[email protected] Lecture 1-5 Power Law Structure Weili Wu Ding-Zhu Du Univ of Texas at Dallas.
Chapter 5 Guillotine Cut (2) Portals Ding-Zhu Du.
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Transcript of Chapter 5 Guillotine Cut (2) Portals Ding-Zhu Du.
![Page 2: Chapter 5 Guillotine Cut (2) Portals Ding-Zhu Du.](https://reader034.fdocuments.us/reader034/viewer/2022051401/5697bf911a28abf838c8e8d0/html5/thumbnails/2.jpg)
Rectilinear Steiner Tree
• Given a set of points in the rectilinear plane, find a minimum length tree interconnecting them.
• Those given points are called terminals.
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Initially
Edge length< RSMT
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Initially
n x n grid2 2
n = # of terminalsL
n
L
n
Ln
2
Total movingLength:
If PTAS exists for grid points, then it exists for general case.
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(1/3-2/3)-cut
Longer edge1/3 2/3
Shorter edge
Longer edge
> 1/3
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Cut line position
n x n grid2 2
L
Cut line alwayspasses throughthe center of a cell.
1 ( assume)
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Depth of (1/3-2/3)-cut
Note that every two parallel cut lines has distanceat least one. Therefore, the smallest rectangle hasarea 1.
After one cut, each resulting rectangle has area Within a factor of 2/3 from the original one.
Hence, depth of cuts < (4 log n)/(log (3/2)) = O(log n) since
(2/3) n > 1depth 4
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(1/3-2/3)-Partition
O(log n)
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Portals
m portals divide a cut segment equally.
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Restriction
A rectilinear Steiner tree T is restricted if
there exists a (1/3-2/3)-partition such thatIf a segment of T passes through a cutLine, it passes at a portal.
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Minimum Restricted RST can be computed in time n 2 bydynamic programming
O(m)
Choices of each cut line = O(n )2
# of subproblems = n 2O(m)24
26
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# of subproblem
Each subproblems can be described by three facts:
1. Position of for edges of a rectangle.
2. Position of portals at each edge.
4. Partition of using portals on the boundary. (In each part of the partition, all portals are connected and every terminal inside of the rectangle is connected to some tree containing a portal. )
O(n )8
O(n )4
2O(m)3. Set of using portals.
2O(m)
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Position of portals
O(n )O(n )2 2
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# of partitions
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1 k
N(k) = # of partitions
N(k) = N(k-1) + N(k-2)N(1) + ··· + N(1)N(k-2) + N(k-1) = N(k-1)N(0) + N(k-2)N(1) + ··· + N(0)N(k-1)
N(0)=1
f(x) = N(0) + N(1)x + N(2)x + ··· + N(k)x + ···2 k
xf(x) = f(x) - 12
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Analysis (idea)
• Consider a MRST T.
• Choose a (1/3-2/3)-partition.
• Modify it into a restricted RST by moving cross-points to portals.
• Estimate the total cost of moving cross-points.
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Choice of (1/3-2/3)-partition
Each cut is chosen to minimize # of cross-points.
(# of cross-points) x (1/3 longer edge length)< (length of T lying in rectangle).
1/3 2/3
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Moving cross-points to portals
Cost = (# of cross-points) x ( edge length/(m+1)) < (3/(m+1)) x (length of T lying in rectangle)
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Moving cost at each level of
(1/3-2/3)-Partition < (3/(m+1)) x (length of T )
O(log n)
Total cost < O(log n)(3 / (m+1)) x (length of T)
Choose m = (1/ε) O(log n). Then 2 = n . O(m) O(1/ε)
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RSMT has (1+ε)-approximation with running Time n .
O(1/ε)
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Thanks, End