Geometric Applications of A Matrix Se~ching Algorithm
Alok Aggarwal 1
Maria M. Klawe 2
Shlomo Moran 1'4
Peter Shor 3
Robert WIlber 2
1. IBM T. J. Watson Center, Yorktown Heights
2. IBM Almaden Research Center, San Jose
3. Math. Sciences Research Institute, Berkeley
4. On leave from The Technion, Haifa, Israel.
I. Introduction
points in the plane. In fact, using some geometric
properties of a convex polygon and the fact that any
point can be the nearest neighbor of at most six
The all-farthest neighbor problem for a set of n
points in the plane, P, is to find for each point Pie P,
other points in the plane, Lee and Preparata (LP78)
another point pj~ P with j # i such that
obtained a O(n) algorithm for the all-nearest neighbor problem on a convex polygon. However, since
d(Pi, pj)
-
max
l<_k<n
d(p i, Pk)
a single point could be the farthest point for all
n-1
other points even if the points form the
where d(Pi, Pj) denotes the Euclidean distance be-
vertices of a convex polygon, the algorithm pro-
tween Pi and pj. The all-nearest neighbor problem
posed by Lee and Preparata cannot be extended to
consists of finding the nearest point for every point
solve the all-farthest neighbor problem in linear
in the set. Shamos and Hoey (SH75) have shown
time.
that O(n log n) is the optimal bound for the allnearest
neighbor
problem, and
Toussaint and
A simple polygon is unimodal if for every vertex
Bhattacharya (TB81) as well as Preparata (Pr77)
have shown that 0(n log n) is also an optimal bound
Pk the function defined by the Euclidean distance
between Pk and the remaining vertices (traversed in
for the all-farthest neighbor problem.
clockwise order) contains only one local maximum.
For any m _> 1, this definition of unimodal polygons
The fl(n log n) bounds given in (SH76, TB81,
can be extended to m-modal polygons, in a natural
Pr77) do not apply when the input set forms the
manner. Somewhat contrary to one's intuition,
vertices of a convex polygon (say, given in clock-
Avis, Toussaint and Bhattacharya (ATB82) have
provided examples of convex polygons in which
wise order) rather than being an arbitrary set of
n/2 vertices have n/4 local maxima in each of their
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distance functions. The counter intuitive fact that
the distance functions can be multimodal has often
resulted either in incorrect algorithms or in increased time complexities for some of them.
Aggarwal and Melville (AM83) have shown that
$00.75 whether a convex polygon is m-modal can be deter-
285
mined in O(n x m) time and Toussaint (To82) has
problem are distinct. These restrictions can be re-
provided a very simple and intuitive algorithm for
moved; we impose t h e m to simplify the definitions
solving the all-farthest neighbor problem for a con-
and algorithms. Let A j denote the j - t h column of A
vex unimodal polygon in O(n) time. Here we show
and
that the all-farthest neighbor problem can be solved
A [i 1. . . . .
for a c o n v e x polygon in O(n) time, regardless of its
modallty. We use the same m e t h o d to speed up several other geometric algorithms by a factor of
log n.
Ai
denote
the
i-th
row
of
A.
ik ; Jl . . . . Jk ] denotes the s u b m a t r i x of
ik and columns
Jl . . . . .
Jk • Given a m a t r i x A, the maximum problem
is to d e t e r m i n e for each i, the c o l u m n j(i) at which
A that is f o r m e d of rows i 1. . . . .
This paper is divided into five sections. Section
A i has its m a x i m u m value. The m a t r i x A is monotone
if for 1 < i 1 < i2 < n, j(il) < j(i2)- A is totally
monotone if e v e r y s u b m a t r i x of A is m o n o t o n e . It is
II discusses a combinatorial problem for matrices
easy to verify that this is equivalent to having e v e r y
under two kinds of constraints -- a " w e a k " con-
2 x 2 s u b m a t r i x of A be m o n o t o n e .
straint and a " s t r o n g " constraint.
