sintan (problem 3.4.5)

Percentage Accurate: 50.8% → 100.0%

Time: 13.3s

Alternatives: 5

Speedup: 67.6×

Specification

Math

\[\begin{array}{l} \\ \frac{x - \sin x}{x - \tan x} \end{array} \]

FPCore

(FPCore (x) :precision binary64 (/ (- x (sin x)) (- x (tan x))))

C

double code(double x) {
	return (x - sin(x)) / (x - tan(x));
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    code = (x - sin(x)) / (x - tan(x))
end function

Java

public static double code(double x) {
	return (x - Math.sin(x)) / (x - Math.tan(x));
}

Python

def code(x):
	return (x - math.sin(x)) / (x - math.tan(x))

Julia

function code(x)
	return Float64(Float64(x - sin(x)) / Float64(x - tan(x)))
end

MATLAB

function tmp = code(x)
	tmp = (x - sin(x)) / (x - tan(x));
end

Wolfram

code[x_] := N[(N[(x - N[Sin[x], $MachinePrecision]), $MachinePrecision] / N[(x - N[Tan[x], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Initial Program: 50.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{x - \sin x}{x - \tan x} \end{array} \]

FPCore

(FPCore (x) :precision binary64 (/ (- x (sin x)) (- x (tan x))))

C

double code(double x) {
	return (x - sin(x)) / (x - tan(x));
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    code = (x - sin(x)) / (x - tan(x))
end function

Java

public static double code(double x) {
	return (x - Math.sin(x)) / (x - Math.tan(x));
}

Python

def code(x):
	return (x - math.sin(x)) / (x - math.tan(x))

Julia

function code(x)
	return Float64(Float64(x - sin(x)) / Float64(x - tan(x)))
end

MATLAB

function tmp = code(x)
	tmp = (x - sin(x)) / (x - tan(x));
end

Wolfram

code[x_] := N[(N[(x - N[Sin[x], $MachinePrecision]), $MachinePrecision] / N[(x - N[Tan[x], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Alternative 1: 100.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 0.031:\\ \;\;\;\;\left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot \left(x \cdot x\right)\right) - 0.5\\ \mathbf{else}:\\ \;\;\;\;\frac{x - \sin x}{x - \tan x}\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (if (<= x 0.031)
   (- (+ (* -0.009642857142857142 (pow x 4.0)) (* 0.225 (* x x))) 0.5)
   (/ (- x (sin x)) (- x (tan x)))))

C

x = abs(x);
double code(double x) {
	double tmp;
	if (x <= 0.031) {
		tmp = ((-0.009642857142857142 * pow(x, 4.0)) + (0.225 * (x * x))) - 0.5;
	} else {
		tmp = (x - sin(x)) / (x - tan(x));
	}
	return tmp;
}

Fortran

NOTE: x should be positive before calling this function
real(8) function code(x)
    real(8), intent (in) :: x
    real(8) :: tmp
    if (x <= 0.031d0) then
        tmp = (((-0.009642857142857142d0) * (x ** 4.0d0)) + (0.225d0 * (x * x))) - 0.5d0
    else
        tmp = (x - sin(x)) / (x - tan(x))
    end if
    code = tmp
end function

Java

x = Math.abs(x);
public static double code(double x) {
	double tmp;
	if (x <= 0.031) {
		tmp = ((-0.009642857142857142 * Math.pow(x, 4.0)) + (0.225 * (x * x))) - 0.5;
	} else {
		tmp = (x - Math.sin(x)) / (x - Math.tan(x));
	}
	return tmp;
}

Python

x = abs(x)
def code(x):
	tmp = 0
	if x <= 0.031:
		tmp = ((-0.009642857142857142 * math.pow(x, 4.0)) + (0.225 * (x * x))) - 0.5
	else:
		tmp = (x - math.sin(x)) / (x - math.tan(x))
	return tmp

Julia

x = abs(x)
function code(x)
	tmp = 0.0
	if (x <= 0.031)
		tmp = Float64(Float64(Float64(-0.009642857142857142 * (x ^ 4.0)) + Float64(0.225 * Float64(x * x))) - 0.5);
	else
		tmp = Float64(Float64(x - sin(x)) / Float64(x - tan(x)));
	end
	return tmp
end

