Difference of squares

Percentage Accurate: 94.1% → 98.6%

Time: 2.0s

Alternatives: 4

Speedup: 0.6×

• 43.3% 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: 94.1% 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: 98.6% accurate, 0.1× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} \mathbf{if}\;a \leq 5 \cdot 10^{+253}:\\ \;\;\;\;\mathsf{fma}\left(a, a, b \cdot \left(-b\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b)
 :precision binary64
 (if (<= a 5e+253) (fma a a (* b (- b))) (* a a)))

C

a = abs(a);
double code(double a, double b) {
	double tmp;
	if (a <= 5e+253) {
		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 <= 5e+253)
		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, 5e+253], N[(a * a + N[(b * (-b)), $MachinePrecision]), $MachinePrecision], N[(a * a), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if a < 4.9999999999999997e253

   1. Initial program 95.9%

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

      1. sqr-neg 95.9%

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

      2. cancel-sign-sub 95.9%

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

      3. fma-def 99.2%

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

   3. Simplified 99.2%

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

   if 4.9999999999999997e253 < a

   1. Initial program 72.7%

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

   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 99.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 5 \cdot 10^{+253}:\\ \;\;\;\;\mathsf{fma}\left(a, a, b \cdot \left(-b\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 2: 75.6% accurate, 0.4× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} \mathbf{if}\;a \cdot a \leq 5.4 \cdot 10^{-138} \lor \neg \left(a \cdot a \leq 108000000\right) \land a \cdot a \leq 2.4 \cdot 10^{+117}:\\ \;\;\;\;b \cdot \left(-b\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b)
 :precision binary64
 (if (or (<= (* a a) 5.4e-138)
         (and (not (<= (* a a) 108000000.0)) (<= (* a a) 2.4e+117)))
   (* b (- b))
   (* a a)))

C

a = abs(a);
double code(double a, double b) {
	double tmp;
	if (((a * a) <= 5.4e-138) || (!((a * a) <= 108000000.0) && ((a * a) <= 2.4e+117))) {
		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) <= 5.4d-138) .or. (.not. ((a * a) <= 108000000.0d0)) .and. ((a * a) <= 2.4d+117)) 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) <= 5.4e-138) || (!((a * a) <= 108000000.0) && ((a * a) <= 2.4e+117))) {
		tmp = b * -b;
	} else {
		tmp = a * a;
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b):
	tmp = 0
	if ((a * a) <= 5.4e-138) or (not ((a * a) <= 108000000.0) and ((a * a) <= 2.4e+117)):
		tmp = b * -b
	else:
		tmp = a * a
	return tmp

Julia

a = abs(a)
function code(a, b)
	tmp = 0.0
	if ((Float64(a * a) <= 5.4e-138) || (!(Float64(a * a) <= 108000000.0) && (Float64(a * a) <= 2.4e+117)))
		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) <= 5.4e-138) || (~(((a * a) <= 108000000.0)) && ((a * a) <= 2.4e+117)))
		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], 5.4e-138], And[N[Not[LessEqual[N[(a * a), $MachinePrecision], 108000000.0]], $MachinePrecision], LessEqual[N[(a * a), $MachinePrecision], 2.4e+117]]], N[(b * (-b)), $MachinePrecision], N[(a * a), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 a a) < 5.40000000000000057e-138 or 1.08e8 < (*.f64 a a) < 2.3999999999999999e117

   1. Initial program 100.0%

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

   2. Taylor expanded in a around 0 87.8%

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

      1. unpow2 87.8%

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

      2. mul-1-neg 87.8%

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

      3. distribute-rgt-neg-in 87.8%

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

   4. Simplified 87.8%

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

   if 5.40000000000000057e-138 < (*.f64 a a) < 1.08e8 or 2.3999999999999999e117 < (*.f64 a a)

   1. Initial program 90.6%

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

   2. Taylor expanded in a around inf 78.2%

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

      1. unpow2 78.2%

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

   4. Simplified 78.2%

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

4. Final simplification 82.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot a \leq 5.4 \cdot 10^{-138} \lor \neg \left(a \cdot a \leq 108000000\right) \land a \cdot a \leq 2.4 \cdot 10^{+117}:\\ \;\;\;\;b \cdot \left(-b\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 3: 96.9% accurate, 0.6× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} \mathbf{if}\;b \cdot b \leq 5 \cdot 10^{+298}:\\ \;\;\;\;a \cdot a - b \cdot b\\ \mathbf{else}:\\ \;\;\;\;b \cdot \left(-b\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b)
 :precision binary64
 (if (<= (* b b) 5e+298) (- (* a a) (* b b)) (* b (- b))))

C

a = abs(a);
double code(double a, double b) {
	double tmp;
	if ((b * b) <= 5e+298) {
		tmp = (a * a) - (b * b);
	} else {
		tmp = b * -b;
	}
	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 ((b * b) <= 5d+298) then
        tmp = (a * a) - (b * b)
    else
        tmp = b * -b
    end if
    code = tmp
end function

Java

a = Math.abs(a);
public static double code(double a, double b) {
	double tmp;
	if ((b * b) <= 5e+298) {
		tmp = (a * a) - (b * b);
	} else {
		tmp = b * -b;
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b):
	tmp = 0
	if (b * b) <= 5e+298:
		tmp = (a * a) - (b * b)
	else:
		tmp = b * -b
	return tmp

Julia

a = abs(a)
function code(a, b)
	tmp = 0.0
	if (Float64(b * b) <= 5e+298)
		tmp = Float64(Float64(a * a) - Float64(b * b));
	else
		tmp = Float64(b * Float64(-b));
	end
	return tmp
end

MATLAB

a = abs(a)
function tmp_2 = code(a, b)
	tmp = 0.0;
	if ((b * b) <= 5e+298)
		tmp = (a * a) - (b * b);
	else
		tmp = b * -b;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_] := If[LessEqual[N[(b * b), $MachinePrecision], 5e+298], N[(N[(a * a), $MachinePrecision] - N[(b * b), $MachinePrecision]), $MachinePrecision], N[(b * (-b)), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 b b) < 5.0000000000000003e298

   1. Initial program 100.0%

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

   if 5.0000000000000003e298 < (*.f64 b b)

   1. Initial program 77.2%

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

   2. Taylor expanded in a around 0 91.2%

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

      1. unpow2 91.2%

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

      2. mul-1-neg 91.2%

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

      3. distribute-rgt-neg-in 91.2%

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

   4. Simplified 91.2%

      \[\leadsto \color{blue}{b \cdot \left(-b\right)} \]
3. Recombined 2 regimes into one program.

4. Final simplification 98.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \cdot b \leq 5 \cdot 10^{+298}:\\ \;\;\;\;a \cdot a - b \cdot b\\ \mathbf{else}:\\ \;\;\;\;b \cdot \left(-b\right)\\ \end{array} \]

Alternative 4: 53.5% 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 94.9%

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

2. Taylor expanded in a around inf 54.7%

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

   1. unpow2 54.7%

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

4. Simplified 54.7%

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

5. Final simplification 54.7%

   \[\leadsto a \cdot a \]

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 2023271 
(FPCore (a b)
  :name "Difference of squares"
  :precision binary64

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

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