The Snowblower Problem
Computer (2006)
- arXiv: cs/0603026
Available from arxiv.org
or
Abstract
We introduce the snowblower problem (SBP), a new optimization problem that is closely related to milling problems and to some material-handling problems. The objective in the SBP is to compute a short tour for the snowblower to follow to remove all the snow from a domain (driveway, sidewalk, etc.). When a snowblower passes over each region along the tour, it displaces snow into a nearby region. The constraint is that if the snow is piled too high, then the snowblower cannot clear the pile. We give an algorithmic study of the SBP. We show that in general, the problem is NP-complete, and we present polynomial-time approximation algorithms for removing snow under various assumptions about the operation of the snowblower. Most commercially-available snowblowers allow the user to control the direction in which the snow is thrown. We differentiate between the cases in which the snow can be thrown in any direction, in any direction except backwards, and only to the right. For all cases, we give constant-factor approximation algorithms; the constants increase as the throw direction becomes more restricted. Our results are also applicable to robotic vacuuming (or lawnmowing) with bounded capacity dust bin and to some versions of material-handling problems, in which the goal is to rearrange cartons on the floor of a warehouse.
Available from arxiv.org
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The Snowblower Problem
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The Snowblower Problem
Esther M. Arkin1, Michael A. Bender2, Joseph S. B. Mitchell1, and Valentin
Polishchuk1
1 Department of Applied Mathematics and Statistics, Stony Brook University, Stony Brook,
NY 11794, USA. {estie,jsbm,kotya}@ams.stonybrook.edu. ⋆⋆⋆
2 Department of Computer Science, Stony Brook University, Stony Brook, NY 11794, USA.
bender@cs.stonybrook.edu †
Summary. We introduce the snowblower problem (SBP), a new optimization problem that is
closely related to milling problems and to some material-handling problems. The objective in
the SBP is to compute a short tour for the snowblower to follow to remove all the snow from
a domain (driveway, sidewalk, etc.). When a snowblower passes over each region along the
tour, it displaces snow into a nearby region. The constraint is that if the snow is piled too high,
then the snowblower cannot clear the pile.
We give an algorithmic study of the SBP. We show that in general, the problem is NP-
complete, and we present polynomial-time approximation algorithms for removing snow un-
der various assumptions about the operation of the snowblower. Most commercially-available
snowblowers allow the user to control the direction in which the snow is thrown. We differ-
entiate between the cases in which the snow can be thrown in any direction, in any direction
except backwards, and only to the right. For all cases, we give constant-factor approximation
algorithms; the constants increase as the throw direction becomes more restricted.
Our results are also applicable to robotic vacuuming (or lawnmowing) with bounded ca-
pacity dust bin and to some versions of material-handling problems, in which the goal is to
rearrange cartons on the floor of a warehouse.
1 Introduction
During a recent major snowstorm in the northeastern USA, one of the authors
used a snowblower to clear an expansive driveway. A snowblower is a “material
shifting machine”, lifting material from one place, and depositing it nearby. There
is a skill in making sure that the deposited piles of snow do not grow higher than
the maximum depth capacity of the snowblower. This experience crystallized into an
algorithmic question we have called the Snowblower Problem (SBP): How does one
optimally use a snowblower to clear a given polygonal region?
⋆⋆⋆ Partially supported by the U.S.-Israel Binational Science Foundation (2000160), NASA
(NAG2-1620), NSF (CCR-0098172, ACI-0328930, CCF-0431030), and Metron Aviation.
† Partially supported by NSF Grants EIA-0112849 and CCR-0208670.
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The Snowblower Problem
Esther M. Arkin1, Michael A. Bender2, Joseph S. B. Mitchell1, and Valentin
Polishchuk1
1 Department of Applied Mathematics and Statistics, Stony Brook University, Stony Brook,
NY 11794, USA. {estie,jsbm,kotya}@ams.stonybrook.edu. ⋆⋆⋆
2 Department of Computer Science, Stony Brook University, Stony Brook, NY 11794, USA.
bender@cs.stonybrook.edu †
Summary. We introduce the snowblower problem (SBP), a new optimization problem that is
closely related to milling problems and to some material-handling problems. The objective in
the SBP is to compute a short tour for the snowblower to follow to remove all the snow from
a domain (driveway, sidewalk, etc.). When a snowblower passes over each region along the
tour, it displaces snow into a nearby region. The constraint is that if the snow is piled too high,
then the snowblower cannot clear the pile.
We give an algorithmic study of the SBP. We show that in general, the problem is NP-
complete, and we present polynomial-time approximation algorithms for removing snow un-
der various assumptions about the operation of the snowblower. Most commercially-available
snowblowers allow the user to control the direction in which the snow is thrown. We differ-
entiate between the cases in which the snow can be thrown in any direction, in any direction
except backwards, and only to the right. For all cases, we give constant-factor approximation
algorithms; the constants increase as the throw direction becomes more restricted.
Our results are also applicable to robotic vacuuming (or lawnmowing) with bounded ca-
pacity dust bin and to some versions of material-handling problems, in which the goal is to
rearrange cartons on the floor of a warehouse.
1 Introduction
During a recent major snowstorm in the northeastern USA, one of the authors
used a snowblower to clear an expansive driveway. A snowblower is a “material
shifting machine”, lifting material from one place, and depositing it nearby. There
is a skill in making sure that the deposited piles of snow do not grow higher than
the maximum depth capacity of the snowblower. This experience crystallized into an
algorithmic question we have called the Snowblower Problem (SBP): How does one
optimally use a snowblower to clear a given polygonal region?
⋆⋆⋆ Partially supported by the U.S.-Israel Binational Science Foundation (2000160), NASA
(NAG2-1620), NSF (CCR-0098172, ACI-0328930, CCF-0431030), and Metron Aviation.
† Partially supported by NSF Grants EIA-0112849 and CCR-0208670.
Page 2
2 Esther M. Arkin, Michael A. Bender, Joseph S. B. Mitchell, Valentin Polishchuk
The SBP shows up also in other contexts: Consider a mobile robot that is
equipped with a device that allows it to pick up a carton and then place the car-
ton down again, possibly on a stack of cartons, in a location just next to it; with each
such operation, the robot shifts a unit of “material”. The SBP models the problem
in which the robot is to move a set of boxes to a specified destination in the most
efficient manner, subject to the constraint that it cannot stack boxes higher than a
capacity bound.
In a third motivating application, consider a robotic lawnmower or vacuum
cleaner that has a catch basin for the clippings, leaves, dust, or other debris. The
goal is to remove the debris from a region, with the constraint that the catch basin
must be emptied (e.g., in the compost pile) whenever it gets full.
The SBP is related to other problems on milling, vehicle routing, and traveling
salesman tours, but there are two important new features: (a) material must be moved
(snow must be thrown), and (b) material may not pile up too high.
While the SBP arises naturally in these other application domains, for the rest of
the paper, we use the terminology of snow removal.
The objective of the SBP is to find the shortest snowblower tour that clears a
domain P , assumed to be initially covered with snow at uniform depth 1. An impor-
tant parameter of the problem is the maximum snow depth D > 1 through which
the snowblower can move. At all times no point of P should have snow of greater
depth than D. The snow is to be moved to points outside of P . We assume that each
point outside P is able to receive arbitrarily much snow (i.e., that the driveway is
surrounded by a “cliff” over which we can toss as much snow as we want).3
Snowblowers offer the user the ability to control the direction in which the snow
is thrown. Some throw directions are preferable over others; e.g., throwing the snow
back into the user’s face is undesirable. However, it can be cumbersome to change
the throw direction too frequently during the course of clearing; also, some variants
(e.g., lawnmowing with clippings) require that the throw direction is fixed. Thus,
we consider three throw models. In the default model throwing the snow backwards
is allowed. In the adjustable-throw model the snow can be thrown only to the left,
right, or forward. In the fixed-throw model the snow is always thrown to the right.
Even though it seems silly to allow the throw direction to be back into one’s face, the
default model is the starting point for the analysis of other models and is equivalent
to the vacuum cleaner problem (discussed later).
Results. In this paper we introduce the snowblower problem, model its variants,
and give the first algorithmic results for its solution. We show that the SBP is NP-
complete for multiply connected domains P . Our main results are constant-factor
approximation algorithms for each of the three throw models, assuming D ≥ 2; refer
to Table 1. The approximation ratio of our algorithms increases as the throw direction
becomes more restricted. We give extensions for clearing polygons with holes, both
where the holes are obstacles and cliffs. Then we discuss how to adapt our algorithms
3 The “cliff” assumption accurately models the capacitated vacuum cleaner problem for
which there is a (central) “dustpan vac” in the baseboard, where a robotic vacuum cleaner
may empty its load [1] and applies also to urban snow removal using snow melters [2].
The SBP shows up also in other contexts: Consider a mobile robot that is
equipped with a device that allows it to pick up a carton and then place the car-
ton down again, possibly on a stack of cartons, in a location just next to it; with each
such operation, the robot shifts a unit of “material”. The SBP models the problem
in which the robot is to move a set of boxes to a specified destination in the most
efficient manner, subject to the constraint that it cannot stack boxes higher than a
capacity bound.
In a third motivating application, consider a robotic lawnmower or vacuum
cleaner that has a catch basin for the clippings, leaves, dust, or other debris. The
goal is to remove the debris from a region, with the constraint that the catch basin
must be emptied (e.g., in the compost pile) whenever it gets full.
The SBP is related to other problems on milling, vehicle routing, and traveling
salesman tours, but there are two important new features: (a) material must be moved
(snow must be thrown), and (b) material may not pile up too high.
While the SBP arises naturally in these other application domains, for the rest of
the paper, we use the terminology of snow removal.
The objective of the SBP is to find the shortest snowblower tour that clears a
domain P , assumed to be initially covered with snow at uniform depth 1. An impor-
tant parameter of the problem is the maximum snow depth D > 1 through which
the snowblower can move. At all times no point of P should have snow of greater
depth than D. The snow is to be moved to points outside of P . We assume that each
point outside P is able to receive arbitrarily much snow (i.e., that the driveway is
surrounded by a “cliff” over which we can toss as much snow as we want).3
Snowblowers offer the user the ability to control the direction in which the snow
is thrown. Some throw directions are preferable over others; e.g., throwing the snow
back into the user’s face is undesirable. However, it can be cumbersome to change
the throw direction too frequently during the course of clearing; also, some variants
(e.g., lawnmowing with clippings) require that the throw direction is fixed. Thus,
we consider three throw models. In the default model throwing the snow backwards
is allowed. In the adjustable-throw model the snow can be thrown only to the left,
right, or forward. In the fixed-throw model the snow is always thrown to the right.
Even though it seems silly to allow the throw direction to be back into one’s face, the
default model is the starting point for the analysis of other models and is equivalent
to the vacuum cleaner problem (discussed later).
Results. In this paper we introduce the snowblower problem, model its variants,
and give the first algorithmic results for its solution. We show that the SBP is NP-
complete for multiply connected domains P . Our main results are constant-factor
approximation algorithms for each of the three throw models, assuming D ≥ 2; refer
to Table 1. The approximation ratio of our algorithms increases as the throw direction
becomes more restricted. We give extensions for clearing polygons with holes, both
where the holes are obstacles and cliffs. Then we discuss how to adapt our algorithms
3 The “cliff” assumption accurately models the capacitated vacuum cleaner problem for
which there is a (central) “dustpan vac” in the baseboard, where a robotic vacuum cleaner
may empty its load [1] and applies also to urban snow removal using snow melters [2].
Page 3
The Snowblower Problem 3
for clearing nonrectilinear polygons and polygons with uneven initial distributions
of snow. We conclude by giving a succinct representation of the snowblower tour, in
which the tour specification is polynomial in the complexity of the input polygon.
