Examples.Basics.BasicTests:f2 from sbv-4.4

Percentage Accurate: 93.7% → 96.2%

Time: 7.3s

Precision: binary64

Cost: 708

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

Math

\[x \cdot x - y \cdot y \]

↓

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

FPCore

(FPCore (x y) :precision binary64 (- (* x x) (* y y)))

↓

(FPCore (x y)
 :precision binary64
 (if (<= (* x x) 4e+294) (- (* x x) (* y y)) (* x x)))

C

double code(double x, double y) {
	return (x * x) - (y * y);
}

↓

double code(double x, double y) {
	double tmp;
	if ((x * x) <= 4e+294) {
		tmp = (x * x) - (y * y);
	} else {
		tmp = x * x;
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = (x * x) - (y * y)
end function

↓

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

Java

public static double code(double x, double y) {
	return (x * x) - (y * y);
}

↓

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

Python

def code(x, y):
	return (x * x) - (y * y)

↓

def code(x, y):
	tmp = 0
	if (x * x) <= 4e+294:
		tmp = (x * x) - (y * y)
	else:
		tmp = x * x
	return tmp

Julia

function code(x, y)
	return Float64(Float64(x * x) - Float64(y * y))
end

↓

function code(x, y)
	tmp = 0.0
	if (Float64(x * x) <= 4e+294)
		tmp = Float64(Float64(x * x) - Float64(y * y));
	else
		tmp = Float64(x * x);
	end
	return tmp
end

MATLAB

function tmp = code(x, y)
	tmp = (x * x) - (y * y);
end

↓

function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((x * x) <= 4e+294)
		tmp = (x * x) - (y * y);
	else
		tmp = x * x;
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := N[(N[(x * x), $MachinePrecision] - N[(y * y), $MachinePrecision]), $MachinePrecision]

↓

code[x_, y_] := If[LessEqual[N[(x * x), $MachinePrecision], 4e+294], N[(N[(x * x), $MachinePrecision] - N[(y * y), $MachinePrecision]), $MachinePrecision], N[(x * x), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

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

   1. Initial program 100.0%

      \[x \cdot x - y \cdot y \]

   if 4.00000000000000027e294 < (*.f64 x x)

   1. Initial program 73.9%

      \[x \cdot x - y \cdot y \]

   2. Taylor expanded in x around inf 92.8%

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

   3. Simplified 92.8%

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

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

3. Recombined 2 regimes into one program.

4. Final simplification 98.0%

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

Alternatives

Alternative 1
Accuracy	96.2%
Cost	708

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

Alternative 2
Accuracy	76.4%
Cost	520

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

Alternative 3
Accuracy	54.0%
Cost	192

\[x \cdot x \]

Reproduce

herbie shell --seed 2023277 
(FPCore (x y)
  :name "Examples.Basics.BasicTests:f2 from sbv-4.4"
  :precision binary64
  (- (* x x) (* y y)))