We show that
each instance of the all-farthest neighbor problem
We n o w show that an instance of the all-farthest
for c o n v e x polygons is an instance of the m a t r i x
neighbor problem on a c o n v e x polygon with n
problem under the strong constraint -- and thus also
vertices can be regarded as an instance of the maxi-
an instance of the matrix problem u n d e r the weak
m u m problem on an n x 2 n -
1 totally m o n o t o n e
constraint. Section HI demonstrates that there is an
matrix.
~2(m log n) lower b o u n d for solving the problem on
c o n v e x polygon in clockwise order.
n x rn matrices subject only to the w e a k constraint.
n x 2n - 1 m a t r i x A as follows. F o r a n y integer u
Consequently, if one only makes w e of the weak
let
constraint in solving the all-farthest neighbor p r o b -
i<j<i+n--
lem (in the m a n n e r used by researchers in the past),
then
then one c a n n o t hope to achieve a linear time sol-
A (i,j) = --j. (The non-positive entries are set so as
ution.
Section IV shows that the m a t r i x problem
Let Pl . . . .
ou
,Pn denote the vertices of a
denote
Define an
((u - 1) m o d n) + 1.
1 then A ( i , j ) = d ( P i ,
A(i,j)=j-n,
and
if
poj).
If j < i
j_>i+n
to m a k e e v e r y value in a r o w distinct.)
If
then
N o w sup-
with the strong constraint can be solved in O ( m )
pose the 2 x 2 s u b m a t r i x A [4 J" ; k, l], with i < j and
time when m _> n. This yields a linear time solution
k < 1, has only positive entries. Then we must have
for the all-farthest neighbor problem on c o n v e x
i < j < k < I < i + n. In this case the vertices Pi, Pj,
polygons. Section V summarizes the i m p r o v e m e n t s
P*k , and Pot are in clockwise order a r o u n d the
that can be obtained over previous algorithms for
polygon. F r o m the triangle inequality one can show
solving geometric optimization and c o m p u t e r aided
design problems. Because of the space constraint,
that d(Pi, P . k ) >_.d(pi, p ° 1) or d(pj, Pel) > d(pj, Pok),
or both. Since the distances are distinct the a b o v e
we only discuss the all-farthest neighbor problem
inequalities
here and provide the details for the other problems
m o n o t o n e . The non-positive entries ensure that all
/
in the final version of the paper.
are
strict.
Thus
A [i, j ; k, 1]
is
other 2 x 2 submatrices of A are also m o n o t o n e .
Thus A is totally m o n o t o n e , and b y solving the
II. The Matrix Problem
m a x i m u m problem on A we can solve the allfarthest neighbor problem for the polygon.
Let A be an n x m m a t r i x with real entries. We
will assume that all entries within a r o w of A are
IIl. An fl(m log n) Lower Bound for the Maximum
distinct. Similarly, we assume that all distances be-
Problem on Arbitrary Monotone Matrices
tween pairs of vertices in the farthest neighbor
286
T h e m a x i m u m p r o b l e m on a m o n o t o n e n x m
assigns a value subject to the condition that it m u s t
m a t r i x A can be solved by the following straightfor-
be possible to position the m a x i m a of the rows in a
w a r d divide and conquer algorithm. Let i = I"n/2"1
w a y consistent with the values fixed so far and the
and in O ( m ) time find the column j at which h i has
rnonotonicity condition.
its m a x i m u m value.
Recursively solve the m a x i -
mum
problem
on
All .....
i - - 1; 1 . . . . .
j'] (when i > 1 and j >
and . 4 I ' i + 1 . . . . .
the
n;j, .... m]
T h e o r e m 3.1: Let A be an n x m m a t r i x , w h e r e n is
submatrices
(when i < n
a p o w e r of 2.
1)
e= 1 orm.
and
L e t h be a positive integer and let
Setf=m+
1-e.
