1 The problem From a given partial order, produce a compatible - - PDF document

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Einführung in die Programmierung Introduction to Programming

• Prof. Dr. Bertrand Meyer

Chair of Softw are Engineering Chair of Softw are Engineering

Lecture 19: Topological sort Part 1: Problem and math basis

by Caran d’Ache

Introduction to Programming, lecture 19: Topological sort

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“Topological sort”

From a given partial order, produce a compatible total order

Introduction to Programming, lecture 19: Topological sort

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compatible total order

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The problem

Partial order: ordering constraints between elements of a set, e.g.

“Remove the dishes before discussing politics” “Walk to Üetliberg before lunch”

From a given partial order, produce a compatible total order

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g

“Take your medicine before lunch” “Finish lunch before removing dishes”

Total order: sequence including all elements of set Compatible: the sequence respects all ordering constraints

Üetliberg, Medicine, Lunch, Dishes, Politics : OK Medicine, Üetliberg, Lunch, Dishes, Politics : OK Politics, Medicine, Lunch, Dishes, Üetliberg : not OK

Why we are doing this!

Very common problem in many different areas Interesting, efficient, non-trivial algorithm Illustration of many algorithmic techniques Illustration of data structures, complexity (big-Oh

notation), and other topics of last lecture

Illustration of software engineering techniques:

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Illustration of software engineering techniques:

from algorithm to component with useful API

Opportunity to learn or rediscover important

mathematical concepts: binary relations (order relations in particular) and their properties

It’s just beautiful!

Today: problem and math basis Next time: detailed algorithm and component

Reading assignment for next Monday

Touch of Class, chapter on topological sort: 17

Introduction to Programming, lecture 19: Topological sort

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Topological sort: example uses

From a dictionary, produce a list of definitions such that no word

• ccurs prior to its definition

Produce a complete schedule for carrying out a set of tasks, subject to ordering constraints (This is done for scheduling

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g maintenance tasks for industrial tasks,

• ften with thousands of constraints)

Produce a version of a class with the features reordered so that no feature call appears before the feature’s declaration

Rectangles with overlap constraints

Constraints: [B, A], [D, A], [A, C], [B, D], [D, C] D C E

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B D A

Rectangles with overlap constraints

Constraints: [B, A], [D, A], [A, C], [B, D], [D, C] Possible solution: D C E B D E A C

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B D A

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The problem

Partial order: ordering constraints between elements of a set, e.g.

“Remove the dishes before discussing politics” “Walk to Üetliberg before lunch”

From a given partial order, produce a compatible total order

Introduction to Programming, lecture 19: Topological sort

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g

“Take your medicine before lunch” “Finish lunch before removing dishes”

Total order: sequence including all elements of set Compatible: the sequence respects all ordering constraints

Üetliberg, Medicine, Lunch, Dishes, Politics : OK Medicine, Üetliberg, Lunch, Dishes, Politics : OK Politics, Medicine, Lunch, Dishes, Üetliberg : not OK

Pictured as a graph

Üetliberg Lunch Dishes Politics

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“Remove the dishes before discussing politics” “Walk to Üetliberg before lunch” “Take your medicine before lunch” “Finish lunch before removing dishes”

Medicine

Sometimes there is no solution

“Introducing recursion requires that

students know about stacks”

“You must discuss abstract data

types before introducing stacks

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types before introducing stacks

“Abstract data types rely on

recursion”

The constraints introduce a cycle

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Overall structure (1)

Given: A type G A set of elements

• f type G

constraints: LIST [TUPLE [G, G ]] class ORDERABLE [G] feature A set of constraints between these elements elements: LIST [G ]

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Required: An enumeration of the elements, in an order compatible with the constraints topsort : LIST [G ] is ... ensure compatible (Result, constraints) end between these elements

Some mathematical background...

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Binary relation on a set

Any property that either holds or doesn’t hold between two elements of a set On a set PERSON of persons, example relations are:

mother : a mother b holds if and only if a is the

mother of b

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father : child : sister : sibling : (brother or sister)

Notation: a r b to express that r holds of a and b.