D 2 or 3 any D ≥ 4
Apx. 6 8
Ref. Theorem 2
D any D ≥ 2
Apx. 4 + 3D/⌊D/2⌋
Ref. Theorem 3
D any D ≥ 2
Apx. 34 + 24D/⌊D/2⌋
Ref. Theorem 4
Table 1: Approximation factors of our algorithms in the default model (left), adjustable-throw
model (middle), and fixed-throw model (right).
Related Work. The SBP is closely related to milling and lawn-mowing prob-
lems, which have been studied extensively in the NC-machining and computational-
geometry literatures; see e.g., [3, 4, 10]. The SBP is also closely related to material-
handling problems, in which the goal is to rearrange a set of objects (e.g., cartons)
within a storage facility; see [8, 9, 13]. The SBP may be considered as an interme-
diate point between the TSP/lawnmowing/milling problems and material-handling
problems. Indeed, for D = ∞, the SBP is that of optimal milling. Unlike most
material-handling problems, the SBP formulation allows the material (snow) to pile
up on a single pixel, and it is this compressibility of the material that distinguishes
the SBP from previously studied material-handling problems. With TSP and related
problems, a pixel only is visited a constant number of times, whereas with material-
handling problems, pixels may have to be visited a number of times exponential in
the input size. For this reason, material-handling problems are not even known to be
in NP [8, 9], in contrast with the SBP. Note that in material handling problems the
objective is to minimize workload (distance traveled while loaded), while in the SBP
(as in the milling/mowing problems) the objective is to minimize total travel distance
(loaded or not).
The SBP is also related to the earth-mover’s distance (EMD), which is the min-
imum amount of work needed to rearrange one distribution (of earth, snow, etc.) to
another; see [7]. In the EMD literature, the question is explored mostly from an exis-
tential point of view, rather than planning the actual process of rearrangement. In the
SBP, we are interested in optimizing the length of the tour, and we do not necessarily
know in advance the final distribution of the snow after it has been removed from P .
Notation. The input is a polygonal domain, P . Since we are mainly concerned
with proving constant factor approximation algorithms, it suffices to consider dis-
tances measured according to the L1 metric. We consider the snowblower to be an
(axis-parallel) unit square that moves horizontally or vertically by unit steps. This
justifies our assumption, in most of our discussion, that P is an integral-orthogonal
simple polygon, which is comprised of a union of pixels – (closed) unit squares with
disjoint interiors and integral coordinates. In Section 5 we remark how our methods
extend to general (nonrectilinear) regions.
We say that two pixels are adjacent or neighbors if they share a side; the degree
of a pixel is the number of its neighbors. For a region R ⊆ P (subset of pixels),
for clearing nonrectilinear polygons and polygons with uneven initial distributions
of snow. We conclude by giving a succinct representation of the snowblower tour, in
which the tour specification is polynomial in the complexity of the input polygon.
D 2 or 3 any D ≥ 4
Apx. 6 8
Ref. Theorem 2
D any D ≥ 2
Apx. 4 + 3D/⌊D/2⌋
Ref. Theorem 3
D any D ≥ 2
Apx. 34 + 24D/⌊D/2⌋
Ref. Theorem 4
Table 1: Approximation factors of our algorithms in the default model (left), adjustable-throw
model (middle), and fixed-throw model (right).
Related Work. The SBP is closely related to milling and lawn-mowing prob-
lems, which have been studied extensively in the NC-machining and computational-
geometry literatures; see e.g., [3, 4, 10]. The SBP is also closely related to material-
handling problems, in which the goal is to rearrange a set of objects (e.g., cartons)
within a storage facility; see [8, 9, 13]. The SBP may be considered as an interme-
diate point between the TSP/lawnmowing/milling problems and material-handling
problems. Indeed, for D = ∞, the SBP is that of optimal milling. Unlike most
material-handling problems, the SBP formulation allows the material (snow) to pile
up on a single pixel, and it is this compressibility of the material that distinguishes
the SBP from previously studied material-handling problems. With TSP and related
problems, a pixel only is visited a constant number of times, whereas with material-
handling problems, pixels may have to be visited a number of times exponential in
the input size. For this reason, material-handling problems are not even known to be
in NP [8, 9], in contrast with the SBP. Note that in material handling problems the
objective is to minimize workload (distance traveled while loaded), while in the SBP
(as in the milling/mowing problems) the objective is to minimize total travel distance
(loaded or not).
The SBP is also related to the earth-mover’s distance (EMD), which is the min-
imum amount of work needed to rearrange one distribution (of earth, snow, etc.) to
another; see [7]. In the EMD literature, the question is explored mostly from an exis-
tential point of view, rather than planning the actual process of rearrangement. In the
SBP, we are interested in optimizing the length of the tour, and we do not necessarily
know in advance the final distribution of the snow after it has been removed from P .
Notation. The input is a polygonal domain, P . Since we are mainly concerned
with proving constant factor approximation algorithms, it suffices to consider dis-
tances measured according to the L1 metric. We consider the snowblower to be an
(axis-parallel) unit square that moves horizontally or vertically by unit steps. This
justifies our assumption, in most of our discussion, that P is an integral-orthogonal
simple polygon, which is comprised of a union of pixels – (closed) unit squares with
disjoint interiors and integral coordinates. In Section 5 we remark how our methods
extend to general (nonrectilinear) regions.
We say that two pixels are adjacent or neighbors if they share a side; the degree
of a pixel is the number of its neighbors. For a region R ⊆ P (subset of pixels),
Page 4
4 Esther M. Arkin, Michael A. Bender, Joseph S. B. Mitchell, Valentin Polishchuk
let GR denote the dual graph of R, having a vertex in the center of each pixel of R
and edges between adjacent pixels. Sometimes when we speak of the region R, we
implicitly mean the dual graph GR. A pixel of degree less than four is a boundary
pixel. For a boundary pixel, a side that is also on the boundary of P is called a
boundary side. The set of boundary sides, ∂P , forms the boundary of P . We treat
∂P as a set of boundary sides, rather than just as a closed curve. We assume that
the elements of ∂P are ordered as they are encountered when the boundary of P is
traversed counterclockwise.
An articulation vertex of a graph G is a vertex whose removal disconnects G.
We assume that GP has no articulation vertices. (Our algorithms can be adapted to
regions having articulation vertices, at a possible increase in approximation ratio.)
Algorithms Overview. Our algorithms proceed by clearing the polygon Voronoi-
cell-by-Voronoi-cell, starting from the Voronoi cell of the garage g — the pixel on
the boundary of P at which the snowblower tour starts and ends. The order of the
boundary sides in ∂P provides a natural order in which to clear the cells. We observe
that the Voronoi cell of each boundary side is a tree of one of two special types,
which we call lines and combs. We show how to clear the trees efficiently in each of
the throw models. We prove that our algorithms give constant-factor approximations
by charging the lengths of the tours produced by the algorithms to two lower bounds,
described in the next section.
2 Preliminaries
Tie Breaking. Our rules for finding V(p) for a pixel p that is equidistant between
two or more boundaries is based on the direction of the shortest path from p to V(p);
horizontal edges are preferred to vertical, going down has higher priority than going
up, going to the right — than going left.
In details, let p be a pixel, equidistant from m boundary sides {e1 . . . em} = E ⊂
∂P . Let r(p), l(p), d(p), u(p) denote the pixels to the left, above, to the right and
below p. First consider the case when p is a boundary pixel. If r(p) /∈ P , then the
right side of p is defined to be V(p); else if l(p) /∈ P , the left side of p is V(p); else
if d(p) /∈ P , the bottom side of p is V(p); otherwise, the top side of p is V(p). If p
is not a boundary pixel, we build the path from p to V(p) as follows. If there exists a
side e ∈ E such that there exist a (shortest) path from p to e, going through r(p) then
V(p) = V(r(p)); else if there exists a side e ∈ E such that there exist a (shortest)
path from p to e, going through l(p) then V(p) = V(l(p)); else if there exists a side
e ∈ E such that there exist a (shortest) path from p to e, going through d(p) then
V(p) = V(d(p)); otherwise V(p) = V(u(p)).
We first describe the rules for finding the closest boundary side for pixels that
are equidistant between two or more boundaries. We give preference to the boundary
sides as follows: horizontal sides are preferred over vertical. The lowest horizontal
side has the highest priority among horizontal sides, and among all lowest horizontal
sides, the leftmost has the highest priority. The leftmost vertical side has the highest
priority among vertical sides, and among all leftmost vertical sides, the highest has
let GR denote the dual graph of R, having a vertex in the center of each pixel of R
and edges between adjacent pixels. Sometimes when we speak of the region R, we
implicitly mean the dual graph GR. A pixel of degree less than four is a boundary
pixel. For a boundary pixel, a side that is also on the boundary of P is called a
boundary side. The set of boundary sides, ∂P , forms the boundary of P . We treat
∂P as a set of boundary sides, rather than just as a closed curve. We assume that
the elements of ∂P are ordered as they are encountered when the boundary of P is
traversed counterclockwise.
An articulation vertex of a graph G is a vertex whose removal disconnects G.
We assume that GP has no articulation vertices. (Our algorithms can be adapted to
regions having articulation vertices, at a possible increase in approximation ratio.)
Algorithms Overview. Our algorithms proceed by clearing the polygon Voronoi-
cell-by-Voronoi-cell, starting from the Voronoi cell of the garage g — the pixel on
the boundary of P at which the snowblower tour starts and ends. The order of the
boundary sides in ∂P provides a natural order in which to clear the cells. We observe
that the Voronoi cell of each boundary side is a tree of one of two special types,
which we call lines and combs. We show how to clear the trees efficiently in each of
the throw models. We prove that our algorithms give constant-factor approximations
by charging the lengths of the tours produced by the algorithms to two lower bounds,
described in the next section.
2 Preliminaries
Tie Breaking. Our rules for finding V(p) for a pixel p that is equidistant between
two or more boundaries is based on the direction of the shortest path from p to V(p);
horizontal edges are preferred to vertical, going down has higher priority than going
up, going to the right — than going left.
In details, let p be a pixel, equidistant from m boundary sides {e1 . . . em} = E ⊂
∂P . Let r(p), l(p), d(p), u(p) denote the pixels to the left, above, to the right and
below p. First consider the case when p is a boundary pixel. If r(p) /∈ P , then the
right side of p is defined to be V(p); else if l(p) /∈ P , the left side of p is V(p); else
if d(p) /∈ P , the bottom side of p is V(p); otherwise, the top side of p is V(p). If p
is not a boundary pixel, we build the path from p to V(p) as follows. If there exists a
side e ∈ E such that there exist a (shortest) path from p to e, going through r(p) then
V(p) = V(r(p)); else if there exists a side e ∈ E such that there exist a (shortest)
path from p to e, going through l(p) then V(p) = V(l(p)); else if there exists a side
e ∈ E such that there exist a (shortest) path from p to e, going through d(p) then
V(p) = V(d(p)); otherwise V(p) = V(u(p)).
We first describe the rules for finding the closest boundary side for pixels that
are equidistant between two or more boundaries. We give preference to the boundary
sides as follows: horizontal sides are preferred over vertical. The lowest horizontal
side has the highest priority among horizontal sides, and among all lowest horizontal
sides, the leftmost has the highest priority. The leftmost vertical side has the highest
priority among vertical sides, and among all leftmost vertical sides, the highest has
Page 5
The Snowblower Problem 5
the highest priority. In fact, any tie-breaking rule can be applied as long as it is
applied consistently. The particular choice of the rule only affects the orientation of
the combs.
