Superellipse on a square budget

Question

What’s better than an ordinary ellipse?   A super ellipse, of course.

And what’s better than a superellipse?   That’s right, a superellipse and a square.

But who can plot both of those luxury items?   Anyone, with a single function you can make at home.

Here are implicit function plots of a superellipse, two squares, and a handy almost-superellipse loop.

$$\require{begingroup} \begingroup \def \K { \kern-.5em } \def \p #1#2{ | {2 \over \large\raise.2ex\pi} #1 \kern.1em | ^ {#2} \! } \begin{matrix} &&&& \small\sf \rlap { In~the~lap~of~luxury } &&& &&&& \small\sf \rlap{ On~a~budget } \\[2ex] \small\sf Loop &&& \large {\raise.2ex 1 \over 2\surd2} &\K = &\K f(x,y) & \K = & \K \p x{2.5} + \p y{2.5} &&& 1 & \K = & \K h \, (x{+}y \, ,x{-}y) \\[1ex] \small\sf Square &&& 1 &\K = &\K g(x,y) & \K = & \K \p x\infty + \p y\infty &&& 0 & \K = & \K h \, (x{+}y \, ,x{-}y) \end{matrix} \kern2em \\ \tiny\strut \endgroup$$

The status- conscious among us may impress themselves by recognizing  $ {\raise.2ex 1 \over 2\surd2} = f(x,y) $  as nothing less than a genuine superellipse, yet the handy-dandy  $\raise-.5ex\strut 1 = h \, (x{+}y \, ,x{-}y) $  is almost identical, with 12 common points and less than .011 of maximum separation.

Counting typographically, $f(x,y)$ and $g(x,y)$ are defined by 19 and 15 raw components, respectively, including fraction lines, decimal points, and everything else after the equals signs.

       Can you define $h(x,y)$ with fewer than 10 raw components?

The only components available are those already present in the definitions of $f$ and $g$, as well as all other digits and constants, along with any other letters as long as they spell out trigonometric functions.   Every letter counts, so, for example, ${ \small\raise .5ex \unicode {8220} } \kern-2mu \sin \kern-3mu { \small\raise.5ex \unicode{8221} }$ would contribute 3 components.

Both loops pass through $( \pm { \Large\pi \over \large 4 } , \pm { \Large\pi \over \large4 } )$ and the squares’ sides are at $x = \pm { \Large\pi \over \large 2 }$ and $y = \pm { \Large\pi \over \large2 }$.   Portions of the Cartesian plane beyond the square may contain other points and curves.   Never mind that $g(x,y)$ is defined casually and without regard to the square’s vertices.

^{Online freebies that helped in preparation and could help in solution:
Function Grapher – Good with explicit plots; no ads.
MathGrapher at eMathHelp – Good with implicit plots, sometimes bad with pop-up ads.}

Answer 1

Answer:

$\cos~x + \cos~y$

Solution:

Approximate the x and y intercepts of the loop as $\frac{\pi}{3}$ (since $2^{-\frac{8}{5}} \simeq \frac{1}{3.03}$). Then we need $h(\pm \frac{\pi}{3}, \pm \frac{\pi}{3}) = 1$, so to turn those $\pm\frac{\pi}{3}$s into something reasonable we probably need $\cos~x$ (there are other ways, of course, but it was hinted that we need trig and also it takes care of both positive and negative very compactly: $\cos~\frac{\pi}{3} = \cos~\frac{-\pi}{3} = \frac{1}{2}$). It's likely to be symmetric in $x$ and $y$ so try out $\cos~x + \cos~y$ for now. We've got the loop; now to check what that looks like when $h = 0$ so we can modify it for the square and woah we're already done.

Answer 2

Non-solution that may yet be of some interest:

If we write

$h(u,v)=2-\cos u-\cos v$

then (exactly) the right thing happens at $h=0$ and we are within the necessary bounds at $h=1$. But the definition is, alas!, 11 symbols long.

[EDITED to add:]

D'oh. Actually the above is the result of a mis-conversion from my MATLAB fiddlings, and doesn't in fact do the right thing at $h=0$. If I do the conversion correctly, what I should have written before was

$h(u,v)=\cos u+\cos v$

... but bmcfluff got there after the wrong version of my answer and before this correction. So it goes :-).