jmc

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, $$\begin{array}{rl}f(-x)& =(-1{)}^{\lfloor -x\rfloor }\sqrt{\frac{1}{4}-{\left(-x-\lfloor -x\rfloor -\frac{1}{2}\right)}^2}\\ & =(-1{)}^{-\lfloor x\rfloor -1}\sqrt{\frac{1}{4}-{\left(-x+\lfloor x\rfloor +1-\frac{1}{2}\right)}^2}\\ & =(-1{)}^{-\lfloor x\rfloor -1}\sqrt{\frac{1}{4}-{\left(-x+\lfloor x\rfloor +\frac{1}{2}\right)}^2}\\ & =-(-1{)}^{-\lfloor x\rfloor }\sqrt{\frac{1}{4}-{\left(x-\lfloor x\rfloor -\frac{1}{2}\right)}^2}\\ & =-f(x).\end{array}$$Therefore, $f(x)$ is an $\boxed{\text{odd}}$ function.

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



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