Difference of squares

Percentage Accurate: 93.9% → 98.2%

Time: 1.9s

Alternatives: 4

Speedup: 0.8×

• 44.4% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

\[\begin{array}{l} \\ a \cdot a - b \cdot b \end{array} \]

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 93.9% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ a \cdot a - b \cdot b \end{array} \]

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 98.2% accurate, 0.1× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} \mathbf{if}\;a \leq 8 \cdot 10^{+221}:\\ \;\;\;\;\mathsf{fma}\left(a, a, b \cdot \left(-b\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b)
 :precision binary64
 (if (<= a 8e+221) (fma a a (* b (- b))) (* a a)))

C

a = abs(a);
double code(double a, double b) {
	double tmp;
	if (a <= 8e+221) {
		tmp = fma(a, a, (b * -b));
	} else {
		tmp = a * a;
	}
	return tmp;
}

Julia

a = abs(a)
function code(a, b)
	tmp = 0.0
	if (a <= 8e+221)
		tmp = fma(a, a, Float64(b * Float64(-b)));
	else
		tmp = Float64(a * a);
	end
	return tmp
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_] := If[LessEqual[a, 8e+221], N[(a * a + N[(b * (-b)), $MachinePrecision]), $MachinePrecision], N[(a * a), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if a < 8.0000000000000004e221

   1. Initial program 95.7%

      \[a \cdot a - b \cdot b \]
   2. Step-by-step derivation

      1. fma-neg 98.3%

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

      2. distribute-rgt-neg-in 98.3%

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

   3. Simplified 98.3%

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

   if 8.0000000000000004e221 < a

   1. Initial program 90.5%

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

   2. Taylor expanded in a around inf 100.0%

      \[\leadsto \color{blue}{{a}^{2}} \]
   3. Step-by-step derivation

      1. unpow2 100.0%

         \[\leadsto \color{blue}{a \cdot a} \]

   4. Simplified 100.0%

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

4. Final simplification 98.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 8 \cdot 10^{+221}:\\ \;\;\;\;\mathsf{fma}\left(a, a, b \cdot \left(-b\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 2: 77.2% accurate, 0.4× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} \mathbf{if}\;a \cdot a \leq 1.8 \cdot 10^{-61} \lor \neg \left(a \cdot a \leq 6500000000000\right) \land a \cdot a \leq 3.8 \cdot 10^{+76}:\\ \;\;\;\;b \cdot \left(-b\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b)
 :precision binary64
 (if (or (<= (* a a) 1.8e-61)
         (and (not (<= (* a a) 6500000000000.0)) (<= (* a a) 3.8e+76)))
   (* b (- b))
   (* a a)))

C

a = abs(a);
double code(double a, double b) {
	double tmp;
	if (((a * a) <= 1.8e-61) || (!((a * a) <= 6500000000000.0) && ((a * a) <= 3.8e+76))) {
		tmp = b * -b;
	} else {
		tmp = a * a;
	}
	return tmp;
}

Fortran

NOTE: a should be positive before calling this function
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8) :: tmp
    if (((a * a) <= 1.8d-61) .or. (.not. ((a * a) <= 6500000000000.0d0)) .and. ((a * a) <= 3.8d+76)) then
        tmp = b * -b
    else
        tmp = a * a
    end if
    code = tmp
end function

Java

a = Math.abs(a);
public static double code(double a, double b) {
	double tmp;
	if (((a * a) <= 1.8e-61) || (!((a * a) <= 6500000000000.0) && ((a * a) <= 3.8e+76))) {
		tmp = b * -b;
	} else {
		tmp = a * a;
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b):
	tmp = 0
	if ((a * a) <= 1.8e-61) or (not ((a * a) <= 6500000000000.0) and ((a * a) <= 3.8e+76)):
		tmp = b * -b
	else:
		tmp = a * a
	return tmp

Julia

a = abs(a)
function code(a, b)
	tmp = 0.0
	if ((Float64(a * a) <= 1.8e-61) || (!(Float64(a * a) <= 6500000000000.0) && (Float64(a * a) <= 3.8e+76)))
		tmp = Float64(b * Float64(-b));
	else
		tmp = Float64(a * a);
	end
	return tmp
end

MATLAB

a = abs(a)
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (((a * a) <= 1.8e-61) || (~(((a * a) <= 6500000000000.0)) && ((a * a) <= 3.8e+76)))
		tmp = b * -b;
	else
		tmp = a * a;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_] := If[Or[LessEqual[N[(a * a), $MachinePrecision], 1.8e-61], And[N[Not[LessEqual[N[(a * a), $MachinePrecision], 6500000000000.0]], $MachinePrecision], LessEqual[N[(a * a), $MachinePrecision], 3.8e+76]]], N[(b * (-b)), $MachinePrecision], N[(a * a), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 a a) < 1.80000000000000007e-61 or 6.5e12 < (*.f64 a a) < 3.80000000000000024e76