Voronoi Cell Structure. For a pixel p ∈ P let V(p) denote the element of
∂P , closest to p. In case of ties, the tie-breaking rule is applied. Inspired by
computational-geometry terminology, we call V(p) the Voronoi side of p. We let
δ(p) denote the length of the path from p to the pixel having V(p) as a side (see
Fig. 7, right). For a boundary side e ∈ ∂P we let Voronoi(e) denote the (possibly,
empty) set of pixels, having e is the Voronoi side: Voronoi(e) = {p ∈ P |V(p) = e}.
We call Voronoi(e) the Voronoi cell of e. The Voronoi cells of the elements of ∂P
form a partition of P , which is called the Voronoi decomposition of P .
A set of pixels L whose dual graph GL is a straight path or a path with one bend,
is called a line. Each line L has a root pixel p, which corresponds to one of the two
leaves of GL, and a base, e ∈ ∂P , which is a side of p.
A (horizontal) comb C is a union of pixels consisting of a set of vertically adjacent
(horizontal) rows of pixels, with all of the rightmost pixels (or all of the leftmost
pixels) in a common column. (A vertical comb is defined similarly; however, by our
tie breaking rules, we need consider only horizontal combs.) A comb is a special type
of histogram polygon [6]. The common vertical column of rightmost/leftmost pixels
is called the handle of comb C, and each of the rows is called a tooth. A leftward
comb has its teeth extending leftwards from the handle; a rightward comb is defined
similarly. The pixel of a tooth that is furthest from the handle is the tip of the tooth.
The topmost row is the wisdom tooth of the comb. The root pixel p of the comb is
either the bottommost or topmost pixel of the handle, and its bottom or top side,
e ∈ ∂P , is the base of the comb. See Fig. 1, left. The union of a leftward comb and
a rightward comb having a common root pixel is called a double-sided comb.
c
∗
∗
∗
∗
∗
∗
∗••••
1
1 2
2
3 3 3
3 3
4 4
5
7 8
8
9
9
10
Fig. 1: Left: a comb. The base is bold. The pixels in the handle are marked with asterisks, the
pixels in the wisdom tooth are marked with bullets. Right: Voronoi cells. The sides of ∂P are
numbered 1 . . . 28 counterclockwise. The pixels in the Voronoi cell of a side are marked with
the corresponding number. Voronoi cell of side 3 is a comb; Voronoi cells of sides 6, 11, 17,
25, 28 are empty; cells of sides 1, 7, 10, 18, 24 are lines, comprised of just one pixel; cells of
the other edges are lines with more than one pixel.
the highest priority. In fact, any tie-breaking rule can be applied as long as it is
applied consistently. The particular choice of the rule only affects the orientation of
the combs.
Voronoi Cell Structure. For a pixel p ∈ P let V(p) denote the element of
∂P , closest to p. In case of ties, the tie-breaking rule is applied. Inspired by
computational-geometry terminology, we call V(p) the Voronoi side of p. We let
δ(p) denote the length of the path from p to the pixel having V(p) as a side (see
Fig. 7, right). For a boundary side e ∈ ∂P we let Voronoi(e) denote the (possibly,
empty) set of pixels, having e is the Voronoi side: Voronoi(e) = {p ∈ P |V(p) = e}.
We call Voronoi(e) the Voronoi cell of e. The Voronoi cells of the elements of ∂P
form a partition of P , which is called the Voronoi decomposition of P .
A set of pixels L whose dual graph GL is a straight path or a path with one bend,
is called a line. Each line L has a root pixel p, which corresponds to one of the two
leaves of GL, and a base, e ∈ ∂P , which is a side of p.
A (horizontal) comb C is a union of pixels consisting of a set of vertically adjacent
(horizontal) rows of pixels, with all of the rightmost pixels (or all of the leftmost
pixels) in a common column. (A vertical comb is defined similarly; however, by our
tie breaking rules, we need consider only horizontal combs.) A comb is a special type
of histogram polygon [6]. The common vertical column of rightmost/leftmost pixels
is called the handle of comb C, and each of the rows is called a tooth. A leftward
comb has its teeth extending leftwards from the handle; a rightward comb is defined
similarly. The pixel of a tooth that is furthest from the handle is the tip of the tooth.
The topmost row is the wisdom tooth of the comb. The root pixel p of the comb is
either the bottommost or topmost pixel of the handle, and its bottom or top side,
e ∈ ∂P , is the base of the comb. See Fig. 1, left. The union of a leftward comb and
a rightward comb having a common root pixel is called a double-sided comb.
c
∗
∗
∗
∗
∗
∗
∗••••
1
1 2
2
3 3 3
3 3
4 4
5
7 8
8
9
9
10
Fig. 1: Left: a comb. The base is bold. The pixels in the handle are marked with asterisks, the
pixels in the wisdom tooth are marked with bullets. Right: Voronoi cells. The sides of ∂P are
numbered 1 . . . 28 counterclockwise. The pixels in the Voronoi cell of a side are marked with
the corresponding number. Voronoi cell of side 3 is a comb; Voronoi cells of sides 6, 11, 17,
25, 28 are empty; cells of sides 1, 7, 10, 18, 24 are lines, comprised of just one pixel; cells of
the other edges are lines with more than one pixel.
Page 6
6 Esther M. Arkin, Michael A. Bender, Joseph S. B. Mitchell, Valentin Polishchuk
An analysis of the structure of the Voronoi partition under our tie breaking rules
gives:
Lemma 1. For a side e ∈ ∂P , the Voronoi cell of e is either a line (whose dual graph
is a straight path), or a comb, or a double-sided comb. By our tie-breaking rule, the
combs may appear only as the Voronoi cells of horizontal edges. The double-sided
combs may appear only as the Voronoi cells of (horizontal) edges of length 1.
Let p be a boundary pixel of P , let e ∈ ∂P be the side of p such that p ∈
Voronoi(e). We denote Voronoi(e) by T (p) or T (e), indicating that it is a unique
tree (a line or a comb) that has p as the root and e as the base.
Lower Bounds. We exhibit two lower bounds on the cost of an optimal tour, one
based on the number of pixels and the other on the Voronoi decomposition of the
domain. At any time let s(R) be the set of pixels of R covered with snow and also,
abusing notation, the amount of snow on these pixels. Let d(R) = 1D
∑
p∈s(R) δ(p) .
Lemma 2. Let R be a subset of P with the snowblower starting from a pixel outside
R. Then s(R) and d(R) are lower bounds on the cost to clear R.
Proof. For the snow lower bound, observe that regionR cannot be cleared with fewer
than s(R) snowblower moves because each pixel of s(R) needs to be visited.
For the distance lower bound, observe that, in order to clear the snow initially
residing on a pixel p, the snowblower has to make at least δ(p) moves. When the
snow from p is carried to the boundary of P and thrown away, the snow from at most
D − 1 other pixels can be thrown away simultaneously. Thus, a region R cannot be
cleared with fewer than d(R) moves. ⊓⊔
NP-Completeness. It is known ( [11, 12]) that the Hamiltonian path problem in
cubic grid graphs is NP-complete. The problem can be straightforwardly reduced to
SBP. If G is a cubic grid graph, construct an (integral orthohedral) domain P such
that G = GP . Since GP is cubic, each pixel p ∈ P is a boundary pixel, thus, the
snowblower can throw the snow away from p upon entering it. Hence, SBP on P is
equivalent to TSP on G, which has optimum less than n + 1 iff G is Hamiltonian
(where n is the number of nodes in G). The reduction works for any D ≥ 1.
The algorithms proposed in this paper show that any domain can be cleared using
a set of moves of cardinality polynomial in the number of pixels in the domain,
assuming D ≥ 2. Thus, we obtain
Theorem 1. If D ≥ 2, the SBP is NP-complete, both in the default model and in the
adjustable throw model, for inputs that are polygonal domains with holes.
3 Approximation Algorithm for the Default Model
In this section we give an 8-approximation algorithm for the case when the snow
can be thrown in all four directions. We first show how to clear a line efficiently
An analysis of the structure of the Voronoi partition under our tie breaking rules
gives:
Lemma 1. For a side e ∈ ∂P , the Voronoi cell of e is either a line (whose dual graph
is a straight path), or a comb, or a double-sided comb. By our tie-breaking rule, the
combs may appear only as the Voronoi cells of horizontal edges. The double-sided
combs may appear only as the Voronoi cells of (horizontal) edges of length 1.
Let p be a boundary pixel of P , let e ∈ ∂P be the side of p such that p ∈
Voronoi(e). We denote Voronoi(e) by T (p) or T (e), indicating that it is a unique
tree (a line or a comb) that has p as the root and e as the base.
Lower Bounds. We exhibit two lower bounds on the cost of an optimal tour, one
based on the number of pixels and the other on the Voronoi decomposition of the
domain. At any time let s(R) be the set of pixels of R covered with snow and also,
abusing notation, the amount of snow on these pixels. Let d(R) = 1D
∑
p∈s(R) δ(p) .
Lemma 2. Let R be a subset of P with the snowblower starting from a pixel outside
R. Then s(R) and d(R) are lower bounds on the cost to clear R.
Proof. For the snow lower bound, observe that regionR cannot be cleared with fewer
than s(R) snowblower moves because each pixel of s(R) needs to be visited.
For the distance lower bound, observe that, in order to clear the snow initially
residing on a pixel p, the snowblower has to make at least δ(p) moves. When the
snow from p is carried to the boundary of P and thrown away, the snow from at most
D − 1 other pixels can be thrown away simultaneously. Thus, a region R cannot be
cleared with fewer than d(R) moves. ⊓⊔
NP-Completeness. It is known ( [11, 12]) that the Hamiltonian path problem in
cubic grid graphs is NP-complete. The problem can be straightforwardly reduced to
SBP. If G is a cubic grid graph, construct an (integral orthohedral) domain P such
that G = GP . Since GP is cubic, each pixel p ∈ P is a boundary pixel, thus, the
snowblower can throw the snow away from p upon entering it. Hence, SBP on P is
equivalent to TSP on G, which has optimum less than n + 1 iff G is Hamiltonian
(where n is the number of nodes in G). The reduction works for any D ≥ 1.
The algorithms proposed in this paper show that any domain can be cleared using
a set of moves of cardinality polynomial in the number of pixels in the domain,
assuming D ≥ 2. Thus, we obtain
Theorem 1. If D ≥ 2, the SBP is NP-complete, both in the default model and in the
adjustable throw model, for inputs that are polygonal domains with holes.
3 Approximation Algorithm for the Default Model
In this section we give an 8-approximation algorithm for the case when the snow
can be thrown in all four directions. We first show how to clear a line efficiently
Page 7
The Snowblower Problem 7
with the operation called line-clearing. We then introduce another operation, the
brush, and show how to clear a comb efficiently with a sequence of line-clearings
and brushes. Finally, we splice the subtours through each line and comb into a larger
tour, clearing the entire domain. The algorithm for the default model, developed in
this section, serves as a basis for the algorithms in the other models.
Clearing a Line. Let L be a line of pixels; let p and e be its root and the base. We
are interested in clearing lines for which the base is a boundary side, i.e., e ∈ ∂P .
Let ℓ = s(L); let the first J pixels of L counting from p be clear. We assume that
p is already clear (J > 0); the snow from it was thrown away through the side e as
the snowblower first entered pixel p. Let L|J denote L with the J pixels clear; let
ℓ− J = kD + r.4 Denote by (L|J)D the first kD pixels of L|J covered with snow;
denote by Lr the last r pixels on L|J . The idea of decomposing L|J into (L|J)D
and Lr is that the snow from (L|J)D is thrown away with k “fully-loaded” throws,
and the snow from Lr is thrown away with (at most one) additional “under-loaded”
throw.