Note: relation names will appear in green

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Example: the before relation

The set of interest:

Tasks = {Politics, Lunch, Medicine, Dishes, Üetliberg}

“Remove the dishes before discussing politics” “Walk to Üetliberg before lunch” “Take your medicine before lunch” “Finish lunch before removing dishes”

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Tasks {Politics, Lunch, Medicine, Dishes, Üetliberg}

The constraining relation:

Dishes before Politics Üetliberg before Lunch Medicine before Lunch Lunch before Dishes

Some special relations on a set X

universal [X ]: holds between any two elements of X id [X ]: holds between every element of X and itself empty [X ]: holds between no elements of X

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Relations: a more precise mathematical view

We consider a relation r on a set P as: A set of pairs in P x P, containing all the pairs [x, y] such that x r y.

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Then x r y simply means that [x, y] ∈ r

See examples on next slide

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Example: the before relation

The set of interest:

“Remove dishes before discussing politics” “Walk to Üetliberg before lunch” “Take your medicine before lunch” “Finish lunch before removing dishes”

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elements = {Politics, Lunch, Medicine, Dishes, Üetliberg} The constraining relation: before = {[Dishes, Politics], [Üetliberg, Lunch], [Medicine, Lunch], [Lunch, Dishes]}

Using ordinary set operators

spouse = wife ∪ husband sibling = sister ∪ brother ∪ id [Person] sister ⊆ sibling

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g father ⊆ ancestor universal [X ] = X x X (cartesian product) empty [X ] = ∅

Possible properties of a relation

(On a set X. All definitions must hold for every a, b, c... ∈ X.) Total*: (a ≠ b) ⇒ ((a r b) ∨ (b r a)) Reflexive: a r a Irreflexive: not (a r a) S t i b b

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Symmetric: a r b ⇒ b r a Antisymmetric: (a r b) ∧ (b r a) ⇒ a = b Asymmetric: not ((a r b) ∧ (b r a)) Transitive: (a r b) ∧ (b r c) ⇒ a r c

*Definition of “total” is specific to this discussion (there is no standard definition). The other terms are standard.

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Examples (on a set of persons)

sibling

Reflexive, symmetric, transitive

sister

Symmetric, irreflexive Reflexive, antisymmetric

family_head

(a family_head b means a is the head

• f b’s family, with one head per family)

mother

Asymmetric irreflexive

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Total: (a ≠ b) ⇒ (a r b) ∨ (b r a) Reflexive: a r a Irreflexive: not (a r a) Symmetric: a r b ⇒ b r a Antisymmetric: (a r b) ∧ (b r a) ⇒ a = b Asymmetric: not ((a r b) ∧ (b r a)) Transitive: (a r b) ∧ (b r c) ⇒ a r c mother

Asymmetric, irreflexive

Total order relation (strict)

Relation is strict total order if:

• Total
• Irreflexive
• Transitive

Total: (a ≠ b) ⇒ ((a r b) ∨ (b r a)) Irreflexive: not (a r a) Symmetric: a r b ⇒ b r a Asymmetric: not ((a r b) ∧ (b r a)) Transitive: (a r b) ∧ (b r c) ⇒ a r c

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Example: “less than” < on natural numbers

0 < 1 0 < 2, 0 < 3, 0 < 4, ... 1 < 2 1 < 3 1 < 4, ... 2 < 3 2 < 4, ...

...

1 2 3 4 5

Theorem

A strict order is asymmetric (total)

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Total order relation (strict)

Relation is strict total order if:

Total Irreflexive Transitive Total: (a ≠ b) ⇒ ((a r b) ∨ (b r a)) Irreflexive: not (a r a) Symmetric: a r b ⇒ b r a Asymmetric: not ((a r b) ∧ (b r a)) Transitive: (a r b) ∧ (b r c) ⇒ a r c

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Theorem: A strict order is asymmetric total

Total order relation (non-strict)

Relation is non-strict total order if: Total Reflexive Transitive Antisymmetric

Total: (a ≠ b) ⇒ ((a r b) ∨ (b r a)) Irreflexive: not (a r a) Symmetric: a r b ⇒ b r a Antisymmetric: (a r b) ∧ (b r a) ⇒ a = b Transitive:(a r b) ∧ (b r c) ⇒ a r c

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Example: “less than or equal” ≤ on natural numbers

0 ≤ 0 0 ≤ 1, 0 ≤ 2, 0 ≤ 3, 0 ≤ 4, ... 1 ≤ 1 0 ≤ 2, 0 ≤ 3, 0 ≤ 4, ... 1 ≤ 2, 1 ≤ 3, 1 ≤ 4, ... 2 ≤ 2 2 ≤ 3, 2 ≤ 4, ...

