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Algorithmic Aspects of Searching in the Past

Christine KupichInstitut für Informatik, Universität Freiburg

Lecture 1: Persistent Data StructuresAdvanced Topics in Algorithms & Data Structures

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Overview

• Motivation

• Example: Natural search trees

• Making data structures partially persistent

• Example: Partially persistent red-black trees

• An application: Point location

• An application: Grounded 2-dimensional range searching

• Making data structures fully persistent

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Motivation

Ephemeral: no mechanism to revert to previous states

A structure is called persistent, if it supports access to multiple versions.

Partially persistent: All versions can be accessed but only the newest version can be

modified.

Fully persistent: All versions can be accessed and modified.

Confluently persistent: Two or more old versions can be combined into one new

version.

Oblivious: The data structure yields no knowledge about the sequence of operations

that have been applied to it other than the final result of the operations.

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Example: Natural search trees

Only partially oblivious!

• Insertion history can sometimes be reconstructed.

• Deleted keys are not visible.

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57 3

1

5

7

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Simple methods for making structures persistent

• Structure-copying method: Make a copy of the data structure each time it is

changed. Yields full persistence at the price of (n) time and space per update to a

structure of size n

• Store a log-file of all updates! In order to access version i, first carry out i updates,

starting with the initial structure, and generate version i. (i) time per access, O(1)

space and time per update

• Hybrid-method: Store the complete sequence of updates and additionally each k-th

version for a suitably chosen k. Result: Any choice of k causes blowup in either

storage space or access time

Are there any better methods?

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Making data structures persistent

Several constructions to make various data structures persistent have been devised,

but no general approach has been taken until the seminal paper by Driscoll,

Sarnak, Sleator and Tarjan, 1986.

They propose methods to make linked data structures partially as well as fully

persistent.

Let’s first have a look at how to make structures partially persistent

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Fat node method - partial persistence

Record all changes made to node fields in the nodes

• Each fat node contains same fields as ephemeral node and a version stamp• Add a modification history to every node: each field in a node contains a list of version-value pairs

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Fat node method - partial persistence

Modifications

• Ephemeral update step i creates new node: create a new fat node with version stamp i and original field values• Ephemeral update step i changes a field: store the field value plus a timestamp

Each node knows what its value was at any previous point in time

Access field f of version iChoose the value with maximum version stamp no greater than i

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Fat node method - analysis

• Time cost per access gives O(log m) slowdown per node (using binary search on the modification history)• Time and Space cost per update step is O(1) (to store the modification along with the timestamp at the end of the modification history)

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Fat node method - Example

A partially persistent search tree. Insertions:5,3,13,15,1,9,7,11,10, followed by deletion of item 13.

5

1-10

3 13

2 3

44

151

56

9

7

77

11

8

10

910

1010

10

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Path-copying method - partial persistence

• Make a copy of the node before changing it to point to the new child. Cascade the change back until root is reached. Restructuring costs O(height_of_tree) per update operation• Every modification creates a new root

• Maintain an array of roots indexed by timestamp.

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Path-copying method - Example

5

1 7

3

0

version 0:

version 1:Insert (2)

version 2:Insert (4)

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Path-copying method - Example

5

1 7

3

0

version 0:

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Path-copying method - partial persistence

5 5

1 1 7

3 3

2

0 1

version 1:Insert (2)

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Path-copying method - partial persistence

5 5 5

1 1 1 7

3 3 3

2 4

0 1 2

version 1:Insert (2)

version 2:Insert (4)

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Node-copying method - partial persistence

Extend each node by a time-stamped modification box (initially empty)

Version before the modification

time t

Versionat/ after time t

k

t: rplp rp

Searching in version j

Follow an entry pointer with largest version number i, i <= j Compare keys and follow newest pointer no greater than j

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Node-copying method - partial persistence

5

1

3

7

version 0

version 1:Insert (2)version 2:Insert (4)

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Node-copying method - partial persistence

5

1

3

2

7

1 lp

version 0:

version 1:Insert (2)

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Node-copying method - partial persistence

5

1

3

2

3

4

7

1 lp

version 1:Insert (2)

version 2:Insert (4)

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Node-copying method - partial persistence

5

1

3

2

3

4

72 rp

1 lp

version 1:Insert (2)version 2:Insert (4)

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Node-copying method - partial persistence

Modification

If modification box empty, fill it.

