1

1. Write down the permutations that are the automorphisms of this graph.

2. Write the cycle structure notation for each of the automorphisms.

3. How many independent sets of order 2 does it have?

2

Two independent sets S and T are equivalent if there is an automorphism of G mapping the vertices of S to those in T.

Given a graph G with automorphism group order gand an independent set S such that G has k automorphisms mapping the independent set S to itself, there will be g/k different independent sets of the graph that correspond to S.

Usually the minimum of these is chosen to be the canonical representative of its equivalence class.

3

1. Which automorphisms map this independent set to itself?

2. Which other independent sets are equivalent to it?

4

4 * 3 = 12

5

Sorting these lexicographically:{0, 3} < {1,4} < {2,5}

6

1. Which automorphisms map this independent set to itself?

2. Which other independent sets are equivalent to it?

72 * 6 = 12

8

Sorting these lexicographically:{0, 2} < {0, 4} < {1, 3} < {1, 5} < {2,4} < {3,5}

9

The independent sets of order two fall into two equivalence classes. We could choose the minimum one in each equivalence class as the representative for its class:

10

For each of these two embeddings, apply clockwise BFS starting at the red vertex with first child the green vertex and direction clockwise.

11

0 (2): 1 21 (3): 0 2 32 (3): 0 4 13 (3): 1 4 54 (3): 2 5 35 (2): 3 4

0 (2): 1 21 (2): 0 2 32 (3): 0 4 13 (3): 1 5 44 (3): 2 3 55 (2): 3 4

Lexicographically smaller.

12

0 (2): 1 21 (3): 0 2 32 (3): 0 4 13 (3): 1 4 54 (3): 2 5 35 (2): 3 4

Lexicographically smaller.

0 (3): 1 2 31 (3): 0 4 52 (2): 0 33 (3): 0 2 44 (3): 1 3 55 (2): 1 4

13

0 (2): 1 21 (3): 0 2 32 (3): 0 4 13 (3): 1 4 54 (3): 2 5 35 (2): 3 4

Lexicographically smaller

0 (2): 1 21 (3): 0 3 22 (3): 0 1 43 (3): 1 5 44 (3): 2 3 55 (2): 3 4

14

0 (2): 1 21 (3): 0 2 32 (3): 0 4 13 (3): 1 4 54 (3): 2 5 35 (2): 3 4

0 (2): 1 21 (3): 0 3 22 (3): 0 1 43 (3): 1 5 44 (3): 2 3 55 (2): 3 4

Face walking: 1 2 0

Face walking:1 2 4 3

15

0 (2): 1 21 (3): 0 2 32 (3): 0 4 13 (3): 1 4 54 (3): 2 5 35 (2): 3 4

0 (2): 1 21 (3): 0 2 32 (3): 0 4 13 (3): 1 4 54 (3): 2 5 35 (2): 3 4

Automorphism since the same.

16

To get the canonical form of an embedding:

For each of the n choices of the root r do For each of the degree(r) choices of the first child f do For each choice d of direction do Do a clockwise BFS with root r first child f and direction d

Choose the lex. smallest as the canonical form.

17

To get automorphisms of an embedding:

For each of the n choices of the root r do For each of the degree(r) choices of the first child f do For each choice d of direction do Do a clockwise BFS with root r first child f and direction d

If the rotation system matches the canonical form the BFI labelling is an automorphism.

Alan Turing Celebration Lecture Biological Evolution as a Form of Learning Leslie Valiant, CC and Applied Math., Harvard Wed. Oct 10, 3:30pm, ECS 124

Living organisms function according to protein circuits. Darwin's theory of evolution suggests that these circuits have evolved through variation guided by natural selection. However, the question of which circuits can so evolve in realistic population sizes and within realistic numbers of generations has remained essentially unaddressed.

