H8xx

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CY3G2 Modern Heuristics
Lecture 8

• Dr. Gillian Walker
• Room 182
• g.c.walker_at_reading.ac.uk

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Problem of last week

• There are four cities located on the vertices of
  a square.
• Design a network of roads such that
• Every city is connected with every other city.
• The total length of road is minimised.

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Solution

• There are many possibilities which satisfy the
  problem
• But each network has a different length.
• The optimal solution isnt obvious and the proof
  of the optimal solution isnt obvious.

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Solution

• Optimal solution
• This is known as Steiners problem and there is
  no known efficient algorithm for solving it.
• However there are many engineering applications
  telephone networking, routing aircraft.

All angles 120
The total length of the network is 1 v3
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Summary

• We have spent the last two weeks studying
  evolutionary algorithms.
• We have seen the advantages to using a population
  of solutions, where we use competition and
  ranking to promote or advance better solutions,
  allowing them to mate and compete for a place in
  new generations of solutions.
• In this lecture we are going to look at how we
  can further develop evolutionary algorithms to
  deal with problem constraints.

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Constraint satisfaction

• Constraint satisfaction problems are problems
  that can be solved by satisfying constraints
• Sudoku
• The 8 queens problem
• In the real world virtually all decision making
  involves constraints.
• What distinguishes different types of problem,
  are the different types of constraints
• Rules
• Data dependencies
• Algebraic expressions

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Feasible and Infeasible

• We have previously discussed two types of
  constraints
• HARD constraints that must be satisfied in order
  to have a feasible solution
• SOFT constraints we hope to accomplish but they
  dont render the solution infeasible.
• From these definitions it becomes apparent that
  constraints divide our search space into
• Areas which contain feasible solutions (satisfy
  all the constraints)
• Areas which contain infeasible solutions
  (solutions which breech one, some or all the hard
  constraints)

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Feasible vs Infeasible solutions
Search space

• Solutions B, D and E are feasible.
• Solutions A, C and F are infeasible.
• In the context of our evolutionary algorithm how
  should we treat the infeasible solutions?

A
E
D
F
C
B
Feasible search space
Infeasible search space
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How to deal with infeasible solutions?

• Having feasible and infeasible solutions can
  affect how we design our evolutionary algorithm.
• What if an infeasible solution scores better with
  our evaluation function than a feasible one?
• We need an evaluation function to treat both
  feasible and infeasible solutions. This is hard.
• In general a solution is to design two evaluation
  functions, evalf and evalu for the feasible and
  infeasible solutions respectively.

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How to design the algorithm?

• How are the functions evalf and evalu related?
  Should we assume that evalf ? evalu (the symbol ?
  is better than so is greater than for
  maximisation problems and smaller than for
  minimisation problems)
• Should we kill off infeasible solutions all
  together? Death Penalty
• Should we repair infeasible solutions by moving
  them to the closest point on the feasible space.
  (Replace them with the closest feasible solution)
• If we repair infeasible solutions should we
  replace the infeasible solution in the population
  or should we only use the repair function for
  evaluations?

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How to design the algorithm?

• As we want to find a feasible solution should we
  penalise infeasible ones?
• Should we change the topology of the search space
  by using decoders that might translate infeasible
  solutions into feasible ones?
• How should we go about finding a feasible
  solution?
• Recent trends concentrate on
• Operators and decoders which impose a restriction
  that any feasible solution is better than any
  infeasible solution.
• Multiobjective optimisation techniques.

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Evalf vs Evalu

• Which is better an infeasible solution that is
  almost the optimum or an unrealistic feasible
  function?
• What is the shortest route from A to B?

Path 1
B
A
Path 2
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Rejecting infeasible solutions

• The death penalty heuristic is popular in
  evolutionary algorithms.
• Kill off all the infeasible solutions at each
  step.
• There is no need to separately evaluate feasible
  and infeasible solutions.
• The death penalty heuristic works well when the
  feasible search space is convex and fills a
  regional proportion of the search space.
• Otherwise the approach has problems
• What if you randomly sampled the search space for
  your initial population and all solutions were
  infeasible?

