On the independence number of minimum distance graphs
Discrete Comput Geom
G. Csizmadia 0
0 Courant Institute , New YorkUniversity, New York,NY 10012 , USA
Let F = F(n) denote the largest number such that any set of n points in the plane with minimum distance 1 has at least F elements, no two of which are at unit distance. Improving a result by Pollack, we show that F(n) > 9n. We also give an O(n log n) time algorithm for selecting at least 9 n elements in a set of n points with minimum distance 1 so that no two selected points are at distance 1.

Theorem. Given n points in the plane with minimum distance 1, we can always choose
at least 9 n o f them so that their minimum distance is greater than 1.
Using the above terminology, the theorem states that F ( n ) > 9 n .
2. Proof of the Theorem
Let X be a set of n points in the plane with minimum distance 1. Let G = G (X) denote
the minimum distance graph associated with X. We will prove that a ( G ) >_ 9 n . It is
sufficient to prove the following lemma:
L e m m a 1. Every minimum distance graph G has an independent set P o f k < 9
vertices such that if m is the number o f vertices in G  P, incident to at least one element
o f P , then k / ( k + m) > ~93"
The main theorem follows from Lemma 1, since we can repeatedly apply the lemma
to remove an independent set and its neighbors with the given property. The union of
these independent sets is an independent set with at least 9 n elements.
The proof uses two auxiliary results (Lemmas 2 and 3), whose proofs are postponed
to the last section.
The main idea of the proof is the following. First we can suppose that all vertices have
degree at least 3. Then if you walk counterclockwise around the boundary of G, then
the direction you are facing must rotate counterclockwise around 360 ~. Only vertices
of degree 3 contribute to this counterclockwise rotation so the boundary must contain
some such vertices. Moreover, the counterclockwise rotations at degree 3 vertices must
outweigh the clockwise rotations at degree 4 and 5 vertices. A careful case analysis of
the possible positions of the vertices and their neighbors gives the desired independent
set P.
P r o o f o f Lemma 1.
We can make the following simple observations:
(a) I f a vertex has degree 2, then it forms an independent set satisfying the conditions
of Lemma 1;
(b) if two nonadjacent vertices of degree 3 share a neighbor, then they form such an
independent set; and
(c) if a vertex of degree 3 shares two neighbors with a nonadjacent vertex of degree
4, then they form such a set; etc.
Let d ( p ) denote the degree of a vertex p of G. By the above observations, we can
suppose that d ( p ) > 3 for all p 6 G.
Define the boundary graph C' as follows. Let rl denote a vertex of G with minimum
ycoordinate. Draw a halfline from rl pointing downward, and turn it counterclockwise
around rl until it hits a neighbor of rl. Call that neighbor r2. Then turn the segment rlr2
counterclockwise around r2 until it hits another neighbor of r2. Call that neighbor r3.
Continue this procedure until we hit a point r i that has been encountered before. Then
ri = ri+t, SOwe have a cycle C' = {ri+l, ri+2 . . . . . ri+t}.
On the IndependenceNumberof MinimumDistanceGraphs
Pi
Definition. Let Pi1, ql, and Pi+l denote the neighbors o f a point Pi E C t with
d ( p i ) = 3 (in clockwise order as is shown in Fig. 1). If Z P i  I Piql = Zql PiPi+l ~ 60~
then Pi is called regular. A vertex o f degree 3 is called irregular if it is not regular.
If Pi is irregular, then at least one of its neighbors, say pi+l, does not have a c o m m o n
neighbor with pi. If d(pi+l) > 4, then call the (Pi, Pi+I) pair a special pair. Note that
an irregular pi may be the member of two special pairs.
L e m m a 2. Let Pi E C be irregular, d(pi+l) > 4, and ZqlPipi+l = c~ > 60 ~ Then
a(pi) + a(pi+l) < 0 (see Fig. 1).
Thus, counterclockwise (positive) turns at irregular vertices are followed by clockwise
(negative) turns o f at least the same degree, unless the irregular vertex is followed by
another vertex of degree 3. In other words, for special pairs (Pi, Pi+I) we have a ( p i )
qa(pi+l) < O.
Corollary. I f Pi is irregular and ~Z.~,ki+=1i l a (Pk) > O, then d (Pi l ) = 3 or d (Pi+ 1) = 3..
Furthermore, t9y~ ~2"_~i,+k=1i_ I a(pk) > 60 ~ then d ( p i  l ) = d(pi+l) = 3.
We may suppose that there are no two vertices pj, pj+2 E C with d ( p j ) = d(pj+2) =
3, for otherwise pj and Pj+2 would have five neighbors altogether, and L e m m a 1 would
follow.
Next we choose a family of pairwise disjoint blocks (strings of consecutive vertices)
in C. This procedure is the heart of the proof.
