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The Zone Theorem
The Cutting Lemma Revisited
Tom Jurgenson
1
The Zone Theorem
2
Definitions reminders
 Is a sub-space of d-1 dimensions.
 Is a partition of
into relatively open convex sets.
 Are 0/1/(d-1)-dimension faces (respectively) in
 Are the connected components of
3
.
also called d-faces
Example d=2 (Plane)
 Vertices are points, edges are segments, and facets are also
segments since 2-1=1.
 Cells are 2-dimesional connected components in
 The blue lines are the Hyperplanes in H
 Their arrangement creates:
 Vertices – in red
 Edges\Facets – blue segments(excluding red)
 Cells – light blue areas.
4
The Zone Theorem - Motivation
 Motivation – we would like to bound the number of faces
that can see any part of a given hyperplane. For that we’ll
need to define what ‘see’ means, and define the set of objects
(faces) that ‘see’ g (the zone).
5
Definition for ‘sees’
 A face F can see hyperplane g of H (set of hyperplanes), if
there are points
such that the open segment
xy does not intersect any hyperplane of H.
 Important Note: It does not matter which point x we
choose, either all of them can see g or none can.
 In this example, the red line represents a segment between
the thick line – g, and the face – the cell. The blue line
exemplifies the fact that no segment connects that cell with
the hyperplane g.
6
Zone Definition
 Zone – The zone of hyperplane g, is the set of the faces of the
arrangement of H that can see g.
 Here is the zone (dark grey area) for hyperplane g (the thick line)
and hyperplane set H (all the thin lines). Each face (vertex, edge
and cell) in the zone of g can see g.
7
The Zone Theorem
The number of faces in the zone of any hyperplane in an
arrangement of n hyperplanes in
is
with the
constant of proportionality depending on d.
we will prove this theorem in an inductive manner
8
Base case, d=2 (The plane)
 Let H be a set of n lines in the plane in general position. We
consider the zone of line g.
 In this case the faces would be: vertices (0-faces), edges (1faces), and the cells are convex polygons that may or may not
be bounded(1-faces).
 The bound for the number of faces in the zone in this case
should be
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Bound for Edges is enough
 The number of vertices is proportional to the number of
edges since in a convex polygon the number of vertices is
equal to the number of edges. And in open cells, the number
of edges is at most twice the number of vertices.
 It is clear that the number of polygons is bounded by the
number of vertices by using Euler’s formula: F=v-e+2.
Therefore proportional to the number of edges.
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Bound for Edges
 Let g be a horizontal line.
 We’ll count the number of edges that see g that are also
above g.
 The number of edges that intersect g that are also above g is
bounded by n. Since each line (hyperplane in H) out of the n
lines can only intersect g in one point.
 The other edges are disjoint from g, and in the next slides we
are going to find a bound for these edges.
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Bound for the disjoint edges
 Let uv be an edge that sees g but is disjoint from it.
 Let
be the line containing uv. And let a be the
intersection of g and h.
 Since u is a vertex - it is the intersection of two lines – so let
be the other line that passes through u. Let b be the
intersection of l and g.
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Right Edge of Line l
 Definition: uv is right edge of l if b is on the right of a. uv is
left edge of l if b is on the left of a.
 We’ll show that for each such
there is at most one
right edge.
 Let’s assume uv is a right edge for l.
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Only one right edge
 Assume by contradiction that there are two right edges for l.
Let uv and xy be the two right edges, and assume that uv lies
below xy.
 By definition xy should see some point of g.
 However, every point to the left of a on g is obscured by the
line l. And every point to the right of a on g is obscured by
the line h.
 This contradiction shows that only one right edge can exist
for each line.
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Only one right edge (2)
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Conclusion for d=2
 Each line in H has at most one right edge, and by symmetry
also at most one left edge. There is also an edge that
intersects g. The same can be argued for edges below g, for
the lower right and lower left edges, and an edge that
intersects. Altogether there are at most 6 edges for each
line in H in the zone.
