facility location problems in the read-only memory with ...igwoa/indo-german-talk.pdfeuclidean...

Euclidean 1-centerFacility location in tree network

Open Problems

Facility location problems in the read-only memorywith constant work-space

Subhas C. Nandy

Indian Statistical Institute, Kolkata, India

withBinay K. Bhattacharya, Minati De and Sasanka Roy

Subhas. C. Nandy Facility location using constant work-space


Open Problems

IntroductionMegiddo’s Algorithm for Constrained-MECConstrained-MEC using constant spaceMegiddo’s Algorithm for MECMEC using constant space

Table of Contents

1 Euclidean 1-centerIntroductionMegiddo’s Algorithm for Constrained-MECConstrained-MEC using constant spaceMegiddo’s Algorithm for MECMEC using constant space

2 Facility location in tree networkCentroid of a TreeWeighted 1-center of a Tree

3 Open Problems



Open Problems


Euclidean 1-center

Input

A set of n points P in R2

Objective

Find c∗ ∈ R2 such thatmaxx∈P

d(c∗, x) =

minc∈R2

maxx∈P

d(c , x)

Compute the center c∗ ofthe Minimum EnclosingCircle (MEC)



Open Problems


Euclidean 1-center

Input


Objective

Find c∗ ∈ R2 such thatmaxx∈P

d(c∗, x) =

minc∈R2

maxx∈P

d(c , x)

Compute the center c∗ ofthe Minimum EnclosingCircle (MEC)

c∗



Open Problems


Literature

Proposed by Sylvester, 1857

Using FVD takes O(n log n) time

O(n) time and O(n) space, Megiddo, 1983

O(n2) time O(1) space in Read-only model,by Asano, Mulzer, Rote and Wang, 2011

Open Problem

Any sub-quadratic time algorithm in read-only model usingconstant space?



Open Problems


Literature

Open Problem


Our Contribution a

aAppeared in FSTTCS 2012

In-place algorithm: O(n) time and O(1) extra space

Read-only Model: O(n1+ε) time and O( 1ε log n) space, where

ε is a positive constant less than 1



Open Problems


Literature

Open Problem


Here we present

Read-only Model: O((n + M) log4 n) time using O(1) space,where M is the time to compute median using O(1) spacea.

aKnown lower bound of M by T.M. Chan(SODA,2009): Ω(n log logS n) timeusing O(S) bits



Open Problems


Read-only Model

Input: Read-only

Constraints during Execution:Limited space for storing temporary results

After Execution:Desired result will be reported



Open Problems


Constant-work-space Read-only Model

Advantages

Less prone to failure

Secured

Multiple process can access same data simultaneously

Can handle big data



Open Problems


Example

Sorting: An O(n2) time

Find the maximum and report

Find the 2nd maximum and report

....

Find the smallest and report



Open Problems


Fundamental Algorithms

Sorting

O(n2

s + n log s) time and O(s) extra-space (by Frederickson,1987)

Time-space product lower bound: Ω(n2) (by Borodin, 1981)

Selection

O(n log3 n) time algorithm using O(1) space (by Munro andPaterson, 1978)

O(n1+ε) time and O( 1ε ) extra-space (by Munro and Raman,

1996)

Lower bound: Ω(n log logS n) time using O(S) bits (by T.M.Chan, 2009)



Open Problems


Constrained MEC

Input


A vertical line L



Open Problems


Constrained MEC

Input


A vertical line L

Objective

Compute the MinimumEnclosing Circle whosecenter m∗ lies on L

m∗



Open Problems


Constrained MEC



Open Problems


Constrained MEC

Arbitrarily pair-up the points



Open Problems


Constrained MEC

Draw the perpendicular bisectors



Open Problems


Constrained MEC

Observe the intersection points on the line L



Open Problems


Constrained MEC

m

Find the median(m) of the intersection points



Open Problems


Constrained MEC

m

Find farthest point(s) from m



Open Problems


Constrained MEC

m

Farthest points are both above and below of m=⇒ m is the result



Open Problems


Constrained MEC

m

All farthest points are in same side of m



Open Problems


Constrained MEC

m

Prune 14 th points



Open Problems


Constrained MEC

Repeat same for the rest of the points



Open Problems


Constrained MEC

1 begin2 while |P| ≥ 4 do3 Arbitrarily pair up the points in P. Let

PAIR = (P[2i − 1],P[2i ]), i = 1, 2, . . . , b |P|2 c be the setof aforesaid disjoint pairs;