Going through the vertices of a negative block in counterclockwise direction yields
a negative turn of at least  6 0 ~ and all positive turns take place in a positive block.
Therefore, since going around the boundary is a 360 ~ turn, there must be more positive
blocks than negative ones.
(i) (a) Assume that a regular vertex ri has a neighbor of degree 5. Then fix such
a neighbor si, and let {ri, si } be called a neutral block. (Note that no two
neutral blocks have nonempty intersection, because si ~ sj unless ri = rj.)
(b) Assume that an irregular vertex ri is an element o f a special pair. For each
such ri fix an si such that (ri, si) forms a special pair. Then {ri, si} is called
a neutral block.
(ii) (a) If a regular vertex ri has no neighbor si with d(si) = 5, then it forms a
positive block.
(b) Each pair of irregular vertices (Pi, P i + I ) forms a positive block.
(iii) (a) Each si with d(si) = 5, which does not belong to any neutral block, forms a
negative block.
(b) Each set of I consecutive vertices Pk+l . . . . . Pk+l, is called a negative block,
whenever d(pi) = 4 (k + 1 < i < k + 1), d ( p t ) ~ 4, d(pk+l+l) 5~ 4,
z k +k+l l a(pi) <_ 60 ~ and neither Pk+l nor pk+t is a member of a neutral
block.
Note that some vertices Pi, . . . . . Pi, may not belong to any of the blocks defined
above, but for them ~~lj=l a ( p # ) < 0 holds.
Clearly, each negative block contributes  6 0 ~ (or less) to Y~pi~c a(pi). Each positive
turn not neutralized by the negative turn, at a following vertex of degree 4 or degree
5, is now represented by a positive block. Moreover, by the corollary, a positive block
contributes at most 60 ~ to ~p,~C a(pi). So ~,p,~C a(pi) >__240 ~ implies
I{P: P is a positive block} I > I{N: N is a negative block}[ + 4.
Let B1 . . . . . Bs denote the disjoint blocks in the order as they appear in C. There
exists a positive block Bj for which Bj+l is either positive or neutral.
(i) If Bj consists of a single regular vertex, then suppose it is P2. It follows that
d ( p l ) < 4. If Bj consists of two irregular vertices, then denote them by Pl and
P2.
(ii) If Bj+l is a positive block, and it is a regular vertex, then we can suppose that
it is pro. If Bj+I consists of two irregular vertices, then denote them by Pm and
Pm+l.
(iii) If Bj+I is a neutral block, then let Pm E Bj+I be the vertex in Bj+l with degree
3. In this case, either d ( p m  l ) > 4 or d(pm+l) >__4.
On the Independence Number of MinimumDistance Graphs
Thus there are eight possible types of (Bj, Bj+I) pairs, but in any case, the degrees of
Pl, P2 . . . . . pm are either:
(a) 4, 3, 3; or
(b) 4 , 3 , < 4 , < 4 . . . . . < 4, < 5 , 3 ; o r
(c) 3 , 3 , < 4 , < 4 . . . . . < 4 , < 5 , 3 ;
and in cases (a) and (b), P2 is regular and so the first and third vertices share two neighbors.
< 4 and < 5 are denoting vertices of degree at most 4 and at most 5, respectively.
This is all we need for the rest of the proof. We distinguish two cases:
(a) Suppose that there exists an h with 3 < h < 15, h < m, such that d(ph) = 3.
There are no negative blocks between Bj and Bj+I, therefore d(pi) < 4 for all i
where 3 < i < m  2. Then:
(i) i f h is even, set P = {Ph, Ph2 . . . . . P 2 } ;
(ii) i f h is odd, set P = {Ph, Ph2 . . . . . P l } ;
and L e m m a 1 follows.
(b) Let m > 16 and d(pi) = 4 (3 < i < 14). There was no negative block between
Bj and Bj+I, therefore Y~314 a(pi) >  6 0 ~
L e m m a 3 . Suppose that there is a sequence pj, P2. . . . . p l 4 i n C s u c h t h a t d ( p 2 ) = 3,
d(pi) = 4 (3 < i < 14), Y~,134 a(pi) >  6 0 ~ and Pl and P3 have at most six neighbors
altogether.
Then we can find k < 9 independent vertices of G such that the number of vertices
incident to at least one of them is at most 3k  1.
Thus, L e m m a 1 follows.
3. Proofs of the Lemmas
[]
C l a i m . Let Po. . . . . P6 be consecutive vertices of C such that d(pi) = 4 (1 < i < 5)
a n d a ( p i ) +a(pi+t) >  6 0 ~ (1 < i < 4). Then:
(i) Pk and Pk+l (k = 1,2, 3, 4) have a common neighbor rk (see Fig. 3).
