graph-theoretic representation of relation composition

graph-theoretic representation of relation composition

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complex

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Complex

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dyadic

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Dyadic

There is another form of representation for 2-adic

relations

that is useful to keep in mind, especially for its ability to render the logic of many complex

formulas

almost instantly understandable to the mind’s eye. This is the representation in terms of

bipartite graphs

(

http://planetmath.org/BipartiteGraph

)

, or

bigraphs

for short.

Here is what

G

and

H

look like in the bigraph picture:

o---------------------------------------o
| . . . . . . . . . . . . . . . . . . . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . . . . . . ./|\. . . . . . . . . . |
| . . . . . . . / | \ . . . . . . .G. . |
| . . . . . . ./. | .\. . . . . . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . . . . . . . . . . . . . . . . . . |
o---------------------------------------o
Figure 9. G = 4:3 + 4:4 + 4:5

o---------------------------------------o
| . . . . . . . . . . . . . . . . . . . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . . . . . .\. | ./. . . . . . . . . |
| . . . . . . . \ | / . . . . . . .H. . |
| . . . . . . . .\|/. . . . . . . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . . . . . . . . . . . . . . . . . . |
o---------------------------------------o
Figure 10. H = 3:4 + 4:4 + 5:4

These graphs may be read to say:

G

puts

4

in relation to

3

,

4

,

5

.

H

puts

3

,

4

,

5

in relation to

4

.

To form the composite relation (

http://planetmath.org/CompositionOfRelations

)

G

∘

H

, one simply follows the bigraph for

G

by the bigraph for

H

, here arranging the bigraphs in order down the page, and then treats any non-empty set of paths of length two between two nodes as being

equivalent

to a single directed edge between those nodes in the composite bigraph for

G

∘

H

.

Here’s how it looks in pictures:

o---------------------------------------o
| . . . . . . . . . . . . . . . . . . . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . . . . . . ./|\. . . . . . . . . . |
| . . . . . . . / | \ . . . . . . .G. . |
| . . . . . . ./. | .\. . . . . . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . . . . . .\. | ./. . . . . . . . . |
| . . . . . . . \ | / . . . . . . .H. . |
| . . . . . . . .\|/. . . . . . . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . . . . . . . . . . . . . . . . . . |
o---------------------------------------o
Figure 11. G Followed By H

o---------------------------------------o
| . . . . . . . . . . . . . . . . . . . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . . . . . . . | . . . . . . . . . . |
| . . . . . . . . | . . . . . . .G o H. |
| . . . . . . . . | . . . . . . . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . . . . . . . . . . . . . . . . . . |
o---------------------------------------o
Figure 12. G Composed With H

Once again we find that

G

∘

H

=

4

:

4

.

To see the utility of the bigraph picture of 2-adic relations, let us devise a slightly more complex example of a

composition

problem, and use it to illustrate the logic of the

matrix multiplication

formula.

Keeping to the same space

X

=

{

1

,

2

,

3

,

4

,

5

,

6

,

7

}

, define the 2-adic relations

M

,

N

⊆

X

×

X

as follows:

M

=

2

:

1

+

2

:

2

+

2

:

3

+

4

:

3

+

4

:

4

+

4

:

5

+

6

:

5

+

6

:

6

+

6

:

7

N

=

1

:

1

+

2

:

1

+

3

:

3

+

4

:

3

+

4

:

5

+

5

:

5

+

6

:

7

+

7

:

7

Here are the bigraph pictures:

o---------------------------------------o
| . . . . . . . . . . . . . . . . . . . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . . ./|\. . ./|\. . ./|\. . . . . . |
| . . . / | \ . / | \ . / | \ . . .M. . |
| . . ./. | .\./. | .\./. | .\. . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . . . . . . . . . . . . . . . . . . |
o---------------------------------------o
Figure 13. Dyadic Relation M

o---------------------------------------o
| . . . . . . . . . . . . . . . . . . . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . | ./. . | ./.\. | . .\. | . . . . |
| . . | / . . | / . \ | . . \ | . .N. . |
| . . |/. . . |/. . .\| . . .\| . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . . . . . . . . . . . . . . . . . . |
o---------------------------------------o
Figure 14. Dyadic Relation N

To form the composite relation

M

∘

N

, one simply follows the bigraph for

M

by the bigraph for

N

, here arranging the bigraphs in order down the page, and then counts any non-empty set of paths of length two between two nodes as being equivalent to a single directed edge between those two nodes in the composite bigraph for

M

∘

N

.