Otherwise, make a copy of the node, using only the latest values, i.e. value in modification box plus the value we want to insert, without using modification box

Cascade this change to the node’s parent

If the node is a root, add the new root to a sorted array of roots

Access time gets O(1) slowdown per node, plus additive O(log m) cost for finding the correct root

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Node-copying method - Example

A partially persistent search tree. Insertions: 5,3,13,15,1,9,7,11,10, followed by deletion of item 13.

5

1-2

22

3

5

13

3-9

15

4

1

513

6

9

7

7

9

11

8

10

9

5

10

1110

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Node-copying method - partial persistence

The amortized costs (time and space) per modification are O(1).Proof: Using the potential technique

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Potential technique

Potential techniqueThe potential is a function of the entire data structure

Definition potential function: A measure of a data structure whose change after an operation corresponds to the time cost of the operation • The initial potential has to be equal to zero and non-negative for all versions• The amortized cost of an operation is the actual cost plus the change in potential• Different potential functions lead to different amortized bounds

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Node-copying method - partial persistence

Definitions

• Live nodes: they form the latest version (reachable from the root of the most recent version), dead otherwise• Full live nodes: live nodes whose modification boxes are full

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Node-copying method - potential paradigm

The potential function f (T): the number of full live nodes in T(initially zero)

The amortized cost of an operation is the actual cost plus the change in potential

Δ f =?

Each modification involves k number of copies, each with a O(1) space and time cost, and one change to a modification box with O(1) time costChange in potential after update operation i: Δ f =

Space: O(k + Δ f), time: O(k + 1 + Δ f)

Hence, a modification takes O(1) amortized space and O(1) amortized time

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Red-black trees

Constraints

• All missing nodes are regarded as black• Any red node has a black parent• From any node, all paths to a missing node contain the same number of black nodes

Depth of an n-node red-black tree is at most 2 log nRoot is colored black

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Red-black trees

Rebalancing transformations - insertion

1.

2.

bubble the violation up the tree

recolor

recolor

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Red-black trees

Rebalancing transformations - insertion

11 rr4.

lr + recolor parent and gran-parent

3.leaving no inconsistency

An insertion requires O(log n) recolorings plus at most 2 rotations

Case 3.

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Red-black trees - partial persistence

A red-black tree can be made partially persistent using the node copying method at an amortized space cost of O(1) per insertion or deletion and a worst-case time cost of O(log n) per access, insertion or deletion.

Each node contains: • a key • 2 pointers for the successors• a color bit and • an extra pointer (version stamp, direction)

Colors are not used in access operations. Old colors can be overwritten

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Red-black trees - partial persistence

An Example: insert E, C, M, O, N

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Red-black trees - partial persistence

An Example: insert E, C, M, O, N

recolor

E

C

1-2

r,b

2r

E

1

r,bInsert C Insert M E

M

3

b

r

E

C

1-2

r,b

2r

O

E

M

3-4

b

r

4r

E

C

1-2

r,b

2r

O

E

M

3-4

b,r,b

r,b4

r

E

C

1-2

r,b

2r,b

Insert O

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Red-black trees - partial persistence

Insert N

O

E

M

3-4

b

r,b4

r

E

C

1-2

r,b

2r,b

Nr

O

E

M

3-4

b

r,b4

r

E

C

1-2

r,b

2r,b

5

RR

Or

N

Mr,b

r

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Or

N

Mr,b

r LR +recolor

Or

N

M

r,b

r

Nr

O

E

M

3-4

b

r,b4

r

E

C

1-2

r,b

2r,b

5

Nr,b

O

E

M

3-5

b

r,b

4r

E

C

1-2

r,b

2r,b

5

Mr

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Application: Grounded 2-Dimensional Range Searching

Given a set of points, and a query triple (a,b,i)

Report the set of points a<x<b and y<i.

a b

i

x

y

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Application: Grounded 2-Dimensional Range Searching

a b

i

To answer a query: Report all points in version i whose x-coordinates are in [a,b]. Query time?

Persistent red-black tree: Space ? , preprocessing time ?

Version i contains every point for which y<i. Use x-coordinates as keys.

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1-Dimensional Range Search

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Application: Planar point location

Suppose that the Euclidian plane is subdivided into polygons by n line segments that intersect only at their endpoints.

