Examples.Basics.ProofTests:f4 from sbv-4.4

Percentage Accurate: 93.8% → 99.9%

Time: 3.5s

Alternatives: 4

Speedup: 0.8×

• 44.1% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

\[\begin{array}{l} \\ \left(x \cdot x + \left(x \cdot 2\right) \cdot y\right) + y \cdot y \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (+ (+ (* x x) (* (* x 2.0) y)) (* y y)))

C

double code(double x, double y) {
	return ((x * x) + ((x * 2.0) * y)) + (y * y);
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = ((x * x) + ((x * 2.0d0) * y)) + (y * y)
end function

Java

public static double code(double x, double y) {
	return ((x * x) + ((x * 2.0) * y)) + (y * y);
}

Python

def code(x, y):
	return ((x * x) + ((x * 2.0) * y)) + (y * y)

Julia

function code(x, y)
	return Float64(Float64(Float64(x * x) + Float64(Float64(x * 2.0) * y)) + Float64(y * y))
end

MATLAB

function tmp = code(x, y)
	tmp = ((x * x) + ((x * 2.0) * y)) + (y * y);
end

Wolfram

code[x_, y_] := N[(N[(N[(x * x), $MachinePrecision] + N[(N[(x * 2.0), $MachinePrecision] * y), $MachinePrecision]), $MachinePrecision] + N[(y * y), $MachinePrecision]), $MachinePrecision]

Initial Program: 93.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \left(x \cdot x + \left(x \cdot 2\right) \cdot y\right) + y \cdot y \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (+ (+ (* x x) (* (* x 2.0) y)) (* y y)))

C

double code(double x, double y) {
	return ((x * x) + ((x * 2.0) * y)) + (y * y);
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = ((x * x) + ((x * 2.0d0) * y)) + (y * y)
end function

Java

public static double code(double x, double y) {
	return ((x * x) + ((x * 2.0) * y)) + (y * y);
}

Python

def code(x, y):
	return ((x * x) + ((x * 2.0) * y)) + (y * y)

Julia

function code(x, y)
	return Float64(Float64(Float64(x * x) + Float64(Float64(x * 2.0) * y)) + Float64(y * y))
end

MATLAB

function tmp = code(x, y)
	tmp = ((x * x) + ((x * 2.0) * y)) + (y * y);
end

Wolfram

code[x_, y_] := N[(N[(N[(x * x), $MachinePrecision] + N[(N[(x * 2.0), $MachinePrecision] * y), $MachinePrecision]), $MachinePrecision] + N[(y * y), $MachinePrecision]), $MachinePrecision]

Alternative 1: 99.9% accurate, 0.1× speedup

Math

\[\begin{array}{l} [x, y] = \mathsf{sort}([x, y])\\ \\ \begin{array}{l} \mathbf{if}\;y \leq 3.6 \cdot 10^{+148}:\\ \;\;\;\;x \cdot \left(x + y \cdot 2\right) + y \cdot y\\ \mathbf{else}:\\ \;\;\;\;{y}^{2}\\ \end{array} \end{array} \]

FPCore

NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (if (<= y 3.6e+148) (+ (* x (+ x (* y 2.0))) (* y y)) (pow y 2.0)))

C

assert(x < y);
double code(double x, double y) {
	double tmp;
	if (y <= 3.6e+148) {
		tmp = (x * (x + (y * 2.0))) + (y * y);
	} else {
		tmp = pow(y, 2.0);
	}
	return tmp;
}

Fortran

NOTE: x and y should be sorted in increasing order before calling this function.
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= 3.6d+148) then
        tmp = (x * (x + (y * 2.0d0))) + (y * y)
    else
        tmp = y ** 2.0d0
    end if
    code = tmp
end function

Java

assert x < y;
public static double code(double x, double y) {
	double tmp;
	if (y <= 3.6e+148) {
		tmp = (x * (x + (y * 2.0))) + (y * y);
	} else {
		tmp = Math.pow(y, 2.0);
	}
	return tmp;
}

