Visibility Graph
Team 10
NakWon Lee, Dongwoo Kim
Robot Motion Planning
• Consider the case of point robot
• The polygons in S are obstacles, and their total
number of edges is denoted by n
• The point robot can touch obstacles, because
obstacles are open set.
• Start position 𝑝𝑠𝑡𝑎𝑟𝑡 . Goal position 𝑝𝑔𝑜𝑎𝑙 .
• Paths do not intersect the interior of any of the
obstacles.
Robot Motion Planning
• Use trapezoidal map 𝑇(𝐶𝑓𝑟𝑒𝑒 )of the free
configuration space 𝐶𝑓𝑟𝑒𝑒 .
• Configuration space is the parameter space of a
robot 𝑅. Denoted by 𝐶(𝑅).
• For a point robot, 𝐶𝑓𝑟𝑒𝑒 𝑅 was simply the empty
space between obstacles.
Configuration Space
Shortest Paths
• Construct the trapezoidal map 𝑇(𝐶𝑓𝑟𝑒𝑒 ) for
configuration space 𝐶𝑓𝑟𝑒𝑒 .
• Construct the load map 𝐺𝑟𝑜𝑎𝑑 .
• Find the trapezoids which have the start point or
end point.
• Find the shortest route in the load map 𝐺𝑟𝑜𝑎𝑑 using
breadth first search algorithm.
Shortest Paths
Shortest Paths
• Shortest path in road map is not a real shortest path
• because some arcs are btw nodes that are far apart, whereas
others are btw nodes that are close to each other.
• It is just minimum number of hop.
• To improve this problem, give each arc a weight
corresponding to the Euclidean length, and use graph
search algorithm that find the shortest path in a
weighted graph.
• Dijkstra's algorithm
• But, it is still not a shortest path.
Shortest Paths
• Think of this path as an elastic
rubber band.
• Fix the endpoints at the start and
goal position.
• Try to tighten the rubber band.
• It will be stopped by the obstacles.
• The new path will follow parts of
the obstacle boundaries and
straight line segments through
open space.
Shortest Paths
• Lemma 15.1 Any shortest path between 𝑝𝑠𝑡𝑎𝑟𝑡 and
𝑝𝑔𝑜𝑎𝑙 among a set 𝑆 of disjoint polygonal obstacles
is a polygonal path whose inner vertices are
vertices of 𝑆
• Inner vertex – A vertex that is not the begin or endpoint
of the path.
Shortest Paths
• Proof.
• Suppose for a contradiction that a shortest path 𝜏 is not
polygonal.
• Point 𝑝 on 𝜏 that lies in the interior of the free space
with the property that no line segment containing 𝑝 is
contained in 𝜏.
𝑝
𝜏
𝑣1
𝑛𝑜𝑡 𝑝
𝑣2
Shortest Paths
• There is a disc of positive radius centered
at 𝑝 that is completely contained in the
free space.
• Part of 𝜏 inside the disk, which is not a
straight line segment, can be shortened
by replacing it with the segment
connecting the points which lie on circle.
• Contradict the theorem “any shortest
path must be locally shortest”.
Shortest Paths
• Now consider a vertex 𝑣 on 𝜏. It cannot lies in the interior of the free
space.
• Consider the disc centered at 𝑣 such that half of the disc is contained in
the free space.
• Can replace the sub path inside the disc with a straight line segment.
• The only possibility left is that 𝑣 is an obstacle vertex.
𝑣
𝑣
short cut
𝜏
𝑣1
𝜏
𝑣2
𝑣1
𝑣2
Shortest Paths
• Construct a road map with this characterization.
• This road map is visibility graph of 𝑆, denoted by
𝐺𝑣𝑖𝑠 (𝑆).
• Its nodes are the vertices of 𝑆.
• There is an edge between vertices 𝑣 and 𝑤 if they
can see each other.
• The segment 𝑣𝑤 dose not intersect the interior of
any obstacle in 𝑆.
Shortest Paths
• Two vertices that can see each other are called
visible.
• The segment connecting visible two vertices is
called a visibility edge.
• Endpoint of the same obstacle edge always see
each other.
• Hence, the obstacle edges form a subset of the
edges of 𝐺𝑣𝑖𝑠 (𝑆).
Shortest Paths
• To make a complete road map to find shortest path,
add start and goal point as vertices to 𝑆.
• So we consider the visibility graph of the set
𝑆 ∗ ≔ 𝑆 ∪ {𝑝𝑠𝑡𝑎𝑟𝑡 , 𝑝𝑔𝑜𝑎𝑙 }.
• Corollary 15.2 The shortest path between 𝑝𝑠𝑡𝑎𝑟𝑡
and 𝑝𝑔𝑜𝑎𝑙 among a set 𝑆 of disjoint polygonal
obstacles consists of arcs of the visibility graph
𝐺𝑣𝑖𝑠 (𝑆 ∗ ), where 𝑆 ∗ ≔ 𝑆 ∪ {𝑝𝑠𝑡𝑎𝑟𝑡 , 𝑝𝑔𝑜𝑎𝑙 }.