Proof ofLemma 2. W e m a y supposethatd(pi+l) = 4 . D e n o t e t h e n e i g h b o r s o f p i + l by
pi, q2, q3, and pi+2 (see Fig. 1). C l e a r l y , / q l piPi+l = ot implies ZpiPi+lq2 >__180~ o~.
Therefore,
Let ro and r5 denote the remaining unlabeled neighbors of Pl and Ps, respectively.
Then:
(ii) rk (k = 1, 2, 3, 4) have at most two neighbors ( s k  l , s'k) different from Pk, Pk+l,
rkl, and rk+ 1 .
Suppose that both r2 and r3 have exactly two such nei"ghbors and ~~42a(pi) >  3 0 ~
Then:
(iii) r2 and r3 have a common neighbor s2 = s~ (s2 r P3); and
(iv) s2 has at most two neighbors different from sl, r2, r3, and s~ (see Fig. 7).
Proof. (i) Let k = 1 and suppose for contradiction that Pt and P2 have no c o m m o n
neighbor. Let us use the notation of Fig. 4. Clearly, Zbpl p2 + Z p l p2c > 180 ~ Zpopl b >
120 ~ and Zcp2P3 > 120 ~ So
This is a contradiction, so C l a i m (i) is true for k = 1. For k = 2, 3, 4, the p r o o f is similar.
(ii) Suppose k = 1 and that Pl and P2 have the c o m m o n neighbor r l , and rl has three
neighbors sl, s2, s3 as is shown in Fig. 5.
Let t~  roplrl and fl = rlP2r2. T h e n t~ + fl < 180 ~ We also know that Z p l r l s l >
180 ~  ~, Zs3rlP2 > 180 ~ /~. But
360 ~ = Z p l r l s l + Zslrls2 + Zs2rls3 + Zs3rlp2 t Zp2rlpt
> (180 ~
~
~
~
~
~
~,
a contradiction. For k = 2, 3, 4, the proof is similar.
b
c
I
(iii) Suppose that s2 # s~ (see Fig. 6). Let Ir2r31 = t and Isis21 = s. Let ot = / r l p2r2,
fl = Zr2P3r3, and y = / r 3 P 4 r 4 . T h e n ct + fl + y < 210 ~ Clearly, t = 2 s i n ( f l / 2 ) and
8 = /s~r2r3 < 360 ~  (Zp2r2sl ]ZsIr2S2 d/r3r2P3 + Zp3r2P2)
Similarly, 8' = / r 2 r a s 2 < 120 ~  fl/2. This implies
So
 2
> s i n ( 9 0 ~
> s i n
t,
t~
/ "
t~
know that Lr3s2r 2 = /3. ThUS if S2 had three neighbors tl, t2, t3, then we would get
360 ~ > 13 + (180 ~  e) + 60 ~ + 60 ~ + (180 ~  e')
Proof o f Lemma 3. Claim (i) implies that all Pi, Pi+l pairs (3 < i < 13) have a c o m
mon neighbor ri (see Fig. 8). Suppose that there is an rj (4 _< j _< 12) which has only
one neighbor different from rjl, rj+l, pj, and pj+l.
Now:
(a) if j is odd, then pick Pj+2, rj, p j  l , Pj3 . . . . . P2; and
(b) if j is even, then pick Pj+2, rj, Pjl, Pj3 . . . . . Pl.
These [ j / 2 J + 2 vertices have altogether 3 ( [ j / 2 J + 2)  1 neighbors.
Thus, we can suppose by Claim (ii) that all rj (4 < j _< 12) has two other neighbors.
~~.314a(pi) >  6 0 ~ impliesthateither ~,a4a(pi) >  3 0 ~ or ~~.~aa(pi) >  3 0 ~
loss o f generality suppose )~.913a(pi) >  3 0 ~
Claim (iii) implies that there exists s9 = s~, sl0 = s'lo, and Sll = S'll (see Fig. 8).
By Claim (iv), sl0 has at most two neighbors different from s9, sll, rio, or rll. Let
On the Independence Number of Minimum Distance Graphs
P = {Sl0, r12, r9, P14, P n , Ps, P6, P4, P2} Then the number o f points incident to at
least one element o f P is at most 26 = 3 99  1, completing the proof o f L e m m a 3. []
Remark. The above proof gives us an algorithm for selecting at least ~93n elements in
a set o f n points with minimum distance 1 so that no two selected points are at distance
1. The algorithm finds at most 9 and at least 1, new independent points in each step.
The minimum distance graph can be constructed in O (n log n) time, and after that,
each step takes at most C time, where C is a large constant. We have to make at most n
steps, therefore the algorithm can be completed in O (n log n) time.
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4. R. Pollack . Increasing the minimum distance of a set of points , J. Combinatorial Theory, SeriesA , 40 , 450 , 1985 .