Lecture 2

Question 1

What new idea is introduced in sGA compared to general EA format?

Answer 1

Parent selection before variation

Question 2

What kind of Parent Selection does sGA have?

Answer 2

Proportionate selection:
Meaning that each parent has its own probability to be selected as a parent.

Question 3

What scoring system is used to determine the parents in sGA?

Answer 3

The individuals with higher fitness have higher probability of being select as parent.

Question 4

What are the three issues with proportionate selection in sGA?

Answer 4

• Required: fitness maximization and non-negative fitness
• Far-above average individuals quickly dominate population
• Similar fitness (convergence) dissolves selection pressure

Question 5

What type of variations are used in sGA?

Answer 5

One point, two point or uniform- crossover.

Question 6

How is the next generation of population selected in sGA?

Answer 6

surivor selection: replace population with offspring

Question 7

describe psudo code for sGA

Answer 7

1 𝑡 ← 0
2 𝑷𝑡← 𝑐𝑟𝑒𝑎𝑡𝑒𝐴𝑛𝑑𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒𝐼𝑛𝑖𝑡𝑖𝑎𝑙𝐼𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙𝑠 𝑛
3 𝒘𝒉𝒊𝒍𝒆 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝐶𝑟𝑖𝑡𝑒𝑟𝑖𝑜𝑛𝑁𝑜𝑡𝑆𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑(𝑷𝑡) 𝒅𝒐
3.1 𝑺𝑡 ← 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑡𝑒𝑆𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛𝑃𝑎𝑟𝑒𝑛𝑡𝑠(𝑷𝑡, 𝑛)
3.2 𝑶𝑡 ← ∅
3.3 𝝅 ← 𝑟𝑎𝑛𝑑𝑜𝑚𝑃𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛(𝑛)
3.4 𝒇𝒐𝒓 𝑖 ∈ {0,1, … , (𝑛/2) − 1} 𝒅𝒐
3.4.1 𝑶𝑡 ← 𝑶𝑡 ∪ 𝑐𝑟𝑜𝑠𝑠𝑜𝑣𝑒𝑟𝐴𝑛𝑑𝑀𝑢𝑡𝑎𝑡𝑖𝑜𝑛(𝑆𝝅2𝑖𝑡 , 𝑆𝝅2𝑖+1𝑡 , 𝑝𝑐, 𝑝𝑚)
3.5 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒𝐴𝑙𝑙𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠(𝑶𝑡)
3.6 𝑷𝑡+1 ← 𝑶𝑡
3.7 𝑡 ← 𝑡 + 1

Question 8

Why do we require a schema?

Answer 8

To investigate the dynamics of sGA so that we can determine if the GA will work well.

Question 9

What is the domain of an Schema?

Answer 9

Is a string (vector) of length l that contains 0,1 or ⋆

Question 10

What does a Schema represent?

Answer 10

Schema represents subset of all ℓ-dimensional binary genotypes

Question 11

What is an other common name for Schema?

Answer 11

a similarity subset:
– Set of genotypes that are similar according to schema

Question 12

What does it mean in schemas when: schema 𝒉 iff 𝒙 ∈ 𝒉

Answer 12

That genotype/solution 𝒙 has matched with schema 𝒉

Question 13

What is the general purpose of Schema Analysis?

Answer 13

– Observe change in average fitness of a schema
in a single generation
– Extrapolate result over multiple generations
– Gives a prediction of the behavior of the average
schema fitness

Question 14

In Schema Analysis, what does 𝜇(𝒉, 𝑡) (mu) denote?

Answer 14

The number of matches of schema 𝒉 in generation t in population P^t.

Question 15

In Schema Analysis, what does 𝜑(𝒉, 𝑡) (phi) denote?

Answer 15

The population-based cumulative fitness of a schema:
The sum of the fitness values of all genotypes in 𝑷^t that matched with schema 𝒉 in generation t.

Question 16

How do we calculate population-based average schema fitness?

