Tight Bounds for Minimax Grid Matching, With Applications to the Average Case Analysis of Algorithms

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Tight Bounds for Minimax Grid Matching, With
Applications to the Average Case Analysis of
Algorithms

• Tom Leighton and Peter Shor
• Presented by Linna Wei

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• Annual ACM Symposium on Theory of
  ComputingProceedings of the eighteenth annual
  ACM symposium on Theory of computing
• 1986

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area

• Wafer-scale integration of systolic arrays
• Two dimensional discrepancy problems
• Testing pseudorandom number generators
• Maximum up-right matching problem (Karp, Luby and
  Marchetti-Spaccamela, 2-dimensional bin packing)

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• P any particular set of N random points.
• L(P) minimum length in the perfect matching
  between points in P to the grid points

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• L(P) is the minimum over all perfect matchings of
  the maximum distance between any pair of matched
  points or simply the minimax matching length for
  P.

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goal

• The expected value of L(P) for random P.

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?

• True for any random point but with high
  probability not true for every grid point.
• With probability less than 1-1/N there is a
  circular region with area in the
  square that does not contain any random points at
  all.
• With probability 1-1/N,

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• Wafer-Scale Integration of Systolic Arrays
• Tom Leighton and Charles E. Leiserson
• IEEE Transactions on Computers
• Vol. c-34, No. 5, 1985
• Goal minimize
• the longest wire
• Observation
• Length of the longest
• wire is the length of the longest sequence of
• dead cells in the snake-like string.

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• For each cell, the probability of it to be live
• or dead is ½.
• K cells are dead, 1/2k
• Any set of 2lgN cells are
• dead, 1/N2
• Less than N sets of 2lgN consecutive cells
• The chances are less than 1/N of having to skip
  more than 2lgN cells in the entire snake-like
  path of length N.

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Related Combinatorial Problems

• The Maximum Up-Right Matching Problem
• Two-Dimensional Discrepancy Problem
• For applications and proof

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The Maximum Up-Right Matching Problem

• An up-right matching
• Is a one-to-one matching
• of pluses to minuses.
• Every plus is either un-
• matched or is matched to a single minus that lies
  above and to the right of the plus.
• Maximum minimizes the number of unmatched points.

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Relationship between minimax grid matching and
maximum up-right matching

• Any very high probability upper bound on L(P) can
  be transformed into a very high probability upper
  bound for by simply multiplying by
  .

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• Consider an up-right matching problem with N
  random pluses and N random minuses. Let d(N) be a
  very high probability upper bound on L(P).
• With very high probability
• Then form a matching on
• by matching the plus
• Identified with grid point
• (i,j) to the minus identified
• With grid point (i2d(N),
• j2d(N))

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• The procedure forms an up-right matching.
• The only pluses not matched by the procedure are
  those identified with grid points in the topmost
  2d(N) rows and the rightmost 2d(N) columns.(less
  than )
• The claimed very high probability bound for
  immediately follows from the very high
  probability bound for L(P). And it is conceivable
  that a symmetric condition is also true.

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Two-Dimensional Discrepancy Problem

• Up-right discrepancy problem
• Consider a square with
• area N that contains N
• random pluses and N
• random minuses, define discrepancy of the region
  R, to be he number of pluses in R less
  the number of minuses in R. R is the up-right
  regions.

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• The maximum value of over all up-right
  regions is precisely the number of unmatched
  pluses in a maximum up-right matching for
  .

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Applications

• Wafer-Scale Integration
• Two-Dimensional Bin Packing
• One-Dimensional Bin Packing
• Dynamic Allocation
• Testing Pseudorandom Number Generatiors

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In Upward Right Matching

• The Average-case Analysis of Some On-line
  Algorithms for Bin Packing
• Peter W. Shor
• Combinatorica, Vol. 6, Issue 2, Pages 179
  200, 1986

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• Proof Sketch To obtain a lower bound of k for
  the number of unmatched points in an upward right
  matching in a square, it suffices to split the
  square into two sections, such that in the lower
  section there are k more points than points,
  and such that
• the boundary
• dividing the
• sections always
• goes down and
• to the right.

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• To make the proof easier, we will rotate the
  square 45 degrees, so now we want a boundary
  which has a slope between -1 and 1. We will then
  construct a boundary where the expected number of
  extra s below it is

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• We will produce the boundary in stages. At each
  stage, the boundary will consists of line
  segments and triangles. If a triangle is on a
  segment of the boundary, then the final boundary
  will pass through the two extreme vertices of the
  triangle and the section between these vertices
  will be contained within the triangle. At each
  stage, we will replace every triangle with either
  a line segment or with two triangles each having
  a quarter of the area of the old triangles. We
  will finally stop the refinement when the
  triangles are so small that they have on the
  average only one point in each.

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• For this triangle which
• has two vertices at
• diagonal vertices of the
• square, its sides have a
• slope of
• Check points and points in it.
• If - points gt points, refine the boundary to
  two smaller triangles so the boundary is below
  the quadrilateral.
• Otherwise, refine the boundary to two smaller
  triangles above the quadrilateral.
• Repeat.

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Properties of triangles
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• At each stage, we are testing r regions each of
  area to decide whether or not
  to include them. Since the expected difference
  between the number of and points a region of
  area A contains is , each stage adds
  on the average extra
  points to the lower region. Thus, after
  stages,
• we have an expected number of
• extra points in the lower region.

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• Proof
• Step 1 convert the minimax grid matching problem
  into a dual discrepancy problem with a
  straightforward application of Halls theorem.
• Step 2 Intuition and Motivation for step 3
• Step 3 Prove the necessary bounds for the
  discrepancy problem.

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Conversion to the Dual Discrepancy Problem

• We define a partition of the square with
  area N into subareas, each with side
  length . In step3 we will prove that
  there is a constant such that with very
  high probability, the discrepancy of every simply
  connected region whose boundary lies along the
  edges of is at most
• where p is the perimeter of the region.

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Decomposition Into Triangles The Intuition
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Formally Bounding the Discrepancy The Proof

• Prove the average discrepancy of a triangle in
  the decomposition of the polygonal region R is
  the expected discrepancy

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• Section 1 In the deterministic section the
  discrepancy of any R satisfying the hypothesis
  can be bounded by the sum of the discrepancy of
  disjoint regions.
• Section 2 In the probabilistic section we
  establish very high probability bounds on the
  discrepancies of the regions in the class, thus
  obtaining an upper bound on the discrepancy of
  any R.

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• The End