We clear line L starting at p by moving all the snow through the base e and
returning back to p. The basic clearing operation is a back throw. In a back throw
the snowblower, entering a pixel u from pixel v, throws u’s snow backward onto
v. Starting from p, the snowblower moves along L away from p until either the
snowblower moves through D pixels covered with snow or the snowblower reaches
the other end of L; this is called the forward pass. Next, the snowblower makes a
U-turn and moves back to p, pushing all the snow in front of it and over e; this is
called the backward pass. A forward and backward pass that clears exactly D units
of snow is called a D-full pass.
Lemma 3. For arbitrary D ≥ 4 the line-clearing cost is at most 2s(L\p)+4d(L|J).
For D = 2, 3 the line-clearing cost is at most 2s(L \ p) + 2d(L|J). If every pass is
D-full, the cost is 4d(L|J) for D ≥ 4 and 2d(L|J) for D = 2, 3.
Proof. The clearing cost is c(L|J) = c((L|J)D)+ c(Lr) =
∑k
i=1 2(J − 1 + iD)+
2(ℓ − 1) = 2kJ + Dk(k + 1) − 2k + 2(ℓ − 1). The snow lower bound of L \ p is
s(L\p) = ℓ−1. The distance lower bound of (L|J)D is d((L|J)D) = 1D
∑kD
i=1(J+
i) = kJ + k(kD + 1)/2.
Thus,
c(L|J) = 2s(L \ p) +
(
2 +
D − 3
J + (Dk + 1)/2
)
d((L|J)D)
If every pass is a D-full pass, then c(Lr) = 0. Therefore, c(L|J) = c((L|J)D) =
(
2 + D−3J+(Dk+1)/2
)
d((L|J)D). ⊓⊔
Clearing a Comb. Let C be a comb with the root p, base e, and handle H of length
H . Let ℓ1 . . . ℓH be the lengths of the teeth of the comb. Since we are interested
4 For ease of presentation, we adapt the following convention. For d ∈ {D, ⌊D/2⌋} and an
integer w we understand the equality w = ad+ b as follows: b and a are the remainder and
the quotient, respectively, of w divided by d.
with the operation called line-clearing. We then introduce another operation, the
brush, and show how to clear a comb efficiently with a sequence of line-clearings
and brushes. Finally, we splice the subtours through each line and comb into a larger
tour, clearing the entire domain. The algorithm for the default model, developed in
this section, serves as a basis for the algorithms in the other models.
Clearing a Line. Let L be a line of pixels; let p and e be its root and the base. We
are interested in clearing lines for which the base is a boundary side, i.e., e ∈ ∂P .
Let ℓ = s(L); let the first J pixels of L counting from p be clear. We assume that
p is already clear (J > 0); the snow from it was thrown away through the side e as
the snowblower first entered pixel p. Let L|J denote L with the J pixels clear; let
ℓ− J = kD + r.4 Denote by (L|J)D the first kD pixels of L|J covered with snow;
denote by Lr the last r pixels on L|J . The idea of decomposing L|J into (L|J)D
and Lr is that the snow from (L|J)D is thrown away with k “fully-loaded” throws,
and the snow from Lr is thrown away with (at most one) additional “under-loaded”
throw.
We clear line L starting at p by moving all the snow through the base e and
returning back to p. The basic clearing operation is a back throw. In a back throw
the snowblower, entering a pixel u from pixel v, throws u’s snow backward onto
v. Starting from p, the snowblower moves along L away from p until either the
snowblower moves through D pixels covered with snow or the snowblower reaches
the other end of L; this is called the forward pass. Next, the snowblower makes a
U-turn and moves back to p, pushing all the snow in front of it and over e; this is
called the backward pass. A forward and backward pass that clears exactly D units
of snow is called a D-full pass.
Lemma 3. For arbitrary D ≥ 4 the line-clearing cost is at most 2s(L\p)+4d(L|J).
For D = 2, 3 the line-clearing cost is at most 2s(L \ p) + 2d(L|J). If every pass is
D-full, the cost is 4d(L|J) for D ≥ 4 and 2d(L|J) for D = 2, 3.
Proof. The clearing cost is c(L|J) = c((L|J)D)+ c(Lr) =
∑k
i=1 2(J − 1 + iD)+
2(ℓ − 1) = 2kJ + Dk(k + 1) − 2k + 2(ℓ − 1). The snow lower bound of L \ p is
s(L\p) = ℓ−1. The distance lower bound of (L|J)D is d((L|J)D) = 1D
∑kD
i=1(J+
i) = kJ + k(kD + 1)/2.
Thus,
c(L|J) = 2s(L \ p) +
(
2 +
D − 3
J + (Dk + 1)/2
)
d((L|J)D)
If every pass is a D-full pass, then c(Lr) = 0. Therefore, c(L|J) = c((L|J)D) =
(
2 + D−3J+(Dk+1)/2
)
d((L|J)D). ⊓⊔
Clearing a Comb. Let C be a comb with the root p, base e, and handle H of length
H . Let ℓ1 . . . ℓH be the lengths of the teeth of the comb. Since we are interested
4 For ease of presentation, we adapt the following convention. For d ∈ {D, ⌊D/2⌋} and an
integer w we understand the equality w = ad+ b as follows: b and a are the remainder and
the quotient, respectively, of w divided by d.
Page 8
8 Esther M. Arkin, Michael A. Bender, Joseph S. B. Mitchell, Valentin Polishchuk
in clearing combs for which the base e is a boundary side (e ∈ ∂P ), we assume
that pixel p is already clear — the snow from it was thrown away through e as the
snowblower first entered p.
Our clearing strategy is as follows. While there exists a line L ⊂ C rooted at p,
such that s(L) ≥ D, we perform as many D-full passes on L as we can. When no
such L remains, we call the comb brush-ready and we use another clearing operation,
the brush, to finish the clearing.
A brush, essentially, is a “capacitated” depth-first-search. Among the teeth of a
brush-ready comb that are not fully cleared, let t be the tooth, furthest from the base.
In a brush, we move the snowblower from p through the handle, turn into t, reach
its tip, U-turn, come back to the handle (pushing the pile of snow), turn onto the
handle, move by the handle back towards p until we reach the next not fully cleared
tooth, turn onto the tooth, and so on. We continue clearing the teeth one-by-one in
this manner until D units of snow have been moved (or all the snow on the comb has
been moved). Then we push the snow to p through the handle and across e. This tour
is called a brush (Fig. 2).
6?
Fig. 2: Left: a brush-ready comb. The snow is shown in light gray. Center: a brush, D = 4;
the part of the brush, traveling through the handle, is bold. Right: the comb after the brush.
Lemma 4. For arbitrary D ≥ 4 the comb C can be cleared at a cost of at most
4s(C \ p) + 4d(C \ p) (at most 4s(C \ p) + 2d(C \ p) for D = 2, 3).
Proof. If s(C \ p) < D, then the cost of clearing is just 2s(C \ p), so suppose,
s(C \p) ≥ D. Let B be the number of brushes used; let B be the set of pixels cleared
by the brushes. For b = 1 . . . B let tb and t′b be the first and the last tooth visited
during the bth brush. For b ∈ {1 . . . B − 1} the bth brush enters at least 2 teeth, so
tb > t′b ≥ tb+1.
Each brush can be decomposed into two parts: the part, traveling through the
teeth, and the part, traveling through the handle (see Fig. 2). Since each tooth is
visited during at most 2 brushes, the length of the first part is at most 4 times the size
of all teeth, that is, 4s(C \ H). The total length of the second part of all brushes is
2
∑B
b=1(tb − 1). Thus, the cost of the “brushing” is
in clearing combs for which the base e is a boundary side (e ∈ ∂P ), we assume
that pixel p is already clear — the snow from it was thrown away through e as the
snowblower first entered p.
Our clearing strategy is as follows. While there exists a line L ⊂ C rooted at p,
such that s(L) ≥ D, we perform as many D-full passes on L as we can. When no
such L remains, we call the comb brush-ready and we use another clearing operation,
the brush, to finish the clearing.
A brush, essentially, is a “capacitated” depth-first-search. Among the teeth of a
brush-ready comb that are not fully cleared, let t be the tooth, furthest from the base.
In a brush, we move the snowblower from p through the handle, turn into t, reach
its tip, U-turn, come back to the handle (pushing the pile of snow), turn onto the
handle, move by the handle back towards p until we reach the next not fully cleared
tooth, turn onto the tooth, and so on. We continue clearing the teeth one-by-one in
this manner until D units of snow have been moved (or all the snow on the comb has
been moved). Then we push the snow to p through the handle and across e. This tour
is called a brush (Fig. 2).
6?
Fig. 2: Left: a brush-ready comb. The snow is shown in light gray. Center: a brush, D = 4;
the part of the brush, traveling through the handle, is bold. Right: the comb after the brush.
Lemma 4. For arbitrary D ≥ 4 the comb C can be cleared at a cost of at most
4s(C \ p) + 4d(C \ p) (at most 4s(C \ p) + 2d(C \ p) for D = 2, 3).
Proof. If s(C \ p) < D, then the cost of clearing is just 2s(C \ p), so suppose,
s(C \p) ≥ D. Let B be the number of brushes used; let B be the set of pixels cleared
by the brushes. For b = 1 . . . B let tb and t′b be the first and the last tooth visited
during the bth brush. For b ∈ {1 . . . B − 1} the bth brush enters at least 2 teeth, so
tb > t′b ≥ tb+1.
Each brush can be decomposed into two parts: the part, traveling through the
teeth, and the part, traveling through the handle (see Fig. 2). Since each tooth is
visited during at most 2 brushes, the length of the first part is at most 4 times the size
of all teeth, that is, 4s(C \ H). The total length of the second part of all brushes is
2
∑B
b=1(tb − 1). Thus, the cost of the “brushing” is
Page 9
The Snowblower Problem 9
c(B) ≤ 2
B
∑
b=1
(tb − 1) + 4s(C \ H) ≤ 2
B
∑
b=2
tb + 4s(C \ p) − 2 (1)
since t1 ≤ H and H ≥ 2 (for otherwise C is a line).
There are exactly D pixels cleared during each brush b ∈ {0 . . . B − 1}, and each
of these pixels is at distance at least tb′ from the base of the comb. Thus, the distance
lower bound of the pixels, cleared during brush b, is at least tb′ . Consequently, the
distance lower bound of B
d(B) ≥
B
∑
b=1
tb′ ≥
B−1
∑
b=1
tb+1 =
B
∑
b=2
tb (2)
From (1) and (2), B can be cleared at a cost of at most 2d(B) + 4s(C \ p).
Let P ⊆ C be the pixels, cleared during the line-clearings. By our strategy, during
each line-clearing, every pass is D-full; thus, by Lemma 3, P can be cleared at a cost
of at most 4d(P) (or 2d(P) if D = 2, 3). Since P and B are snow-disjoint and
P ∪ B = C \ p, the lemma follows. ⊓⊔
The above analysis is also valid in the case when the handle is initially clear. This
is the case when the second side of a double-sided comb is being cleared. Thus, a
double-sided comb can be cleared within the same bounds on the cost of clearing.
Clearing the Domain. Now that we have defined the operations which allow us to
clear efficiently lines and combs, we are ready to present the algorithm for clearing
the domain.
Theorem 2. For arbitrary D ≥ 4 (resp., D = 2, 3) an 8-approximate (resp., 6-
approximate) tour can be found in polynomial time.
Proof. Let p1, . . . , pM be the boundary pixels of P as they are encountered when
going around the boundary of P counterclockwise starting from g = p1; let
e1, . . . , eM ∈ ∂P be the boundary sides of p1, . . . , pM such that ei = V e(pi), i =
1 . . .M . The polygonP can be decomposed into disjoint trees T (p1), . . . , T (pM ) =
T (e1), . . . , T (eM ) with the bases e1 . . . , eM , where each tree T (ei) is either a line
or a comb.
