < 4.9999999999999997e304

   1. Initial program 100.0%

      \[a \cdot a - b \cdot b \]

   2. Simplified 100.0%

      \[\leadsto \color{blue}{\mathsf{fma}\left(a, a, b \cdot \left(-b\right)\right)} \]
      Step-by-step derivation

      [Start] 100.0%	\[ a \cdot a - b \cdot b \]
      fma-neg [=>] 100.0%	\[ \color{blue}{\mathsf{fma}\left(a, a, -b \cdot b\right)} \]
      distribute-rgt-neg-in [=>] 100.0%	\[ \mathsf{fma}\left(a, a, \color{blue}{b \cdot \left(-b\right)}\right) \]

   if 4.9999999999999997e304 < (*.f64 a a)

   1. Initial program 85.5%

      \[a \cdot a - b \cdot b \]

   2. Taylor expanded in a around inf 96.8%

      \[\leadsto \color{blue}{{a}^{2}} \]

   3. Simplified 96.8%

      \[\leadsto \color{blue}{a \cdot a} \]
      Step-by-step derivation

      [Start] 96.8%	\[ {a}^{2} \]
      unpow2 [=>] 96.8%	\[ \color{blue}{a \cdot a} \]

3. Recombined 2 regimes into one program.

4. Final simplification 99.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot a \leq 5 \cdot 10^{+304}:\\ \;\;\;\;\mathsf{fma}\left(a, a, b \cdot \left(-b\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	96.8%
Cost	7044

\[\begin{array}{l} \mathbf{if}\;a \cdot a \leq 5 \cdot 10^{+304}:\\ \;\;\;\;\mathsf{fma}\left(a, a, b \cdot \left(-b\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 2
Accuracy	77.3%
Cost	1037

\[\begin{array}{l} \mathbf{if}\;a \cdot a \leq 6.8 \cdot 10^{-64} \lor \neg \left(a \cdot a \leq 0.15\right) \land a \cdot a \leq 8.6 \cdot 10^{+130}:\\ \;\;\;\;b \cdot \left(-b\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 3
Accuracy	96.8%
Cost	708

\[\begin{array}{l} \mathbf{if}\;a \cdot a \leq 10^{+306}:\\ \;\;\;\;a \cdot a - b \cdot b\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 4
Accuracy	53.8%
Cost	192

\[a \cdot a \]

Reproduce

herbie shell --seed 2023243 
(FPCore (a b)
  :name "Difference of squares"
  :precision binary64

  :herbie-target
  (* (+ a b) (- a b))

  (- (* a a) (* b b)))