[PDF] Artificial intelligence 1: informed search





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Schéma bilan des brassages génétiques lors de la méiose GÈNES

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Artificial intelligence 1: informed search

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Schéma bilan des brassages génétiques lors de la méiose GÈNES INDÉPENDANTS prophase I Crossing over M GILLOT Conforme au programme SVT de 2011



[PDF] Schéma dun crossing-over et du brassage intra-chromosomique

Schéma d'un crossing-over et du brassage intra-chromosomique entre 2 gènes situés sur la même paire de chromosome (d'après http://www ac-grenoble fr/)



[PDF] Schéma de méiose Correction 2n = 4 (2 paires de chromosomes) 3

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Artificial intelligence 1: informed search Page

The Fundamental Theorem

Page2 GA

Why does it work?

1.Schema Theorem

2.Search Spaces asHypercubes

Page3

Schema

A Schemais a similarity template describing a

subset of strings with similarities at certain string positions. e.g.For a binary alphabet {0, 1},we motivate a schema by appending a special symbol *, or , producing a ternary alphabet {0, 1, *}that allows us to build schemata. We can think of it as a pattern matching device: a schema matches a particular string if at every location in the schema a 1 matches a 1 in the string, or a 0 matches a 0, or a * matches either. Page

Notation: String, Population

4 Consider strings to be constructed over the binary alphabet

V={0, 1};

Stringsas capital letters

Individual characters by lowercase letters subscripted by their position.

Example:

A = 0111000 may be represented symbolically as:

A= a1a2a3a4a5a6a7

airepresents a gene(binary feature or detector) aivalue represents an allele A(t) represents a population of strings at time (or generation) t. Page

Notation: Schema

5

Consider a schema Htaken from the three-letter

alphabet:

V={ 0, 1, *};

* asterisk is a which matches either a 0or a 1at a particular position. Page6

Schema Matching

A bit stringmatches a particular schemataif that

bit string can be constructed from the schemata by replacing the symbol with the appropriate bit value. e.g.

H = *11*0**

StringA = 0111000

String Ais an example of the schema H

because the string alleles aimatch schema positions hiat the fixed positions

2, 3and 5.

Page Page8

Order of Schema:

o(H) ±is the number of fixed positions present in the template Understanding the building blocks of future solutions

Schema Properties

Schema Order:

o(011*1**) = 4

Schema Defining Length:

į(H) = 5-1 = 4

011*1**

Schema Order:

o(0******) = 1

0******

Schema Defining Length: į(H) = 0, because

there is only one fixed position

Defining Length of Schema:

į(H) ±is the distance between the first

and last specific string position Page9 They provide the basic means for analyzing the net effect of reproductionand genetic operatorson the building blocks contained within the population. Understanding the building blocks of future solutions

Schema Properties

Schemata and their propertiesserve as notational

devices for rigorously discussing and classifying string similarities. Page Page

Effect of Reproductionon Schemata

11 Suppose at time t, there are mexamples of a particular schema Hin population A(t) During reproduction, a string Aigets copied according to its fitness with probability pi= After picking a non-overlapping population of size nwith replacement from the population A(t), we may write the reproductive schema growth equation as: f fi jf

HfntHmtHm)(*),()1,(

),(tHmm f(H) is the average fitness of the strings representing schema Hat time t.

Page12

If we recognise that the average fitness of the entire population as we may express the reproductive schema growth equation as: After picking a non-overlapping population of size nwith replacement from the population A(t), we may write the reproductive schema growth equation as: n ff jf

HfntHmtHm)(*),()1,(

f

HftHmtHm)(*),()1,(

Simplification

Effect of Reproductionon Schemata

Page13

Reproductive schema growth equation:

f

HftHmtHm)(*),()1,(

A particular schema grows as the ratio of the average fitness of the schema to the average fitness of the population. Schemata with fitness values abovethe population average will receive an increasingnumber of samples in the next generation. Schemata with fitness values belowthe population average will receive a decreasingnumber of samples. Allthe schemata in a population grow or decay according to their schema averages under the operation of reproduction alone.

