In Evolutionary programming, recombination is

1.	In Evolutionary programming, recombination is
A.	doesnot use recombination to produce offspring. it only uses mutation
B.	uses recombination such as cross over to produce offspring
C.	uses various recombination operators
D.	none of the mentioned
Answer» B. uses recombination such as cross over to produce offspring

Discussion

No Comment Found

Related MCQs

What are normally the two best measurement units for an evolutionary algorithm?1. Number of evaluations2. Elapsed time3. CPU Time4. Number of generations
There are also other operators, more linguistic in nature, called that can be applied to fuzzy set theory.
Which of the following neural networks uses supervised learning? (A) Multilayer perceptron (B) Self organizing feature map (C) Hopfield network
Given U = {1,2,3,4,5,6,7} A = {(3, 0.7), (5, 1), (6, 0.8)} then A will be: (where ~ complement)
If A and B are two fuzzy sets with membership functions A(x) = {0.6, 0.5, 0.1, 0.7, 0.8} B(x) = {0.9, 0.2, 0.6, 0.8, 0.5}Then the value of (A B) (x) will be
Compute the value of adding the following two fuzzy integers:A = {(0.3,1), (0.6,2), (1,3), (0.7,4), (0.2,5)} B = {(0.5,11), (1,12), (0.5,13)}Where fuzzy addition is defined as A+B(z) = maxx+y=z (min( A(x), B(x))) Then, f(A+B) is equal to
Choose the correct statement1. A fuzzy set is a crisp set but the reverse is not true2. If A,B and C are three fuzzy sets defined over the same universe of discourse such that A B and B C and A C3. Membership function defines the fuzziness in a fuzzy set irrespecive of the elements in the set, which are discrete or continuous
Consider a fuzzy set A defined on the interval X = [0, 10] of integers by the membership Junction A(x) = x / (x+2) Then the cut corresponding to = 0.5 will be
A fuzzy set whose membership function has at least one element x in the universe whose membership value is unity is called
Which selection strategy works with negative fitness value?

In Evolutionary programming, recombination is

Discussion

No Comment Found

Related MCQs

Reply to Comment