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

Algorithms

Tom Leighton and Peter Shor

Presented by Linna Wei

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

• 1986

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)

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

between points in P to the grid points

• 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.

goal

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

?

• 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,

2

2

(log )N

( ) ( lg )L P N

• 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.

• 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.

Related Combinatorial Problems

• The Maximum Up-Right Matching Problem

• Two-Dimensional Discrepancy Problem

• For applications and proof

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.

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 .

( )U P

( )N

• 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 (i+2d(N),

j+2d(N))

( ) ( ) ( ) ( )L P d N and L P d N

P

• 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.

4 ( )d N N

( ) ( ( ))U P O Nd N

( )U P

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.

( )R

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

( )R

P P P

Applications

• Wafer-Scale Integration

• Two-Dimensional Bin Packing

• One-Dimensional Bin Packing

• Dynamic Allocation

• Testing Pseudorandom Number Generatiors

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

1/2 3/4( ) ( (log ) )U P n n

• 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.

• 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

1/2 3/4( (log ) )n n

• 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.

• 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 > + 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.

1/2((log ) )n

Properties of triangles

• 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.

1/2 2(log )c n r

( )nA1/2 1/4( (log ) )n n

(log )n1/2 3/4( (log ) )n n

• Proof:

• Step 1: convert the minimax grid matching problem into a dual discrepancy problem with a straightforward application of Hall’s theorem.

• Step 2: Intuition and Motivation for step 3

• Step 3: Prove the necessary bounds for the discrepancy problem.

3/4( ) (log )L P O N

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.

3/2log

N

N3/4log N

1c

3/4logcp N

Decomposition Into Triangles – The Intuition

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 ( ( ))Area T

• 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.

2 log p

The End