 There are n lines in H so the number of edges is at most 6n.
Linear in n.
 We already argued that the total number of faces is linear in
the number of edges therefore the number of faces in the
zone when d=2 is:
as needed.
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Inductive Step from d-1 to d
 Assume that the total number of faces of a zone in
is
 Mark
as the max. number of (d-1) faces in the zone in
an arrangement of n hyperplanes in
.
 Let H be an arrangement of hyperplanes and g a base
hyperplane s.t
is attained for them.
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Motivation: number of blue facets
 We color one hyperplane in H red, let it be




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. The rest
n-1 hyperplanes in H we color blue.
A Blue Facet is a facet that lies in a blue hyperplane.
Since every facet has a chance of
becoming blue
(choose any blue hyperplane out of n blue hyperplanes), the
expected number of blue facets is
Now we bound the expected number of blue facets in a
different way and use it to estimate
Example in the next slide…
Blue and Red Facets
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Bound for Expected Blue Facets
 Consider the arrangement of blue hyperplanes only. Since
there are n-1 such hyperplanes the number of blue facets is
bounded by
 In the next stage we add the red hyperplane h and see how it
affects the number of blue facets in the zone of g.
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The increase in number of blue facets
caused by h
 The red hyperplane h may split an existing blue facet F into
two new facets – F1 and F2. If both new facets can see g then
we get an increase in the number of facets.
 Claim:
is visible from
(we’ll prove it later)
 If we intersect all blue hyperplanes with g and with h we get
a d-1 dimensional arrangement in which
is a facet in
a zone of the d-2 dimensional hyperplane
. Using the
induction assumption we get
such “splits”. The
total number of facets becomes:
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Split Example
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Proof for the above claim
 Reminder: We wanted to show the following claim:
is visible from
 Proof: Let C be a cell of the zone in the arrangement of blue
hyperplanes having F on the boundary.
 Because F1 and F2 see g we get that
sees
and
sees
 The interior of
is contained in C and we get that
the intersection of
with the hyperplane h
contains a segment witnessing the visibility of
from
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Split Example (II)
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Conclusion for facets
 On the one hand we calculated the expected number of blue
facets:
 On the other hand we found an upper bound for the number
of blue facets:
 We combine the two and get:
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Conclusion for facets (II)
 And as we can see:
as needed.
 Next we are going to discuss d-k faces and not just d-1 faces (or
facets)
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Expected number of d-k faces
 Denote




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the max. possible number of j-faces in the
zone for an arrangement of n hyperplanes of dimension d.
Let H be an arrangement of n hyperplanes where
is attained.
Again a random hyperplane h in H is colored red and the rest
blue.
Reminder: a d-k face is the intersection of k hyperplanes.
A d-k face is blue if its relative interior is disjoint from the
red hyperplane h.
Expected number of d-k faces (2)
 The probability for a d-k face to be blue is
:
[For the selection of first blue hyperplane, then the second on
so on until we select the k-th hyperplane]
 As in the facet case we get that the expected number of blue
d-k faces in the required zone is:
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Upper bound for d-k faces
 By adding the red hyperplane h the number of blue d-k faces
can increase by at most
by the inductive
hypothesis.
 Using the same arguments in lower dimensions as the facets
case we get to the conclusion that:
 And in the next slide we’ll show:
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Algebraic proof
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Bound for edges and vertices
 For the case k=d-1 (edges) by using this method we only get
the bound:
 So the number of edges and vertices must be bound
separately:
 Vertices: Each vertex is contained in some3-face of the zone.
Within such 3-face, the number of vertices is at most 3 times
the number of 2-faces, because the 3-face is a 3D convex
polyhedron. Since H is simple each 2-face is contained in a
bounded number of 3-face. It follows that the total number of
vertices is at most proportional to
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Final details
 Edges are proportional to vertices.
