### Trying to draw the Amoeba


With Mathematica, it's possible to graph $e^{-k x} + e^{-k y} = 1,e^{-k x} - e^{-k y} = 1$ and $e^{-k x} + e^{-k y} = -1$ to get the amoeba of **1 + x + y** when **k = 1**. Then by plotting when $k \to \infty$, these graphs converge to the tropical polynomial 


<http://www.freeimagehosting.net/uploads/7a0cc0b6f3.jpg>


However, I meant to draw **Min(1,x,y)** which I fixed by shifting my coordinates **x,y** by the vector **(1,1)**. This is a bit ad-hoc and I am probably not understanding how constant functions "tropicalize".


### Main question


I would like to "fatten" **Min[x, y, 1, x + y + 1]** into its amoeba, so I thought the right curve should be **1 + x + y + xy**. However, my "neck" is disappearing in the scaling limit. *How should I scale the coefficients correctly to get my amoeba?*


Following Zeb's suggestion (but a few second before he posted it) I came up with this imge
<http://www.freeimagehosting.net/uploads/b612de3bf9.gif>


However, this "dequantization" procedure doesn't always produce the whole tropical curve. Here's the curve I drew to "requantize" **Min[1, x , y , x+ y + 1, -2 + 2x , 2y]**. A line has to be missing b/c of the zero tension condition, as in [Tropical Mathematics](http://arxiv.org/abs/math/0408099) by David Speyer and Bernd Sturmfels.


<http://www.freeimagehosting.net/uploads/50219e2142.gif>


Here is the code. You have to draw 4 different versions of the curve to get all the absolute values. Maybe this should somehow involve complex phases as well.



```
q[x_] := E^( x)
f[a_, b_, c_] := c + a + b + c a b + (1/c^2) a^2 + b^2;
{x0, y0} = { -1, -3};
L = 5;
k = 8;
ContourPlot[
    { f[q[k x], q[k y], q[k ]] == 0, f[-q[k x], q[k y], q[k ]] == 0, 
    f[q[k x], -q[k y], q[k ]] == 0, f[-q[k x], -q[k y], q[k ]] == 0}, 
    {x, x0, x0 + L}, {y, y0, y0 + L}
]

```

Ideally, I want to take any tropical curve and fatten it into its amoeba. Tropical conics and cubics seem the best starting point.




---


In anticipation of comments, by "amoeba" here I think I mean the boundary of the 2D region which is usually called "amoeba".



How to Tropicalize a Polynomial in Two Variables?
For the first amoeba you mentioned, I think your equations should be $e^{-kx}\pm e^{−ky}=\pm e^{-k}$, not $e^{-kx}\pm e^{−ky}=\pm e^{0}$.


For the main question, I think you should be using an equation like $e^{-k}\pm e^{-kx}\pm e^{-ky} \pm e^{-k(x+y+1)} = 0$... so really you want a curve like $c+x+y+cxy$, where when you rescale $x$ and $y$ by raising them both to a power, you also rescale the coefficient $c$ by raising it to the same power.


Edit: Ok, for the second problem you are having, I think this is coming up because plugging in different signs of $x$ and $y$ doesn't give you all the different sign possibilities for your detropicalized polynomial.


So, if you want to get the tropical curve $Min(1,x,y,1+x+y,2x-2,2y)$, you want to use all of the equations $e^{-k}\pm e^{-kx}\pm e^{-ky} \pm e^{-k(x+y+1)} \pm e^{-k(2x-2)} \pm e^{-k(2y)} = 0$.


In fact, I think you don't need to use all $32$ of those equations, you just need to use enough of them that every pair of terms have opposite signs in one of your equations, such as the following three:


$e^{-k}+ e^{-kx}+ e^{-ky} - e^{-k(x+y+1)} - e^{-k(2x-2)} - e^{-k(2y)} = 0$
$e^{-k}- e^{-kx}+ e^{-ky} - e^{-k(x+y+1)} + e^{-k(2x-2)} - e^{-k(2y)} = 0$
$e^{-k}+ e^{-kx}- e^{-ky} - e^{-k(x+y+1)} - e^{-k(2x-2)} + e^{-k(2y)} = 0$


In the limit this will give you the right amoeba, but I'm cheating a bit, because really we should be doing fancy stuff with logarithms of complex numbers.


