(pow (* z0 (floor z1)) 2)

Percentage Accurate: 99.8% → 99.8%
Time: 8.7s
Alternatives: 1
Speedup: 1.0×

Specification

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\[{\left(z0 \cdot \left\lfloor z1\right\rfloor \right)}^{2} \]
(FPCore (z0 z1)
  :precision binary64
  (pow (* z0 (floor z1)) 2))
double code(double z0, double z1) {
	return pow((z0 * floor(z1)), 2.0);
}

public static double code(double z0, double z1) {
	return Math.pow((z0 * Math.floor(z1)), 2.0);
}

def code(z0, z1):
	return math.pow((z0 * math.floor(z1)), 2.0)

function code(z0, z1)
	return Float64(z0 * floor(z1)) ^ 2.0
end

function tmp = code(z0, z1)
	tmp = (z0 * floor(z1)) ^ 2.0;
end

code[z0_, z1_] := N[Power[N[(z0 * N[Floor[z1], $MachinePrecision]), $MachinePrecision], 2], $MachinePrecision]

{\left(z0 \cdot \left\lfloor z1\right\rfloor \right)}^{2}

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.

Accuracy vs Speed?

Herbie found 1 alternatives:

AlternativeAccuracySpeedup
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.

Initial Program: 99.8% accurate, 1.0× speedup?

\[{\left(z0 \cdot \left\lfloor z1\right\rfloor \right)}^{2} \]
(FPCore (z0 z1)
  :precision binary64
  (pow (* z0 (floor z1)) 2))
double code(double z0, double z1) {
	return pow((z0 * floor(z1)), 2.0);
}

public static double code(double z0, double z1) {
	return Math.pow((z0 * Math.floor(z1)), 2.0);
}

def code(z0, z1):
	return math.pow((z0 * math.floor(z1)), 2.0)

function code(z0, z1)
	return Float64(z0 * floor(z1)) ^ 2.0
end

function tmp = code(z0, z1)
	tmp = (z0 * floor(z1)) ^ 2.0;
end

code[z0_, z1_] := N[Power[N[(z0 * N[Floor[z1], $MachinePrecision]), $MachinePrecision], 2], $MachinePrecision]

{\left(z0 \cdot \left\lfloor z1\right\rfloor \right)}^{2}

Reproduce

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herbie shell --seed 2025277 -o generate:taylor -o generate:evaluate
(FPCore (z0 z1)
  :name "(pow (* z0 (floor z1)) 2)"
  :precision binary64
  (pow (* z0 (floor z1)) 2))