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Applications of Genetic Algorithms. Amy Hoover. Actual GA Applications. Software Bug Repair http://www.genetic-programming.org/hc2009/1-Forrest/Forrest-Presentation.pdf Weapon Design in Video Games gar.eecs.ucf.edu Radar coverage. Traveling Salesman Problem.

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actual ga applications
Actual GA Applications
  • Software Bug Repair
    • http://www.genetic-programming.org/hc2009/1-Forrest/Forrest-Presentation.pdf
  • Weapon Design in Video Games
    • gar.eecs.ucf.edu
  • Radar coverage
traveling salesman problem
Traveling Salesman Problem
  • Problem: You want to mine the most gold in the least amount of time.
  • What’s the best route?

B

C

A

E

D

traveling salesman problem1
Traveling Salesman Problem
  • Problem: You want to mine the most gold in the least amount of time.
  • What’s the best route?

4min

4min

B

C

A

5min

2min

2min

7min

3min

E

6min

D

what s the best route
What’s the best route?

4min

4min

B

C

A

5min

2min

2min

7min

3min

E

6min

D

Path 1: ABCDE

How long would this take?

Path 2: ABDEC

How long would this take?

what s the best route1
What’s the best route?

4min

4min

B

C

A

2min

5min

2min

7min

3min

E

6min

D

Path 1: ABCDE

A-B = 4min, B-C = 4min, C-D = 3 min, D-E = 6 min

4 + 4 + 3 + 6 = 17 minutes to mine all the gold!

Path 2: ABDEC

(Class) How long would this take

what s the best route2
What’s the best route?

4min

4min

B

C

A

5min

2min

2min

7min

3min

E

6min

D

Path 1: ABCDE

A-B = 4min, B-C = 4min, C-D = 3 min, D-E = 6 min

4 + 4 + 3 + 6 = 17 minutes to mine all the gold!

Path 2: ABDEC

A-B = 4min, B-D = 5min, D-E = 6 min, E-C = 7 min

4 + 5 + 6 + 7 = 22 minutes to mine all the gold

Which path is better?

problem
Problem
  • Search space grows enormously with each visited node
    • nodeCount!
traveling salesman and gas
Traveling Salesman and GAs
  • Example GA
    • http://dandeblasio.com/cot4810/
  • How can we encode this using GAs?
traveling salesman and gas1
Traveling Salesman and GAs
  • How can we encode this using GAs?
    • Need chromosome!
traveling salesman and gas2
Traveling Salesman and GAs
  • How can we encode this using GAs?
    • Need chromosome!
      • Path = ABCDE
    • Then what?
traveling salesman and gas3
Traveling Salesman and GAs
  • How can we encode this using GAs?
    • Need chromosome!
      • Path = ABCDE
    • Then what?
      • Randomize the initial population!
        • Path 1 = ABCDE
        • Path 2 = ABCED
        • Path 3 = ABECD
        • Path 4 = AEBCD
        • Path 5 = EABCD

    • Then what?
traveling salesman and gas4
Traveling Salesman and GAs
  • How can we encode this using GAs?
    • Calculate the fitness of each individual
        • Path 1 = ABCDE = ?
        • Path 2 = ABCED = ?
        • Path 3 = ABECD = ?
        • Path 4 = AEBCD = ?
        • Path 5 = EABCD = ?
traveling salesman and gas5
Traveling Salesman and GAs
  • How can we encode this using GAs?
    • Calculate the fitness of each individual
    • Keep in mind we want to get the most gold in the LEAST amount of time so we can go do other fun stuff
        • Path 1 = ABCDE = 17 minutes
        • Path 2 = ABCED = 22 minutes
        • Fitness = 100 – 22 = 78
        • Path 3 = ABECD = 4 +2 + 7 + 3 = 16 minutes
        • Path 4 = AEBCD = 2 + 2 +4 + 3 = 11 minutes
        • Fitness = 100 - 11 = 89
        • Path 5 = EABCD = 2 + 4 + 4 + 3 = 13 minutes
traveling salesman and gas6
Traveling Salesman and GAs
  • How can we encode this using GAs?
    • Calculate the fitness of each individual
    • Keep in mind we want to get the most mithril in the LEAST amount of time so we can go do other fun stuff
        • Path 1 = ABCDE = 17 minutes
        • Path 2 = ABCED = 22 minutes
        • Path 3 = ABECD = 4 +2 + 7 + 3 = 16 minutes
        • Path 4 = AEBCD = 2 + 2 +4 + 3 = 11 minutes
        • Path 5 = EABCD = 2 + 4 + 4 + 3 = 13 minutes
traveling salesman and gas7
Traveling Salesman and GAs
  • Who should reproduce?
        • Path 1 = ABCDE = 17 minutes
        • Path 2 = ABCED = 22 minutes
        • Path 3 = ABECD = 4 +2 + 7 + 3 = 16 minutes
        • Path 4 = AEBCD = 2 + 2 +4 + 3 = 11 minutes
        • Path 5 = EABCD = 2 + 4 + 4 + 3 = 13 minutes
slide17
Problems Arise With Crossover (A doesn’t have 2 sets of gold!)
    • “ABCDEF” “ABCFAE”
    • “CDBFAE” “CDBDEF”
  • 2 Options
    • Modify Our Representation
    • Modify Our Crossover Operator
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