Suppose that up to
m a x ( m -- 2, 0) cells h a v e already been queried (i.e.,
j < m). T h e time required b y this algorithm is given
h a v e been assigned fixed values).
by the r e c u r r e n c e
Also, suppose
that no cells h a v e been queried in A f, that a n y cells
f(n,m)
< m + max
that h a v e been queried in A e h a v e been set to h, and
If(rn/21 - l,j) +
1)},
l<_/<_m f ( L n / 2 J , m - j +
that all other queried cells h a v e been set to values
less t h a n 1.
with f ( 0 , m) = f ( n , l ) = 0.
Solving the recurrence,
T h e n an a d v e r s a r y c a n a n s w e r a n y
queries for the remaining cells, in a w a y consistent
we h a v e f ( n , m) = O ( m log n) . (All logarithms in
with the m o n o t o n i c i t y condition, so that in o r d e r to
this section are base 2.) The best previously k n o w n
d e t e r m i n e the positions of the m a x i m a in each r o w
algorithms for the all-farthest neighbor p r o b l e m on
a total of at least ( 1 / 4 ) ( m - 1)(1 + log n) cells m u s t
c o n v e x polygons and for the problems described in
be queried (including those that w e r e initially que-
section V all contain a step that is essentially this
ried), and so that the m a x i m u m value in each r o w
divide and conquer procedure. N o t e that this algo-
is at least h.
r i t h m w o r k s for a r b i t r a r y m o n o t o n e matrices; the
m u c h stronger p r o p e r t y of total m o n o t o n i c i t y is not
Proof:
t a k e n a d v a n t a g e of.
w h e n m = 1 the claimed lower b o u n d is 0 a n d there
In this section we show that
a n y algorithm that solves the m a x i m u m p r o b l e m
We will a s s u m e t h r o u g h o u t that rn > 2, for
is nothing to prove.
W h e n we say that the a d v e r -
for a r b i t r a r y m o n o t o n e matrices must have a worst
sary sets a cell to a low value that m e a n s that the
case time of fl(m log n). So a n y i m p r o v e m e n t on the
cell is set to s o m e previously unused positive value
simple divide and conquer algorithm for the m a t r i -
less t h a n 1.
N o t e that if m - 2
or f e w e r queries
ces corresponding to the applications m u s t m a k e es-
h a v e been m a d e t h e n there are at least t w o columns
sential use of the fact that these matrices are totally
with no queries, say A "il a n d A h. The a d v e r s a r y can
monotone.
a n s w e r future queries in these columns either b y
setting all cells in A "il to h + 1 and all cells in A h to
T h e lower bound is on the n u m b e r of cells of the
low values or b y setting all cells in A la to low values
m a t r i x t h a t m u s t be queried b y a n y algorithm for
and all ceils in A N to h + 1. Either the m a x i m a will
the m a x i m u m
matrices.
all be in A jl or t h e y will all be in A h, and in either
Thus the b o u n d applies even if all additional oper-
case the m a x i m u m value in each r o w will be greater
ations (such as comparisons) are free.
than h . Thus w h e n m - 2 or f e w e r queries h a v e
that
when
(1/4)(m-
problem
n
is
a
on m o n o t o n e
power
1)(1 + Iog n) queries
of
2
must
We p r o v e
at
been m a d e the positions of the m a x i m a have not yet
least
been d e t e r m i n e d .
be made,
f r o m which it follows that for a r b i t r a r y n at least
( 1 / 4 ) ( m - 1) log n queries are required. T h e proof
We use induction on n.
uses an a d v e r s a r y a r g u m e n t . The value of each cell
of the m a t r i x is regarded as being i n d e t e r m i n a t e
Basis step: Suppose n < 4. By the observation
unt~ it is first queried, at which point an a d v e r s a r y
above, at least r n -
287
1 queries m u s t be m a d e , and
when
n < 4, ( 1 / 4 ) ( m - 1)(1 + log n) < m -
1. So
t h e c l a i m is t r u e in this case.
Induction step:
c o l u m n s of .4 c o n t a i n c -
1 q u e r i e d cells, a n d t h e
last m - c c o l u m n s c o n t a i n m - c q u e r i e d cells.