...

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Total order relation (strict)

Relation is strict total order if:

Total Irreflexive Transitive Total: (a ≠ b) ⇒ ((a r b) ∨ (b r a)) Irreflexive: not (a r a) Symmetric: a r b ⇒ b r a Antisymmetric: (a r b) ∧ (b r a) ⇒ a = b Transitive:(a r b) ∧ (b r c) ⇒ a r c

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Partial order relation (strict)

Relation is strict partial order if:

Total Irreflexive Transitive Total: (a ≠ b) ⇒ ((a r b) ∨ (b r a)) Irreflexive: not (a r a) Symmetric: a r b ⇒ b r a Antisymmetric: (a r b) ∧ (b r a) ⇒ a = b Transitive:(a r b) ∧ (b r c) ⇒ a r c

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[1, 2]

1 2

b a

[0, 1]

y x

1 2 3

[3, 0]

c

[4, 2]

d Example: relation between points in a plane: p q if both xp < xq yp < yq <

Theorems

A strict order is asymmetric (total) partial

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A total order is a partial order (“partial” order really means possibly partial)

Example partial order

[1, 2]

1 2

b a

[0, 1]

y

[3, 0] [4, 2]

d

p q if both xp < xq yp < yq

<

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Here the following hold: c a No link between a and c, b and c: a c nor e.g. neither

1 2 3 c

a < b a < d c < d < <

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Possible topological sorts

1 2

b a

y

d

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a < b a < d c < d

Topological sort understood

< Here the relation is: {[a, b], [a, d], [c, d]}

1 2

b a

y

d One of the solutions is: b d

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We are looking for a total order relation t such that

⊆ t

< a, b, c, d a < b a < d c < d

Final statement of topological sort problem From a given partial order, produce a compatible total order where:

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A partial order p is compatible with a total order t if and only if p ⊆ t

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From constraints to partial orders

Is a relation defined by a set of constraints, such as constraints = {[Dishes, Politics], [Üetliberg, Lunch], [Medicine, Lunch], [Lunch, Dishes]}

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always a partial order?

Üetliberg Medicine Lunch Dishes Politics

Powers and transitive closure of a relation

r i +1 = r i ; r where ; is composition

Ü P

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Transitive closure r + = r 1 ∪ r 2 ∪ ... always transitive

M L D P r 1 r 2 r 3

Reflexive transitive closure

r i +1 = r i ; r where ; is composition

Ü P

r 0 = id [X ] where X is the underlying set

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Transitive closure r + = r 1 ∪ r 2 ∪ ... always transitive

M L D P r 1 r 2 r 3 r 0

Reflexive transitive closure: r ∗ = r 0 ∪ r 1 ∪ r 2 ∪ ... always transitive and reflexive

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Acyclic relation

r + ∩ id [X ] = ∅

Ü P

A relation r on a set X is acyclic if and only if:

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before + id [X]

M L D P

Acyclic relations and partial orders

Theorems:

Any (strict) order relation is acyclic. A relation is acyclic if and only if its transitive

closure is a (strict) order.

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(Also: if and only if its reflexive transitive closure is a nonstrict partial order)

From constraints to partial orders

The partial order of interest is before + before = {[Dishes, Politics], [Üetliberg, Lunch], [Medicine, Lunch], [Lunch, Dishes]}

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Üetliberg Medicine Lunch Dishes Politics

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Back to software...

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The basic algorithm idea

p q r s t u v w

• Introduction to Programming, lecture 19: Topological sort

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topsort

u

• What we have seen

The topological sort problem and its applications Mathematical background:

Relations as sets of pairs Properties of relations Order relations: partial/total, strict/nonstrict Transitive, reflexive-transitive closures

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,

The relation between acyclic and order relations The basic idea of topological sort

Next: how to do it for

Efficient operation (O (m+n) for m constraints & n

items)

Good software engineering: effective API

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Reading assignment for next Monday

Touch of Class, chapter on topological sort: 17

Introduction to Programming, lecture 19: Topological sort

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End of lecture 19

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