Q7Soft Computing
Question
2 marks
Q.7. Consider the following binary strings (101010110) and (100110100). Write the new strings using a single-point crossover operator and by choosing the 3rd site at random and mutating the 4th bit of the string.
Answer
Parents: 101010110 and 100110100. Crossover at site 3 (first 3 bits fixed, remainder swapped) gives offspring 101|110100 = 101110100 and 100|010110 = 100010110. Mutating the 4th bit of each (flipping it) gives final strings 101010100 and 100110110.