L e t n b e a p o w e r of 2 g r e a t e r
Let
L
be
one
of
the
two
submatrices
t h a n o r e q u a l to 4 a n d a s s u m e t h a t t h e t h e o r e m is
All .....
r 1 -- 1; 1 . . . . .
c]
t r u e f o r all p o w e r s of 2 less t h a n n. W e s h o w t h a t
A[rl,...,r
2-
c]
t h e t h e o r e m is t r u e f o r n. T h e s e q u e n c e of q u e r i e s
f e w e s t q u e r i e d cells. L h a s c c o l u m n s a n d a t m o s t
1; 1 . . . . .
or
, whichever
has
the
is d i v i d e d i n t o t w o stages. T h e first s t a g e lasts until
/(c - l)/2J
a t o t a l of m - 1 cells h a v e b e e n q u e r i e d ( i n c l u d i n g
b e t h e i n d e x of t h e r o w of A c o n t a i n i n g t h e first r o w
t h o s e cells t h a t h a d b e e n q u e r i e d at t h e s t a r t ) . S i n c e
of L (i.e., k 1 is e q u a l to e i t h e r 1 o r r l ) , a n d let k 2 b e
a t least m - I q u e r i e s m u s t b e m a d e , w e d o r e a c h
t h e i n d e x of t h e r o w of A c o n t a i n i n g t h e last r o w of
5 m a x ( c - 2, 0) q u e r i e d cells. L e t k 1
t h e e n d of t h e first s t a g e . A n y q u e r y m a d e a f t e r t h e
L (i.e., k 2 is e q u a l t o e i t h e r r 1 - 1 o r r 2 - 1 ). S i m i -
(m-l)th
larly,
query
is in
the
second
stage.
For
let
submatrix
R
be
either
1 < i < 3, let r~ = (in~4) + 1. T h e rules f o r a n s w e r -
Aft,
ing a q u e r y t o cell A ( i , j ) d u r i n g t h e first s t a g e a r e
A [r3,...,
as f o l l o w s .
q u e r i e d cells. L e t k 3 b e t h e i n d e x of t h e r o w of A
. . . , r 3 -- 1; c . . . . .
n; c . . . . .
rn]
or
m ] , whichever has the fewest
c o n t a i n i n g t h e first r o w of R, a n d let k 4 b e t h e i n d e x
la)
If i < r 2 a n d j =
1 t h e n if e = 1 set A ( i , j ) t o h,
of t h e r o w of A c o n t a i n i n g t h e l a s t r o w of R. R h a s
m - c + 1
o t h e r w i s e set A (i, j ) t o h + 1.
columns
and
contains
at
most
L (m - c ) / 2 J _< m a x ( m - c - 1, 0) q u e r i e d cells.
m t h e n if e = m s e t A ( i , j ) t o h,
l b ) If i _> r 2 a n d j =
• o t h e r w i s e set A (i, j ) t o h + 1.
2)
If i < r 2 a n d
j#
N o t e t h a t L satisfies t h e c o n d i t i o n s of t h e t h e o -
1 o r if i > r 2 a n d j # m ,
set
r e m , w i t h p a r a m e t e r s h t a n d e t, w h e r e e ~ = 1 a n d
hr=hife=
A (i, l ) t o a l o w v a l u e .
landh'-h+
life=m.
Similarly, R
satisfies t h e c o n d i t i o n s of t h e t h e o r e m , w i t h p a r a m W h e n t h e first s t a g e e n d s , e x a c t l y m - 1 cells of
A have been queried and the values fixed are cons i s t e n t w i t h all m a x i m a
b e i n g in c o l u m n
1 and
in r o w s 1 t h r o u g h r 2 - 1
all m a x i m a
in r o w s r 2
t h r o u g h n b e i n g in c o l u m n m. Q u e r i e d cells in c o l umns 2 through m -
e t e r s h " a n d e", w h e r e e " = m - c + 1 a n d h " = h if
e=mandh'=h+
queries
life=
of cells w i t h i n
1. I n t h e s e c o n d s t a g e
L or R are handled
by
recursively applying the adversary strategy to the
t w o s u b m a t r i c e s , a n d q u e r i e s t o cells o u t s i d e of L
a n d R a r e a n s w e r e d in s u c h a w a y t h a t t h e y i m p o s e
1 all h a v e v a l u e s less t h a n 1.
n o c o n s t r a i n t s u p o n t h e p o s i t i o n s of t h e m a x i m a
w i t h i n L o r R.
