geometric solutions for the ip-lookup and packet classification problem

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Geometric Solutions for theIP-Lookup and Packet Classification Problem

(Lecture 12: The IP-LookUp & Packet Classification Problem, Part II)Advanced Algorithms & Data Structures

Prof. Dr. Th. OttmannAlbert-Ludwigs-Universität Freiburg, Germany

ottmann@informatik.uni-freiburg.de

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Overview

The IP-table lookup problem

Geometric interpretation of the problem

Solution using Priority Search Trees

Solution using augmented range trees

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Dstn Addr

--------

---- ----

--------

Dstn-prefix Next Hop

Forwarding Table

Forwarding EngineHEADER

Lookup in an IP Router

Next-Hop-Computation

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0 224

232-1

128.9.0.0/16

65.0.0.0

142.12.0.0/19

65.0.0.0/8

65.255.255.255

Destination IP Prefix

Outgoing Port

(Next-Hop)

65.0.0.0/ 8 3

128.9.0.0/16 1

142.12.0.0/19 7

IP prefix: 0-32 bits

128.9.16.14

Example Forwarding Table

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128.9.16.0/21128.9.172.0/21

128.9.176.0/24

0 232-1

128.9.0.0/16142.12.0.0/19

65.0.0.0/8

Nested intervalls

Sets of intervalls corresponding to sets of prefixes of IP-addressesare nested:Any two intervalls are either disjoint or one is contained in the other!

Overlaps are impossible!

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128.9.16.0/21128.9.172.0/21

128.9.176.0/24

Routing lookup: Find the longest matching prefix (the most specific route) among all prefixes that match the destination address.Geometric view: For a given point p, find the smallest interval stabbed by p.

0 232-1

128.9.0.0/16142.12.0.0/19

65.0.0.0/8

128.9.16.14

LMP-Matching

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Overview

The IP-table lookup problem

Geometric interpretation of the problem

Solution using Priority Search Trees

Solution using augmented range trees

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Geometric view of IP-table lookup

Find a representation of a (dynamic) set of intervals such that the following operations can be carried out efficiently:

• Insertion of an interval

• Deletion of an interval

• For a given point p: Find the smallest interval stabbed by p (LMP-query)

Assumptions:

Set of intervals is a nested set.

Every newly inserted interval does not overlap with any interval in the set.

Conflict test: For a given interval test whether or not it overlaps (is in conflict) with any interval already in the set.

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Representation of intervals by points

A

B

C

D

A

B

CD

Interval (l, r) is mapped to point (r, l) below the main diagonal.

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Stabbing queries

A

B

C

D

A

BC

D

p lies in (l, r) ⇔ (l, r) is stabbed by p ⇔ ( l ≤ p und p ≤ r )

All points representing intervals stabbed by p lie to theright and below point (p, p)

p

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Geometric interpretation of interval-inclusion

A

B

A

B

A ⊒ B Interval A = (lA, rA) includes interval B = (lB, rB) iff. point A is right below point B A B ⋟

B ⊑ A ⇔ A ⊒ B ⇔ point B left above point A ⇔ B ≼ A

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Disjoint intervals

B

A B

A

X and Y represent disjoint intervals⇔ There is no point p on the diagonal, s.t. p ≼ X und p ≼ Y

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Overlapping intervals

C

p

D

C

D

X and Y represent overlapping intervals⇔ There is a point p on the diagonal, s.t. p ≼ X and p ≼ Y , but X and Y are incomparable with resp. to ≼

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Nested intervals

B

C

A

B

p

A

D

C

X is smallest intervall, that contains point p⇔ p ≼ X and for all Y with p ≼ Y : X ≼ Y

Observation: Sets of nested intervals stabbed by a point are totally ordered with resp. to ≼ Points closer to the diagonal represent smaller intervals.

D

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LMP-Queries

B

C

A

B

p

A

D

C

D

The smallest interval containing a given point p is represented bythe point closest to the diagonal in the region right below p.(This is the topmost leftmost point in the region!)

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Overview

The IP-table lookup problem

Geometric interpretation of the problem

Solution using Priority Search Trees

Solution using augmented range trees

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Priority Search TreesPriority Search trees are a 1.5-dim structure for the storage of points, they support the following operations :

Insertion of a point Deletion of a point South-grounded range queries

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x

x

x´ = ∞

Executable in O(log n) time.

LMP-Query

Find topmost leftmost (orleftmost topmost) point inthe range (x, ∞ , x)

p

x

x

Well defined, if intervalls are nested,otherwise not!

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15,1

40

4,2

9

33,4

28

1

1,3

1

6,10

1317,35

302,4

2

17 28

17,9

17

Finding topmost leftmost points in PST

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Finding leftmost topmost points in PST

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Pairwise distinct search keys

A

BC

D

A

BC

D

(x, y) ↦ (2W x + 2W – 1 – y, y) = ((x, -y), y)All points have pairwise distinct x-coordinates.Points corresponding to intervals with equal left endpointare l.t.r. sorted according to increasing lengths.LMP: Find point with smallest x-coordinate in a given x-range below a threshold!

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2,2

6

6,4

2

7,3

7

2

6

Possibilities for the full dynamization of priority search trees: No rigid skeleton, but growing or shrinking with the point set.

Insertion (5,3)

Dynamic PST

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Balanced trees as skeletons of PSTs

q

x´

p

x

1 2

3

x

x´

1

2 3

Rotation preserves the x-order,but may destroy the y-order of points.

