Question 3578 – Theory-of-Computation

Consider a=0⁺1⁺ and b=2⁺3⁺

Hence (0⁺1⁺ | 2⁺3⁺)* = (0⁺1⁺ + 2⁺3⁺)* = (a+b)*
Now since the min string in 0⁺1⁺ is “01” hence “a” has min length two.
Similarly the min string in 2⁺3⁺ is “23” , hence “b”has min length two.
We require 4 length strings. ⁺
We can generate 4 length strings using “a” only, or using “b” only or using both.⁺
Using “a” one time only → 0⁺1⁺ only: {0001, 0011, 0111}
using “b” one time only → 2⁺3⁺ only : {2223, 2233, 2333}
Using two time, i.e., aa, ab, ba, bb
aa→ 0101
ab→ 0123
ba→ 2301
bb→ 2323
‘Hence total 10 strings {0001, 0011, 0111, 2223, 2233, 2333, 0101, 0123, 2301, 2323}