Difference of squares

Percentage Accurate: 92.9% → 97.9%

Time: 2.0s

Alternatives: 4

Speedup: 0.8×

• 44.3% 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: 92.9% 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: 97.9% accurate, 0.1× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} \mathbf{if}\;a \leq 1.4 \cdot 10^{+236}:\\ \;\;\;\;\mathsf{fma}\left(a, a, b \cdot \left(-b\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b)
 :precision binary64
 (if (<= a 1.4e+236) (fma a a (* b (- b))) (* a a)))

C

a = abs(a);
double code(double a, double b) {
	double tmp;
	if (a <= 1.4e+236) {
		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 <= 1.4e+236)
		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, 1.4e+236], N[(a * a + N[(b * (-b)), $MachinePrecision]), $MachinePrecision], N[(a * a), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if a < 1.39999999999999996e236

   1. Initial program 94.2%

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

      1. sqr-neg 94.2%

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

      2. cancel-sign-sub 94.2%

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

      3. fma-def 97.5%

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

   3. Simplified 97.5%

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

   if 1.39999999999999996e236 < a

   1. Initial program 73.3%

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

   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 97.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 1.4 \cdot 10^{+236}:\\ \;\;\;\;\mathsf{fma}\left(a, a, b \cdot \left(-b\right)\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 2: 76.8% accurate, 0.4× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} \mathbf{if}\;a \cdot a \leq 4.6 \cdot 10^{-134} \lor \neg \left(a \cdot a \leq 1.7 \cdot 10^{-49}\right) \land a \cdot a \leq 100000000:\\ \;\;\;\;b \cdot \left(-b\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b)
 :precision binary64
 (if (or (<= (* a a) 4.6e-134)
         (and (not (<= (* a a) 1.7e-49)) (<= (* a a) 100000000.0)))
   (* b (- b))
   (* a a)))

C

a = abs(a);
double code(double a, double b) {
	double tmp;
	if (((a * a) <= 4.6e-134) || (!((a * a) <= 1.7e-49) && ((a * a) <= 100000000.0))) {
		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) <= 4.6d-134) .or. (.not. ((a * a) <= 1.7d-49)) .and. ((a * a) <= 100000000.0d0)) 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) <= 4.6e-134) || (!((a * a) <= 1.7e-49) && ((a * a) <= 100000000.0))) {
		tmp = b * -b;
	} else {
		tmp = a * a;
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b):
	tmp = 0
	if ((a * a) <= 4.6e-134) or (not ((a * a) <= 1.7e-49) and ((a * a) <= 100000000.0)):
		tmp = b * -b
	else:
		tmp = a * a
	return tmp

Julia

a = abs(a)
function code(a, b)
	tmp = 0.0
	if ((Float64(a * a) <= 4.6e-134) || (!(Float64(a * a) <= 1.7e-49) && (Float64(a * a) <= 100000000.0)))
		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) <= 4.6e-134) || (~(((a * a) <= 1.7e-49)) && ((a * a) <= 100000000.0)))
		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], 4.6e-134], And[N[Not[LessEqual[N[(a * a), $MachinePrecision], 1.7e-49]], $MachinePrecision], LessEqual[N[(a * a), $MachinePrecision], 100000000.0]]], N[(b * (-b)), $MachinePrecision], N[(a * a), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if (*.f64 a a) < 4.6000000000000001e-134 or 1.70000000000000002e-49 < (*.f64 a a) < 1e8

   1. Initial program 100.0%

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

   2. Taylor expanded in a around 0 88.4%

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

      1. unpow2 88.4%

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

      2. mul-1-neg 88.4%

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

      3. distribute-rgt-neg-in 88.4%

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

   4. Simplified 88.4%

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

   if 4.6000000000000001e-134 < (*.f64 a a) < 1.70000000000000002e-49 or 1e8 < (*.f64 a a)

   1. Initial program 87.0%

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

   2. Taylor expanded in a around inf 79.4%

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

      1. unpow2 79.4%

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

   4. Simplified 79.4%

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

4. Final simplification 83.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \cdot a \leq 4.6 \cdot 10^{-134} \lor \neg \left(a \cdot a \leq 1.7 \cdot 10^{-49}\right) \land a \cdot a \leq 100000000:\\ \;\;\;\;b \cdot \left(-b\right)\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 3: 96.3% accurate, 0.8× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} \mathbf{if}\;a \leq 6.2 \cdot 10^{+150}:\\ \;\;\;\;a \cdot a - b \cdot b\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b)
 :precision binary64
 (if (<= a 6.2e+150) (- (* a a) (* b b)) (* a a)))

C

a = abs(a);
double code(double a, double b) {
	double tmp;
	if (a <= 6.2e+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 <= 6.2d+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 <= 6.2e+150) {
		tmp = (a * a) - (b * b);
	} else {
		tmp = a * a;
	}
	return tmp;
}

Python

a = abs(a)
def code(a, b):
	tmp = 0
	if a <= 6.2e+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 <= 6.2e+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 <= 6.2e+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, 6.2e+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 < 6.20000000000000028e150

   1. Initial program 94.8%

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

   if 6.20000000000000028e150 < a

   1. Initial program 76.0%

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

   2. Taylor expanded in a around inf 96.0%

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

      1. unpow2 96.0%

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

   4. Simplified 96.0%

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

4. Final simplification 94.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 6.2 \cdot 10^{+150}:\\ \;\;\;\;a \cdot a - b \cdot b\\ \mathbf{else}:\\ \;\;\;\;a \cdot a\\ \end{array} \]

Alternative 4: 53.6% 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 93.0%

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

2. Taylor expanded in a around inf 52.6%

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

   1. unpow2 52.6%

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

4. Simplified 52.6%

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

5. Final simplification 52.6%

   \[\leadsto a \cdot a \]

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 2023297 
(FPCore (a b)
  :name "Difference of squares"
  :precision binary64

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

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