4 Intermediate-Computation-for-Constrained-MEC; (*It returns a point m on L and a side D *)

5 (* Pruning step *)6 forall the pair of points (P[2i ],P[2i + 1]) ∈ PAIR do7 if The bisector line Li defined by the pair

(P[2i ],P[2i + 1]) intersect to the opposite side of Dfrom m

8 then9 Discard one of P[2i ] and P[2i + 1] from P which lies

on the side of m with respect to the bisector line Li ;10

11 (* Finally, when |P| < 4 *) compute the constrained MEC inbrute force manner and decide on which side of L the center ofunconstrained MEC lies.



Open Problems


Constrained MEC

How to keep track the prunedelements?

pair and the feasible region

a

b

p

q



Open Problems


Dominance Relation

Definition

For a pair of points p, q ∈ P, pis said to dominate q withrespect to a feasible region U,if their perpendicular bisectorb(p, q) does not intersect thefeasible region U, and both qand U lie on the same side ofb(p, q).

a

b

p

q



Open Problems


Dominance Relation

Property

p dominates q with respect toa feasible region U if and onlyif from any point x ∈ U,d(p, x) > d(q, x).

a

b

p

q



Open Problems


Dominance Relation

Lemma

If p dominates q and qdominates r with respect to afeasible region U, then pdominates r with respect tothe feasible region U.

Proof.

x ∈ U, d(p, x) > d(q, x) andd(q, x) > d(r , x) =>d(p, x) > d(r , x)

a

b

p

q



Open Problems


Pairing Scheme

p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 p11 p12 p13 p14 p15 p16 p17

Input Array: P[]

b = +∞

a = −∞

The feasibleregionU = [a, b] on L



Open Problems


Pairing Scheme


Input Array: P[]

b

a = −∞




Open Problems


Pairing Scheme


Input Array: P[]

b

a




Open Problems


Pairing Scheme

At the beginning of k-th phase

U = [a, b]

only one element is valid (i.e dominant) from each block ofconsecutive 2k−1 elements, namelyBki = P[i .2k−1 + 1, i .2k−1 + 2, . . . , (i + 1).2k−1],

i = 1, 2, . . . , d n2k−1 e.

We refer the only valid element of Bki as valid(Bk

i )



Open Problems


Pairing Scheme

At the end of k-th phase

U = [a, b] (modified)

only one element is valid (i.e dominant) from each block ofconsecutive 2k elements, namelyBk+1i = P[i .2k + 1, i .2k + 2, . . . , (i + 1).2k ], i = 1, 2, . . . , d n

2ke.



Open Problems


Pairing Scheme

Formed-Pair in the k-th phase

valid(Bk2i−1) and valid(Bk

2i ) form a pair Pairki fori = 1, 2, . . . , d n

2k−1 e

Valid Pair

A pair Pairki is valid pair with respect to U = [a, b] if thecorresponding perpendicular bisector intersects [a, b] on L.



Open Problems


In the k-th phase

Recognize the only valid element from a block Bki

Lemma

Let p, q ∈ Bki having perpendicular bisector b(p, q), then any one

of the following happens

If b(p, q) intersects outside [a, b] on L, then one of p, q is notvalid(Bk

i ).

If b(p, q) intersects [a, b] on L, then none of p and q arevalid(Bk

i ).