Shortest Paths
Visibility Graph
Definition:
• The visibility graph of s and t and the obstacle set
is a graph whose vertices are s and t the obstacle
vertices, and vertices v and w are joined by an edge
if v and w are either mutually visible or if (v, w) is
an edge of some obstacle.
Visibility Graph
Visible Vertices
Visible Vertices
Visible
Definition
Two points p and q are mutually visible if the open
line segment joining them doesn't intersect the
interior of any obstacle.
Search Tree
Computing the Visibility Graph
• VisibleVertices: 𝑂(𝑛𝑙𝑜𝑔𝑛) – (sorting)
• Operation on blanced search tree : 𝑂(𝑙𝑜𝑔𝑛)
• For every vertices with Visible Vertices.
• VisibilityGraph – 𝑂(𝑛2 𝑙𝑜𝑔𝑛)
Computing the Visibility Graph
• Give an 𝑂(𝑛2 ) procedure for constructing.
• 𝑛 is the number of line segments in the plane.
• No three vertices are collinear??
• The algorithm is not output sensitive.
• An 𝑂(𝑛 log 𝑛 + 𝑘) algorithm (𝑘 is the number of
edges in visibility graph) is existent but quite
complicated.
Computing the Visibility Graph
• The text’s algorithm operates by doing sweep one
vertex at a time.
• New algorithm does the sweep for all vertices
simultaneously.
• When we use the arrangement, angular sort can be
performed for all vertices in 𝑂(𝑛2 ) time.
• If we build the entire arrangement, this sorting
algorithm will involve 𝑂(𝑛2 ) space.
• However it can be implemented in 𝑂(𝑛) space
using an algorithm called topological plane sweep.
Computing the Visibility Graph
• First, recall the algorithm for computing trapezoidal
maps.
• Shoot a bullet up and down from every vertex until
it hits its first line segment.
• Angle 𝜃 continuously sweeps out all slopes from
− ∞ to +∞.
• All bullet lines attached to all the vertices begin to
turn slowly counterclockwise.
Computing the Visibility Graph
• The question is what are the significant event
points, and what happens with each event?
Computing the Visibility Graph
• It is useful to view the problem both in its primal
and dual form.
• 2𝑛 (𝑛 is the number of segments) segment end
points 𝑣 = (𝑣𝑎 , 𝑣𝑏 ), its dual line 𝑣 ∗ : 𝑦 = 𝑣𝑎 𝑥 − 𝑣𝑏 .
• Significant event occurs whenever a bullet path in
the primal plane jumps from one line segment to
another.
• This occurs when 𝜃 reaches the slope of the line
joining two visible endpoints 𝑣 and 𝑤.
Computing the Visibility Graph
event
event
event
Computing the Visibility Graph
• To keep track of which endpoints are visible and
which are not is complicated.
• Instead we will take events to be all angles 𝜃
between two endpoints, whether they are visible
or not.
• By duality, the slope of such an event will
correspond to the 𝑎-coordinate of the intersection
of dual lines 𝑣 ∗ and 𝑤 ∗ in the dual arrangement.
Dual Arrangement
x-coordinate is the slope of dual segment
Dual of point D
Dual of point C
Find Events
• By sweeping the arrangement of the 2𝑛 dual lines
from left-to-right, we will enumerate all the slope
events in angular order.
Find Events
Angular order
• By using topological plane sweep,
event do not need to be sorted.
Topological Plane Sweep
• Provides a way to sweep an
arrangement of lines using
a “flexible” sweeping line.
• Because events do not
need to be sorted, we can
avoid 𝑂(log 𝑛) factor.
Topological Plane Sweep
• Upper horizon tree
• Lower horizon tree
Topological Plane Sweep
• Data structure
• 𝐸[1: 𝑛] – array of line equation. 𝐸 𝑖 = (𝑎𝑖 , 𝑏𝑖 ), if the 𝑖th line
of 𝐻, 𝑙𝑖 , is 𝑦 = 𝑎𝑖 𝑥 + 𝑏𝑖 .
• 𝐻𝑇𝑈[1: 𝑛] – an array representing the upper horizon tree.
𝐻𝑇𝑈[𝑖] is a pair (𝜆𝑖 , 𝜌𝑖 ) of indices indicating the lines that
delimit the segment of 𝑙𝑖 in the upper horizon tree to the left
and to the right.
• 𝐻𝑇𝐿 1: 𝑛 – represents the lower horizon tree.
• 𝐼 – a set of integers, represented as a stack. If 𝑖 is in 𝐼, then 𝑐𝑖
and 𝑐𝑖+1 share a common right endpoint.
• 𝑀[1: 𝑛] – an array holding the current sequence of indices
that form the lines 𝑚1 , 𝑚2 , ⋯ , 𝑚𝑛 of the cut.
• 𝑁 1: 𝑛 – a list of pairs of indices indicating the lines
delimiting each edge of the cut. 𝑁[𝑖] thus encodes the
endpoints of the edge on 𝑀[𝑖]. The same convention as that
above is used for missing endpoints.
