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# 07.2 Holland's Genetic Algorithms Schema Theorem

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One of the fewest Evolutionary algorithms with proof about the Expected number of parents for a certain Schema. The slides have been updated with a better proof, however, the proof still have some problems... I seriously believe that we need a topology stochastic process to really understand what is going on in Genetic Algorithms. This quite tough because of mixing of topology and probability to define a realistic model of populations in Genetic Algorithms.

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### 07.2 Holland's Genetic Algorithms Schema Theorem

1. 1. Artiﬁcial Intelligence Holland’s GA Schema Theorem Andres Mendez-Vazquez April 7, 2015 1 / 37
2. 2. Outline 1 Introduction Schema Deﬁnition Properties of Schemas 2 Probability of a Schema Probability of an individual is in schema H Surviving Under Gene wise Mutation Surviving Under Single Point Crossover The Schema Theorem A More General Version Problems with the Schema Theorem 2 / 37
3. 3. Outline 1 Introduction Schema Deﬁnition Properties of Schemas 2 Probability of a Schema Probability of an individual is in schema H Surviving Under Gene wise Mutation Surviving Under Single Point Crossover The Schema Theorem A More General Version Problems with the Schema Theorem 3 / 37
4. 4. Introduction Consider the Canonical GA Binary alphabet. Fixed length individuals of equal length, l. Fitness Proportional Selection. Single Point Crossover (1X). Gene wise mutation i.e. mutate gene by gene. Deﬁnition 1 - Schema H A schema H is a template that identiﬁes a subset of strings with similarities at certain string positions. Schemata are a special case of a natural open set of a product topology. 4 / 37
5. 5. Introduction Consider the Canonical GA Binary alphabet. Fixed length individuals of equal length, l. Fitness Proportional Selection. Single Point Crossover (1X). Gene wise mutation i.e. mutate gene by gene. Deﬁnition 1 - Schema H A schema H is a template that identiﬁes a subset of strings with similarities at certain string positions. Schemata are a special case of a natural open set of a product topology. 4 / 37
6. 6. Introduction Consider the Canonical GA Binary alphabet. Fixed length individuals of equal length, l. Fitness Proportional Selection. Single Point Crossover (1X). Gene wise mutation i.e. mutate gene by gene. Deﬁnition 1 - Schema H A schema H is a template that identiﬁes a subset of strings with similarities at certain string positions. Schemata are a special case of a natural open set of a product topology. 4 / 37
7. 7. Introduction Consider the Canonical GA Binary alphabet. Fixed length individuals of equal length, l. Fitness Proportional Selection. Single Point Crossover (1X). Gene wise mutation i.e. mutate gene by gene. Deﬁnition 1 - Schema H A schema H is a template that identiﬁes a subset of strings with similarities at certain string positions. Schemata are a special case of a natural open set of a product topology. 4 / 37
8. 8. Introduction Consider the Canonical GA Binary alphabet. Fixed length individuals of equal length, l. Fitness Proportional Selection. Single Point Crossover (1X). Gene wise mutation i.e. mutate gene by gene. Deﬁnition 1 - Schema H A schema H is a template that identiﬁes a subset of strings with similarities at certain string positions. Schemata are a special case of a natural open set of a product topology. 4 / 37
9. 9. Introduction Consider the Canonical GA Binary alphabet. Fixed length individuals of equal length, l. Fitness Proportional Selection. Single Point Crossover (1X). Gene wise mutation i.e. mutate gene by gene. Deﬁnition 1 - Schema H A schema H is a template that identiﬁes a subset of strings with similarities at certain string positions. Schemata are a special case of a natural open set of a product topology. 4 / 37
10. 10. Introduction Consider the Canonical GA Binary alphabet. Fixed length individuals of equal length, l. Fitness Proportional Selection. Single Point Crossover (1X). Gene wise mutation i.e. mutate gene by gene. Deﬁnition 1 - Schema H A schema H is a template that identiﬁes a subset of strings with similarities at certain string positions. Schemata are a special case of a natural open set of a product topology. 4 / 37