Open Problems


In the k-th phase

Recognize the valid element from a block Bki

p1 p2 p3 p4 p5 p6

Candidate

TestingPair p1 p2



Open Problems


In the k-th phase


p1 p2 p3 p4 p5 p6

Candidate

TestingPair p1 p2

p1



Open Problems


In the k-th phase


p1 p2 p3 p4 p5 p6

Candidate

TestingPair p1 p3

p1



Open Problems


In the k-th phase


p1 p2 p3 p4 p5 p6

Candidate

TestingPair p1 p4

p1



Open Problems


In the k-th phase


p1 p2 p3 p4 p5 p6

Candidate

TestingPair p4 p5

p4



Open Problems


In the k-th phase


The valid(Bki ) can be identified in O(|Bk

i |) time using O(1)extra-space.

Lemma

We can enumerate all the valid elements of k-th phase in O(n)time using O(1) extra-space.



Open Problems


In the k-th phase

In an iteration

Consider the intersection points on L of the perpendicularbisectors of the valid pairs

Find the median m among these intersection points

Decide on which side of m the m∗ lies and update U



Open Problems


In the k-th phase

In an iteration




After this iteration

Number of valid pairs decreases by 14



Open Problems


In the k-th phase

In an iteration




After at most log n iterations, none of Pairki are valid pairs, wherei = 1, 2, . . . , d n

2k−1 e



Open Problems


Time and Space Complexity of the Constrained MEC

Lemma

The Constrained MEC can be computed in O((n + M) log2 n) timeusing O(1) extra-space, where M is the time needed to computethe median of n elements in read-only memory when O(1) space isprovided.



Open Problems


Megiddo’s Algorithm for MEC



Open Problems



Arbitrarily pair up the points



Open Problems



Consider their perpendicular bisectors and find the median slope Sm



Open Problems



Pair up the bisector Li , Lj such that slope(Li ) < Sm < slope(Lj)



Open Problems



Consider the intersection points of the paired bisectors (tuple)



Open Problems



Find their median x-coordinate and evoke Decide-on-a-Line



Open Problems



Find median y -coordinate and evoke Decide-on-a-Line



Open Problems



Prune n16 elements



Open Problems



Algorithm 1: MEC(P)

Input: An array P[1, . . . , n] containing a set P of n points in R2.Output: The center π∗ of the minimum enclosing circle of the

points in P.1 begin2 while |P| ≥ 16 do3 Arbitrarily pair up the points in P. Let

PAIR = (P[2i − 1],P[2i ]), i = 1, 2, . . . , b |P|2 c be the setof aforesaid disjoint pairs;

4 Intermediate-Computation-for-MEC; (It returns aquadrant Quad defined by two orthogonal lines LH & LV )

5 (* Pruning step *)6 forall the pair of points (P[2i ],P[2i + 1]) ∈ PAIR do7 if The bisector line Li defined by the pair

(P[2i ],P[2i + 1]) does not intersect the quadrant Quad8 then9 Discard one of P[2i ] and P[2i + 1] from P which

lies on the side of the quadrant Quad with respectto the bisector line Li ;

10

11 (* Finally, when |P| < 16 *) compute by brute force manner.



Open Problems


Decide-on-a-Line using constant space

m∗

Find the Constrained-MEC center m∗



Open Problems



m∗ c∗

Find all the farthest points F



Open Problems



m∗

t1

t2

b1

b2

Convex hull of F contains m∗



Open Problems



m∗

t1

t2

b1

b2

Convex hull of F does not contain m∗



Open Problems



Lemma

Decide-on-a-Line(L) can be computed in O((n + M) log2 n) timeusing O(1) extra-space, where M is the time needed to computethe median of n elements given in a read-only memory using O(1)extra-space.



Open Problems


MEC using constant space

How to keep track of the pruned elements?

Intersection of Feasible Region USubhas. C. Nandy Facility location using constant work-space


Open Problems


Intersection of Feasible Region U of constant complexity

After each iteration we will keep a feasible triangle



Open Problems


Time and Space Complexity

Single iteration

Evokes at most four Constrained MEC

Theorem

The Euclidean 1-center of a set of points in R2 given in aread-only array can be found in O((n + M) log4 n) time using O(1)extra-space, where M is the time needed to compute the median ofn elements in read-only memory when O(1) space is provided.