A f t e r t h e first s t a g e is c o m p l e t e d a c o l u m n c a n d
t w o s u b m a t r i c e s L a n d R a r e s e l e c t e d as f o l l o w s .
T h e r u l e s f o r a n s w e r i n g a q u e r y of cell A (id) a r e
F o r 0 _< j < m, let sj b e t h e n u m b e r of q u e r i e d cells
in c o l u m n s 1 t h r o u g h j of A (s o = 0).
Let c be the
as f o l l o w s .
s m a l l e s t i n t e g e r in [ 1, m ] s u c h t h a t sc = c - 1 ( t h e r e
is s u c h a n i n t e g e r b e c a u s e S,n = m - 1 ). U s i n g t h e
1) If i < k 1 a n d j = 1, o r if i > k 4 a n d j = m, t h e n s e t
A ( i , j ) t o h + 1.
a c t t h a t t h e sj's a r e a n o n d e c r e a s i n g s e q u e n c e of int e g e r s it is e a s y t o s h o w b y i n d u c t i o n t h a t f o r a l l j in
2) I f k 2 < i < k
3andj=cthensetA(i,j)
toh+2.
[0, c - 1 ], sj > j. I n p a r t i c u l a r , s c_ 1 > c - 1. S i n c e
Sc-1 <so,
we
conclude
that
Sc-l = sc = c -
T h e r e f o r e A c h a s n o q u e r i e d cells, t h e first c -
1.
3 a ) If A (i, j ) is in s u b m a t r i x L t h e n fix t h e v a l u e f o r
1
t h a t cell b y a p p l y i n g t h e a d v e r s a r y s t r a t e g y
288
recursively to L, using the p a r a m e t e r s e' and
Also, the m a x i m u m value in each r o w is at least h.
h'.
[]
3b) If ,4 (i, j ) is in s u b m a t r i x R then fix the value for
that cell b y applying the a d v e r s a r y strategy
W h e n A has no initial queries the conditions of
recursively to R , using the p a r a m e t e r s e" and
the t h e o r e m are obviously m e t for a n y h > 1 a n d e
hH .
equal to either 1 or m, so w e h a v e the desired
fl(m log n) l o w e r bound.
4) For all other queries set `4 (i, j) to a low value.
IV. A Linear T i m e Algorithm for The M a x i m u m
Rule
1 ensures that the m a x i m a in rows
1
Problem on Totally Monotone Matrices
through k 1 - 1 are in .41 and that the m a x i m a in
r o w s k 4 + 1 through n are in A m . Rules 2 ensures
that the m a x i m a in rows k 2 + 1 through k 3 - 1 are
in
.4 c.
`4[k I . . . . .
and
The
k2; c + 1. . . . .
the
`4[k3 . . . . .
values
cells
in
m ] are all less than h',
values
k4; 1 . . . . .
of
of
c-
cells
in
1] are all less t h a n h".
By a s s u m p t i o n the recursively applied strategy will
result in a m a x i m u m value of at least h p in each r o w
of L a n d of at least h" in each r o w of R. Thus the
m a x i m a for rows k 1 through k 2 of A will all be in
s u b m a t r i x L, and the m a x i m a for rows k 3 through
k 4 will all be in s u b m a t r i x R. Thus, as claimed, no
e x t e r n a l constraints a r e placed upon the positions
of the m a x i m a within L and R, so the recursive use
H e r e we s h o w that b y m a k i n g use of the strict
constraints imposed b y total m o n o t o n i c i t y we can
solve the m a x i m u m p r o b l e m in linear time.