11. 11. Introduction Deﬁnition 2 If A denotes the alphabet of genes, then A ∪ ∗ is the schema alphabet, where * is the ‘wild card’ symbol matching any gene value. Example: For A ∈ {0, 1, ∗} where ∗ ∈ {0, 1}. 5 / 37
12. 12. Example The Schema H = [0 1 ∗ 1 *] generates the following individuals 0 1 * 1 * 0 1 0 1 0 0 1 0 1 1 0 1 1 1 0 0 1 1 1 1 Not all schemas say the same Schema [ 1 ∗ ∗ ∗ ∗ ∗ ∗] has less information than [ 0 1 ∗ ∗ 1 1 0]. It is more!!! [ 1 ∗ ∗ ∗ ∗ ∗ 0] span the entire length of an individual, but [ 1 ∗ 1 ∗ ∗ ∗ ∗] does not. 6 / 37
13. 13. Example The Schema H = [0 1 ∗ 1 *] generates the following individuals 0 1 * 1 * 0 1 0 1 0 0 1 0 1 1 0 1 1 1 0 0 1 1 1 1 Not all schemas say the same Schema [ 1 ∗ ∗ ∗ ∗ ∗ ∗] has less information than [ 0 1 ∗ ∗ 1 1 0]. It is more!!! [ 1 ∗ ∗ ∗ ∗ ∗ 0] span the entire length of an individual, but [ 1 ∗ 1 ∗ ∗ ∗ ∗] does not. 6 / 37
14. 14. Example The Schema H = [0 1 ∗ 1 *] generates the following individuals 0 1 * 1 * 0 1 0 1 0 0 1 0 1 1 0 1 1 1 0 0 1 1 1 1 Not all schemas say the same Schema [ 1 ∗ ∗ ∗ ∗ ∗ ∗] has less information than [ 0 1 ∗ ∗ 1 1 0]. It is more!!! [ 1 ∗ ∗ ∗ ∗ ∗ 0] span the entire length of an individual, but [ 1 ∗ 1 ∗ ∗ ∗ ∗] does not. 6 / 37
15. 15. Outline 1 Introduction Schema Deﬁnition Properties of Schemas 2 Probability of a Schema Probability of an individual is in schema H Surviving Under Gene wise Mutation Surviving Under Single Point Crossover The Schema Theorem A More General Version Problems with the Schema Theorem 7 / 37
16. 16. Schema Order and Length Deﬁnition 3 - Schema Order o (H) Schema order, o, is the number of non “*” genes in schema H. Example: o(***11*1)=3. Deﬁnition 3 – Schema Deﬁning Length, δ (H). Schema Deﬁning Length, δ(H), is the distance between ﬁrst and last non “*” gene in schema H. Example: δ(***11*1)=7-4=3. Notes Given an alphabet A with|A| = k, then there are (k + 1)l possible schemas of length l. 8 / 37
17. 17. Schema Order and Length Deﬁnition 3 - Schema Order o (H) Schema order, o, is the number of non “*” genes in schema H. Example: o(***11*1)=3. Deﬁnition 3 – Schema Deﬁning Length, δ (H). Schema Deﬁning Length, δ(H), is the distance between ﬁrst and last non “*” gene in schema H. Example: δ(***11*1)=7-4=3. Notes Given an alphabet A with|A| = k, then there are (k + 1)l possible schemas of length l. 8 / 37
18. 18. Schema Order and Length Deﬁnition 3 - Schema Order o (H) Schema order, o, is the number of non “*” genes in schema H. Example: o(***11*1)=3. Deﬁnition 3 – Schema Deﬁning Length, δ (H). Schema Deﬁning Length, δ(H), is the distance between ﬁrst and last non “*” gene in schema H. Example: δ(***11*1)=7-4=3. Notes Given an alphabet A with|A| = k, then there are (k + 1)l possible schemas of length l. 8 / 37
19. 19. Outline 1 Introduction Schema Deﬁnition Properties of Schemas 2 Probability of a Schema Probability of an individual is in schema H Surviving Under Gene wise Mutation Surviving Under Single Point Crossover The Schema Theorem A More General Version Problems with the Schema Theorem 9 / 37
20. 20. Probabilities of belonging to a Schema H What do we want? The probability that individual h is from schema H: P (h ∈ H) We need the following probabilities Pdistruption(H, 1X) = probability of schema being disrupted due to crossover. Pdisruption (H, mutation) =probability of schema being disrupted due to mutation Pcrossover (H survive) 10 / 37
21. 21. Probabilities of belonging to a Schema H What do we want? The probability that individual h is from schema H: P (h ∈ H) We need the following probabilities Pdistruption(H, 1X) = probability of schema being disrupted due to crossover. Pdisruption (H, mutation) =probability of schema being disrupted due to mutation Pcrossover (H survive) 10 / 37
22. 22. Probabilities of belonging to a Schema H What do we want? The probability that individual h is from schema H: P (h ∈ H) We need the following probabilities Pdistruption(H, 1X) = probability of schema being disrupted due to crossover. Pdisruption (H, mutation) =probability of schema being disrupted due to mutation Pcrossover (H survive) 10 / 37