Our algorithm clears P tree-by-tree starting with T (e1) = T (g). By Lem-
mas 3 and 4, for i = 1 . . .M , the tree T (pi) \ pi can be cleared at a cost of at
most 4s(T (pi) \ pi) + 4d(T (pi) \ pi) starting from pi and returning to pi. Since
⋃M
1 T (pi) \ pi = P \ {p1 . . . pM}, the interior of P can be cleared at a cost
of at most c(P \ {p1 . . . pM}) = 4s(P \ {p1 . . . pM}) + 4d(P \ {p1 . . . pM}) ≤
4s(P \ g) + 4d(P \ g)− 4M + 4. Finally, the tours clearing the interior of P can be
spliced into a tour, clearing P at a cost of at most 2M . Since the optimum is at least
s(P \ g) and is at least d(P \ g), the theorem follows. ⊓⊔
4 Other Models
In this section we give approximation algorithms for the case when the throw
direction is restricted. Specifically, we first consider the adjustable throw direction
c(B) ≤ 2
B
∑
b=1
(tb − 1) + 4s(C \ H) ≤ 2
B
∑
b=2
tb + 4s(C \ p) − 2 (1)
since t1 ≤ H and H ≥ 2 (for otherwise C is a line).
There are exactly D pixels cleared during each brush b ∈ {0 . . . B − 1}, and each
of these pixels is at distance at least tb′ from the base of the comb. Thus, the distance
lower bound of the pixels, cleared during brush b, is at least tb′ . Consequently, the
distance lower bound of B
d(B) ≥
B
∑
b=1
tb′ ≥
B−1
∑
b=1
tb+1 =
B
∑
b=2
tb (2)
From (1) and (2), B can be cleared at a cost of at most 2d(B) + 4s(C \ p).
Let P ⊆ C be the pixels, cleared during the line-clearings. By our strategy, during
each line-clearing, every pass is D-full; thus, by Lemma 3, P can be cleared at a cost
of at most 4d(P) (or 2d(P) if D = 2, 3). Since P and B are snow-disjoint and
P ∪ B = C \ p, the lemma follows. ⊓⊔
The above analysis is also valid in the case when the handle is initially clear. This
is the case when the second side of a double-sided comb is being cleared. Thus, a
double-sided comb can be cleared within the same bounds on the cost of clearing.
Clearing the Domain. Now that we have defined the operations which allow us to
clear efficiently lines and combs, we are ready to present the algorithm for clearing
the domain.
Theorem 2. For arbitrary D ≥ 4 (resp., D = 2, 3) an 8-approximate (resp., 6-
approximate) tour can be found in polynomial time.
Proof. Let p1, . . . , pM be the boundary pixels of P as they are encountered when
going around the boundary of P counterclockwise starting from g = p1; let
e1, . . . , eM ∈ ∂P be the boundary sides of p1, . . . , pM such that ei = V e(pi), i =
1 . . .M . The polygonP can be decomposed into disjoint trees T (p1), . . . , T (pM ) =
T (e1), . . . , T (eM ) with the bases e1 . . . , eM , where each tree T (ei) is either a line
or a comb.
Our algorithm clears P tree-by-tree starting with T (e1) = T (g). By Lem-
mas 3 and 4, for i = 1 . . .M , the tree T (pi) \ pi can be cleared at a cost of at
most 4s(T (pi) \ pi) + 4d(T (pi) \ pi) starting from pi and returning to pi. Since
⋃M
1 T (pi) \ pi = P \ {p1 . . . pM}, the interior of P can be cleared at a cost
of at most c(P \ {p1 . . . pM}) = 4s(P \ {p1 . . . pM}) + 4d(P \ {p1 . . . pM}) ≤
4s(P \ g) + 4d(P \ g)− 4M + 4. Finally, the tours clearing the interior of P can be
spliced into a tour, clearing P at a cost of at most 2M . Since the optimum is at least
s(P \ g) and is at least d(P \ g), the theorem follows. ⊓⊔
4 Other Models
In this section we give approximation algorithms for the case when the throw
direction is restricted. Specifically, we first consider the adjustable throw direction
Page 10
10 Esther M. Arkin, Michael A. Bender, Joseph S. B. Mitchell, Valentin Polishchuk
formulation. This is a convenient case for the snowblower operator who does not
want the snow thrown in his face. We then consider the fixed throw direction model,
which assumes that the snow is always thrown to the right.
We remark that the low approximation factors of the algorithms for the default
model, presented in the previous section, were due to a very conservative clearing:
the snow from every pixel p ∈ P was thrown through the Voronoi side V(p). Un-
fortunately, it seems hard to preserve this appealing property if throwing back is
forbidden. The reason is that the comb in the Voronoi cell Voronoi(e) of a boundary
side e ∈ ∂P often has a “staircase”-shaped boundary (see Fig. 8); clearing the first
“stair” in the staircase cannot be done without throwing the snow onto a pixel of
Voronoi(e′), where e′ 6= e is another boundary side. That’s why the approximation
factors of the algorithms in this section are higher than those in the previous one.
Adjustable Throw Direction
In the adjustable-throw model the snow cannot be thrown backward but can be
thrown in the three other directions. To give a constant-factor approximation algo-
rithm for this case, we show how to emulate line-clearings and brushes avoiding back
throws (Fig. 3). The approximation ratios increase slightly in comparison with the
default model.
Line-clearing. We can emulate a (half of a) pass by a sequence of moves, each with
throwing the snow to the left, forward or to the right (Fig. 3, left and center). Thus,
the line-clearing may be executed in the same way as it was done if the back throws
were allowed. The only difference is that now the snow is moved to the base when
the snow from only ⌊D/2⌋ pixels (as opposed to D pixels) of the line is gathered.
Lemma 5. The line-clearing cost is at most 3D/⌊D/2⌋d(L|J) + 2s(L \ p). If every
pass is ⌊D/2⌋-full, the cost is 3D/⌊D/2⌋d(L|J).
Proof. Let ℓ − J = k′ ⌊D/2⌋ + r′. Let (L|J)⌊D/2⌋ be the first k′⌊D/2⌋ pixels of
L|J , let Lr′ be its last r′ pixels. Then the cost of the clearing of L|J is c(L|J) =
c((L|J)⌊D/2⌋)+c(Lr′) =
∑k′
i=1 2(J+i⌊D/2⌋)+2ℓ = 2k′J+⌊D/2⌋k′(k′+1)+2ℓ.
The lower bounds are given by s(L \ p) = ℓ− 1 and
d((L|J)⌊ D2 ⌋) =
1
D
k′⌊D2 ⌋
∑
i=1
(J + i) = ⌊
D
2 ⌋
D
[
k′J + k
′(k′⌊D2 ⌋ + 1)
2
]
(3)
Thus,
c(L|J) ≤ D
⌊D2 ⌋
2 +
2 +
⌊
D
2
⌋
k′ − k′
k′J + ⌊
D
2 ⌋
2 k′2 + k
′
2
d(L \ p) + 2s(L \ p) ⊓⊔
Brush. Brush also does not change too much from the default case. The difference
is the same as with the line-clearing: now, instead of clearing D pixels with a brush,
formulation. This is a convenient case for the snowblower operator who does not
want the snow thrown in his face. We then consider the fixed throw direction model,
which assumes that the snow is always thrown to the right.
We remark that the low approximation factors of the algorithms for the default
model, presented in the previous section, were due to a very conservative clearing:
the snow from every pixel p ∈ P was thrown through the Voronoi side V(p). Un-
fortunately, it seems hard to preserve this appealing property if throwing back is
forbidden. The reason is that the comb in the Voronoi cell Voronoi(e) of a boundary
side e ∈ ∂P often has a “staircase”-shaped boundary (see Fig. 8); clearing the first
“stair” in the staircase cannot be done without throwing the snow onto a pixel of
Voronoi(e′), where e′ 6= e is another boundary side. That’s why the approximation
factors of the algorithms in this section are higher than those in the previous one.
Adjustable Throw Direction
In the adjustable-throw model the snow cannot be thrown backward but can be
thrown in the three other directions. To give a constant-factor approximation algo-
rithm for this case, we show how to emulate line-clearings and brushes avoiding back
throws (Fig. 3). The approximation ratios increase slightly in comparison with the
default model.
Line-clearing. We can emulate a (half of a) pass by a sequence of moves, each with
throwing the snow to the left, forward or to the right (Fig. 3, left and center). Thus,
the line-clearing may be executed in the same way as it was done if the back throws
were allowed. The only difference is that now the snow is moved to the base when
the snow from only ⌊D/2⌋ pixels (as opposed to D pixels) of the line is gathered.
Lemma 5. The line-clearing cost is at most 3D/⌊D/2⌋d(L|J) + 2s(L \ p). If every
pass is ⌊D/2⌋-full, the cost is 3D/⌊D/2⌋d(L|J).
Proof. Let ℓ − J = k′ ⌊D/2⌋ + r′. Let (L|J)⌊D/2⌋ be the first k′⌊D/2⌋ pixels of
L|J , let Lr′ be its last r′ pixels. Then the cost of the clearing of L|J is c(L|J) =
c((L|J)⌊D/2⌋)+c(Lr′) =
∑k′
i=1 2(J+i⌊D/2⌋)+2ℓ = 2k′J+⌊D/2⌋k′(k′+1)+2ℓ.
The lower bounds are given by s(L \ p) = ℓ− 1 and
d((L|J)⌊ D2 ⌋) =
1
D
k′⌊D2 ⌋
∑
i=1
(J + i) = ⌊
D
2 ⌋
D
[
k′J + k
′(k′⌊D2 ⌋ + 1)
2
]
(3)
Thus,
c(L|J) ≤ D
⌊D2 ⌋
2 +
2 +
⌊
D
2
⌋
k′ − k′
k′J + ⌊
D
2 ⌋
2 k′2 + k
′
2
d(L \ p) + 2s(L \ p) ⊓⊔
Brush. Brush also does not change too much from the default case. The difference
is the same as with the line-clearing: now, instead of clearing D pixels with a brush,
Page 11
The Snowblower Problem 11
s
6?→
s
@
s s
→
6?
@ bdP
6
-
Fig. 3: Emulating line-clearing and brush. The (possible) snow locations are in light gray; s is
the snowblower. Left: forward and backward passes in the default model; there are D units of
snow on the checked pixel. Center: the passes emulation; there is (at most) 2⌊D/2⌋ units of
snow on the checked pixel. Right: the snow to be cleared during a brush is in light gray; there
is ⌊D/2⌋ light gray pixels. Together with each gray pixel, the snow from at most 1 other pixel
is moved — thus, the brush is feasible.
only ⌊D/2⌋ pixels are (considered) cleared (Fig. 3, right). The definition of a brush-
ready comb is changed too — now we require that there is less than ⌊D/2⌋ pixels
covered with snow on each tooth of such a comb.
Lemma 6. A comb can be cleared at a cost of 3D/⌊D/2⌋d(C \ p) + 4s(C \ p).
Proof. In comparison with the default model (Lemma 4) several observations are in
place. The number of brushes may go up; we still denote it by B. The cost of the
brushes 1 . . . B − 1 does not change. If the Bth brush has to enter the first tooth,
there may be 2 more moves needed to return to the root of the comb (see Fig. 3,
right); hence, the total cost of the brushing (1) may go up by 2. The distance lower
bound (2) goes down by D/⌊D/2⌋. The rest of the proof is identical to the proof of
Lemma 4 (with Lemma 5 used in place of Lemma 3). ⊓⊔
Observe that in fact the snow can be removed from more than ⌊D/2⌋ pixels during
a brush; we just ignore it for now in our analysis. Note that a double-sided comb can
also be cleared in the described way.
