Math FPCore C Julia Wolfram TeX \[\left(\left(x \cdot y + \frac{z \cdot t}{16}\right) - \frac{a \cdot b}{4}\right) + c
\]
↓
\[\begin{array}{l}
t_1 := \left(\frac{z \cdot t}{16} + x \cdot y\right) - \frac{b \cdot a}{4}\\
\mathbf{if}\;t_1 \leq \infty:\\
\;\;\;\;c + t_1\\
\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(t \cdot 0.0625, z, -0.25 \cdot \left(b \cdot a\right)\right)\\
\end{array}
\]
(FPCore (x y z t a b c)
:precision binary64
(+ (- (+ (* x y) (/ (* z t) 16.0)) (/ (* a b) 4.0)) c)) ↓
(FPCore (x y z t a b c)
:precision binary64
(let* ((t_1 (- (+ (/ (* z t) 16.0) (* x y)) (/ (* b a) 4.0))))
(if (<= t_1 INFINITY) (+ c t_1) (fma (* t 0.0625) z (* -0.25 (* b a)))))) double code(double x, double y, double z, double t, double a, double b, double c) {
return (((x * y) + ((z * t) / 16.0)) - ((a * b) / 4.0)) + c;
}
↓
double code(double x, double y, double z, double t, double a, double b, double c) {
double t_1 = (((z * t) / 16.0) + (x * y)) - ((b * a) / 4.0);
double tmp;
if (t_1 <= ((double) INFINITY)) {
tmp = c + t_1;
} else {
tmp = fma((t * 0.0625), z, (-0.25 * (b * a)));
}
return tmp;
}
function code(x, y, z, t, a, b, c)
return Float64(Float64(Float64(Float64(x * y) + Float64(Float64(z * t) / 16.0)) - Float64(Float64(a * b) / 4.0)) + c)
end
↓
function code(x, y, z, t, a, b, c)
t_1 = Float64(Float64(Float64(Float64(z * t) / 16.0) + Float64(x * y)) - Float64(Float64(b * a) / 4.0))
tmp = 0.0
if (t_1 <= Inf)
tmp = Float64(c + t_1);
else
tmp = fma(Float64(t * 0.0625), z, Float64(-0.25 * Float64(b * a)));
end
return tmp
end
code[x_, y_, z_, t_, a_, b_, c_] := N[(N[(N[(N[(x * y), $MachinePrecision] + N[(N[(z * t), $MachinePrecision] / 16.0), $MachinePrecision]), $MachinePrecision] - N[(N[(a * b), $MachinePrecision] / 4.0), $MachinePrecision]), $MachinePrecision] + c), $MachinePrecision]
↓
code[x_, y_, z_, t_, a_, b_, c_] := Block[{t$95$1 = N[(N[(N[(N[(z * t), $MachinePrecision] / 16.0), $MachinePrecision] + N[(x * y), $MachinePrecision]), $MachinePrecision] - N[(N[(b * a), $MachinePrecision] / 4.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$1, Infinity], N[(c + t$95$1), $MachinePrecision], N[(N[(t * 0.0625), $MachinePrecision] * z + N[(-0.25 * N[(b * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]
\left(\left(x \cdot y + \frac{z \cdot t}{16}\right) - \frac{a \cdot b}{4}\right) + c
↓
\begin{array}{l}
t_1 := \left(\frac{z \cdot t}{16} + x \cdot y\right) - \frac{b \cdot a}{4}\\
\mathbf{if}\;t_1 \leq \infty:\\
\;\;\;\;c + t_1\\
\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(t \cdot 0.0625, z, -0.25 \cdot \left(b \cdot a\right)\right)\\
\end{array}
Alternatives Alternative 1 Accuracy 98.8% Cost 8004
\[\begin{array}{l}
t_1 := \left(\frac{z \cdot t}{16} + x \cdot y\right) - \frac{b \cdot a}{4}\\
\mathbf{if}\;t_1 \leq \infty:\\
\;\;\;\;c + t_1\\