23. 23. Probability of Disruption Consider now The CGA using ﬁtness proportionate parent selection on-point crossover (1X) bitwise mutation with probability Pm Genotypes of length l The Schema could be disrupted if the cross over falls between the ends Pdistruption(H, 1X) = δ(H) (l − 1) (1) 0 1 0 0 1 0 11 / 37
24. 24. Probability of Disruption Consider now The CGA using ﬁtness proportionate parent selection on-point crossover (1X) bitwise mutation with probability Pm Genotypes of length l The Schema could be disrupted if the cross over falls between the ends Pdistruption(H, 1X) = δ(H) (l − 1) (1) 0 1 0 0 1 0 11 / 37
25. 25. Probability of Disruption Consider now The CGA using ﬁtness proportionate parent selection on-point crossover (1X) bitwise mutation with probability Pm Genotypes of length l The Schema could be disrupted if the cross over falls between the ends Pdistruption(H, 1X) = δ(H) (l − 1) (1) 0 1 0 0 1 0 11 / 37
26. 26. Probability of Disruption Consider now The CGA using ﬁtness proportionate parent selection on-point crossover (1X) bitwise mutation with probability Pm Genotypes of length l The Schema could be disrupted if the cross over falls between the ends Pdistruption(H, 1X) = δ(H) (l − 1) (1) 0 1 0 0 1 0 11 / 37
27. 27. Probability of Disruption Consider now The CGA using ﬁtness proportionate parent selection on-point crossover (1X) bitwise mutation with probability Pm Genotypes of length l The Schema could be disrupted if the cross over falls between the ends Pdistruption(H, 1X) = δ(H) (l − 1) (1) 0 1 0 0 1 0 11 / 37
28. 28. Probability of Disruption Consider now The CGA using ﬁtness proportionate parent selection on-point crossover (1X) bitwise mutation with probability Pm Genotypes of length l The Schema could be disrupted if the cross over falls between the ends Pdistruption(H, 1X) = δ(H) (l − 1) (1) 0 1 0 0 1 0 11 / 37
29. 29. Why? Given that you have δ(H) = the distance between ﬁrst and last non “*” last position in Genotype - ﬁrst position in Genotype = l − 1 Case I δ(H) = 1, when the positions of the non “*” are next to each other Case II δ(H) = l − 1, when the positions of the non “*” are in the extremes 12 / 37
30. 30. Why? Given that you have δ(H) = the distance between ﬁrst and last non “*” last position in Genotype - ﬁrst position in Genotype = l − 1 Case I δ(H) = 1, when the positions of the non “*” are next to each other Case II δ(H) = l − 1, when the positions of the non “*” are in the extremes 12 / 37
31. 31. Why? Given that you have δ(H) = the distance between ﬁrst and last non “*” last position in Genotype - ﬁrst position in Genotype = l − 1 Case I δ(H) = 1, when the positions of the non “*” are next to each other Case II δ(H) = l − 1, when the positions of the non “*” are in the extremes 12 / 37
32. 32. Why? Given that you have δ(H) = the distance between ﬁrst and last non “*” last position in Genotype - ﬁrst position in Genotype = l − 1 Case I δ(H) = 1, when the positions of the non “*” are next to each other Case II δ(H) = l − 1, when the positions of the non “*” are in the extremes 12 / 37
33. 33. Remarks about Mutation Observation about Mutation Mutation is applied gene by gene. In order for schema H to survive, all non * genes in the schema much remain unchanged. Thus Probability of not changing a gene 1 − Pm (Pm probability of mutation). Probability of requiring that all o(H) non * genes survive, (1 − Pm)o(H) . Typically the probability of applying the mutation operator, pm 1. The probability that the mutation disrupt the schema H Pdisruption (H, mutation) = 1 − (1 − Pm)o(H) ≈ o (H) Pm (2) After ignoring high terms in the polynomial!!! 13 / 37
34. 34. Remarks about Mutation Observation about Mutation Mutation is applied gene by gene. In order for schema H to survive, all non * genes in the schema much remain unchanged. Thus Probability of not changing a gene 1 − Pm (Pm probability of mutation). Probability of requiring that all o(H) non * genes survive, (1 − Pm)o(H) . Typically the probability of applying the mutation operator, pm 1. The probability that the mutation disrupt the schema H Pdisruption (H, mutation) = 1 − (1 − Pm)o(H) ≈ o (H) Pm (2) After ignoring high terms in the polynomial!!! 13 / 37
