Is
\[f(x) = (-1)^{\lfloor x \rfloor} \sqrt{\frac{1}{4} - \left( x - \lfloor x \rfloor - \frac{1}{2} \right)^2}\]an even function, odd function, or neither?

Enter "odd", "even", or "neither".
If $x$ is an integer, then $x = \lfloor x \rfloor,$ so
\[\frac{1}{4} - \left( x - \frac{1}{2} - \lfloor x \rfloor \right)^2 = \frac{1}{4} - \frac{1}{4} = 0,\]which means $f(x) = 0.$

Otherwise, $\lfloor x \rfloor < x < \lfloor x \rfloor + 1,$ so
\[-\lfloor x \rfloor - 1 < -x < -\lfloor x \rfloor,\]which means $\lfloor -x \rfloor = -\lfloor x \rfloor - 1.$  Hence,
\begin{align*}
f(-x) &= (-1)^{\lfloor -x \rfloor} \sqrt{\frac{1}{4} - \left( -x - \lfloor -x \rfloor - \frac{1}{2} \right)^2} \\
&= (-1)^{-\lfloor x \rfloor - 1} \sqrt{\frac{1}{4} - \left( -x + \lfloor x \rfloor + 1 - \frac{1}{2} \right)^2} \\
&= (-1)^{-\lfloor x \rfloor - 1} \sqrt{\frac{1}{4} - \left( -x + \lfloor x \rfloor + \frac{1}{2} \right)^2} \\
&= -(-1)^{-\lfloor x \rfloor} \sqrt{\frac{1}{4} - \left( x - \lfloor x \rfloor - \frac{1}{2} \right)^2} \\
&= -f(x).
\end{align*}Therefore, $f(x)$ is an $\boxed{\text{odd}}$ function.

The graph of $y = f(x)$ is the following:

[asy]
unitsize(2.5 cm);

draw(arc((1/2,0),1/2,0,180),red);
draw(arc((3/2,0),1/2,180,360),red);
draw(arc((5/2,0),1/2,90,180),red);
draw(arc((-1/2,0),1/2,180,360),red);
draw(arc((-3/2,0),1/2,0,180),red);
draw(arc((-5/2,0),1/2,270,360),red);
draw((-2.5,0)--(2.5,0));
draw((0,-1/2)--(0,1/2));

label("$\dots$", (2.7,0));
label("$\dots$", (-2.7,0));

dot("$(\frac{1}{2},0)$", (1/2,0), S);
dot("$(\frac{3}{2},0)$", (3/2,0), N);
dot("$(-\frac{1}{2},0)$", (-1/2,0), N);
dot("$(-\frac{3}{2},0)$", (-3/2,0), S);
[/asy]

The graph consists of semicircles centered at the half integers, with radii $\frac{1}{2}.$