 In conclusion, for every k=1…d-2 we get
 For vertices and edges we get
 In conclusion we get that the zone has
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Zones in other arrangements
 The max. complexity of a zone can be investigated for
objects other than hyperplanes, and that leads to many
related problems that we won’t discuss here.
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Cutting Lemma Revisited
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General Plan
 First, we’ll prove the cutting lemma for the planar case with
a tight bound. We already saw a proof for a bound that was
not tight that uses random sampling, and a different proof
that is tight but does not use random sampling.
 The following proof is both tight and uses random
sampling in a way that makes it easy to be generalized
to higher dimensions.
 But first we’ll do a review for some cutting-lemma related
definitions.
35
Review of Definitions (I)
 A generalized triangle is the intersection of 3 half planes.
 A cutting of the plane into generalized triangles, is the
subdivision of the plane to a disjoint set generalized triangles
that their union covers the entire plane.
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Review of Definitions (II)
 A
cutting of the plane with a set of n of lines H – is a
cutting of the plane into generalized triangles
s.t the interior of each such
is intersected
by at most
lines from H.
 Note – the triangles’ edges may or may not include lines
from H.
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½ cutting example
 In the next example we have a ½ cutting. The red lines are
lines in H, the black lines are the lines that create the
triangulation.
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The Cutting Lemma
 For every set H of n lines in the plane and every r>1 there
exists a 1/r cutting for H the size
 In other words, there is a subdivision of the plane into
generalized triangles s.t the interior of each triangle is
intersected by at most n/r lines of H.
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The sampler
 Goal – sample line. Once the lines are sampled their
intersection (possibly with some modifications) would result
in a triangulation.
 In the previous lecture when we wanted to select a subgroup
S of H, we picked |S| lines out of H with repetitions. This
procedure may result with a subgroup of size smaller than
|S|.
 In order to simplify the calculation, we’ll choose S by
independent Bernoulli trials. We fix the probability of p=s/n,
and include a line from H in S using that probability.
 For this point on, let s=|S|
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Why sampling and triangulation does
not work?
 In a previous lecture we already saw a sampling algorithm
that outputs a triangulation that with probability close to 1
non of the triangles in the outputted triangulation is
intersected by more than
where is some
constant (Weak Cutting Lemma)
 We will demonstrate that a similar statement with a
is not generally true. So our method would have to include
more than just sampling in order to achieve the desired
triangulation.
 To see that we’ll examine a simpler case than the plane – 1dimensional situation.
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The 1-D case
 H is a set of n points.
 A generalized triangle in this case is the segment between two
points.
 We would like to count the longest consecutive count of
unselected points of H. Let k be our target number for such a
count.
 In example below, n=30 and s=15 so p=0.5. The black dots
are selected points (In this example there are exactly 15
selected dots, but it doesn’t have to be this way)
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k consecutive unselected points
 We’ll show that for a fixed k it is very likely that k
consecutive unselected points show up in a sequence of n
points, where n is sufficiently large
 For simplicity assume n is divisible by k and divide our n
points in to blocks the length of k. Notice that this is an even
more restrictive case than a general consecutive count.
 In each such a block there is a probability of
for the entire block to be unselected. Therefore the
probability of not obtaining any block as entirely unselected
is
and this probability is
exponentially small for k proportional to
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Why is it exponentially small?
 p is a constant, let p=0.5. The probability becomes:
 Now set k=0.5logn:
 And that exponentially small as n goes to infinity.
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Conclusion of 1D
 A sequence of k that is proportional to log of n is very likely
to appear using this sampling method. This is why this
method is not good enough to get the 1/r cutting we require.
 Of course, in the 1D case we can define a sampler that selects
every n/s point. However it is not clear how to transform
this selection to the plane (Where the lines are not ordered).
 
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2nd attempt – Two level decomposition
 General plan: Instead of trying to improve the sampling
algorithm to generate a 1/r cutting from the start:
1. we’ll take a random sample (obtained as before with
probability p=s/n independent Bernoulli trials)
2. We just saw that this is not good enough, so we’ll take
every triangle that intersects more lines than required and
divide it so each new triangle will be good for our 1/r
cutting.