Clearing the Domain. As in the default case (Theorem 2),
Theorem 3. A (4 + 3D/⌊D/2⌋)-approximate tour can be found in polynomial time.
Comment on the Parity of D. We remark that if D is even, the cost of the clearing is
the same as it would be if the snowblower were able to move through snow of depth
D + 1 (the slight increase of 6/(D− 1) in the approximation factor would be due to
the decrease of the distance lower bound).
Fixed Throw Direction
In reality, changing the throw direction requires some effort. In particular, a snow
plower does not change the direction of snow displacement at all. In this section we
s
6?→
s
@
s s
→
6?
@ bdP
6
-
Fig. 3: Emulating line-clearing and brush. The (possible) snow locations are in light gray; s is
the snowblower. Left: forward and backward passes in the default model; there are D units of
snow on the checked pixel. Center: the passes emulation; there is (at most) 2⌊D/2⌋ units of
snow on the checked pixel. Right: the snow to be cleared during a brush is in light gray; there
is ⌊D/2⌋ light gray pixels. Together with each gray pixel, the snow from at most 1 other pixel
is moved — thus, the brush is feasible.
only ⌊D/2⌋ pixels are (considered) cleared (Fig. 3, right). The definition of a brush-
ready comb is changed too — now we require that there is less than ⌊D/2⌋ pixels
covered with snow on each tooth of such a comb.
Lemma 6. A comb can be cleared at a cost of 3D/⌊D/2⌋d(C \ p) + 4s(C \ p).
Proof. In comparison with the default model (Lemma 4) several observations are in
place. The number of brushes may go up; we still denote it by B. The cost of the
brushes 1 . . . B − 1 does not change. If the Bth brush has to enter the first tooth,
there may be 2 more moves needed to return to the root of the comb (see Fig. 3,
right); hence, the total cost of the brushing (1) may go up by 2. The distance lower
bound (2) goes down by D/⌊D/2⌋. The rest of the proof is identical to the proof of
Lemma 4 (with Lemma 5 used in place of Lemma 3). ⊓⊔
Observe that in fact the snow can be removed from more than ⌊D/2⌋ pixels during
a brush; we just ignore it for now in our analysis. Note that a double-sided comb can
also be cleared in the described way.
Clearing the Domain. As in the default case (Theorem 2),
Theorem 3. A (4 + 3D/⌊D/2⌋)-approximate tour can be found in polynomial time.
Comment on the Parity of D. We remark that if D is even, the cost of the clearing is
the same as it would be if the snowblower were able to move through snow of depth
D + 1 (the slight increase of 6/(D− 1) in the approximation factor would be due to
the decrease of the distance lower bound).
Fixed Throw Direction
In reality, changing the throw direction requires some effort. In particular, a snow
plower does not change the direction of snow displacement at all. In this section we
Page 12
12 Esther M. Arkin, Michael A. Bender, Joseph S. B. Mitchell, Valentin Polishchuk
consider the fixed throw direction model, i.e., the case of the snowblower which
can only throw the snow to the right. We exploit the same idea as in the previous
subsection — reducing the problem in the fixed throw direction model to the problem
in the default model. All we need is to show how to emulate line-clearing and brush.
In what follows we retain the notation from the previous section.
Lemma 7. The line-clearing cost is at most 24D/⌊D/2⌋d(L|J) + 25s(L \ p). If
every pass is ⌊D/2⌋-full, the cost is 24D/⌊D/2⌋d(L|J).
Proof. We first consider clearing a line whose dual graph is embedded as a single
straight line segment and whose base is perpendicular to the segment; we describe
the line-clearing, assuming that the line is vertical. Next, we extend the solution to
the case when the base is parallel to the edges of the dual graph; this can only be a
horizontal line — the first tooth in a (double-)comb. Finally, we consider clearing an
L-shaped line; this can only by a tooth together with the (part of the) handle.
A Line L with GL⊥e. As in the adjustable-throw case (see Fig. 3, left and center),
to clear L we will need to use the pixels to the right of L to throw the snow onto.
Let p′ be the pixel, following p in ∂P . Before the line-clearing is begun, it will be
convenient to have p′ clear. Thus, the first thing we do upon entering L (through p) is
clearing p′. Together with returning the snowblower to p it takes 2 or 4 moves (Fig. 4,
left); we call these moves the double-base setup.
e
s-
e
s-ff
6
s
s
Fig. 4: Left: the double-base setup. Right: before the forward pass the snow below the snow-
blower is cleared on both lines.
Then, the following invariant is maintained during line-clearing. If the snow-
blower is at a pixel q ∈ L before starting the forward pass, all pixels on L from p
to q are clear, along with the pixels to the right of them (Fig. 4, right). The invariant
holds in the beginning of the line-clearing and our line-clearing strategy respects it.
Each back throw is emulated with 5 moves (Fig. 5, left). After moving up by
⌊D/2⌋ pixels (and thus, gathering 2⌊D/2⌋ units of snow on these ⌊D/2⌋ pixels), the
snowblower U-turns and moves towards p “pushing” the snow in front of it; a push
is emulated with 11 moves (Fig. 6).
The above observations already show that the cost of line-clearing increases only
by a multiplicative constant in comparison with the adjustable-throw case. A more
consider the fixed throw direction model, i.e., the case of the snowblower which
can only throw the snow to the right. We exploit the same idea as in the previous
subsection — reducing the problem in the fixed throw direction model to the problem
in the default model. All we need is to show how to emulate line-clearing and brush.
In what follows we retain the notation from the previous section.
Lemma 7. The line-clearing cost is at most 24D/⌊D/2⌋d(L|J) + 25s(L \ p). If
every pass is ⌊D/2⌋-full, the cost is 24D/⌊D/2⌋d(L|J).
Proof. We first consider clearing a line whose dual graph is embedded as a single
straight line segment and whose base is perpendicular to the segment; we describe
the line-clearing, assuming that the line is vertical. Next, we extend the solution to
the case when the base is parallel to the edges of the dual graph; this can only be a
horizontal line — the first tooth in a (double-)comb. Finally, we consider clearing an
L-shaped line; this can only by a tooth together with the (part of the) handle.
A Line L with GL⊥e. As in the adjustable-throw case (see Fig. 3, left and center),
to clear L we will need to use the pixels to the right of L to throw the snow onto.
Let p′ be the pixel, following p in ∂P . Before the line-clearing is begun, it will be
convenient to have p′ clear. Thus, the first thing we do upon entering L (through p) is
clearing p′. Together with returning the snowblower to p it takes 2 or 4 moves (Fig. 4,
left); we call these moves the double-base setup.
e
s-
e
s-ff
6
s
s
Fig. 4: Left: the double-base setup. Right: before the forward pass the snow below the snow-
blower is cleared on both lines.
Then, the following invariant is maintained during line-clearing. If the snow-
blower is at a pixel q ∈ L before starting the forward pass, all pixels on L from p
to q are clear, along with the pixels to the right of them (Fig. 4, right). The invariant
holds in the beginning of the line-clearing and our line-clearing strategy respects it.
Each back throw is emulated with 5 moves (Fig. 5, left). After moving up by
⌊D/2⌋ pixels (and thus, gathering 2⌊D/2⌋ units of snow on these ⌊D/2⌋ pixels), the
snowblower U-turns and moves towards p “pushing” the snow in front of it; a push
is emulated with 11 moves (Fig. 6).
The above observations already show that the cost of line-clearing increases only
by a multiplicative constant in comparison with the adjustable-throw case. A more
Page 13
The Snowblower Problem 13
careful look at the Figs. 5, 6 reveals that: (1) in the push emulation the first two moves
are the opposites of the last two, thus, all 4 moves may be omitted – consequently,
a push may be emulated by a sequence of only 7 moves; (2) if the boundary side,
following e, is vertical, the last push, throwing the snow away from P , may require 9
moves (Fig. 5, right); and, (3) when emulating the last back throw in a forward pass,
the last 2 of the 5 moves (the move up and the move to the right in Fig. 5, left) can
be omitted – indeed, during the push emulation, the snowblower may as well start
to the right of the snow (see Fig. 6). Thus, a line L|J can be cleared at a cost of
c(L|J) ≤ 4 +
∑k′
i=1
(
J − 1 + (i− 1)
⌊
D
2
⌋
+ 5
⌊
D
2
⌋
+ 7(J + i
⌊
D
2
⌋
− 1)
)
+J +
5r′ + 7(ℓ− 1).
s
s
→66 2
e
ff? s
6
Fig. 5: Left: emulating back throw. Right: pushing the 2 ⌊D/2⌋ units of snow away from P
and returning the snowblower to p may require 9 moves.
2
2
2
2
2
2 2 2
→ → → →
s s s
s
s2 2
?
2
?
64 4
Fig. 6: Emulating pushing the snow in front of the snowblower.
A Line L with GL||e. Consider a horizontal line, extending to the left of the base;
such a line may represent the first tooth of a comb. The double-base can be cleared
with 8 or 12 moves (see Fig. 9), the root can be cleared with 3 moves (see Fig. 10,
left) instead of 9 moves (see Fig. 5, right); the rest of the clearing does not change.
Consider now a horizontal line extending to the right of the base (see Fig. 10,
right); such a line may appear as the first tooth in a double-sided comb. The double-
base for such a line can be cleared with 3 moves (see Fig. 10, right); the rest of the
clearing is the same as for the vertical line.
L-shaped Line. An L-shaped line L consists of a vertical and a horizontal segment.
Each of the segments can be cleared as described above.
Thus, any line L|J can be cleared at a cost of at most c(L|J) ≤ 12 +
∑k′
i=1
(
J − 1 + (i− 1)
⌊
D
2
⌋
+ 5
⌊
D
2
⌋
+ 7(J + i
⌊
D
2
⌋
− 1)
)
+ J + 5r′ + 7(ℓ− 1).
careful look at the Figs. 5, 6 reveals that: (1) in the push emulation the first two moves
are the opposites of the last two, thus, all 4 moves may be omitted – consequently,
a push may be emulated by a sequence of only 7 moves; (2) if the boundary side,
following e, is vertical, the last push, throwing the snow away from P , may require 9
moves (Fig. 5, right); and, (3) when emulating the last back throw in a forward pass,
the last 2 of the 5 moves (the move up and the move to the right in Fig. 5, left) can
be omitted – indeed, during the push emulation, the snowblower may as well start
to the right of the snow (see Fig. 6). Thus, a line L|J can be cleared at a cost of
c(L|J) ≤ 4 +
∑k′
i=1
(
J − 1 + (i− 1)
⌊
D
2
⌋
+ 5
⌊
D
2
⌋
+ 7(J + i
⌊
D
2
⌋
− 1)
)
+J +
5r′ + 7(ℓ− 1).
s
s
→66 2
e
ff? s
6
Fig. 5: Left: emulating back throw. Right: pushing the 2 ⌊D/2⌋ units of snow away from P
and returning the snowblower to p may require 9 moves.
2
2
2
2
2
2 2 2
→ → → →
s s s
s
s2 2
?
2
?
64 4
Fig. 6: Emulating pushing the snow in front of the snowblower.
A Line L with GL||e. Consider a horizontal line, extending to the left of the base;
such a line may represent the first tooth of a comb. The double-base can be cleared
with 8 or 12 moves (see Fig. 9), the root can be cleared with 3 moves (see Fig. 10,
left) instead of 9 moves (see Fig. 5, right); the rest of the clearing does not change.
Consider now a horizontal line extending to the right of the base (see Fig. 10,
right); such a line may appear as the first tooth in a double-sided comb. The double-
base for such a line can be cleared with 3 moves (see Fig. 10, right); the rest of the
clearing is the same as for the vertical line.