Python

[x, y] = sort([x, y])
def code(x, y):
	tmp = 0
	if y <= 3.6e+148:
		tmp = (x * (x + (y * 2.0))) + (y * y)
	else:
		tmp = math.pow(y, 2.0)
	return tmp

Julia

x, y = sort([x, y])
function code(x, y)
	tmp = 0.0
	if (y <= 3.6e+148)
		tmp = Float64(Float64(x * Float64(x + Float64(y * 2.0))) + Float64(y * y));
	else
		tmp = y ^ 2.0;
	end
	return tmp
end

MATLAB

x, y = num2cell(sort([x, y])){:}
function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 3.6e+148)
		tmp = (x * (x + (y * 2.0))) + (y * y);
	else
		tmp = y ^ 2.0;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := If[LessEqual[y, 3.6e+148], N[(N[(x * N[(x + N[(y * 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(y * y), $MachinePrecision]), $MachinePrecision], N[Power[y, 2.0], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 3.60000000000000006e148

   1. Initial program 93.0%

      \[\left(x \cdot x + \left(x \cdot 2\right) \cdot y\right) + y \cdot y \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. +-commutative 93.0%

         \[\leadsto \color{blue}{\left(\left(x \cdot 2\right) \cdot y + x \cdot x\right)} + y \cdot y \]

      2. associate-*l* 93.0%

         \[\leadsto \left(\color{blue}{x \cdot \left(2 \cdot y\right)} + x \cdot x\right) + y \cdot y \]

      3. distribute-lft-out 97.7%

         \[\leadsto \color{blue}{x \cdot \left(2 \cdot y + x\right)} + y \cdot y \]

   4. Applied egg-rr 97.7%

      \[\leadsto \color{blue}{x \cdot \left(2 \cdot y + x\right)} + y \cdot y \]

   if 3.60000000000000006e148 < y

   1. Initial program 80.5%

      \[\left(x \cdot x + \left(x \cdot 2\right) \cdot y\right) + y \cdot y \]
   2. Add Preprocessing

   3. Taylor expanded in x around 0 100.0%

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

4. Final simplification 98.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 3.6 \cdot 10^{+148}:\\ \;\;\;\;x \cdot \left(x + y \cdot 2\right) + y \cdot y\\ \mathbf{else}:\\ \;\;\;\;{y}^{2}\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 98.8% accurate, 0.8× speedup

Math

\[\begin{array}{l} [x, y] = \mathsf{sort}([x, y])\\ \\ \begin{array}{l} \mathbf{if}\;y \leq 2.3 \cdot 10^{+194}:\\ \;\;\;\;x \cdot \left(x + y \cdot 2\right) + y \cdot y\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(y + x \cdot 2\right)\\ \end{array} \end{array} \]

FPCore

NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y)
 :precision binary64
 (if (<= y 2.3e+194) (+ (* x (+ x (* y 2.0))) (* y y)) (* y (+ y (* x 2.0)))))

C

assert(x < y);
double code(double x, double y) {
	double tmp;
	if (y <= 2.3e+194) {
		tmp = (x * (x + (y * 2.0))) + (y * y);
	} else {
		tmp = y * (y + (x * 2.0));
	}
	return tmp;
}

Fortran

NOTE: x and y should be sorted in increasing order before calling this function.
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if (y <= 2.3d+194) then
        tmp = (x * (x + (y * 2.0d0))) + (y * y)
    else
        tmp = y * (y + (x * 2.0d0))
    end if
    code = tmp
end function

Java

assert x < y;
public static double code(double x, double y) {
	double tmp;
	if (y <= 2.3e+194) {
		tmp = (x * (x + (y * 2.0))) + (y * y);
	} else {
		tmp = y * (y + (x * 2.0));
	}
	return tmp;
}

Python

[x, y] = sort([x, y])
def code(x, y):
	tmp = 0
	if y <= 2.3e+194:
		tmp = (x * (x + (y * 2.0))) + (y * y)
	else:
		tmp = y * (y + (x * 2.0))
	return tmp

