LARGEST EMPTY.pdf - Library(ISI Kolkata) - Indian Statistical Institute

J. Chaudhuri et al. / Journal of Algorithms 46 (2003) 54–78 77
5. Conclusion
In this paper, we have considered the problem of locating the largest empty rectangle,
of any arbitrary orientation, among a set of points. An algorithmic technique is proposed
to solve this problem which inspects all the PMERs present in the plane. One may hope
for a faster algorithm without considering all the PMERs.
Acknowledgments
The authors acknowledge Prof. Micha Sharir for suggesting the example in Fig. 5, which
proves that the upper bound of the number of PMERs is Ω(n 3 ). We thank the referees of
the paper for valuable suggestions which improved the presentation of the paper.
Appendix A. Area calculation
Let us consider a set of MERs denoted by six-tuples {p i ,p j ,p k ,p l ,φ,ψ}. ThePMER
is a member in this set which has largest area. Let the grid angle corresponding to the
PMER be θ.Thevalueofθ is obtained as follows:
At grid angle θ, the lines corresponding to the north and south boundaries of the PMER
will be (y −y i ) = m∗(x −x i ) and (y −y k ) = m∗(x −x k ), respectively, where m = tan(θ).
The lines at the east and west boundaries will be (y − y l ) = (−1/m) ∗ (x − x l ) and
(y − y j ) = (−1/m)∗ (x − x j ), respectively. So we have the coordinates of its four corners
as follows:
( m(yl − y i ) + (m 2 x i + x l )
north-east:
m 2 , m(x l − x i ) + (m 2 )
y l + y i )
+ 1
m 2 ,
+ 1
( m(yj − y i ) + (m 2 x i + x j )
north-west:
m 2 , m(x j − x i ) + (m 2 )
y j + y i )
+ 1
m 2 ,
+ 1
( m(yl − y k ) + (m 2 x k + x l )
south-east:
m 2 , m(x l − x k ) + (m 2 )
y l + y k )
+ 1
m 2 ,
+ 1
( m(yj − y k ) + (m 2 x k + x j )
south-west:
m 2 , m(x j − x k ) + (m 2 )
y j + y k )
+ 1
m 2 .
+ 1
The area of the rectangle is
A θ = (m(y l − y j ) + (x l − x j ))((y k − y i ) + m(x i − x k ))
m 2 , φ θ ψ.
+ 1
This is a unimodal function in 0 θ π/2. Its maximum value can be obtained by solving
δ
δθ A θ = 0. Now we have
δ
δθ A 1 [
θ = (yk
m 2 − y i )(y l − y j ) + (x i − x k )(x l − x j ) ]
+ 1
+ 2m [ (x i − x k )(y l − y j ) + (y k − y i )(x l − x j ) ]
− m 2[ (y l − y j )(y k − y i ) + (x l − x j )(x i − x k ) ] , where m = tan(θ).