L-shaped Line. An L-shaped line L consists of a vertical and a horizontal segment.
Each of the segments can be cleared as described above.
Thus, any line L|J can be cleared at a cost of at most c(L|J) ≤ 12 +
∑k′
i=1
(
J − 1 + (i− 1)
⌊
D
2
⌋
+ 5
⌊
D
2
⌋
+ 7(J + i
⌊
D
2
⌋
− 1)
)
+ J + 5r′ + 7(ℓ− 1).
Page 14
14 Esther M. Arkin, Michael A. Bender, Joseph S. B. Mitchell, Valentin Polishchuk
Since the snow and distance (3) lower bounds do not change, the lemma follows.
⊓⊔
Lemma 8. A comb can be cleared at a cost of 34s(C \ p) + 24D/⌊D/2⌋d(C \ p).
Proof. Brush in the fixed throw direction model can be described easily using anal-
ogy with: a) brush in default and adjustable-throw models and b) line-clearing in
fixed-throw model. As in the adjustable-throw model, we prepare to clear ⌊D/2⌋
pixels during each brush. Same as with line-clearing, we setup the double-base for
the comb with at most 12 moves; also, 9 moves per brush may be needed to push
the snow away from P through the base. Back throw and push can be emulated with
5 and 7 moves (Fig. 5, left and Fig. 6). Thus, if the cost of a brush (1) in the de-
fault model was, say, c, the cost of the brush in the fixed-throw model is at most
7c+9. Since any brush starts with the double-base setup, c ≥ 6; this, in turn, implies
7c+9 ≤ (51/6)c. Hence, the cost of clearingB increases by at most a factor of 51/6.
By Lemma 7, the cost of clearing P , c(P) ≤ 24D/⌊D/2⌋d(P). The snow and
distance lower bounds do not change in comparison with the adjustable-throw case.
The lemma now follows from simple arithmetic. ⊓⊔
As in the default and adjustable-throw models (Theorems 2,3),
Theorem 4. A (34+24D/⌊D/2⌋)-approximate tour can be found in polynomial time.
5 Extensions
Polygons with Holes. Our methods extend to the case in which P is a polygonal
domain with holes. There are two natural ways that holes may arise in the model.
In the first, the holes represent obstacles for the snowblower (e.g., walls of build-
ings that border the driveway). No snow can be thrown onto such holes; the holes’
boundaries serve as walls for the motion of the snowblower and for the deposition of
snow. Our algorithm for the default model extends immediately to this variation. The
SBP in restricted-throw models, however, may become infeasible; it is not always
possible for the snowblower to enter a thin (one-pixel-width) “channel” between two
walls.
In the second variation, the holes’ boundaries are assumed to be the same “cliffs”
as the polygon’s outer boundary. It is in fact this version of the problem that we
proved to be NP-complete. With some modifications our algorithms work for this
variation as well; we sketch here the necessary modifications. The boundary, ∂P ,
of the domain now consists of the boundary pixel sides both on the outer boundary
of the polygon, denoted ∂oP , and on the boundaries of the holes. As in the simple
polygon case, we build the Voronoi decomposition of the domain and prepare to
clear it Voronoi-cell-by-Voronoi-cell. The only problem with it now is that there is no
readily available Hamiltonian cycle through the elements of ∂P . To build a suitable
cycle we find a spanning tree, T, in the graph having a node for each hole and a node
for ∂oP and having an edge between two holes H1 and H2 if there are boundary
Since the snow and distance (3) lower bounds do not change, the lemma follows.
⊓⊔
Lemma 8. A comb can be cleared at a cost of 34s(C \ p) + 24D/⌊D/2⌋d(C \ p).
Proof. Brush in the fixed throw direction model can be described easily using anal-
ogy with: a) brush in default and adjustable-throw models and b) line-clearing in
fixed-throw model. As in the adjustable-throw model, we prepare to clear ⌊D/2⌋
pixels during each brush. Same as with line-clearing, we setup the double-base for
the comb with at most 12 moves; also, 9 moves per brush may be needed to push
the snow away from P through the base. Back throw and push can be emulated with
5 and 7 moves (Fig. 5, left and Fig. 6). Thus, if the cost of a brush (1) in the de-
fault model was, say, c, the cost of the brush in the fixed-throw model is at most
7c+9. Since any brush starts with the double-base setup, c ≥ 6; this, in turn, implies
7c+9 ≤ (51/6)c. Hence, the cost of clearingB increases by at most a factor of 51/6.
By Lemma 7, the cost of clearing P , c(P) ≤ 24D/⌊D/2⌋d(P). The snow and
distance lower bounds do not change in comparison with the adjustable-throw case.
The lemma now follows from simple arithmetic. ⊓⊔
As in the default and adjustable-throw models (Theorems 2,3),
Theorem 4. A (34+24D/⌊D/2⌋)-approximate tour can be found in polynomial time.
5 Extensions
Polygons with Holes. Our methods extend to the case in which P is a polygonal
domain with holes. There are two natural ways that holes may arise in the model.
In the first, the holes represent obstacles for the snowblower (e.g., walls of build-
ings that border the driveway). No snow can be thrown onto such holes; the holes’
boundaries serve as walls for the motion of the snowblower and for the deposition of
snow. Our algorithm for the default model extends immediately to this variation. The
SBP in restricted-throw models, however, may become infeasible; it is not always
possible for the snowblower to enter a thin (one-pixel-width) “channel” between two
walls.
In the second variation, the holes’ boundaries are assumed to be the same “cliffs”
as the polygon’s outer boundary. It is in fact this version of the problem that we
proved to be NP-complete. With some modifications our algorithms work for this
variation as well; we sketch here the necessary modifications. The boundary, ∂P ,
of the domain now consists of the boundary pixel sides both on the outer boundary
of the polygon, denoted ∂oP , and on the boundaries of the holes. As in the simple
polygon case, we build the Voronoi decomposition of the domain and prepare to
clear it Voronoi-cell-by-Voronoi-cell. The only problem with it now is that there is no
readily available Hamiltonian cycle through the elements of ∂P . To build a suitable
cycle we find a spanning tree, T, in the graph having a node for each hole and a node
for ∂oP and having an edge between two holes H1 and H2 if there are boundary
Page 15
The Snowblower Problem 15
pixel sides, e1 ∈ ∂H1 and e2 ∈ ∂H2, such that there exist adjacent pixels p1 ∈
Voronoi(e1) and p2 ∈ Voronoi(e2). Then, after Voronoi(e1) is cleared, we direct the
snowblower to start clearing Voronoi(e2).5 After that, the snowblower continues to
clear the Voronoi cells of the boundary pixel sides of H2 until another edge of T,
connecting H2 to a hole H3, is encountered (if H2 is a leaf of T, H3 = H1), and so
on. The cost of clearing the Voronoi cells of the sides in ∂P does not change, and
the cost of traversing T is at most twice the size of P ; thus, the proposed algorithm
remains a constant-factor approximation algorithm.
Nonrectilinear Polygonal Domains. Our discussion so far has assumed that P is
integral-orthogonal; in this case, the snowblower makes axis-parallel movements. If
P is rectilinear, but not necessarily integral, we proceed as in [3]: first, the bound-
ary of P is traversed once, and then our algorithms are applied to the remaining
part, P ′, of the domain P . Every time the snow is thrown away from P ′, a certain
length (which depends on the throw model) may need to be added to the cost of the
tour; thus, the approximation factors of our algorithms may increase by an additive
constant.
We can also extend our methods to general nonrectilinear domains. Since the
snowblower is not allowed to move outside the domain, care must be taken about
specifying which portion of the domain is actually clearable. This portion can be
found by traversing the boundary of the domain; then, the accessible portion can be
cleared as described above.
Implicit Representation of the Tour. As in [3, 4], we make the distinction be-
tween explicit and implicit representations of the domains and snowblower tours. As
mentioned in the introduction, we assume that a domain is given as the union of pix-
els; this way the size of the input to the problem is O(N), where N is the number
of pixels in the domain. The size of the description of the snowblower tour produced
by our algorithm is polynomial in N , i.e., polynomial in the input size. Instead, the
domain may be given in polygonal representation, as a list of coordinates of its n
vertices; the size of such a representation is O(n logW ), where W is the largest
coordinate in the input. In principle, N may be Ω(W ), and hence the length of the
description of the snowblower tour may appear to be exponential in the size of the
input. To avoid such a blow-up in the size of the tour, it is possible to represent the
tour succinctly so that the size of the representation is polynomial in n logW .
Indeed, our algorithms produce tours, comprised of line-clearings and brushes.
Our Voronoi decomposition of the domain is the discretized version of the Voronoi
diagram of the edges of the domain; the latter is O(n) in size and can be found
efficiently ( [5]). Given the diagram, it is easy to constrain the (axis-parallel) motion
of the snowblower to stay within a Voronoi cell of an edge: when “in doubt”, i.e.,
when the snowblower is about to enter a pixel, intersected by an edge of the Voronoi
diagram, it can be decided “in place”, in constant time (based on the tie breaking
rules), whether entering the pixel will place the snowblower into the Voronoi cell of
another side of P .
5 The clearing starts from creating a path of width 2 from p2 to e2; the path “bridges” the
holes H1 and H2.
pixel sides, e1 ∈ ∂H1 and e2 ∈ ∂H2, such that there exist adjacent pixels p1 ∈
Voronoi(e1) and p2 ∈ Voronoi(e2). Then, after Voronoi(e1) is cleared, we direct the
snowblower to start clearing Voronoi(e2).5 After that, the snowblower continues to
clear the Voronoi cells of the boundary pixel sides of H2 until another edge of T,
connecting H2 to a hole H3, is encountered (if H2 is a leaf of T, H3 = H1), and so
on. The cost of clearing the Voronoi cells of the sides in ∂P does not change, and
the cost of traversing T is at most twice the size of P ; thus, the proposed algorithm
remains a constant-factor approximation algorithm.
Nonrectilinear Polygonal Domains. Our discussion so far has assumed that P is
integral-orthogonal; in this case, the snowblower makes axis-parallel movements. If
P is rectilinear, but not necessarily integral, we proceed as in [3]: first, the bound-
ary of P is traversed once, and then our algorithms are applied to the remaining
part, P ′, of the domain P . Every time the snow is thrown away from P ′, a certain
length (which depends on the throw model) may need to be added to the cost of the
tour; thus, the approximation factors of our algorithms may increase by an additive
constant.
We can also extend our methods to general nonrectilinear domains. Since the
snowblower is not allowed to move outside the domain, care must be taken about
specifying which portion of the domain is actually clearable. This portion can be
found by traversing the boundary of the domain; then, the accessible portion can be
cleared as described above.
Implicit Representation of the Tour. As in [3, 4], we make the distinction be-
tween explicit and implicit representations of the domains and snowblower tours. As
mentioned in the introduction, we assume that a domain is given as the union of pix-
els; this way the size of the input to the problem is O(N), where N is the number
of pixels in the domain. The size of the description of the snowblower tour produced
by our algorithm is polynomial in N , i.e., polynomial in the input size. Instead, the
domain may be given in polygonal representation, as a list of coordinates of its n
vertices; the size of such a representation is O(n logW ), where W is the largest
coordinate in the input. In principle, N may be Ω(W ), and hence the length of the
description of the snowblower tour may appear to be exponential in the size of the
input. To avoid such a blow-up in the size of the tour, it is possible to represent the
tour succinctly so that the size of the representation is polynomial in n logW .
Indeed, our algorithms produce tours, comprised of line-clearings and brushes.