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Repairing infeasible solutions Pros

• Repairing infeasible solutions is common in
  evolutionary computation, especially in
  combinatorial optimisation problems (TSP,
  knapsack problem)
• In these cases it is easy to repair infeasible
  solutions to make them feasible.
• The repaired version is used either
• In evaluation
• Replaces the infeasible solution in the
  population.
• The process of repairing infeasible solutions is
  related to a combination of learning and
  evolution (Baldwin effect).
• Learning in this case is as a result of a local
  search to find the nearest feasible solution. The
  improvement is translated to the individual.
• A local search is equivalent to the learning
  during a single generation.

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Repairing infeasible solutions Cons

• The weakness of this approach lies in the problem
  dependence.
• Different repair procedures have to be designed
  for each individual problem.
• There is no standard heuristic, you could use
• A greedy heuristic
• A random approach.
• In some cases repairing infeasible solutions
  becomes as complicated as solving the problem
  itself! (Often in nonlinear transporting
  problems, scheduling and timetabling.)

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Replacing individuals by their repaired versions

• Replacing infeasible solutions with their
  repaired counterpart relates to Lamarckian
  evolution.
• Lamarckian assumed that a person improves during
  their lifetime and that these improvements are
  coded back into the genetics, and can be passed
  to the next generation.
• The most common replacement strategy was reported
  by Orvosh and Davis and is known as the 5 rule.
  In combinatorial mathematics problems the best
  results were achieved when 5 of the repaired
  solutions replaced the original solutions.
• In other algorithms (GENOCOP III) 15 was
  optimum. It is dependent on the problem.

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Penalizing infeasible solutions

• The most common way to deal with infeasible
  solutions is to penalize for their infeasibility,
  either by
• Awarding a penalty for the infeasibility
• Introducing a cost for repairing a solution.
• The relationship between the infeasible solution
  and the feasible part of the search space plays a
  significant role in how the penalizing occurs.
• An appropriate penalty function should consider
  the ease of repairing a function as well as the
  quality of the repaired version.

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Decoders

• Another reasonable heuristic for dealing with the
  issue of feasibility is to use specialized
  representations and variation operators to
  maintain the feasibility of the solutions in the
  population.
• One possibility is a decoder where the solution
  encodes to an instruction about how to get to a
  solution, rather than being the solution.
• Think of a knapsack problem which was interpreted
  as Take an item if possible. This would always
  lead to a feasible solution.
• Each decoder imposes a relationship between a
  feasible solution and a decoded solution.

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Decoders

• It is important that several conditions are
  satisfied
• For each solution there is an encoded solution.
• For each encoded solution there is a feasible
  solution.
• All solutions should be represented in the same
  number of encodings.
• For NLP, Fourier transforms or wavelet transforms
  could be used.

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Problem of the Week

• There are five houses, each of a different colour
  and inhabited by women of different
  nationalities, with different pets, favourite
  drinks and cars. Moreover
• The Englishwoman lives in the red house.
• The Spaniard owns the dog.
• The woman in the green house drinks cocoa.
• The Ukrainian drinks eggnog.
• The green house is immediately to the right of
  the ivory house.
• The owner of the Toyota car also owns snails.
• The owner of the Ford lives in the yellow house.

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Problem of the week

• The man in the middle house drinks milk.
• The Norwegian lives in the first house on the
  left.
• The woman who owns the Chevrolet lives in the
  house next to the house where the woman owns a
  fox.
• The Ford owners house is next to the house where
  the horse is kept.
• The Mercedes-Benz owner drinks orange juice.
• The Japanese drives a Volkswagen.
• The Norwegian lives next to the blue house.
• The question is .... Who owns the zebra?
  Furthermore, who drinks water?