Open Problems


Similar prune-and-search algorithm

Linear programming in R3

minx1,x2,x3

d1x1 + d2x2 + d3x3

subject to: a′ix1 + b′ix2 + c ′i x3 ≥ βi , i ∈ I = 1, 2, . . . n.

can be computed in O((n + M) log4 n) time using O(1)extra-space, where M is the time needed to compute the median ofn elements in read-only memory when O(1) space is provided.



Open Problems

Centroid of a TreeWeighted 1-center of a Tree

Table of Contents



3 Open Problems



Open Problems


Centroid of a Tree

Centroid of a tree

A vertex v of the tree T such that the size of the largest subtreeof v is minimum among all vertices of T .

Available Results

Can be computed in O(n) time using O(n) space.

In read-only environment, it is easy to compute in O(n2) timeusing O(1) extra space.

Our Result

A linear time algorithm using O(1) extra space in read-onlyenvironment.



Open Problems


Centroid of a Tree

Data Structure

The tree T is stored as a DCEL in read-only memory, such that foreach vertex v the following operations can be performed in O(1)time.

Parent(v)

FirstChild(v)

NextChild(u, v) - Child of u which is next to v

We useTm(t): the subtree of the node t rooted at one of its neighbors m.



Open Problems


Centroid of a Tree

Algorithm

1. Start from an arbitraryvertex t ∈ V

2. Compute a neighbor m oft having maximumnumber of nodes in thesubtree Tm(t).

3. If |Tm(t)| ≤ bn2c, thenreport t as the centroid.Otherwise, the centroidlies in Tm(t).

t′t

This subtree is denoted as ∆

This subtree is denoted as Tm(t)

This subtree is denoted as Tt′(t)

m



Open Problems


Centroid of a Tree

We now concentrate of Step 2 of the algorithm.

Invariant

Maintain three variables t, t ′

and Size satisfying

Initially t = root, t ′ = ∅and Size = 0

If t ′ 6= ∅ then t and t ′ areadjacent vertices andSize = |Tt′(t)|.

t′t

This subtree is denoted as ∆

This subtree is denoted as Tm(t)

This subtree is denoted as Tt′(t)

m

Note: During the search t ′ is the parent of t



Open Problems


Centroid of a Tree

Compute m such that |Tm(t)| = maxv∈N(t)

|Tv (t)| as follows:

Initially, t ′ is the predecessor of t, Size = |Tt′(t)|, andSIZE = MaxSize = |Tt(t ′)| computed in the earlier step.

Use two pointers φ1 and φ2 to count the size of two subtreesof T in a parallel manner.

Use the proceduresParent(u), FirstChild(u) and NextChild(u, v)to compute the size of the subtrees pointed by φ1 and φ2.



Open Problems


Centroid of a Tree

When the counting of a pointer in one subtree is finished, itstarts with an unprocessed subtree immediately.

Each time, when the counting of a subtree by one of thepointers is finished, the count is added in a counter TOTAL.

When the task of one of the pointers is finished, and it doesnot find any unprocessed subtree of T , then the process stopswithout completing the task of the other pointer in which itwas working.

Its count is available by computing SIZE − TOTAL.

During this processthe node m ∈ N(t) having maximum |Tm(t)|, andMaxSize = |Tm(t)| is maintained.



Open Problems


Centroid of a Tree

If MaxSize ≤ bn2c, report m as centroid.

Else recurse with t ′ = t, t = m, Size = n −MaxSize andSIZE = MaxSize.

The recursion continues in the subtrees rooted at m. One of itsneighbors is t, whose elements are already visited in the previousiteration, and the count is Tm(t)



Open Problems


Centroid of a Tree

Result

In each iteration, at most half of the elements of T are visitedby each of φ1 and φ2.

In an iteration, either answer is reported;or the process starts with the unfinished subtree, which is oflargest size among the subtrees of t.

The total time complexity of an iteration is 2× Size.

Since in each iteration, a unfinished subtree is processed,the charge goes to the subtrees which are completelyprocessed.