For
each i, let j(i) be such that A(i, j(i)) is the m a x i m u m
e l e m e n t in A i. T h e k e y c o m p o n e n t of the algorithm
is the subroutine
REDUCE . It takes as input a n
n x m totally m o n o t o n e m a t r i x A, with m > n. T h e
value r e t u r n e d b y REDUCE is an n x n s u b m a t r i x
of A, C, with the p r o p e r t y that, for 1 _< i __. n, subm a t r i x C contains c o l u m n A Ai). R E D U C E
does a
constant a m o u n t of w o r k per c o m p a r i s o n , a n d does
at m o s t 2 m O(m)
n-
1 comparisons, so runs in t i m e
.
of the a d v e r s a r y strategy in rules (3a) and (3b) is
Say t h a t an e l e m e n t A (i, j) is killed if, using the
valid.
results of a n y c o m p a r i s o n s m a d e so far a n d the total
S u b m a t r i c e s L and R each h a v e n[4 rows.
assumption,
at
least
(1/4)(c-
By
1)(I + l o g ( n / 4 ) )
queries are needed to l o c a t e the m a x i m a within L
m o n o t o n i c i t y of A, it can be s h o w n that A(i, j) is not
the m a x i m u m e l e m e n t in Ai, i.e., j ~ j(i). A c o l u m n
is killed if all of its elements are killed.
and at least ( 1 / 4 ) ( , , - c)(1 + l o g ( n / 4 ) ) queries are
L e m m a 4.1: L e t A be a totally m o n o t o n e m a t r i x . If
needed to locate the m a x i m a within R. In addition,
A (r, Jl) > A (r, J2)
at the end of the first stage t h e r e are at least
|A(i, j2): 1 < i < r } are killed. On the other hand,
if
A ( r , A ) < A(r, j2)
then
the
entries
in
{A(i, jl): r _< i < n } are killed.
(m - 1 ) / 2 queries in .4 outside of L and R .
So the
total n u m b e r of queries needed to find the m a x i m a
then
the
entries
in
in A is at least
Proof: T h e first claim follows f r o m the fact that
(m -
1
1)
1
n
+ ~ - ( c -- 1)(1 + log -~--) +
n
1
- ~ ( m -- c)(1 + log -~-) - ~ - ( m -- 1)(1 + log n).
A [i, r ; Jl, J2 ] is m o n o t o n e for all I < i < r .
Simi-
larly, the second claim follows f r o m the fact that
A [r, i ; Jl, J2 ] is m o n o t o n e for all r < i _< n . []
289
L e t t h e index of C b e t h e l a r g e s t k s u c h t h a t f o r
b r a n c h e s of t h e c a s e s t a t e m e n t a r e e x e c u t e d .
A
all 1 < j < k a n d 1 < i < j , e l e m e n t C(i, j ) is killed.
c o l u m n is d e l e t e d o n l y in t h e last t w o cases, a n d
N o t e t h a t e v e r y m a t r i x has i n d e x a t least 1.
s i n c e a t o t a l of m - n c o l u m n s a r e d e l e t e d w e h a v e
b + c = m - n. T h e i n d e x i n c r e a s e s in t h e first c a s e
a n d d e c r e a s e s in t h e last case, a n d is u n c h a n g e d in
T h e a l g o r i t h m R E D U C E is as f o l l o w s .
t h e s e c o n d case. S i n c e t h e i n d e x s t a r t s a t 1 a n d e n d s
R E D UCE(A)
n o h i g h e r t h a n n t h e net i n c r e a s e in t h e i n d e x is
C ,~--A ; k.,--1
a -- c < n -- 1. C o m b i n i n g t h e s e t w o f a c t s , w e h a v e
~ h i l e C has m o r e t h a n n c o l u m n s do
timet=a
+ b + c < a + 2b + c < 2 m - n - 1 .
[]
case
C(k, k) > C(k, k + 1) a n d k < n:
We
k ~ - - k + 1.
now
describe
solves the maximum
C(k, k) > C(k, k + 1) a n d k = n:
MAXCOMPUTE,
which
problem on an n x m totally
m o n o t o n e m a t r i x , w h e r e m >_ n.
D e l e t e c o l u m n C k +l .
C ( k , k ) < C ( k , k + 1):
M A X C O M P UTE(A )
D e l e t e c o l u m n Ck; k,~- k - 1 .