Given such a polygonal subdivision and an on-line sequence of query points in the plane, the planar point location problem, is to determine for each query point the polygon containing it.

Measure an algorithm by three parameters:

1) The preprocessing time.

2) The space required for the data structure.

3) The time per query.

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Planar point location - example

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Solving planar point location (Cont.)

Dobkin-Lipton:

Partition the plane into vertical slabs by drawing a vertical line through each endpoint.

Within each slab the lines are totally ordered.

Allocate a search tree per slab containing the lines and with each line associate the polygon above it.

Allocate another search tree on the x-coordinates of the vertical lines

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Planar point location -- example

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Solving planar point location (Cont.)

To answer a query:

first find the appropriate slab

then search the slab to find the polygon

Query time is O(log n)

How about the space ?

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Planar point location -- bad example

Total # lines O(n), and number of lines in each slab is O(n).

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Planar point location & persistence

So how do we improve the space bound ?

Key observation: The lists of the lines in adjacent slabs are very similar.

Create the search tree for the first slab.

Then obtain the next one by deleting the lines that end at the corresponding vertex and adding the lines that start at that vertex

How many insertions/deletions are there all together ?

2n (One insertion and one deletion per segment)

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Planar point location & persistence (cont)

Updates should be persistent since we need all search trees at the end.

Partial persistence is enough.

Well, we already have the path copying method, lets use it.What do we get ?

O(n log n) space and O(n log n) preprocessing time.

Using the node-copying method, we can improve the space bound to O(n).

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Making data structures fully persistent

• With this type of persistence the versions don't form a simple linear path, they form

a version tree (since you can also update in the past). Lack of linear ordering.

• Impose a total ordering on the versions (version list)

• The version list defines a preorder on the version tree (for navigation): for any

version i, the descendants of i in the version tree occur consecutively in the version

list, starting with i.

0

1

2

3

4

5

6 7

version list:

A version tree

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Making data structures fully persistent

iA iC

iG iA iM

iI

0

1

2

3

4

6

7 8

iK

dE

10

11

iM9

dM

iO

5

12

iE

Search tree versions:

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Full persistence

It must be possible to:

• perform insertions in the version list and

• given two versions i and j, determine whether i precedes or follows j in the version

list

This list order problem has been addressed by Dietz and Sleator

• order queries are answered in O(1) worst case time with an O(1) amortized time

bound for insertion

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Fat node method - full persistence

• Each fat node contains same fields as ephemeral node plus space for extra fields

(each with a field name and a version stamp)

• Each field in a node contains a list of version-value pairs

Access

• Versions are compared with respect to their position in the version list, not with

respect to their numeric values

• Access a field in version i: search for the version stamp rightmost in version list,

but not to the right of i

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Fat node method - Example

iA iC

iG iA iM

iI

0

1

2

3

4

6

7 8

iK

dE

10

11

iM9

dM

iO

5

12

iE

Version list: 1,6,7,10,11,2,8,9,3,4,5,12

E

A C

A

G

K

M M

I

O

1-10,12

1-10, 12

A fully persistent search tree

11 6 2 7 9

3

4

12

5

10

112

8

2

33

2

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Fat node method - full persistence

Update operation i

• Add i to the version list

• Update step creates new node: create new fat node with original field values (stamp i)

• Update step changes a field f: we have to guarantee that the new value of f will be used only in version i

Time cost per access and update step

• O(log m), provided each set of field values is stored in a search tree, ordered by version stamp

Space cost

• Worst-case space cost per update step is O(1)

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Applications

Partially persistent balanced search trees

• give a simple solution to the planar point location problem, the grounded 2-dimensional range searching problem, …

• can be used as a substitute for Chazelle‘s hive graph (geometric retrieval)

Fully persistent data structures can be used for

• the binary dispatching problem (OO – languages: find for a invocation the most specific applicable method)

• text editing

Oblivious data structures

• cryptography

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References

• J. R. Driscoll, N. Sarnak, D. D. Sleator, and R. E. Tarjan: Making data structures

persistent. Journal of Computer and System Sciences, 38:86-124, 1989. Final

version.

• N. Sarnak, R. E. Tarjan. Planar Point Location Using Persistent Search Trees:

Communications of the ACM,29:669 – 679, July 1986.

• D. Micciancio: Oblivious Data Structures: Applications to Cryptography.1997.