Answer 16

Cumulative fitness over number of matches for a certain schema:
HAT! 𝜑(𝒉, 𝑡) = 𝜑(𝒉, 𝑡) / 𝜇(𝒉, 𝑡)

Question 17

What does ☆ mean in Shema analysis?

Answer 17

A unique schema of length ℓ, that only includes ⋆.
☆ = ⋆⋆..⋆⋆

(Basically: it includes all solutions. )

Question 18

How can you express proportionate selection probability using Schema Analysis?

Answer 18

The fitness of that Individual over the cumulative fitness of the total population.
p = fitness(P^t_i) / 𝜑(☆, 𝑡)

Question 19

In Schema Analysis, what does 𝜓(𝒉, 𝑡) (psi) mean?

Answer 19

The probability of a selection being matched by 𝒉 is 𝜓(𝒉, 𝑡). Which is denoted by :
𝜑(𝒉, 𝑡) / 𝜑(☆, 𝑡)

Question 20

What is the expected number of copies of instances of schema 𝒉 in the next generation without variation? (equation)

Answer 20

𝑛 𝜓(𝒉, 𝑡) = 𝜇(𝒉, 𝑡) HAT 𝜑(𝒉,𝑡) / HAT 𝜑(☆,𝑡)

Question 21

What can you conclude in Schema Analysis if population-based average schema
fitness remains larger than the average population fitness?

Answer 21

then # matches of 𝒉
grows exponentially

Question 22

How do we denote Cumulative Population fitness?

Answer 22

𝜑(☆, 𝑡)

Question 23

How do we denote population-based average schema fitness?

Answer 23

HAT 𝜑(𝒉, 𝑡)

Question 24

How do we denote average
.population fitness

Answer 24

HAT 𝜑(☆, 𝑡)

Question 25

What does 𝜖(𝒉) denote?

Answer 25

probability of schema disruption when applying variation.

Question 26

What does disruption mean in schema analysis?

Answer 26

at least one parent ** matches** a schema
but the offspring do not match that schema

Question 27

What does the schama growth equation tell us?

Answer 27

The expected number of copies of instances of schema 𝒉 in the next generation with variation!

Question 28

What is the schema growth equation?

Answer 28

𝛦[𝜇(𝒉, 𝑡 + 1) ≥ (1 − 𝜖(𝒉)) * 𝜇(𝒉, 𝑡) * HAT𝜑(𝒉, t) / HAT 𝜑(☆, 𝑡)

Question 29

When the expected number of times a schema appears in
the population increases, what do we expect to happen to the probability of schema disruption?

Answer 29

It becomes as small as possble. Because we maximize the (1 − 𝜖(𝒉)) term in our formula.

Question 30

Why is the problem that the average population fitness catches up with the schema fitness, not really a problem?

Answer 30

By the time this happens, most genotypes have already converged to match that highly-fit schema.

Question 31

When does the probability of disruption increase in Schema Analysis?

Answer 31

– with defining length of schema under study (i.e., distance
between outer fixed bits)
– and with order of schema (number of fixed bits in schema)

Question 32

Why do we need multiple runs of the same algorithm?

Answer 32

The random initial population and the randomness in variationa and mutations means that this algorithm is stochastic. A singular run can also lead to local optimalisations, instead of the optimal solution.

Question 33

What kind of selection is Tournament selection?

Answer 33

Rank/dominate based

Question 34

What is the advantage of rank/domination based selection?

Answer 34

It removes the dependency on the actual fitness value.

Question 35

What does a Tournament Selection with replacement mean?

Answer 35

Where individuals can be parents to multiple offspring. Due to the fixed number of tournaments, this can mean that some individuals do not get the chance to create offspring.

Question 36

What does a Tournament Selection without replacement mean?

Answer 36

Every individual becomes a parent, before individuals are assigned a second time to become a parent. This makes sure all individuals in a population are concidered during selection.

Question 37

How often is a solution considered in tournament selection without replacement?

Answer 37

(sn)/N times.
Where s is the size of each tournament, n is the amount of resulting solutions, and N is the population set.

