Angle Bisector Algorithm and Modified Dynamic Programming Algorithm for
Dubins Traveling Salesman Problem
Abstract
A Dubin’s Travelling Salesman Problem (DTSP) of finding a minimum length
tour through a given set of points is considered. DTSP has a Dubins
vehicle, which is capable of moving only forward with constant speed. In
this paper, first, a worst case upper bound is obtained on DTSP tour
length by assuming DTSP tour sequence same as Euclidean Travelling
Salesman Problem (ETSP) tour sequence. It is noted that, in the worst
case, \emph{any algorithm that uses of ETSP tour
sequence} is a constant factor approximation algorithm for DTSP. Next,
two new algorithms are introduced, viz., Angle Bisector Algorithm (ABA)
and Modified Dynamic Programming Algorithm (MDPA). In ABA, ETSP tour
sequence is used as DTSP tour sequence and orientation angle at each
point $i_k$ are calculated by using angle bisector of the relative
angle formed between the rays $i_{k}i_{k-1}$ and
$i_ki_{k+1}$. In MDPA, tour sequence and orientation angles are
computed in an integrated manner. It is shown that the ABA and MDPA are
constant factor approximation algorithms and ABA provides an improved
upper bound as compared to Alternating Algorithm (AA)
\cite{savla2008traveling}. Through numerical
simulations, we show that ABA provides an improved tour length compared
to AA, Single Vehicle Algorithm (SVA)
\cite{rathinam2007resource} and Optimized Heading
Algorithm (OHA) \cite{babel2020new,manyam2018tightly}
when the Euclidean distance between any two points in the given set of
points is at least $4\rho$ where
$\rho$ is the minimum turning radius. The time
complexity of ABA is comparable with AA and SVA and is better than OHA.
Also we show that MDPA provides an improved tour length compared to AA
and SVA and is comparable with OHA when there is no constraint on
Euclidean distance between the points. In particular, ABA gives a tour
length which is at most $4\%$ more than the ETSP tour
length when the Euclidean distance between any two points in the given
set of points is at least $4\rho$.