Julia

x, y = sort([x, y])
function code(x, y)
	tmp = 0.0
	if (y <= 2.3e+194)
		tmp = Float64(Float64(x * Float64(x + Float64(y * 2.0))) + Float64(y * y));
	else
		tmp = Float64(y * Float64(y + Float64(x * 2.0)));
	end
	return tmp
end

MATLAB

x, y = num2cell(sort([x, y])){:}
function tmp_2 = code(x, y)
	tmp = 0.0;
	if (y <= 2.3e+194)
		tmp = (x * (x + (y * 2.0))) + (y * y);
	else
		tmp = y * (y + (x * 2.0));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := If[LessEqual[y, 2.3e+194], N[(N[(x * N[(x + N[(y * 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(y * y), $MachinePrecision]), $MachinePrecision], N[(y * N[(y + N[(x * 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if y < 2.30000000000000005e194

   1. Initial program 91.7%

      \[\left(x \cdot x + \left(x \cdot 2\right) \cdot y\right) + y \cdot y \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. +-commutative 91.7%

         \[\leadsto \color{blue}{\left(\left(x \cdot 2\right) \cdot y + x \cdot x\right)} + y \cdot y \]

      2. associate-*l* 91.7%

         \[\leadsto \left(\color{blue}{x \cdot \left(2 \cdot y\right)} + x \cdot x\right) + y \cdot y \]

      3. distribute-lft-out 97.8%

         \[\leadsto \color{blue}{x \cdot \left(2 \cdot y + x\right)} + y \cdot y \]

   4. Applied egg-rr 97.8%

      \[\leadsto \color{blue}{x \cdot \left(2 \cdot y + x\right)} + y \cdot y \]

   if 2.30000000000000005e194 < y

   1. Initial program 85.7%

      \[\left(x \cdot x + \left(x \cdot 2\right) \cdot y\right) + y \cdot y \]
   2. Add Preprocessing

   3. Taylor expanded in x around 0 85.7%

      \[\leadsto \color{blue}{2 \cdot \left(x \cdot y\right)} + y \cdot y \]
   4. Step-by-step derivation

      1. associate-*r* 85.7%

         \[\leadsto \color{blue}{\left(2 \cdot x\right) \cdot y} + y \cdot y \]

      2. distribute-rgt-out 96.4%

         \[\leadsto \color{blue}{y \cdot \left(2 \cdot x + y\right)} \]

   5. Applied egg-rr 96.4%

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

4. Final simplification 97.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 2.3 \cdot 10^{+194}:\\ \;\;\;\;x \cdot \left(x + y \cdot 2\right) + y \cdot y\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(y + x \cdot 2\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 55.9% accurate, 1.9× speedup

Math

\[\begin{array}{l} [x, y] = \mathsf{sort}([x, y])\\ \\ y \cdot \left(y + x \cdot 2\right) \end{array} \]

FPCore

NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y) :precision binary64 (* y (+ y (* x 2.0))))

C

assert(x < y);
double code(double x, double y) {
	return y * (y + (x * 2.0));
}

Fortran

NOTE: x and y should be sorted in increasing order before calling this function.
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = y * (y + (x * 2.0d0))
end function

Java

assert x < y;
public static double code(double x, double y) {
	return y * (y + (x * 2.0));
}

Python

[x, y] = sort([x, y])
def code(x, y):
	return y * (y + (x * 2.0))

Julia

x, y = sort([x, y])
function code(x, y)
	return Float64(y * Float64(y + Float64(x * 2.0)))
end

MATLAB

x, y = num2cell(sort([x, y])){:}
function tmp = code(x, y)
	tmp = y * (y + (x * 2.0));
end

Wolfram

NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := N[(y * N[(y + N[(x * 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 91.0%

   \[\left(x \cdot x + \left(x \cdot 2\right) \cdot y\right) + y \cdot y \]
2. Add Preprocessing