We suggest that computational learning theory offers the framework for investigating this question, of how circuits can come into being adaptively from experience, without a designer. We formulate evolution as a form of learning from examples. The targets of the learning process are the functions of highest fitness. The examples are the experiences. The learning process is constrained so that the feedback from the experiences is Darwinian. We formulate a notion of evolvability that distinguishes function classes that are evolvable with polynomially bounded resources from those that are not. The dilemma is that if the function class, say for the expression levels of proteins in terms of each other, is too restrictive, then it will not support biology, while if it is too expressive then no evolution algorithm will exist to navigate it.

CSC 482/582: Project proposal slides deadline extended to Oct. 11.

You should be working on a literature review and planning your methodology.

Assignment #1 has been posted: due Thurs. Oct. 18 at the beginning of class.Assignments may be handed in up to 5 business days after the official due date with a 2% penalty for each business day that the assignment is overdue.

Graph Isomorphism

The graph isomorphism problem has no known polynomial time algorithm which works for an arbitrary graph.

Canonical form: If two graphs are isomorphic, their canonical forms must be the same, otherwise, they must be different.

For trees and planar graphs, a canonical form can be computed in polynomial time. 21

The ( ) Canonical Form for a Tree

22

Label each vertex with the string ( ). While more than two vertices remain do

Locate all the leaves. Remove each leaf placing its label

within the ( and ) of the parent so that within the ( ) of the parent, the strings

representing the children are sorted.End while

If there is one vertex left, the canonical form of the tree is the label of this vertex.

If there are two vertices u and v left, the canonical form of T is label(u) concatenated with label(v) where label(u) ≤ label(v).

23

( )

( )

( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( )

( )

( )

( )

( ) ( )( )

( )

Label each vertex with ( ) .

24

( )

( )

( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( )

( )

( )

( )

( ) ( )( )

( )

Identify the leaves.25

(( ))

(( )( )( ))

( )

(( )( )( ))

(( )( )) (( )( ))

(( ))

Put labels of leaves inside ( ) of their neighbours making sure strings are in sorted order.

26

(( ))

(( )( )( ))

( )

(( )( )( ))

(( )( )) (( )( ))

(( ))

27

(( ))

(( )( )( ))

( )

(( )( )( ))

(( )( )) (( )( ))

(( ))

Identify the leaves.28

( (( )) ( ) ( ) ( )) ( (( )( )) (( )( )) ( ) )

((( )( )( )))

Put labels of leaves inside ( ) of their neighbours making sure strings are in sorted order.

29

( (( )) ( ) ( ) ( )) ( (( )( )) (( )( )) ( ) )

((( )( )( )))

30

( (( )) ( ) ( ) ( )) ( (( )( )) (( )( )) ( ) )

((( )( )( )))

31

( ( (( )( )) (( )( )) ( ) )

( (( )) ( ) ( ) ( ))

(( )( )( )) )

CANONICAL FORM FOR THIS TREE:

( ( (( )( )) (( )( )) ( ) ) ( (( )) ( ) ( ) ( ))(( )( )( )) )

32

( ( (( )( )) (( )( )) ( ) )

( (( )) ( ) ( ) ( ))

(( )( )( )) )

To reconstruct the tree:

Factor what is in each ( ) into minimal well-parenthesized strings.

33

( (( )( )) (( )( )) ( ) )

To reconstruct the tree: Add a leaf for each well-parenthesized string w with label w.

( (( )) ( ) ( ) ( ))

(( )( )( ))

34

( (( )( )) (( )( )) ( ) )

( (( )) ( ) ( ) ( ) )

( ( ) ( ) ( ) )

Factor what is in each ( ) into minimal well-parenthesized strings.

35

( (( )( )) (( )( )) ( ) )( (( )) ( ) ( ) ( ) )

( ( ) ( ) ( ) )

Add a leaf for each one.36

With the labels on them.

( ( ) )

( )

( )

( )

( ) ( ) ( )

(( )( )) (( )( ))

( )

37

( ( ) )

( )

( )

( )

( ) ( ) ( )

(( )( )) (( )( ))

( )

38

( ) ( ) ( ) ( ) ( )

39

40