35. 35. Remarks about Mutation Observation about Mutation Mutation is applied gene by gene. In order for schema H to survive, all non * genes in the schema much remain unchanged. Thus Probability of not changing a gene 1 − Pm (Pm probability of mutation). Probability of requiring that all o(H) non * genes survive, (1 − Pm)o(H) . Typically the probability of applying the mutation operator, pm 1. The probability that the mutation disrupt the schema H Pdisruption (H, mutation) = 1 − (1 − Pm)o(H) ≈ o (H) Pm (2) After ignoring high terms in the polynomial!!! 13 / 37
36. 36. Remarks about Mutation Observation about Mutation Mutation is applied gene by gene. In order for schema H to survive, all non * genes in the schema much remain unchanged. Thus Probability of not changing a gene 1 − Pm (Pm probability of mutation). Probability of requiring that all o(H) non * genes survive, (1 − Pm)o(H) . Typically the probability of applying the mutation operator, pm 1. The probability that the mutation disrupt the schema H Pdisruption (H, mutation) = 1 − (1 − Pm)o(H) ≈ o (H) Pm (2) After ignoring high terms in the polynomial!!! 13 / 37
37. 37. Remarks about Mutation Observation about Mutation Mutation is applied gene by gene. In order for schema H to survive, all non * genes in the schema much remain unchanged. Thus Probability of not changing a gene 1 − Pm (Pm probability of mutation). Probability of requiring that all o(H) non * genes survive, (1 − Pm)o(H) . Typically the probability of applying the mutation operator, pm 1. The probability that the mutation disrupt the schema H Pdisruption (H, mutation) = 1 − (1 − Pm)o(H) ≈ o (H) Pm (2) After ignoring high terms in the polynomial!!! 13 / 37
38. 38. Remarks about Mutation Observation about Mutation Mutation is applied gene by gene. In order for schema H to survive, all non * genes in the schema much remain unchanged. Thus Probability of not changing a gene 1 − Pm (Pm probability of mutation). Probability of requiring that all o(H) non * genes survive, (1 − Pm)o(H) . Typically the probability of applying the mutation operator, pm 1. The probability that the mutation disrupt the schema H Pdisruption (H, mutation) = 1 − (1 − Pm)o(H) ≈ o (H) Pm (2) After ignoring high terms in the polynomial!!! 13 / 37
39. 39. Outline 1 Introduction Schema Deﬁnition Properties of Schemas 2 Probability of a Schema Probability of an individual is in schema H Surviving Under Gene wise Mutation Surviving Under Single Point Crossover The Schema Theorem A More General Version Problems with the Schema Theorem 14 / 37
40. 40. Gene wise Mutation Lemma 1 Under gene wise mutation (Applied Gene by Gene), the (lower bound) probability of an order o(H) schema H surviving at generation (No Disruption) t is, 1 − o (H) Pm (3) 15 / 37
41. 41. Probability of an individual is sampled from schema H Consider the Following 1 Probability of selection depends on 1 Number of instances of schema H in the population. 2 Average ﬁtness of schema H relative to the average ﬁtness of all individuals in the population. Thus, we have the following probability P (h ∈ H) = PUniform (h in Population) × Mean Fitness Ratio 16 / 37
42. 42. Probability of an individual is sampled from schema H Consider the Following 1 Probability of selection depends on 1 Number of instances of schema H in the population. 2 Average ﬁtness of schema H relative to the average ﬁtness of all individuals in the population. Thus, we have the following probability P (h ∈ H) = PUniform (h in Population) × Mean Fitness Ratio 16 / 37
43. 43. Probability of an individual is sampled from schema H Consider the Following 1 Probability of selection depends on 1 Number of instances of schema H in the population. 2 Average ﬁtness of schema H relative to the average ﬁtness of all individuals in the population. Thus, we have the following probability P (h ∈ H) = PUniform (h in Population) × Mean Fitness Ratio 16 / 37
44. 44. Probability of an individual is sampled from schema H Consider the Following 1 Probability of selection depends on 1 Number of instances of schema H in the population. 2 Average ﬁtness of schema H relative to the average ﬁtness of all individuals in the population. Thus, we have the following probability P (h ∈ H) = PUniform (h in Population) × Mean Fitness Ratio 16 / 37