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Balanced trees as skeletons of PSTs

q

x´

p

x

1 2

3

x

x´

1

2 3

px occurs in subtree 1:Insert p into subtree 1;Pull up a point from subtree 2 or 3.

2 log n steps for the trickle down process!

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Balanced trees as skeletons of PSTs

q

x´

p

x

1 2

3

x

x´

1

2 3

px occurs in subtree 2:Compare py with the priorities stored at the roots ofsubtree 1 and 2;Insert a point into subtree 2 or 3;Pull up a point from subtree 2 or 3.

2 log n steps for the trickle down process!

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Conflict detection (Insertion)

x

vu

y

new (query) interval

(1) There is an interval (x, y) stabbed by u that has a right endpoint y < v

x y

vu

(2) The range (u, v) contains a left endpoint x of an interval (x, y) that has a right endpoint not in this range, i.e. y > v

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Conflict detection (Insertion)

x

vu

y

new (query) interval

(1) There is an interval (x, y) stabbed by u that has a right endpoint y < v

Points represent intervals sortedaccording to their right endpoints

(u, u)

v

(y, x)

u

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Conflict detection (Insertion)x y

vu

(2) The range (u, v) contains a left endpoint x of an interval (x, y) that has a right endpoint not in this range, i.e. y > v

Map interval (x, y) to point (x, -y). Store points in PST

v

(x, -y)

u

-v-y < -v ⇔ y > v

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Overview

The IP-table lookup problem

Geometric interpretation of the problem

Solution using Priority Search Trees

Solution using augmented range trees

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1-dim Range Tree: Summary

Let P be a set of n points in 1-dim space.

P can be stored in a balanced binary leaf-search tree such that the following holds:

Construction time: O(n log n)

Space requirement: O(n)

Insertion of a point: O(log n) time

Deletion of a point: O(log n) time

1-dim-range-query: Reporting all k points falling into a given query range can be carried out in time O(long n + k).

The performance of 1-dim range trees does not depend on the chosen balancing scheme!

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MinXinRectangle Queries

Problem: Given a set P of points that changes under insertions and deletions,

construct a data structure to store P that can be updated in O(lg n) time and that can

find the point with minimal x-coordinate in a given range below a given

threshold in O(log n) time.

l r

y0

MinXinRectangle(l, r, y0)

Assumption: All points have pairwise different x-coordinates

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MinXinRectangle Queries

lr

y0

MinXinRectangle(l, r, y0)

Assumption: All points have pairwise different x-coordinates

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Min-augmented

Range

Tree

(2, 12) (3, 4) (4, 11)

(5, 3) (8, 5)

(11, 21)

(14, 7)

(21, 8)(15, 2) (17, 30)

2 4

3

11

8

5

14

17

15 21

2

2

2

3

4 3

3

3

Two data structures in one:

Leaf-search tree on x-coordinates of points

Min-tournament tree on y-coordinates of points

82

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MinXinRectangle(l, r, y0)

l r

Split node

Search for the boundary values l, r.Find the leftmost umbrella node witha min-field ≤ y0.

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MinXinRectangle(l, r, y0)

l r

Split node

Search for the boundary values l, r.Find the leftmost umbrella node with a min-field ≤ y0.

Proceed to the left son of the current node, if its min-field is ≤ y0,

and to the right son, otherwise. Return the point at the leaf.

MinXinRectangle(l, r, y0) can be found in time O(height of tree).

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UpdatesInsert operation

Insert node as into a standard binary leaf search tree.

Adjust min-fields of every ancestor of the new node by playing a min tournament for each node and its sibling along the search path.

Delete operation: Similar

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Maintaining

min-fields

Under

Rotations

s1

s2s3

s4 s5

s1

s3s5

s4

min{s5, s3}

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Min-augmented

Range

Trees–Summary

Theorem: There exists a data structure to represent a dynamic set S of n points in the plane with the following properties:The data structure allows updates and to answer MinXinRectangle(l, r, y0) queries in O(log n) time. The data structure occupies O(n) space.

Note: The data structure can be based on an arbitrary scheme of balanced binary leaf search trees.

Min-augmented range trees can be used (instaed of PSTs) to solve the dynamic version of the IP lookup problem without changing the time complexity of the lookup and update operations.

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Conflict detection (Insertion)

x

vu

y

new (query) interval

(1) There is an interval (x, y) stabbed by u that has a right endpoint y < v

Points represent intervals sortedaccording to their right endpoints

(u, u)

v

(y, x)

u

Choose a min-augmented rangetree andcheck whether there is at leastone point in the range [u,v] below the threshold u.

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Conflict detection (Insertion)x y

vu

(2) The range (u, v) contains a left endpoint x of an interval (x, y) that has a right endpoint not in this range, i.e. y > v

Map interval [x, y] to point (x, y). Store points in max-augmented range tree and check whether there is a point(x, y) in the range [u, v] with a right endpoint y > v.

v

(x, y)

u

v

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• Implement and test new solutions with real (benchmark) data.

• Find a simple, efficient solution for the general, higher-dimensional case.

• Consider the case of bursty and clustered updates.

• Investigate the effect of using other data structures, likefully dynamic segment trees,relaxed balanced search trees, relaxed balanced PST, ….

• Find an efficient method for conflict detection in 2 (or more) dimensions.

Open problems