MATLAB

x = abs(x)
function tmp_2 = code(x)
	tmp = 0.0;
	if (x <= 0.031)
		tmp = ((-0.009642857142857142 * (x ^ 4.0)) + (0.225 * (x * x))) - 0.5;
	else
		tmp = (x - sin(x)) / (x - tan(x));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := If[LessEqual[x, 0.031], N[(N[(N[(-0.009642857142857142 * N[Power[x, 4.0], $MachinePrecision]), $MachinePrecision] + N[(0.225 * N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - 0.5), $MachinePrecision], N[(N[(x - N[Sin[x], $MachinePrecision]), $MachinePrecision] / N[(x - N[Tan[x], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < 0.031

   1. Initial program 38.2%

      \[\frac{x - \sin x}{x - \tan x} \]

   2. Taylor expanded in x around 0 63.3%

      \[\leadsto \color{blue}{\left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot {x}^{2}\right) - 0.5} \]
   3. Step-by-step derivation

      1. unpow2 63.3%

         \[\leadsto \left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot \color{blue}{\left(x \cdot x\right)}\right) - 0.5 \]

   4. Applied egg-rr 63.3%

      \[\leadsto \left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot \color{blue}{\left(x \cdot x\right)}\right) - 0.5 \]

   if 0.031 < x

   1. Initial program 99.9%

      \[\frac{x - \sin x}{x - \tan x} \]
3. Recombined 2 regimes into one program.

4. Final simplification 70.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 0.031:\\ \;\;\;\;\left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot \left(x \cdot x\right)\right) - 0.5\\ \mathbf{else}:\\ \;\;\;\;\frac{x - \sin x}{x - \tan x}\\ \end{array} \]

Alternative 2: 98.9% accurate, 1.8× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 3:\\ \;\;\;\;\left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot \left(x \cdot x\right)\right) - 0.5\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x)
 :precision binary64
 (if (<= x 3.0)
   (- (+ (* -0.009642857142857142 (pow x 4.0)) (* 0.225 (* x x))) 0.5)
   1.0))

C

x = abs(x);
double code(double x) {
	double tmp;
	if (x <= 3.0) {
		tmp = ((-0.009642857142857142 * pow(x, 4.0)) + (0.225 * (x * x))) - 0.5;
	} else {
		tmp = 1.0;
	}
	return tmp;
}

Fortran

NOTE: x should be positive before calling this function
real(8) function code(x)
    real(8), intent (in) :: x
    real(8) :: tmp
    if (x <= 3.0d0) then
        tmp = (((-0.009642857142857142d0) * (x ** 4.0d0)) + (0.225d0 * (x * x))) - 0.5d0
    else
        tmp = 1.0d0
    end if
    code = tmp
end function

Java

x = Math.abs(x);
public static double code(double x) {
	double tmp;
	if (x <= 3.0) {
		tmp = ((-0.009642857142857142 * Math.pow(x, 4.0)) + (0.225 * (x * x))) - 0.5;
	} else {
		tmp = 1.0;
	}
	return tmp;
}

Python

x = abs(x)
def code(x):
	tmp = 0
	if x <= 3.0:
		tmp = ((-0.009642857142857142 * math.pow(x, 4.0)) + (0.225 * (x * x))) - 0.5
	else:
		tmp = 1.0
	return tmp

Julia

x = abs(x)
function code(x)
	tmp = 0.0
	if (x <= 3.0)
		tmp = Float64(Float64(Float64(-0.009642857142857142 * (x ^ 4.0)) + Float64(0.225 * Float64(x * x))) - 0.5);
	else
		tmp = 1.0;
	end
	return tmp
end