B~-- R E D U C E ( A )
endcase
if n =
return(C)
1 lhenreturn
C.-- B[2, 4 . . . . . 2 L n / 2 j ; 1,2 . . . . . n ]
MAXCOMPUTE(C)
T h e i n v a r i a n t m a i n t a i n e d is t h a t k is t h e i n d e x of C.
Also, only killed columns are deleted.
Using the bounds due to the known positions
It is e a s y t o
of t h e m a x i m a in t h e e v e n r o w s of B, f i n d t h e
see t h a t t h e s e c o n d i t i o n s h o l d . T h e i n v a r i a n t h o l d s
m a x i m a in t h e o d d r o w s of B.
i n i t i a l l y b e c a u s e t h e i n d e x of C a t t h e s t a r t is 1. If
C ( k , k ) > C ( k , k + 1) t h e n b y L e m m a
4.1 all ele-
T h e o r e m 4.3: W h e n n < m M A X C O M P U T E
m e n t s of C k'+l in r o w s 1 t h r o u g h k a r e k i l l e d . T h u s
the
if k < n t h e i n d e x of C i n c r e a s e s b y 1, a n d if k = n
n x m m a t r i x in t i m e O(m).
maximum
problem
on
a totally
solves
monotone
c o l u m n C k + l is killed, a n d t h e i n d e x of C r e m a i n s
the same.
Proof:
If C(k, k ) < C ( k , k + 1) t h e n b y L e m m a
Let
f(n,m)
be
the
time
taken
by
4.1 all e l e m e n t s of C k in c o l u m n s k t h r o u g h n a r e
MAXCOMPUTE
k i l l e d , a n d s i n c e t h e e l e m e n t s of C k in c o l u m n s 1
r e m 4.2 w e k n o w t h a t t h e c a l l t o R E D U C E
t h r o u g h k - 1 w e r e a l r e a d y k i l l e d , C k is k i l l e d .
t i m e O(m) a n d t h a t b y f i n d i n g t h e m a x i m a in t h e
In
t h a t c a s e t h e i n d e x of C d e c r e a s e s b y 1.
T h e o r e m 4.2: I n O(m) c o m p a r i s o n s , a l g o r i t h m RE-
on a n n x m m a t r i x .
From Theotakes
r o w s of t h e n x n m a t r i x
B we have found the
m a x i m a in t h e r o w s of A.
T h e a s s i g n m e n t of t h e
e v e n r o w s of B t o C is r e a l l y j u s t t h e m a n i p u l a t i o n
D UCE r e d u c e s t h e m a x i m u m p r o b l e m f o r a n n x m
of a list of r o w s , a n d c a n b e d o n e in O(n) t i m e . C is
totally monotone matrix to the maximum
an n/2 x n totally monotone matrix so the recur-
problem
for a n n x n t o t a l l y m o n o t o n e m a t r i x .
sive call t o M A X C O M P U T E
Proof: REDUCE terminates when C has n columns,
so t h e o u t p u t is a n n x n s u b m a t r i x of A.
That C
t a k e s t i m e f ( n / 2 , n) .
T h e last s t e p c a n b e d o n e in O(n) t i m e .
T h u s , for
s u i t a b l e c o n s t a n t s c 1 a n d c2, t h e t i m e is
c o n t a i n s all c o l u m n s of A t h a t h a v e a m a x i m u m
f ( n , m) < cln + c2m + f(--~, n),
v a l u e for s o m e r o w of A f o l l o w s f r o m t h e a b o v e
d i s c u s s i o n . F o r t h e t i m e a n a l y s i s , let a, b, a n d c d e note, respectively, the number
w h i c h h a s t h e s o l u t i o n f(n, m) < 2 ( c 1 + c2)n + c2m.
of t i m e s t h e t h r e e
S i n c e m >_ n. this is O(m). []
290
We have also obtained tight bounds for the
Finally, we discuss an application of our m a t r i x
m a x i m u m problem on totally m o n o t o n e matrices in
search algorithm to a wire routing problem. Con-
the case w h e r e m < n, which we state here without
sider the problem of connecting n corresponding
proof.