3. Taylor expanded in x around 0 56.6%

   \[\leadsto \color{blue}{2 \cdot \left(x \cdot y\right)} + y \cdot y \]
4. Step-by-step derivation

   1. associate-*r* 56.6%

      \[\leadsto \color{blue}{\left(2 \cdot x\right) \cdot y} + y \cdot y \]

   2. distribute-rgt-out 59.7%

      \[\leadsto \color{blue}{y \cdot \left(2 \cdot x + y\right)} \]

5. Applied egg-rr 59.7%

   \[\leadsto \color{blue}{y \cdot \left(2 \cdot x + y\right)} \]

6. Final simplification 59.7%

   \[\leadsto y \cdot \left(y + x \cdot 2\right) \]
7. Add Preprocessing

Alternative 4: 14.6% accurate, 2.6× speedup

Math

\[\begin{array}{l} [x, y] = \mathsf{sort}([x, y])\\ \\ 2 \cdot \left(y \cdot x\right) \end{array} \]

FPCore

NOTE: x and y should be sorted in increasing order before calling this function.
(FPCore (x y) :precision binary64 (* 2.0 (* y x)))

C

assert(x < y);
double code(double x, double y) {
	return 2.0 * (y * x);
}

Fortran

NOTE: x and y should be sorted in increasing order before calling this function.
real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = 2.0d0 * (y * x)
end function

Java

assert x < y;
public static double code(double x, double y) {
	return 2.0 * (y * x);
}

Python

[x, y] = sort([x, y])
def code(x, y):
	return 2.0 * (y * x)

Julia

x, y = sort([x, y])
function code(x, y)
	return Float64(2.0 * Float64(y * x))
end

MATLAB

x, y = num2cell(sort([x, y])){:}
function tmp = code(x, y)
	tmp = 2.0 * (y * x);
end

Wolfram

NOTE: x and y should be sorted in increasing order before calling this function.
code[x_, y_] := N[(2.0 * N[(y * x), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 91.0%

   \[\left(x \cdot x + \left(x \cdot 2\right) \cdot y\right) + y \cdot y \]
2. Add Preprocessing

3. Taylor expanded in x around 0 56.6%

   \[\leadsto \color{blue}{2 \cdot \left(x \cdot y\right)} + y \cdot y \]

4. Taylor expanded in x around inf 18.4%

   \[\leadsto \color{blue}{2 \cdot \left(x \cdot y\right)} \]

5. Final simplification 18.4%

   \[\leadsto 2 \cdot \left(y \cdot x\right) \]
6. Add Preprocessing

Developer target: 93.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ x \cdot x + \left(y \cdot y + \left(x \cdot y\right) \cdot 2\right) \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (+ (* x x) (+ (* y y) (* (* x y) 2.0))))

C

double code(double x, double y) {
	return (x * x) + ((y * y) + ((x * y) * 2.0));
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = (x * x) + ((y * y) + ((x * y) * 2.0d0))
end function

Java

public static double code(double x, double y) {
	return (x * x) + ((y * y) + ((x * y) * 2.0));
}

Python

def code(x, y):
	return (x * x) + ((y * y) + ((x * y) * 2.0))

Julia

function code(x, y)
	return Float64(Float64(x * x) + Float64(Float64(y * y) + Float64(Float64(x * y) * 2.0)))
end

MATLAB

function tmp = code(x, y)
	tmp = (x * x) + ((y * y) + ((x * y) * 2.0));
end

Wolfram

code[x_, y_] := N[(N[(x * x), $MachinePrecision] + N[(N[(y * y), $MachinePrecision] + N[(N[(x * y), $MachinePrecision] * 2.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Reproduce

herbie shell --seed 2024073 
(FPCore (x y)
  :name "Examples.Basics.ProofTests:f4 from sbv-4.4"
  :precision binary64

  :alt
  (+ (* x x) (+ (* y y) (* (* x y) 2.0)))

  (+ (+ (* x x) (* (* x 2.0) y)) (* y y)))