Topological Plane Sweep
If segment is the rightmost on 𝑙𝑖 ,
set 𝜌𝑖 = 0
If segment is the leftmost on 𝑙𝑖 ,
set 𝜆𝑖 = −1
Topological Plane Sweep
• Algorithm
Initialization:
• 1. Sort the lines of the arrangement by slope.
• 2. Find the leftmost and the rightmost intersection point of the lines. Let the
two points be (𝑥𝑙 , 𝑦𝑙 ) and (𝑥𝑟 , 𝑦𝑟 ).
• 3. Create vertical lines 𝑥 = 𝑥𝑙 − 1, 𝑥 = 𝑥𝑟 + 1 as left boundary and right
boundary. Determine the intersection points of lines 𝑙1 , ⋯ , 𝑙𝑛 with the
boundaries.
• 4. Create upper horizon tree: Insert 𝑙1 , ⋯ , 𝑙𝑛 . Assume 𝑙1 , ⋯ , 𝑙𝑘 have been
inserted. These lines form an upper bay. To insert 𝑙𝑘+1 , begin at its endpoint on
the left boundary. Walk in counter clockwise order around the bay till we find
the intersection point of 𝑙𝑘+1 with an edge.
• 5. Create lower horizon tree similarly by starting the travers at endpoints on the
right boundary.
• 6. Initialize 𝑁: Let 𝐻𝑇𝑈[𝑖] = (−1, 𝑟) and 𝐻𝑇𝐿[𝑖] = (−1, 𝑠). If 𝑙𝑟 intersects 𝑙𝑖 to
the left of the intersection point of 𝑙𝑠 and 𝑙𝑖 then the right delimiting line of 𝑒𝑖 is
𝑟. Otherwise, the right delimiting line of 𝑒𝑖 is 𝑠.
• 7. Initialize 𝐼 by scanning 𝑁.
Topological Plane Sweep
• Algorithm
Elementary Step
• While 𝐼 = Λ
• 1. Pop 𝑖 from 𝐼
• 2. Swap 𝑀[𝑖], 𝑀[𝑖 + 1] /*lines are going to cross, after the
elementary step*/
• 3. 𝑁 𝑖 . 𝜆 ← 𝑁 𝑖 + 1 . 𝜌, 𝑁 𝑖 + 1 . 𝜆 ← 𝑁[𝑖]. 𝜌 /*the point of
elementary step becomes the left endpoint of the two new cut
edges */
• 4. Update 𝐻𝑇𝑈, 𝐻𝑇𝐿.
• 5. 𝑁 𝑖 . 𝜌 ← 𝑐𝑙𝑜𝑠𝑒𝑟 𝐻𝑇𝑈 𝑀 𝑖 . 𝜌, 𝐻𝑇𝐿 𝑀 𝑖 . 𝜌, 𝑁 𝑖 + 1 . 𝜌 ←
𝑐𝑙𝑜𝑠𝑒𝑟 𝐻𝑇𝑈 𝑀 𝑖 + 1 . 𝜌, 𝐻𝑇𝐿 𝑀 𝑖 + 1 . 𝜌 /* find the new right
endpoints */
• 6. If 𝑁[𝑖 + 1]. 𝜌 = 𝑀[𝑖 + 2] then push 𝑖 + 1 into 𝐼. If 𝑁[𝑖]. 𝜌 =
𝑀[𝑖 − 1] then push 𝑖 − 1 into 𝐼. /* push valid vertices formed after
sweeping into I if there is any */
What Happens at Each Event
• For each vertex 𝑣, there are two bullet paths
growing from 𝑣 along the line with slope 𝜃.
• Let 𝑓(𝑣) and 𝑏(𝑣) denote the line segments that
forward bullet path and backward bullet path hit.
• If either path does not hit any segment then we
store a special null value.
• Each slope is determined by exactly two lines.
What Happens at Each Event
• Possible scenarios
What Happens at Each Event
• Same segment
• If 𝑣 and 𝑤 are endpoints of the same segment, then
they are visible, and we add the edge (𝑣, 𝑤) to the
visibility graph.
• How can I check this?
What Happens at Each Event
• Invisible
• Consider the distance
from 𝑣 to 𝑤.
• Compute the contact
point of the bullet path
shot from 𝑣 in direction
𝜃 with segment 𝑓(𝑣) or
𝑏(𝑣).
• If 𝑣𝑤 is longer than 𝑣𝑝,
𝑣 and 𝑤 are not visible.
Calculate the distance of 𝑣𝑤
𝑝
Calculate the distance of 𝑣𝑝
What Happens at Each Event
• Segment entry
• If we are entering the segment, then we set 𝑓(𝑣) or
𝑏(𝑣) to this segment.
• 𝑣𝑤 can be a visibility edge.
What Happens at Each Event
• Segment exit
• The bullet path will need to
shoot out and find the next
segment that it hits.
• It is available in 𝑂(1) time.
• Since we are sweeping over 𝑤
at the same time that we are
sweeping over 𝑣.
• We know that the bullet
extension from 𝑤 hits 𝑓 𝑤 .
• So, new 𝑓(𝑣) is same as 𝑓(𝑤).
𝑓(𝑤)
Thank You
Q&A