Question 38

How many tournaments are needed to select n solutions in a generation for Tournament Selection?

Answer 38

n solutions = n tournaments

You only select the best solution each time.

Question 39

Does high selection pressure result in faster or slower convergence?

Answer 39

faster

Question 40

Is it easy to exert high selection pressure in tournament selection?

Answer 40

Yes. By selecting only the best solution per tournament, it increases the pressure that only the best are selected.

Question 41

What is the formula for the Tournament Selection Takeover Time
?

Answer 41

t ≈ log(𝑛 log(𝑛)) /
log(𝑠)
where 𝑛 is the number of tournaments per generation (often equal to pop size)
and 𝑠 is tournament-size

Question 42

Why is population size almost irrelevant for the efficiency in tournament selection?

Answer 42

The takeover time scales logrithmic, hence have less impact on.

Question 43

What does the P+O setting mean in Tournament selection?

Answer 43

That the tournament will select the best one out of two individuals and their offspring.

Furthermore, the best solution gets 2 copies in next gen

Question 44

Why is the P+O setting favorable in tournament selection?

Answer 44

**Ensures elitism: **
Best solution found so far does not get lost (due to variation)

Question 45

Why does the best ind. get 2 copies in a P+O setting?

Answer 45

This makes convergens to (local) optimum faster.

Question 46

What is the biggest downside of high(er) mutation probability?

Answer 46

It slows down overall convergence.

Question 47

Does increasing tournament selection size also increase selection presure?

Answer 47

Yes, a individual now needs to be (relatively) better to more individuals in order to be the best.

Question 48

How does a uniqueness constraint on genes influence the efficiency of sGA’s? and how do we solve this?

Answer 48

Because the genetype now consists of unique genes, crossover is not (necessarily) possible anymore. This breaks sGA.

We solve this by introducing a repair operators.

Question 49

What is the def. of a permutation?

Answer 49

the action of changing the arrangement, especially the linear order, of a set of items.

Question 50

What is a repair operator and how does it work?

Answer 50

An extra operation for your algorithm that makes sure that permutation problems stay in the feasible solution space.

It detects clashes during crossover and replaces problems with unused numbers. Both the clashes and unused numbers are picked in random order.

Question 51

What is a permutation-specific operator?

Answer 51

An improved version of a repair operator, which aims to remove clashes by the means of new variators, specifically for permutations.

Question 52

Explain Davis’s order crossover (OX)

Answer 52

• 2 parent permutations, 1 offspring
• Copy a part from the first parent.
• Read the rest of the second parent and omit elements that are already present in offspring
• after cut, wrap around to the start.

Question 53

Explain Partially Mapped Crossover (PMX)

Answer 53

– 2 parent permutations, 1 offspring
– Copy part from first parent to offspring
– Now the second parent will look at the indexes of the already copied genes, and sees what its own genes are.
- if these genes don’t match (which is very likely) it will place its own gene from that index, at the index of what the offsprings gene is.

9 0 7 2 8 6 5 3 1 4
2 5 7 9 4 6 0 3 8 1
→
. . . 2 8 6 5 . . .
(now: 9 needs to go
where 2 was)
→
9 . . 2 8 6 5 . . .
(now: 4 needs to go
where 8 was)

Question 54

Explain Edge crossover (EX)

Answer 54

– 2 parent permutations, 1 offspring
- First,for each element list elements it is connected to (neighbours) in either parent.
- Pick a random element as your starter
- Then, pick the next element from the adjacency list from the previous selected element. Pick the one with the shortest adjecency list. (make sure you omit elements that already have been used.)
- if no neighbours are still available, pick a random element to continue with.
- Repeat until all elements are present in offpsring.

Question 55

Explain Bean’s generic solution to permutations

Answer 55

• 2 parents 1 offspring.
• For each parent, assign a (random) real value (0,1) to each gene to indicate it’s index in that genotype.
• Now the standard crossovers are available to use on the default genotype, to ensure all elements are present.
• after crossover, you can sort the elements in the genotype based on their index value