   1. Initial program 100.0%

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

   2. Taylor expanded in a around 0 87.4%

      \[\leadsto \color{blue}{-1 \cdot {b}^{2}} \]
   3. Step-by-step derivation

      1. unpow2 87.4%

         \[\leadsto -1 \cdot \color{blue}{\left(b \cdot b\right)} \]

      2. mul-1-neg 87.4%

         \[\leadsto \color{blue}{-b \cdot b} \]

      3. distribute-rgt-neg-in 87.4%

         \[\leadsto \color{blue}{b \cdot \left(-b\right)} \]

   4. Simplified 87.4%

      \[\leadsto \color{blue}{b \cdot \left(-b\right)} \]

   if 1.80000000000000007e-61 < (*.f64 a a) < 6.5e12 or 3.80000000000000024e76 < (*.f64 a a)

   1. Initial program 91.1%

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

   2. Taylor expanded in a around inf 83.5%

      \[\leadsto \color{blue}{{a}^{2}} \]
   3. Step-by-step derivation

      1. unpow2 83.5%

         \[\leadsto \color{blue}{a \cdot a} \]

   4. Simplified 83.5%

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

4. Final simplification 85.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot a \leq 1.8 \cdot 10^{-61} \lor \neg \left(a \cdot a \leq 6500000000000\right) \land a \cdot a \leq 3.8 \cdot 10^{+76}:\\ \;\;\;\;b \cdot \left(-b\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 3: 96.6% accurate, 0.8× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} \mathbf{if}\;a \leq 5.9 \cdot 10^{+150}:\\ \;\;\;\;a \cdot a - b \cdot b\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b)
 :precision binary64
 (if (<= a 5.9e+150) (- (* a a) (* b b)) (* a a)))

C

a = abs(a);
double code(double a, double b) {
	double tmp;
	if (a <= 5.9e+150) {
		tmp = (a * a) - (b * b);
	} else {
		tmp = a * a;
	}
	return tmp;
}

Fortran

NOTE: a should be positive before calling this function
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8) :: tmp
    if (a <= 5.9d+150) then
        tmp = (a * a) - (b * b)
    else
        tmp = a * a
    end if
    code = tmp
end function

Java

a = Math.abs(a);
public static double code(double a, double b) {
	double tmp;
	if (a <= 5.9e+150) {
		tmp = (a * a) - (b * b);
	} else {
		tmp = a * a;
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b):
	tmp = 0
	if a <= 5.9e+150:
		tmp = (a * a) - (b * b)
	else:
		tmp = a * a
	return tmp

Julia

a = abs(a)
function code(a, b)
	tmp = 0.0
	if (a <= 5.9e+150)
		tmp = Float64(Float64(a * a) - Float64(b * b));
	else
		tmp = Float64(a * a);
	end
	return tmp
end

MATLAB

a = abs(a)
function tmp_2 = code(a, b)
	tmp = 0.0;
	if (a <= 5.9e+150)
		tmp = (a * a) - (b * b);
	else
		tmp = a * a;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_] := If[LessEqual[a, 5.9e+150], N[(N[(a * a), $MachinePrecision] - N[(b * b), $MachinePrecision]), $MachinePrecision], N[(a * a), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if a < 5.90000000000000023e150

   1. Initial program 96.2%

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

   if 5.90000000000000023e150 < a

   1. Initial program 90.9%

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

   2. Taylor expanded in a around inf 95.5%

      \[\leadsto \color{blue}{{a}^{2}} \]
   3. Step-by-step derivation

      1. unpow2 95.5%

         \[\leadsto \color{blue}{a \cdot a} \]

   4. Simplified 95.5%

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

4. Final simplification 96.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 5.9 \cdot 10^{+150}:\\ \;\;\;\;a \cdot a - b \cdot b\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 4: 53.4% accurate, 2.3× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ a \cdot a \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b) :precision binary64 (* a a))

C

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

Fortran

NOTE: a should be positive before calling this function
real(8) function code(a, b)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    code = a * a
end function

Java

a = Math.abs(a);
public static double code(double a, double b) {
	return a * a;
}

Python

a = abs(a)
def code(a, b):
	return a * a

Julia

a = abs(a)
function code(a, b)
	return Float64(a * a)
end

MATLAB

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

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_] := N[(a * a), $MachinePrecision]

Derivation

1. Initial program 95.3%

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

2. Taylor expanded in a around inf 55.8%

   \[\leadsto \color{blue}{{a}^{2}} \]
3. Step-by-step derivation

   1. unpow2 55.8%

      \[\leadsto \color{blue}{a \cdot a} \]

4. Simplified 55.8%

   \[\leadsto \color{blue}{a \cdot a} \]

5. Final simplification 55.8%

   \[\leadsto a \cdot a \]

Developer target: 100.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \left(a + b\right) \cdot \left(a - b\right) \end{array} \]

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Reproduce

herbie shell --seed 2023230 
(FPCore (a b)
  :name "Difference of squares"
  :precision binary64

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

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