\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(t \cdot 0.0625, z, -0.25 \cdot \left(b \cdot a\right)\right)\\
\end{array}
\]
Alternative 2 Accuracy 99.0% Cost 19904
\[\mathsf{fma}\left(x, y, \mathsf{fma}\left(b, -0.25 \cdot a, \mathsf{fma}\left(z, \frac{t}{16}, c\right)\right)\right)
\]
Alternative 3 Accuracy 98.9% Cost 13632
\[\mathsf{fma}\left(t, \frac{z}{16}, \mathsf{fma}\left(x, y, c - b \cdot \frac{a}{4}\right)\right)
\]
Alternative 4 Accuracy 98.6% Cost 7360
\[\mathsf{fma}\left(t \cdot 0.0625, z, x \cdot y + \left(c - b \cdot \frac{a}{4}\right)\right)
\]
Alternative 5 Accuracy 98.5% Cost 2116
\[\begin{array}{l}
t_1 := \left(\frac{z \cdot t}{16} + x \cdot y\right) - \frac{b \cdot a}{4}\\
\mathbf{if}\;t_1 \leq \infty:\\
\;\;\;\;c + t_1\\
\mathbf{else}:\\
\;\;\;\;b \cdot \left(-0.25 \cdot a\right)\\
\end{array}
\]
Alternative 6 Accuracy 66.3% Cost 1616
\[\begin{array}{l}
t_1 := 0.0625 \cdot \left(z \cdot t\right)\\
t_2 := t_1 + x \cdot y\\
t_3 := c + b \cdot \left(-0.25 \cdot a\right)\\
\mathbf{if}\;b \cdot a \leq -1 \cdot 10^{+162}:\\
\;\;\;\;t_3\\
\mathbf{elif}\;b \cdot a \leq -2 \cdot 10^{-178}:\\
\;\;\;\;t_2\\
\mathbf{elif}\;b \cdot a \leq -4 \cdot 10^{-263}:\\
\;\;\;\;c + t_1\\
\mathbf{elif}\;b \cdot a \leq 2 \cdot 10^{+164}:\\
\;\;\;\;t_2\\
\mathbf{else}:\\
\;\;\;\;t_3\\
\end{array}
\]
Alternative 7 Accuracy 36.1% Cost 1508
\[\begin{array}{l}
t_1 := 0.0625 \cdot \left(z \cdot t\right)\\
t_2 := b \cdot \left(-0.25 \cdot a\right)\\
\mathbf{if}\;y \leq -1.9 \cdot 10^{-113}:\\
\;\;\;\;x \cdot y\\
\mathbf{elif}\;y \leq -3.5 \cdot 10^{-264}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;y \leq -4.5 \cdot 10^{-302}:\\
\;\;\;\;t_2\\
\mathbf{elif}\;y \leq 1.9 \cdot 10^{-251}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;y \leq 2.1 \cdot 10^{-213}:\\
\;\;\;\;c\\
\mathbf{elif}\;y \leq 1.1 \cdot 10^{-124}:\\
\;\;\;\;t_2\\
\mathbf{elif}\;y \leq 1.1 \cdot 10^{-21}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;y \leq 11500000000000:\\
\;\;\;\;t_2\\
\mathbf{elif}\;y \leq 3.7 \cdot 10^{+183}:\\
\;\;\;\;t_1\\
\mathbf{else}:\\
\;\;\;\;x \cdot y\\
\end{array}
\]
Alternative 8 Accuracy 57.0% Cost 1504
\[\begin{array}{l}
t_1 := c + 0.0625 \cdot \left(z \cdot t\right)\\
t_2 := c + x \cdot y\\
t_3 := c + b \cdot \left(-0.25 \cdot a\right)\\
\mathbf{if}\;y \leq -1.9 \cdot 10^{-113}:\\
\;\;\;\;t_2\\
\mathbf{elif}\;y \leq -1.5 \cdot 10^{-262}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;y \leq -4.2 \cdot 10^{-301}:\\
\;\;\;\;t_3\\
\mathbf{elif}\;y \leq 3.8 \cdot 10^{-230}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;y \leq 3.2 \cdot 10^{-124}:\\
\;\;\;\;t_3\\
\mathbf{elif}\;y \leq 3.3 \cdot 10^{-22}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;y \leq 54000000000000:\\
\;\;\;\;t_3\\
\mathbf{elif}\;y \leq 3.7 \cdot 10^{+183}:\\
\;\;\;\;t_1\\
\mathbf{else}:\\
\;\;\;\;t_2\\
\end{array}
\]
Alternative 9 Accuracy 88.5% Cost 1353
\[\begin{array}{l}
