\section {10.2. Characterization of convex tracings}
Convex tracings have a rather straightforward definition, which we choose to present below in the language of car trips.
\definition{\Defi}{A polygonal tracing is said to be {\it convex} if it consists of a single tour around a convex polygon, traversed in such a way that the polygon lies always to the same side of the car.}
This definition is quite natural, but it is somewhat hard to use in the context of polygonal tracings. The rest of this section will be devoted to giving a couple of alternative characterizations of convex tracings, of a more ``local'' character. These characterizations are based only in the machinery we have developed in the previous chapter, without explicit reference to the idea of convexity.
Convex tracings satisfy many interesting properties. Some of the most important are listed below:
\indent $\PP0$. The car never passes twice through the same point or faces twice along the same line.
\hangindent 40pt after0
In more precise terms: for any point $p$ of the plane there is at most one turn $T$ and one {\it moving} instant $t$ such that $\po T(t)=p$; for any line $\lscr$, there is at most one turn $T$ and one {\it turning} instant $t$ such that $\ta T(t)=\lscr$. Note that there are infinitely many instants when the car sits still while doing a turn, or faces along the same line while moving, but these instants do not count as far as $\PP0$ is concerned.
\indent $\PP1$. The line of the car never passes between two points of the tracing.
\indent $\PP1\pr$. The car never lies between two tangents of the tracing.
\hangindent 40pt after0
Property $\PP1$ says that the line along which the car is facing never passes simultaneously to the left of some point of the tracing and to the right of some other point. In a similar vein, $\PP1\pr$ means the car never lies simultaneously to the left of some tangent of the tracing and to the right of some other tangent.
\figspace 3cm
\indent $\PP2$. The car moves always forwards or always backwards.
\indent $\PP2\pr$. The car turns always to the left or always to the right.
\indent $\PP3$. The car never faces twice towards the same point.
\indent $\PP3\pr$. The car never peeks twice out of the same line.
\hangindent 40pt after0
Property $\PP3$ means that for any point $x$ of the plane, there is at most one tour $T$ and one {\it turning} instant $t$ such that $T(t)$ is facing towards $x$. Note that during any move there are infinitely many instants where the car is facing towards the same point, but these do not count as far as $\PP3$ is concerned. In a similar way, $\PP3\pr$ means that for any line $\lscr$ there is at most one tour $T$ and one instant $t$ such that $T(t)$ is on $\lscr$ and facing to the right of it.
\figspace 3cm
These properties are not independent of each other; indeed, for any polygonal tracing,
{\def\eqq{\;\;eqv\;\;}
$$ \PP0 \and \PP1\eqq \PP0 \and \PP1\pr\eqq \PP2 \and \PP3\eqq \PP2\pr \and \PP3\pr,\eqno(1)$$}
and each of these pairs implies the tracing is convex (if not empty). We will prove (1) in the following sequence:
\indent $\hbox{\it Convexity} \implies \PP0 \and \PP1\and \PP1\pr$ (theorem {\Thevii});
\indent $\PP0 \and \PP1\implies \PP2 \and \PP2\pr \and \PP3\and \PP3\pr$ (theorem {\Theviii});
\indent $\PP2 \and \PP3\implies \PP0 \and \PP1$ (theorem {\Theix});
\indent $\PP0 \and \PP1\implies\hbox {\it convexity}$ (theorem {\Thex});
\indent $\PP2\pr \and \PP3\pr\implies \PP0 \and \PP1\pr$ (theorem {\Thexi});
\indent $\PP0 \and \PP1\pr\implies\hbox {\it convexity}$ (theorem {\Thexii}).
It is not hard to prove that $\PP3$ remains true in the limiting case when the point goes to infinity. That is, {\it in a convex tracing the car faces in each direction at most once}. Indeed, we could have replaced $\PP3$ by this weaker property, and it would still be the case that $\PP2$ and $\PP3$ implies convexity.
