Intersections
Two Convex Polygons
R. Inkulu
http://www.iitg.ac.in/rinkulu/
(Intersections: Convex Polygons)
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Problem Description
Given two convex polygons P1 and P2 of sizes n1 and n2 respectively,
compute P = P1 ∩ P2 .
(Intersections: Convex Polygons)
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Outline
1
Prepara-Hoey’s Algorithm
2
O’Rourke’s Algorithm
(Intersections: Convex Polygons)
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Plane-Sweep Algorithm
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• Sort upper and lower hull vertices of P1 based on their x-coordi (resp.
P2 ) and save in list L1 (resp. L2 )
• Merge L1 and L2 to obtain L; with each vertex save info such as which
polygon it belongs to etc.,
• Computer P while traversing L:
compute the intersection of two quadrilaterals in each slab, and
remove a vertex from P whenever three vertices are collinear.
(Intersections: Convex Polygons)
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Analysis
• O(n1 + n2 ) time.
• O(n1 + n2 ) space.
(Intersections: Convex Polygons)
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Outline
1
Prepara-Hoey’s Algorithm
2
O’Rourke’s Algorithm
(Intersections: Convex Polygons)
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Characterizing sickles
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• Two successive intersection points on P together form a sickle.
• Removing all the sickles yield the required.
• Intersection points’ always occur in pairs.
(Intersections: Convex Polygons)
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Removing sickels
1,13,28
2,15,30
0’,12,27
11,26
14,29
16
3,17
10,25
0’’,18
9,24
4,19
5,20
6,21
8,23
7,22
• While traversing boundaries of P1 and P2 simultaneously in CCW order,
remove sickles, while maintaining P.
• How many pointer advances are required in total?
(Intersections: Convex Polygons)
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Making a pointer reach an intersection point v
ri
v
w
Let the pointers together advance for n1 + n2 steps:
• boundary of one polygon is guaranteed to be completely traversed (from
the listed cases).
• ensures one of the pointers’ reachng some intersection point (whenever
P 6= φ).
(Intersections: Convex Polygons)
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Placing both the pointers at the same intersection point
cases (b), (d)
ri
v
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case (c)
br
case (a)
v
rm
br
w
rt
case (d)
bj
w
bs
case (b)
bs
(a) bptr is on (bs − br ]; move
bptr till v
(b) bptr is on (br − bs ];
move rptr till w
all these cases are listed exhaustively later.
• The pointer advance schedule requires to be designed to place both the
pointers at the same intersection point without causing any deadlocks.
(Intersections: Convex Polygons)
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Identifying sickles in succession
13,28
12,27
15,30
11,26
14,29
16
17
10,25
18
9,24
19
5,20
6,21
8,23
7,22
• Once both the pointers pass through the same intersection point, all the
sickles will be constructed in succession (due to cases mentioned above)
by a single traversal of P1 and P2 .
(Intersections: Convex Polygons)
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Pointer advancing approach
P
advance either
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advance
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(c)
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advance
(d)
bptr and rptr are moving in CCW direction; polygonal chains are shown as lines; solid arrows
represent the current edge to which bptr and rptr refer to.
• Do not advance on a boundary whose current edge may contain a yet to
be found intersection.
(Intersections: Convex Polygons)
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Algorithm
• Find all the intersection points (and sickles) of P1 and P2 while
advancing pointers over P1 and P2 together for 2(n1 + n2 ).
(implicitly) first brings both the pointers to the same intersection point.
• If no intersection has been found, determine whether P1 ⊂ P2 or
P2 ⊂ P1 or P1 ∩ P2 = φ.
(Intersections: Convex Polygons)
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Running Example
1,13,28
2,15,30
0’,12,27
11,26
14,29
16
3,17
10,25
0’’,18
9,24
4,19
5,20
6,21
8,23
7,22
(Intersections: Convex Polygons)
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Analysis
• 4(n1 + n2 ) time.
placing rptr/bptr at an intersection point v: n1 + n2
finding all the sickles (including bringing the other point to v): n1 + n2
two polygon inclusion tests: 2(n1 + n2 )
• O(n1 + n2 ) space.
(Intersections: Convex Polygons)
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