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Partial Ordering. Lecture 11: Oct 17. (based on slides in MIT 6.042). Representing Relations as Graphs. Given a set A, a binary relation R on A is a subset of pairs of A. For example, in a social network, the set A = set of people. - PowerPoint PPT Presentation
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Partial Ordering
Lecture 11: Oct 17 (based on slides in MIT 6.042)
Representing Relations as Graphs
Given a set A, a binary relation R on A is a subset of pairs of A.
For example, in a social network, the set A = set of people.
John
relation R = friendship, which is a symmetric relationship.
Mary
Tom
DavidSam
A symmetric relationship can be represented by an
undirected graph.
For equivalence relationship, the graph has a special structure.
Equivalence Relationship
For example, let A = set of positive integers, and
relationship R be “the same remainder when divided by 3”.
1
4
7
2
5
8
3
6
9
Each connected component is a complete subgraph.
Asymmetric relations is represented by a directed graph.
Asymmetric Relationship
Miami
Chicago
LA
BostonNew York
For example, set A = set of basketball teams
relation R = whether team a beats team b
The resulting directed graph has no special structure.
Asymmetric “Ordering” Relationship
For some asymmetric relations, there is an “ordering” between the objects.
For example, the set A = all subsets of {x,y,z},
the relation R = whether X is a proper subset of Y
• Asymmetry
• Transitivity
No directed cycles.
Partial Order
A partial order is a binary relation which satisfies the following properties.
(1) Asymmetry: aRb implies (bRa) for all a,b A
(2) Transitivity: aRb and bRc implies aRc for all a,b,c A.
a
b
a
b
c
Example of Partial Order
Directed graph representation Partial order representation
When there is a directed path from X to Y, it is
understood that it implies there is a relation (X,Y) by
transitivity.
Example of Partial Order
Directed graph representation Partial order representation
1
2
3
4
1
2
3
4
Let A = set of positive integers
relation R = “is less than”
a divides b iff ka = b for some k
2
10
3 5
15
30
More Example: Divide
1 {1}
2 {1,2}
3 {1,3}
5 {1,5}
15{1,3,5,15}
10 {1,2,5,10}
30 {1,2,3,5,10,15,30}
Subset from Divide
The divide relation
can also be captured
by the subset relation.
Definition. A binary relation, R, on set A,
is a strict partial order iff
it is transitive and
asymmetric.
Strict Partial Order
In a strict partial ordering, can we have aRa?
Weak partial order is the same as
strict partial order except
aRa for all a A (reflexivity)
NO, because of asymmetry.
e.g. “less than equal”,
“subset”, etc.
Total Order
Definition. A pair (a,b) is incomparable if neither aRb nor bRa
A total order is a partial order with no incomparable pairs.
That is, in a total order,
either aRb or bRa for all ab A
e.g. ({x,y},{y,z}) is incomparable,
({x},{z}) is incomparable, etc.
1. ≤ on the Integers
2. < on the Reals
3. on Sets (subset)
4. on Sets (proper subset)
5. Divide
6. Properly divide
Some Partial Orders
Which are total?
Which are strict?
1,2
2,4,6
subject c is a direct prerequisite for subject d
c→d
Subject Prerequisites
18.01 → 6.042 → 6.046 → 6.840
18.01 is indirect prereq. of 6.840
18.01 → 6.042 → 6.046 → 6.840
To find a feasible ordering of the courses
we need to start with a course with no prerequisite.
Partial order representation.
Directed graph representation.
d is minimal: no “smaller” element
d is minimum: d is “smaller” than everything
. ( )c d c d
.c d d c
Minimal vs Minimum
minimum means "smallest” – a prereq. for every subject
Directed Acyclic Graph
Does a partial order always have a minimal element?
18.01 → 6.042 → 6.046 → 6.840
The directed graph representation of a partial order is a directed acyclic graph.
violate asymmetry
A directed cycle wouldDirected graph representation.
Existence of a Minimal Element
What is a minimal element in the directed acyclic graph?
A vertex with indegree zero.
Does a directed acyclic graph always have an indegree zero vertex?
Start from any vertex v. Always go backward.
Since there is no directed cycle,
there must be a indegree zero vertex.
Fact. There is a vertex with indegree zero in a directed acyclic graph.
Fact. There is a minimal element in a partial order.
