Logistic regression 2

?

Percentage Accurate: 79.9% → 90.0%
Time: 9.6s
Precision: binary64
Cost: 32772

?

\[\log \left(1 + e^{x}\right) - x \cdot y \]
↓
\[\begin{array}{l} \mathbf{if}\;\log \left(1 + e^{x}\right) - x \cdot y \leq 10^{+305}:\\ \;\;\;\;\mathsf{fma}\left(x, -y, \mathsf{log1p}\left(e^{x}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(0.5 - y\right) + \log 2\\ \end{array} \]
(FPCore (x y) :precision binary64 (- (log (+ 1.0 (exp x))) (* x y)))
↓
(FPCore (x y)
 :precision binary64
 (if (<= (- (log (+ 1.0 (exp x))) (* x y)) 1e+305)
   (fma x (- y) (log1p (exp x)))
   (+ (* x (- 0.5 y)) (log 2.0))))
double code(double x, double y) {
	return log((1.0 + exp(x))) - (x * y);
}
↓
double code(double x, double y) {
	double tmp;
	if ((log((1.0 + exp(x))) - (x * y)) <= 1e+305) {
		tmp = fma(x, -y, log1p(exp(x)));
	} else {
		tmp = (x * (0.5 - y)) + log(2.0);
	}
	return tmp;
}

function code(x, y)
	return Float64(log(Float64(1.0 + exp(x))) - Float64(x * y))
end
↓
function code(x, y)
	tmp = 0.0
	if (Float64(log(Float64(1.0 + exp(x))) - Float64(x * y)) <= 1e+305)
		tmp = fma(x, Float64(-y), log1p(exp(x)));
	else
		tmp = Float64(Float64(x * Float64(0.5 - y)) + log(2.0));
	end
	return tmp
end

code[x_, y_] := N[(N[Log[N[(1.0 + N[Exp[x], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] - N[(x * y), $MachinePrecision]), $MachinePrecision]
↓
code[x_, y_] := If[LessEqual[N[(N[Log[N[(1.0 + N[Exp[x], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] - N[(x * y), $MachinePrecision]), $MachinePrecision], 1e+305], N[(x * (-y) + N[Log[1 + N[Exp[x], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[(x * N[(0.5 - y), $MachinePrecision]), $MachinePrecision] + N[Log[2.0], $MachinePrecision]), $MachinePrecision]]

\log \left(1 + e^{x}\right) - x \cdot y
↓
\begin{array}{l}
\mathbf{if}\;\log \left(1 + e^{x}\right) - x \cdot y \leq 10^{+305}:\\
\;\;\;\;\mathsf{fma}\left(x, -y, \mathsf{log1p}\left(e^{x}\right)\right)\\

\mathbf{else}:\\
\;\;\;\;x \cdot \left(0.5 - y\right) + \log 2\\


\end{array}

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Herbie found 10 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Bogosity?

Bogosity

Target

Original79.9%
Target99.9%
Herbie90.0%
\[\begin{array}{l} \mathbf{if}\;x \leq 0:\\ \;\;\;\;\log \left(1 + e^{x}\right) - x \cdot y\\ \mathbf{else}:\\ \;\;\;\;\log \left(1 + e^{-x}\right) - \left(-x\right) \cdot \left(1 - y\right)\\ \end{array} \]

Derivation?

  1. Split input into 2 regimes

  2. if (-.f64 (log.f64 (+.f64 1 (exp.f64 x))) (*.f64 x y)) < 9.9999999999999994e304

    1. Initial program 99.9%

      \[\log \left(1 + e^{x}\right) - x \cdot y \]

    2. Simplified100.0%

      \[\leadsto \color{blue}{\mathsf{fma}\left(x, -y, \mathsf{log1p}\left(e^{x}\right)\right)} \]
      Step-by-step derivation

      [Start]99.9%

      \[ \log \left(1 + e^{x}\right) - x \cdot y \]

      sub-neg [=>]99.9%

      \[ \color{blue}{\log \left(1 + e^{x}\right) + \left(-x \cdot y\right)} \]

      +-commutative [=>]99.9%

      \[ \color{blue}{\left(-x \cdot y\right) + \log \left(1 + e^{x}\right)} \]

      distribute-rgt-neg-in [=>]99.9%

      \[ \color{blue}{x \cdot \left(-y\right)} + \log \left(1 + e^{x}\right) \]

      fma-def [=>]100.0%

      \[ \color{blue}{\mathsf{fma}\left(x, -y, \log \left(1 + e^{x}\right)\right)} \]

      log1p-def [=>]100.0%

      \[ \mathsf{fma}\left(x, -y, \color{blue}{\mathsf{log1p}\left(e^{x}\right)}\right) \]

    if 9.9999999999999994e304 < (-.f64 (log.f64 (+.f64 1 (exp.f64 x))) (*.f64 x y))