45. 45. Then Finally P (h ∈ H) = (Number of individuals matching schema H at generation t) (Population Size) × (Mean ﬁtness of individuals matching schema H) (Mean ﬁtness of individuals in the population) (4) 17 / 37
46. 46. Then Finally P (h ∈ H) = m (H, t) f (H, t) Mf (t) (5) where M is the population size and m(H, t) is the number of instances of schema H at generation t. Lemma 2 Under ﬁtness proportional selection the expected number of instances of schema H at time t is E [m (H, t + 1)] = M · P (h ∈ H) = m (H, t) f (H, t) f (t) (6) 18 / 37
47. 47. Then Finally P (h ∈ H) = m (H, t) f (H, t) Mf (t) (5) where M is the population size and m(H, t) is the number of instances of schema H at generation t. Lemma 2 Under ﬁtness proportional selection the expected number of instances of schema H at time t is E [m (H, t + 1)] = M · P (h ∈ H) = m (H, t) f (H, t) f (t) (6) 18 / 37
48. 48. Why? Note the following M independent samples (Same Probability) are taken to create the next generation of parents Thus m (H, t + 1) = Ih1 + Ih2 + ... + IhM Remark: The indicator random variable of ONE for these samples!!! Then E [m (H, t + 1)] = E [Ih1 ] + E [Ih2 ] + ... + E [IhM ] 19 / 37
49. 49. Why? Note the following M independent samples (Same Probability) are taken to create the next generation of parents Thus m (H, t + 1) = Ih1 + Ih2 + ... + IhM Remark: The indicator random variable of ONE for these samples!!! Then E [m (H, t + 1)] = E [Ih1 ] + E [Ih2 ] + ... + E [IhM ] 19 / 37
50. 50. Why? Note the following M independent samples (Same Probability) are taken to create the next generation of parents Thus m (H, t + 1) = Ih1 + Ih2 + ... + IhM Remark: The indicator random variable of ONE for these samples!!! Then E [m (H, t + 1)] = E [Ih1 ] + E [Ih2 ] + ... + E [IhM ] 19 / 37
51. 51. Finally But, M samples are taken to create the next generation of parents E [m (H, t + 1)] = P (h1 ∈ H) + P (h2 ∈ H) + ... + P (hM ∈ H) Remember the Lemma 5.1 in Cormen’s Book Finally, because P (h1 ∈ H) = P (h2 ∈ H) = ... = P (hM ∈ H) E [m (H, t + 1)] = M × P (h ∈ H) QED!!! 20 / 37
52. 52. Finally But, M samples are taken to create the next generation of parents E [m (H, t + 1)] = P (h1 ∈ H) + P (h2 ∈ H) + ... + P (hM ∈ H) Remember the Lemma 5.1 in Cormen’s Book Finally, because P (h1 ∈ H) = P (h2 ∈ H) = ... = P (hM ∈ H) E [m (H, t + 1)] = M × P (h ∈ H) QED!!! 20 / 37
53. 53. Outline 1 Introduction Schema Deﬁnition Properties of Schemas 2 Probability of a Schema Probability of an individual is in schema H Surviving Under Gene wise Mutation Surviving Under Single Point Crossover The Schema Theorem A More General Version Problems with the Schema Theorem 21 / 37
54. 54. Search Operators – Single point crossover Observations Crossover was the ﬁrst of two search operators introduced to modify the distribution of schema in the population. Holland concentrated on modeling the lower bound alone. 22 / 37
55. 55. Search Operators – Single point crossover Observations Crossover was the ﬁrst of two search operators introduced to modify the distribution of schema in the population. Holland concentrated on modeling the lower bound alone. 22 / 37
56. 56. Crossover Consider the following Generated Individual h = 1 0 1 | 1 1 0 0 H1 = * 0 1 | * * * 0 H2 = * 0 1 | * * * * Crossover Remarks 1 Schema H1 will naturally be broken by the location of the crossover operator unless the second parent is able to ‘repair’ the disrupted gene. 2 Schema H2 emerges unaﬀected and is therefore independent of the second parent. 3 Thus, Schema with long deﬁning length are more likely to be disrupted by single point crossover than schema using short deﬁning lengths. 23 / 37
57. 57. Crossover Consider the following Generated Individual h = 1 0 1 | 1 1 0 0 H1 = * 0 1 | * * * 0 H2 = * 0 1 | * * * * Crossover Remarks 1 Schema H1 will naturally be broken by the location of the crossover operator unless the second parent is able to ‘repair’ the disrupted gene. 2 Schema H2 emerges unaﬀected and is therefore independent of the second parent. 3 Thus, Schema with long deﬁning length are more likely to be disrupted by single point crossover than schema using short deﬁning lengths. 23 / 37