Does a partial order always have a minimal element?
v
18.02
18.01
6.046
6.840
18.03
6.001
6.034
6.003
8.01
8.02
6.002
6.004
6.033
6.857
6.042
Course Schedule
Partial order representation.
Antichain: Set of subjects with no prereqs among them
-- can be taken in any order. (said to be incomparable)
Antichain
In the directed graph representation, an antichain corresponds
to a set of vertices with no edges between them.
18.02
18.01
6.046
6.840
18.03
6.001
6.034
6.003
8.01
8.02
6.002
6.004
6.033
6.857
6.042
may have other
antichains
Some Antichains
Chain: Set of successive prereqs
-- must be taken in order.
(subjects said to be comparable)
Chain
In the directed graph representation,
a chain corresponds to a directed path.
18.02
18.01
6.046
6.840
18.03
6.001
6.034
6.003
8.01
8.02
6.002
6.004
6.033
6.857
6.042
Some Chains
18.02
18.01
6.046
6.840
18.03
6.001
6.034
6.003
8.01
8.02
6.002
6.004
6.033
6.857
6.042
Maximum Length Chain
also known as a critical pathHow many terms to graduate?
6 terms are necessary
and sufficient (but may need to take many courses per term...)
18.02
18.01
6.046
6.840
18.03
6.001
6.034
6.003
8.01
8.02
6.002
6.004
6.033
6.857
6.042
Minimum Length Schedule
Finding minimum schedule =
Partition into minimum
number of antichains
Graph Colouring
minimum length schedule >= maximum length chain
Can we always find a schedule with length = maximum length chain?
This is a graph colouring problem.
Minimum colouring
Maximum path length
In the directed graph representation, an antichain corresponds
to a set of vertices with no edges between them.
An antichain can be
coloured by the same
colour
Graph Colouring
Theorem. In a directed acyclic graph,
minimum colouring <= maximum path length.
1. Let k be the maximum path length in G.
2. Let M be the set of vertices with indegree 0.
3. Consider G-M.
4. The maximum path length in G-M is k-1.
5. By induction, G-M can be coloured using k-1 colours.
6. Use a new colour for vertices in M.
7. We are done.
M
G-M
Corollary. Every partial ordering can be partitioned into k antichains,
where k is equal to the maximum chain length.
Dilworth’s Theorem
Dilworth’s Theorem. For all t > 0, every partially
ordered set with n elements must have either a chain
of size greater than t or an antichain of size at least
n/t.
Corollary. Every partial ordering can be partitioned into k antichains,
where k is equal to the maximum chain length.
(Proof of Dilworth’s theorem)
1. If the maximum chain length is at most t,
2. then the partial ordering can be partitioned into t antichains.
3. Therefore, there is an antichain with at least n/t elements.
Application of Dilworth’s Theorem
For example, if S = (6, 4, 7, 9, 1, 2, 5, 3, 8),
then 647 and 7253 are both subsequences of S.
An increasing subsequence of S is a subsequence
of whose successive elements get larger;
A decreasing subsequence is defined similarly.
Let S be a sequence of n different numbers.
A subsequence of S is a sequence that
can be obtained by deleting elements of S.
A longest increasing subsequence is 1238,
and a longest decreasing subsequence is 641.
Application of Dilworth’s Theorem
Can you find a sequence of 9 numbers with
no increasing subsequence of length 4 and
no deceasing subsequence of length 4?
3, 2, 1, 6, 5, 4, 9, 8, 7
Can you do better?
Claim. For any sequence of 9 numbers,
there must be an increasing sequence of length 3,
or a decreasing sequence of length 3.
Claim. For any sequence of 9 numbers,
there must be an increasing sequence of length 3,
or a decreasing sequence of length 3.
Application of Dilworth’s Theorem
Given any sequence a1, a2,…, a9
Define a partial ordering as follow.
ai aj iff i < j and ai < aj
6 4 7 9 1 2 5 3 8
•A chain = a path = an increasing subsequence.
•An antichain = a set of vertices with no edges = a decreasing
subsequence.
Dilworth’s Theorem. For all t > 0, every partially
ordered set with n elements must have either a chain
of size greater than t or an antichain of size at least
n/t.
Application of Dilworth’s Theorem
6 4 7 9 1 2 5 3 8