t_1 := 0.0625 \cdot \left(z \cdot t\right)\\
\mathbf{if}\;b \cdot a \leq -1 \cdot 10^{+162} \lor \neg \left(b \cdot a \leq 2 \cdot 10^{+49}\right):\\
\;\;\;\;\left(c + t_1\right) - \left(b \cdot a\right) \cdot 0.25\\
\mathbf{else}:\\
\;\;\;\;c + \left(t_1 + x \cdot y\right)\\
\end{array}
\]
Alternative 10 Accuracy 86.1% Cost 1225
\[\begin{array}{l}
\mathbf{if}\;b \cdot a \leq -1 \cdot 10^{+162} \lor \neg \left(b \cdot a \leq 2 \cdot 10^{+164}\right):\\
\;\;\;\;c + b \cdot \left(-0.25 \cdot a\right)\\
\mathbf{else}:\\
\;\;\;\;c + \left(0.0625 \cdot \left(z \cdot t\right) + x \cdot y\right)\\
\end{array}
\]
Alternative 11 Accuracy 85.7% Cost 1225
\[\begin{array}{l}
t_1 := 0.0625 \cdot \left(z \cdot t\right)\\
\mathbf{if}\;b \cdot a \leq -1 \cdot 10^{+162} \lor \neg \left(b \cdot a \leq 2 \cdot 10^{+49}\right):\\
\;\;\;\;t_1 - \left(b \cdot a\right) \cdot 0.25\\
\mathbf{else}:\\
\;\;\;\;c + \left(t_1 + x \cdot y\right)\\
\end{array}
\]
Alternative 12 Accuracy 86.6% Cost 1224
\[\begin{array}{l}
t_1 := \left(b \cdot a\right) \cdot 0.25\\
t_2 := 0.0625 \cdot \left(z \cdot t\right)\\
\mathbf{if}\;b \cdot a \leq -1 \cdot 10^{+162}:\\
\;\;\;\;\left(c + x \cdot y\right) - t_1\\
\mathbf{elif}\;b \cdot a \leq 2 \cdot 10^{+49}:\\
\;\;\;\;c + \left(t_2 + x \cdot y\right)\\
\mathbf{else}:\\
\;\;\;\;t_2 - t_1\\
\end{array}
\]
Alternative 13 Accuracy 35.4% Cost 1112
\[\begin{array}{l}
t_1 := 0.0625 \cdot \left(z \cdot t\right)\\
\mathbf{if}\;y \leq -1.3 \cdot 10^{-113}:\\
\;\;\;\;x \cdot y\\
\mathbf{elif}\;y \leq 2.4 \cdot 10^{-251}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;y \leq 8.2 \cdot 10^{-125}:\\
\;\;\;\;c\\
\mathbf{elif}\;y \leq 2.9 \cdot 10^{-22}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;y \leq 0.0135:\\
\;\;\;\;c\\
\mathbf{elif}\;y \leq 3.7 \cdot 10^{+183}:\\
\;\;\;\;t_1\\
\mathbf{else}:\\
\;\;\;\;x \cdot y\\
\end{array}
\]
Alternative 14 Accuracy 52.0% Cost 848
\[\begin{array}{l}
t_1 := c + x \cdot y\\
t_2 := 0.0625 \cdot \left(z \cdot t\right)\\
\mathbf{if}\;t \leq -9 \cdot 10^{-29}:\\
\;\;\;\;t_2\\
\mathbf{elif}\;t \leq 6.1 \cdot 10^{-69}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;t \leq 8.8 \cdot 10^{-26}:\\
\;\;\;\;b \cdot \left(-0.25 \cdot a\right)\\
\mathbf{elif}\;t \leq 1.75 \cdot 10^{+190}:\\
\;\;\;\;t_1\\
\mathbf{else}:\\
\;\;\;\;t_2\\
\end{array}
\]
Alternative 15 Accuracy 56.9% Cost 713
\[\begin{array}{l}
\mathbf{if}\;y \leq -1.9 \cdot 10^{-113} \lor \neg \left(y \leq 6.4 \cdot 10^{+183}\right):\\
\;\;\;\;c + x \cdot y\\
\mathbf{else}:\\
\;\;\;\;c + 0.0625 \cdot \left(z \cdot t\right)\\
\end{array}
\]
Alternative 16 Accuracy 34.2% Cost 456
\[\begin{array}{l}
\mathbf{if}\;y \leq -1.5 \cdot 10^{-182}:\\
\;\;\;\;x \cdot y\\
\mathbf{elif}\;y \leq 3.4 \cdot 10^{+57}:\\
\;\;\;\;c\\
\mathbf{else}:\\
\;\;\;\;x \cdot y\\
\end{array}
\]
Alternative 17 Accuracy 22.6% Cost 64
\[c
\]