Numeric.Integration.TanhSinh:everywhere from integration-0.2.1

Percentage Accurate: 93.8% → 99.9%

Time: 5.3s

Precision: binary64

Cost: 708

• 24.5% of points produce a very large (infinite) output. You may want to add a precondition.

Math

\[x \cdot \left(1 + y \cdot y\right) \]

↓

\[\begin{array}{l} \mathbf{if}\;y \cdot y \leq 4 \cdot 10^{+136}:\\ \;\;\;\;x \cdot \left(y \cdot y + 1\right)\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(y \cdot x\right)\\ \end{array} \]

FPCore

(FPCore (x y) :precision binary64 (* x (+ 1.0 (* y y))))

↓

(FPCore (x y)
 :precision binary64
 (if (<= (* y y) 4e+136) (* x (+ (* y y) 1.0)) (* y (* y x))))

C

double code(double x, double y) {
	return x * (1.0 + (y * y));
}

↓

double code(double x, double y) {
	double tmp;
	if ((y * y) <= 4e+136) {
		tmp = x * ((y * y) + 1.0);
	} else {
		tmp = y * (y * x);
	}
	return tmp;
}

Fortran

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    code = x * (1.0d0 + (y * y))
end function

↓

real(8) function code(x, y)
    real(8), intent (in) :: x
    real(8), intent (in) :: y
    real(8) :: tmp
    if ((y * y) <= 4d+136) then
        tmp = x * ((y * y) + 1.0d0)
    else
        tmp = y * (y * x)
    end if
    code = tmp
end function

Java

public static double code(double x, double y) {
	return x * (1.0 + (y * y));
}

↓

public static double code(double x, double y) {
	double tmp;
	if ((y * y) <= 4e+136) {
		tmp = x * ((y * y) + 1.0);
	} else {
		tmp = y * (y * x);
	}
	return tmp;
}

Python

def code(x, y):
	return x * (1.0 + (y * y))

↓

def code(x, y):
	tmp = 0
	if (y * y) <= 4e+136:
		tmp = x * ((y * y) + 1.0)
	else:
		tmp = y * (y * x)
	return tmp

Julia

function code(x, y)
	return Float64(x * Float64(1.0 + Float64(y * y)))
end

↓

function code(x, y)
	tmp = 0.0
	if (Float64(y * y) <= 4e+136)
		tmp = Float64(x * Float64(Float64(y * y) + 1.0));
	else
		tmp = Float64(y * Float64(y * x));
	end
	return tmp
end

MATLAB

function tmp = code(x, y)
	tmp = x * (1.0 + (y * y));
end

↓

function tmp_2 = code(x, y)
	tmp = 0.0;
	if ((y * y) <= 4e+136)
		tmp = x * ((y * y) + 1.0);
	else
		tmp = y * (y * x);
	end
	tmp_2 = tmp;
end

Wolfram

code[x_, y_] := N[(x * N[(1.0 + N[(y * y), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

↓

code[x_, y_] := If[LessEqual[N[(y * y), $MachinePrecision], 4e+136], N[(x * N[(N[(y * y), $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision], N[(y * N[(y * x), $MachinePrecision]), $MachinePrecision]]

Target

Original	93.8%
Target	99.9%
Herbie	99.9%

\[x + \left(x \cdot y\right) \cdot y \]

Derivation

1. Split input into 2 regimes

2. if (*.f64 y y) < 4.00000000000000023e136

   1. Initial program 99.9%

      \[x \cdot \left(1 + y \cdot y\right) \]

   if 4.00000000000000023e136 < (*.f64 y y)

   1. Initial program 82.9%

      \[x \cdot \left(1 + y \cdot y\right) \]

   2. Taylor expanded in y around inf 82.9%

      \[\leadsto \color{blue}{{y}^{2} \cdot x} \]

   3. Simplified 99.8%

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

      [Start] 82.9%	\[ {y}^{2} \cdot x \]
      unpow2 [=>] 82.9%	\[ \color{blue}{\left(y \cdot y\right)} \cdot x \]
      associate-*l* [=>] 99.8%	\[ \color{blue}{y \cdot \left(y \cdot x\right)} \]

3. Recombined 2 regimes into one program.

4. Final simplification 99.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;y \cdot y \leq 4 \cdot 10^{+136}:\\ \;\;\;\;x \cdot \left(y \cdot y + 1\right)\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(y \cdot x\right)\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	99.9%
Cost	708

\[\begin{array}{l} \mathbf{if}\;y \cdot y \leq 4 \cdot 10^{+136}:\\ \;\;\;\;x \cdot \left(y \cdot y + 1\right)\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(y \cdot x\right)\\ \end{array} \]

Alternative 2
Accuracy	92.9%
Cost	580

\[\begin{array}{l} \mathbf{if}\;y \cdot y \leq 1:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(y \cdot y\right)\\ \end{array} \]

Alternative 3
Accuracy	99.0%
Cost	580

\[\begin{array}{l} \mathbf{if}\;y \cdot y \leq 0.05:\\ \;\;\;\;x\\ \mathbf{else}:\\ \;\;\;\;y \cdot \left(y \cdot x\right)\\ \end{array} \]

Alternative 4
Accuracy	99.9%
Cost	448

\[x + y \cdot \left(y \cdot x\right) \]

Alternative 5
Accuracy	52.1%
Cost	64

\[x \]

Reproduce

herbie shell --seed 2023271 
(FPCore (x y)
  :name "Numeric.Integration.TanhSinh:everywhere from integration-0.2.1"
  :precision binary64

  :herbie-target
  (+ x (* (* x y) y))

  (* x (+ 1.0 (* y y))))