Effect of Reproductionon Schemata

Page14

Reproductive schema growth equation:

f

HftHmtHm)(*),()1,(

Suppose we assume that a particular schema Hremains above average an amount with a cconstant. Under this assumption, we can write:

QuantitaveEffect of Reproductionon Schemata

fc ),(*)1()(*),()1,(tHmcf fcftHmtHm Starting at t=0, and assuming a stationary value of c, we obtain the equation: tcHmtHm)1(*)0,(),( Reproductionallocates exponentially increasing (decreasing) numbers of trials to above (below) average schema. Page Reproductioncan allocate exponentially increasing and decreasing numbers of schemata to future generations in parallel. Many, many different schemata are sampled in parallelaccording to the same rule through the use of nsimple reproduction operations. However, reproductiondoes notpromote exploration of new regions of the search space.

This is where crossoversteps in.

QuantitaveEffect of Reproductionon Schemata

tcHmtHm)1(*)0,(),( Page Consider a particular string of length l= 7 and two representative schemata within that string:

Effect of Crossoveron Schemata

A = 0111000

H1= *1****0

H2= ***10**

Recall:Crossover Operation

crossover proceeds with the random selection of a mate; Random selection of a crossover site, and the exchange of substrings from the beginning of the string to the crossover site inclusively with the corresponding substring of the chosen mate. Page Consider a particular string of length l= 7 and two representative schemata within that string:

Effect of Crossoveron Schemata

Assuming that we have the following randomly chosen crossover site: 3

A = 0 1 1 |1 0 0 0

H1= * 1 * |* * * 0

H2= * * * |1 0 * *

A = 0 1 1 1 0 0 0

H1= * 1 * * * * 0

H2= * * * 1 0 * *

Page

Effect of Crossoveron Schemata

Assuming that we have the following randomly chosen crossover site: 3

A = 0 1 1 |1 0 0 0

H1= * 1 * |* * * 0

H2= * * * |1 0 * *

H1isdestroyed. Defining length = 5

H2will survive. Defining length = 1

H1is less likely to survive crossover than schema H2because on average the crossover site is more likely to fall between the extreme fixed positions. Page

Effect of Crossoveron Schemata

A = 0 1 1 |1 0 0 0

H1= * 1 * |* * * 0

H2= * * * |1 0 * *

H1is less likely to survive crossover than schema H2because on average the crossover site is more likely to fall between the extreme fixed positions. crossover site is selected uniformly at random among the l-1=7-1 = 6 possible sites, then H1is destroyed with probability pdand survives with probability ps.

H1isdestroyed. Defining length = 5

H2will survive. Defining length = 1

6 5 )1( )( l Hpd 6

11 dspp

H1 Page

Effect of Crossoveron Schemata

A = 0 1 1 |1 0 0 0

H1= * 1 * |* * * 0

H2= * * * |1 0 * *

H1is less likely to survive crossover than schema H2because on average the crossover site is more likely to fall between the extreme fixed positions. If the crossover site is selected uniformlyat random among the l-1=7-1 = 6 possible sites. Similarly, you can calculate the probability of destruction and survival for H2as follows:

H1isdestroyed. Defining length = 5

H2will survive. Defining length = 1

6 1 )1( )( l Hpd 6

51 dspp

H2 Page

Lower Bound on CrossoverSurvival Probability

To generalise, a schema survives when the cross over site falls outside the defining length. The survival probability under simple crossover is ps )1( )(1 l Hps

Lower Bound on Crossover

Survival Probability

Page

Lower Bound on CrossoverSurvival Probability

To generalise, a schema survives when the cross over site falls outside the defining length. The survival probability under simple crossover is ps )1( )(1 l Hps If we consider the probability of performing a crossover operation to be pc, )1( )(1quotesdbs_dbs28.pdfusesText_34
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