58. 58. Crossover Consider the following Generated Individual h = 1 0 1 | 1 1 0 0 H1 = * 0 1 | * * * 0 H2 = * 0 1 | * * * * Crossover Remarks 1 Schema H1 will naturally be broken by the location of the crossover operator unless the second parent is able to ‘repair’ the disrupted gene. 2 Schema H2 emerges unaﬀected and is therefore independent of the second parent. 3 Thus, Schema with long deﬁning length are more likely to be disrupted by single point crossover than schema using short deﬁning lengths. 23 / 37
59. 59. Crossover Consider the following Generated Individual h = 1 0 1 | 1 1 0 0 H1 = * 0 1 | * * * 0 H2 = * 0 1 | * * * * Crossover Remarks 1 Schema H1 will naturally be broken by the location of the crossover operator unless the second parent is able to ‘repair’ the disrupted gene. 2 Schema H2 emerges unaﬀected and is therefore independent of the second parent. 3 Thus, Schema with long deﬁning length are more likely to be disrupted by single point crossover than schema using short deﬁning lengths. 23 / 37
60. 60. Now, we have Lemma 3 Under single point crossover, the (lower bound) probability of schema H surviving at generation t is, Pcrossover (H survive) =1 − Pcrossover (H does not survive) =1 − pc δ(H) l − 1 Pdiﬀ (H, t) Where Pdiﬀ (H, t) is the probability that the second parent does not match schema H. pc is the a priori selected threshold of applying crossover. 24 / 37
61. 61. Now, we have Lemma 3 Under single point crossover, the (lower bound) probability of schema H surviving at generation t is, Pcrossover (H survive) =1 − Pcrossover (H does not survive) =1 − pc δ(H) l − 1 Pdiﬀ (H, t) Where Pdiﬀ (H, t) is the probability that the second parent does not match schema H. pc is the a priori selected threshold of applying crossover. 24 / 37
62. 62. Now, we have Lemma 3 Under single point crossover, the (lower bound) probability of schema H surviving at generation t is, Pcrossover (H survive) =1 − Pcrossover (H does not survive) =1 − pc δ(H) l − 1 Pdiﬀ (H, t) Where Pdiﬀ (H, t) is the probability that the second parent does not match schema H. pc is the a priori selected threshold of applying crossover. 24 / 37
63. 63. How? We can see the following Pcrossover (H does not survive) = Pc × Pdistruption(H, 1X) × Pdiﬀ (H, t) After all Pc is used to decide if the crossover will happen. The second parent could come from the same schema, and yes!!! We do not have a disruption!!! Then Pcrossover (H does not survive) = Pc × δ(H) l − 1 × Pdiﬀ (H, t) 25 / 37
64. 64. How? We can see the following Pcrossover (H does not survive) = Pc × Pdistruption(H, 1X) × Pdiﬀ (H, t) After all Pc is used to decide if the crossover will happen. The second parent could come from the same schema, and yes!!! We do not have a disruption!!! Then Pcrossover (H does not survive) = Pc × δ(H) l − 1 × Pdiﬀ (H, t) 25 / 37
65. 65. How? We can see the following Pcrossover (H does not survive) = Pc × Pdistruption(H, 1X) × Pdiﬀ (H, t) After all Pc is used to decide if the crossover will happen. The second parent could come from the same schema, and yes!!! We do not have a disruption!!! Then Pcrossover (H does not survive) = Pc × δ(H) l − 1 × Pdiﬀ (H, t) 25 / 37
66. 66. How? We can see the following Pcrossover (H does not survive) = Pc × Pdistruption(H, 1X) × Pdiﬀ (H, t) After all Pc is used to decide if the crossover will happen. The second parent could come from the same schema, and yes!!! We do not have a disruption!!! Then Pcrossover (H does not survive) = Pc × δ(H) l − 1 × Pdiﬀ (H, t) 25 / 37
67. 67. In addition Worst case lower bound Pdiﬀ (H, t) = 1 (7) 26 / 37
68. 68. Outline 1 Introduction Schema Deﬁnition Properties of Schemas 2 Probability of a Schema Probability of an individual is in schema H Surviving Under Gene wise Mutation Surviving Under Single Point Crossover The Schema Theorem A More General Version Problems with the Schema Theorem 27 / 37
69. 69. The Schema Theorem The Schema Theorem The expected number of schema H at generation t + 1 when using a canonical GA with proportional selection, single point crossover and gene wise mutation (where the latter are applied at rates pc and Pm) is, E [m (H, t + 1)] ≥ m (H, t) f (H, t) f (t) 1 − pc δ(H) l − 1 Pdiﬀ (H, t) − o (H) Pm (8) 28 / 37