\theorem{\Thevii}{A convex tracing satisfies $\PP0$, $\PP1$ and $\PP1\pr$.}
\proof{Immediate.}
\theorem{\Theviii}{If a polygonal tracing satisfies $\PP0$ and $\PP1$, then it contains at most one tour, and satisfies also $\PP2$, $\PP2\pr$, $\PP3$, and $\PP3\pr$.}
\proof{First, note that $\PP0$ forbids null moves and turns, for during them the car stays for infinitely many instants on the same point and facing into the same tangent, respectively.
\mfig 4cm
\pp Suppose now a tour of the tracing contained both forward and backward moves; then there would be a forward move $[si\to si\pr)$ followed by a backward one $[s{i+1}\to s{i+1}\pr)$. But then, for any state $r$ in the intervening turn $(si\pr\to s{i+1})$, the tangent $\ta r$ would pass between the points $\po si$ and $\po s{i+1}\pr$ of the tracing, contradicting $\PP1$.
\endmfig
\pp Suppose now a tour contained both left and right turns; there would be a left turn $[s{i-1}\pr\to si)$ followed by a right one $[si\pr\to s{i+1})$. Since the moves are all done in the same way, the points $\po s{i-1}$ and $\po s{i+1}\pr$ would lie on opposite sides of the tangent of the move $[si\to si\pr)$, again contradicting $\PP1$.
\pp Therefore, all moves and turns on the same tour are performed in the same way. Now suppose the tracing contains two distinct tours $R$ and $S$. Each contains at least three non-colinear points, and therefore we can always find a line $\lscr$ that passes between two points of $R$ and also between two points of $S$. By lemma {\Theiii}, there are two moving states $r1$ and $s1$ that peek to the right of $\lscr$, and two others $r2$ and $s2$ that peek to the left of it, all with distinct positions by $\PP0$.
\figspace 3cm
\pp It is easy to check that, no matter how they are arranged along $\lscr$, one of the four tangents will pass between two of the four points, contradicting $\PP1$. We conclude therefore that the tracing has at most one tour $T$, and satisfies $\PP2$ and $\PP2\pr$.
\pp Let's now suppose there are two distinct turning instants $a$ and $b$ such that $T(a)$ and $T(b)$ both face towards some point $x$. These two states cannot have the same tangent, because of $\PP0$; therefore, there is some line $\lscr$ that passes through $x$ all of whose points (except $x$) lie between the two tangents at $a$ and $b$.
\figspace 3cm
\pp The points $T(a)$ and $T(b)$ must be on opposite sides of $\lscr$, and by lemma {\Thei} the car must pass twice through $\lscr$, once when going from $T(a)$ to $T(b)$ and a second time when returning from $T(b)$ to $T(a)$. Because of $\PP1\pr$, it cannot cross the line anywhere except on $x$; but then it will pass twice through the same point, contradicting $\PP0$.
\pp To conclude, let us suppose there are two moving instants $a$ and $b$ when $T$ is peeking out of some line $\lscr$. The two states $T(a)$ and $T(b)$ have different positions because of $\PP0$; let $x$ be any point on $\lscr$ between them. By lemma {\Theii}, there are two distinct turning instants (one when going from $T(a)$ to $T(b)$ and the other when returning from $T(b)$ to $T(a)$) when the line of the car passes through the point $x$. The tangents at those instants must coincide with $\lscr$ or its opposite, otherwise they will pass between $\po T(a)$ and $\po T(b)$.
\figspace 3cm
\pp Because of $\PP0$, those two tangents must be opposite to each other. But then the moves to which $T(a)$ and $T(b)$ belong lie on the same side of $\lscr$, and therefore in opposite sides of the two tangents, contradicting $\PP1\pr$.}
Therefore, $\PP0$ through P$3\pr$ are valid for convex tracings. What is more important, these properties characterize convextity, as the next two theorems show:
\theorem{\Theix}{If a tracing satisfies $\PP2$ and $\PP3$, then it satisfies also $\PP0$ and $\PP1$.}
\mfig 3cm
\proof{First, notice that, because of $\PP3$, in each tour of the tracing all turns are made in the same way: null turns are automatically forbidden, and two consecutive turns in opposite directions always contain two states that face towards the same point.