MATLAB

x = abs(x)
function tmp_2 = code(x)
	tmp = 0.0;
	if (x <= 3.0)
		tmp = ((-0.009642857142857142 * (x ^ 4.0)) + (0.225 * (x * x))) - 0.5;
	else
		tmp = 1.0;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := If[LessEqual[x, 3.0], N[(N[(N[(-0.009642857142857142 * N[Power[x, 4.0], $MachinePrecision]), $MachinePrecision] + N[(0.225 * N[(x * x), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - 0.5), $MachinePrecision], 1.0]

Derivation

1. Split input into 2 regimes

2. if x < 3

   1. Initial program 38.2%

      \[\frac{x - \sin x}{x - \tan x} \]

   2. Taylor expanded in x around 0 63.3%

      \[\leadsto \color{blue}{\left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot {x}^{2}\right) - 0.5} \]
   3. Step-by-step derivation

      1. unpow2 63.3%

         \[\leadsto \left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot \color{blue}{\left(x \cdot x\right)}\right) - 0.5 \]

   4. Applied egg-rr 63.3%

      \[\leadsto \left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot \color{blue}{\left(x \cdot x\right)}\right) - 0.5 \]

   if 3 < x

   1. Initial program 99.9%

      \[\frac{x - \sin x}{x - \tan x} \]

   2. Taylor expanded in x around inf 98.2%

      \[\leadsto \color{blue}{1} \]
3. Recombined 2 regimes into one program.

4. Final simplification 69.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 3:\\ \;\;\;\;\left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot \left(x \cdot x\right)\right) - 0.5\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \]

Alternative 3: 98.8% accurate, 22.8× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 2.6:\\ \;\;\;\;0.225 \cdot \left(x \cdot x\right) - 0.5\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x) :precision binary64 (if (<= x 2.6) (- (* 0.225 (* x x)) 0.5) 1.0))

C

x = abs(x);
double code(double x) {
	double tmp;
	if (x <= 2.6) {
		tmp = (0.225 * (x * x)) - 0.5;
	} else {
		tmp = 1.0;
	}
	return tmp;
}

Fortran

NOTE: x should be positive before calling this function
real(8) function code(x)
    real(8), intent (in) :: x
    real(8) :: tmp
    if (x <= 2.6d0) then
        tmp = (0.225d0 * (x * x)) - 0.5d0
    else
        tmp = 1.0d0
    end if
    code = tmp
end function

Java

x = Math.abs(x);
public static double code(double x) {
	double tmp;
	if (x <= 2.6) {
		tmp = (0.225 * (x * x)) - 0.5;
	} else {
		tmp = 1.0;
	}
	return tmp;
}

Python

x = abs(x)
def code(x):
	tmp = 0
	if x <= 2.6:
		tmp = (0.225 * (x * x)) - 0.5
	else:
		tmp = 1.0
	return tmp

Julia

x = abs(x)
function code(x)
	tmp = 0.0
	if (x <= 2.6)
		tmp = Float64(Float64(0.225 * Float64(x * x)) - 0.5);
	else
		tmp = 1.0;
	end
	return tmp
end

MATLAB

x = abs(x)
function tmp_2 = code(x)
	tmp = 0.0;
	if (x <= 2.6)
		tmp = (0.225 * (x * x)) - 0.5;
	else
		tmp = 1.0;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := If[LessEqual[x, 2.6], N[(N[(0.225 * N[(x * x), $MachinePrecision]), $MachinePrecision] - 0.5), $MachinePrecision], 1.0]

Derivation

1. Split input into 2 regimes

2. if x < 2.60000000000000009

   1. Initial program 38.2%

      \[\frac{x - \sin x}{x - \tan x} \]

   2. Taylor expanded in x around 0 64.5%

      \[\leadsto \color{blue}{0.225 \cdot {x}^{2} - 0.5} \]
   3. Step-by-step derivation

      1. unpow2 63.3%

         \[\leadsto \left(-0.009642857142857142 \cdot {x}^{4} + 0.225 \cdot \color{blue}{\left(x \cdot x\right)}\right) - 0.5 \]

   4. Applied egg-rr 64.5%

      \[\leadsto 0.225 \cdot \color{blue}{\left(x \cdot x\right)} - 0.5 \]

   if 2.60000000000000009 < x

   1. Initial program 99.9%

      \[\frac{x - \sin x}{x - \tan x} \]

   2. Taylor expanded in x around inf 98.2%

      \[\leadsto \color{blue}{1} \]
3. Recombined 2 regimes into one program.

