I am trying to understand duality explained in the book Computational Geometry Algorithms and Applications, 3rd Ed – de Berg et al. Unfortunately, I have some problem of solving a question in this book. In the book they explain how a point p = ($p_x$ ,$p_y$) can be transformed by a duality transform to $p* := (y= p_x x – p_y)$ .
The duality transform got the following properties:
**Observation 8.3** Let p be a point in the plane and let l be a non-vertical line in the plane. The duality transform o → o∗ has the following properties: 1. It is incidence preserving: p ∈ l if and only if l∗ ∈ p∗ . 2. It is order preserving: p lies above l if and only if l∗ lies above p∗ .
This is the question I am trying to answer:
The dual transform of Section 8.2 has minus signs. Suppose we change
them to plus signs, so the dual of a point ($p_x$ , $p_y$ ) is the
line y = $p_x$ x + $p_y$ ,
and the dual of the line $y = mx + b$ is the point (m, b). Is this dual
transform incidence and order preserving?
The problem is that I don't know how a line would look like if it is defined by $p* := (y= p_x x + p_y)$. I hope that someone could help me solve this question.
Best Answer
The dual transform you propose does not preserve incidence.
Now it is not surprising that the transform is not order-preserving either. I leave it up to you to find an example.
All is not lost, though. Interestingly, there is a plane duality where the point $p = (p_x,p_y)$ is sent to the line $y = p_x\:x + p_y$. However, this duality is not equal to its inverse: to go back from the dual plane to the primal plane, we have to send the dual point $q = (q_x,q_y)$ to the line $y = -q_x\:x + q_y$ in the primal plane. \begin{array}{|ccc|} \hline \text{primal:} & & \text{dual:}\\[2ex] (a,b) & \quad\longleftrightarrow\quad & y = ax + b\\ y = cx + d & \quad\longleftrightarrow\quad & (-c,d)\\ \hline \end{array} This duality does indeed preserve incidence. (I don't know whether it is order preserving; I don't find that particularly interesting. Perhaps you can work it out for yourself if you really want to know.) You can see that it fixes the above example: we reflect the point $\ell^* = (-1,2)$ in the $y$-axis and do indeed obtain a point on the line $p^*$. More generally, it is easily seen that the following are equivalent:
Now how did I find this alternative duality? I could have just tried to make the algebra work (and I'm a little embarrassed now that I didn't do that), but instead I tried to gain some geometric intuition for the problem. Both of these dualities are special cases of duality in projective geometry.
Now, I may choose an identification of the Euclidean plane with a subset of the projective plane $\mathbb{P}^2$. For this I choose to work with the affine hyperplane $\mathcal A = \{(x,y,z) \in \mathbb{R}^3 \: : \: z = -1\}$. This is illustrated in the following example:
Dualising in projective geometry is easy: we send every point and every line to its orthogonal complement. Thus, lines become points and vice versa, and it is easy to see that incidence is preserved (if $p \subseteq \ell$, then $\ell^\perp \subseteq p^\perp$ by elementary linear algebra).
I tried to find the projective interpretation of the duality given in the book by De Berg et al. Note that the point $(0,0)$ is sent to the line $y = 0$; therefore we have a point that is incident with its dual. This is a problem in the projective setting, for no one-dimensional subspace of $\mathbb{R}^3$ is contained in its orthogonal complement. We therefore have to identify the dual plane with some affine hyperplane $\mathcal B \subseteq \mathbb{R}^3$ different from $\mathcal A$. However, there is not much choice left for us: every vertical line in the primal plane must go to a point at infinity. If we embed a vertical line into the projective plane, we see that its dual is a one-dimensional subspace of the $xz$-plane. Therefore $\mathcal B$ must be parallel to $xz$-plane! This doesn't leave a lot of choice, although we still have some options. Let us go with $\mathcal B = \{(x,y,z) \in \mathbb{R}^3 \: : \: y = -1\}$, for reasons of symmetry. (Recall that $\mathcal A$ is the plane $z = -1$, so it seams natural to go with the plane $y = -1$ here.)
As it turns out, this gives us precisely the duality from the book! Let us work out in detail what happens:
Now, how did I find the formula that fixes your proposed duality? I simply replaced the plane $\mathcal B = \{(x,y,z)\in\mathbb{R}^3 \: : \: y = -1\}$ by the plane $\mathcal B' = \{(x,y,z) \in \mathbb{R}^3 \: : \: y = 1\}$ and worked out the duality (embedding the primal plane into the projective plane, taking orthogonal complements, and projecting onto $\mathcal B'$). Since $\mathcal B'$ is the reflection of $\mathcal B$ in the $xz$-plane, I wouldn't be surprised to learn that this duality is order-reversing. But then again, I haven't checked...
Note that this recipe can be used to find other dualities as well. For instance, replace $\mathcal A$ by the plane $\mathcal A' = \{(x,y,z)\in\mathbb{R}^3 \: : \: z = 1\}$. Now if we use $\mathcal A'$ and $\mathcal B'$, we get a duality where the point $p = (p_x,p_y)$ is sent to the line $y = -p_x\: x - p_y$ and vice versa. This duality is again symmetrical, in the sense that the same formula works for going from the primal to the dual plane and vice versa. Perhaps the symmetric dualities are precisely the ones that preserve order? Maybe I'm being too optimistic now. Nevertheless, there is still much about plane duality left for you and me to discover!