70. 70. Proof We use the following quantities Pcrossover (H survive) = 1 − pc δ(H) l−1 Pdiﬀ (H, t) ≤ 1 Pno−disruption (H, mutation) = 1 − o (H) Pm ≤ 1 Then, we have that E [m (H, t + 1)] =M × P (h ∈ H) =M m (H, t) f (H, t) Mf (t) = m (H, t) f (H, t) f (t) ≥ m (H, t) f (H, t) f (t) × 1 − pc δ(H) l − 1 Pdiﬀ (H, t) × [1 − o (H) Pm] 29 / 37
71. 71. Proof We use the following quantities Pcrossover (H survive) = 1 − pc δ(H) l−1 Pdiﬀ (H, t) ≤ 1 Pno−disruption (H, mutation) = 1 − o (H) Pm ≤ 1 Then, we have that E [m (H, t + 1)] =M × P (h ∈ H) =M m (H, t) f (H, t) Mf (t) = m (H, t) f (H, t) f (t) ≥ m (H, t) f (H, t) f (t) × 1 − pc δ(H) l − 1 Pdiﬀ (H, t) × [1 − o (H) Pm] 29 / 37
72. 72. Proof We use the following quantities Pcrossover (H survive) = 1 − pc δ(H) l−1 Pdiﬀ (H, t) ≤ 1 Pno−disruption (H, mutation) = 1 − o (H) Pm ≤ 1 Then, we have that E [m (H, t + 1)] =M × P (h ∈ H) =M m (H, t) f (H, t) Mf (t) = m (H, t) f (H, t) f (t) ≥ m (H, t) f (H, t) f (t) × 1 − pc δ(H) l − 1 Pdiﬀ (H, t) × [1 − o (H) Pm] 29 / 37
73. 73. Thus We have the following E [m (H, t + 1)] ≥ m (H, t) f (H, t) f (t) 1 − pc δ(H) l − 1 Pdiﬀ (H, t) − o (H) Pm + ... pc δ(H) l − 1 Pdiﬀ (H, t)o (H) Pm ≥ m (H, t) f (H, t) f (t) 1 − pc δ(H) l − 1 Pdiﬀ (H, t) − o (H) Pm The las inequality is possible because pc δ(H) l−1 Pdiﬀ (H, t)o (H) Pm ≥ 0 30 / 37
74. 74. Remarks Observations The theorem is described in terms of expectation, thus strictly speaking is only true for the case of a population with an inﬁnite number of members. What about a ﬁnite population? In the case of ﬁnite population sizes the signiﬁcance of population drift plays an increasingly important role. 31 / 37
75. 75. Remarks Observations The theorem is described in terms of expectation, thus strictly speaking is only true for the case of a population with an inﬁnite number of members. What about a ﬁnite population? In the case of ﬁnite population sizes the signiﬁcance of population drift plays an increasingly important role. 31 / 37
76. 76. Remarks Observations The theorem is described in terms of expectation, thus strictly speaking is only true for the case of a population with an inﬁnite number of members. What about a ﬁnite population? In the case of ﬁnite population sizes the signiﬁcance of population drift plays an increasingly important role. 31 / 37
77. 77. Outline 1 Introduction Schema Deﬁnition Properties of Schemas 2 Probability of a Schema Probability of an individual is in schema H Surviving Under Gene wise Mutation Surviving Under Single Point Crossover The Schema Theorem A More General Version Problems with the Schema Theorem 32 / 37
78. 78. More General Version More General Version E [m (H, t + 1)] ≥ m (H, t) α (H, t) {1 − β(H, t)} (9) Where α(H, t)is the “selection coeﬃcient” β(H, t) is the “transcription error.” This allows to say that H survives if α(H, t) ≥ 1 − β (H, t) or m (H, t) f (H, t) f (t) ≥ 1 − pc δ(H) l − 1 Pdiﬀ (H, t) − o (H) Pm 33 / 37
79. 79. More General Version More General Version E [m (H, t + 1)] ≥ m (H, t) α (H, t) {1 − β(H, t)} (9) Where α(H, t)is the “selection coeﬃcient” β(H, t) is the “transcription error.” This allows to say that H survives if α(H, t) ≥ 1 − β (H, t) or m (H, t) f (H, t) f (t) ≥ 1 − pc δ(H) l − 1 Pdiﬀ (H, t) − o (H) Pm 33 / 37
80. 80. More General Version More General Version E [m (H, t + 1)] ≥ m (H, t) α (H, t) {1 − β(H, t)} (9) Where α(H, t)is the “selection coeﬃcient” β(H, t) is the “transcription error.” This allows to say that H survives if α(H, t) ≥ 1 − β (H, t) or m (H, t) f (H, t) f (t) ≥ 1 − pc δ(H) l − 1 Pdiﬀ (H, t) − o (H) Pm 33 / 37
81. 81. More General Version More General Version E [m (H, t + 1)] ≥ m (H, t) α (H, t) {1 − β(H, t)} (9) Where α(H, t)is the “selection coeﬃcient” β(H, t) is the “transcription error.” This allows to say that H survives if α(H, t) ≥ 1 − β (H, t) or m (H, t) f (H, t) f (t) ≥ 1 − pc δ(H) l − 1 Pdiﬀ (H, t) − o (H) Pm 33 / 37
82. 82. Observation Observation This is the basis for the observation that short (deﬁning length), low order schema of above average population ﬁtness will be favored by canonical GAs, or the Building Block Hypothesis. 34 / 37