Thus, the total time complexity becomes O(n).



Open Problems


Weighted 1-center of a Tree

Weighted 1-center of a tree

Given a tree T having each vertex attached with positive weights,a point π on an edge e∗ of T is said to be the weighted 1-center if

∆w (π) = minp∈T

∆w (p)

where ∆w (p) = maxv∈V

w(v)× d(p, v), and

d(p, v) is the distance of the vertex v from the point p along theedges of T .



Open Problems


Weighted 1-center of a Tree

Overview of the Algorithm

Identify an edge e∗ on which the 1-center lies

Next, compute the point π (1-center) on e∗.

An important result

- If u is a fixed vertex in T , and

- v ′ be a vertex in T satisfyingw(v ′)d(v ′, u) = max

v∈Vw(v)d(v , u) that lies in Tt(u), then

- the 1-center of T lies in Tt(u+).

Here Tt(u+) is the subtree of T containing Tt(u) ∪ u ∪ (u, t)



Open Problems


Finding e∗

1. Initialize T ′ = T .

2. Repeat the following steps until T ′ is an edge.

Compute the centroid in T ′. Let it be u.Identify the vertex v ′ traversing all the subtrees of u in thetree T completely.Suppose v ′ ∈ Tt(u). Set T ′ = T ′

t (u+).



Open Problems


Finding e∗

Since, in each iteration, half of the vertices are prunned, to setT ′, the number of iteration is O(log n).

In order to compute v ′, the entire tree T is scanned. Thus,the time complexity of each iteration is O(n).

Result

The overall time complexity of identifying e∗ is O(n log n).The space complexity is O(1).

An important property

At most two internal vertices of T become the leaves of T ′.This enables us to encode T ′ using only four variables.



Open Problems


Computing the 1-center on the edge e∗ = [u∗, v ∗]

Query(m) — for a query point m ∈ e∗

Test whether π = m or π ∈ [u∗,m] or π ∈ [m, v∗].

Process

For a point x ∈ [a, b], definef1(x, v) = w(v)× (d(u∗, v) + d(x, u∗)) for all v ∈ Tu∗ (v∗),f2(x, v) = w(v)× (d(v∗, v) + d(x, v∗)) for all v ∈ Tv∗ (u∗).

u∗

v∗

v(∈ Tv(u))

x

v(∈ Tu(v))

Find f1(k1,m) = maxv∈Tu∗ (v∗)

f1(m, v), and f2(k2,m) = maxv∈Tv∗ (u∗)

f1(m, v)

If f1(k1,m) = f2(k2,m), then π = mElse if f1(k1,m) > f2(k2,m) then π ∈ [a,m], else π ∈ [m, b].

Time complexity of Query(m): O(n)



Open Problems


Computing m

Dominance

For a pair of vertices p, q ∈ Tu∗(v∗), p dominates q if f1(p, x) andf1(q, x) do not intersect in [a, b], and f1(q, x) < f1(p, x).



Open Problems


Computing m

Process

Pair up vertices of V1 = Tu∗(v∗).

If one of them dominates the other, then prune the dominatedone.Otherwise compute the point of intersection x ∈ [a, b] off1(p, x) and f1(q, x).

Similarly, pair up vertices of V2 = Tv∗(u∗),and compute the point of intersetion of f1(p, x) and f1(q, x)for the undominated pairs (p, q) ∈ V2.

Compute the median m of these intersection points inread-only memory.



Open Problems


Complexity

Result

Time complexity : O((n + M) log2 n), andSpace: O(1), wherewhere M is the time needed for computing the median of nelements in read-only environment using O(1) space.



Open Problems

Table of Contents



3 Open Problems



Open Problems

Open Problem

Farthest Pair of Points

Given a set P of n points in R2, find the farthest pair of points insub-quadratic time in the constant-work-space read-only model.

Maximum Empty Circle

Given a set P of n points in R2, find the the maximum emptycircle (whose center is inside the convex hull of P) in sub-quadratictime in the constant-work-space read-only model.



Open Problems

Thank You!


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