\endmfig
\mfig 6cm
\pp Now suppose there were two distinct tours $R$ and $S$. As in the proof of theorem {\Theviii}, there should be some line $\lscr$ that passes between two points of $R$ and also between two points of $S$. By lemma {\Theiii}, there should be two states $r1$ of $R$ and $s1$ of $S$ peeking to the left of $\lscr$, and two others $r2$ and $s2$ peeking to the right of it. Then a point $x$ on $\lscr$ sufficiently far away in the direction of $\di \lscr$ would be to the right of $r1$ and $s1$ and to the left of $r2$ and $s2$. By lemma {\Theiv}, for two turning instants $c$ and $c\pr$ we would have $R(c)$ and $S(c\pr)$ facing towards $x$, contradicting $\PP3$. So, there is at most one tour.
\endmfig
\pp Let us now prove that $\PP1$ holds. Assume first all turns are to the left, and suppose there is some turning state $P(s)$ whose tangent passes between two points of the tour. Then by lemma {\Theiii} there would be a moving state $P(c)$ peeking to the left of $\ta P(s)$. If $c\pr$ is the end of the corresponding move, then $P(c\pr)$ lies to the left of $P(s)$; there would be an instant $d\in[c\pr\sp s)$ when $P(d)$ is to the left of $\ta P(s)$ but facing towards it.
\figspace 3cm
\pp Then if we were to choose $x$ on the forward ray of $P(s)$ and sufficiently far away, $x$ would lie to the right of $P(c)$ and to the left of $P(d)$; by lemma {\Theiv}, there would be a turning instant $e\in(c\sp d)$ when $P(e)$ faces towards $x$. But $P(s)$ also faces towards $x$, a contradiciton. A similar argument holds if all turns are to the right, so we conclude $\PP1$ holds.
\mfig 4cm
\pp Finally, let's show $\PP0$ holds. Clearly, $\PP3$ forbids two turning instants with same tangent. Suppose there were two moves $[si\to si\pr)$ and $[sj\to sj\pr)$ passing through the same point $x$. If the tour is moving backwards, then $si\pr$ and $sj\pr$ are distinct turning states that face towards point $x$. If the tour moves forward, then the turns ending at $si$ and $sj$ contain two states that face towards some point in a neighborhood of $x$ (a special argument is needed for the case when the two moves have opposite orientations, but we will leave the details to the reader). Either case, $\PP3$ is contradicted.}
\endmfig
\theorem{\Thex}{A nonempty tracing that satisfies $\PP0$ and $\PP1$ is convex.}
\proof{By theorem {\Theviii}, the tracing contains a single tour $\langle s0\, s0\pr\, s1\, s1\pr\, \ldots\, s{n-1}\, s{n-1}\pr\rangle$. The path traversed by the car is the polygon $\s P=\langle \po s0, \po s1, \ldots, \po s{n-1}\rangle$, which is simple because of $\PP0$. If the car always turns counterclockwise, then each move $[si\to si\pr)$, and therefore all points of the tracing, will lie to the left of the tangent supporting the next move $[s{i+1}\to s{i+1}\pr)$. So, the polygon $\s P$ is contained in the intersection of the left half-planes of the lines that support its edges; this means it is a convex polygon.
\pp A similar argument holds for the case when all turns are to the right.}
\theorem{\Thexi}{If a tracing satisfies $\PP2\pr$ and $\PP3\pr$, then it also satisfies $\PP0$ and $\PP1\pr$.}
\theorem{\Thexii}{A nonempty tracing that satisfies properties $\PP0$ and $\PP1\pr$ is convex.}
The proofs of these theorems will become trivial after we have studied the concept of geometric duality, so we will not bother to give them here. Of course, both theorems can also be proved directly, by ``dualizing'' the proofs of theorems {\Theix} and {\Thex}.
In conclusion, $\PP0$ and $\PP1$ completely characterize convex tracings, and the same is true of $\PP2$ and $\PP3$, $\PP0$ and $\PP1\pr$, and $\PP2\pr$ and $\PP3\pr$. Note that they are almost ``local'' properties and can be expressed using only the language of polygonal tracings, as we wanted.