Here’s how it looks in pictures:

o---------------------------------------o
| . . . . . . . . . . . . . . . . . . . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . . ./|\. . ./|\. . ./|\. . . . . . |
| . . . / | \ . / | \ . / | \ . . .M. . |
| . . ./. | .\./. | .\./. | .\. . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . | ./. . | ./ \. | . .\. | . . . . |
| . . | / . . | / . \ | . . \ | . .N. . |
| . . |/. . . |/. . .\| . . .\| . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . . . . . . . . . . . . . . . . . . |
o---------------------------------------o
Figure 15. M Followed By N

o---------------------------------------o
| . . . . . . . . . . . . . . . . . . . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . . ./.\. . ./.\. . ./.\. . . . . . |
| . . . / . \ . / . \ . / . \ . .M o N. |
| . . ./. . .\./. . .\./. . .\. . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . . . . . . . . . . . . . . . . . . |
o---------------------------------------o
Figure 16. M Composed With N

Let us hark back to that mysterious matrix multiplication formula, and see how it appears in the light of the bigraph representation.

The

coefficient

of the composition

M

∘

N

between

i

and

j

in

X

is given as follows:

(

M

∘

N

)

i

⁢

j

=

∑

k

M

i

⁢

k

⁢

N

k

⁢

j

.

Graphically interpreted, this is a

sum over paths

. Starting at the node

i

,

M

i

⁢

k

being 1 indicates that there is an edge in the bigraph of

M

from node

i

to node

k

, and

N

k

⁢

j

being 1 indicates that there is an edge in the bigraph of

N

from node

k

to node

j

. So the

∑

k

ranges over all possible intermediaries

k

, ascending from 0 to 1 just as soon as there happens to be some path of length two between nodes

i

and

j

.

It is instructive at this point to compute the other possible composition that can be formed from

M

and

N

, namely, the composition

N

∘

M

, that takes

M

and

N

in the opposite order. Here is the graphic

computation

:

o---------------------------------------o
| . . . . . . . . . . . . . . . . . . . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . | ./. . | ./.\. | . .\. | . . . . |
| . . | / . . | / . \ | . . \ | . .N. . |
| . . |/. . . |/. . .\| . . .\| . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . . ./|\. . ./|\. . ./|\. . . . . . |
| . . . / | \ . / | \ . / | \ . . .M. . |
| . . ./. | .\./. | .\./. | .\. . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . . . . . . . . . . . . . . . . . . |
o---------------------------------------o
Figure 17. N Followed By M

o---------------------------------------o
| . . . . . . . . . . . . . . . . . . . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . . . . . . . . . . . . . . . . . . |
| . . . . . . . . . . . . . . . .N o M. |
| . . . . . . . . . . . . . . . . . . . |
| . . o . o . o . o . o . o . o . .X. . |
| . . 1 . 2 . 3 . 4 . 5 . 6 . 7 . . . . |
| . . . . . . . . . . . . . . . . . . . |
o---------------------------------------o
Figure 18. N Composed With M

In sum,

N

∘

M

=

0

. This example affords

sufficient

evidence that

relational composition

, just like its kindred, matrix multiplication, is a non-commutative algebraic operation.

Title

graph-theoretic representation of relation composition

Canonical name

GraphtheoreticRepresentationOfRelationComposition

Date of creation

2013-03-22 17:50:36

Last modified on

2013-03-22 17:50:36

Owner

Jon Awbrey (15246)

Last modified by

Jon Awbrey (15246)

Numerical id

9

Author

Jon Awbrey (15246)

Entry type

Example

Classification

msc 68P15

Classification

msc 08A02

Classification

msc 05C65

Classification

msc 05B30

Classification

msc 03E20

Classification

msc 05B20

Classification

msc 03B10

Classification

msc 68R01

Related topic

AlgebraicRepresentationOfRelationComposition

Related topic

GeometricRepresentationOfRelationComposition

Related topic

MatrixRepresentationOfRelationComposition

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