 Finally, we’ll show that the total number of triangles is
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But first, more definitions
 Let T be the collection of triangles after triangulating the
random sample.
 Let
denote the set of lines of H that intersect the
triangle
and
number of lines that
intersect.
 Let the excess of
be
 Note: if
so the triangle
can be included in the 1/r cutting. If
we subdivide it
to a collection of finer triangles.
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Subdivision of triangles with excess
greater than 1
 We consider the arrangement of
(instead of H), and
we construct a
cutting for it. (How? We’ll see later...)
 We intersect this triangulation with . This process may
produce triangles, but also quadrilaterals, pentagons, and
hexagons. Each of these convex polygons is subdivided into
triangles.
 Each triangle in the
cutting is intersected by at most
triangles, and therefore are valid for the 1/r cutting.
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Example:
Up – the process of subdividing a triangle with excess
greater than 1.
Left – An example for a hexagon created by combining
the two triangulations.
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But How? The suboptimal cutting
lemma.
 As of now, we can’t subdivide
because the cutting lemma is
not valid yet. So let’s use a weaker claim:
 The Suboptimal Cutting Lemma
for every finite collection of lines and any u>1, there exists a
1/u cutting consisting of at most:
triangles, where K is a suitable constant.
 We’ll prove the suboptimal cutting lemma later.
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Using the suboptimal cutting lemma
 If we use the suboptimal cutting lemma for producing the
cuttings, we can estimate that the number of triangles
in the 1/r cutting is bounded by:
 Note that we get one if the excess is less or equal to 1. If the
excess is greater than 1 we get the bound of the yet unproven
suboptimal cutting lemma times 4.
 The times 4 is for the triangulation of a
possible hexagon as shown here:
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Using the suboptimal cutting lemma (II)
We have two goals now:
1. Prove the suboptimal cutting lemma.
2. Show that although we typically have triangles
as large as about log(r) the expected number of triangles in
T with excess t or larger decreases exponentially as a
function of t.

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We’ll prove the first claim by demonstrating a triangulation
method that achieves that bound – the vertical
decomposition (next slide). The second claim would be
included in the proof of the Cutting Lemma.
Vertical decomposition - reminder
 We erect vertical segments upwards and downwards from
each vertex in the arrangement of S and extends them until
they meet another line (or all the way to infinity)
 We get trapezoids, but those can be split into 2 triangles.
 Let Trpz(S) denote the set of generalized trapezoids in the
vertical decomposition of S.
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Proposition:
Trapezoids with large excess are rare
Proposition: Let H be a fixed set of n lines in general
position, let p=r/n, where
, let S be a
random sample drawn from H by independent Bernoulli
trials with success probability of p, and let
be a real
parameter. Let
denote the set of trapezoids in
with excess at least t
Then the expected number of trapezoids in
is bounded as follows:
For a suitable constant C.
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Notes for this proposition
 If r is not smaller than n/2 than we can easily find a cutting,
as we saw in the cutting lemma lecture it’s only gets
interesting when r is proportional to logn.
 Reminder: How to construct such a cutting? If r is
proportional to n one can select all lines in the H, as shown
before this creates the desired subdivision.
 We use this limitation to find a finer cutting only because this
is needed in the proof of this proposition (but we’ll get there
much later)
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Proof of the suboptimal cutting lemma
 Set
for a sufficiently large constant
A.
 Assume that the trapezoid proposition holds – and choose a
sample S accordingly.
 Activate the proposition twice – once for
and once
for
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First case
 For
by using the proposition:
for some constant B.
 Also note that
 Finally:
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Second case
 For
by using the proposition:
for some constant C. We’ll group the constants together:
 Since:
there exists A large enough s.t
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Proof of the Suboptimal Cutting
Lemma – Conclusion
 From these two cases and the linearity of expectation we get:
 There exist a sample were both parts of the sum are smaller
than 2/3. Notice that the second part is at least 1 and
therefore 0.