    1. Initial program 31.7%

      \[\log \left(1 + e^{x}\right) - x \cdot y \]

    2. Simplified31.7%

      \[\leadsto \color{blue}{\mathsf{log1p}\left(e^{x}\right) - x \cdot y} \]
      Step-by-step derivation

      [Start]31.7%

      \[ \log \left(1 + e^{x}\right) - x \cdot y \]

      log1p-def [=>]31.7%

      \[ \color{blue}{\mathsf{log1p}\left(e^{x}\right)} - x \cdot y \]

    3. Taylor expanded in x around 0 70.7%

      \[\leadsto \color{blue}{\left(0.5 - y\right) \cdot x + \log 2} \]
  3. Recombined 2 regimes into one program.

  4. Final simplification91.4%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\log \left(1 + e^{x}\right) - x \cdot y \leq 10^{+305}:\\ \;\;\;\;\mathsf{fma}\left(x, -y, \mathsf{log1p}\left(e^{x}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(0.5 - y\right) + \log 2\\ \end{array} \]

Alternatives

Alternative 1
Accuracy90.0%
Cost32772
\[\begin{array}{l} \mathbf{if}\;\log \left(1 + e^{x}\right) - x \cdot y \leq 10^{+305}:\\ \;\;\;\;\mathsf{fma}\left(x, -y, \mathsf{log1p}\left(e^{x}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(0.5 - y\right) + \log 2\\ \end{array} \]
Alternative 2
Accuracy90.0%
Cost26436
\[\begin{array}{l} \mathbf{if}\;\log \left(1 + e^{x}\right) - x \cdot y \leq 10^{+305}:\\ \;\;\;\;\mathsf{log1p}\left(e^{x}\right) - x \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(0.5 - y\right) + \log 2\\ \end{array} \]
Alternative 3
Accuracy76.6%
Cost6984
\[\begin{array}{l} \mathbf{if}\;x \leq -7.2 \cdot 10^{-6}:\\ \;\;\;\;-x \cdot y\\ \mathbf{elif}\;x \leq 1.7 \cdot 10^{-47}:\\ \;\;\;\;\log 2 + x \cdot 0.5\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(0.5 - y\right)\\ \end{array} \]
Alternative 4
Accuracy89.5%
Cost6980
\[\begin{array}{l} \mathbf{if}\;x \leq -1.4:\\ \;\;\;\;-x \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(0.5 - y\right) + \log 2\\ \end{array} \]
Alternative 5
Accuracy87.6%
Cost6852
\[\begin{array}{l} \mathbf{if}\;x \leq -52:\\ \;\;\;\;-x \cdot y\\ \mathbf{else}:\\ \;\;\;\;\log 2 - x \cdot y\\ \end{array} \]
Alternative 6
Accuracy76.5%
Cost6728
\[\begin{array}{l} \mathbf{if}\;x \leq -3.9 \cdot 10^{-6}:\\ \;\;\;\;-x \cdot y\\ \mathbf{elif}\;x \leq 1.02 \cdot 10^{-45}:\\ \;\;\;\;\log 2\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(0.5 - y\right)\\ \end{array} \]
Alternative 7
Accuracy53.1%
Cost580
\[\begin{array}{l} \mathbf{if}\;x \leq -2 \cdot 10^{-310}:\\ \;\;\;\;-x \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot 0.5 - x \cdot y\\ \end{array} \]
Alternative 8
Accuracy53.1%
Cost452
\[\begin{array}{l} \mathbf{if}\;x \leq -2 \cdot 10^{-310}:\\ \;\;\;\;-x \cdot y\\ \mathbf{else}:\\ \;\;\;\;x \cdot \left(0.5 - y\right)\\ \end{array} \]
Alternative 9
Accuracy50.9%
Cost256
\[-x \cdot y \]
Alternative 10
Accuracy5.3%
Cost192
\[x \cdot 0.5 \]

Reproduce?

herbie shell --seed 2023271 
(FPCore (x y)
  :name "Logistic regression 2"
  :precision binary64

  :herbie-target
  (if (<= x 0.0) (- (log (+ 1.0 (exp x))) (* x y)) (- (log (+ 1.0 (exp (- x)))) (* (- x) (- 1.0 y))))

  (- (log (+ 1.0 (exp x))) (* x y)))