L2 is dcfl and L1 is Cfl
soln for L2 Let E = { aibj | i ≠ j and 2i ≠ j }
Split it as unions of 3 languages.
{ aibj | i > j } ∪ { aibj | i < j and 2i > j } ∪ { aibj | 2i < j }
G1 = { aibj | i > j }
G2 = { aibj | i < j and 2i > j }
G3 = { aibj | 2i < j }
G1 and G3 are quite easy to figure out.
for G1
S → aSb | aS1
S1 → aS1 |ε
for G3
S → aSbb | S1b
S1 → S1b | ε
Now tricky part is G2.
S → aSb | aS1b
S1 → aS1bb | aS2bb
S2 → ε
note- L2 is given as problem 2.24 in shipser book.