Top-rated recipes tagged "genetic_algorithms" but not "back" and "neural_networks"http://code.activestate.com/recipes/tags/-back+genetic_algorithms-neural_networks/top/2011-11-27T06:45:00-08:00ActiveState Code RecipesEvolutionary Algorithm (Generation of Prime Numbers) (Python)
2011-11-27T06:45:00-08:00Alexander James Wallarhttp://code.activestate.com/recipes/users/4179768/http://code.activestate.com/recipes/577964-evolutionary-algorithm-generation-of-prime-numbers/
<p style="color: grey">
Python
recipe 577964
by <a href="/recipes/users/4179768/">Alexander James Wallar</a>
(<a href="/recipes/tags/algorithm/">algorithm</a>, <a href="/recipes/tags/example/">example</a>, <a href="/recipes/tags/genetic/">genetic</a>, <a href="/recipes/tags/genetic_algorithm/">genetic_algorithm</a>, <a href="/recipes/tags/genetic_algorithms/">genetic_algorithms</a>, <a href="/recipes/tags/list/">list</a>, <a href="/recipes/tags/number/">number</a>, <a href="/recipes/tags/of/">of</a>, <a href="/recipes/tags/prime/">prime</a>, <a href="/recipes/tags/primelist/">primelist</a>, <a href="/recipes/tags/primes/">primes</a>, <a href="/recipes/tags/theory/">theory</a>).
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<p>This is an evolutionary algorithm that returns a random list of prime numbers. This code is highly inefficient for a reason. This algorithm is more of a proof of concept that if a prime was a heritable trait, it would not be a desired one. </p>
<p>Parameters:</p>
<p>isPrime --> n: number to check if it is prime
allPrimes --> n: size of list of random primes, m: the primes in the list will be between 0 and m</p>
Partition Problem and natural selection (Python)
2009-10-27T00:27:14-07:00Jai Vikram Singh Vermahttp://code.activestate.com/recipes/users/4171450/http://code.activestate.com/recipes/576937-partition-problem-and-natural-selection/
<p style="color: grey">
Python
recipe 576937
by <a href="/recipes/users/4171450/">Jai Vikram Singh Verma</a>
(<a href="/recipes/tags/approximate_solution/">approximate_solution</a>, <a href="/recipes/tags/complete/">complete</a>, <a href="/recipes/tags/crossover/">crossover</a>, <a href="/recipes/tags/easiest_hard_problem/">easiest_hard_problem</a>, <a href="/recipes/tags/generations/">generations</a>, <a href="/recipes/tags/genetic_algorithms/">genetic_algorithms</a>, <a href="/recipes/tags/natural_selection/">natural_selection</a>, <a href="/recipes/tags/np/">np</a>, <a href="/recipes/tags/np_complete/">np_complete</a>, <a href="/recipes/tags/np_hard/">np_hard</a>, <a href="/recipes/tags/optimal_solution/">optimal_solution</a>, <a href="/recipes/tags/partition_problem/">partition_problem</a>).
Revision 4.
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<p>Partition problem
From Wikipedia, the free encyclopedia</p>
<p>In computer science, the partition problem is an NP-complete problem. The problem is to decide whether a given multiset of integers can be partitioned into two "halves" that have the same sum. More precisely, given a multiset S of integers, is there a way to partition S into two subsets S1 and S2 such that the sum of the numbers in S1 equals the sum of the numbers in S2? The subsets S1 and S2 must form a partition in the sense that they are disjoint and they cover S. The optimization version asks for the "best" partition, and can be stated as: Find a partition into two subsets S1,S2 such that max(sum(S_1), sum(S_2)) is minimized (sometimes with the additional constraint that the sizes of the two sets in the partition must be equal, or differ by at most 1).</p>
<p>The partition problem is equivalent to the following special case of the subset sum problem: given a set S of integers, is there a subset S1 of S that sums to exactly t /2 where t is the sum of all elements of S? (The equivalence can be seen by defining S2 to be the difference S − S1.) Therefore, the pseudo-polynomial time dynamic programming solution to subset sum applies to the partition problem as well.</p>
<p>Although the partition problem is NP-complete, there are heuristics that solve the problem in many instances, either optimally or approximately. For this reason, it has been called the "The Easiest Hard Problem" by Brian Hayes.</p>