Our Voronoi decomposition of the domain is the discretized version of the Voronoi
diagram of the edges of the domain; the latter is O(n) in size and can be found
efficiently ( [5]). Given the diagram, it is easy to constrain the (axis-parallel) motion
of the snowblower to stay within a Voronoi cell of an edge: when “in doubt”, i.e.,
when the snowblower is about to enter a pixel, intersected by an edge of the Voronoi
diagram, it can be decided “in place”, in constant time (based on the tie breaking
rules), whether entering the pixel will place the snowblower into the Voronoi cell of
another side of P .
5 The clearing starts from creating a path of width 2 from p2 to e2; the path “bridges” the
holes H1 and H2.
Page 16
16 Esther M. Arkin, Michael A. Bender, Joseph S. B. Mitchell, Valentin Polishchuk
The clearing of a Voronoi face in the Voronoi diagram is done tree-by-tree: first
a (double-)comb is cleared (if present), then a set of lines (if present), then the other
comb (if present). Thus, finding a short representation of a tour boils down to exhibit-
ing succinct representations of the tours through a comb and through a set of lines
with adjacent bases comprising a boundary edge of P . The descriptions of these tours
given in Sections 3 and 4 provide such representations.
The Vacuum Cleaner Problem. Consider the following problem. The floor – a
polygonal domain, possibly with holes – is covered with dust and debris. The house
is equipped with a central vacuum system, and certain places on the boundary of
the floor (the baseboard) are connected to the “dustpan vac” – a dust dump location
of infinite capacity [1]. The robotic vacuum cleaner has a dust/debris capacity D
and must be emptied to a dump location whenever full. The described problem is
equivalent to the SBP in default throw model, and actually provided the motivation
to study the SBP with throwing backwards allowed.
Nonuniform Depth of Snow. Our algorithms generalize straightforwardly to the
case in which some pixels of the domain initially contain more than one unit of
snow. For a problem instance to be feasible it is required that there is less than D
(less than ⌊D/2⌋ in restricted-throw direction models) units of snow on each pixel.
The approximation ratios in this case depend (linearly) on D (or, in general, on the
ratio of D to the minimum initial depth of snow on P ).
Capacitated Disposal Region. Another generalization of the problem is to con-
sider the dump locations to have finite capacities. If instead of “cliffs” at the bound-
ary of P , there is a finite capacity (maximum depth) associated with each point in
the complement of P , the SBP more accurately models some material handling prob-
lems, but also becomes considerably more difficult. The snow lower bound still ap-
plies, the distance lower bound transforms to a lower bound based on a minimum-
cost matching between the pixels in P and the pixels in the complement of P . This
problem represents a computational problem related to “earth-mover distance” [7]
and is beyond the scope of this paper.
Open Problems. The complexity of the SBP in simple polygons and the complex-
ity of the SBP in the fixed-throw model are open. We also do not have an algorithm
for the case of holes-obstacles and restricted throw direction; the hardness of this
version is also open.
References
1. http://www.centralvacuumstores.com/vacpan.htm.
2. http://www.plowsunlimited.com/snow melters.htm.
3. E. M. Arkin, S. P. Fekete, and J. S. B. Mitchell. Approximation algorithms for lawn
mowing and milling. Comput. Geom. Theory Appl., 17:25–50, 2000.
4. E. M. Arkin, M. Held, and C. L. Smith. Optimization problems related to zigzag pocket
machining. Algorithmica, 26(2):197–236, 2000.
5. F. Aurenhammer. Voronoi diagrams: A survey of a fundamental geometric data structure.
ACM Comput. Surv., 23(3):345–405, Sept. 1991.
The clearing of a Voronoi face in the Voronoi diagram is done tree-by-tree: first
a (double-)comb is cleared (if present), then a set of lines (if present), then the other
comb (if present). Thus, finding a short representation of a tour boils down to exhibit-
ing succinct representations of the tours through a comb and through a set of lines
with adjacent bases comprising a boundary edge of P . The descriptions of these tours
given in Sections 3 and 4 provide such representations.
The Vacuum Cleaner Problem. Consider the following problem. The floor – a
polygonal domain, possibly with holes – is covered with dust and debris. The house
is equipped with a central vacuum system, and certain places on the boundary of
the floor (the baseboard) are connected to the “dustpan vac” – a dust dump location
of infinite capacity [1]. The robotic vacuum cleaner has a dust/debris capacity D
and must be emptied to a dump location whenever full. The described problem is
equivalent to the SBP in default throw model, and actually provided the motivation
to study the SBP with throwing backwards allowed.
Nonuniform Depth of Snow. Our algorithms generalize straightforwardly to the
case in which some pixels of the domain initially contain more than one unit of
snow. For a problem instance to be feasible it is required that there is less than D
(less than ⌊D/2⌋ in restricted-throw direction models) units of snow on each pixel.
The approximation ratios in this case depend (linearly) on D (or, in general, on the
ratio of D to the minimum initial depth of snow on P ).
Capacitated Disposal Region. Another generalization of the problem is to con-
sider the dump locations to have finite capacities. If instead of “cliffs” at the bound-
ary of P , there is a finite capacity (maximum depth) associated with each point in
the complement of P , the SBP more accurately models some material handling prob-
lems, but also becomes considerably more difficult. The snow lower bound still ap-
plies, the distance lower bound transforms to a lower bound based on a minimum-
cost matching between the pixels in P and the pixels in the complement of P . This
problem represents a computational problem related to “earth-mover distance” [7]
and is beyond the scope of this paper.
Open Problems. The complexity of the SBP in simple polygons and the complex-
ity of the SBP in the fixed-throw model are open. We also do not have an algorithm
for the case of holes-obstacles and restricted throw direction; the hardness of this
version is also open.
References
1. http://www.centralvacuumstores.com/vacpan.htm.
2. http://www.plowsunlimited.com/snow melters.htm.
3. E. M. Arkin, S. P. Fekete, and J. S. B. Mitchell. Approximation algorithms for lawn
mowing and milling. Comput. Geom. Theory Appl., 17:25–50, 2000.
4. E. M. Arkin, M. Held, and C. L. Smith. Optimization problems related to zigzag pocket
machining. Algorithmica, 26(2):197–236, 2000.
5. F. Aurenhammer. Voronoi diagrams: A survey of a fundamental geometric data structure.
ACM Comput. Surv., 23(3):345–405, Sept. 1991.
Page 17
The Snowblower Problem 17
6. F. Chin, J. Snoeyink, and C. A. Wang. Finding the medial axis of a simple polygon in
linear time. Discrete Comput. Geom., 21(3):405–420, 1999.
7. S. Cohen and L. Guibas. The earth mover’s distance: Lower bounds and invariance under
translation. Technical Report CS-TR-97-1597, 1997.
8. J. Culberson. Sokoban is PSPACE-complete. In N. S. E. Lodi, L. Pagli, editor, Proc.
Internat. Conf. Fun with Algorithms, pages 65–76, Elba, Italy, June 1998. Carelton Sci-
entific.
9. E. D. Demaine, M. L. Demaine, M. Hoffmann, and J. O’Rourke. Pushing blocks is hard.
Comput. Geom. Theory Appl., 26(1).
10. M. Held. On the Computational Geometry of Pocket Machining, volume 500 of Lecture
Notes Comput. Sci. Springer-Verlag, June 1991.
11. A. Itai, C. H. Papadimitriou, and J. L. Szwarcfiter. Hamilton paths in grid graphs. SIAM
J. Comput., 11:676–686, 1982.
12. C. H. Papadimitriou and U. V. Vazirani. On two geometric problems related to the travel-
ing salesman problem. J. Algorithms, 5:231–246, 1984.
13. V. Polishchuk. The box mover problem. In Proceedings of the 16th Canadian Conference
on Computational Geometry (CCCG’04), pages 36–39, 2004.
6. F. Chin, J. Snoeyink, and C. A. Wang. Finding the medial axis of a simple polygon in
linear time. Discrete Comput. Geom., 21(3):405–420, 1999.
7. S. Cohen and L. Guibas. The earth mover’s distance: Lower bounds and invariance under
translation. Technical Report CS-TR-97-1597, 1997.
8. J. Culberson. Sokoban is PSPACE-complete. In N. S. E. Lodi, L. Pagli, editor, Proc.
Internat. Conf. Fun with Algorithms, pages 65–76, Elba, Italy, June 1998. Carelton Sci-
entific.
9. E. D. Demaine, M. L. Demaine, M. Hoffmann, and J. O’Rourke. Pushing blocks is hard.
Comput. Geom. Theory Appl., 26(1).
10. M. Held. On the Computational Geometry of Pocket Machining, volume 500 of Lecture
Notes Comput. Sci. Springer-Verlag, June 1991.
11. A. Itai, C. H. Papadimitriou, and J. L. Szwarcfiter. Hamilton paths in grid graphs. SIAM
J. Comput., 11:676–686, 1982.
12. C. H. Papadimitriou and U. V. Vazirani. On two geometric problems related to the travel-
ing salesman problem. J. Algorithms, 5:231–246, 1984.
13. V. Polishchuk. The box mover problem. In Proceedings of the 16th Canadian Conference
on Computational Geometry (CCCG’04), pages 36–39, 2004.
Page 18
18 Esther M. Arkin, Michael A. Bender, Joseph S. B. Mitchell, Valentin Polishchuk
Appendix: Figures
Tie
Hor. edge
Vert. edge
down
up
left
right
left
right
left
right
up
down
up
down
@
,,
,
XXX
HHH
p
e
Fig. 7: Left: the tie breaking tree; on a tie, the upper edge of the tree is followed. Right: an
example; by the tie breaking rules, V(p) = e, the bold edge; δ(p) = 4.
E
p
a
b
c
Fig. 8: Part of the comb of Voronoi(E). V e(p) ⊆ E. V e(a), V e(b), V e(c) * E, so the
snow form a must be thrown onto Voronoi(E′) 6= Voronoi(E). Nevertheless, observe that
δ(a) = δ(p) − 1 – one could try to make use of it.
Appendix: Figures
Tie
Hor. edge
Vert. edge
down
up
left
right
left
right
left
right
up
down
up
down
@
,,
,
XXX
HHH
p
e
Fig. 7: Left: the tie breaking tree; on a tie, the upper edge of the tree is followed. Right: an
example; by the tie breaking rules, V(p) = e, the bold edge; δ(p) = 4.
E
p
a
b
c
Fig. 8: Part of the comb of Voronoi(E). V e(p) ⊆ E. V e(a), V e(b), V e(c) * E, so the
snow form a must be thrown onto Voronoi(E′) 6= Voronoi(E). Nevertheless, observe that
δ(a) = δ(p) − 1 – one could try to make use of it.
Page 19
The Snowblower Problem 19
s
s s
→ →
-- ffff
s s
→ →
-- ffff
s
Fig. 9: Setting up the double-base for clearing a horizontal line extending to the left of its
base. Depending on the direction of the edge adjacent to the base from the right, there are 8
(above) or 12 (below) moves necessary.
s
×
s
s
Fig. 10: Left: clearing the root of a horizontal, extending to the left, line with 3 moves. There
is 2 ⌊D/2⌋ units of snow on the checked pixel. Right: Setting up the double-base for clearing
a horizontal line extending to the right of its base.
s
s s
→ →
-- ffff
s s
→ →
-- ffff
s
Fig. 9: Setting up the double-base for clearing a horizontal line extending to the left of its
base. Depending on the direction of the edge adjacent to the base from the right, there are 8
(above) or 12 (below) moves necessary.
s
×
s
s
Fig. 10: Left: clearing the root of a horizontal, extending to the left, line with 3 moves. There
is 2 ⌊D/2⌋ units of snow on the checked pixel. Right: Setting up the double-base for clearing
a horizontal line extending to the right of its base.
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