 So there exists a sample S with both
and
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Proof of the Suboptimal Cutting
Lemma – Conclusion (II)
 That means that there exists a 1/u cutting into
triangles.
 Suboptimal Cutting Lemma proof is complete.
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Proof of the Cutting Lemma (I)
 Reminder: To produce a 1/r cutting we pick a sample S with
probability p=r/n as described before. We let T be the
collection of triangles of the vertical decomposition of S
(actually – trapezoids, but that has no effect). For every
triangle t in T with excess greater than 1 we refine the
triangulation with a
cutting. We’ll estimate the
expected number of triangles:
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Proof of the Cutting Lemma (II)
 As
and
 Since each part of the sum is at least 1. Also group by
order
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‘s
Proof of the Cutting Lemma (III)
 Notice that: if
then
so
 Also, the summing over triangles with
is less then summing over
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Proof of the Cutting Lemma (IV)
 Using the suboptimal cutting lemma and the trapezoid proposition
for some M large enough
 The ratio test shows that the series converges to some
number. So we get that the total number of triangles is
as needed.
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Just one more
Now we just need to prove the trapezoid proposition, for
that we’ll need some more definitions:

is the set of all trapezoids that can ever appear in the vertical
decomposition for some subset S of H (including the empty
set).

is the set of lines of H incident to at least one vertex
of . Notice that
(all the possible cases, up to
symmetry are shown in the next slide)
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D cases
 The following shows that
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Some Properties of Vertical
Decomposition
 C0: As we saw:
. Moreover, any subset S of H is
the defining set for at most a constant number of trapezoids
in Reg.
 C1: For any
we have
(the
defining set must be present) and
(no
intersecting lines may be present).
 C2: For any
and any
s.t
and
we have
 C3: For every
we have
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Notes about the properties
 C3: Think of the iterative process of adding one vertical line
at a time. Each splits an existing region in two. Since the lines
are in general position the number of intersections is about
 C2 is the most interesting - it means that the vertical
decomposition is defined “locally”:
is present in
whenever it is not
excluded for simple local reasons.
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Proof of the Trapezoid Proposition
 First case
:
From C3 we get
is the sum of independent random variables.
and so we get:
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where |S|
First case continued:
For now we have:
for
some constant M independent of t.
On the other hand
by
combining these two and setting C=2M we get:
as required.
This concludes the case of
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Second case
 For now we assume
. Let
be a random sample
as defined before with probability p. From C1 and C2 we get:
(a trapezoid appears iff all
his defining lines are selected into S, and no line that
intersects it are selected into S).
 Let the set of all possible trapezoids with excess at least t:
 The expected number of trapezoids can be written as:
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Second case continued:
 We now define p’=p/t and since t>1 we get: p’<p
 Let S’ be an a sample drawn of H with probability p’
 From what we saw earlier:
 Also:
 And finally:
The last transition holds from limiting the sum.
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Second case continued (II):
Now be expanding with p:
Where R is defined:
As we’ll see in the next slide R is bounded by:
Finally we derive:
for a sufficiently large constant C (and than choosing the max
between it and the first case) the proposition is proved!
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Lower bound for R
 And since
 For every real p:
 Since p’<p<0.5:
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and
we get:
Lower bound for R (II)
 By the last slide we see that:
 Since R is defined:
it will always be
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Cutting Lemma conclusion
 Since the trapezoid decomposition lemma is proved the
entire proof is done.
 As mentioned before this Lemma can be generalized to
higher dimensions, and the proof we have just presented can
also be generalized to prove it (but it won’t be covered now).
 Cutting Lemma for Higher Dimensions: Let d>0 be a fixed
integer, let H be a set of n hyperplanes in
and let r be a
parameter
.Then there exists a 1/r cutting for H the
size of
generalized simplexes s.t the interior of each simplex
is intersected by at most n/r hyperplanes of H.
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