4. Final simplification 70.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 2.6:\\ \;\;\;\;0.225 \cdot \left(x \cdot x\right) - 0.5\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \]

Alternative 4: 98.5% accurate, 67.6× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ \begin{array}{l} \mathbf{if}\;x \leq 1.6:\\ \;\;\;\;-0.5\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x) :precision binary64 (if (<= x 1.6) -0.5 1.0))

C

x = abs(x);
double code(double x) {
	double tmp;
	if (x <= 1.6) {
		tmp = -0.5;
	} else {
		tmp = 1.0;
	}
	return tmp;
}

Fortran

NOTE: x should be positive before calling this function
real(8) function code(x)
    real(8), intent (in) :: x
    real(8) :: tmp
    if (x <= 1.6d0) then
        tmp = -0.5d0
    else
        tmp = 1.0d0
    end if
    code = tmp
end function

Java

x = Math.abs(x);
public static double code(double x) {
	double tmp;
	if (x <= 1.6) {
		tmp = -0.5;
	} else {
		tmp = 1.0;
	}
	return tmp;
}

Python

x = abs(x)
def code(x):
	tmp = 0
	if x <= 1.6:
		tmp = -0.5
	else:
		tmp = 1.0
	return tmp

Julia

x = abs(x)
function code(x)
	tmp = 0.0
	if (x <= 1.6)
		tmp = -0.5;
	else
		tmp = 1.0;
	end
	return tmp
end

MATLAB

x = abs(x)
function tmp_2 = code(x)
	tmp = 0.0;
	if (x <= 1.6)
		tmp = -0.5;
	else
		tmp = 1.0;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := If[LessEqual[x, 1.6], -0.5, 1.0]

Derivation

1. Split input into 2 regimes

2. if x < 1.6000000000000001

   1. Initial program 38.2%

      \[\frac{x - \sin x}{x - \tan x} \]

   2. Taylor expanded in x around 0 63.2%

      \[\leadsto \color{blue}{-0.5} \]

   if 1.6000000000000001 < x

   1. Initial program 99.9%

      \[\frac{x - \sin x}{x - \tan x} \]

   2. Taylor expanded in x around inf 98.2%

      \[\leadsto \color{blue}{1} \]
3. Recombined 2 regimes into one program.

4. Final simplification 69.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq 1.6:\\ \;\;\;\;-0.5\\ \mathbf{else}:\\ \;\;\;\;1\\ \end{array} \]

Alternative 5: 50.3% accurate, 207.0× speedup

Math

\[\begin{array}{l} x = |x|\\ \\ -0.5 \end{array} \]

FPCore

NOTE: x should be positive before calling this function
(FPCore (x) :precision binary64 -0.5)

C

x = abs(x);
double code(double x) {
	return -0.5;
}

Fortran

NOTE: x should be positive before calling this function
real(8) function code(x)
    real(8), intent (in) :: x
    code = -0.5d0
end function

Java

x = Math.abs(x);
public static double code(double x) {
	return -0.5;
}

Python

x = abs(x)
def code(x):
	return -0.5

Julia

x = abs(x)
function code(x)
	return -0.5
end

MATLAB

x = abs(x)
function tmp = code(x)
	tmp = -0.5;
end

Wolfram

NOTE: x should be positive before calling this function
code[x_] := -0.5

Derivation

1. Initial program 49.6%

   \[\frac{x - \sin x}{x - \tan x} \]

2. Taylor expanded in x around 0 51.9%

   \[\leadsto \color{blue}{-0.5} \]

3. Final simplification 51.9%

   \[\leadsto -0.5 \]

Reproduce

herbie shell --seed 2023305 
(FPCore (x)
  :name "sintan (problem 3.4.5)"
  :precision binary64
  (/ (- x (sin x)) (- x (tan x))))