terminals {Pi} and {Qi} with wires. The P/s are arranged in order b y index along a horizontal row,
Theorem 4.4: When n > m the necessary and suffi-
and the Qi's are also arranged in order along a lower
cient time required to solve the m a x i m u m problem
horizontal row. The connecting wires must be sep-
on
arated by some m i n i m u m distance, and often there
a
totally
monotone
n × m
matrix
is
O(m(1 + log(n/m))) .
are other constraints. For example, the wires m a y
be constrained to lie on a rectilinear grid or m a y be
V. Applications of the Matrix Searching Algorithm
constrained to consist of horizontal, vertical, or 45
degree angle segments.
Lee and Preparata have shown (LP78) that the
Assume that the Qi's are
along the x axis with Q1 at the origin. Define the
they
offset to be the x coordinate of P1 and call the y coordinate of the Pi's the separation. Given the design
crucially use the fact that any point in the plane can
rules, the separation problem is t o find, for some
be the nearest neighbor of at most six other points.
fixed offset, the m i n i m u m separation that permits a
Like the all-farthest neighbor problem, we can also
legal wiring. The optimal offset problem is to find
solve the all-nearest neighbor in linear time without
the offset that allows the m i n i m u m separation.
even using this fact. F u r t h e r m o r e , we can solve the
Dolce et al. (DKSSU81) found a linear time algo-
following problem in linear time. Given t w o c o n v e x
polygons, for every vertex in one polygon, find its
r i t h m for the m i n i m u m separation problem in the
farthest (or its nearest) vertex in the other polygon.
rectilinear integer grid.
Our algorithm improves the previous best k n o w n
s h o w e a that for a very general class of design rules
solution for this problem by a factor of log n .
the m i n i m u m separation problem can be solved in
all-nearest neighbor problem for a c o n v e x polygon
can
be solved in linear time.
However,
case where the wires are constrained to lie on a
Seigel and Dolev (SDS1)
O(n log n) time. T h e y also obtained linear bounds
Boyce et al. (BDDG82) have shown that given a
c o n v e x n-gon, the m a x i m u m area (or m a x i m u m
perimeter)
inscribed
k-gon
O(kn log n + n log2n) time.
can
be
found
in
Since the diameter
problem for a c o n v e x polygon can be regarded as
finding the m a x i m u m perimeter k-gon when k = 2,
it is not surprising that our linear time solution for
the m a t r i x problem can be used to r e d u c e the time
complexity
of
Boyce
et
al.'s
algorithm
to
O(kn + n log n). In particular, we can find a maximum
area
(or m a x i m u m
perimeter)
inscribed
quadrilateral in O(n log n) time. In a similar manner, Aggarwal, Chang and Yap ( A C Y 8 5 ) have
shown that the minimum area circumscribing k-gon
can be found in O(n 2 log k log n) time and we can
reduce the time c o m p l e x i t y of their algorithm to
O(n 2 log k).
for m o r e general constraints than those used by
Dolev et al. (DKSSU81), such as w h e r e the wires lie
on a quarter integer grid and consist of segments at
angles that are multiples of 45 degrees.
However,
for some natural design rules, such as a wiring
scheme that permits arbitrarily shaped wires, with
a m i n i m u m separation of 1 unit, Seigel and Dolev
w e r e not able to do any b e t t e r than their generic
O(n log n) algorithm. The O(n log n) algorithm was
based
upon
a
certain
matrix
being
totally
m o n o t o n e , so we can provide a linear time algorithm for all design rules in the class they defined, in
particular for the case w h e r e the wires can have arbitrary shapes.
Also, they showed that in such
schemes the optimal offset problem can always be
solved in O(n log2n) time, and we can r e m o v e one
of the log n factors in that algorithm.
291
Acknowledgements: The authors thank Ashok K.
Chandra, Don Coppersmith, S. Rao Kosaraju, Ravi
Nair, and Martin Tompa for stimulating discussions.
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