83. 83. Outline 1 Introduction Schema Deﬁnition Properties of Schemas 2 Probability of a Schema Probability of an individual is in schema H Surviving Under Gene wise Mutation Surviving Under Single Point Crossover The Schema Theorem A More General Version Problems with the Schema Theorem 35 / 37
84. 84. Problems Problem 1 Only the worst-case scenario is considered. No positive eﬀects of the search operators are considered. This has lead to the development of Exact Schema Theorems. Problem 2 The theorem concentrates on the number of schema surviving not which schema survive. Such considerations have been addressed by the utilization of Markov chains to provide models of behavior associated with speciﬁc individuals in the population. 36 / 37
85. 85. Problems Problem 1 Only the worst-case scenario is considered. No positive eﬀects of the search operators are considered. This has lead to the development of Exact Schema Theorems. Problem 2 The theorem concentrates on the number of schema surviving not which schema survive. Such considerations have been addressed by the utilization of Markov chains to provide models of behavior associated with speciﬁc individuals in the population. 36 / 37
86. 86. Problems Problem 1 Only the worst-case scenario is considered. No positive eﬀects of the search operators are considered. This has lead to the development of Exact Schema Theorems. Problem 2 The theorem concentrates on the number of schema surviving not which schema survive. Such considerations have been addressed by the utilization of Markov chains to provide models of behavior associated with speciﬁc individuals in the population. 36 / 37
87. 87. Problems Problem 1 Only the worst-case scenario is considered. No positive eﬀects of the search operators are considered. This has lead to the development of Exact Schema Theorems. Problem 2 The theorem concentrates on the number of schema surviving not which schema survive. Such considerations have been addressed by the utilization of Markov chains to provide models of behavior associated with speciﬁc individuals in the population. 36 / 37
88. 88. Problems Problem 1 Only the worst-case scenario is considered. No positive eﬀects of the search operators are considered. This has lead to the development of Exact Schema Theorems. Problem 2 The theorem concentrates on the number of schema surviving not which schema survive. Such considerations have been addressed by the utilization of Markov chains to provide models of behavior associated with speciﬁc individuals in the population. 36 / 37
89. 89. Problems Problem 1 Only the worst-case scenario is considered. No positive eﬀects of the search operators are considered. This has lead to the development of Exact Schema Theorems. Problem 2 The theorem concentrates on the number of schema surviving not which schema survive. Such considerations have been addressed by the utilization of Markov chains to provide models of behavior associated with speciﬁc individuals in the population. 36 / 37
90. 90. Problems Problem 3 Claims of “exponential increases” in ﬁt schema i.e., if the expectation operator of Schema Theorem is ignored and the eﬀects of crossover and mutation discounted, the following result was popularized by Goldberg, m(H, t + 1)≥(1 + c)m(H, t) where c is the constant by which ﬁt schema are always ﬁtter than the population average. PROBLEM!!! Unfortunately, this is rather misleading as the average population ﬁtness will tend to increase with t, thus population and ﬁtness of remaining schema will tend to converge with increasing ‘time’. 37 / 37
91. 91. Problems Problem 3 Claims of “exponential increases” in ﬁt schema i.e., if the expectation operator of Schema Theorem is ignored and the eﬀects of crossover and mutation discounted, the following result was popularized by Goldberg, m(H, t + 1)≥(1 + c)m(H, t) where c is the constant by which ﬁt schema are always ﬁtter than the population average. PROBLEM!!! Unfortunately, this is rather misleading as the average population ﬁtness will tend to increase with t, thus population and ﬁtness of remaining schema will tend to converge with increasing ‘time’. 37 / 37