Difference of squares

Percentage Accurate: 93.7% → 96.2%

Time: 7.2s

Precision: binary64

Cost: 708

• 43.1% of points produce a very large (infinite) output. You may want to add a precondition.

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;a \cdot a \leq 4 \cdot 10^{+294}:\\ \;\;\;\;a \cdot a - b \cdot b\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

FPCore

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

↓

(FPCore (a b)
 :precision binary64
 (if (<= (* a a) 4e+294) (- (* 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) <= 4e+294) {
		tmp = (a * a) - (b * b);
	} else {
		tmp = a * a;
	}
	return tmp;
}

Fortran

real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = (a * a) - (b * b)
end function

↓

real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8) :: tmp
    if ((a * a) <= 4d+294) then
        tmp = (a * a) - (b * b)
    else
        tmp = a * a
    end if
    code = tmp
end function

Java

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

↓

public static double code(double a, double b) {
	double tmp;
	if ((a * a) <= 4e+294) {
		tmp = (a * a) - (b * b);
	} else {
		tmp = a * a;
	}
	return tmp;
}

Python

def code(a, b):
	return (a * a) - (b * b)

↓

def code(a, b):
	tmp = 0
	if (a * a) <= 4e+294:
		tmp = (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) <= 4e+294)
		tmp = Float64(Float64(a * a) - Float64(b * b));
	else
		tmp = Float64(a * a);
	end
	return tmp
end

MATLAB

function tmp = code(a, b)
	tmp = (a * a) - (b * b);
end

↓

function tmp_2 = code(a, b)
	tmp = 0.0;
	if ((a * a) <= 4e+294)
		tmp = (a * a) - (b * b);
	else
		tmp = a * a;
	end
	tmp_2 = 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], 4e+294], N[(N[(a * a), $MachinePrecision] - N[(b * b), $MachinePrecision]), $MachinePrecision], N[(a * a), $MachinePrecision]]

Target

Original	93.7%
Target	100.0%
Herbie	96.2%

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

Derivation

1. Split input into 2 regimes

2. if (*.f64 a a) < 4.00000000000000027e294

   1. Initial program 100.0%

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

   if 4.00000000000000027e294 < (*.f64 a a)

   1. Initial program 73.9%

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

   2. Taylor expanded in a around inf 92.8%

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

   3. Simplified 92.8%

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

      [Start] 92.8%	\[ {a}^{2} \]
      unpow2 [=>] 92.8%	\[ \color{blue}{a \cdot a} \]

3. Recombined 2 regimes into one program.

4. Final simplification 98.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot a \leq 4 \cdot 10^{+294}:\\ \;\;\;\;a \cdot a - b \cdot b\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	96.2%
Cost	708

\[\begin{array}{l} \mathbf{if}\;a \cdot a \leq 4 \cdot 10^{+294}:\\ \;\;\;\;a \cdot a - b \cdot b\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 2
Accuracy	76.4%
Cost	520

\[\begin{array}{l} \mathbf{if}\;a \leq -1.8 \cdot 10^{+63}:\\ \;\;\;\;a \cdot a\\ \mathbf{elif}\;a \leq 1.12 \cdot 10^{-72}:\\ \;\;\;\;b \cdot \left(-b\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 3
Accuracy	54.0%
Cost	192

\[a \cdot a \]

Reproduce

herbie shell --seed 2023277 
(FPCore (a b)
  :name "Difference of squares"
  :precision binary64

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

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